ARS MATHEMATICA CONTEMPORANEA
VOLUME 25, NUMBER 2
2025
AVAILABLE AT: https://amc-journal.eu
ISSN: 1855-3974
The publication is partially supported by the Slovenian Research Agency from the Call for
co-financing of scientific periodical publications.

ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.01
https://doi.org/10.26493/1855-3974.3085.3ea
(Also available at http://amc-journal.eu)
On regular graphs with Šoltés vertices*
Nino Bašić †
FAMNIT, University of Primorska, Koper, Slovenia and
IAM, University of Primorska, Koper, Slovenia and
Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia
Martin Knor ‡
Slovak University of Technology in Bratislava, Slovakia
Riste Škrekovski
FAMNIT, University of Primorska, Koper, Slovenia and
Faculty of Mathematics and Physics, University of Ljubljana, Slovenia and
Faculty of Information Studies, Novo mesto, Slovenia and
Rudolfovo - Science and Technology Centre Novo Mesto, Slovenia
Received 13 March 2023, accepted 4 March 2024, published online 10 March 2025
Abstract
Let W (G) be the Wiener index of a graph G. We say that a vertex v ∈ V (G) is a Šoltés
vertex in G if W (G − v) = W (G), i.e. the Wiener index does not change if the vertex
v is removed. In 1991, Šoltés posed the problem of identifying all connected graphs G
with the property that all vertices of G are Šoltés vertices. The only such graph known to
this day is C11. As the original problem appears to be too challenging, several relaxations
were studied: one may look for graphs with at least k Šoltes vertices; or one may look
for α-Šoltés graphs, i.e. graphs where the ratio between the number of Šoltés vertices and
the order of the graph is at least α. Note that the original problem is, in fact, to find all
1-Šoltés graphs. We intuitively believe that every 1-Šoltés graph has to be regular and has
to possess a high degree of symmetry. Therefore, we are interested in regular graphs that
contain one or more Šoltés vertices. In this paper, we present several partial results. For
every r ≥ 1 we describe a construction of an infinite family of cubic 2-connected graphs
with at least 2r Šoltés vertices. Moreover, we report that a computer search on publicly
*The second and third authors acknowledge partial support of the Slovenian research agency ARRS; program
P1-0383 and ARRS project J1-3002, and the annual work program of Rudolfovo.
†Corresponding author. The author is supported in part by the Slovenian Research Agency (research program
P1-0294 and research projects J1-1691, N1-0140, and J1-2481).
‡The author acknowledges partial support by Slovak research grants VEGA 1/0069/23, VEGA 1/0011/25,
APVV-23-0076 and APVV-22-0005.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.01
available collections of vertex-transitive graphs did not reveal any 1-Šoltés graph. We are
only able to provide examples of large 13 -Šoltés graphs that are obtained by truncating
certain cubic vertex-transitive graphs. This leads us to believe that no 1-Šoltés graph other
than C11 exists.
Keywords: Šoltés problem, Wiener index, regular graph, cubic graph, Cayley graph, Šoltés vertex.
Math. Subj. Class. (2020): 05C12, 05C90, 20B25
1 Introduction
All graphs under consideration in this paper are simple and undirected. The Wiener index
of a graph G, denoted by W (G), is defined as
W (G) =
1
2
∑
u∈V (G)
∑
v∈V (G)
dG(u, v) =
∑
{u,v}⊆V (G)
dG(u, v), (1.1)
where dG(u, v) is the distance between vertices u and v (i.e. the length of a shortest path
between u and v). If the graph G is disconnected, we may take W (G) = ∞. The Wiener
index was introduced in 1947 [16] and has been extensively studied ever since. For a recent
survey on the Wiener index see [9]. The transmission of a vertex v in a graph G, denoted by
wG(v), is defined as wG(v) =
∑
u∈V (G) dG(u, v). Note that (1.1) can also be expressed
as W (G) = 12
∑
u∈V (G) wG(u).
Let v ∈ V (G). The graph obtaned from G by removing a vertex v is denoted by G−v.
If we remove a vertex v from a graph G, any of the following scenarios may occur:
(a) the Wiener index decreases (e.g. W (K7) = 21 and W (K7 − v) = W (K6) = 15);
(b) the Wiener index increases (e.g. W (Wh9) = 56 and W (Wh9−v0) = W (C8) = 64,
where Whn denotes the wheel graph on n vertices and v0 is the central vertex of the
wheel);
(c) the Wiener index does not change (e.g. W (Wh8) = 42 and W (Wh8 − v0) =
W (C7) = 42, where v0 is the central vertex of Wh8).
We say that a vertex v ∈ V (G) is a Šoltés vertex of G if W (G) = W (G − v), i.e. the
Wiener index of G does not change if the vertex v is removed. If G is disconnected, with
two non-trivial components or with at least three components, then every vertex is a Šoltés
vertex. Therefore, it is natural to require that G is connected. Let
S(G) = {v ∈ V (G) | W (G) = W (G− v)}
and let 0 ≤ α ≤ 1. We say that a graph G is an α-Šoltés graph if |S(G)| ≥ α|V (G)|,
i.e. the ratio between the number of Šoltés vertices of G and the order of G is at least α.
For example, the graph in Figure 1 is the smallest cubic 13 -Šoltés graph. Note that G is an
1-Šoltés graph if every vertex in G is a Šoltés vertex. The only 1-Šoltés graph known to
this day is the cycle on 11 vertices, C11. In this paper, Šoltés graph is simply the synonym
for 1-Šoltés graph.
E-mail addresses: nino.basic@famnit.upr.si (Nino Bašić), martin.knor@stuba.sk (Martin Knor),
riste.skrekovski@fmf.uni-lj.si (Riste Škrekovski)
N. Bašić et al.: On regular graphs with Šoltés vertices 3
The Šoltés problem [15] was forgotten for nearly three decades. It was revived and
popularised in 2018 by Knor et al. [7]. They considered a relaxation of the original prob-
lem: one may look for graphs with a prescribed number of Šoltés vertices. They showed
that there exists a unicyclic graph on n vertices with at least one Šoltés vertex for every
n ≥ 9. They also showed that there exists a unicyclic graph with a cycle of length c and
at least one Šoltés vertex for every c ≥ 5, and that every graph is an induced subgraph of
some larger graph with a Šoltés vertex. They have further shown that a Šoltés vertex in
a graph may have a prescribed degree [8]. Namely, they proved that for any d ≥ 3 there
exist infinitely many graphs with a Šoltés vertex of degree d. Necessary conditions for the
existence of Šoltés vertices in Cartesian products of graphs were also considered [8]. In
2021, Bok et al. [3] showed that for every k ≥ 1 there exist infinitely many cactus graphs
with exactly k distinct Šoltés vertices. In 2021, Hu et al. [6] studied a variation of the prob-
lem and showed that there exist infinitely many graphs where the Wiener index remains the
same even if r ≥ 2 distinct vertices are removed from the graph.
Akhmejanova et al. [1] considered another possible relaxation of the problem: Do there
exist graphs with a given percentage of Šoltés vertices? They constructed two infinite
families of graphs with a relatively high proportion of Šoltés vertices. Their first family
comprises graphs B(k), k ≥ 2, where B(k) is a 2k5k+6 -Šoltés graph on 5k + 6 vertices.
Two vertices of B(k) are of degree k + 1, while the remaining vertices are of degree 2.
The percentage of Šoltés vertices is below 25 , but tends to
2
5 as k goes to infinity. They
also introduced a two-parametric infinite family L(k,m), m ≥ 7 and k ≥ m−3
m−6 . Here,
the percentage of Šoltés vertices is below 12 , but tends to
1
2 as k goes to infinity for a fixed
m. These graphs contain at least one leaf, at least km vertices of degree 2, and a vertex of
degree km+ 1.
In the present paper we focus on regular graphs. Our intuition lead us to believe that
the solutions to the original Šoltés problem should be graphs having all vertices of the same
degree.
Conjecture 1.1. If G is a Šoltés graph, then G is regular.
For a general regular graph G, the values W (G − u) and W (G − v) might be signifi-
cantly different for two different vertices u, v ∈ V (G); it may happen that removal of one
vertex increases the Wiener index, while removal of the other vertex descreases the Wiener
index. However, W (G−u) and W (G−v) are equal if vertices u and v belong to the same
vertex orbit. Therefore, we believe that a Šoltés graph is likely to be vertex transitive.
Conjecture 1.2. If G is a Šoltés graph, then G is vertex transitive.
Figure 1: The smallest cubic 13 -Šoltés graph has 24 vertices. Its Šoltés vertices are coloured
red.
4 Ars Math. Contemp. 25 (2025) #P2.01
Among truncations of cubic vertex-transitive graphs we found several 13 -Šoltés graphs;
see Section 4. Interestingly, all our examples are in fact Cayley graphs and this leads us to
pose the following conjecture.
Conjecture 1.3. If G is a Šoltés graph, then G is a Cayley graph.
It is not hard to obtain small examples of regular graphs with Šoltés vertices. We used
the geng [11] software to generate small k-regular graphs (for k = 3, 4 and 5). Let Rr
denote the class of all r-regular graphs and let Rrn denote the set of r-regular graphs on n
vertices. Let N(G, k) be the number of graphs in the class G that contain exactly k Šoltés
vertices.
Table 1 shows the numbers of (non-isomorphic) cubic graphs of orders n ≤ 24 that
contain Šoltés vertices. We can see that cubic graphs of order 12 or less do not contain
Šoltés vertices. There are plenty of examples with one Šoltés vertex. Cubic graphs with two
Šoltés vertices first appear at order n = 14; there are three such graphs (see Figure 2(a)–
(c)). Examples with three and four Šoltés vertices appear at order n = 16; there is one
cubic graph with three and two cubic graphs with four Šoltés vertices (see Figure 2(d)–
(f)). At order n = 18, there are no graphs with three Šoltés vertices, however there is
only one graph with four Šoltés vertices (see Figure 2(g)). Numbers of 4-regular and 5-
regular graphs with respect to their number of Šoltés vertices are given in Tables 2 and 3,
respectively.
N(R3
n
, k)
n |R3
n
| k = 1 k = 2 k = 3 k = 4 k = 5 k = 6 k = 7 k = 8
≤ 12 112 - - - - - - - -
14 509 4 3 - - - - - -
16 4060 108 37 1 2 - - - -
18 41301 1014 200 - 1 - - - -
20 510489 13460 1076 6 13 - - - -
22 7319447 194432 9610 151 52 - 1 - -
24 117940535 3161124 130087 2596 333 2 3 - 1
Table 1: The numbers of (non-isomorphic) cubic graphs with Šoltés vertices. There are no
graphs with Šoltés vertices for orders up to 12. The column labeled |R3n| gives the total
number of cubic graphs of order n. Symbol ‘-’ is a replacement for 0 (i.e. no such graph
exists). Each of the next columns gives the numbers of graphs with k Šoltés vertices for
k = 1, 2, . . . , 8. A blue-coloured number means that we have provided drawings of these
graphs (see Figures 1 and 2).
In the next two sections we construct an infinite family of cubic 2-connected graphs
with at least 2r, r ≥ 1, Šoltés vertices. It recently came to our attention that Dobrynin in-
dependently found an infinite family of cubic graphs with four Šoltés vertices [4]. However,
our method is more general.
N. Bašić et al.: On regular graphs with Šoltés vertices 5
N(R4n, k)
n |R4n| k = 1 k = 2 k = 3 k = 4
≤ 12 1894 - - - -
13 10778 - 1 - -
14 88168 30 6 - -
15 805491 265 85 - 5
16 8037418 2191 472 - -
17 86221634 14430 2097 4 1
Table 2: The numbers of (non-isomorphic) quartic graphs of orders up to 17 with Šoltés
vertices. Naming conventions used in Table 1 also apply here.
N(R5n, k)
n |R5n| k = 1 k = 2 k = 3 k = 4
≤ 12 7912 - - - -
14 3459383 8 3 - -
Table 3: The numbers of (non-isomorphic) quintic graphs of order up to 14 with Šoltés
vertices. Naming conventions used in Table 1 also apply here.
(a) (b) (c) (d) (e)
(f) (g) (h)
Figure 2: Examples of small regular graphs with two or more Šoltés vertices: (a) to (c) are
the three cubic graphs of order 14 with two Šoltés vertices; (d) and (e) are the two cubic
graphs of order 16 with four Šoltés vertices; (f) is the only cubic graph of order 16 with
three Šoltés vertices; (g) is the only cubic graph of order 18 with four Šoltés vertices; (h)
is the only quartic graph of order 13 with two Šoltés vertices. Šoltés vertices are coloured
red.
6 Ars Math. Contemp. 25 (2025) #P2.01
2 Cubic 2-connected graphs with two Šoltés vertices
In the present section we prove the following result.
Theorem 2.1. There exist infinitely many cubic 2-connected graphs G which contain at
least two Šoltés vertices.
We prove Theorem 2.1 by a sequence of lemmas. We start by giving several definitions.
First, we define a graph Gt on 8t+ 8 vertices, where t ≥ 1. Take 2t copies of the diamond
graph (i.e. K4 − e) and connect their degree-2 vertices, so that a ring of 2t copies of
K4 − e is formed. Add a disjoint 4-cycle to that graph. Then subdivide one of the edges
that connects two consecutive diamonds by two vertices, denote them by z1 and z2, and
connect z1 and z2 with two opposite vertices of the 4-cycle. Add a leaf to each remaning
degree-2 vertex of the 4-cycle. Denote the resulting graph by Gt. For an illustration,
see G3 in Figure 3. Note that Gt has exactly two leaves, denote them by v1 and v2, and
all the remaining vertices have degree 3. Denote by u1 and u2 the two vertices at the
longest distance from v1. This distance is dG(v1, u1) = dG(v1, u2) = 3t + 3 and also
dG(v2, u1) = dG(v2, u2) = 3t + 3. Observe that Gt has an automorphism (a symmetry)
fixing both v1 and v2, while interchanging u1 with u2.
u1
u2
z1
z2
v1
v2
Figure 3: The graph G3.
Now, we determine f(t) = W (Gt − u1)−W (Gt). We compute f(t) by summing the
contributions of all vertices (first the contribution of quadruples of vertices of all copies of
K4 − e, and then the contribution of the vertices which are not in any copy of K4 − e). As
the calculation is long and tedious, we present just the result
f(t) = 16t3 − 8t2 − 26t− 14, (2.1)
which was checked by a computer. Actually, the exact value of f(t) is not important here.
The crucial property is that for t big enough, f(t) is positive. In fact, limt→∞ f(t) = ∞;
see also Section 3, where a lower bound for f(t) is given.
To motivate the above definition, we briefly describe the main idea of the proof. We
attach trees T1 and T2 to vertices v1 and v2 of Gt, and then we add edges to them so that
the resulting graph H will be cubic and 2-connected. See Figure 4 for an example. This
will be done in three phases.
N. Bašić et al.: On regular graphs with Šoltés vertices 7
P1) In the first phase, we will construct trees T1 and T2 to guarantee that vertices u1
and u2 are indeed Šoltés vertices. The vertices of T1 and T2 can be partitioned into
several layers based on their distance to v1 and v2, respectively. The resulting graph
will be denoted by Q.
P2) In the second phase, we will add the ‘red’ edges, whose endpoints are in two consec-
utive layers, to graph Q. The resulting graph will be denoted by R. The purpose of
the second phase is to make sure that the sum of free valencies is even within each
layer, making the next phase possible. Note that although the final graph H is cubic,
the intermediate graphs, Q and R, are subcubic. The free valency of a vertex v in a
subcubic graph G is 3− degG(v).
P3) In the last phase, we will add blue edges to R in order to to obtain cycles and paths,
so that the resulting graph H will be cubic and 2-connected. The endpoints of ‘blue’
edges will reside in the same layer of the forest T1 ∪ T2. Adding ‘red’ and ‘blue’
edges has no influence on Šoltésness of u1 and u2.
u1
u2
v1
v2
Figure 4: A graph H with two Šoltés vertices, namely u1 and u2, that contains G3.
Let us consider the resulting graph H . Observe that if x ∈ V (H) \ V (Gt), then we
have wH(x) − wH−u1(x) = dH(u1, x). Clearly, every (u1, x)-path contains one of the
vertices from {v1, v2}. Hence, for calculating W (H) −W (H − u1), only the distance of
the new vertices x to {v1, v2} matters.
We need to find suitable trees T1 and T2 rooted at v1 and v2, respectively. Each of
these trees will have q vertices and their depth, say d, will be determined later. What next
properties do T1 and T2 need to have? Let ℓi be the number of vertices of the forest T1∪T2
at distance i, 1 ≤ i ≤ d, from {v1, v2}. Since the resulting graph will be cubic and 2-
connected, we have 2 ≤ ℓ1 ≤ 4, 2 ≤ ℓ2 ≤ 8, 2 ≤ ℓ3 ≤ 24, . . . and for the last value ℓd
we have 1 ≤ ℓd ≤ 2d+1. The trees attached to v1 and v2 may be paths in which case we
get ℓ1 = ℓ2 = · · · = ℓq = 2 (since each of T1 and T2 have exactly q vertices). In this case
the transmission of vj in Tj is biggest possible, 1 ≤ j ≤ 2. In the other extremal situation
the transmission of vj in Tj is smallest possible, 1 ≤ j ≤ 2, which results in ℓi = 2i+1,
1 ≤ i < d. If x ∈ V (H)\V (Gt) is at distance i from {v1, v2}, then dH(u1, x) = 3t+3+i.
8 Ars Math. Contemp. 25 (2025) #P2.01
Denote by D the sum of distances from all vertices of V (H) \ V (Gt) to u1. Then
D =
d
∑
i=1
(3t+ 3 + i)ℓi. (2.2)
Observe that we need to find a finite sequence (ℓ1, ℓ2, . . . , ℓd) so that f(t) = D. As the
resulting graph H has to be cubic, we need to add an even number of vertices, 2q, to the
graph Gt. What are the bounds for D?
First, we determine the lower bound; let us denote it by Dm. This bound will be
obtained when ℓi attains the maximum possible value of 2i+1 for every 1 ≤ i ≤ d− 1. In
other words, we are attaching complete binary trees to v1 and v2. Let a = ⌊log2(2q + 3)⌋.
Recall that the depth of a complete binary tree with n vertices is ⌊log2(n)⌋ and note that in
our case a− 1 = d. Then
ℓ1 = 4,
ℓ2 = 8,
...
ℓa−2 = 2
a−1,
ℓa−1 = 2q −
a−2
∑
i=1
ℓi = 2q − 2a + 4.
The above sequence will be called the short sequence and denoted by Lm. Using the
formula
∑a
i=1 ix
i−1 = (axa+1 − (a+ 1)xa + 1)/(x− 1)2, we get
Dm =
a−2
∑
i=1
(3t+ 3 + i)2i+1 + (3t+ a+ 2)(2q − 2a + 4)
= (3t+ 1)2q +
a−2
∑
i=1
(i+ 2)2i+1 + (a+ 1)(2q − 2a + 4) + 2 · 21 + 1 · 20 − 5
= (3t+ 1)2q +
a
∑
i=1
i2i−1 + (a+ 1)(2q − 2a + 4)− 5
= (3t+ 1)2q + a2a+1 − (a+ 1)2a + 1 + (a+ 1)2q − (a+ 1)2a + 4(a+ 1)− 5
= (3t+ 1)2q + a2a+1 − (a+ 1)2a+1 + (a+ 1)2q + 4a
= (3t+ 1)2q − 2a+1 + (a+ 1)2q + 4a.
(2.3)
Now we find the upper bound Dm for D. In this case ℓ1 = ℓ2 = · · · = ℓq = 2; this
sequence will be called the long sequence and denoted by Lm. Therefore,
Dm = 2(3t+ 4) + 2(3t+ 5) + · · ·+ 2(3t+ 3 + q)
= (3t+ 3)2q + 2
q
∑
i=1
i
= (3t+ 1)2q + q2 + 5q.
(2.4)
Note that Dm and Dm are functions of q and t.
N. Bašić et al.: On regular graphs with Šoltés vertices 9
t 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
qmin 9 17 27 38 50 64 78 94 110 128 146 165 185 205 227 249
qmax 9 21 38 59 85 116 152 192 238 288 344 405 471 542 617 698
t 19 20 21 22 23 24 25 26 27 28 29 30 31
qmin 272 295 319 344 370 396 422 450 478 506 535 565 595
qmax 785 876 973 1075 1182 1294 1411 1533 1661 1795 1933 2077 2225
Table 4: The minimum and maximum value of q which satisfy the condition of Lemma 2.2.
Lemma 2.2. Let t ≥ 3. Then there exists q, such that Dm ≤ f(t) ≤ Dm.
Proof. Observe that if q ∼ 4t
√
t then we have Dm ≤ f(t) ≤ Dm for large enough t.
For small values of t we computed the minimum and maximum value of q that satisfies the
condition of the lemma; see Table 4.
Note that the value q in Lemma 2.2 is uniquely determined only for t = 3. For larger
values of t we get a range of options. Any q between qmin and qmax can be used. Moreover,
different values of q lead to non-isomorphic graphs H .
Now, we show that for every integer D, Dm ≤ D ≤ Dm, there exists a finite sequence
(ℓ1, ℓ2, . . . , ℓd) which realises D. Moreover, the graph Gt can be extended to H by at-
taching trees T1 and T2 that have ℓi vertices at distance i from {v1, v2}, such that H is a
2-connected cubic graph.
Let us define an operation M that modifies one such sequence L = (ℓ1, ℓ2, . . . , ℓd).
Definition 2.3. Let L = (ℓ1, ℓ2, . . . , ℓd). Set ℓd+1 = 0. Let i be the smallest value such
that ℓi ≥ 3 and either
(i) 2(ℓi − 1) > ℓi+1 + 1 or
(ii) ℓi = ℓi+1 = 3.
We say that M(L) = (ℓ′1, ℓ′2, . . . , ℓ′d) is a modification of sequence L if ℓ′j = ℓj for all j,
1 ≤ j ≤ d+ 1, except for j ∈ {i, i+ 1}, for which ℓ′i = ℓi − 1 and ℓ′i+1 = ℓi+1 + 1.
Lemma 2.4. Let t ≥ 1 and q ≥ 0. For every D, Dm ≤ D ≤ Dm, there exists a finite
sequence L = (ℓ1, ℓ2, . . . , ℓd), such that
(i)
∑d
i=1 ℓi = 2q;
(ii) 2 ≤ ℓi ≤ 2i+1 for i < d and 1 ≤ ℓd ≤ 2d+1;
(iii) ℓi+1 ≤ 2ℓi.
Namely, L = MD−Dm(Lm).
Proof. We start with the short sequence Lm = (4, 8, 16, . . . ). It clearly satisfies condi-
tions (i) to (iii). We have already seen that Lm realises Dm. This established the base of
induction.
Every other D can be realised by a sequence that is obtained from Lm by iteratively
applying operation M. Assume that after (D − 1) −Dm steps we obtained the sequence
L = (ℓ1, ℓ2, . . . , ℓd) which realises D−1. Obviously, M(L) realises D. It is easy to check
that conditions (i) to (iii) are satisfied for M(L).
10 Ars Math. Contemp. 25 (2025) #P2.01
Next, we prove two additional properties that hold when operation M is iteratively
applied on Lm.
Lemma 2.5. Let t ≥ 1 and q ≥ 0. Let L = (ℓ1, ℓ2, . . . , ℓd) be obtained from Lm by
iteratively applying operation M. Then the following holds:
(i) If (ii) of Definition 2.3 applies, then ℓj = 2 for all j < i.
(ii) If index i in Definition 2.3 is such that ℓi+1 = 0, then ℓi ≥ 4.
Proof. (i): Observe that if part (ii) of Definition 2.3 applies and i ≥ 2, then ℓi−1 = 2, since
otherwise i − 1 satisfies the assumption of the definition, thus i is not the smallest such
value. Moreover, if there is k, k < i, with ℓk ≥ 3, then choose the largest possible k with
this property. Then ℓk+1 = ℓk+2 = · · · = ℓi−1 = 2. Hence, k satisfies the assumption
of the definition, a contradiction. It means that if part (ii) of the definition applies, then
ℓi−1 = ℓi−2 = · · · = ℓ1 = 2.
(ii): If ℓi+1 = 0 then clearly ℓi ≥ 3. Suppose that ℓi = 3. Then ℓi−1 = 2, since
otherwise the assumptions apply to i − 1. As 2q is even, there must be k < i such that
ℓk ≥ 3. Let k be the largest possible value with this property. Then the assumptions apply
to k, a contradiction. Thus, if ℓi+1 = 0 then ℓi ≥ 4.
Note that part (ii) of Lemma 2.5 means that in the sequence L′ = M(L), ℓ′i ≥ 3 and
ℓ′i+1 = 1. The corresponding graph H will thus have a single vertex of degree 3 in the final
layer.
Example 2.6. For an illustration, the sequence of sequences
Lm, M(Lm), M2(Lm), M3(Lm), . . .
for 2q = 20 is
(4, 8, 8), (4, 5, 6, 5), (3, 4, 5, 6, 2), (2, 3, 4, 7, 4), (2, 2, 3, 5, 7, 1),
(4, 7, 9), (4, 4, 7, 5), (3, 4, 4, 7, 2), (2, 3, 4, 6, 5), (2, 2, 3, 5, 6, 2),
(4, 6, 10), (3, 5, 7, 5), (3, 3, 5, 7, 2), (2, 3, 4, 5, 6), (2, 2, 3, 4, 7, 2),
(4, 6, 9, 1), (3, 5, 6, 6), (2, 4, 5, 7, 2), (2, 3, 4, 4, 7), (2, 2, 3, 4, 6, 3),
(4, 6, 8, 2), (3, 4, 7, 6), (2, 4, 5, 6, 3), (2, 3, 3, 5, 7), (2, 2, 3, 4, 5, 4),
(4, 5, 9, 2), (3, 4, 6, 7), (2, 4, 4, 7, 3), (2, 2, 4, 5, 7), etc.
(4, 5, 8, 3), (3, 4, 5, 8), (2, 3, 5, 7, 3), (2, 2, 4, 5, 6, 1),
(4, 5, 7, 4), (3, 4, 5, 7, 1), (2, 3, 5, 6, 4), (2, 2, 4, 4, 7, 1),
There are altogether 67 sequences since Dm −Dm = 66 when 2q = 20.
Lemma 2.7. Let t ≥ 3. There exist rooted trees T1 and T2 such that vertices u1 and u2
are Šoltés vertices in the graph Q obtained from Gt by attaching T1 and T2 to vertices v1
and v2.
Proof. By Lemma 2.2, there exists q such that Dm ≤ f(t) ≤ Dm. From Lemma 2.4, we
obtain the sequence L, which gives us the appropriate number of vertices in every layer of
the forest T = T1 ∪ T2. This ensures that v1 and v2 are Šoltés vertices in Q.
N. Bašić et al.: On regular graphs with Šoltés vertices 11
Now, we construct a graph Q containing Gt and realising L. Let Tj be the tree rooted
at vj , 1 ≤ j ≤ 2. Then T1 will have ⌈ℓi/2⌉ vertices at distance i from v1 and T2 will have
⌊ℓi/2⌋ vertices at distance i from v2. Observe that it is possible to construct both T1 and
T2. We have two possibilities.
Case 1: ℓi = 2ℓi−1. Then either ℓi = 2i+1, ℓi−1 = 2i, . . . , ℓ1 = 4 and on the first i levels
both T1 and T2 are complete binary trees of height i; or ℓ1 = ℓ2 = · · · = ℓi−1 = 2 and
ℓi = 4, which means that both T1 and T2 contain one vertex at levels 1, 2, . . . , i − 1 and
two vertices at level i.
Case 2: ℓi ≤ 2ℓi−1 − 1. If ℓi−1 is even then we can construct i-th level of both T1 and T2,
and at least one vertex of level i− 1 of T2 will have degree less than 3. On the other hand,
if ℓi−1 is odd then T2 has only (ℓi−1 − 1)/2 vertices at level i− 1. However, it has ⌊ℓi/2⌋
vertices at level i and ⌊ℓi/2⌋ ≤ ⌊(2ℓi−1 − 1)/2⌋ = ℓi−1 − 1 = 2⌊ℓi−1/2⌋, so in T2, the
number of vertices at level i is at most twice the number of vertices at level i − 1. In this
case, at least one vertex at level i − 1 in T1 will have degree less than 3. This concludes
Case 2.
We construct the trees T1 and T2 so that at each level we minimise the number of
vertices of degree 3. Observe that then there is no level in which there are vertices of
degree 1 and also vertices of degree 3.
Consider v1 and v2 as vertices of level 0, and set ℓ0 = 2. We plan to add edges within
levels (i.e. ‘blue’ edges) to create a cubic graph, but sometimes we must also add edges
between consecutive levels (i.e. ‘red’ edges). First, we add necessary edges connecting
vertices of different levels. For every i ≥ 1, if ∑ij=0 ℓj is odd then add an edge joining a
vertex (of degree ≤ 2) of (i− 1)-th level with a vertex of i-th level.
Lemma 2.8. It is possible to add edges to Q as described above, so that the resulting graph
has no parallel edges and it is subcubic.
Proof. We add the red edges step by step starting with level 1, together with creating the
trees T1 and T2. And we show that at each level it is possible to add a required edge. We
distinguish two cases:
Case 1: There is no red edge between levels i − 2 and i − 1. If 2ℓi−1 = ℓi, then either
(a) ℓ1 = ℓ2 = · · · = ℓi−1 = 2 and ℓi = 4, or (b) ℓj = 2j+1 for all j ≤ i. In both
subcases
∑i
j=0 ℓj is even, and no red edge is added between levels i − 1 and i. Suppose
that 2ℓi−1 > ℓi. Then there is a vertex at (i − 1)-st level, say x, whose degree is less than
3. If 2ℓi > ℓi+1, then there is a vertex at i-th level, say y, whose degree is also less than 3,
and we can add the edge xy. (Observe that if we add these additional edges together with
the construction of trees T1 and T2, then we do not create parallel edges. The only problem
occures when x is the unique vertex at level i− 1 in Tj and y is also in Tj . But then either
ℓi−1 = 3 and j = 2, in which case x can be chosen in T1, or ℓi−1 = 2 and ℓi = 3 in which
case x can be chosen in T2 and y in T1.) On the other hand, if 2ℓi = ℓi+1 then (recall that
2ℓi−1 > ℓi) ℓ1 = · · · = ℓi−1 = ℓi = 2 and ℓi+1 = 4, so
∑i
j=0 ℓj is even, and no red edge
is added between levels i− 1 and i.
Case 2: There is a red edge between levels i− 2 and i− 1. Then
∑i−1
j=0 ℓj is odd. Assume
that we also have to add a red edge between levels i − 1 and i. Then ∑ij=0 ℓj is also odd
which means that ℓi is even and that 2ℓi−1 > ℓi, as shown in Case 1. Hence, 2ℓi−1 > ℓi+1,
12 Ars Math. Contemp. 25 (2025) #P2.01
so there is a vertex at (i − 1)-st level, say x, whose degree is less than 3. As 2ℓi > ℓi+1,
there is a vertex at i-th level, say y, whose degree is also less than 3, and we can add the
edge xy. (Multiple edges can be avoided analogously as in Case 1.)
We remark that we did not precisely specify how to choose the vertices x and y, when
a red edge is add between levels i− 1 and i in case there are several possibilities. Here are
a few simple rules to follow when choosing x or y at level i:
(i) if there are at least three leaves (in T1∪T2) at level i, then do not choose these leaves;
(ii) if there is eactly one leaf w at level i, w ∈ V (Tj), then there must be a protected
degree-2 vertex at level i in T3−j (i.e. the protected vertex shall not be chosen);
(iii) if there are two leaves w and w′ at level i (one of them is in T1 and the other in T2, as
we will prove later), then one degree-2 vertex from T1 and one degree-2 vertex from
T2 has to be protected;
(iv) if there are no leaves at level i, we have no constraints.
The above rules will be fully justified later, when we will consider 2-connectivity of the
resulting graph H . Denote by R the graph obtained after adding red edges, as described
above. We have the following statement.
Lemma 2.9. In each level of R, the sum of free valencies is even.
Proof. Let 1 ≤ i ≤ d. We prove the statement for level i. So denote by a1, a2, . . . , aℓi the
vertices at i-th level. Our task is to show that
∑ℓi
j=1(3−degR(aj)) is even. We distinguish
two cases, with two subcases each.
Case 1:
∑i
j=0 ℓj is odd. Then there are ℓi + 1 edges between levels ℓi−1 and ℓi in R.
If ℓi+1 is odd, then
∑i+1
j=0 ℓj is even, so there are ℓi+1 edges between levels i and i + 1.
Hence,
∑ℓi
j=1(3− degR(aj)) = 3ℓi − ℓi − 1− ℓi+1 is even. On the other hand, if ℓi+1 is
even, then
∑i+1
j=0 ℓj is odd, so there are ℓi+1 + 1 edges between levels i and i+ 1. Hence,
∑ℓi
j=1(3− degR(aj)) = 3ℓi − ℓi − 1− ℓi+1 − 1 is even.
Case 2:
∑i
j=0 ℓj is even. Then there are ℓi edges between levels ℓi−1 and ℓi in R. If
ℓi+1 is odd, then
∑i+1
j=0 ℓj is odd, so there are ℓi+1 + 1 edges between levels i and i + 1.
Hence,
∑ℓi
j=1(3− degR(aj)) = 3ℓi − ℓi − ℓi+1 − 1 is even. On the other hand, if ℓi+1 is
even, then
∑i+1
j=0 ℓj is also even, so there are ℓi+1 edges between levels i and i+1. Hence,
∑ℓi
j=1(3− degR(aj)) = 3ℓi − ℓi − ℓi+1 is even.
By Lemma 2.9, the sum of free valencies is even at each level of R. This means that, in
general, after we add some edges connecting vertices within level i, and when we do that
for all i, 1 ≤ i ≤ d, the resulting graph H will be cubic. We now describe how to add these
‘blue’ edges, so that H will be 2-connected, and how to resolve the cases when a level has
small number of remaining degree-2 vertices.
Observation. H will be 2-connected if for every leaf x of T , say x ∈ V (Tk), where
1 ≤ k ≤ 2 and x is a vertex at level i, there is a path, say P , containing only the vertices of
level i and connecting x with a vertex, say y, of T3−k.
N. Bašić et al.: On regular graphs with Šoltés vertices 13
We refer to the above as the 2-connectivity condition. The reason is that P can be
completed to a cycle using a (vk, x)-path in Tk and a (v3−k, y)-path in T3−k. Since all
cycles constructed in this way contain three vertices of Gt, the resulting graph H will be
2-connected. We remark that in one special case the path P will contain vertices of levels
i and i+1, but it will still be possible to complete P to a cycle containing three vertices of
Gt. Thus, our attention will be focused on the leaves of T .
Lemma 2.10. It is possible to add edges to R so that the resulting graph H will be cubic
and 2-connected.
Proof. First, we consider the d-th (i.e., the last) level. Note that all the vertices at level d
are leaves of T . Since
∑d
j=1 ℓj = 2q is even, each vertex at level d has degree 1 in R. We
distinguish three cases.
Case 1: ℓd ≥ 3. In this case, we add to graph R a cycle passing through all the vertices
of level d. Then the vertices of level d will have degree 3 and they will satisfy the 2-
connectivity condition, since ⌈ℓd/2⌉ vertices of level d are in T1 and ⌊ℓd/d⌋ of them are
in T2.
Case 2: ℓd = 1. Then ℓd−1 ≥ 3, as already shown. In this case, we replace the sequence
L = (ℓ1, ℓ2, . . . , ℓd−1, 1) by L∗ = (ℓ1, ℓ2, . . . , ℓd−1, 3) and we find a 2-connected cubic
graph H∗ realizing L∗. In this graph, the pendant vertices of level d are connected to
three different vertices of level d− 1 in T , since at each level we minimised the number of
degree-3 vertices. Since
∑d
j=1 ℓj is even, there are no other edges connecting vertices of
level d− 1 with those of level d in R. Hence, we add a 3-cycle as described in Case 1, and
then contract the three vertices at level d to a single vertex. If H∗ is cubic and 2-connected,
then so is the resulting graph H .
Case 3: ℓd = 2. In this case, we simultaneously resolve the problem for levels d and d− 1.
Since
∑d−1
j=1 ℓj is even, all vertices of level d− 1 have degree 1 except for two, which have
degree 2. (Recall that T was constructed so that at each level the number of vertices of
degree 3 was minimised.)
If ℓd−1 = 2, then connect both vertices of level d− 1 with both vertices of level d and
add an edge connecting the vertices of level d. Then the vertices of levels d− 1 and d have
degree 3 and they satisfy the 2-connectivity condition.
If ℓd−1 ≥ 3, then pick a vertex of degree 1 at level d − 1, say x, and join it to both
vertices of level d. Then x has degree 3, but since it is a leaf of T , it does not satisfy
the 2-connectivity condition in the strict sense. Nevertheless, there is a cycle in H which
contains edges of Gt, a path connecting v1 with a vertex of level d in T1, a path connecting
v2 with a vertex of level d in T2, and the two edges connecting x with the vertices of level
d, which is sufficient. Then add to H the edge connecting vertices of level d and add a path
passing through all vertices of level d− 1 except x, and starting/ending in the two degree-2
vertices. This resolves the problem for levels d and d− 1. This concludes Case 3.
We now turn to level i, 1 ≤ i < d. In case ℓd = 2, we assume i < d− 1. Then vertices of
level i are connected to vertices of levels i− 1 and i+1 using only the edges of R, and we
now add only edges connecting vertices within level i, i.e. the blue edges.
In some cases, we specify positions of red edges that were added to T to form R, to
justify the four rules for choosing vertices x and y in the process of creating R.
14 Ars Math. Contemp. 25 (2025) #P2.01
If there is no leaf at level i, then all vertices of this level have degrees 2 and 3 in R. By
Lemma 2.9, there is an even number of degree-2 vertices. Thus, we can add a collection
of independent edges so that all vertices of level i will have degree 3. Since there were no
leaves, the vertices of level i satisfy the 2-connectivity condition.
Now suppose that there are leaves at level i. Since we minimised the number of vertices
of degree 3 when constructing T , there are no vertices of degree 3 in level i. Consequently,
each vertex of level i is connected to at most one vertex in level i+ 1 in T . Hence, T1 and
T2 have, respectively, ⌈ℓi/2⌉ − ⌈ℓi+1/2⌉ and ⌊ℓi/2⌋ − ⌊ℓi+1/2⌋ leaves at level i. Denote
k = (⌈ℓi/2⌉−⌈ℓi+1/2⌉)−(⌊ℓi/2⌋−⌊ℓi+1/2⌋). Since k = (⌈ℓi/2⌉−⌊ℓi/2⌋)−(⌈ℓi+1/2⌉−
⌊ℓi+1/2⌋), we have
−1 ≤ k ≤ 1. (2.5)
This means that the numbers of leaves at level i in T1 and T2 differ by at most one, and also
that there are no degree-3 vertices at level i in T . Moreover, since i < d, level i contains
two vertices, say b1 and b2, such that b1 ∈ V (T1), b2 ∈ V (T2) and b1, b2 are not leaves in
T . (Recall that if i = d− 1 and ℓd = 1, then then we solve this case for L∗ where ℓd = 3,
and afterwards we provide the contraction of vertices at level d, see Case 2 above.) Then
degT (b1) = degT (b2) = 2. We distinguish three cases.
Case 1: T has at least 3 leaves at level i. If E(R) \ E(T ) contains an edge connecting
levels i − 1 and i, then this edge will terminate at b1, and if E(R) \ E(T ) contains an
edge connecting levels i and i + 1, then this edge will start at b2. (Note that if we create
T and R simultaneously, level by level, then we can form b1 and b2, so that we do not get
parallel edges. In the worst case we relabel b1 and b2, so that b1 ∈ V (T2) and b2 ∈ V (T1).)
This leaves the leaves untouched. Then we add a cycle passing through all leaves of level
i and add a collection of independent edges so that all vertices of level i become degree-3
vertices. Since, at level i, at least one leaf is in T1 and at least one is in T2, the vertices at
level i satisfy the 2-connectivity condition.
Case 2: T has exactly two leaves at level i. Denote these vertices by a1 and a2. As
mentioned above, we may assume that a1 ∈ V (T1) and a2 ∈ V (T2). If E(R) \ E(T )
contains an edge connecting levels i − 1 and i, then this edge will terminate at a1, and if
E(R) \ E(T ) contains an edge connecting levels i and i + 1, then this edge will start at
a2. (Again, not to create parallel edges, the red edge between levels i − 1 and i may be
connected to a2 instead of a1, and then possible red edge between levels i and i + 1 will
start at a1.) Then add edges a1b2, a2b1, and a collection of independent edges so that all
vertices of level i become degree-3 vertices. Due to the presence of edges a1b2 and a2b1,
the vertices at level i satisfy the 2-connectivity condition.
Case 3: T has exactly one leaf at level i. Denote this vertex by a. Without loss of generality,
assume that a ∈ V (T1). If E(R) \ E(T ) contains an edge connecting levels i − 1 and i,
then this edge will terminate at b1, and if E(R) \E(T ) contains an edge connecting levels
i and i+1, then this edge will start at a. (Not to create parallel edges, the red edge between
levels i− 1 and i may be connected to a instead of b1, and then possible red edge between
levels i and i + 1 will start at b1.) Then add the edge ab2, and a collection of independent
edges so that all vertices of level i become degree-3 vertices. Due to the presence of edge
ab2, vertices at level i satisfy the 2-connectivity condition.
N. Bašić et al.: On regular graphs with Šoltés vertices 15
3 Cubic 2-connected graphs with 2r Šoltés vertices
Now we generalise Theorem 2.1 to higher amount of Šoltés vertices.
Theorem 3.1. Let r ≥ 1. There exist infinitely many cubic 2-connected graphs G which
contain at least 2r Šoltés vertices.
Proof. We reconsider the graph Gt from the proof of Theorem 2.1. This graph consists of
a chain of 2t diamonds attached to vertices z1 and z2 of a graph on 8 vertices. Denote this
graph on 8 vertices by F .
We construct Gt,r. Take a binary tree B of depth r−1. This tree has 2r−1 vertices, out
of which 2r−1 are leaves. Denote these leaves by a1, a2, . . . , a2r−1 . Let B′ be a copy of B.
To distinguish endvertices of B′ from those of B, put to the endvertices of B′ dashes. Now
take 2r−1 chains of 2t diamonds and identify the ends (the vertices of degree 2) of k-th
chain with ak and a′k, respectively. Finally, join the roots of B and B
′ (i.e., the vertices of
degree 2) by edges to z1 and z2.
u1
u2
z1
z2
v1
v2
Figure 5: The graph G3,2. Edges of binary trees of depth 1 are red.
Denote by Gt,r the resulting graph, see Figure 5 for G3,2. Then Gt,r has 8t2r−1 +
2(2r−1− 1)+8 vertices. Moreover, all central vertices of 2r−1 chains of diamonds belong
to the same orbit of Gt,r. Observe that there are 2r such vertices. Let u1 and u2 be central
vertices of one of the chains of diamonds. If we show that limt→∞(W (Gt,r − u1) −
W (Gt,r)) = ∞, we can complete Gt,r analogously as Gt was completed to H in the proof
of Theorem 2.1, to obtain a cubic 2−connected graph with at least 2r Šoltés vertices.
Thus, it remains to show that W (Gt,r − u1) − W (Gt,r) tends to infinity as t → ∞.
Observe that W (Gt,r − u1)−W (Gt,r) equals
∑
x,y∈V (Gt,r)\{u1}
(
dGt,r−u1(x, y)− dGt,r (x, y)
)
− wGt,r (u1).
We first estimate wGt,r (u1) from above. For small i, there are at most 4 vertices at
distance i from u1. For bigger i the amount of vertices at distance i grows, but it cannot
exceed 4 ·2r+8 since there are 2r−1 chains attached to F and F itself has 8 vertices. Thus,
16 Ars Math. Contemp. 25 (2025) #P2.01
wGt,r (u1) ≤ (2r+2 + 8)
∑1+3t+2r+3t+3
i=1 i. And if we held r constant, wGt,r (u1) can be
bounded from above by a quadratic polynomial in t.
Now we estimate
∑
x,y∈V (Gt,r)\{u1}
(
dGt,r−u1(x, y) − dGt,r (x, y)
)
from below. For
every x, y ∈ V (Gt,r)\{u1} we have dGt,r−u1(x, y) ≥ dGt,r (x, y), since Gt,r has all paths
which exist in Gt,r −u1. However, it suffices to consider only x, y being in the same chain
of diamonds as u1. Observe that the distance from u2 to a neighbour of u1 ( ̸= u2) is 2 in
Gt,r, but it is at least 6t in Gt,r − u1 (with equality if r = 1, i.e. if Gt,r = Gt). So this
distance is increased at least by 6t − 2. The distance from u2 to the second neighbour of
u1 is increased at least by 6t − 4, etc. However, we should consider also a neighbour of
u2 ( ̸= u1). For this vertex the distances are increased at least by 6t−4, 6t−6, . . . Summing
up,
D ≥
3t−1
∑
j=1
j
∑
i=1
2i = 2
3t−1
∑
j=1
(
j + 1
2
)
= 2
(
3t+ 1
3
)
.
Consequently, W (Gt,r−u1)−W (Gt,r) is bounded from below by a cubic polynomial
(in t) with leading coefficient 9. Thus, limt→∞(W (Gt,r − u1) − W (Gt,r)) = ∞ as
required.
4 Concluding remarks and further work
We believe that if there exists another Šoltés graph in addition to C11, it is likely to be
vertex-transitive or has a low number of vertex orbits. Vertices of the same orbit are either
all Šoltés vertices or none of them is.
Holt and Royle [5] have constructed a census of all vertex-transitive graphs with less
than 48 vertices; these graphs can be obtained from their Zenodo repository [14] in the
graph6 format [10]. The repository contains 100 720 391 graphs in total, 100 716 591 of
which are connected [12]. The computer search revealed that the only Šoltés graph among
them is the well-known C11.
We also examined the census of cubic vertex-transitive graphs by Potočnik, Spiga and
Verret [13]. Their census contains all (connected) cubic vertex-transitive graphs on up to
1280 vertices; there are 111 360 such graphs. CVT(n, i) denotes the i-th graph of order
n in the census. No Šoltés graph has been found, but the search revealed that there exist
graphs that are 13 -Šoltés, i.e.
1
3 of all vertices are Šoltés vertices. We found 7 cubic
1
3 -
Šoltés graphs; all of them are trunctations of certain cubic vertex-transitive graphs. In
this paper, the truncation of a graph G is denoted by Tr(G). Note that the truncation of
a vertex-transitive graph is not necessarily a vertex-transitive graph; in the case of cubic
graphs, there may be up to 3 vertex orbits. When doing the computer search, we have to
check the Šoltés property for one vertex from each orbit only. Here is the list of cubic
vertex-transitive graphs G, such that Tr(G) is a 13 -Šoltés graph:
CVT(384, 805), CVT(600, 259), CVT(768, 3650), CVT(1000, 302),
CVT(1056, 538), CVT(1056, 511), CVT(1280, 967).
Interestingly, all these graphs are Cayley graphs. Several properties of these graphs are
listed in the Appendix. The graph CVT(768, 3650) is the only non-bipartite example,
while the rest are bipartite. Girths of these graph are values from the set {4, 6, 8, 10, 12}.
We were able to identify seven such graphs. However, we believe that there could exist
many more.
N. Bašić et al.: On regular graphs with Šoltés vertices 17
Problem 4.1. Find an infinite family of cubic vertex-trainsitive graphs {Gi}∞i=1, such that
Tr(Gi) is a 13 -Šoltés graph for all i ≥ 1.
Moreover, we also found an example of a 4-regular 13 -Šoltés graph, namely the graph
L(CVT(324, 104)). It has order 486 and is the line graph of CVT(324, 104), which is a
Cayley graph. More data can be found in the Appendix.
Of course, 13 -Šoltés is still a long way from being Šoltés. The next conjecture is ad-
ditionally reinforced by the fact that there are no Šoltés graphs among vertex-transitive
graphs with less than 48 vertices.
Conjecture 4.2. The cycle on eleven vertices, C11, is the only Šoltés graph.
ORCID iDs
Nino Bašić https://orcid.org/0000-0002-6555-8668
Martin Knor https://orcid.org/0000-0003-3555-3994
Riste Škrekovski https://orcid.org/0000-0001-6851-3214
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N. Bašić et al.: On regular graphs with Šoltés vertices 19
A Appendix
There are 7 cubic vertex-transitive graphs G on up to 1280 vertices, such that Tr(G) is a
1
3 -Šoltés graph. Since all these graph are Cayley graphs, we give the generating set for the
Cayley graph. Note that the group itself (its permutation representation) is given implicitly
by these generators; however, we also give the group’s ID from GAP’s library of small
groups [2]. SmallGroup(n, k) is the k-th group of order n from that library. We also
calculated girth, diameter and tested all graphs for bipartiteness.
• CVT(384, 805):
Group([(2, 4)(6, 12, 17, 19, 18, 14)(7, 10, 20)(8, 15, 16, 9, 11, 13),
(1, 4)(2, 3)(5, 13)(6, 18)(7, 17)(8, 15)(9, 12)(10, 14)(11, 19)(16, 20)])
∼= SmallGroup(384, 5781)
girth: 6; diameter: 10; bipartite? True
• CVT(600, 259):
Group([(1, 2)(3, 4)(5, 6)(7, 9)(8, 10)(13, 14)(16, 17),
(1, 3)(5, 7)(6, 8)(9, 12)(10, 13)(11, 14)(15, 16),
(1, 4)(2, 3)(5, 6)(8, 11)(9, 10)(13, 14)(15, 17)])
∼= SmallGroup(600, 103)
girth: 10; diameter: 13; bipartite? True
• CVT(768, 3650):
Group([(2, 3)(4, 7)(5, 6)(10, 12),
(2, 4)(3, 5)(6, 7)(9, 10)(11, 12),
(1, 2)(3, 6)(4, 8)(5, 7)])
∼= SmallGroup(768, 1090104)
girth: 8; diameter: 16; bipartite? False
• CVT(1000, 302):
Group([(4, 5)(6, 7)(9, 14)(10, 18)(11, 21)(12, 23)(13, 25)(15, 24)(16, 27)(17, 29)
(19, 26)(20, 31)(22, 28)(30, 32),
(1, 2)(3, 4)(5, 6)(8, 9)(10, 15)(11, 19)(12, 18)(13, 21)(14, 23)(16, 17)
(20, 27)(22, 29)(24, 31)(25, 32)(26, 28),
(1, 2)(8, 17)(9, 27)(10, 11)(12, 26)(13, 24)(14, 22)(15, 18)(16, 30)(20, 32)
(21, 28)(23, 31)(25, 29)])
∼= SmallGroup(1000, 105)
girth: 10; diameter: 15; bipartite? True
20 Ars Math. Contemp. 25 (2025) #P2.01
• CVT(1056, 538):
Group([(2, 3)(4, 5)(6, 8)(7, 10)(9, 11)(13, 14)(15, 16)(17, 18)(19, 20)(21, 22),
(1, 2)(4, 6)(7, 9)(12, 13)(14, 15)(16, 17)(18, 19)(20, 21),
(4, 7)(6, 9)(10, 11)(12, 13)(14, 15)(16, 17)(18, 19)(20, 21)])
∼= SmallGroup(1056, 493)
girth: 4; diameter: 22; bipartite? True
• CVT(1056, 511):
Group([(2, 3)(12, 13)(14, 15)(16, 17)(18, 19)(20, 21),
(1, 2)(4, 5)(6, 7)(8, 9)(10, 11)(12, 14)(15, 16)(17, 18)(19, 20)(21, 22),
(5, 6)(7, 8)(9, 10)(12, 13)(14, 15)(16, 17)(18, 19)(20, 21)])
∼= SmallGroup(1056, 468)
girth: 4; diameter: 22; bipartite? True
• CVT(1280, 967):
Group([(1, 2)(3, 5)(7, 9)(8, 11)(10, 15)(12, 19)(13, 20)(14, 22)(16, 25)(17, 27)
(18, 29)(21, 32)(23, 34)(24, 30)(26, 31)(28, 37),
(1, 3)(4, 5)(6, 7)(8, 12)(9, 13)(10, 16)(11, 17)(15, 23)(18, 30)(20, 31)
(21, 27)(22, 25)(24, 28)(26, 36)(29, 33)(35, 37),
(1, 4)(2, 3)(6, 8)(7, 10)(9, 14)(11, 18)(12, 20)(13, 21)(15, 24)(16, 26)
(17, 28)(19, 29)(22, 33)(23, 35)(25, 30)(27, 36)(31, 34)(32, 37)])
∼= SmallGroup(1280, 81752)
girth: 12; diameter: 16; bipartite? True
There exists one cubic vertex-transitive graph G on up to 1280 vertices, such that L(G) is
a 13 -Šoltés graph.
• CVT(324, 104):
Group([(2, 9)(3, 4)(5, 7)(6, 8),
(1, 5)(2, 6)(3, 9)(4, 7),
(2, 9)(3, 6)(4, 8)])
∼= SmallGroup(324, 39)
girth: 4; diameter: 12; bipartite: True
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.02
https://doi.org/10.26493/1855-3974.3205.3d6
(Also available at http://amc-journal.eu)
Indicated total domination game*
Michael A. Henning †
Department of Mathematics and Applied Mathematics, University of Johannesburg,
Auckland Park, 2006 South Africa
Douglas F. Rall
Professor Emeritus of Mathematics, Furman University, Greenville, SC, USA
Received 23 August 2023, accepted 21 March 2024, published online 10 March 2025
Abstract
A vertex u in a graph G totally dominates a vertex v if u is adjacent to v in G. A
total dominating set of G is a set S of vertices of G such that every vertex of G is totally
dominated by a vertex in S. The indicated total domination game is played on a graph
G by two players, Dominator and Staller, who take turns making a move. In each of his
moves, Dominator indicates a vertex v of the graph that has not been totally dominated in
the previous moves, and Staller chooses (or selects) any vertex adjacent to v that has not yet
been played, and adds it to a set D that is being built during the game. The game ends when
every vertex is totally dominated, that is, when D is a total dominating set of G. The goal
of Dominator is to minimize the size of D, while Staller wants just the opposite. Providing
that both players are playing optimally with respect to their goals, the size of the resulting
set D is the indicated total domination number of G, denoted by γit(G). In this paper we
present several results on indicated total domination game. Among other results we prove
that the indicated total domination number of a graph is bounded below by the well studied
upper total domination number.
Keywords: Total domination game, indicated total domination game.
Math. Subj. Class. (2020): 05C65, 05C69
*The authors wish to thank the reviewers for their careful and thorough reading of the paper, and helpful
comments and suggestions that led to an improved version of the paper.
†Corresponding author. Research of the first author was supported in part by the South African National
Research Foundation, Grant Numbers 132588 and 129265, and the University of Johannesburg.
E-mail addresses: mahenning@uj.ac.za (Michael A. Henning), doug.rall@furman.edu (Douglas F. Rall)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.02
1 Introduction
In 2010 Brešar, Klavžar, and Rall [4] published the seminal paper on the domination game
which belongs to the growing family of competitive optimization graph games. Domination
games played on graphs are now very well studied in the literature. The subsequent rapid
growth by the scientific community of research on domination games played on graphs in-
spired the recent book entitled “Domination games played on graphs” by Brešar, Henning,
Klavžar, and Rall [2], which presented the state of the art results at the time and shows that
the area is rich for further research. In this paper, we study the total version of the indicated
domination game.
A neighbor of a vertex v in G is a vertex that is adjacent to v. The open neighborhood of
v in G is the set of neighbors of v, denoted NG(v). Thus, NG(v) = {u ∈ V : uv ∈ E(G)}.
The closed neighborhood of v is the set NG[v] = {v} ∪NG(v). For a set of vertices S ⊆
V (G) and a vertex v belonging to the set S, the open S-private neighborhood is defined
by pn(v, S) = {w ∈ V : NG(w) ∩ S = {v}}. The S-external private neighborhood of
v is the set epn(v, S) = pn(v, S) \ S, and the open S-internal private neighborhood is
the set ipn(v, S) = pn(v, S) ∩ S. We note that pn(v, S) = epn(v, S) ∪ ipn(v, S). The
closed neighborhood of a set S ⊆ V (G) is the set ∪NG[v] where the union is taken over
all vertices v ∈ S.
A vertex of degree 1 in a graph is called a leaf and its neighbor a support vertex. An
isolated vertex in G is a vertex of degree 0 in G. An isolate-free graph is a graph which
contains no isolated vertex. A trivial graph is the graph of order 1, and a nontrivial graph
has order at least 2. The join of two graphs G and H , denoted G⊕H , is constructed from
their disjoint union by adding edges making every vertex in G adjacent to every vertex in
H . The subgraph of G induced by a set S ⊆ V (G) is denoted by G[S]. For an integer
k ≥ 1, we let [k] = {1, . . . , k} and [k]0 = {0, 1, . . . , k}.
A vertex u in a graph G dominates a vertex v if u = v or u is adjacent to v in G. A
dominating set of G is a set S of vertices of G such that every vertex in G is dominated by
a vertex in S. A vertex u in a graph G totally dominates a vertex v if u is adjacent to v in G.
A total dominating set, abbreviated TD-set, of G is a set S of vertices of G such that every
vertex of G is totally dominated by a vertex in S, that is, every vertex in G has a neighbor in
S. The total domination number of G, denoted γt(G), is the minimum cardinality of a TD-
set in G, while the upper total domination number of G, denoted Γt(G), is the maximum
cardinality of a minimal TD-set in G. A minimal TD-set of cardinality Γt(G) we call a
Γt-set of G. For other graph theory terminology not defined herein, the reader is referred to
[15], and for other recent books on domination in graphs, we refer the reader to [13, 14, 18].
1.1 Domination games in graphs
The domination game, as introduced in [4], is played on a graph G by two players: Domi-
nator and Staller. They alternate taking moves in which they select a vertex of G. A move
is legal if the selected vertex dominates at least one vertex which is not already dominated
by previously played vertices. The game ends when there are no legal moves, so when
the set of played vertices is a dominating set of G. The goal of Dominator is to finish the
game with minimum number of moves, while the aim of Staller is to maximize the number
of moves. If both players play optimally, then the number of moves played on G is the
invariant (see [2, 5]) the game domination number γg(G), which is the number of moves
M. A. Henning and D. F. Rall: Indicated total domination game 3
on G if Dominator starts the game. We refer to the domination game when Dominator
makes the first move in the game as the D-game (standing for ‘Dominator start game’).
The total domination game was introduced in [16] as follows. Given a graph G, two
players, called Dominator and Staller, take turns choosing a vertex from G. Each vertex
chosen must totally dominate at least one vertex not totally dominated by the set of vertices
previously chosen. Such a chosen vertex is called a legal move. The game ends when there
is no legal move available. Dominator wishes to minimize the number of vertices selected,
while the goal of Staller is just the opposite. The game total domination number, γtg(G),
of G is the number of moves played on G when both players play optimally and Dominator
starts the game. As before, we refer to the total domination game when Dominator makes
the first move in the game as the D-game. The total domination game is now extensively
studied in the literature; see, for example, [2, 17].
A sequence of vertices in a graph G is a total dominating sequence if every vertex v
in the sequence totally dominates at least one vertex that was not totally dominated by any
vertex that precedes v in the sequence, and at the end all vertices of G are totally dominated.
While the length of a shortest such sequence is the total domination number γt(G) of G,
a total dominating sequence of maximum length is the Grundy total domination number,
γtgr(G), of G.
1.2 Open-open irredundant sets and induced matchings
Fellows, Fricke, Hedetniemi, and Jacobs [10] unified the study of private neighbors of
vertices by forming a cohesive theory of private neighbors in graphs, yielding the so-called
private neighbor cube. We mention here two such parameters defined in [10].
If each vertex in a set S of vertices in a graph G has a private neighbor inside S other
than itself, then the subgraph G[S] induced by the set S consists of disjoint copies of the
complete graph K2 on two vertices, that is, G[S] is a 1-regular subgraph of G. The set
of edges in such an induced subgraph is called an induced matching by Cameron [6] and
Faudree, Gyárfás, Schelp, and Tuza [9] in 1989, and a strong matching by Golumbic and
Laskar [11] in 1993. The maximum order of an induced matching in G is the induced
matching number, denoted α′ind(G), of G. Induced matchings in graphs are now very well
studied in the literature.
If each vertex in a set S of vertices in a graph G has a private neighbor other than itself,
either inside S or outside S, then such a set is coined an open-open irredundant set in [10].
Such a set satisfies the condition that for every vertex v ∈ S, the set N(v) \N(S \ {v}) is
not empty. The open-open irredundance number, denoted OOIR(G), of G is the maximum
cardinality of an open-open irredundant set in G.
1.3 Indicated total domination game
Grzesik [12] proposed and first studied the indicated coloring game which is a combinato-
rial game played on a graph G by two players, and a fixed set C of colors. In each round
of the game Ann indicates an uncolored vertex, and Ben colors it using a color from C,
obeying just the proper coloring rule. The goal of Ann is to achieve a proper coloring of
the whole graph, while Ben is trying to prevent this. The minimum cardinality of the set of
colors C for which Ann has a winning strategy is called the indicated chromatic number,
χi(G), of a graph G.
4 Ars Math. Contemp. 25 (2025) #P2.02
Recently, Brešar, Bujtás, Iršič, Rall and Tuza [1] introduced and studied the indicated
domination game inspired by the indicated coloring game. The indicated domination game
is played on a graph G by two players, Dominator and Staller, who take turns making
a move. In each of his moves, Dominator indicates a vertex v of the graph that has not
been dominated in the previous moves, and Staller chooses (or selects) any vertex from
the closed neighborhood of v that has not yet been played, and adds it to a set D that is
being built during the game. The game ends when there is no undominated vertex left,
that is, when D is a dominating set. The goal of Dominator is to minimize the size of D,
while Staller wants just the opposite. Providing that both players are playing optimally with
respect to their goals, the size of the resulting set D is the indicated domination number of
G, and is denoted by γi(G).
In this paper, we study the total version of the indicated domination game. The indi-
cated total domination game is played on an isolate-free graph G by two players, Domina-
tor and Staller, who take turns making a move. In each of his moves, Dominator indicates a
vertex v of the graph that has not been totally dominated in the previous moves, and Staller
chooses (or selects) any vertex from the open neighborhood of v, and adds it to a set D that
is being built during the game. The game ends when every vertex is totally dominated, that
is, when D is a TD-set. The goal of Dominator is to minimize the size of D, while Staller
wants just the opposite. Providing that both players are playing optimally with respect to
their goals, the size of the resulting set D is the indicated total domination number of G,
and is denoted by γit(G). Upon completion of the game, the set D of vertices chosen by
Staller is a TD-set of G (possibly, D = V (G)). By considering the game tree (which we
do not define here) for the indicated total domination game, identical arguments as in [2,
Chapter 1.2] show that the indicated total domination number is well-defined. Throughout
this paper we only consider isolate-free graphs.
2 The continuation principle
A partially total dominated graph is a graph together with a declaration that some vertices
are already totally dominated. Since Dominator only indicates vertices of the graph that
have not yet been totally dominated, he is not permitted to indicate any vertices that are
declared already totally dominated. Adopting the notation in [17], given a graph G and
a subset S of vertices of G, we denote by G|S the partially total dominated graph with
S as the set of declared vertices already totally dominated. Thus, γit(G|S) is the number
of moves needed to finish the game on G|S when Dominator and Staller both are playing
optimally.
A key lemma in both the domination game and the total domination game (and other
variants of certain domination games played on graphs) is the so-called Continuation Prin-
ciple first presented by Kinnersley, West, and Zamani in [19]. The Continuation Principle
is a powerful tool for obtaining good upper bounds on game domination parameters.
We show that the Continuation Principle unfortunately does not hold for the indicated
total domination game. That is, we show that if G is a graph and A,B ⊆ V (G) with
B ⊆ A, then it is not necessarily true that γit(G|A) ≤ γ
i
t(G|B). Perhaps the simplest
counterexample is to let G be the graph obtained from a 6-cycle v1v2 . . . v6v1 by adding
two new vertices u1 and u4, and adding the edges u1v1, u4v4 and v1v4. The graph G is
illustrated in Figure 1.
M. A. Henning and D. F. Rall: Indicated total domination game 5
v2 v3
v5v6
v1u1 v4 u4
Figure 1: A graph G.
In the graph G shown in Figure 1, let B = ∅ and let A = {u1, u4}. Suppose the
indicated total domination game is played on G|B. We note that G|B = G, and so the
game is played on the graph G. In this game Dominator would first indicate vertex u1,
and Staller would be compelled to select vertex v1. At this point the vertices in the set
{u1, v2, v4, v6} are totally dominated. Dominator as his second move indicates vertex u4
to which Staller must choose vertex v4 and the game is complete. Thus, γ
i
t(G|B) ≤ 2.
Since γit(G|B) ≥ γt(G) = 2, we infer that γ
i
t(G|B) = 2.
Now let the indicated total domination game be played on G|A. If Dominator indicates
vertex v1, then Staller can choose vertex u1. This leaves vertices v2, v3, v4, v5 and v6 to be
totally dominated, and that will require at least two more choices by Staller. If Dominator
indicates vertex v4, then by symmetry Staller can guarantee that at least three vertices are
played (starting with the vertex u4). If Dominator indicates vertex v2, then Staller can
choose vertex v1. This leaves vertices v1, v3 and v5 to be totally dominated, and Staller can
guarantee that at least two additional vertices are played. All other choices for Dominator to
indicate follow from symmetry. Therefore, γit(G|A) ≥ 3. As observed earlier, γ
i
t(G|B) =
2. Hence, γit(G|B) < γ
i
t(G|A).
Since the Continuation Principle does not hold for the indicated total domination game,
this indicates that obtaining good upper bounds on the indicated total domination number
is likely to be challenging.
3 The upper total domination number
In this section we show that the indicated total domination number of an isolate-free graph
is at least the upper total domination of the graph. A fundamental property of minimal
TD-sets was established by Cockayne, Dawes, and Hedetniemi [7] in 1980.
Lemma 3.1 ([7]). A TD-set S in a graph G is a minimal TD-set if and only if every vertex
v ∈ S has an open S-external private neighbor or an open S-internal private neighbor,
that is, if and only if |epn(v, S)| ≥ 1 or |ipn(v, S)| ≥ 1.
As an application of Lemma 3.1, we can prove that the indicated total domination
number of a graph is at least its upper total domination number.
Proposition 3.2. If G is an isolate-free graph, then Γt(G) ≤ γ
i
t(G).
6 Ars Math. Contemp. 25 (2025) #P2.02
Proof. Let S be a Γt-set of G, and so S is a minimal TD-set of G of cardinality Γt(G). By
Lemma 3.1, |epn(v, S)| ≥ 1 or |ipn(v, S)| ≥ 1 for every vertex v ∈ S, and so, pn(v, S) =
epn(v, S) ∪ ipn(v, S) ̸= ∅. Let p = Γt(G) and define a partition (V1, V2, . . . , Vp) of
the vertex set V (G) into p sets as follows. For i ∈ [p], let pn(vi, S) ⊆ Vi ⊆ NG(vi).
Staller’s strategy is to always select a vertex from S to totally dominate the vertex indicated
by Dominator. More precisely, whenever Dominator indicates a vertex v to be totally
dominated, Staller identifies the set Vi that contains the vertex v for some i ∈ [p] and
selects the vertex vi ∈ S. Since pn(w, S) ̸= ∅ for every vertex w ∈ S, this guarantees
that upon completion of the game Staller selects all vertices in the set S, implying that
γit(G) ≥ |S| = Γt(G).
The upper total domination number of a path Pn of order n is established in [8].
Proposition 3.3 ([8]). For n ≥ 2 an integer, Γt(Pn) = 2⌊
n+1
3
⌋.
We determine next the indicated total domination number of a path Pn of order n,
and give a strategy for Dominator to play on a path. Recall that for k ≥ 1, we let [k] =
{1, . . . , k} and [k]0 = {0} ∪ [k].
Theorem 3.4. For n ≥ 2, γit(Pn) = Γt(Pn).
Proof. Let T be a path v1v2 . . . vn of order n ≥ 2. If n = 2, then it is immediate that
γit(T ) = Γt(T ) = 2. Hence we may assume that n ≥ 3. By Proposition 3.2, γ
i
t(T ) ≥
Γt(T ). Hence it suffices for us to prove that γ
i
t(T ) ≤ Γt(T ) from which we infer that
γit(Pn) = Γt(Pn). We consider three cases.
Case 1. n ≡ 0 (mod 3). Thus, n = 3k for some k ≥ 1. Dominator’s strategy is
to indicate on his (i + 1)st move the vertex v3i+1 for i ∈ [k − 1]0. Thus on his first k
moves, Dominator indicates the vertices v1, v4, . . . , v3k−2 in turn. This forces Staller to
play the vertex v2 on her first move, and if k ≥ 2, then she is required to play either the
vertex v3i or v3i+2 on her (i+ 1)st move for all i ∈ [k − 1]. On each of her moves, Staller
totally dominates two new vertices. Hence after the first k moves of Staller, exactly 2k
vertices on the path T are totally dominated. For the remaining k vertices on T that are not
yet totally dominated, Dominator simply indicates each such vertex on consecutive moves,
thus guaranteeing that the game is complete after at most k additional moves of Staller.
Thus, the game is finished in at most 2k moves, and so γit(T ) ≤ 2k = 2⌊
n+1
3
⌋ = Γt(Pn).
Case 2. n ≡ 1 (mod 3). Thus, n = 3k + 1 for some k ≥ 1. Dominator’s strategy
is to indicate on his (i + 1)st move the vertex v3i+1 for i ∈ [k]0. Thus on his first k + 1
moves, Dominator indicates the vertices v1, v4, . . . , v3k+1 in turn. This forces Staller to
play the vertex v2 on her first move, the vertex v3i or v3i+2 on her (i + 1)st move for all
i ∈ [k − 1], and the vertex v3k on her (k + 1)st move. On each of her moves, Staller
totally dominates two new vertices. Hence after the first k + 1 moves of Staller, exactly
2(k + 1) vertices on the path T are totally dominated. For the remaining k − 1 vertices on
T that are not yet totally dominated, Dominator indicates each such vertex on consecutive
moves, thus guaranteeing that the game is complete after at most k − 1 additional moves
of Staller. Thus, the game is finished in at most (k + 1) + (k − 1) = 2k moves, and so
γit(T ) ≤ 2k = 2⌊
n+1
3
⌋ = Γt(Pn).
M. A. Henning and D. F. Rall: Indicated total domination game 7
Case 3. n ≡ 2 (mod 3). Thus, n = 3k + 2 for some k ≥ 1. Dominator’s strategy is to
indicate on his ith move the vertex v3i−2 for i ∈ [k]. Thus on his first k moves, Dominator
indicates the vertices v1, v4, . . . , v3k−2 in turn. This forces Staller to play the vertex v2 on
her first move, and to play the vertex v3i−3 or v3i−1 on her ith move for i ∈ [k] \ {1}. On
each of her moves, Staller totally dominates two new vertices. Hence after the first k moves
of Staller, exactly 2k vertices on the path T are totally dominated. For the remaining k+2
vertices on T that are not yet totally dominated, Dominator indicates each such vertex
on consecutive moves, thus guaranteeing that the game is complete after at most k + 2
additional moves of Staller. Thus, the game is finished in at most k + (k + 2) = 2k + 2
moves, and so γit(T ) ≤ 2k + 2 = 2⌊
n+1
3
⌋ = Γt(Pn).
A natural problem is to extend the path result in Theorem 3.4 and to determine if the
indicated total domination number is equal to the upper total domination in the class of
trees. We state this formally as follows.
Question 3.5. Is it true that if T is a nontrivial tree, then Γt(T ) = γ
i
t(T )?
The following result gives a partial result in this direction.
Proposition 3.6. If T is a nontrivial tree in which every vertex is a leaf or a support vertex,
then γit(T ) = Γt(T ).
Proof. Let T be a tree of order n ≥ 2 in which every vertex is a leaf or a support vertex.
If n = 2, then the result is immediate. Hence we may assume that n ≥ 3. If T is a
star K1,n−1, then it is straightforward to check that Γt(T ) = γ
i
t(T ) = 2. Hence, we may
assume that T has diameter at least 3, implying that T contains at least two support vertices.
Every TD-set of T contains the set of support vertices of T in order to totally dominate all
leaves in T . Moreover, every minimal TD-set of T is unique and consists of the support
vertices of T , implying that Γt(T ) = s where here s ≥ 2 denotes the number of support
vertices in T . We show next that γit(T ) = s. Let v1, v2, . . . , vs be the support vertices in
T , and let ui be an arbitrary leaf neighbor of vi for i ∈ [s]. On Dominator’s ith move, he
indicates the leaf ui for i ∈ [s]. In order to totally dominate the leaf ui, Staller is required
to play the vertex vi on her ith move. The resulting set of s vertices played by Staller is
the set of support vertices in T , which is a TD-set of T and completes the game. Hence,
Dominator can guarantee that the indicated total domination game finishes in at most s
moves, that is, γit(T ) ≤ s = Γt(T ). By Proposition 3.2, Γt(G) ≤ γ
i
t(G). Consequently,
Γt(G) = γ
i
t(G).
We show next that there exists connected graphs G satisfying Γt(G) < γ
i
t(G). For this
purpose, for a given graph G and an integer k ≥ 2, the kth-power graph of G, denoted
Gk, is the graph with the same vertex set as G and where two vertices u and v are adjacent
in Gk if dG(u, v) ≤ k. We show in the following result that the kth power of a cycle of
order 2k + 3 has upper total domination number less than its indicated total domination
number.
Proposition 3.7. If k ≥ 2 and G = Ck2k+3, then Γt(G) < γ
i
t(G).
8 Ars Math. Contemp. 25 (2025) #P2.02
Proof. Let C be the cycle C2k+3 with V (C) = {v1, . . . , v2k+3} and E(C) = {vivi+1 : i ∈
[2k+3]} where the indices are computed modulo 2k+3, and consider the kth power G =
Ck of the cycle C. First we claim that no minimal TD-set of G contains two consecutive
vertices from the original cycle C. Without loss of generality suppose, to the contrary, that
{v1, v2} is a subset of a minimal TD-set A of G. We note that NG({v1, v2}) = V (G) \
{vk+3}. This implies that A contains a neighbor, vj , of vk+3 in G. Thus, {v1, v2, vj} ⊆ A,
where 3 ≤ j ≤ 2k + 3 and j ̸= k + 3. If j = k + 4, then since dC(v1, vk+4) = k and
dC(v2, vk+4) = k + 1, we infer that N(v2) \ N(A \ {v2}) = ∅, and so epn(v2, A) =
ipn(v2, A) = ∅. If j = k + 2, then since dC(v1, vk+2) = k + 1 and dC(v2, vk+2) = k,
we infer that N(v1) \N(A \ {v1}) = ∅, and so epn(v1, A) = ipn(v1, A) = ∅. If k + 5 ≤
j ≤ 2k + 3, then N(v1) \ N(A \ {v1}) = ∅, and so epn(v1, A) = ipn(v1, A) = ∅. If
3 ≤ j ≤ k + 1, then N(v2) \N(A \ {v2}) = ∅, and so epn(v2, A) = ipn(v2, A) = ∅. In
all cases, we contradict the minimality of the TD-set A as given by Lemma 3.1. Hence, no
minimal TD-set of G contains two consecutive vertices from the original cycle C. However
if 1 ≤ i < j ≤ 2k + 3 and k ≥ dC(vi, vj) > 1, then {vi, vj} is a minimal TD-set
of G = Ck and, in fact, every minimal TD-set of G is formed this way. We infer that
Γt(G) = 2.
When the indicated total domination game is played on G we may assume by symmetry
that v1 is the first vertex chosen by Staller. The vertices v1, vk+2 and vk+3 are the only three
vertices in G not totally dominated by the vertex v1. If Dominator now indicates v1 or vk+2
on his second move, then Staller can select v2 which leaves vk+3 not totally dominated.
On the other hand, if Dominator now indicates vk+3 on his second move, then Staller can
choose v2k+3, which leaves vk+2 not totally dominated. Therefore, Staller can force at least
three vertices to be played upon completion of the game, implying that γit(G) = 3.
By Proposition 3.7, there exists connected graphs G of arbitrarily large minimum de-
gree satisfying Γt(G) < γ
i
t(G). We show in the next section that there exist connected
graphs G such that γit(G) can be arbitrarily larger than Γt(G).
4 The open-open irredundance number
Let G be an isolate-free graph. From the definition of an open-open irredundant set, if
S is a minimal TD-set in G, then S is an open-open irredundant set. In particular, every
Γt-set of G is an open-open irredundant set of G. Hence as observed in [10], it holds that
Γt(G) ≤ OOIR(G). By Proposition 3.2, we have Γt(G) ≤ γ
i
t(G). Hence it is a natural
question to ask whether γit(G) ≤ OOIR(G). We show in this section that in general there
is no relation between the indicated total domination number, γit(G), of a graph G and the
open-open irredundance number, OOIR(G), of G.
Let G be the family of graphs constructed as follows. Let H be the graph obtained
from the join of two vertex disjoint copies Q1 : u1v1w1x1 and Q2 : u2v2w2x2 of a path
P4 of order 4, and so H = P4 ⊕ P4. For k ≥ 1, let H1, . . . , Hk be k vertex disjoint
copies of H . Further, let Qi,1 : ui,1vi,1wi,1xi,1 and Qi,2 : ui,2vi,2wi,2xi,2 be the paths in
Hi corresponding to the paths Q1 and Q2 in H for i ∈ [k]. Let G1 = H1, and for k ≥ 2
let Gk be obtained from the disjoint union of the graphs H1, . . . , Hk by adding the edges
wi,2wi+1,1 for all i ∈ [k] where addition is taken modulo k. The graph Gk is illustrated in
Figure 2. Let G = {Gk : k ≥ 1}.
M. A. Henning and D. F. Rall: Indicated total domination game 9
x1,1 x2,1 x3,1 xk,1x1,2 x2,2 x3,2 xk,2
w1,1 w2,1 w3,1 wk,1w1,2 w2,2 w3,2 wk,2
v1,1 v2,1 v3,1 vk,1v1,2 v2,2 v3,2 vk,2
u1,1 u2,1 u3,1 uk,1u1,2 u2,2 u3,2 uk,2
Figure 2: The graph Gk in the family G.
Proposition 4.1. For k ≥ 1, we have γit(Gk) = 3k, Γt(Gk) = 2k and OOIR(Gk) = 2k.
Proof. For k ≥ 1, consider the graph G = Gk ∈ G. Let the indicated total domination
game be played on G. We provide a strategy for Staller that will ensure at least 3k vertices
are chosen. By symmetry we may assume that in his first move Dominator indicates a ver-
tex that belongs to the path Q1,1. Staller responds by playing x1,2, which totally dominates
all vertices on the path Q1,1 and the vertex w1,2. Furthermore, later in the game Staller will
only play vertices from the path Q1,2 when Dominator indicates a vertex in Q1,2 that is not
yet totally dominated. This implies that at least three vertices from Q1,2 will be played by
Staller since both v1,2 and w1,2 are support vertices in the path Q1,2.
As play progresses and Dominator first indicates a vertex in Hj for some j ∈ [k] \ {1},
Staller will choose a vertex in a manner similar to how she responded to Dominator’s first
play. That is, if Dominator indicates a vertex that belongs to the path Qj,1, then Staller
plays the vertex xj,2. In this case when Dominator indicates any further vertices in Qj,2,
Staller will choose a vertex that belongs to the path Qj,2. On the other hand, if the first
vertex Dominator indicates from Hj is a vertex that belongs to the path Qj,2, then Staller
chooses the vertex xj,1 and ensures that two more additional vertices from Qj,1 are chosen
in the remainder of the game. This strategy ensures that at least three vertices will be
chosen by Staller in each of the k copies of H in the graph Gk. Dominator can prevent
more than three vertices being chosen from Hj , for each j ∈ [k], by playing as follows.
The first vertex he indicates in Hj should be xj,1. A short analysis shows that it is then to
Staller’s advantage to select either xj,2 or uj,2. By indicating xj,2 and then uj,2 on his next
two moves Dominator ensures that at most three vertices will be selected from Hj . We
therefore infer that Staller has a strategy to play exactly three vertices from every copy of
H = P4 ⊕ P4 in Gk, implying that γ
i
t(Gk) = 3k.
Let D be a Γt-set of Gk, and so D is a minimal TD-set of Gk of maximum cardinality.
Let Di = D ∩ V (Hi) for all i ∈ [k]. We show that |Di| ≤ 2. If Di contains a vertex
from Qi,1 and a vertex from Qi,2, then two such vertices form a set that totally dominates
all vertices of Hi, and by the minimality of the set D we infer that |Di| = 2. If Di con-
tains vertices from exactly one of Qi,1 and Qi,2, say from Qi,1, then in order to totally
dominate the vertices ui,1 and xi,1 the set Di contains the vertices vi,1 and wi,1, respec-
tively. However, the set {vi,1, wi,1} totally dominates all vertices of Hi, and once again
by the minimality of the set D we infer that |Di| = 2. This is true for all i ∈ [k], and so
Γt(Gk) ≤ 2k. Since the set {x1,1, x1,2, . . . , xk,1, xk,2}, for example, is a minimal TD-set
of Gk, we note that Γt(Gk) ≥ 2k. Consequently, Γt(Gk) = 2k. Similarly, any open-open
10 Ars Math. Contemp. 25 (2025) #P2.02
irredundant set in Gk can contain at most two vertices from Hi for each i ∈ [k], and so
OOIR(Gk) ≤ 2k. As observed earlier, 2k = Γt(G) ≤ OOIR(G) ≤ 2k. Consequently,
OOIR(G) = 2k.
By Proposition 4.1, there exist connected graphs G such that γit(G) can be arbitrarily
larger than Γt(G) and OOIR(G). We show next that there exist connected graphs G such
that OOIR(G) can be arbitrarily larger than γit(G). Let k be a positive integer at least 5,
and let Fk,1 and Fk,2 be two disjoint copies of the complete graph Kk, where V (Fk,1) =
{u1, u2, . . . , uk} and V (Fk,2) = {v1, v2, . . . , vk}. Let Fk be obtained from the disjoint
union of Fk,1 and Fk,2 by adding the k − 1 edges uivi for i ∈ [k − 1]. Thus, Fk is
obtained from the prism Kk □K2 of a complete graph Kk by removing one of the added
edges between a pair of corresponding vertices in the copies of the complete graph, that is,
Fk ∼= (Kk □K2) − ukvk. The graph Fk is illustrated in Figure 3, where for clarity we
omit the edges in the complete graphs Fk,1 and Fk,2.
u1 u2 u3 uk−1 uk
v1 v2 v3 vk−1 vk
Kk
Kk
Figure 3: The graph Fk = (Kk □K2)− ukvk.
Proposition 4.2. For k ≥ 5, we have γit(Fk) = 4 and OOIR(Fk) = k − 1.
Proof. The set {u1, u2, . . . , uk−1} is an open-open irredundant set of Fk, which implies
that OOIR(Fn) ≥ k − 1. Suppose A is an open-open irredundant subset of V (Fn) such
that |A| ≥ k. If |A∩{u1, u2, . . . , uk}| ≥ 1 and |A∩{v1, v2, . . . , vk}| ≥ 1, then neither of
these intersections has cardinality more than 2 and thus |A| ≤ 4, which is a contradiction.
We may thus assume that A = {u1, u2, . . . , uk}. This is also a contradiction since N(uk)\
N(A \ {uk}) = ∅. Therefore, OOIR(Fk) = k − 1.
We show next that γit(Fk) = 4. Let the indicated total domination game be played
on the graph Fk. Suppose that Dominator indicates uk on his first move. Staller can only
choose a vertex from {u1, u2, . . . , uk−1} to totally dominate uk. Without loss of general-
ity we may assume Staller plays u1, and this vertex totally dominates {v1, u2, u3, . . . , uk}.
Suppose that Dominator indicates vk on his second move. Staller could choose any vertex
from {v1, v2, . . . , vk−1} to totally dominate vk. If she chooses v1, then the game ends. Any
other choice by Staller, say vj where 2 ≤ j ≤ k − 1, leaves just two vertices, namely, u1
and vj , not totally dominated. Thus, at most two more vertices can be chosen by Staller.
Regardless of which of these two vertices are indicated by Dominator on his third move,
Staller can choose the appropriate vertex from {uk, vk} and thus make sure the game ends
with four vertices being chosen. Suppose next that Dominator indicates vj on his second
move, where 2 ≤ j ≤ k−1. In this case, Staller chooses the vertex vk, thereby totally dom-
inating all vertices except for u1 and vk. If Dominator indicates u1, then Staller chooses the
vertex uk, while if Dominator indicates vk, then Staller chooses the vertex v2. Regardless
of what vertex Dominator indicates, Staller can force four vertices to be chosen before the
game ends.
M. A. Henning and D. F. Rall: Indicated total domination game 11
Suppose that Dominator indicates a vertex different from uk and vk. By symmetry
and renaming vertices if necessary, we may assume that Dominator indicates vertex u1.
In this case, Staller chooses the vertex uk, thereby totally dominating vertices in the set
{u1, u2, . . . , uk−1}. Dominator must then indicate a vertex from the set {v1, v2, . . . , vk, uk}
on his second move. However, regardless of what vertex Dominator indicates, Staller can
force four vertices to be chosen before the game ends. Therefore, γit(Fk) = 4.
By Proposition 4.2, there exist connected graphs G such that OOIR(G) can be arbi-
trarily larger than γit(G).
5 The induced matching number
The following relation between the open-open irredundance number and the induced match-
ing number is given in [11].
Theorem 5.1 ([11]). If G is an isolate-free graph, then 2α′ind(G) ≤ OOIR(G). Moreover
if G is a bipartite graph, then 2α′ind(G) = OOIR(G).
As observed earlier, Γt(G) ≤ OOIR(G) holds for all isolate-free graphs G. Hence
as a consequence of Theorem 5.1, if G is an isolate-free bipartite graph, then Γt(G) ≤
2α′ind(G). Recall that by Proposition 3.2, if G is an isolate-free graph, then Γt(G) ≤
γit(G). A natural problem is to determine if the parameters γ
i
t(G) and 2α
′
ind(G) are related.
Let G = {Gk : k ≥ 1} be the family of graphs constructed earlier (see Figure 2 for
an illustration of the graph Gk that belongs to the family G). For k ≥ 1 if Gk ∈ G,
then by Proposition 4.1 we have γit(Gk) = 3k. Since each of the k copies of the graph
H = P4 ⊕P4 used to build the graph Gk is dense and has induced matching number equal
to 1, we observe that α′ind(Gk) = k. This yields the following result.
Proposition 5.2. For k ≥ 1, we have γit(Gk) = 3k and 2α
′
ind(Gk) = 2k.
Moreover for k ≥ 1, let Bk be the graph obtained from k vertex disjoint copies of
K3 by selecting one vertex from each copy of K3 and identifying these vertices into one
new vertex v, and then adding a new vertex u and adding the edge uv. The graph B4, for
example, is illustrated in Figure 4(a). Dominator can force the game to be completed after
two moves by indicating the vertex u on his first move, thereby forcing Staller to play the
vertex v on her first move. All vertices are now totally dominated, except for the vertex
v. Hence, after Dominator indicates the vertex v on his second move, Staller plays any
neighbor of v on her second move, and the game is over, and so γit(Bk) = 2. However, the
k edges in the triangles that do not contain the vertex v form an induced matching in Bk,
implying that α′ind(Bk) = k.
Proposition 5.3. For k ≥ 1 we have γit(Bk) = 2 and α
′
ind(Bk) = k.
By Proposition 5.2, there exist connected graphs G such that γit(G) can be arbitrarily
larger than 2α′ind(G), while by Proposition 5.3, there exist connected graphs H such that
2α′ind(H) can be arbitrarily larger than γ
i
t(H). Thus, in general there is no relation between
the indicated total domination number, γit(G), of an isolate-free graph G and twice the
induced matching number, 2α′ind(G), of G. However, we pose the following question.
Question 5.4. Is it true that if G is an isolate-free bipartite graph, then γit(G) ≤ 2α
′
ind(G)?
12 Ars Math. Contemp. 25 (2025) #P2.02
u
v
u
v
v1 v2 v3 v4
(a) B4 (b) J4
Figure 4: The graphs B4 and J4.
For k ≥ 1, let Jk be the graph obtained from k vertex disjoint copies of C4 by selecting
one vertex from each copy of C4 and identifying these vertices into one new vertex v, and
then adding a new vertex u and adding the edge uv. Let Q1, . . . , Qk be the k copies of
C4 in the graph Jk, and so each cycle Qi contains the vertex v for i ∈ [k]. Let vi be the
vertex in Qi that is not adjacent to v for i ∈ [k]. The graph J4, for example, is illustrated in
Figure 4(b). Dominator can force the game to be completed after k+1 moves by indicating
the vertex vi on his ith move for i ∈ [k], and then indicating the vertex u on his (k + 1)st
move. This strategy of Dominator forces Staller to play the vertex v and one neighbor of
the vertex vi for all i ∈ [k], thereby producing a TD-set in the graph Jk, implying that
γit(Jk) ≤ k + 1. (One can readily show that Staller has a strategy to prolong the game by
at least k + 1 moves, implying that γit(Jk) = k + 1.) However, selecting k edges of Jk,
one edge incident with each of the vertices vi for i ∈ [k], produces an induced matching in
Jk, implying that α
′
ind(Jk) = k. Hence, 2α
′
ind(Jk)− γ
i
t(Jk) = 2k− (k+1) = k− 1. We
state this formally as follows.
Proposition 5.5. There exist connected bipartite graphs G such that 2α′ind(G) can be
arbitrarily larger than γit(G).
The following result gives a partial answer to Question 5.4 in the case when T is a tree
that contains a maximum induced matching satisfying certain properties.
Proposition 5.6. If a nontrivial tree T contains a maximum induced matching M such that
one end of every edge in M is a leaf of T , then γit(T ) ≤ 2α
′
ind(T ).
Proof. Let T be a nontrivial tree of order n that contains a maximum induced matching
M such that one end of every edge in M is a leaf of T . If n = 2, then T = K2 and
γit(T ) = 2 = 2α
′
ind(T ). Hence, we may assume that n ≥ 3. We define V (M) as the set
of vertices that are incident with an edge of M . Let M = {e1, . . . , ek} and let ei = uiwi
for i ∈ [k]. By supposition, one of ui and wi is a leaf for all i ∈ [k]. Renaming vertices if
necessary, we may assume that ui is a leaf for all i ∈ [k]. Let U = {u1, . . . , uk} and let
W = {w1, . . . , wk}, and so V (M) = U ∪ W . Let X be the boundary of the set V (M)
in the tree T , and so X is the set of vertices not in the set V (M) that have a neighbor in
V (M). We note that since every vertex in the set U is a leaf (with a neighbor in W ), the
set W totally dominates the boundary X (and totally dominates the set U ). Let Y be the
set of vertices not totally dominated by the set W in the tree T . We proceed further with
the following claim.
Claim 5.7. If Y = ∅, then γit(T ) ≤ 2α
′
ind(T ).
M. A. Henning and D. F. Rall: Indicated total domination game 13
Proof of Claim 5.7. Suppose that Y = ∅, implying that V (T ) = U ∪W ∪X and that W
is a dominating set of T . In this case, Dominator’s strategy is to indicate on his ith move
the leaf ui for i ∈ [k]. Thus on his first k moves, Dominator indicates the leaves u1, . . . , uk
in turn. This forces Staller to play the unique neighbor of ui, namely vertex wi, on her
ith move for all i ∈ [k]. After these k moves of Staller have been played, the vertices
w1, . . . , wk have been added to the set that will grow to a TD-set upon completion of the
game. Thus at this stage of the game, W is the set of vertices that have been played. By
our earlier observations, the set W totally dominates the set U ∪X . Thus the only vertices
not yet totally dominated in the game are those vertices that belong to the set W .
Dominator now ensures that the current set of played vertices, namely W , can be ex-
tended to a TD-set of the tree T by adding to the set at most k vertices. This goal he
readily achieves by indicating the vertices in W in turn that are not yet totally dominated.
More precisely, Dominator indicates the vertex w1 on his (k + 1)st move. After Staller’s
reply, if the game is not yet over, then Dominator indicates on his next move the vertex wj
with smallest subscript j that has not yet been totally dominated. Continuing in this way,
Dominator has a strategy to complete the game in at most 2k = 2α′ind(T ) moves, and so
γit(T ) ≤ 2k = 2α
′
ind(T ).
By Claim 5.7, we may assume that |Y | ≥ 1, for otherwise the desired result follows.
If Y is not an independent set, then we could add to M an arbitrary edge that belongs to
the induced subtree T [Y ] of T , contradicting the maximality of the induced matching M .
Hence, Y is an independent set, and so all neighbors of vertices in Y belong to the set X .
Dominator now employs an opening game strategy in the game that ensures that all
vertices in Y are totally dominated as follows. On his first more, he indicates an arbitrary
vertex y1 ∈ Y . Staller must reply by playing a neighbor of y1, which as observed earlier
belongs to the set X . Let x1 be the vertex played by Staller in response to Dominator
indicating the vertex y1, and so x1 ∈ X . If a vertex in Y is not totally dominated by the
vertex x1, then Dominator indicates a vertex y2 ∈ Y that is not adjacent to x1. Let x2
be the vertex played by Staller in response to Dominator indicating the vertex y2, and so
x2 ∈ X and x1 ̸= x2. Continuing in this way, Dominator indicates on each of his next
moves a vertex in Y not yet totally dominated by the set of vertices played to date, thereby
forcing Staller to respond by playing a vertex in X that totally dominates the indicated
vertex. Suppose that this process takes r moves to reach the situation when all vertices of
Y are totally dominated. Further, suppose yi is the vertex indicated by Dominator on his
ith move and that xi is the vertex played by Staller on her ith move for i ∈ [r]. We call the
vertex yi the partner of the vertex xi for i ∈ [r]. This completes the opening game strategy
of Dominator.
Let X1 = {x1, . . . , xr} ⊆ X be the resulting set of vertices played by Staller on her
first r moves. The set X1 totally dominates the set Y . Let Y1 = {y1, . . . , yr} be the set
of vertices indicated by Dominator on his first r moves, and so xiyi is an edge of T (and
yi is the partner of xi) for i ∈ [r]. Let W1 be the set of all vertices in W that are (totally)
dominated by the set X1.
Claim 5.8. |X1| ≤ |W1|.
Proof of Claim 5.8. Suppose, to the contrary, that |X1| > |W1|. Then there must exist
w ∈ W1 such that |NT (w) ∩ X1| ≥ 2. Renaming vertices if necessary, we may assume
that x1, x2 ∈ NT (w)∩X1. But then, the induced matching obtained from M by removing
14 Ars Math. Contemp. 25 (2025) #P2.02
the edge uw, where u ∈ U is the neighbor of w, and adding the edges x1y1 and x2y2 has
cardinality |M |+ 1, a contradiction to the maximality of M .
By Claim 5.8, |X1| ≤ |W1|. Let X2 = X \ X1 and let W2 = W \ W1. Further, let
Ui be the set of neighbors of vertices in Wi that belong to the set U for i ∈ [2]. Hence, the
induced subgraph T [Ui ∪Wi] is a disjoint union of |Wi| copies of K2, where each copy of
K2 consists of a vertex in Wi and its unique neighbor that belongs to the set U for i ∈ [2].
In his middle game strategy, Dominator ensures that all vertices in W are played as
follows. Dominator indicate on his (r + i)th move the leaf ui for i ∈ [k]. Thus on his next
k moves immediately following the first r moves in his opening game strategy, Dominator
indicates the leaves u1, . . . , uk in turn. This forces Staller to play the unique neighbor of
ui, namely vertex wi, on her (r + i)th move for all i ∈ [k]. After these r + k moves
of Staller have been played, the vertices w1, . . . , wk have been added to the set of played
vertices, yielding the current set X1∪W of vertices that have been played to date by Staller.
The set W of vertices played to date in the game totally dominates the set U ∪X . Recall
that Dominator’s opening game strategy ensures that the set X1 totally dominates the set
Y ∪W1. Thus, Dominator middle game strategy adds k new vertices to the set of vertices
played by Staller, in addition to the r vertices she played in the opening game phase of
the game. Upon completion of his middle game strategy, the only vertices not yet totally
dominated in the game are those vertices that belong to the set W2.
In his end game strategy, Dominator ensures that all vertices in W2 are totally domi-
nated as follows. Dominator orders the vertices in W2 sequentially. Let w2,1, w2,2, . . . , w2,q
be the resulting ordering of the vertices in W2 by Dominator, where q = |W2|. On
each of his subsequent moves, Dominator indicates the vertex w2,j with smallest sub-
script j that has not yet been totally dominated where j ∈ [q]. Thus, Dominator’s end
game strategy adds q new vertices to the set of vertices played by Staller, in addition
to the r vertices she played in the opening game phase of the game and the k vertices
she played in the middle game phase of the game. Upon completion of Dominator’s
end game strategy, at most r + k + q vertices are played by Staller, and the resulting
set of played vertices is a TD-set of the tree T . Recall that |W1| ≥ |X1| = r, and
so q = |W2| = |W | − |W1| ≤ k − r. Therefore, Dominator has a strategy to com-
plete the game in at most (r + q) + k ≤ 2k = 2|M | = 2α′ind(T ) moves, and so
γit(T ) ≤ 2k = 2α
′
ind(T ).
6 The game total domination number
We show in this section that in general there is no relation between the indicated total dom-
ination number, γit(G), of an isolate-free graph G and the game total domination number,
γtg(G), of G.
Proposition 6.1. If k ≥ 3 and Tk is the tree obtained from a star K1,k by subdividing
every edge once, then γit(Tk) = Γt(Tk) = 2k and γtg(Tk) = k + 1.
Proof. For k ≥ 3, let T = Tk be the tree obtained from a star K1,k by subdividing ev-
ery edge exactly once. Let v be the central vertex of T (of degree k), and let NT (v) =
{x1, x2, . . . , xk}. Further, let yi be the leaf neighbor of xi for i ∈ [k]. The tree Tk is
illustrated in Figure 5.
The set of leaves together with the set of support vertices in Tk is a minimal TD-set of
maximum cardinality, and thus Γt(Tk) = 2k.
M. A. Henning and D. F. Rall: Indicated total domination game 15
v
x1
x2
xk
y1
y2
yk
Figure 5: The tree Tk in the proof of Proposition 6.1.
When the total domination game is played on Tk, Dominator plays the vertex v as
his first move. The set of legal moves remaining is the set of support vertices, namely
{x1, x2, . . . , xk} and each of these vertices must be played during the game in order to
totally dominate the leaves of Tk. This strategy of Dominator shows that γtg(Tk) ≤ k+ 1.
In any play of the total domination game on Tk exactly k moves will be required to totally
dominate the set of leaves, namely the set {y1, . . . , yk}. Furthermore, either v or y1 must
be played to totally dominate the vertex x1. This shows that γtg(Tk) ≥ k + 1. Therefore,
γtg(Tk) = k + 1.
By Proposition 3.2 we have γit(Tk) ≥ Γt(Tk) = 2k. By Proposition 5.6 we readily
infer that γit(Tk) ≤ 2k. Consequently, γ
i
t(Tk) = 2k.
By Proposition 6.1, there exist connected graphs G such that γit(G) can be arbitrar-
ily larger than γtg(G). Using the result of Proposition 6.1 and the fact that Γt(G) ≤
OOIR(G), we see that there exist trees T such that OOIR(T ) is arbitrarily larger than
γtg(T ). We show next that there exist connected graphs G such that γtg(G) can be arbi-
trarily larger than both γit(G) and OOIR(G).
Proposition 6.2. For k ≥ 4 an even integer, if Tk is the tree obtained from a star K1,k by
subdividing every edge three times, then γit(Tk) ≤ 2k+2 = OOIR(Tk) and γtg(Tk) ≥
5
2
k.
Proof. For k ≥ 4 an even integer, let Tk be the tree obtained from a star K1,k by subdivid-
ing every edge exactly three times. Let v be the central vertex of Tk (of degree k), and let
Qi : vuiviwixi be the k paths of length 4 emanating from v in Tk for i ∈ [k]. The tree Tk
is illustrated in Figure 6.
v
u1
u2
uk
v1
v2
vk
w1
w2
wk
x1
x2
xk
Figure 6: The tree Tk in the proof of Proposition 6.2.
16 Ars Math. Contemp. 25 (2025) #P2.02
First we show that OOIR(Tk) = 2(k + 1). Let M be an induced matching in Tk of (max-
imum) cardinality α′ind(Tk). By the maximality of M , the induced matching M contains
an edge from the set {uivi, viwi, wixi} for all i ∈ [k]. If the edge uivi or the edge viwi
belong to the induced matching M , then we can replace such an edge in M with the edge
wixi. Hence, we may choose the induced matching M to contain the edge wixi for all
i ∈ [k]. No additional edge incident with vi or wi can be added to these k edges without
violating the requirement that M is an induced matching. Hence, M contains exactly one
additional edge, namely an edge incident with the vertex v. Hence, the set
M = {vu1} ∪
k⋃
i=1
{wixi},
for example, is an induced matching in Tk of maximum cardinality, and so α
′
ind(Tk) =
|M | = k + 1. By Theorem 5.1, we therefore infer that OOIR(Tk) = 2(k + 1).
We show next that γtg(Tk) ≥
5
2
k. Let the total domination D-game be played on Tk.
We provide a strategy for Staller that forces at least 5
2
k moves to be made in the game.
We note that when the game has ended, the vertex wi has been played in order to totally
dominate the leaf xi for each i ∈ [k]. Further, at least one vertex from {vi, xi} will have
been played in order to totally dominate the vertex wi for each i ∈ [k]. Thus at least
two vertices are played from every set {vi, wi, xi} for all i ∈ [k]. Staller’s strategy is
to play vertex ui, for as many values of i ∈ [k] as possible, whenever no other vertex
from {ui, vi, wi, xi} has been played. Since k is even, by following this strategy she can
ensure that at least k/2 of the vertices in {u1, . . . , uk} will be played. For each j ∈ [k]
for which Staller played vertex uj , as observed earlier the vertex wj and at least one of vj
and xj will be played in the course of the game, implying that at least three vertices from
the set {uj , vj , wj , xj} are played. Therefore upon completion of the total domination D-
game, Staller guarantees that at least three moves are played from at least k/2 of the sets
{uj , vj , wj , xj}, and from the remainder of the sets {uj , vj , wj , xj} at least two moves are
played. Therefore, when the game is complete at least 3 × k
2
+ 2 × k
2
= 5
2
k moves were
made, implying that γtg(Tk) ≥
5
2
k.
Finally, let the indicated total domination game be played on Tk. We give a strategy for
Dominator that ensures at most 2k + 2 vertices are chosen. In his first k moves Dominator
indicates - in any order - the vertices in {x1, . . . , xk}, thereby forcing Staller to respond by
playing all k support vertices, namely the vertices in the set {w1, . . . , wk}. After Staller’s
first k moves the result is that the set ∪ki=1{vi, xi} is therefore totally dominated. Now
Dominator indicates (in order) the vertices in the sequence u1, . . . , uk. If Staller responds
by playing v1, . . . , vk, respectively, then every vertex except for vertex v has been totally
dominated. As a result 2k + 1 vertices will be chosen by Staller. On the other hand, if
Staller plays v1, . . . , vj , respectively, for some 1 ≤ j < k and then plays vertex v when
Dominator indicates the vertex uj+1, exactly one vertex from {vi, xi} will be played by
Staller for each i ∈ [k] \ [j] when the game ends. In this case k+ j+1+(k− j) = 2k+1
vertices will be chosen upon completion of the game. Finally, if Staller plays the vertex v
when Dominator indicates vertex u1, then exactly one vertex from {vi, xi} will be played
by Staller (in order to totally dominate the vertex wi) for each i ∈ [k] and one vertex
from {u1, . . . , uk} will be played by Staller when Dominator indicates the vertex v. In this
case a total of k + 1 + k + 1 = 2k + 2 vertices will be chosen by Staller. This strategy
by Dominator ensures that at most 2k + 2 vertices will be played by Staller. Therefore,
γit(Tk) ≤ 2k + 2.
M. A. Henning and D. F. Rall: Indicated total domination game 17
By Proposition 6.2, there therefore exist connected graphs G such that γtg(G) can be
arbitrarily larger than γit(G).
We also remark that when the total domination game (D-game) is played on the graph
Fk defined earlier immediately before the statement of Proposition 4.2 where k ≥ 5, then
Dominator can ensure that both vertices u1 and v1 are played in the first three moves.
Indeed, Dominator plays the vertex u1 in his first move. Staller can force the game to last
at least three moves by playing vertex uk on her first move. Therefore, γtg(Fk) = 3. Recall
that by Proposition 4.2, we have γit(Fk) = 4. Hence for k ≥ 5, the graph Fk is another
example of a graph satisfying γtg(Fk) < γ
i
t(Fk).
7 The Grundy total domination number
Recall that the length of a longest total dominating sequence in G is the Grundy total
domination number, γtgr(G), of G. The definition of the indicated total domination game
implies that the sequence of vertices selected by Staller in the indicated total domination
game is a total dominating sequence. From this we infer that γit(G) ≤ γ
t
gr(G). Let A be
an open-open irredundant set in G of cardinality OOIR(G) and construct a sequence S
using the vertices of A in any order. If S is not a total dominating sequence, then it can
be extended to form one. This shows that OOIR(G) ≤ γtgr(G). We state this formally as
follows.
Obervation 7.1. If G is an isolate-free graph, then γit(G) ≤ γ
t
gr(G) and OOIR(G) ≤
γtgr(G).
As a consequence of Proposition 3.3 and the following result given in [3], we infer that
there exist trees T such that γtgr(T ) is arbitrarily larger than γ
i
t(T ).
Theorem 7.2 ([3]). If T is a tree of order n, then γtgr(T ) = n if and only if T has a perfect
matching.
8 Summary and concluding remarks
The Hasse diagram in Figure 7 shows the relationships between the invariants studied in this
paper for graphs with no isolated vertices. The invariants γit, OOIR and γtg are pairwise in-
comparable. Indeed, Propositions 4.1 and 4.2 show that the differences γit(G)−OOIR(G)
and OOIR(G)−γit(G) can be arbitrarily large; Propositions 6.1 and 6.2 show that γ
i
t(G)−
γtg(G) and γtg(G)− γ
i
t(G) can be arbitrarily large; Propositions 6.1 and 6.2 together with
the fact that every minimal TD-set is open-open irredundant show that OOIR(G)−γtg(G)
and γtg(G) − OOIR(G) can be arbitrarily large. Consider the graph Bk defined in Sec-
tion 5 for each positive integer k. We see from Proposition 5.3 that γit(Bk) = 2 and
2α′ind(Bk) = 2k. It is also easy to verify that γtg(Bk) = γt(Bk) = Γt(Bk) = 2. Thus,
2α′ind can be arbitrarily larger than γ
i
t, γtg, γt and Γt. On the other hand, if G is the corona
of a complete graph of order k, then 2α′ind(G) = 2, γ
i
t(G) = γt(G) = Γt(G) = k and
γtg(G) = k+1, which shows that 2α
′
ind can be arbitrarily smaller than these four invariants
as well. Therefore, 2α′ind is incomparable with each of γ
i
t, γtg, γt and Γt.
18 Ars Math. Contemp. 25 (2025) #P2.02
2α′indγt
Γt
γit OOIR γtg
γtgr
Figure 7: Relations between the invariants studied in this paper.
ORCID iDs
Michael A. Henning https://orcid.org/0000-0001-8185-067X
Douglas F. Rall https://orcid.org/0000-0002-5482-756X
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.03
https://doi.org/10.26493/1855-3974.2944.9cd
(Also available at http://amc-journal.eu)
Edge-transitive core-free Nest graphs*
István Kovács †
UP IAM, University of Primorska, Muzejski trg 2 and
UP FAMNIT, University of Primorska, Glagoljaška ulica 8,
SI-6000 Koper, Slovenia
Received 21 August 2022, accepted 24 July 2023, published online 11 March 2025
Abstract
A finite simple graph Γ is called a Nest graph if it is regular of valency 6 and admits an
automorphism ρ with two orbits of the same length such that at least one of the subgraphs
induced by these orbits is a cycle. We say that Γ is core-free if no non-trivial subgroup
of the group generated by ρ is normal in Aut(Γ). In this paper we show that, if Γ is
edge-transitive and core-free, then it is isomorphic to one of the following graphs: the
complement of the Petersen graph, the Hamming graph H(2, 4), the Shrikhande graph and
a certain normal 2-cover of K3,3 by Z42.
Keywords: Bicriculant, edge-transitive, primitive permutation group.
Math. Subj. Class. (2020): 05C25, 20B25
1 Introduction
All groups in this paper will be finite and all graphs will be finite and simple. A graph
admitting an automorphism with two orbits of the same length is called a bicirculant. Sym-
metry properties of bicirculants have attracted considerable attention (see, e.g., [1, 5, 7, 16,
22, 23, 25, 29]). Following [17], for an integer d ≥ 3, we denote by F(d) the family of reg-
ular graphs having valency d and admitting an automorphism with two orbits of the same
length such that at least one of the subgraphs induced by these orbits is a cycle. Jajcay et
al. [12] initiated the investigation of the edge-transitive graphs in the classes F(d), d ≥ 6.
The families F(d) with 3 ≤ d ≤ 5 were studied under different names. The graphs in
F(3) were introduced by Watkins [27] under the name generalised Petersen graphs, the
*The author is grateful to the reviewers for pointing out a few inaccuracies and also for the valuable comments
and suggestions, which improved the presentation.
†Partially supported by the Slovenian Research Agency (research program P1-0285, research projects N1-
0062, J1-9108, J1-1695, J1-2451, N1-0140 and N1-0208).
E-mail address: istvan.kovacs@upr.si (István Kovács)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.03
graphs in F(4) by Wilson [28] under the name Rose Window graphs, and the graphs in
F(5) by Arroyo et al. [2] under the name Tabačjn graphs. The automorphism groups of
these graphs form the subject of the papers [9, 10, 15, 18], and the question which of them
are edge-transitive has been answered in [2, 10, 15].
Jajcay et al. [12] asked whether there exist edge-transitive graphs in F(d) for d ≥ 6.
Following [26], they call the graphs in F(6) Nest graphs. Several infinite families of edge-
transitive Nest graphs were exhibited, which turn out to have interesting properties (e.g.,
half-arc-transitivity). However, no edge-transitive graph of valency larger than 6 was found.
Recently, it was proved by the author and Ruff [17] that the complement of the Petersen
graph is the only edge-transitive graph in F(d) with d ≥ 6, which has twice an odd number
of vertices. The main result of [12] is the classification of the edge-transitive Nest graphs
of girth 3 (see [12, Theorem 3.6]), and the task to classify all edge-transitive Nest graphs
was posed as [12, Problem 1.2]. In what follows, the Nest graphs will be described via
their representation due to [12, Construction 3.1], which goes as follows. Let n ≥ 4 and
let a, b, c, k ∈ Zn such that each of them is distinct from 0 (the zero element of Zn), the
elements a, b and c are pairwise distinct, and in the case when n is even, k ̸= n/2. Then
the Nest graph N (n; a, b, c; k) is defined to have vertex set {ui : i ∈ Zn} ∪ {vi : i ∈ Zn},
and three types of edges such as
• {ui, ui+1} for i ∈ Zn (rim edges),
• {vi, vi+k} for i ∈ Zn (hub edges),
• {ui, vi}, {ui, vi+a}, {ui, vi+b} and {ui, vi+c} for i ∈ Zn (spoke edges),
where the sums in the subscripts are computed in Zn. It is easy to see that the permutation
ρ of V (Γ), defined as ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1), is an automorphism of
Γ with orbits {ui : i ∈ Zn} and {vi : i ∈ Zn}, and the subgraph induced by the former
orbit is a cycle. It is not hard to show that all the graphs N (n; a, b, c; k) comprise the whole
family F(6).
In the case of both the Rose Window and the Tabačjn graphs, the classification of the
edge-transitive graphs was obtained in two main steps. The so called core-free graphs were
found first and the rest was retrieved from the core-free graphs using covering techniques
(see [2, 15]). Here is the formal definition of a core-free Nest graph.
Definition 1.1. Let Γ = N (n; a, b, c; k) be a Nest graph and ρ be the permutation of V (Γ)
defined as ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1). Then Γ is core-free if no non-trivial
subgroup of ⟨ρ⟩ (the group generated by ρ) is normal in Aut(Γ).
Remark 1.2. The term “core-free” comes from group theory. For a subgroup A ≤ B, the
core of A in B is the largest normal subgroup of B contained in A. In the case when A has
trivial core, it is also called core-free. In this context, Definition 1.1 can be rephrased by
saying that Γ is core-free if and only if ⟨ρ⟩ is core-free in Aut(Γ).
Our goal in this paper is to determine the edge-transitive core-free Nest graphs. For
an explanation why this task is more subtle than in the case of Rose Window and Tabačjn
graphs, we refer to [12, page 9]. The edge-transitive non-core-free Nest graphs are handled
in the paper [14].
The main result of this paper is the following theorem.
I. Kovács: Edge-transitive core-free Nest graphs 3
Theorem 1.3. If N (n; a, b, c; k) is an edge-transitive core-free graph, then it is isomorphic
to one of the following graphs:
N (5; 1, 2, 3; 2), N (8; 1, 3, 4; 3), N (8; 1, 2, 5; 3) and N (12; 2, 4, 8; 5).
Remark 1.4. The fact that each of the Nest graphs in Theorem 1.3 is core-free was
mentioned by Jajcay et al., see [12, page 9]. The first three of them are well-known
strongly regular graphs. The Nest graph N (5; 1, 2, 3; 2) is the complement of the Pe-
tersen graph, N (8; 1, 3, 4; 3) is the Hamming graph H(2, 4), and N (8; 1, 2, 5; 3) is the
Shrikhande graph. The fourth Nest graph N (12; 2, 4, 8; 5) is not strongly-regular, it can be
described as a normal 2-cover of the complete bipartite graph K3,3 by Z42 (for the definition
of a normal 2-cover, see the 2nd paragraph of Subsection 2.1).
The paper is organised as follows. Section 2 contains the needed results from graph and
group theory. In Section 3 we review some results about Nest graphs obtained in [12, 17].
Section 4 is devoted to the Nest graphs in the form N (2m; 2,m, 2 +m; 1), m is odd. The
main result (Proposition 4.1) is a characterisation, which was mentioned in [12] without a
proof, and which is needed for us in the proof Theorem 1.3. The latter proof is presented
in Section 5.
2 Preliminaries
2.1 Graph theory
Given a graph Γ, let V (Γ), E(Γ), A(Γ) and Aut(Γ) denote its vertex set, edge set, arc
set and automorphism group, respectively. The number |V (Γ)| is called the order of Γ.
The set of vertices adjacent with a given vertex v is denoted by Γ(v). If G ≤ Aut(Γ) and
v ∈ V (Γ), then the stabiliser of v in G is denoted by Gv , the orbit of v under G by vG, and
the set of all G-orbits by Orb(G, V (Γ)). If B ⊆ V (Γ), then the setwise stabiliser of B in
G is denoted by G{B}. If G is transitive on V (Γ), then Γ is said to be G-vertex-transitive,
and Γ is simply called vertex-transitive when it is Aut(Γ)-vertex-transitive; (G-)edge- and
(G-)arc-transitive graphs are defined correspondingly.
Let π be an arbitrary partition of V (Γ) and for a vertex v ∈ V (Γ), let π(v) denote the
class containing v. The quotient graph of Γ with respect to π, denoted by Γ/π, is defined
to have vertex set π, and edges {π(u), π(v)}, where {u, v} ∈ E(Γ) such that π(u) ̸= π(v).
Now, if there exists a constant r such that
∀{u, v} ∈ E(Γ) : π(u) ̸= π(v) and |Γ(u) ∩ π(v)| = r,
then Γ is called an r-cover of Γ/π. The term cover will also be used instead of 1-cover. In
the special case when π = Orb(N,V (Γ)) for an intransitive normal subgroup N◁Aut(Γ),
Γ/N will also be written for Γ/π and when Γ is also an r-cover (cover, respectively) of
Γ/N , then the term normal r-cover (normal cover, respectively) will also be used. It is
well-known that this is always the case when Γ is edge-transitive. More precisely, if Γ is a
G-edge-transitive graph, Γ is regular with valency κ, and N ◁G is intransitive, then Γ is a
normal r-cover of Γ/N for some r and r divides κ.
A graph admitting a regular cyclic group of automorphisms is called a circulant. A
recursive classification of finite arc-transitive circulants was obtained independently by
Kovács [13] and Li [19]. The paper [13] also provides an explicit characterisation (see
4 Ars Math. Contemp. 25 (2025) #P2.03
[13, Theorem 4]), which was rediscovered recently by Li et al. [20]. The characterisation
presented below follows from the proof of [13, Theorem 4] or from [20, Theorem 1.1].
In what follows, given a cyclic group C and a divisor d of |C|, we denote by Cd the
unique subgroup of C of order d.
Theorem 2.1 ([13]). Let Γ be a connected arc-transitive graph of order n and of valency
κ and suppose that C ≤ Aut(Γ) is a regular cyclic subgroup. Then one of the following
holds.
(a) Γ is the complete graph.
(b) C is normal in Aut(Γ).
(c) B = Orb(Cd, V (Γ)) is a block system for Aut(Γ) for some divisor d of gcd(n, κ),
d > 1. Γ is a normal d-cover of Γ/B, and Γ/B is a connected arc-transitive circulant
of valence κ/d.
(d) B1 = Orb(Cd, V (Γ)) and B2 = Orb(Cn/d, V (Γ)) are block systems for Aut(Γ) for
some divisor d of n such that d > 3, gcd(d, n/d) = 1 and d − 1 divides κ. Γ/B1 is
a connected arc-transitive circulant of valency κ/(d− 1), Γ/B2 ∼= Kd, and
Aut(Γ) = G1 ×G2, (2.1)
where Cd ≤ G1, G1 ∼= Sd, Cn/d < G2, and G2 ∼= Aut(Γ/B1).
Remark 2.2. Although not used later, it is worth mentioning that the graph Γ in part (c) is
isomorphic to the lexicographical product Γ/B[Kd], where Kd is the edgeless graph on d
vertices, and the graph Γ in part (d) is isomorphic to the tensor (direct) product Kd×Γ/B1
(see [13, 20]).
In the rest of the section we restrict ourselves to arc-transitive circulants of small va-
lency.
Lemma 2.3 ([3, part (ii) of Corollary 1.3]). Let Γ be a connected arc-transitive graph of
order n and of valency κ, where κ = 3 or 4, and suppose that C ≤ Aut(Γ) is a regular
cyclic subgroup. Then one of the following holds.
(1) Γ is isomorphic to one of the graphs: K4,K5,K3,3 and K5,5 − 5K2.
(2) κ = 4 and C is normal in Aut(Γ).
(3) κ = 4, n is even, B = Orb(C2, V (Γ)) is a block system for Aut(Γ), and Γ is a
normal 2-cover of Γ/B, which is a cycle.
Lemma 2.4. Let Γ be a connected arc-transitive graph of order n > 14 and of valency
6, and suppose that C ≤ Aut(Γ) is a regular cyclic subgroup. Then Aut(Γ) contains a
normal subgroup N such that one of the following holds.
(1) N = C, or
(2) n ≡ 4(mod 8) and N = Cn/4, or
(3) N ∼= Zℓ3 for ℓ ≥ 2 and C3 < N .
I. Kovács: Edge-transitive core-free Nest graphs 5
Proof. Γ belongs to one of the families (a) – (d) in Theorem 2.1.
Family (a): This case cannot occur as n > 14.
Family (b): Part (1) follows.
Family (c): In this case Orb(Cd, V (Γ)) is a block system for Aut(Γ), where d = 2 or
d = 3 or d = 6. Let B = Orb(Cd, V (Γ)).
If d = 2, then Γ/B has valency 3. It follows from Lemma 2.3 that n ≤ 12, but this is
excluded.
If d = 3, then choose N to be the Sylow 3-subgroup of the kernel of the action of
Aut(Γ) on B. It is clear that C3 ≤ N . The quotient graph Γ/B is a cycle. Using this,
one can see that a Sylow 3-subgroup of a vertex stabilizer in Aut(Γ) is contained in N , in
particular, N ̸= C3. It follows that N ∼= Zℓ3 for some ℓ ≥ 2. Also, N is characteristic
in the latter kernel, which implies that N ◁ Aut(Γ). Finally, Orb(N,V (Γ)) = B =
Orb(C3, V (Γ)), and so C3 < N , i.e., part (3) holds.
If d = 6, then Γ ∼= K6,6. This contradicts the assumption that n > 14.
Family (d): In this case it follows from the assumption that n > 14 that Orb(C4, V (Γ))
and Orb(Cn/4, V (Γ)) are block systems for ] Aut(Γ) and n ≡ 4(mod 8). Furthermore,
Aut(Γ) = G1 ×G2,
where C4 < G1, G1 ∼= S4, Cn/4 < G2 and G2 ∼= Aut(Γ/B1), where B1 = Orb(C4, V (Γ)).
The graph Γ/B1 is connected of valency 2, hence it is a cycle of length n/4. It follows that
Cn/4 is characteristic in G2, and as G2 ◁Aut(Γ), part (2) follows.
2.2 Group theory
Our terminology and notation are standard and we follow the books [8, 11]. The socle of
a group G, denoted by soc(G), is the subgroup generated by the set of all minimal normal
subgroups (see [8, page 111]). The group G is called almost simple if soc(G) = T , where
T is a non-abelian simple group. In this case G is embedded in Aut(T ) so that its socle is
embedded via the inner automorphisms of T , and we also write T ≤ G ≤ Aut(T ) (see [8,
page 126]).
Our proof of Theorem 1.3 relies on the classification of primitive groups containing a
cyclic subgroup with two orbits due to Müller [24]. Here we need only the special case
when the cyclic subgroup is semiregular.
Theorem 2.5 ([24, Theorem 3.3]). Let G be a primitive permutation group of degree 2n
containing an element with two orbits of the same length. Then one of the following holds,
where G0 denotes the stabiliser of a point in G.
(1) (Affine action) Zm2 ◁ G ≤ AGL(m, 2), where n = 2
m−1. Furthermore, one of the
following holds.
(a) n = 2, and G0 = GL(2, 2).
(b) n = 2, and G0 = GL(1, 4).
(c) n = 4, and G0 = GL(3, 2).
(d) n = 8, and G0 is one of the following groups: Z5 : Z4, ΓL(1, 16), (Z3 × Z3) :
Z4, ΣL(2, 4), ΓL(2, 4), A6, GL(4, 2), (S3 × S3) : Z2, S5, S6 and A7.
(2) (Almost simple action) G is an almost simple group and one of the following holds.
6 Ars Math. Contemp. 25 (2025) #P2.03
(a) n ≥ 3, soc(G) = A2n, and A2n ≤ G ≤ S2n in its natural action.
(b) n = 5, soc(G) = A5, and A5 ≤ G ≤ S5 in its action on the set of 2-subsets of
{1, 2, 3, 4, 5}.
(c) n = (qd−1)/2(q−1), soc(G) = PSL(d, q), and PSL(d, q) ≤ G ≤ PΓL(d, q)
for some odd prime power q and even number d ≥ 2 such that (d, q) ̸= (2, 3).
(d) n = 6 and soc(G) = G = M12.
(e) n = 11, soc(G) = M22, and M22 ≤ G ≤ Aut(M22).
(f) n = 12 and soc(G) = G = M24.
If G is a group in one of the families (a) – (f) in part (2) above, then it follows from
[8, Theorem 4.3B] that soc(G) is the unique minimal normal subgroup of G. Therefore,
we have the following corollary.
Corollary 2.6. Let G be a primitive permutation group in one of the families (a) – (f) in
part (2) of Theorem 2.5, and let N ◁G, N ̸= 1. Then N is also primitive.
For a transitive permutation group G ≤ Sym(Ω), the subdegrees of G are the lengths
of the orbits of a point stabiliser Gω , ω ∈ Ω. Since G is transitive, it follows that the
subdegrees do not depend on the choice of ω (see [8, page 72]). The number of orbits
of Gω is called the rank of G. The actions of a group G on sets Ω and Ω′ are said to be
equivalent if there is a bijection φ : Ω → Ω′ such that
∀ω ∈ Ω, ∀g ∈ G : φ(ωg) = (φ(ω))g.
Now, suppose that G is a group in one of the families (a) – (f) in part (2) of Theorem 2.5.
If G is in family (a), then G is clearly 2-transitive. If 2n ̸= 6, then the action is unique up
to equivalency. If 2n = 6, then G admits two inequivalent actions. Suppose that G ∼= S6.
Consider the action of PGL(2, 5) on the projective line PG(1, 5), denote this line by Ω.
The group PGL(2, 5) ∼= S5 and it has index 6 in the symmetric group Sym(Ω). The action
of Sym(Ω) on the set of right cosets of PG(2, 5) is inequivalent with its natural action on
Ω. If G ∼= A6, the above construction can be repeated with considering PSL(2, 5) instead
of PGL(2, 5). Note that, if H is a stabilizer of a point in G with respect to either of the two
inequivalent actions, then with respect to the other action, H acts transitively.
If G is in family (b), then the action is unique up to equivalence and the subdegrees are
1, 3 and 6.
Let G be in family (c). The semiregular cyclic subgroup of G with two orbits is con-
tained in a regular cyclic group, called the Singer subgroup of PGL(d, q) (see [11, Chap-
ter 2, Theorem 7.3]). In this case the action is unique up to equivalence if and only if d = 2.
If d ≥ 4, then the action of G is equivalent to either its natural action on the set of points
of the projective geometry PG(d − 1, q), or to its natural action on the set of hyperplanes
of PG(d− 1, q). In both actions G is 2-transitive.
Let G = M12 in the family (d). Then G has two inequivalent actions. These ac-
tions can be described using the action of the Mathieu group M24 on the Steiner System
S(5, 8, 24) and the fact the setwise stabilizer of a dodecad in M24 is isomorphic to M12 (see
[8, pages 207–208]). Note that, if H is a stabilizer of a point in G with respect to either of
the two inequivalent actions, then with respect to the other action, H acts transitively.
Finally, if G is in the families (e) – (f), then the action is unique up to equivalence and
G is 2-transitive (this can also be read off from [6]). All this information is summarised in
the lemma below.
I. Kovács: Edge-transitive core-free Nest graphs 7
Lemma 2.7. Let G be a primitive permutation group in one of the families (a) – (f) in part
(2) of Theorem 2.5.
(1) G is 2-transitive, unless G belongs to family (b). In the latter case the subdegrees
are 1, 3 and 6.
(2) The action of G is unique up to equivalence, unless one of the following holds.
(i) G ∼= A6 or S6 or M12, G admits two inequivalent faithful actions, and if H is a
stabilizer of a point in G with respect to either of the two inequivalent actions,
then with respect to the other action, H acts transitively.
(ii) G is in family (c) and d ≥ 4. In the latter case G admits two inequivalent
faithful actions, namely, the natural actions on the set of points and the set of
hyperplanes, respectively, of the projective geometry PG(d− 1, q).
The following result about G-arc-transitive bicirculants can be found in Devillers et
al. [7]. The proof works also for the edge-transitive bicirculants, in fact, it is an easy
consequence of Theorem 2.5.
Proposition 2.8 ([7, part (1) of Proposition 4.2]). Let Γ be a G-edge-transitive bicirculant
such that G is a primitive group. Then Γ is one of the following graphs:
(1) The complete graph, and G is one of the 2-transitive groups described in part (2) of
Theorem 2.5.
(2) The Petersen graph or its complement, and A5 ≤ G ≤ S5.
(3) The Hamming graph H(2, 4) or its complement, and G is a rank 3 subgroup of
AGL(4, 2).
(4) The Clebsch graph or its complement, and G is a rank 3 subgroup of AGL(4, 2).
Using the computational result that there exists no edge-transitive graph in F(d) with
7 ≤ d ≤ 10 and of order at most 100, one can easily deduce which of the graphs in the
families (1) – (4) above belongs also to the family F(d) for some d ≥ 3.
Corollary 2.9. Let Γ ∈ F(d) be a G-edge-transitive graph for some d ≥ 3. If G is
primitive on V (Γ), then Γ is isomorphic to one of the graphs: K6, the Petersen graph and
its complement, and the Hamming graph H(2, 4).
Finally, we also need a result of Lucchini [21] about core-free cyclic subgroups (this
serves as a key tool in [2, 15] as well). For the definition of a core-free subgroup, see the
remark following Definition 1.1.
Theorem 2.10 ([21]). If C is a core-free cyclic proper subgroup of a group G, then |C|2 <
|G|.
3 Nest graphs
In this section we review some previous results about Nest graphs, which were obtained in
[12, 17].
8 Ars Math. Contemp. 25 (2025) #P2.03
Lemma 3.1 ([12, Lemma 3.2]). Let Γ = N (n; a, b, c; k) and suppose that c = a + b (in
Zn). Then Γ is edge-transitive if and only if it is also arc-transitive.
The next result establishes some obvious isomorphisms.
Lemma 3.2 ([12, Lemma 3.3]). The graph N (n; a, b, c; k) is isomorphic to
N (n; a′, b′, c′; k), where {a, b, c} = {a′, b′, c′}, as well as to any of the graphs:
N (n; a, b, c;−k), N (n;−a,−b,−c; k) and N (n;−a, b− a, c− a; k).
The graphs in the next lemma will be further studied in the next section.
Lemma 3.3 ([12, Lemma 3.4]). If m ≥ 3 is an odd integer, then the graph N (2m; 2,m,
2+m; 1) is arc-transitive having vertex stabilisers of order 12. Furthermore, the stabiliser
of u0 in Aut(Γ) is the dihedral group D6 of order 12 generated by the involutions φ and η
defined by
uφi =
{
u−i if i is even,
v−i+1 if i is odd
and vφi =
{
u−i+1 if i is even,
v−i+2 if i is odd,
and uηi = ui and v
η
i = vi+m for every i ∈ Zn.
Suppose that Γ is a G-edge-transitive Nest graph. In the next two lemmas we consider
block systems for G. A block system B is said to be minimal if it is non-trivial, and no
non-trivial block for G is contained properly in a block of B (by non-trivial we mean that
the block is neither a singleton subset nor the whole vertex set). We say that B is normal if
B = Orb(N,V (Γ)) for some N ◁G. Furthermore, we say that B is cyclic if any block in
B is contained in either {ui : i ∈ Zn} or {vi : i ∈ Zn}.
Lemma 3.4 ([17, Lemma 4.1]). Let Γ be a G-edge-transitive Nest graph of order 2n such
that
C < G, C = ⟨ρ⟩, and ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1),
and let B be a cyclic block system for G with blocks of size d, d < n/2. Then the following
hold.
(1) The kernel of the action of G on B is equal to Cd (the subgroup of C of order d).
(2) Γ is a normal cover of Γ/B.
(3) Γ/B is a Ḡ-edge-transitive Nest graph of order 2n/d, where Ḡ is the image of G
induced by its action on B.
Remark 3.5. Suppose that the graph Γ in the lemma above is given as Γ = N (n; a, b, c; k)
for a, b, c, k ∈ Zn. Note that, then Γ/B ∼= N (n/d; f(a), f(b), f(c); f(k)), where f is the
homomorphism from Zn to Zn/d such that f(1) = 1.
Lemma 3.6 ([17, Lemma 4.2]). Let Γ be a G-edge-transitive Nest graph of order 2n such
that
C < G, C = ⟨ρ⟩, and ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1),
and let B be a non-cyclic block system for G with blocks of size d. Then the following hold.
I. Kovács: Edge-transitive core-free Nest graphs 9
(1) The number d is even and any block in B is a union of two Cd/2-orbits.
(2) The group C acts transitively on B and the kernel of the action of C on B is equal to
Cd/2.
(3) If d > 2 and B is minimal, then B is normal.
4 A property of the graphs N (2m; 2,m, 2 + m; 1), m is odd
In this section we give the following characterisation of the Nest graphs in the title. As we
said in the introduction, this was mentioned already in [12] without a proof.
Proposition 4.1. Let Γ = N (n; a, b, c; k) be an edge-transitive graph such that n > 8
and suppose that there exists a non-identity automorphism of Γ, which fixes all vertices ui,
i ∈ Zn. Then Γ ∼= N (2m; 2,m, 2 +m; 1) for some odd number m.
We prove first an auxiliary lemma.
Lemma 4.2. Let Γ = N (2m; a,m, a+m; k), where m > 2, a = 2 or m− 2, and k = 1
or m−1. Then Γ is edge-transitive if and only if m is odd and Γ ∼= N (2m; 2,m, 2+m; 1).
Proof. The “if” part follows from Lemma 3.3.
For the “only if” part, assume that Γ is edge-transitive. If a = m−2, then −a = 2+m,
−m = m and −(a+m) = 2. By Lemma 3.2, we find that
Γ ∼= Γ′ := N (2m; 2,m, 2 +m; k),
where k = 1 or m− 1. We have to show that m is odd and k = 1.
Assume to the contrary that m is even or k = m−1. Moreover, let us choose m so that
it is the smallest number for which this happens. A quick check with the computer algebra
package MAGMA [4] shows that m ≥ 12.
Define the binary relation ∼ on V (Γ′) by letting u ∼ v whenever u = v or |Γ′(u) ∩
Γ′(v)| = 4 for any u, v ∈ V (Γ′). Suppose that u0 ∼ ui, i ̸= 0. If ui is adjacent with u1 or
u−1, then i = 2 or −2, and
∣
∣{v0, v2, vm, v2+m} ∩ {vi, v2+i, vm+i, v2+m+i}
∣
∣ = 3.
because n = 2m ≥ 24. As m ≥ 12, this cannot happen and the common neighbors of
u0 and ui are the vertices v0, v2, vm and v2+m. This yields that {0, 2,m, 2 + m} + i =
{0, 2,m, 2 + m} holds in Zn. Using that m ≥ 12, we find that i = m. Note that this
implies that ui ∼ uj if and only if i = j or i = j +m.
A similar argument shows that vi ∼ vj if and only if i = j or i = j +m.
Now suppose that u0 ∼ vi. Then both u1, u−1 are adjacent with vi, and it is not hard
to show that i = 1 or i = m + 1. This implies that ui ∼ vj if and only if j − i = 1 or
j − i = m+ 1. All these show that ∼ is an equivalence relation whose classes are the sets
Bi := {ui, ui+m, vi+1, vi+1+m}, i ∈ Zn.
Clearly, ∼ is invariant under Aut(Γ′), so the classes above form a block system for
Aut(Γ′). This block system will be denoted by B. Part of the graph with k = m − 1 is
shown in Figure 1.
10 Ars Math. Contemp. 25 (2025) #P2.03
ui−2 ui−1 ui ui+1 ui+2
ui−2+m ui−1+m ui+m ui+1+m ui+2+m
vi−1 vi vi+1 vi+2 vi+3
vi−1+m vi+m vi+1+m vi+2+m vi+3+m
Figure 1: The Nest graph N (2m; 2,m, 2 +m;m− 1).
Let K be the kernel of the action of Aut(Γ′) on B. We prove next that K is faithful
on every block. Suppose that g ∈ K fixes pointwise the block Bi for some i ∈ Zn. Any
pair of vertices in Bi+1 is contained in a unique 4-cycle intersecting Bi at two vertices (see
Figure 1). This means that g maps any pair of Bi+1 to itself, implying that g fixes pointwise
Bi+1. Repeating the argument, we conclude that g fixes pointwise each block Bi, i.e., g is
the identity automorphism.
Let n = 2m, C = ⟨ρ⟩, where ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1). Using the
facts that K is faithful on every block Bi and the quotient graph Γ/B is an n/2-cycle whose
automorphism group is isomorphic to the dihedral group Dn/2 of order n, we obtain the
bound
|Aut(Γ′)| ≤ |K| · n = 24n.
Thus |C|2 = n2 ≥ |Aut(Γ′)| because n = 2m ≥ 24. By Theorem 2.10, C has a non-
trivial core in Aut(Γ′), let this core be denoted by N .
Assume first that |N | is even. Then as C2 is characteristic in N and N ◁ Aut(Γ′), we
obtain that C2 ◁ Aut(Γ′). Thus Orb(C2, V (Γ)) is a block system for Aut(Γ′). But this
is impossible because u0 has one neighbour from the orbit {u1, u1+m} and two from the
orbit {v0, vm}.
Let |N | be odd and choose an odd prime divisor p of |N |. It follows as above that
Cp ◁Aut(Γ
′). Clearly, p divides m.
Assume that m = p or 2p. If m = p, then by our initial assumptions, k = p−1. But this
means that the edge {v0, vk} is contained in a Cp-orbit, contradicting that Cp ◁ Aut(Γ′)
and Γ′ is edge-transitive. If m = 2p, then u0 has one neighbour from the Cp-orbit {u4i+1 :
0 ≤ i ≤ p − 1} and two from the Cp-orbit {v4i : 0 ≤ i ≤ p − 1} (namely, v0 and
v2+m = v2+2p), which is a contradiction again.
Let m > 2p. By Lemma 3.4(3) and the remark after the lemma,
Γ′/Cp ∼= N (2m/p; f(2), f(m), f(2 +m), f(k)),
I. Kovács: Edge-transitive core-free Nest graphs 11
where f is the homomorphism from Z2m to Z2m/p such that f(1) = 1. Since m > 2p and
p is odd, it follows that
f(2) = 2, f(m) = m/p and f(2 +m) = 2 +m/p.
Furthermore, f(k) = 1 if k = 1 and f(k) = m/p − 1 if k = m − 1. By the minimality
of m, we see that m/p is odd and f(k) = 1. This, however, contradicts that m is even or
k = m− 1.
Proof of Proposition 4.1. Let H and N be the setwise and the pointwise stabiliser, respec-
tively, of the set {ui : i ∈ Zn} in Aut(Γ). Then N ̸= 1 and N ◁ H . It follows that the
N -orbits contained in V := {vi : i ∈ Zn} form a block system for the action of H on
V , implying that Orb(N,V ) = Orb(Cd, V ) for some d > 1. Since N ≤ Aut(Γ)u0 , it
follows that every element in N maps {v0, va, vb, vc} to itself, and therefore, the latter set
is a union of some orbits under N , hence some orbits under Cd as well. This yields that d
is even, so that n = 2m, ρm ∈ Cd, and we may write w.l.o.g. that
a < m, b = m, c = a+m, and k < m.
Let η be the permutation of the vertex set acting as
uηi = ui and v
η
i = vi+m (i ∈ Zn).
It is easy to check that η ∈ Aut(Γ).
Note that, by Lemma 3.1, Γ is arc-transitive, so Aut(Γ)u0 is transitive on Γ(u0). Let
s = |Γ(v0) ∩ Γ(vm)|. It is easy to see that s ≥ 4. Define the graph ∆ as follows:
V (∆) = Γ(u0) and E(∆) = {{w,w
′} : |Γ(w) ∩ Γ(w′)| = s}.
Note that ∆ is vertex-transitive, in particular, it is regular.
Assume for the moment that u1 and u−1 are adjacent in ∆. This means |Γ(u1) ∩
Γ(u−1)| = s ≥ 4. Since Γ(u1) ∩ Γ(u−1) ∩ {ui : i ∈ Zn} = {u0}, we conclude that
|{1, 1 + a, 1 +m, 1 + a+m} ∩ {−1,−1 + a,−1 +m,−1 + a+m}| ≥ 3. (4.1)
At least one of 1 and 1+m is in the intersection. If it is 1, then 1 = −1+a or −1+a+m
because n > 4. As a < m, we find that a = 2. Similarly, if 1 +m is in the intersection,
then 1 + m = −1 + a or −1 + a + m, and we find again that a = 2. Now, substituting
a = 2 in (5.3), a contradiction arises because n > 8.
Therefore, u1 and u−1 are not adjacent in ∆. Using also that η ∈ Aut(Γ), see above,
we obtain that u1 must be adjacent with v0 or va.
Assume first that u1 and v0 are adjacent. Then Γ(u1) ∩ Γ(v0) = {u0, u2, vk, v−k},
hence 2 ∈ {0, n−a,m, n−a+m} and k ∈ {1, 1+a, 1+m, 1+a+m}. Since a, k < m
and n > 4, we find in turn that a = m− 2, and k = 1 or m− 1.
Now, if u1 and va are adjacent, then Γ(u1) ∩ Γ(va) = {u0, u2, va+k, va−k}, whence
2 ∈ {a, 0, a + m,m} and a + k ∈ {1, 1 + a, 1 + m, 1 + a + m}. Since a, k < m and
n > 4, we find in turn that a = 2, and k = 1 or m− 1.
To sum up, a = 2 or m − 2 and k = 1 or m − 1, and so the proposition follows from
Lemma 4.2.
12 Ars Math. Contemp. 25 (2025) #P2.03
Corollary 4.3. Let Γ be a G-edge-transitive Nest graph of order 2n such that n > 8 and
C < G, C = ⟨ρ⟩, and ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1),
and suppose that Orb(Cn/2, V (Γ)) is a block system for G. Then Cn/2 ◁G.
Proof. Let K be the kernel of the action of G on the block system Orb(Cn/2, V (Γ)), and
let K∗ and (Cn/2)∗ denote the image of K and Cn/2, respectively, induced by their action
on U := {ui : i ∈ Zn}. Since the subgraph of Γ induced by U is a cycle of length n > 8,
it follows that (Cn/2)∗ is characteristic in K∗. Therefore, if K is faithful on U , then Cn/2
is characteristic in K and as K ◁G, we obtain that Cn/2 ◁G, as required.
If K is not faithful on U , then by Proposition 4.1, n = 2m, m is odd, and Γ ∼=
Γ′ := N (2m; 2,m, 2 +m; 1). Consider the group ⟨C,φ, η⟩, where φ and η are defined in
Lemma 3.3. This is transitive on V (Γ′) and also contains the stabiliser of u0 in Aut(Γ′),
therefore, Aut(Γ′) = ⟨C,φ, η⟩. A straightforward computation shows that φρ2φ = ρ−2
and ηρ2 = ρ2η, and hence Cn/2 ◁ Aut(Γ′). All these show that Cn/2 ◁ G holds in this
case as well.
5 Proof Theorem 1.3
Throughout this section we keep the following notation.
Hypothesis 5.1.
Γ = N (n; a, b, c; k) is a Nest graph of order 2n, n ≥ 4,
C = ⟨ρ⟩, where ρ = (u0, u1, . . . , un−1)(v0, v1, . . . , vn−1),
G ≤ Aut(Γ) such that ρ ∈ G, G acts transitively on E(Γ), and coreG(C) = 1.
Instead of Theorem 1.3 we show the following slightly more stronger theorem. The
proof will be given in the end of the section.
Theorem 5.2. Assuming Hypothesis 5.1, Γ is isomorphic to one of the graphs:
N (5; 1, 2, 3; 2), N (8; 1, 3, 4; 3), N (8; 1, 2, 5; 3) and N (12; 2, 4, 8; 5).
We start with a computational result, which we retrieved from [12, Table 1] with the
help of MAGMA [4]. Here we use the obvious facts that C is also core-free in Aut(Γ) and
that Aut(Γ) is primitive whenever so is G.
Lemma 5.3. Assuming Hypothesis 5.1, if n ≤ 50, then the following hold.
(1) Γ is isomorphic to one of the graphs:
N (5; 1, 2, 3; 2), N (8; 1, 3, 4; 3), N (8; 1, 2, 5; 3) and N (12; 2, 4, 8; 5).
(2) G is either primitive and Γ ∼= N (5; 1, 2, 3; 2) or N (8; 1, 3, 4; 3); or for N :=
soc(Aut(Γ)), N ∼= Z22 if n = 8 and N
∼= Z42 if n = 12, and the N -orbits have
length 4.
The existence of a non-trivial non-cyclic block system is established next.
Lemma 5.4. Assuming Hypothesis 5.1, suppose that n > 8. Then G admits a non-trivial
non-cyclic block system.
I. Kovács: Edge-transitive core-free Nest graphs 13
Proof. Observe that, if G is primitive, then Corollary 2.9 shows that Γ ∼= N (5; 1, 2, 3; 2)
or N (8; 1, 3, 4; 3). As n > 8, G is imprimitive.
Let B be a non-trivial block system with blocks of size d. If B is cyclic, then by
Lemma 3.4(1) and Corollary 4.3, Cd ◁G, where Cd is the subgroup of C of order d. This
contradicts our assumption that coreG(C) = 1, so B is non-cyclic.
In the next two lemmas we study non-cyclic block systems with blocks of size 2.
Lemma 5.5. Assuming Hypothesis 5.1, suppose that n > 50 and B is a non-cyclic block
system for G with blocks of size 2. Then Γ/B has valency 12.
Proof. Let K be the kernel of the action of G on B, and for a subgroup X ≤ G, denote by
X̄ the image of X induced by its action on B. For a block B ∈ B, we write B = {uB , vB},
where uB ∈ {ui : i ∈ Zn} and by vB ∈ {vi : i ∈ Zn}, and define the permutation τ of
V (Γ) as
τ :=
∏
B∈B
(uB vB). (5.1)
Observe that τ commutes with any element of G.
Now define the graph Γ′ as
V (Γ′) := V (Γ) and E(Γ′) := {{u0, u1}
x : x ∈ ⟨G, τ⟩} (5.2)
Then E(Γ) = {{u0, u1}x : x ∈ G} ⊆ E(Γ′). Also, ⟨τ,G⟩ ≤ Aut(Γ′), hence Γ′ is both
vertex- and edge-transitive. Since τ commutes with every element of G, it follows that
E(Γ′) = E(Γ) ∪ E(Γ)τ and E(Γ) = E(Γ)τ or E(Γ) ∩ E(Γ)τ = ∅. Notice that
Γ/B = Γ′/B,
hence we are done if show that Γ′/B has valency 12.
Denote by d and d′ the valency of Γ′ and Γ′/B, respectively. Now, d = |E(Γ′)|/n =
6 |E(Γ
′)|
|E(Γ)| . This shows that d = 6 if E(Γ) = E(Γ
′), and d = 12 otherwise.
Assume for the moment that d = 6, i.e., E(Γ) = E(Γ′). In this case τ ∈ K, hence B
is normal and Γ′ is a normal r-cover of Γ′/B and r = 1 or r = 2.
If r = 2, then d′ = 3. As n > 50, this is impossible due to Lemma 2.3. Here we
use the facts that Γ′/B is edge-transitive and C̄ is regular on V (Γ′/B). It is well-known
that an edge-transitive circulant graph is also arc-transitive. Thus r = 1 and d′ = 6. As
n > 50, Lemma 2.4 can be applied to Γ′/B and C̄. This says that Aut(Γ′/B) has a normal
subgroup N such that
(1) N = C̄, or
(2) n ≡ 4(mod 8) and N = C̄n/4, or
(3) N ∼= Zℓ3 for ℓ ≥ 2 and C̄3 ≤ N .
In case (1), N < Ḡ, hence KC ◁ G. The condition r = 1 yields that K = ⟨τ⟩. Thus
KC is abelian and ⟨x2 : x ∈ KC⟩ = Cn if n is odd and Cn/2 if n is even. Using that
the latter group is characteristic in KC and KC ◁ G, we obtain that coreG(C) ̸= 1, a
contradiction.
14 Ars Math. Contemp. 25 (2025) #P2.03
In case (2), N < Ḡ, hence KCn/4 ◁ G. Since KC is abelian, it follows that Cn/4 is
characteristic in KC, implying that Cn/4 ◁G, a contradiction.
In case (3), C̄3 ≤ Ḡ ∩ N . Since Ḡ ∩ N ◁ Ḡ, it follows that G contains a normal
subgroup M such that M = ⟨τ⟩ × S, where S ∼= Zℓ
′
3 for some ℓ
′ ≥ 1. Thus S is normal
in G, and we obtain that Orb(S, V (Γ)) is a non-trivial cyclic block system for G. This
contradicts Lemma 5.4, and we conclude that d = 12.
The graph Γ′ is a normal r-cover of Γ′/B, where r = 1 or r = 2, and we have
d′ = d/r = 12/r. If r = 2, then u0 is adjacent with 6 vertices that are contained in the set
{ui : i ∈ Zn}. It follows from the definition of Γ′ that this impossible. Thus r = 1, and so
d′ = 12.
Lemma 5.6. Assuming Hypothesis 5.1, suppose that n > 50 and B is a non-cyclic block
system for G with blocks of size 2. Then there is a normal non-cyclic block system for G
with blocks of size 4.
Proof. Let K be the kernel of the action of G on B, and for a subgroup X ≤ G, denote
by X̄ the image of X induced by its action on B. For the sake of simplicity we write Γ̄ for
Γ/B.
By Lemma 5.5, Γ̄ has valency 12. This implies that K = 1. As C̄ ≤ Aut(Γ̄) and it
is regular on V (Γ̄), Theorem 2.1 can be applied to Γ̄ and C̄. As n > 50, Γ̄ cannot be the
complete graph. Also, if C̄◁Aut(Γ̄), then C◁G because K = 1. This is also impossible,
hence Γ̄ is in one of the families (c) and (d) of Theorem 2.1.
Case 1: Γ̄ is in family (c).
In this case Orb(C̄d, V (Γ̄)) is a block system for Aut(Γ̄), hence for Ḡ as well, where
d ∈ {2, 3, 4, 6}. Let N be the unique subgroup of G for which N̄ is the kernel of the
action of Ḡ on Orb(C̄d, V (Γ̄)). Note that N ◁ G and N ∼= N̄ because K = 1. Let
B′ = Orb(N,V (Γ)). It follows that B′ is non-cyclic and it has blocks of size 2d.
Let d = 2. Then B′ is normal with blocks of size 4, so the conclusion of the lemma
holds.
Let d = 3. Then the Sylow 3-subgroup of N̄ is normal in Ḡ. It follows in turn that, the
Sylow 3-subgroup of N is normal in G, the orbits of the latter subgroup form a non-trivial
cyclic block system for G. This contradicts Lemma 5.4.
Let d = 4. Then Γ̄ has valency 3. It follows from Lemma 2.3 that n ≤ 6, but this is
excluded.
Finally, let d = 6. Let τ be the permutation of V (Γ) defined in (5.1) and Γ′ be the graph
defined in (5.2). Let ∆ be the subgraph of Γ′ induced by the set uN0 ∪ u
N
1 . It is not hard to
show that ∆ is a bipartite graph, it has valency 6, and it is also edge-transitive. Moreover,
if B ∈ B such that B ⊂ uN1 , then
|∆(u0) ∩B| = 1. (5.3)
Since ⟨τ⟩ × C6 ≤ Aut(∆), it follows that ∆ is uniquely determined by ∆(u0). It follows
from the definition of Γ′ that |Γ′(u0) ∩ {ui : i ∈ Zn}| = 4. Therefore, replacing uN0 ∪ u
N
1
with uN0 ∪ u
N
n−1 if necessary, we may assume w.l.o.g. that |∆(u0) ∩ {ui : i ∈ Zn}| ≤ 2.
This together with (5.3) show that there are 6 possibilities for ∆. Letting B = {u0, vb}
and n = 6l, the biparts of ∆ are {uxl, vxl+b : x ∈ Z6} and {uxl+1, vxl+b+1 : x ∈ Z6}.
Using (5.3), ∆ can be described as follows. There exists i ∈ Z6 such that ∆(u0) ∩ {ui :
I. Kovács: Edge-transitive core-free Nest graphs 15
i ∈ Zn} = {u1, uil+1}. Then as τ ∈ Aut(∆), ∆(v0) ∩ {vi : i ∈ Zn} = {vb+1, vil+b+1},
and for every j ∈ Z6,
∆(ujl) = {vxl+b+1 : x ∈ Z6, x ̸= j, j + i} ∪ {ujl+1, u(j+i)l+1},
∆(vjl+b) = {uxl+b+1 : x ∈ Z6, x ̸= i, j + i} ∪ {vjl+b+1, v(j+i)l+b+1}.
A computation with MAGMA [4] shows that none of these 6 graphs is edge-transitive.
Case 2: Γ̄ is in family (d).
We finish the proof by showing this case does not occur. Theorem 2.1 shows that
B1 := Orb(C̄d, V (Γ̄)) and B2 := Orb(C̄n/d, , V (Γ̄)) are blocks for Ḡ for some divisor d
of n such that d ∈ {4, 5, 7} and gcd(d, n/d) = 1. Furthermore,
C̄d × C̄n/d < Ḡ ≤ Aut(Γ
′) = G1 ×G2,
where C̄d ≤ G1, G1 ∼= Sd, C̄n/d < G2 and G2 ∼= Aut(Γ′/B1).
Let d = 4. Then Γ̄/B1 has valency 4. Using also that n/d is odd and that n > 20, it
follows from Lemma 2.3 that C̄n/d ◁ Ḡ, hence Cn/d ◁G, a contradiction.
Let d = 5. Then Γ̄/B1 has valency 3, hence n ≤ 30 by Lemma 2.3, which is excluded.
Finally, let d = 7. Then Γ̄/B1 is a cycle of length n/7, implying that C̄n/d ◁ Ḡ, so
Cn/d ◁G, a contradiction.
Before the proof of Theorem 5.2 we need two more lemmas dealing with non-cyclic
block systems with blocks of size at least 4.
Lemma 5.7. Assuming Hypothesis 5.1, suppose that n > 50 and B is a minimal non-cyclic
block system for G with blocks of size at least 4, and let B ∈ B be any block. Then the
permutation group of B induced by G{B} is an affine group.
Proof. For a subgroup X ≤ G{B}, denote by X∗ the image of X induced by its action
on B. As B is minimal, (G{B})∗ is a primitive permutation group. Also, (C{B})∗ is a
semiregular cyclic subgroup of (G{B})∗ with 2 orbits, hence Theorem 2.5 can be applied
to (G{B})∗. This shows that (G{B})∗ is either an affine group or it is one of the groups
in the families (a) – (f) in part (2) of Theorem 2.5. Assume that the latter case occurs. We
derive in three steps that this leads to a contradiction. Let K be the kernel of the action of
G on B.
Step 1: K acts faithfully on every block in B.
Since K ◁G{B}, it follows that K∗ ◁ (G{B})∗. By Corollary 2.6, K∗ is primitive and
belongs to the same family as (G{B})∗.
Assume to the contrary that K is not faithful on every block. Using the connectedness
of Γ, it is easy to show that there are blocks B,B′ in B with the following properties: The
kernel of the action of K on B is non-trivial on B′, and Γ has an edge {w,w′} such that
w ∈ B and w′ ∈ B′. Denote by N the latter kernel. Now as N ◁ K and K is primitive
on B′, N is transitive on B′. Thus the orbit (w′)N = B′, and so w is adjacent with any
vertex in B′. Since B is normal, it follows that the subgraph of Γ induced by B ∪ B′ is
isomorphic to the complete bipartite graph Km,m, where m = |B|. On the other hand,
m ≥ 6, showing that Γ ∼= K6,6, a contradiction.
Denote by B and B′ the blocks containing u0 and u1, respectively.
16 Ars Math. Contemp. 25 (2025) #P2.03
Step 2: The action of K on B is equivalent with its action on B′.
Assume to the contrary that the actions are inequivalent. Then K is a group described
in one of the cases (i) and (ii) of Lemma 2.7(2). If K is described in Lemma 2.7(2)(i), then
|B| = 6 or 12, and the stabilizer Ku0 acts transitively on the block B
′. This implies that
every vertex in B is adjacent with every vertex in B′, hence |B| = 6 and Γ ∼= K6,6. This
is impossible.
Let K be a group described in Lemma 2.7(2)(ii). Then K belongs to family (c) in
Theorem 2.5(2) with d ≥ 4, and the elements in B and B′ correspond to the points and the
hyperplanes of the projective geometry PG(d − 1, q), respectively. The set B′ splits into
two Ku0 -orbits of lengths
(qd−1 − 1)/(q − 1) and q(qd−1 − 1)/(q − 1).
The first orbit consists of the hyperplanes of PG(d−1, q) through the point represented by
u0, and the second orbit consists of the remaining hyperplanes. Clearly, the minimum of
these numbers is bounded above by the valency of Γ, implying that qd−1 − 1 ≤ 6(q − 1),
and hence qd−2 < 6. This is impossible because d ≥ 4.
Step 3: coreG(C) ̸= 1.
Since K acts equivalently on B and B′, it follows that Ku0 = Kv for some vertex
v ∈ B′ (see [8, Lemma 1.6B]). Define the binary relation ∼ on V (Γ) by letting u ∼ v if
and only if Ku = Kv . It is not hard to show, using that K ◁G, that ∼ is a G-congruence
(see [8, Exercise 1.5.4]), and so there is a block for G containing u0 and v. Also, as K is
not regular, this block is non-trivial, and this shows that v ̸= u1.
By Lemma 2.7(1), K is 2-transitive on B′, unless |B′| = 10, K = A5 or S5, and it has
subdegrees 1, 3 and 6.
Assume first that K is 2-transitive on B′. Then the orbit u
Ku0
1 = u
Kv
1 = |B
′| − 1 and
each vertex in uKv1 is adjacent with u0. Hence u0 has |B
′| − 1 neighbours in B′. On the
other hand, as B is normal, this number divides 6, so |B| = |B′| = 4, contradicting that
G∗{B} is an almost simple group.
We are left with the case that |B′| = 10, K = A5 or S5, and it has subdegrees 1, 3 and
6. Consequently, u0 has 3 or 6 neighbours in B′.
If u0 has 6 neighbours, then it is clear that n = 10, which is excluded.
Now assume that u0 has 3 neighbours in B′. In this case Γ is a normal 3-cover of a cycle
of length n/5. Since Γ/B is a cycle of length n/5, it follows that |G| ≤ |K| · 2n/5 = 48n.
Using that n > 50, Theorem 2.10 shows that coreG(C) ̸= 1.
Lemma 5.8. Assuming Hypothesis 5.1, suppose that n > 50 and B is a minimal non-cyclic
block system for G with blocks of size at least 4. Then the blocks have size 4.
Proof. Let K be the kernel of the action of G on B, and let B ∈ B be the block containing
u0. Denote by (G{B})∗ the permutation group of B induced by G{B}. By Lemma 5.7,
(G{B})
∗ is an affine group, and thus it is one of the groups in the families (a) – (d) in part
(1) of Theorem 2.5. In particular, |B| ∈ {4, 8, 16}. Assume to the contrary that |B| > 4.
By Lemma 3.6, B is normal, hence Γ is a normal r-cover of Γ/B for r ∈ {1, 2, 3}.
Case 1: r = 1.
I. Kovács: Edge-transitive core-free Nest graphs 17
In this case K is regular on every block, in particular, K ∼= Z32 or Z
4
2. On the other
hand, by Lemma 3.6(2), C|B|/2 < K, a contradiction.
Case 2: r = 2.
In this case Γ/B has valency 3. It follows from Lemma 2.3 that n ≤ 48, but this is
excluded.
Case 3: r = 3.
Then Γ/B is a cycle of length 2n/|B|. This implies that the action of Gu0 on Γ(u0)
admits a block system consisting of two blocks of size 3. Consequently, the restriction of
Gu0 to Γ(u0) is a {2, 3}-group. This together with the fact that Γ is connected yield that
Gu0 is also a {2, 3}-group. Now checking the stabilisers in part (1) of Theorem 2.5, we
find that |B| = 16 and
(G{B})
∗
u0
∼= (Z3 × Z3) : Z4 or (S3 × S3) : Z2 (5.4)
Assume for the moment K is not faithful on B. Then there exist adjacent blocks B′ and
B′′ such that the kernel of the action of K on B′ is non-trivial on B′′. Denote this kernel
by L. The L-orbits contained in B′′ have the same size, which is equal to 2s for some
1 ≤ s ≤ 4. On the other hand, for w ∈ B′, the set Γ(w)∩B′′ is L-invariant, implying that
|Γ(w) ∩B′′| is equal to some power of 2, a contradiction. Thus K is faithful on B.
The group K contains a normal subgroup E such that E ∼= Z42. Note that B =
Orb(E, V (Γ)). Let P be the Sylow 3-subgroup of G{B}. Since Γ/B is a cycle, it follows
that P ≤ K. This also shows that P ∼= Z23. Also, C8 ≤ K, and in view of (5.4), we obtain
that |(G{B})∗ : K| ≤ 2 and if the index is equal to 2, then (G{B})∗u0
∼= (S3 × S3) : Z2. A
direct check by MAGMA [4] shows that in the latter case (G{B})∗ has a unique subgroup
of index 2 containing an element of order 8, which is also primitive. All these show that K
is primitive on B.
Denote by ∆ be the subgraph of Γ induced by uE0 ∪ u
E
1 . Using that E ∼= Z
4
2 acting
regularly on both uE0 and u
E
1 , it is not hard to show that ∆ is the union of four 3-dimensional
cube Q3. If ∆1 is a component of ∆, then |V (∆1) ∩ B| = 4 (note that B = uE0 ) and
V (∆1) ∩ B is a block for K. This, however, contradicts the fact that K is primitive on
B.
We are ready to settle Theorem 5.2, and therefore, Theorem 1.3 as well.
Proof of Theorem 5.2. In view of Lemma 5.3, we may assume that n > 50. It follows
from Lemmas 5.4 – 5.8 that G admits a normal non-cyclic block system with blocks of size
4. Denote this block system by B. Let K be the kernel of the action of G on B, and for a
subgroup X ≤ G, denote by X̄ the image of X induced by its action on B. As B is normal,
Γ is a normal r-cover of Γ/B for some r ∈ {1, 2, 3}. We exclude below all possibilities
case-by-case.
Case 1: r = 1.
In this case |K| = 4 and K ∩ C = C2. If K ∼= Z4, then C2 is characteristic in K, and
therefore, it is normal in G. This is impossible because coreG(C) = 1, hence K ∼= Z22
The graph Γ/B is edge-transitive, it has valency 6, and C̄ is regular on V (Γ/B). As
n > 50, Lemma 2.4 can be applied to Γ/B and C̄. It follows that Aut(Γ/B) has a normal
subgroup N such that
18 Ars Math. Contemp. 25 (2025) #P2.03
(1) N = C̄, or
(2) n ≡ 4(mod 8) and N = C̄n/4, or
(3) N ∼= Zℓ3 for ℓ ≥ 2 and C̄3 ≤ N .
In case (1), we obtain that KC ◁ G, whereas in case (2), KCn/4 ◁ G. In either
case, |KC : C| = 2, and therefore, for the derived subgroup (KC)′, (KC)′ ≤ C. Thus
(KC)′ ≤ coreG(C), and so (KC)′ = 1, i.e., KC is an abelian group. Then we obtain
that Cn/2 ◁ G in case (1), and Cn/4 ◁ G in case (2). None of these is possible because
coreG(C) = 1.
In case (3), G contains a normal subgroup M such that KC3 ≤ M and M/K ∼= Zℓ
′
3 for
some ℓ′ ≥ 1. Then M can be written as M = KS where C3 ≤ S and S ∼= Zℓ
′
3 . As C2 ≤
K, we obtain that C3 commutes with K, and so C3 ≤ O3(M), where O3(M) denotes
the largest normal 3-subgroup of M . As O3(M) is characteristic in M , O3(M) ◁ G.
This yields that Orb(O3(M), V (Γ)) is a non-trivial cyclic block system, a contradiction to
Lemma 5.4.
Case 2: r = 2.
In this case Γ/B has valency 3, hence n ≤ 12 by Lemma 2.3, which is excluded.
Case 3: r = 3.
The group K is faithful on every block of B. This can be shown by copying the ar-
gument that has been used in Case 3 in the proof of Lemma 5.8. Since Γ/B is a cycle of
length n/2, it follows that |G| ≤ |K| · n = 24n. Using that n > 50, Theorem 2.10 shows
that coreG(C) ̸= 1, a contradiction.
ORCID iDs
István Kovács https://orcid.org/0009-0001-5199-2850
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.04
https://doi.org/10.26493/1855-3974.3130.d46
(Also available at http://amc-journal.eu)
On optimal λ-separable packings in the plane
Károly Bezdek *
Department of Mathematics and Statistics, University of Calgary, Canada and
Department of Mathematics, University of Pannonia, Veszprém, Hungary
Zsolt Lángi †
Department of Algebra and Geometry, Budapest University of Technology and Economics
and HUN-REN Alfréd Rényi Institute of Mathematics, Budapest, Hungary
Received 17 May 2023, accepted 8 April 2024, published online 12 March 2025
Abstract
Let P be a packing of circular disks of radius ρ > 0 in the Euclidean, spherical, or
hyperbolic plane. Let 0 ≤ λ ≤ ρ. We say that P is a λ-separable packing of circular
disks of radius ρ if the family P ′ of disks concentric to the disks of P having radius λ
form a totally separable packing, i.e., any two disks of P ′ can be separated by a line which
is disjoint from the interior of every disk of F ′. This notion bridges packings of circular
disks of radius ρ (with λ = 0) and totally separable packings of circular disks of radius ρ
(with λ = ρ). In this note we extend several theorems on the density, tightness, and contact
numbers of disk packings and totally separable disk packings to λ-separable packings of
circular disks of radius ρ in the Euclidean, spherical, and hyperbolic plane. In particular,
our upper bounds (resp., lower bounds) for the density (resp., tightness) of λ-separable
packings of unit disks in the Euclidean plane are sharp for all 0 ≤ λ ≤ 1 with the extremal
values achieved by λ-separable lattice packings of unit disks. On the other hand, the bounds
of similar results in the spherical and hyperbolic planes are not sharp for all 0 ≤ λ ≤ ρ
although they do not seem to be far from the relevant optimal bounds either. The proofs
use local analytic and elementary geometry and are based on the so-called refined Molnár
decomposition, which is obtained from the underlying Delaunay decomposition and as such
might be of independent interest.
Keywords: Euclidean, spherical and hyperbolic plane, λ-separable packing, density, tightness, con-
tact number, refined Molnár decomposition.
Math. Subj. Class. (2020): 52A55, 52A40, 52C15.
*Corresponding author. Partially supported by a Natural Sciences and Engineering Research Council of
Canada Discovery Grant.
†Partially supported by the ERC Advanced Grant “ERMiD”, and the NKFIH grant K147544.
E-mail addresses: kbezdek@ucalgary.ca (Károly Bezdek), zlangi@math.bme.hu (Zsolt Lángi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.04
1 Introduction
Let M ∈ {E2,H2, S2} denote the Euclidean, hyperbolic, or spherical plane, i.e., one of
the simply connected complete 2-dimensional Riemannian manifolds of constant sectional
curvature. Since simply connected complete space forms, the sectional curvature of which
have the same sign are similar, we may assume without loss of generality that the sectional
curvature of M is 0,−1, or 1.
Recall that finding the densest packing of congruent (circular) disks in M is a classical
problem of discrete geometry that has been investigated in great details with the basic
results published in L. Fejes Tóth’s classical book [11]. The concept of totally separable
packings is more recent. It was introduced by G. Fejes Tóth and L. Fejes Tóth in [8] and
attracted significant attention. In this paper we introduce a new family of packings called
λ-separable packings that bridge packings and totally separable packings. Before defining
it, we recall some notions from the theory of packings.
A countable family P of convex bodies in the Euclidean plane E2 is called a packing
if any two elements are non-overlapping, i.e. their interiors are disjoint. A packing P is
called translative if all elements in P are translates of a given planar convex body, and a
lattice packing, if it is translative, and the translation vectors form a lattice. A packing
P is called totally separable if for any two elements of P there is a straight line in R2
that separates them, and does not intersect the interior of any element of P . An often
investigated concept regarding packings is their density, which measure the area of the
fraction of the plane covered by the elements of the packing. Namely, if P is a packing in
E
2, then its (upper) density is defined as the quantity
lim sup
τ→∞
area
(
⋃
K∈P(τB) ∩K
)
τ2π
,
where area(·) denotes area, and τB is the closed Euclidean disk of radius τ , centered at
the origin o. It is worth noting that the notion of packing and totally separable packing
can be naturally extended to the spherical and hyperbolic planes. By the compactness of
the spherical plane S2, the density of a packing can be defined in a straightforward way,
without using limits, whereas it has been observed by Böröczky [5] that the notion of
density of sphere packings in hyperbolic space has to to be introduced with respect to well-
defined underlying cell decompositions of the hyperbolic space (such as the hyperbolic
refined Molnár decomposition introduced in the Appendix A).
Definition 1.1. Let 0 ≤ λ ≤ ρ. Let P be a packing of (circular) disks of radius ρ in
M ∈ {E2,H2, S2}. We say that P is a λ-separable packing of disks of radius ρ if the
family P ′ of disks concentric to the disks of P having radius λ form a totally separable
packing in M, i.e., any two disks of P ′ can be separated by a line in M which is disjoint
from the interior of every disk of P ′.
Remark 1.2. Clearly, a packing of disks of radius ρ in M is a 0-separable packing of disks
of radius ρ in M ∈ {E2,H2, S2}. On the other hand, a totally separable packing of disks
of radius ρ in M is a ρ-separable packing of disks of radius ρ in M ∈ {E2,H2, S2}.
In this note we are going to investigate λ-separable packings of disks of radius ρ in
M ∈ {E2,H2, S2} from the point of view of density, tightness, and contact numbers.
Before stating our main theorems, we introduce some notation that we are going to use in
the formulation of all our results, and also in their proofs.
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 3
Let
xs1(y) :=
1
2
arcsin
cosλ sin2 y
√
sin2 y − sin2 λ
, if 0 < λ <
π
4
, arcsin tanλ < y <
π
2
, (1.1)
xs2(y) :=
π
2
− 1
2
arcsin
cosλ sin2 y
√
sin2 y − sin2 λ
, if 0 < λ <
π
4
, arcsin tanλ < y <
π
2
, (1.2)
xh(y) :=
1
2
arcsinh
coshλ sinh2 y
√
sinh2 y − sinh2 λ
, if 0 < λ < y, (1.3)
xe(y) :=
y2
2
√
y2 − λ2
, if 0 < λ < y. (1.4)
Furthermore, for any 0 < λ < π4 , arcsin tanλ < y <
π
2 and i = 1, 2 let T
s
i (y)
denote the spherical isosceles triangle of edge lengths 2y, 2xsi (y), 2x
s
i (y), and let R
s
i (y)
denote the circumradius of T si (y). Similarly, for any 0 < λ < y, let T
h(y) denote the
hyperbolic isosceles triangle with edge lengths 2y, 2xh(y), 2xh(y), and let Rh(y) denote
the circumradius of Th(y). We denote by T e(y) the Euclidean isosceles triangle with
edge lengths 2y, 2xe(y), 2xe(y) and its circumradius by Re(y). In addition, we denote
the regular spherical, hyperbolic, Euclidean triangle of edge length 2ρ by T sreg(ρ),T
h
reg(ρ)
and T ereg(ρ), respectively, and the circumradii of these triangles by R
s
reg(ρ), R
h
reg(ρ), and
Rereg(ρ), respectively.
Remark 1.3. It is an elementary computation to check that xs1 as a function of y is strictly
decreasing over the closed interval S1 := [arcsin tanλ, arcsin(
√
2 sinλ)] and strictly in-
creasing over S2 := [arcsin(
√
2 sinλ), π/2], with xs1(arcsin tanλ) = x
s
1(π/2) =
π
4 and
xs1(arcsin(
√
2 sinλ)) = λ. Since xs2(y) =
π
2 − xs1(y), similar statements hold for xs2. For
k = 1, 2 and j = 1, 2, we denote the inverse of the restriction of xsj to Sk by x
s
j |−1Sk . We
remark that arcsin tanλ < arcsin(
√
2 sinλ) < π2 holds for all λ ∈ (0, π/4). Further-
more, xh(y) is a strictly decreasing function of y on H1 := (λ, arcsinh(
√
2 sinhλ)], and
strictly increasing on H2 := [arcsinh(
√
2 sinhλ),+∞), with xh(y) → +∞ as y → λ+
or y → +∞. We denote the inverse of xh on the two intervals by xh|−1H1 and x
h|−1H2 ,
respectively.
For any 0 < λ ≤ arcsin 35 , let
yss(λ) := arcsin
√
3 + 5 sin2 λ−
√
9− 34 sin2 λ+ 25 sin4 λ
8
and (1.5)
ysb(λ) := arcsin
√
3 + 5 sin2 λ+
√
9− 34 sin2 λ+ 25 sin4 λ
8
, (1.6)
and for any λ > 0, let
yhs (λ) := arcsinh
√
5 sinh2 λ− 3 +
√
25 sinh4 λ+ 34 sinh2 λ+ 9
8
. (1.7)
4 Ars Math. Contemp. 25 (2025) #P2.04
Finally, for any 0 < λ < π2 , we set
ysmin(λ) := arcsin


1
6
√
6A1/3 − 216
(
− 109 L4 + 23L2
)
A1/3
+ 60L2

 , (1.8)
where L := sinλ and A := 100L6 − 36L4 + 12
√
−375L12 + 750L10 − 471L8 + 96L6.
Similarly, for any 0 < λ, we set
yhmin(λ)
:= arcsinh
√
5L2
3
+
2
3
√
10L4 + 6L2 cos
(
1
3
arccos
(
(25L2 + 9)L
4
√
2(5L2 + 3)3/2
)
− 2π
3
)
,
(1.9)
where L := sinhλ.
An elementary computation shows that the above expressions exist on the required
intervals.
1.1 Euclidean results
One of the fundamental problems in the theory of packings is finding the maximum density
for certain subfamilies of packings.
Definition 1.4. Let B = {x ∈ E2 : ||x|| ≤ 1} denote the unit disk centered at the origin
of E2, where || · || stands for the Euclidean norm of the corresponding vector (resp., point)
in E2. Let δλ(B) (resp., δ
∗
λ(B)) denote the largest density of λ-separable packings of unit
disks (resp., λ-separable lattice packings of unit disks) in E2, i.e., the largest fraction of E2
covered by λ-separable packings of unit disks (resp., λ-separable lattice packings of unit
disks) in E2.
It was proved in [8] that δ1(B) = δ
∗
1(B) =
π
4 . On the other hand, it is well known [11]
that δ0(B) = δ
∗
0(B) =
π√
12
. The following theorem extends these results to λ-separable
packings as follows.
Theorem 1.5. Let B = {x ∈ E2 : ||x|| ≤ 1}. Then
δλ(B) = δ
∗
λ(B) =
{
π√
12
, if 0 ≤ λ ≤
√
3
2 ,
π
4λ , if
√
3
2 ≤ λ ≤ 1.
Furthermore, for 0 ≤ λ ≤
√
3
2 and
√
3
2 ≤ λ ≤ 1 the lattice packing F of unit disks whose
Delaunay triangles are T ereg(1) and T
e(
√
2− 2
√
1− λ2), respectively, is a λ-separable
packing with density δλ(B) (cf. Figure 1).
Our next topic is the tightness of λ-separable packings.
Definition 1.6. Let γλ(B) (resp., γ
∗
λ(B)) denote the smallest r > 0 such that there exists
a λ-separable unit disk packing B +X (resp., a λ-separable lattice packing B + Λ of unit
disks, where Λ denotes a 2-dimensional lattice in E2) satisfying E2 = rB + X (resp.,
E
2 = rB + Λ). We call γλ(B) (resp., γ
∗
λ(B)) the λ-separable tightness constant (resp.,
λ-separable lattice tightness constant) of B. In particular, if P := B +X is a packing of
unit disks in E2, then the smallest r > 0 satisfying E2 = rB +X is called the tightness of
P .
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 5
Figure 1: A densest λ-separable packing of unit disks in E2 with λ = 0.93. The unit disks
and the sides of the Delaunay triangles are denoted by solid lines. The concentric disks of
radius λ and the lines separating them are drawn with dotted and dashed lines, respectively.
The bases of the Delaunay triangles are horizontal, and their length is greater than 2. The
legs of the Delaunay triangles are of length 2.
The notion of tightness constant goes back to Rogers (see [20]) and it is often called the
simultaneous packing and covering constant. Closely related notions were introduced by
Ryskov [16] as well as L. Fejes Tóth [10]. It was noted in [10] that γ0(B) = γ
∗
0 (B) =
2√
3
.
The following theorem is an extension of that to λ-separable packings.
Theorem 1.7. Let B = {x ∈ E2 : ||x|| ≤ 1}. Then
γλ(B) = γ
∗
λ(B) =







2√
3
, if 0 ≤ λ ≤
√
3
2 ,√
2−2
√
1−λ2
λ , if
√
3
2 ≤ λ ≤ 2
√
2
3 ,
3
√
3λ
4 , if
2
√
2
3 ≤ λ ≤ 1.
Furthermore, the tightness of a λ-separable packing F of unit disks is γλ(B) if and only if
the Delaunay triangles defined by F tile E2 and are all congruent to
T ereg(1), T
e
(√
2− 2
√
1− λ2
)
and T e
(
√
3
2
λ
)
,
respectively (see Figure 2).
According to [13] the number of touching pairs in a packing of n > 1 unit disks in E2
(called the contact number of the given unit disk packing) is at most ⌊3n −
√
12n− 3⌋,
where ⌊·⌋ denotes the lower integer part of the given real. On the other hand, it is proved
in [4] (see also [1]) that the contact number of an arbitrary totally separable packing of
n > 1 unit disks in E2 is at most ⌊2n − 2√n⌋. Both upper bounds are sharp and one can
characterize the extremal configurations (see [14] and [1]). For a comprehensive survey
on the contact numbers, see [2]. The problem of finding a sharp upper bound for the
contact numbers of λ-separable packings of n > 1 unit disks in E2 seems to be a difficult
question. We have only the following result, which is obtained using the proof technique
of Theorem 4 of Eppstein [7] combined with Theorem 1.5. To state it, we first define a
necessary concept in Definition 1.8.
6 Ars Math. Contemp. 25 (2025) #P2.04
(a) (b) (c)
T T
T
Figure 2: The Delaunay triangle T of three λ-separable unit disks in E2 for λ = 0.5 <
√
3
2
(case (a)), λ = 0, 9 ∈
(√
3
2 ,
2
√
2
3
)
(case (b)) and λ = 0.97 > 2
√
2
3 (case (c)). The
disks of radius λ centered at the vertices of the triangles, and the lines separating them
are drawn by dotted and dashed lines, respectively. The triangle T in case (a) is a regular
triangle of edge length 2, the one in case (b) is an isosceles triangle of edge lengths 2, 2 and
2
√
2− 2
√
1− λ2, and the one in case (c) is an isosceles triangle with edge lengths 3λ√
2
,
3λ√
2
and
√
6λ.
Definition 1.8. Let cλ(n,B) denote the largest number of touching pairs of unit disks in a
λ-separable packing of n > 1 unit disks in E2.
Theorem 1.9.
(i) Let 0 ≤ λ ≤
√
3
2 . Then cλ(n,B) = ⌊3n−
√
12n− 3⌋ holds for all n > 1.
(ii) Let
√
3
2 < λ ≤ 1. Then ⌊2n− 2
√
n⌋ ≤ cλ(n,B) ≤ 2n−
√
πλ
√
n+O(1) holds for
n > 1, where 1.6494... =
√
π
√
3
2 <
√
πλ ≤ √π = 1.7724....
Theorem 1.9 is the motivation behind Problem 1.10.
Problem 1.10. Let
√
3
2 < λ ≤ 1. Then prove or disprove that cλ(n,B) = ⌊2n − 2
√
n⌋
holds for all n > 1.
Remark 1.11. Recall the following more general question of Swanepoel [18]: Is it true
that the number of touching pairs in any packing of n > 1 unit disks in E2 having no
touching triplets (i.e., whose contact graph is triangle-free) is at most ⌊2n − 2√n⌋? We
note that if
√
3
2 < λ ≤ 1, then clearly in any λ-separable packing of n > 1 unit disks in E2
there are no touching triplets. Thus, if the answer to Swanepoel’s question is affirmative,
then so is the answer to Problem 1.10.
1.2 Spherical results
In the next two theorems we deal with λ-separable packings of spherical caps of radius ρ
in S2. Clearly, if such a packing contains at least three caps, then λ ≤ π4 , and we will
see that in this case the inequality λ ≤ π2 − ρ is also satisfied. Totally separable packings
of congruent spherical caps have been investigated in the recent paper [3]. In particular,
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 7
[3, Theorem 1.10] provides an upper bound on the density of a totally separable packing
of congruent spherical caps on S2. The next theorem extends Part (i) of Theorem 1.10 in
[3] to the density of λ-separable packings of spherical caps of radius ρ in S2, introduced in
Definition 1.12.
Definition 1.12. Let δsλ(ρ) denote the largest density of λ-separable packings of spherical
caps of radius ρ in S2, i.e., the largest fraction of S2 covered by a λ-separable packing of
spherical caps of radius ρ in S2.
Theorem 1.13. Let 0 ≤ λ ≤ ρ < π2 with ρ > 0 and λ ≤ π2 − ρ. Then
δsλ(ρ) ≤





δ(T s1 (x
s
1|−1S1 (ρ))), if λ ≤ ρ ≤ min
{
yss(λ),
π
4
}
,
δ(T sreg(ρ)), if y
s
s(λ) < ρ ≤ ysb(λ),
δ(T s2 (ρ)), if
π
4 < ρ, and ρ < y
s
s(λ) or ρ > x
s
b(λ).
(1.10)
Furthermore, in the above three cases we have equality if and only if there is a λ-separable
packing F of spherical caps of radius ρ whose Delaunay triangles tile S2 and are congruent
to T s1 (x
s
1|−1S1 (ρ)), T
s
reg(ρ), or T
s
2 (ρ), respectively.
Our next topic is the tightness of λ-separable packings of spherical caps on S2.
Definition 1.14. Let γsλ(ρ) denote the smallest r > 0 such that there exists a λ-separable
packing P of spherical caps of radius ρ in S2 having the property that the spherical caps of
radius r concentric to the spherical caps of P cover S2.
The investigation of the tightness of packings of spherical caps of radius ρ was started
by L. Fejes Tóth in [9], where an upper bound was derived, which was sharp for certain
values of ρ. The following theorem extends these investigations to λ-separable packings of
spherical caps of radius ρ in S2 as follows.
Theorem 1.15. Let 0 ≤ λ ≤ ρ < π2 with ρ > 0 and λ ≤ π2 − ρ. Then
γsλ(ρ) ≥









Rs1(y
s
min(λ)), if λ ≤ ρ ≤ xs1(ysmin(λ)),
Rs1(x
s
1|−1S1 (ρ)), if x
s
1(y
s
min(λ)) < ρ ≤ min
{
ys(λ),
π
4
}
,
Rsreg(ρ), if y
s
s(λ) < ρ ≤ ysb(λ),
Rs2(ρ), if
π
4 < ρ, and ρ < y
s
s(λ) or ρ > x
s
b(λ).
(1.11)
Furthermore, in the above four cases we have equality if and only if there is a λ-separable
packing F of spherical caps of radius ρ whose Delaunay triangles tile S2 and are congruent
to T s1 (y
s
min(λ)), T
s
1 (x
s
1|−1S1 (ρ)), T
s
reg(ρ), or T
s
2 (ρ), respectively.
Remark 1.16. Using the characterization of tilings of S2 with congruent spherical (isosce-
les) triangles (see [6] and [19]), one may find pairs (λ, ρ) with 0 < λ < ρ where our
estimates are sharp. In particular, an elementary computation shows that the tilings of S2
generated by the isosceles triangles H16 and H20 in [19] are the Delaunay decomposi-
tions of λ-separable packings of spherical caps of radius ρ with both maximal density and
minimal tightness. The values of λ and ρ are
ρ =
1
2
arcsin
(√
2
√
2− 2
)
≈ 0.57186, λ = arcsin
(
2 sin π8
√
1 +
√
2
√
4 +
√
2
)
≈ 0.53644
8 Ars Math. Contemp. 25 (2025) #P2.04
for H16,
ρ =
1
2
arcsin
(
√
1 + 2 cos π5
1 + cos π5
)
≈ 0.55357,
λ = arcsin
(
2√
5
sin
π
10
√
1 + 2 cos
π
5
)
≈ 0.46365
for H20.
In addition, for n = 4, 6, 12, let Fn denote the family of n spherical caps of radius ρn
with ρ4 = arcsin
√
2
3 , ρ6 =
π
4 , ρ12 = arcsin
1
2 sin 2π
5
, where the centers of the caps are the
vertices of a regular tetrahedron, octahedron or icosahedron inscribed in S2, respectively.
The Delaunay triangles of Fn are regular triangles of edge length 2ρn, and thus, they are λ-
separable families of maximal density and minimal tightness for any value of λ satisfying
yss(λ) < ρ ≤ ysb(λ).
1.3 Hyperbolic results
Finally, in the next two theorems we deal with λ-separable packings of hyperbolic disks of
radius ρ in H2.
Definition 1.17. Let δhλ(ρ) denote the largest density of λ-separable packings of hyperbolic
disks of radius ρ in the cells of the hyperbolic refined Molnár decompositions in H2.
Our result for the density of λ-separable hyperbolic disks is as follows.
Theorem 1.18. Let 0 ≤ λ ≤ ρ, where 0 < ρ. Then
δhλ(ρ) ≤
{
δ(Th(xh|−1H1(ρ))), if λ ≤ ρ ≤ y
h
s (λ),
δ(T sreg(ρ)), if y
h
s (λ) < ρ.
(1.12)
Furthermore, in the above two cases we have equality if there is a λ-separable packing of
hyperbolic disks of radius ρ whose Delaunay triangles are congruent to Th(xh|−1H1(ρ)) or
Threg(ρ), respectively.
Next, we introduce the tightness of a λ-separable family of hyperbolic disks, and state
our result about it.
Definition 1.19. Let γhλ(ρ) denote the smallest r > 0 such that there exists a λ-separable
packing P of hyperbolic disks of radius ρ in H2 having the property that the hyperbolic
disks of radius r concentric to the hyperbolic disks of P cover H2.
Theorem 1.20. Let 0 ≤ λ ≤ ρ, where 0 < ρ. Then
γsλ(ρ) ≥





Rh(yhmin(λ)), if λ ≤ ρ ≤ xh(yhmin(λ)),
Rh(xh|−1H1(ρ)), if x
h(yhmin(λ)) < ρ ≤ yhs (λ),
Rhreg(ρ), if y
h
s (λ) < ρ.
(1.13)
Furthermore, in the above three cases we have equality if there is a λ-separable packing
of hyperbolic disks of radius ρ whose Delaunay triangles tile H2 and are congruent to
Th(yhmin(λ)), T
h(xh|−1H1(ρ)), or T
h
reg(ρ), respectively.
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 9
1.4 Preliminaries
In the rest of the paper we prove the theorems stated. The proofs use local analytic and ele-
mentary geometry and are based on the so-called refined Molnár decompositions. We note
that the Euclidean Molnár decomposition was introduced by Molnár in [15], the spherical
refined analogue of which was introduced and applied in [3]. For the sake of completeness,
the Appendix A of this paper gives a description of the hyperbolic refined Molnár decom-
position, the method of which extends to the Euclidean as well as the spherical plane.
In the proofs, for any two points p, q in M with M ∈ {E2,H2, S2}, which are assumed
not to be antipodal if M = S2, we denote the unique shortest geodesic segment connect-
ing them by [p, q]. The distance of p from q is labelled by d(p, q), and conv(X) (resp.,
area(X)) stands for the convex hull (resp., the area) of the set X ⊂ M. If M = S2,
conv(X) denotes the intersection of all closed hemispheres of S2 containing X , or if no
such hemisphere exists, then we set conv(X) = S2. Finally for any set S ⊆ M with
M ∈ {E2,H2, S2}, cl(S) denotes the closure of S.
2 Proofs of Theorems 1.13 and 1.15
In the proof we use the refined Molnár decomposition defined by a finite point set X of
S
2, as in the proof of Theorem 2 of [3]. We note that even though in the definition of this
decomposition we assumed that no closed hemisphere contains X , the same argument can
be applied to define it if there is a closed hemisphere of S2 containing X . In this case
the union of the cells of the refined Molnár decomposition is the spherical convex hull
conv(X) of X .
We prove our statement in a more general setting. Namely, we say that a packing F
of spherical caps of radius ρ is R-locally λ-separable if any triple of caps in F whose
centers are at pairwise spherical distances at most R are λ-separable. In the proof we
assume that F is a (2Rρ)-locally λ-separable packing of spherical caps of radius ρ, where
Rρ = arcsin(
√
2 sin ρ) (if ρ > π4 , then Rρ does not exist, in this case we simply assume
that F is λ-separable). In addition, we assume that F is (2Rρ)-saturated in S2, i.e., every
point p of S2 is at distance at most 2Rρ from the center of a cap in F , as otherwise the
spherical cap of center p and radius ρ can be added to Fm preserving its (2Rρ)-separability.
This implies that the circumradius of any cell in the Delaunay decomposition of the set X
of the centers of the elements of F is at most 2Rρ.
Here we note that Rρ is the circumradius of a regular spherical quadrangle with edge
length 2ρ.
We decompose the cells of the M -decomposition of conv(X) into two types of cells P
as follows:
(i) P has circumradius at most Rρ, in this case we say that P is type 1, or
(ii) it is of the form P = cl (conv{v, ci, cj} \ conv{v′, ci, cj}), where ci, cj ∈ C, v is
the circumcenter of a Delaunay cell with ci and cj as vertices and with circumradius
at least Rρ, and d(v
′, ci) = d(v
′, cj). In this case we say that P is type 2.
The above defined cell decomposition is called the refined M -decomposition of the
packing.
10 Ars Math. Contemp. 25 (2025) #P2.04
We observe that since the sides of P are of length at least 2ρ, any type 1 spherical
polygon P is either a triangle containing its circumcenter, or a regular spherical quadrangle
with edge length 2ρ; the latter quadrangle can be decomposed into two triangles containing
their circumcenter.
Let P be a cell of the refined M -decomposition. If the sum of the internal angles of P
is denoted by ϕ, then we define the density of F in P as
δ(P ) =
(1− cos ρ)ϕ
area(P )
.
Clearly, to prove Theorem 1.13 (respectively, Theorem 1.15 ), it is sufficient to prove that
for the density of F in any cell (respectively, circumradius of any cell) of the decomposition
is less than or equal to the value promised in the theorem.
If P is a type 2 cell of the decomposition, then this statement readily follows from [3,
Lemma 3], which we quote for completeness, without proof.
Lemma 2.1. Let Q be an isosceles spherical triangle, with vertices q1, q2, p, where p is the
apex. Let the length of the legs of Q be x, and that of the base be y, where 2ρ ≤ y < π
and x2 < y <
π
2 . For i = 1, 2, let Si denote the spherical cap of radius ρ, centered at pi.
Let f(x, y) denote the density of the packing {S1, S2} in Q as a function of x and y. Then
f(x, y) is a strictly decreasing function of both x and y.
Finally, if P is a type 1 cell of the decomposition, then the assertion follows from
Lemmas 2.2 and 2.3.
Lemma 2.2. Let T = conv{a, b, c} ⊂ S2 be a spherical triangle with edge lengths at
least 2ρ that contains its circumcenter. For ∗ ∈ {a, b, c}, let C∗ denote the closed spherical
cap of radius λ ≤ ρ centered at ∗. Assume that there are lines La and Lb such that La
separates Ca from Cb and Cc, and Lb separates Cb from Ca and Cc. Let δ(T ) denote the
density of {Cx, Cy, Cz} with respect to T . Then 0 ≤ λ < π4 , λ ≤ ρ ≤ π2 − ρ, and we have
the following.
δ(T ) ≤





δ(T s1 (x
s
1|−1S1 (ρ))), if λ ≤ ρ ≤ min
{
yss(λ),
π
4
}
,
δ(T sreg(ρ)), if y
s
s(λ) < ρ ≤ ysb(λ),
δ(T s2 (ρ)), if
π
4 < ρ, and ρ < y
s
s(λ) or ρ > x
s
b(λ).
(2.1)
Furthermore, in the above three cases we have equality if and only if T is congruent to
T s1 (x
s
1|−1S1 (ρ)), T
s
reg(ρ), or T
s
2 (ρ), respectively.
Lemma 2.3. Let T = conv{a, b, c} ⊂ S2 be a spherical triangle with edge lengths at
least 2ρ that contains its circumcenter. For ∗ ∈ {a, b, c}, let C∗ denote the closed spherical
cap of radius λ ≤ ρ centered at ∗. Assume that there are lines La and Lb such that La
separates Ca from Cb and Cc, and Lb separates Cb from Ca and Cc. Let R(T ) denote the
circumradius of T . Then 0 ≤ λ < π4 , λ ≤ ρ ≤ π2 − λ, and we have the following.
R(T ) ≥









Rs1(y
s
min(λ)), if λ ≤ ρ ≤ xs1(ysmin(λ)),
Rs1(x
s
1|−1S1 (ρ)), if x
s
1(y
s
min(λ)) < ρ ≤ min
{
ys(λ),
π
4
}
,
Rsreg(ρ), if y
s
s(λ) < ρ ≤ ysb(λ),
Rs2(ρ), if
π
4 < ρ, and ρ < y
s
s(λ) or ρ > x
s
b(λ).
(2.2)
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 11
Furthermore, in the above four cases we have equality if and only if T is congruent to
T s1 (y
s
min), T
s
1 (x
s
1|−1S1 (ρ)), T
s
reg(ρ), or T
s
2 (ρ), respectively.
To prove these lemmas, we prove a preliminary lemma which is valid in any plane of
constant curvature.
Lemma 2.4. Let T be a triangle in M ∈ {E2,H2, S2} with vertices a, b, c and containing
its circumcenter. Let Ca, Cb, Cc be disks of radius ρ > 0 centered at a, b, c, respectively.
Assume that a line L separates Ca and Cb from Cc. Then the line passing through the
midpoints of [a, c] and [b, c] separates Ca from Cb and Cc.
Proof. Let L̄ be the line passing through the midpoints of [a, c] and [b, c]. Clearly, it is
sufficient to prove Lemma 2.4 under the additional assumption that there is no ρ′ > ρ such
that some line L′ separates C ′a and C
′
b from C
′
c, where C
′
a, C
′
b and C
′
c are disks of radius
ρ′ > ρ centered at a, b, c, respectively. Thus, we may assume that L cannot be perturbed
in such a way that it is disjoint from all of Ca, Cb, Cc, implying that we have one of the
following.
(a*) All of Ca, Cb, Cc touches L.
(b*) Two of Ca, Cb, Cc touches L at the same point.
(c*) M = S2, and two of Ca, Cb, Cc touches L at two antipodal points.
In case (a*), L passes through the midpoints of [a, c] and [b, c], i.e L̄ = L. In case (b*),
the two disks touch L from opposite sides; that is, one of them is Cc, and we may assume
that the other one is Ca. Then the distance of b from L is strictly larger than ρ, implying
that L strictly separates the midpoint of [b, c] and b. Thus, the midpoint of [c, b] is closer to
a than to c and b, from which we readily have that T does not contain its circumcenter; a
contradiction.
Finally, in case (c*) we may assume that the two disks touching L are Ca and Cb. Then,
by the symmetry of the configuration one can modify L in such a way that the perturbed
line touches Ca and Cb at points which are not antipodal, and thus, there is a line disjoint
from all of Ca, Cb, Cc that separates Ca, Cb from Cc; a contradiction. □
Proof of Lemmas 2.2 and 2.3. Without loss of generality, we may assume that λ > 0.
First, by Girard’s theorem relating the spherical excess of a triangle to its area, δ(T ) is
maximal if and only if area(T ) is minimal among the triangles satisfying the conditions in
Lemma 2.2.
Note that since any side of T is of length at least 2ρ, and any vertex of T is at distance
at least λ from a line passing through the midpoints of two sides, by compactness, both area
and circumradius are minimized for some triangles in the family of triangles satisfying the
conditions in the lemmas. Without loss of generality, we may assume that Tδ has maximal
density (minimal area) and Tγ has minimal circumradius in this family. For the moment,
we deal only with Lemma 2.2, and assume that T = Tδ .
Let G denote the Lexell circle of T generated by c, that is, let G be the circle containing
−a,−b, c. Let H denote the open hemisphere that contains c in its interior, and a, b in its
boundary. Let G0 = H ∩ G. It is well known that if d ∈ H , then area(conv{a, b, d}) =
area(conv{a, b, c}) if and only if d ∈ G0. Let L denote the bisector of the segment [a, b],
and let c0 denote the intersection point of C0 and L. Without loss of generality, we may
12 Ars Math. Contemp. 25 (2025) #P2.04
assume that c is not farther from a than from b. Now we consider the possible position of
all points d ∈ C0 with the property that T ′ = conv{a, b, d} contains its circumcenter o′,
its edge lengths are at least 2ρ, and there is a line that separates Ca from Cb and Cc, and a
line that separates Cb from Ca and Cc.
First, the property that T ′ contains o′ is satisfied if and only if for any {x, y, z} =
{a, b, d}, the distance of the midpoint of [x, y] from z is not less than from x or y. This
condition holds if and only if d ∈ G1 for some closed arc G1 in G0 symmetric to L.
Similarly, there is a closed arc G2 ⊂ G0 symmetric to L such that T ′ has edge lengths at
least 2ρ if and only if d ∈ G2.
Finally, let L′a denote the line through the midpoints of [a, b] and [a, c], and note that
it separates Ca from Cb and Cc. Let L
′
b denote the reflected copy of L
′
a to L, and note
that it separates Cb from Ca and Cc. Furthermore, for any point d lying on the arc of C
connecting c and its reflected copy to L, the disk of radius λ and center d is separated from
Ca by L
′
a, and from Cb by L
′
b. More generally, there is a closed arc G3 in G0, symmetric
to L, such that Ca and Cd is separated from Cb by a line and Cb and Cd are separated from
Ca by another line if and only if d ∈ G3. We remark that the fact that T satisfies the above
three properties implies that G1∩G2∩G3 is not empty. Since this set is a closed arc in G0
symmetric to L, it follows that conv{a, b, c0} contains its circumcenter, has edge lengths
at least 2ρ, and it satisfies the separability properties described in the lemma. Thus, T has
minimal area only if T is symmetric to L and it satisfies at least one of the following.
(a**) Its circumcenter ō lies on [a, b].
(b**) d(a, c) = d(b, c) = 2ρ.
(c**) The lines L′a and L
′
b touch Ca, Cb and Cc.
Consider the case that T satisfies (a**) but it does not satisfy (b**) or (c**). Then
slightly moving all of a, b, c towards ō by the same quantity yields a triangle that has strictly
smaller area and still satisfies the conditions in the lemma; a contradiction. Thus, we have
that T satisfies (b**) or (c**). If T satisfies (b**) but it does not satisfy (c**) and the
distance of a and b is greater than 2ρ, then we may decrease the angle at c while keeping
the distance of c from a and b fixed to obtain a similar contradiction. Hence, T satisfies
(c**), or T is a regular triangle of edge length 2ρ.
We observe that the same consideration can be carried out under the assumption that
T = Tγ , in which the circumcircle of T plays the role of the Lexell circle. Thus, we have
that any triangle T with minimal circumradius and satisfying the conditions in Lemma 2.3
either satisfies (c**), or it is a regular triangle of edge length 2ρ. It is also worth noting
that any isosceles triangle satisfying the required conditions has sides of lengths at least
2ρ. Thus, it is easy to show that if T sreg(ρ) satisfies the conditions of the lemmas, then this
triangle has minimal area and minimal circumradius.
From now on, we assume that T is a triangle satisfying (c**) and the conditions of the
lemmas, and investigate the properties of T .
Let the angles of T be 2α, β, β and its edge lengths be 2x, 2x, 2y. It can be shown that
in that case the triangle with angles 2α, π − β, π − β and edge lengths 2y, π − 2x, π − 2x
also satisfies (c**).
Step 1: computing the edge lengths of the triangles satisfying (c**).
First, we note that λ ≤ π4 holds since there are two separating great circles that divide
S
2 into four lunes.
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 13
Let T be an isosceles spherical triangle with edges of lengths 2x, 2x, 2y (0 < x < y <
π
2 ), apex p, and vertices q1, q2 on its base. Assume that the line L through the midpoints
m,m′ of [q,q2] and [p, q1], respectively, touches and separates the spherical caps of radius λ
centered at p, q1. Let the orthogonal projections of p and q1 on L be p̄ and q̄1, respectively.
Let w ,z and t be the lengths of the arcs [m,m′], [p̄,m′], and [m, p], respectively. By the
spherical Pythagorean Theorem, we have
cos(2x) = cos y cos t,
cosx = cosλ cos z,
cos y = cosλ cos(w − z),
cos t = cosλ cos(w + z),
From this we obtain that
1 = cos2 w + sin2 w =
(cos y + cos t)
2
4 cos2 λ cos2 z
+
(cos y − cos t)2
4 cos2 λ sin2 z
=
(cos y + cos t)
2
4 cos2 x
+
(cos y − cos t)2
4(cos2 λ− cos2 x) ,
which yields that
4 cos2 x(cos2 λ− cos2 x) cos2 y
=
(
cos2 y + cos t cos y
)2
(cos2 λ− cos2 x) +
(
cos2 y − cos t cos y
)2
cos2 x
=
(
cos2 y + cos(2x)
)2
(cos2 λ− cos2 x) +
(
cos2 y − cos(2x)
)2
cos2 x
=
(
cos2 y + cos(2x)
)2
cos2 λ− 4 cos2 y cos(2x) cos2 x.
Since cos2 λ− cos2 x+ cos(2x) = cos2 λ− sin2 x, from this it follows that
4 cos2 x(cos2 λ− sin2 x) cos2 y = cos4 y cos2 λ+ 2 cos2 y cos(2x) cos2 λ
+ cos2(2x) cos2 λ.
Thus,
−4 cos2 x sin2 x cos2 y = cos4 y cos2 λ− 2 cos2 y cos2 λ+ cos2(2x) cos2 λ, implying
sin2(2x)(cos2 λ− cos2 y) = sin4 y cos2 λ.
As 0 < λ ≤ π4 , this yields that
sin(2x) =
cosλ sin2 y
√
cos2 λ− cos2 y
. (2.3)
Thus, for any 0 < λ ≤ π4 , there is a triangle T satisfying (c) if and only if arcsin tanλ ≤
y < π2 , and in this case the solutions, up to congruence, are T
s
1 (y) and T
s
2 (y).
14 Ars Math. Contemp. 25 (2025) #P2.04
Step 2: Finding the values of y such that T s1 (y) or T
s
2 (y) contains its circumcenter.
Note that if the height of an isosceles triangle T corresponding to its base is at least π2 ,
then T contains its circumcenter; in other words, T s2 (y) contains its circumcenter for all
values of y. On the other hand, the triangle T s1 (y) contains its circumcenter if and only if
the height corresponding to its base is of length at least y. Let m denote the length of this
height. Then, by the spherical Pythagorean theorem, we have cosm =
cos 2xs
1
(y)
cos y , implying
that T1(y) contains its circumcenter if and only if cos 2x
s
1(y) ≤ cos2 y. Since xs1(y) ≤ π4
and by (2.3), the latter inequality is equivalent to the inequality
√
1− cos
2 λ sin4 y
√
sin2 y − sin2 λ
≤
(
1− sin2 y
)2
.
Since 0 < λ < π4 , an elementary computation shows that this inequality is satisfied if and
only if 0 ≤ sin2 y ≤ 2 sin2 λ, implying that arcsin tanλ ≤ y ≤ arcsin(
√
2 sinλ). We
note that arcsin(
√
2 sinλ) ≥ arcsin tanλ is satisfied for all 0 ≤ λ ≤ π4 .
Step 3: Finding the relations between the lengths of the sides of T s1 (y), T
s
2 (y).
First, we consider xs1(y) and investigate the inequality x
s
1(y) ≥ y. Then we have
y ≤ π4 , and we need to solve the inequality
sin(2y) ≤ sin(2xs1) =
cosλ sin2 y
√
sin2 y − sin2 λ
under the condition that 0 ≤ xs1, y ≤ π4 . In this case both sides are nonnegative, and we
can write the inequality as
4 sin2 y(1− sin2 y) ≤ (1− sin
2 λ) sin4 y
sin2 y − sin2 λ
.
Multiplying both sides with the denominator and simplifying, we obtain that
4(1− sin2 y)(sin2 y − sin2 λ) ≤ (1− sin2 λ) sin2 y,
and by algebraic transformations we obtain the inequality
4 sin4 y − (3 + 5 sin2 λ) sin2 y + 4 sin2 λ ≥ 0.
For sin2 y this inequality is a quadratic inequality, the discriminant of which is 9−34 sin2 λ+
25 sin4 λ. On the interval λ ∈ [0, π/4], this expression is nonnegative if and only if
0 ≤ λ ≤ arcsin(3/5). On the other hand, since by our conditions π4 ≥ y ≥ arcsin tanλ,
we have sinλ ≤ 1√
3
< 35 . Thus, x
s
1(y) ≥ y implies that 0 ≤ λ ≤ arcsin 1√3 , and the
discriminant of the above inequality is positive. In this case the two roots are
yss(λ) = arcsin
√
3 + 5 sin2 λ−
√
9− 34 sin2 λ+ 25 sin4 λ
8
,
ysb(λ) = arcsin
√
3 + 5 sin2 λ+
√
9− 34 sin2 λ+ 25 sin4 λ
8
,
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 15
where the above expressions can be shown to exist. Furthermore, numeric computations
show that we have arcsin tanλ ≤ yss(λ) ≤ ysb(λ) ≤ π2 , ysb(λ) ≥ π4 for all values of λ, and
yss(λ) ≤ π4 if and only if 0 ≤ λ ≤ arcsin 1√3 .
The investigation of the inequality xs2(y) ≤ y can be done in a similar way, and we
obtain the following.
• If 0 ≤ λ ≤ arcsin 1√
3
and arcsin tanλ ≤ y ≤ ys, then y ≤ x1(y) ≤ x2(y).
• If 0 ≤ λ ≤ arcsin 35 and ys ≤ y ≤ yb, then x1(y) ≤ y ≤ x2(y).
• Otherwise x1(y) ≤ x2(y) ≤ y.
Note that the function yss is strictly increasing whereas y
s
b is strictly decreasing on its
domain (see Figure 3).
Figure 3: The union of the graphs of the functions yss and y
s
b decomposes the region con-
sisting of the points with the possible values of (λ, y) into three connected components,
described in the previous list, which determine the relative position of y, xs1(y) and x
s
2(y).
The region of the possible points is bounded by black curves. The red curve is the union of
the graphs of yss and y
s
b . The green segment is the curve y =
π
4 .
Remark 2.5. Note that by the computations in Step 3 , T1(y)
s or T s2 (y) is a regular triangle
of edge length 2y if and only if y = xs1(y) or y = x
s
2(y). This yields that T
s
reg(ρ) satisfies
the conditions of the lemmas if and only if yss(λ) ≤ ρ ≤ ysb(λ).
Step 4: Computing the areas and the circumradii of T s1 (y) and T
s
2 (y).
For i = 1, 2, let Rsi (y) denote the the circumradius of T
s
i (y). Let the angle of T
s
1 (y) at
p, q1 and q2 be denoted by 2α, β, β, respectively, and note that this implies that the angles
of T s2 (y) at p, −q1, −q2 are 2α, π− β, π− β, respectively. Then, by Girard’s theorem, for
area(T s1 (y)) we have
cos
area(T s1 (y))
2
= sin(γ + δ).
16 Ars Math. Contemp. 25 (2025) #P2.04
Now, applying the spherical laws of sines and cosines, we obtain that
sinα =
√
sin2 y − sin2 λ
sin y cosλ
, cosα = tanλ cot y, sinβ =
tanλ
sin y
and
cosβ =
√
sin2 y − tan2 λ
sin y
.
Using trigonometric identities, from this we obtain
cos
area(T s1 (y))
2
=
√
sin2 y − sin2 λ
√
sin2 y cos2 λ− sin2 λ+ sin2 λ cos y
sin2 y cos2 λ
,
and a similar computation yields
cos
area(T s2 (y))
2
=
−
√
sin2 y − sin2 λ
√
sin2 y cos2 λ− sin2 λ+ sin2 λ cos y
sin2 y cos2 λ
.
Next, the circumradius R of a spherical triangle with angles α, β, γ satisfies
cotR =
√
sin(α− P ) sin(β − P ) sin(γ − P )
sinP
,
where P = 12 (α+ β + γ − π) (see [17]). Thus, Rs1(y) satisfies
cotRs1(y) =
√
sin
(
α+ β − π2
)
sin2
(
π
2 − α
)
sin
(
α+ β − π2
) =
cosα cos(β − α)
√
− cos(β − α) cos(β + α)
.
By our formulas for sinα, cosα, sinβ, cosβ and using trigonometric identities, we obtain
that
− cos(β − α) cos(β + α) = tan2 λ.
Substituting back and using the addition formula for the cosine of the difference of two
angles, we obtain
cotRs1(y) =
sinλ
cos2 λ
(
cos2 y
sin3 y
√
sin2 y cos2 λ− sin2 λ+ cos y
sin3 y
√
sin2 y − sin2 λ
)
,
and a very similar computation yields
cotRs2(y) =
sinλ
cos2 λ
(
−cos
2 y
sin3 y
√
sin2 y cos2 λ− sin2 λ+ cos y
sin3 y
√
sin2 y − sin2 λ
)
.
Step 5: Investigating the monotonicity properties of the areas and the circumradii of
T s1 (y) and T
s
2 (y).
Here we present the computation for T s1 (y).
Denoting sinλ and sin y by L and Y , respectively, we obtain that the value of
cos
area(T s1 (y))
2
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 17
can be written as
f(Y ) =
√
Y 2 − L2
√
Y 2(1− L2)− L2 + L2
√
1− Y 2
Y 2
√
1− L2
.
Then we have
f ′(Y ) =
L2 (A−B)√
Y 2 − L2
√
Y 2(1− L2)− L2
√
1− Y 2(1− L2)
,
where
A = (2Y 2 − 2L2 − Y 2L2)
√
1− Y 2 andB = (2− Y 2)
√
Y 2 − L2
√
Y 2(1− L2)− L2
are positive for any 0 < L < 1√
2
and L√
1−L2 < Y < 1. On the other hand,
A2 −B2 = −Y 6(1− L2)(Y 2 − 2L2),
implying that area(T s1 (y)) is strictly decreasing on the interval
y ∈ [arctan sinλ, arcsin(
√
2 sinλ)],
and it is strictly increasing on y ∈
[
arcsin(
√
2 sinλ), π2
]
. Using a similar computation we
have that area(T s2 (y)) is strictly increasing on its whole domain.
Next, we present the computation for Rs1(y). Using the notation L = sinλ and Y =
sin y, we have that cotRs1(y) is equal to
g(Y ) =
1− Y 2
Y 3
√
Y 2(1− L2)− L2 +
√
1− Y 2
Y 3
√
Y 2 − L2.
We find the maximum of this expression for L√
1−L2 < Y < 1, where L > 0 is fixed.
To do it, we intend to examine first the condition when the squares of the derivatives of the
first and the second members of g(Y ) are equal. With the notation Z = Y 2 − 5L23 , this
leads to the cubic equation Z3 +
(
2L2 − 10L43
)
Z + L
4
3 − 2527L6 = 0. The discriminant
of this equation is negative on the interval L ∈
(
0, 1√
2
)
, implying that it has one real root.
This root yields the unique positive solution sin ysmin(λ)) for Y . Numeric computations
show that for any 0 < λ ≤ π4 ,
arcsin tanλ ≤ ysmin(λ) ≤ arcsin(
√
2 sinλ),
and for any 0 < λ ≤ arcsin 35 , we have xs1(ysmin(λ)) < yss(λ) ≤ ysmin(λ), with equality
on the right if and only if λ = arcsin 1√
3
. In particular, checking the sign of g′(Y ), we
obtain that Rs1(y) is strictly decreasing on the interval [arcsin tanλ, y
s
min(λ)] and strictly
increasing on [ysmin(λ), π/2).
A similar computation shows that Rs2(y) is strictly increasing on the interval
[
arcsin tanλ,
π
2
)
.
18 Ars Math. Contemp. 25 (2025) #P2.04
Step 6: A case analysis to prove the lemmas.
We describe the analysis to find the minimum area triangles from amongst T s1 (y) and
T s2 (Y ) satisfying the conditions of Lemma 2.2. To finish the proof of Lemma 2.3, we can
use a similar argument.
Case 1: 0 < ρ ≤ π4 . Then x2(y) ≥ ρ is satisfied for all values of y, and thus, T s2 (y)
satisfies the conditions if and only if y ≥ max{ρ, arcsin tanλ}.
Subcase 1.1: λ ≤ ρ ≤ yss(λ). By Step 2, T s1 (y) contains its circumcenter of and only
if y ≤ arcsin(
√
2 sinλ), and T s2 (y) contains it for all values of y. Furthermore, the sides
of T s1 (y) are at least 2ρ if and only if y ≥ ρ, and x1(y) ≥ ρ. By Step 3, these inequalities
are equivalent to y ≥ max{ρ, arcsin tanλ}, and y ≤ xs1|−1S1 (ρ) or y ≥ x
s
1|−1S2 (ρ). Since
y ≤ arcsin(
√
2 sinλ), y ≥ xs1|−1S2 (ρ) is not satisfied for any value of y. Thus, we need to
find the minimum area triangle from amongst the T s1 (y) with y ≥ max{ρ, arcsin tanλ}
and y ≤ xs1|−1S1 (ρ), and the T
s
2 (y) with y ≥ max{ρ, arcsin tanλ}. Observe that by the
monotonicity properties of the function xs1|−1S1 and the fact that y
s
s(λ) < arcsin(
√
2 sinλ),
we have that
xs1|−1S1 (ρ) ≥ x
s
1|−1S1 (y
s
s(λ)) = y
s
s(λ) ≥ max{arcsin tanλ, ρ}.
Thus, by Step 5, among the above triangles T s1 (x
s
1|−1S1 (ρ)) has minimal area.
Subcase 1.2: yss(λ) < ρ ≤ π4 . Here T sreg(ρ) satisfies the conditions, and thus, this
triangle has minimal area.
Case 2: π4 < ρ <
π
2−λ. Note that in this case no triangle T s1 (y) satisfies the conditions.
We also note that ysb(λ) <
π
2 − λ for any value of λ.
Subcase 2.1: λ ≤ arcsin 35 and yss(λ) ≤ ρ ≤ ysb(λ). (Here we note that if λ ≤
arcsin 1√
3
, then yss(λ) ≤ ρ holds for all ρ ≥ π4 .)
Like in Subcase 1.2, T sreg(ρ) satisfies the conditions, and thus, this triangle has minimal
area.
Subcase 2.2: if λ > arcsin 35 , or arcsin
1√
3
≤ λ ≤ arcsin 35 and ρ < ys(λ), or
0 ≤ λ < arcsin 35 and ρ > yb(λ). Then T sreg(ρ) does not satisfy the conditions.
The triangle T s2 (y) satisfies the conditions if and only if y ≥ max{arcsin tanλ, ρ} and
xs2|−1S1 (ρ) ≤ y ≤ x
s
2|−1S2 (ρ). By our conditions, x
s
2(ρ) < ρ, implying that ρ < x
s
2|−1S1 (ρ)
or ρ > xs2|−1S2 (ρ) if ρ ≤ arcsin(
√
2 sinλ) or ρ ≥ arcsin(
√
2 sinλ), respectively. Since
ρ ≤ Π2 − λ, and yb(λ) = arcsin(
√
2 sinλ) = arcsin(
√
2 sinλ) if λ = arcsin 1√
3
, we have
that ρ > xs2|−1S2 (ρ) if and only if 0 ≤ λ < arcsin
1√
3
and ρ > ysb(λ), and ρ < x
s
2|−1S2 (ρ)
otherwise (see Figure 4). In both cases, by Step 5, the solution is T s2 (ρ).
Case 3: ρ > π2 − λ. In this case xs2(y) < ρ for all values of y, implying that there is no
triangle satisfying the conditions. □
3 Proofs of Theorems 1.5, 1.7, 1.18, 1.20
The proofs in question are straightforward modifications of the proofs of Theorems 1.13
and 1.15 , and are based on the use of the refined Molnár decomposition. In particular,
let M ∈ {E2,H2}. Like in Section 2, we assume that F is a (2Rρ)-locally λ-separable
packing of spherical caps of radius ρ in M, where Rρ = arcsinh(
√
2 sinh ρ) if M = H2,
and Rρ =
√
2ρ if M = E2. In addition, we assume that F is (2Rρ)-saturated in M, i.e.,
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 19
Figure 4: An illustration for Subcase 2.2. The boundary of the region of the possible
parameter values is drawn with black color. The red curve is the union of the graphs of yss
and ysb . The green curves are the segment y =
π
4 , and the curve y = arcsin(
√
2 sin(λ)).
Observe that the last curve passes through the intersection of the segment y = π2 − λ and
the curve y = yb(λ).
every point p is at distance at most 2Rρ from the center of a cap in F . Thus, the radius of
any circumdisk of the cells in the Delaunay decomposition of the set X of the centers of
the elements of F is at most 2Rρ. Then the refined Molnár decomposition of M defined by
X consists of two types of cells P :
(i) P has circumradius at most Rρ, in this case we say that P is type 1, or
(ii) it is of the form P = cl (conv{v, ci, cj} \ conv{v′, ci, cj}), where ci, cj ∈ X , v is
the circumcenter of a Delaunay cell with ci and cj as vertices and with circumradius
at least Rρ, and d(v
′, ci) = d(v
′, cj). In this case we say that P is type 2.
Like in the spherical case, any type 1 cell is either a triangle containing its circumcenter,
or it can be decomposed into two triangles containing their circumcenters. Thus, we assume
that any type 1 cell is a triangle containing its circumcenter.
To prove the theorems for type 2 cells, we need the following lemma, which is a variant
of [3, Lemma 3].
Lemma 3.1. Let Q be an isosceles triangle in M, with vertices q1, q2, p, where p is the
apex. Let the length of the legs of Q be x, and that of the base be y, where ρ ≤ y2 < x . For
i = 1, 2, let Si denote the disk of radius ρ, centered at qi. Let
f(x, y) =
area(S1 ∪ S2)
area(Q)
denote the density of the packing {S1, S2} in Q as a function of x and y. Then f(x, y) is a
strictly decreasing function of both x and y.
20 Ars Math. Contemp. 25 (2025) #P2.04
This lemma is straightforward to prove if M = E2, using an elementary computation.
If M = H2, then a slight modification of the proof of Lemma 3 of [3] can be applied, which
we leave to the reader. Thus, it is sufficient to prove the assertion for type 1 triangles.
First, we prove Theorem 1.5 for type 1 triangles. Assume that M = E2 and ρ = 1.
Let T be any type 1 triangle. Since the sides of T has lengths at least 2, it follows that
area(T ) ≥
√
3
4 . Similarly, since the Euclidean disks centered at the vertices of T are a
λ-separable system, we have that at least two heights of T are at least 2λ, implying that
area(T ) ≥ λ2 . Thus, area(T ) ≥ max
{√
3
4 ,
λ
2
}
. On the other hand, if 0 ≤ λ ≤
√
3
2 ,
then the packing of unit disks whose Delaunay cells are regular triangles is a λ-separable
packing, and if
√
3
2 < λ ≤ 1, then the same is true for the packing of unit disks whose
Delaunay cells are isosceles triangles whose legs are of length 2 and the length of the
corresponding heights is 2λ, i.e. if they are congruent to T e(
√
2− 2
√
1− λ2). This proves
Theorem 1.5. To prove Theorems 1.7, 1.18 and 1.20, it is sufficient to prove the following
lemmas.
Lemma 3.2. Let T be a non-obtuse triangle in E2 with edge lengths at least 2, and having
two heights of lengths at least 2λ. Let R(T ) denote the circumradius of T . Then
R(T ) ≥







2√
3
, if 0 ≤ λ ≤
√
3
2 ,√
2−2
√
1−λ2
λ , if
√
3
2 ≤ λ ≤ 2
√
2
3 ,
3
√
3λ
4 , if
2
√
2
3 ≤ λ ≤ 1.
(3.1)
Furthermore, we have equality in one of the three cases if and only if T is congruent to
T 2reg(1), or T
e(
√
(2− 2
√
1− λ2)), or T e
(√
3
2λ
)
, respectively.
Lemma 3.3. Let T = conv{a, b, c} ⊂ H2 be a hyperbolic triangle with edge lengths
at least 2ρ that contains its circumcenter. For x ∈ {a, b, c}, let Cx denote the closed
hyperbolic disk of radius λ ≤ ρ centered at x. Assume that there are lines La and Lb such
that La separates Ca from Cb and Cc, and Lb separates Cb from Ca and Cc. Let δ(T )
denote the density of {Cx, Cy, Cz} with respect to T . Then we have the following.
δ(T ) ≤
{
δ(Th(xh|−1H1(ρ))), if λ ≤ ρ ≤ y
h
s (λ),
δ(T sreg(ρ)), if y
s
s(λ) < ρ.
(3.2)
Furthermore, in the above two cases we have equality if and only if T is congruent to
Th(xh|−1H1(ρ)) or T
h
reg(ρ), respectively.
Lemma 3.4. Let T = conv{a, b, c} ⊂ H2 be a hyperbolic triangle with edge lengths
at least 2ρ that contains its circumcenter. For x ∈ {a, b, c}, let Cx denote the closed
hyperbolic disk of radius λ ≤ ρ centered at x. Assume that there are lines La and Lb such
that La separates Ca from Cb and Cc, and Lb separates Cb from Ca and Cc. Let R(T )
denote the circumradius of T . Then we have the following.
R(T ) ≥





Rh(yhmin(λ)), if λ ≤ ρ ≤ xh(ysmin(λ)),
Rh(xh|−1H1(ρ)), if x
h(yhmin(λ)) < ρ ≤ yhs (λ),
Rhreg(ρ), if y
s
s(λ) < ρ.
(3.3)
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 21
Furthermore, in the above three cases we have equality if and only if T is congruent to
Th(yhmin(λ)), T
h(xh|−1H1(ρ)), or T
h
reg(ρ), respectively.
The proof of these lemmas follows the proof of Lemmas 2.2 and 2.3, and we only
sketch them.
Proof of Lemmas 3.2, 3.3 and 3.4. First, by the well known formula relating the angle de-
fect of a hyperbolic triangle to its area, if M = H2, δ(T ) is maximal if and only if area(T )
is minimal among the triangles satisfying the conditions in Lemma 3.3.
Furthermore, similarly like in the previous section, we observe that both area and cir-
cumradius is minimized for some triangle satisfying the conditions in the lemmas.
Let M = H, and let T be a minimum area triangle satisfying the conditions. Let H
denote the open hyperbolic half plane containing c in its interior and a, b on its boundary.
Recall the well known fact that for any d ∈ H a hyperbolic triangle T ′ = conv{a, bd}
satisfies area(T ′) = area(T ) if and only if d lies on the hypercycle through c generated
by the midpoints of [a, c] and [b, c]. Let G0 denote this hypercycle, and note that it is
symmetric to the bisector L of [a, b]. Applying the argument in the previous section, we
obtain there is a hypercycle arc G′ in G0 symmetric to c0 such that for any d ∈ C0, the
triangle conv{a, b, d} satisfies the conditions in Lemma 3.3 if and only if d ∈ G′. The same
argument can be applied for Lemmas 3.2 and 3.4 in which the circumcircle of T plays the
role of G0.
From this, following the argument in the proof of Lemmas 2.2 and 2.3, we obtain that
if T has minimal circumradius or area, then T is either
(a’) a regular triangle of edge length 2ρ, or
(b’) d(a, b) = d(b, c) and the line through the midpoints of [a, c] and [a, b] (respectively,
the midpoints of [b, c] and [a, b]) touches the three disks of radius λ centered at the
vertices of T .
Step 1: computing the edge lengths of the triangles satisfying (b’).
Following the argument in Section 2, we obtain that for a hyperbolic triangle T ,
sinh(2x) =
coshλ sinh2 y
√
cosh2 y − cosh2 λ
=
coshλ sinh2 y
√
sinh2 y − sinh2 λ
,
which is defined for any y > λ, yielding that the only hyperbolic triangle satisfying (b’),
up to congruence, is Th(y). In E2, we can similarly obtain that for any y > λ, up to
congruence, there is a unique isosceles triangle satisfying (b’), namely T e(y).
Step 2: Checking if Th(y) or T e(y) contains its circumcenter.
Applying the idea of the proof in Section 2 yields that Th(y) contains its circumcenter
if and only if arcsinh tanhλ ≤ y ≤ arcsinh(
√
2 sinhλ). Similarly, T e(y) contains its
circumcenter if and only if λ < y ≤
√
2λ.
Step 3: Finding the relations between the lengths of the sides of Th(y) and T e(y).
In H2, a consideration like in the spherical case leads to the fact that the inequality
xh(y) ≥ y is satisfied if and only if
y ≤ yhs (λ) = arcsinh
√
5 sinh2 λ− 3 +
√
25 sinh4 λ+ 34 sinh2 λ+ 9
8
.
22 Ars Math. Contemp. 25 (2025) #P2.04
We note that like in the spherical case, we have that the hyperbolic disks of radius λ cen-
tered at the vertices of Threg(ρ) are totally separable if and only if ρ ≥ yhs (λ). In this
case Threg(ρ) has minimal area and minimal circumradius among the triangles satisfying
the conditions of the lemmas.
In the Euclidean plane, the inequality xe(y) ≥ y is satisfied if and only if λ < y ≤ 2√
3
λ.
Here, T ereg(1) satisfies the conditions if and only if λ ≤
√
3
2 . In this case T
e
reg(1) has
minimal circumradius.
Step 4: Computing area(Th(y)), Ry(y) and Re(y).
A slight modification of the computation in the spherical case shows that
cosh
area(Th(y))
2
=
√
sinh2 y − sinh2 λ
√
sinh2 y cosh2 λ− sinh2 λ+ sinh2 λ cosh y
sinh2 y cosh2 λ
, and
cothRh(y)
=
sinhλ
cosh2 λ
(
cosh2 y
sinh3 y
√
sinh2 y cosh2 λ− sinh2 λ+ cosh y
sinh3 y
√
sinh2 y − sinh2 λ
)
.
Here we used the formula cothR =
√
sin(α+P ) sin(β+P ) sin(γ+P )
sinP for the circumradius of
a hyperbolic triangle with angles α, β, γ, where P = 12 (π − α − β − γ) [17]. Similarly,
for the circumradius Re(y) of T e(y), we obtain
Re(y) =
y3
2λ
√
y2 − λ2
.
Step 5: Investigating the monotonicity properties of area(Th(y)), Rh(y) and Re(y).
A simple computation following the proof in the spherical case shows that area(Th(y))
is strictly decreasing on
[
arcsinh tanhλ, arcsinh(
√
2 sinhλ)
]
, and strictly increasing on
[
arcsinh(
√
2 sinhλ),∞
)
.
We sketch the computation for Rh(y). Denoting sinhλ and sinh y by L and Y , we
obtain that cothRh(y) can be written as
h(Y ) =
(Y 2 + 1)
√
Y 2(L2 + 1)− L2
Y 3
+
√
Y 2 + 1
√
Y 2 − L2
Y 3
.
We examine first the condition when the squares of the derivatives of the first and the
second members of h(Y ) are equal. With the notation Z = Y 2 − 5L23 , this leads to the
cubic equation Z3−
(
2L2 + 10L
4
3
)
Z− 2527L6− L
4
3 = 0. The discriminant of this equation
is positive, implying that it has three real roots. Solving it, we obtain that Y has to satisfy
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 23
one of the following:
Y 2 =
5L2
3
+
2
3
√
10L4 + 6L2 cos
(
1
3
arccos
(
(25L2 + 9)L
4
√
2(5L2 + 3)3/2
))
,
Y 2 =
5L2
3
+
2
3
√
10L4 + 6L2 cos
(
1
3
arccos
(
(25L2 + 9)L
4
√
2(5L2 + 3)3/2
)
− 2π
3
)
,
Y 2 =
5L2
3
+
2
3
√
10L4 + 6L2 cos
(
1
3
arccos
(
(25L2 + 9)L
4
√
2(5L2 + 3)3/2
)
− 4π
3
)
A numeric computation shows that for every L > 0 all three expressions exist, and the third
expression is negative whereas the first two expressions are positive. On the other hand,
an elementary computation shows that the derivatives of the first and the second member
of f(Y ) cannot be zero simultaneously for L > 0, and recall that we examined only if the
two derivatives are equal in absolute value. Since the above expressions are continuous
functions of L at which the two derivatives are equal in absolute value identically, it is
sufficient to test whether the derivatives have the same or opposite signs for one arbitrary
value L > 0. Testing it we obtain that h′(Y ) = 0 if and only if Y 2 is equal to the second
expression. Thus, by Y > 0, we obtain that for any λ > 0, there is a unique stationary
point of the function cothRh(y), namely
yhmin(λ)
= arcsinh
√
5L2
3
+
2
3
√
10L4 + 6L2 cos
(
1
3
arccos
(
(25L2 + 9)L
4
√
2(5L2 + 3)3/2
)
− 2π
3
)
> λ,
where L = sinhλ. A simple computation yields that for any L > 0, limY→L+0 f
′(Y ) =
∞ and limY→∞ Y 2 · f ′(Y ) = −1 < 0, showing that Rh(y) is strictly decreasing on the
interval (λ, yhmin(λ)], and strictly increasing on the interval [y
h
min(λ),∞).
We note that numeric computations show that λ < xh(yhmin(λ)) < y
h
s (λ) < y
h
min(λ) <
arcsinh(
√
2λ) for all λ > 0 (see Figure 5).
In the Euclidean plane, examining the sign of the derivative of Re(y) shows that Re(y)
is strictly decreasing on
(
λ,
√
2
3λ
]
, and strictly increasing on
[√
2
3λ,∞
)
.
Step 6: A case analysis to prove the lemmas.
An investigation similar to the one in Section 2 finishes the proof. □
4 Proof of Theorem 1.9
Theorems of Harborth [13] and of Heitmann and Radin [14] imply in a straightforward way
that cλ(n,B) = ⌊3n−
√
12n− 3⌋ holds for all n > 1 and 0 ≤ λ ≤
√
3
2 .
Now, let
√
3
2 < λ ≤ 1. Then ⌊2n− 2
√
n⌋ ≤ cλ(n,B) follows in a straightforward way
from Theorem 11 in [1]. So, we are left to show that cλ(n,B) ≤ 2n −
√
πλ
√
n + O(1)
holds for n > 1.
First, recall that the proof of Theorem 1.5 in Section 3 gives a proof of
Lemma 4.1. Let
√
3
2 < λ ≤ 1 and let P be a λ-separable packing of unit disks in E2. Then
the density of P in each cell of the refined Molnár decomposition is at most π4λ .
24 Ars Math. Contemp. 25 (2025) #P2.04
Figure 5: An illustration for the curves λ (yellow), xh(yhmin(λ)) (red), y
h
s (λ) (blue),
yhmin(λ) (green) and arcsinh(
√
2 sinhλ) (black) for a hyperbolic triangle.
Second, we use the proof technique of Theorem 4 of Eppstein [7] as well as Lemma 4.1
for proving Lemma 4.3. In order to state it we need
Definition 4.2. Let P be a packing of unit disks in E2. Then the plane graph Gc(P) whose
vertices are the center points of the unit disks of P and whose edges are the line segments
connecting two vertices of Gc(P) if and only if the corresponding two unit disks of P are
tangent, is called the contact graph of P .
Lemma 4.3. Let
√
3
2 < λ ≤ 1 and let P be a λ-separable packing of n > 1 unit disks in
E
2. Then in the contact graph Gc(P), the number of vertex-face incidences on the outer
face of Gc(P) is at least 2
√
πλ
√
n−O(1).
Proof. If the outer face of Gc(P) has at least 2
√
πλ
√
n−O(1) vertex-face incidences, then
we are done. So, assume that the outer face has less than 2
√
πλ
√
n vertex-face incidences.
Then the perimeter of the outer face is less than 4
√
πλ
√
n. Now, let A be the complement
of the outer face of Gc(P) in E2. Notice that if a unit disk of P has its center in the
interior int(A) of A, then the open unit disk belongs to int(A). Furthermore, the cells of
the refined Molnár decomposition of the center points of the unit disks of P generate a
decomposition of A. Next, Lemma 4.1 together with the property that the outer face has
less than 2
√
πλ
√
n vertex-face incidences imply that
area(A) ≥ (n− 2
√
πλ
√
n)π
π
4λ
= 4λ(n− 2
√
πλ
√
n). (4.1)
K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 25
Finally, (4.1) and the isoperimetric inequality applied to A yield for the perimeter
per(A) of A that
per(A) ≥
√
4π area(A) ≥ 2
√
π
√
4λ
√
n− 2
√
πλ
√
n = 4
√
πλ
√
n− 2
√
πλ
√
n
= 4
√
πλ
√
n−O(1),
finishing the proof of Lemma 4.3. □
Third, we need Lemma 5 of [7] stated as follows.
Lemma 4.4. Let G be a triangle-free plane graph on n vertices in which one face has k
vertex-face incidences. Then G has at most 2n− k2 − 2 edges.
Finally, notice that if
√
3
2 < λ ≤ 1 and P is a λ-separable packing of n > 1 unit
disks in E2, then Gc(P) is triangle-free. Thus, Lemma 4.3 proves the existence of a large
face in Gc(P), and plugging the size of this face as the variable k in Lemma 4.4 shows
that cλ(n,B) ≤ 2n −
√
πλ
√
n + O(1) holds for n > 1. This completes the proof of
Theorem 1.9.
ORCID iDs
Károly Bezdek https://orcid.org/0000-0003-3097-0430
Zsolt Lángi https://orcid.org/0000-0002-5999-5343
References
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(2021), 714–729, doi:10.1112/mtk.12102, https://doi.org/10.1112/mtk.12102.
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Geometry, Springer, Berlin; János Bolyai Math. Soc., Budapest, volume 27 of Bolyai Soc.
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[8] G. Fejes Tóth and L. Fejes Tóth, On totally separable domains, Acta Math. Acad. Sci.
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BF02566097.
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the Sphere, and in Space, volume 360 of Grundlehren der mathematischen Wissenschaften,
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978-3-031-21800-2.
[12] B. Grünbaum and G. C. Shephard, Tilings and Patterns, W. H. Freeman and Company, New
York, 1st edition, 1987.
[13] H. Harborth, Lösung zu Problem 664A, Elem. Math. 29 (1974), 14–15.
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(1977), 195–203.
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K. Bezdek and Z. Lángi: On optimal λ-separable packings in the plane 27
A Appendix: The hyperbolic Molnár decomposition
This Appendix introduces a new decomposition technique in H2, whose Euclidean ana-
logue has been discovered by Molnár [15]. It is obtained from the Delaunay decomposi-
tion. For preciseness, we recall some elementary properties of a decomposition [12]. First,
we call a family F of countably many mutually nonoverlapping closed sets whose union
covers H2 a decomposition or tiling of S2. The elements of this family are called tiles
or cells. In case of a convex decomposition the tiles are assumed to be convex, implying
that they are convex polygons. A decomposition is called locally finite if any point has a
neighborhood intersecting only finitely many tiles, and, if the tiles are polygons, it is called
edge-to-edge, if any edge of a tile coincides with exactly one edge of another tile. If F is
edge-to-edge, the edges and the vertices of the tiles are called the edges and the vertices of
F , respectively.
Consider a discrete, countable point system X = {p1, . . . , pk, . . .} in H2. It is well
known that then there is a unique decomposition of conv(X) into convex polygonal tiles
with the following properties:
(1) the tiling is locally finite and edge-to-edge;
(2) the vertices of every tile are points of X;
(3) the circumdisk of every tile contains no point of X in its interior, and only the vertices
of the tile on its boundary.
This convex, locally finite and edge-to-edge tiling is called the Delaunay decomposition, or
D-decomposition defined by X . Now, by means of Lemma A.1 we define another locally
finite, edge-to-edge decomposition, which we call hyperbolic Molnár decomposition in
short, M -decomposition.
Let F be a cell of the D-decomposition, and let us denote the circumdisk of F by
CF ⊂ H2, and the center of CF by oF . If F does not contain oF , then there is a unique
side of F that separates it from F in CF . We call this side a separating side of F . If [pi, pj ]
is a separating side of F , then we call the polygonal curve [pi, o] ∪ [oF , pj ] the bridge
associated to the separating side [pi, pj ] of F (cf. Figure 6).
p
pi
j
oF
F
Figure 6: An illustration for the M -decomposition of a point system on H2. In the figure,
hyperbolic segments are represented with straight line segments; the edges of the cells, the
circumcircles of the cells and the separating sides are denoted by solid, dotted and dashed
lines, respectively.
Our main lemma is the following.
28 Ars Math. Contemp. 25 (2025) #P2.04
Lemma A.1. If we replace all separating sides of a D-decomposition by the corresponding
bridges, we obtain a locally finite, edge-to-edge decomposition of H2.
Proof. As in [15], the proof is based on showing the following two statements.
(a) Bridges may intersect only at their endpoints.
(b) A bridge may intersect a side in the M -decomposition only at its endpoints.
To show (a), recall that for any cell F , the points of X closest to oF are exactly the
vertices of F . Thus, if [oF , pi] and [oF ′ , pj ] are components of bridges with pi ̸= pj and
oF ̸= oF ′ , then the bisector of the segment [pi, pj ] separates [oF , pi] and [oF ′ , pj ], which,
since oF ̸= oF ′ , yields that the components are disjoint.
Now, we show (b), and let [pi, pj ] be a separating side of the cell F . First, observe that
by the definition of separating side, the triangle T with vertices pi, pj and oF intersects
finitely many cells of the D-decomposition at a point different from [pi, pj ], each of which
is different from F . Let F ′ be the cell adjacent to F and having [pi, pj ] as a side. Note that
both oF and oF ′ lie on the bisector of [pi, pj ] such that if m is the midpoint of this arc, then
oF ∈ [oF ′ ,m], implying that the radius of CF ′ is strictly greater than that of CF . If, apart
from [pi, pj ], F
′ contains T in its interior, then (b) clearly holds for the bridge associated to
[pi, pj ]. On the other hand, if F
′ does not contain T in its interior apart from [pi, pj ], then
there is a separating side [pi′ , pj′ ] of F
′, removed during the construction, and this side is
the unique side containing any point of the bridge apart from pi and pj . Thus, no side of F
′
in the M -decomposition contains an interior point of the bridge. Furthermore, observe that
if any side in the M -decomposition contains an interior point of [pi, oF ] ∪ [oF , pj ], then it
also contains an interior point of [pi′ , oF ′ ] ∪ [oF ′ , pj′ ]. Consequently, we may repeat the
argument with F ′ playing the role of F , and since only finitely many cells may intersect T
at a point different from [pi, pj ], and each is different from F , we conclude that [pi, oF ] ∪
[oF , pj ] may intersect any side of the M -decomposition only at pi or pj . □
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.05
https://doi.org/10.26493/1855-3974.3023.c45
(Also available at http://amc-journal.eu)
Upper embeddability of graphs and products of
transpositions associated with edges*
Shuhei Tsujie †
Department of Mathematics, Hokkaido University of Education,
Asahikawa, Hokkaido 070-8621, Japan
Ryo Uchiumi
Department of Mathematics, Graduate School of Science, Osaka University,
Toyonaka, Osaka 560-0043, Japan
Received 4 December 2022, accepted 12 April 2024, published online 13 March 2025
Abstract
Given a graph, we associate each edge with the transposition which exchanges the
endvertices. Fixing a linear order on the edge set, we obtain a permutation of the vertices.
Dénes proved that the permutation is a full cyclic permutation for any linear order if and
only if the graph is a tree.
In this article, we characterize graphs having a linear order such that the associated
permutation is a full cyclic permutation in terms of graph embeddings. Moreover, we give
a counterexample for Eden’s question about an edge ordering whose associated permutation
is the identity.
Keywords: Full cyclic permutation ordering, upper-embeddable graph, 2-cell embedding, rotation
system.
Math. Subj. Class. (2020): 05C25, 05C10, 57M15
*The authors wish to thank Professor Sachiko Saito for her valuable comments about topics of topology. The
authors would also like to express their deepest appreciation to the anonymous referee for careful reading and
pointing out a deficiency in a proof.
†Corresponding author.
E-mail addresses: tsujie.shuhei@a.hokkyodai.ac.jp (Shuhei Tsujie), uchiumi.ryou.1xu@ecs.osaka-u.ac.jp
(Ryo Uchiumi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.05
1
2
3
4
5
e1
e2
e3
e4
e5
e6
Figure 1: A butterfly graph. Figure 2: A dumbbell graph.
1 Introduction
In this article, a graph G stands for a connected multigraph G = (VG, EG, rG), where
• VG is a finite set of vertices,
• EG is a finite set of edges,
• rG is a map from EG to
(
VG
2
)
, the collection of subsets of VG consisting of 2 ele-
ments.
Note that rG(e) represents the endvertices of the edge e. Also note that loops are not
allowed.
Let n be a positive integer and suppose that VG = [n] := {1, 2, . . . , n}. When an edge
e ∈ EG satisfies rG(e) = {u, v}, we associate the transposition τe := (u v) ∈ Sn with the
edge e, where Sn denotes the symmetric group of degree n.
An edge ordering of a graph G is a linear order ≤ω on EG, denoted as a sequence
ω = (e1, . . . , em) in which ei <ω ej if and only if i < j. Given an edge ordering
ω = (e1, . . . , em), we associate the product πω := τem · · · τe1 ∈ Sn.
Definition 1.1. A permutation σ ∈ Sn is called a full cyclic permutation if σ is a cyclic
permutation of length n. An edge ordering ω = (e1, . . . , em) of a graph G is a full cyclic
permutation ordering if the corresponding permutation πω = τem · · · τe1 is a full cyclic
permutation.
Dénes proved the following theorem to state a connection between labeled trees and
factorization of a full cyclic permutation into transpositions.
Theorem 1.2 (Dénes [1]. See also [7, Section 2] and [11, Lemma 2.1]). Given a graph G,
the following are equivalent.
(i) Any edge ordering of G is a full cyclic permutation ordering.
(ii) G is a tree.
Note that Theorem 1.2 plays an important role in the studies of the chromatic symmetric
functions and the chromatic operator for trees [3, 11]. Also, note that recently the second
author [16] studied an analogue of Theorem 1.2 for signed graphs and the hyperoctahedral
group.
It is a natural question to ask what graphs admit a full cyclic permutation ordering. For
example, let G be the butterfly graph pictured in Figure 1 and define the edge ordering ω
by ω := (e1, e2, e3, e4, e5, e6). Then
πω = τe6τe5τe4τe3τe2τe1 = (3 5)(1 3)(4 5)(3 4)(2 3)(1 2) = (1 3 2 5 4).
S. Tsujie and R. Uchiumi: Upper embeddability of graphs and products of transpositions . . . 3
Therefore ω is a full cyclic permutation ordering of the butterfly graph G.
Let β(G) := |EG|−|VG|+1 denote the Betti number (also called the circuit rank) of the
connected graphG. When a graphG has a full cyclic permutation ordering, considering the
signature of a full cyclic permutation ordering, one can show that the Betti number β(G) is
even. However, the converse is false. For example, the dumbbell graph (Figure 2) has no
full cyclic permutation orderings although its Betti number is 2.
We regard a graph as a topological space by identifying each edge with the unit interval
[0, 1] and gluing them at vertices. Then β(G) coincides with the Betti number of G as
a topological space. In this article, we will show that having a full cyclic permutation
ordering is a topological property as follows.
Let Σ be an orientable closed surface and ι : G→ Σ an embedding. We call a connected
component of the complement of the image of ι a face. An embedding ι is a 2-cell embed-
ding if every face is homeomorphic to an open disk. For any 2-cell embedding ι : G → Σ,
|VG| − |EG|+ fι = 2− 2gΣ,
where fι denotes the number of faces of the 2-cell embedding ι and gΣ the genus of Σ.
Then
2gΣ + fι = β(G) + 1.
Therefore maximizing the genus gΣ is equivalent to minimizing the number of faces fι.
Hence the maximum genus γmax(G), the maximum of genus gΣ such that there exists a
2-cell embedding G→ Σ, satisfies
γmax(G) ≤
⌊
β(G)
2
⌋
,
where ⌊ ⌋ denotes the floor function. If the equality holds, then we say that G is upper
embeddable.
Upper embeddable graphs are well-studied objects by many researchers [4, 5, 8, 9, 10,
12, 13, 14, 15, 17, 18]. Jungerman and Xuong gave a combinatorial characterization of
upper embeddability independently.
Theorem 1.3 (Jungerman [5, Theorem 2], Xuong [18, Theorem A]). A connected graph
G with even (odd) Betti number is upper embeddable if and only if there exists a spanning
tree T of G such that all (all but one) connected components of G \ T consists of an even
number of edges.
Here is the main theorem of this article.
Theorem 1.4. Given a graph G, the following are equivalent.
(1) G has a full cyclic permutation ordering.
(2) There exists a 2-cell embedding ι : G→ Σ such that fι = 1.
(3) The Betti number β(G) is even and G is upper embeddable, that is, β(G) =
2γmax(G).
(4) There exists a spanning tree T of G such that every connected components of
G \ T consists of an even number of edges.
4 Ars Math. Contemp. 25 (2025) #P2.05
1
2
3
4
5 1
2
34 5
Figure 3: A 2-cell embedding of the butterfly graph into a torus with exactly one face.
Note that the conditions (2), (3), and (4) are equivalent by the definition of upper em-
beddability and Theorem 1.3. Figure 3 shows how the butterfly graph can be embedded
into a torus with exactly one face.
The organization of this article as follows. In Section 2, we will prove that (3) implies
(1) and give an example of constructing a full cyclic permutation ordering. In Section 3,
we will review the relation of 2-cell embeddings and rotation systems and we will prove
that (1) implies (2). Combining the proofs in Section 2 and Section 3, we will complete the
proof of Theorem 1.4.
In Section 4, we will study another extreme condition, that is, edge orderings ω such
that πω is the identity permutation, which Eden [2] studied. Eden gave necessary condi-
tions for such orderings and asked whether the condition is also sufficient. We will give a
counterexample for this question.
2 Proof that (3) implies (1)
First, we introduce the following lemma.
Lemma 2.1. Let π be a full cyclic permutation in Sn. If the distinct numbers u, v, w appear
in π in this cyclic order, then the product (u v)(v w)π is a full cyclic permutation.
Proof. By the assumption, we can write
π = (u a1 · · · ar v b1 · · · bs w c1 · · · ct),
where ai, bi, ci denote distinct numbers in [n] \ {u, v, w}. Then we obtain
(u v)(v w)π = (u v)(v w)(u a1 · · · ar v b1 · · · bs w c1 · · · ct)
= (u a1 · · · ar w c1 · · · ct v b1 · · · bs),
which is a full cyclic permutation.
We say that two edges e and e′ are adjacent if rG(e) ∩ rG(e
′) ̸= ∅, that is, they have a
common endvertex. Note that the case rG(e) = rG(e
′) is allowed.
Lemma 2.2. Let e and e′ be two adjacent edges in a graph G. If G \ {e, e′} has a full
cyclic permutation ordering, then G has a full cyclic permutation ordering.
Proof. Let ω′ = (e1, . . . , em) be a full cyclic permutation ordering of G \ {e, e
′}. If
rG(e) = rG(e
′), then ω := (e1, . . . , em, e
′, e) is a full cyclic permutation ordering of G
since πω = πω′ .
S. Tsujie and R. Uchiumi: Upper embeddability of graphs and products of transpositions . . . 5
1
2 3
4
5
W5
1
2 3
4
5
T
1
2 3
4
5
W5 \ T
Figure 4: Wheel graph W5 and its spanning tree T .
Now, suppose that rG(e) = {u, v} and rG(e
′) = {v, w} with u ̸= w. If the cycle
order of u, v, w in πω′ is u, v, w, then ω := (e1, . . . , em, e
′, e) is a full cyclic permutation
ordering of G by Lemma 2.1. In a symmetrical manner, if the cycle order is w, v, u, then
let ω := (e1, . . . , em, e, e
′).
The following lemma is required.
Lemma 2.3 (Xoung [17, Lemma 3]). Suppose that G is upper embeddable and β(G) is
even. If G is not a tree, then there exist two adjacent edges e and e′ such that G \ {e, e′} is
upper embeddable.
Proof that (3) implies (1) in Theorem 1.4. We will show that G has a full cyclic permuta-
tion ordering by induction on the Betti number β(G). When β(G) = 0, G is a tree and has
a full cyclic permutation ordering by Dénes’ theorem (Theorem 1.2).
Assume β(G) > 0. By Lemma 2.3, there exist two adjacent edges e and e′ such that
G′ := G \ {e, e′} is upper embeddable. By the induction hypothesis, G′ has a full cyclic
permutation ordering. By Lemma 2.2, G has a full cyclic permutation ordering.
Example 2.4. Consider the wheel graph W5 pictured in Figure 4. The Betti number of
W5 is 4. Let T be the spanning tree of W5 consisting of the edges 12, 23, 34, 45. Then
W5 \ T is connected and consisting of 4 edges. Therefore W5 satisfies the condition (4)
in Theorem 1.4 and hence has a full cyclic permutation ordering. We will construct a full
cyclic permutation ordering following by the proof of Lemma 2.2.
We partition the edges of W5 \T into two adjacent pairs {25, 35} and {14, 15}. Define
the edge ordering ω1 of T by ω1 := (12, 23, 34, 45). By Dénes’ theorem (Theorem 1.2),
πω1 is a full cyclic permutation. Indeed we have
πω1 = (4 5)(3 4)(2 3)(1 2) = (2 1 5 4 3).
Next we define the edge ordering ω2 of T ∪ {25, 35}. Following the proof above, define
ω2 by ω2 := ω1 ∗ (35, 25), where ∗ denotes the concatenation. Then
πω2 = (2 5)(3 5)(2 1 5 4 3) = (4 2 1 3 5).
Similarly, define ω3 by ω3 := ω2 ∗ (15, 14). Then
πω3 = (1 4)(1 5)(4 2 1 3 5) = (4 2 5 1 3).
Thus we obtain a full cyclic permutation ordering ω3 = (12, 23, 34, 45, 35, 25, 15, 14) of
W5.
6 Ars Math. Contemp. 25 (2025) #P2.05
3 Proof that (1) implies (2)
Let IG(v) denote the set of edges of G incident to a vertex v ∈ VG. A rotation system of
G is a collection ρ = (ρv)v∈VG consisting of cyclic orders ρv on IG(v), where a cyclic
order on IG(v) is an equivalence class of linear orders on IG(v) obtained by identifying
(e1, e2, . . . , es) with its circular shift (e2, . . . , es, e1), denoted by [e1, . . . , es].
Every embedding of G on an orientable closed surface defines a rotation system with
the clockwise ordering for each vertex. Conversely, from a rotation system, we can obtain
a 2-cell embedding of G on an orientable closed surface as follows.
Define DG by
DG := { (e, u) ∈ EG × VG | u ∈ rG(e) } .
We call an element of DG a dart. When rG(e) = {u, v}, the dart (e, u) shows an orien-
tation of the edge e from the source u to the target v. Define the involution α on DG by
α(e, u) := (e, v).
Given a rotation system ρ = (ρv)v∈VG , we will define bijections σ and φ from DG
to itself. Suppose that ρv = [e1, . . . , es]. Define σ by σ(ei, v) := (ei+1, v), where we
consider es+1 = e1. Let φ := σ ◦ α.
For every dart d, the target of d coincides with the source of φ(d). Therefore each orbit
in DG/⟨φ⟩ determines a closed walk on G and we can make a polygon whose sides are
formed by the darts in the orbit. Gluing the sides of the polygons obtained from the orbits
in DG/⟨φ⟩ by the involution α, we obtain an embedding of G on a closed surface. One
can show that this surface is actually orientable (See [6, Subsection 3.2]) and hence this
embedding is the desired 2-cell embedding.
Theorem 3.1 ([6, Theorem 3.2.4]). Given a graph G, there exists a one-to-one corre-
spondence between rotation systems of G and 2-cell embeddings of G on oriented closed
surfaces up to orientation-preserving homeomorphism.
Note that, from the construction, the number of the faces of the embedding correspond-
ing to a rotation system is equal to the number of the orbits in DG/⟨φ⟩.
Let ω be an edge ordering of G. For each v ∈ VG, let ωv denote the linear order on
IG(v) induced by ω. Moreover, let ρω,v be the cyclic order on IG(v) determined by ωv .
Thus we obtain the rotation system ρω := (ρω,v)v∈VG from an edge ordering ω and hence
the corresponding bijections σω and φω = σω ◦ α.
Lemma 3.2. Let ω = (e1, . . . , em) be an edge ordering of a graph G. For each v ∈ VG,
let fv denote the minimal edge in IG(v) with respect to ω. Define a map Ψ: VG/⟨πω⟩ →
DG/⟨φω⟩ by
Ψ([v]) := [(fv, v)],
where the brackets denote equivalence classes. Then Ψ is a bijection.
Proof. First, we will show that the map Ψ is well-defined. It is sufficient to show that
[(fv, v)] = [(fπω(v), πω(v))] for each v ∈ VG.
Fix v ∈ VG. The edge ordering ω = (e1, . . . , em) defines the set Tv as follows.
Tv := { ei ∈ EG | (τiτi−1 · · · τ1)(v) ̸= (τi−1 · · · τ1)(v) } ,
S. Tsujie and R. Uchiumi: Upper embeddability of graphs and products of transpositions . . . 7
where we agree with (τi−1 · · · τ1)(v) = v if i = 1. Suppose that Tv = { ej1 , ej2 , . . . , ejs }
with j1 < j2 < · · · < js. Then (ej1 , . . . , ejs) is a trail from v to πω(v). Let v0 := v and
for k ∈ {1, . . . , s} define vk recursively as the endvertex of ejk other than vk−1. Note that
vs = πω(v).
From the definition of Tv , for each k ∈ {1, . . . , s− 1}, ejk+1 covers ejk in IG(vk) with
respect to ω. Therefore φω(ejk , vk−1) = (ejk+1 , vk). Since ejs is the maximal element in
IG(vs) with respect to the order ω, we have φω(ejs , vs−1) = (fvs , vs) = (fπω(v), πω(v)).
Thus Ψ is well-defined.
Next, to prove the surjectivity, take an orbit W ∈ DG/⟨φω⟩. Let f be the minimal
element in
{ e ∈ EG | (e, v) ∈W for some v ∈ VG }
with respect to ω. Suppose (f, v) ∈ W . Assume f is not minimal in IG(v) with respect to
ωv . Then f
′ := σ−1ω (f) is less than f in IG(v) with respect to ω. Let rG(f
′) = {v, v′}.
Then φω(f
′, v′) = (f, v) and hence (f ′, v′) ∈ W . This contradicts to the minimality of f .
Therefore f is minimal in IG(v) and hence f = fv . Hence Ψ([v]) = [(f, v)] =W .
Finally, we prove the injectivity. Let u, v ∈ VG and suppose that Ψ([u]) = Ψ([v]).
Then we have φsω(fu, u) = (fv, v) for some s ∈ Z. Without loss of generality, we can
assume that s > 0.
Recall the edges of G are ordered by ω = (e1, . . . , em). We can write the edge fu as
fu = ej0 with some j0 ∈ {1, . . . ,m}. Moreover we can obtain the walk (ej0 , ej1 , . . . , ejs)
by (ejk , vk) := φω(ejk−1 , vk−1) for k ∈ {1, . . . , s}, where v0 := u. Note that (ej0 , v0) =
(fu, u) and (ejs , vs) = (fv, v). Suppose that
{ k ∈ {1, . . . , s} | jk−1 ≥ jk } = {p1, . . . , pt}
with p1 < · · · < pt = s. Then π
i
ω(v0) = vjpi for i ∈ {1, . . . , t}. In particular, π
t
ω(u) =
πtω(v0) = vjpt = vs = v. Thus [u] = [v] and hence f is injective.
Now we are ready to prove that (1) implies (2).
Proof that (1) implies (2) in Theorem 1.4. Let ω be a full cyclic permutation ordering of
G. Then the number of faces of the 2-cell embedding corresponding to the rotation system
ρω is equal to |DG/⟨φω⟩| = |VG/⟨πω⟩| = 1.
4 Identity permutation ordering
In this section, a graph is not necessarily connected. We say that an edge ordering ω of a
graph G is an identity permutation ordering if πω = ε, where ε denotes the identity permu-
tation. Every edgeless graph vacuously has an identity permutation ordering. The minimal
example of a non-trivial graph having an identity permutation ordering is the 2-cycle C2.
Eden [2] studied simple graphs that have an identity permutation ordering and mentioned
the complete graph K4 is the minimal example. Figure 5 shows identity permutation or-
derings of C2 and K4.
Eden gave necessary conditions (without proof) for simple connected graphs having an
identity permutation ordering as follows.
8 Ars Math. Contemp. 25 (2025) #P2.05
1 2
e1
e2
1
2 3
4
e1
e3
e5
e6
e4
e2
Figure 5: The 2-cycleC2 and the complete graphK4 have an identity permutation ordering.
Proposition 4.1 (Eden [2, page 130]). Let G be a simple connected graph on n vertices
with m edges. If G has an identity permutation ordering, then the following conditions
hold.
(1) m is even.
(2) There exist a set C consisting of closed trails and a map ψ : VG → C such that the
following conditions hold.
(i) ψ is bijective.
(ii) Every v ∈ VG belongs to the closed trail ψ(v).
(iii) The sum of the number of edges of closed trails in C is 2m.
(iv) Each edge of G belongs to exactly two closed trails in C.
Proof. Suppose that ω = (e1, . . . , em) be an identity permutation ordering. Since the
identity permutation is an even permutation, we have m is even. Let Ψ: VG → DG/⟨φω⟩
be the bijection considered in Lemma 3.2. Note that VG = VG/⟨πω⟩ since πω = ε.
For each v ∈ VG, there exists no dart d ∈ DG such that both d and α(d) belong
to Ψ(v) since πω is the identity. Thus, forgetting the direction of each dart in Ψ(v), we
obtain the closed trail ψ(v) ⊆ EG. Letting W := { ψ(v) | v ∈ VG }, we have a surjection
ψ : VG → W .
We will prove ψ is injective. Assume that there exist distinct vertices u, v such that
ψ(u) = ψ(v). Let ω′ = (f1, . . . , fr) be the induced order of ω on ψ(u). Since πω′(u) = u,
the edges f1 and fr are incident to u. Also, f1 and fr are incident to v by the same reason.
Hence f1 and fr are parallel edges between u and v. This contradicts that G is simple.
Therefore ψ is injective and hence bijective.
By the definition of maps Ψ and ψ, every v ∈ VG belongs to ψ(v). Moreover,
∑
v∈VG
|ψ(v)| =
∑
v∈VG
|Ψ(v)| = |DG| = 2m.
For any e ∈ EG, the two darts on e belongs distinct orbits Ψ(u) and Ψ(v). Then e belongs
to ψ(u) and ψ(v) and the other trails do not contain e. Thus the map ψ : VG → C has the
desired properties.
Example 4.2. For the complete graph K4 in Figure 5, the following map ψ satisfies the
conditions in Proposition 4.1.
ψ(1) = {e1, e4, e5}, ψ(2) = {e1, e3, e6}, ψ(3) = {e2, e4, e6}, ψ(4) = {e2, e3, e5}.
S. Tsujie and R. Uchiumi: Upper embeddability of graphs and products of transpositions . . . 9
v1
v2
v3
v4
v5
v6
v7
v8
v9
v10
v11
v12
Figure 6: A counterexample for Eden’s question.
Eden asked whether the necessary conditions in Proposition 4.1 are also sufficient. We
will give a counterexample for this question. Let G be the graph on 12 vertices with 20
edges pictured in Figure 6. Define a map ψ by
ψ(v1) = v1v2v10v1, ψ(v2) = v2v9v10v2, ψ(v3) = v3v10v4v3,
ψ(v4) = v4v10v11v4, ψ(v5) = v5v6v12v5, ψ(v6) = v6v9v12v6,
ψ(v7) = v7v8v12v7, ψ(v8) = v8v11v12v8, ψ(v9) = v9v2v1v10v9,
ψ(v10) = v10v3v4v11v10, ψ(v11) = v11v8v7v12v11, ψ(v12) = v12v5v6v9v12.
The conditions in Proposition 4.1 are satisfied. Suppose thatG has an identity permuta-
tion ordering ω. By Lemma 3.2 there exists a 2-cell embedding ι with fι = |DG/⟨φω⟩| =
|VG| = 12 faces. Then the genus gι satisfies
2− 2gι = |VG| − |EG|+ fι = 12− 20 + 12 = 4.
Therefore gι = −1, which is a contradiction.
Problem 4.3. Characterize graphs that have an identity permutation ordering.
ORCID iDs
Shuhei Tsujie https://orcid.org/0000-0001-5805-910X
Ryo Uchiumi https://orcid.org/0000-0002-4956-7177
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.06
https://doi.org/10.26493/1855-3974.3312.bb7
(Also available at http://amc-journal.eu)
Regular and semi-regular representations of
groups by posets
Jonathan A. Barmak *
Universidad de Buenos Aires, Facultad de Ciencias Exactas y Naturales,
Departamento de Matemática and
CONICET-Universidad de Buenos Aires, Instituto de Investigaciones Matemáticas Luis A.
Santaló (IMAS). Buenos Aires, Argentina.
Received 7 February 2024, accepted 14 April 2024, published online 13 March 2025
Abstract
By a result of Babai, with finitely many exceptions, every group G admits a semi-
regular poset representation with three orbits, that is, a poset P with automorphism group
Aut(P ) ≃ G such that the action of Aut(P ) on the underlying set is free and with three
orbits. Among finite groups, only the trivial group and Z2 have a regular poset repre-
sentation (i.e. semi-regular with one orbit), however many infinite groups admit such a
representation. In this paper we study non-necessarily finite groups which have a regular
representation or a semi-regular representation with two orbits. We prove that if G admits
a Cayley graph which is locally the Cayley graph of a free group, then it has a semi-regular
representation of height 1 with two orbits. In this case we will see that any extension of
the integers by G admits a regular representation. Applications are given to finite simple
groups, hyperbolic groups, random groups and indicable groups.
Keywords: Automorphism group of posets, Cayley graph, Dehn presentation, simple groups, random
groups.
Math. Subj. Class. (2020): 05E18, 06A11, 20B25, 20B27
1 Introduction
A representation of a group G by a poset is a poset P whose automorphism group Aut(P )
is isomorphic to G, together with an isomorphism G → Aut(P ). In other words, it is a
faithful action of G on P such that every automorphism of P is induced by the action. A
representation of G by P is called semi-regular if the action of G on the underlying set of
*Researcher of CONICET. Partially supported by grant PICT 2019-2338, PICT-2017-2806, PIP
11220170100357CO, UBACyT 20020190100099BA, UBACyT 20020160100081BA.
E-mail address: jbarmak@dm.uba.ar (Jonathan A. Barmak)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.06
P is semi-regular (free), that is the stabilizer of each point x ∈ P is trivial. In this case, the
orbit of each point is in bijection with G. A representation is regular if it is semi-regular
with a unique orbit (i.e. transitive).
Birkhoff proved that every finite group G has a semi-regular poset representation with
|G|+ 1 orbits [9], and Frucht then proved that d+ 2 orbits suffice, where d is the cardinal-
ity of any generator set of G [13]. Babai proved in [4, Corollary 0.14], [5, Corollary 4.3],
the surprising result that with finitely many exceptions every group (non-necessarily finite)
admits a semi-regular poset representation with three orbits. Babai’s proof uses his charac-
terization of groups admitting a digraphical regular representation (defined below), which
in turn analyzes properties of generator sets of different classes of groups. In [7] we gave
a short self-contained proof that every finitely generated group admits a semi-regular poset
representation with four orbits.
Note that if a group G admits a regular representation by a poset P , then P is homoge-
neous (i.e. given two points in P , there is an automorphism of P mapping one to the other).
If G is finite, this means that P is discrete (different points are not comparable), so G is a
symmetric group. Since the action is free, G must be trivial or Z2. The classification of
infinite groups admitting a regular poset representation, however, is a non-trivial problem.
Regular representations of groups by totally ordered sets have been studied in [15, 36]. If
a group admits one such representation, it is isomorphic to a subgroup of R.
Problem. Classify those (finite, infinite) groups G which admit a semi-regular poset rep-
resentation with exactly two orbits and those (infinite) groups that admit a regular poset
representation.
Similar classification problems for finite or infinite groups have been addressed and
solved in the context of graphs. Finitely generated groups admitting a graphical regular
representation have been characterized after a long series of articles of different authors
[16, 20, 22, 23, 24, 30, 34, 35, 40]. A graphical regular representation of a group G is
a graph Γ with Aut(Γ) ≃ G and such that Aut(Γ) acts regularly on the vertex set. On
the other hand the proof of the main result in [3] shows that with finitely many exceptions
every finitely generated group admits a graphical semi-regular representation with two
orbits. Other classification problems studied include digraphical regular representations
[4, 5], graphical and digraphical semi-regular representations with more orbits [12], regular
and semi-regular representations by tournaments with two orbits [6, 17], oriented regular
representations [32].
Among semi-regular poset representations with two orbits, there are some which are
perhaps easier to describe: those which are given by posets of height 1 (i.e. the longest
chain has two elements). Because of their relevance in this work and their resemblance
to Cayley graphs, they will receive the name of Cayley (poset) representations. For finite
groups every semi-regular representation with two orbits is of this type.
Recall that the girth g(Γ) of a graph Γ is the length of its shortest cycle (infinite if the
graph is a forest). When S = {sx}x∈X is a generator set of a groupG, the girth g(Γ(G,S))
of the Cayley graph can be interpreted as the shortest non-trivial word in the free group
F (X) generated by X which is trivial in G when evaluated in the sx. In Section 3 we
prove the following
Theorem 1.1. Let G be a group generated by two elements x, y. Suppose that for every
non-trivial word w in the free group of rank 2, with length smaller than or equal to 21,
w(x, y) ̸= e ∈ G, that is g(Γ(G, {x, y})) > 21. Then G admits a Cayley representation.
J. A. Barmak: Regular and semi-regular representations of groups by posets 3
Using ideas from Weigel, Dixon, Pyber, Seress and Shalev we will prove that, with
finitely many exceptions, every finite simple group admits a Cayley representation (Corol-
lary 4.4). A second application of Theorem 1.1 concerns random groups (in Gromov’s
density model, and in the Arzhantseva and Ol’shanskii few relators model). Using results
by Ollivier, and by Arzhantseva and Ol’shanskii regarding Dehn presentations and sixth
groups we will show that for two generators a random group has a Cayley representation
(Corollaries 5.6 and 5.8). In the density model we require d < 15 .
In Section 6 we establish a connection between Cayley and regular representations by
proving that if G is a group different from Z2 which admits a Cayley representation, then
every extension of the integers by G has a regular poset representation (Theorem 6.1).
With similar ideas to those used in the proof of that result we will prove that with finitely
many exceptions every indicable group admits a semi-regular representation with two or-
bits, though not necessarily a Cayley representation.
2 Cayley, Haar and digraphical regular representations
We begin with the study of semi-regular representations with two orbits. Note first that for
a group G, the fact that there exists a poset P with Aut(P ) ≃ G and the action of Aut(P )
on P having two orbits, does not imply that G admits a semi-regular poset representation
with two orbits. For example Z22 is the automorphism group of the poset with underlying
set {0, 1, 0′, 1′} where 0, 1 < 0′, 1′, but Z22 does not admit a semi-regular representation
with two orbits.
We turn to the special case of representations of height 1. Let G be a group and let
S ⊆ G be a non-empty subset. We define the Cayley poset P (G,S) as follows. The
underlying set is a union of two copies G, G′ = {g′| g ∈ G} of G. If g ∈ G and h ∈ gS,
then g < h′. No other pair of different points are comparable. The regular action of G on
each copy of G, by left multiplication, gives an action L : G → Aut(P (G,S)) of G on
P (G,S). If there are no other automorphisms in P than those induced by (in the image of)
L, P (G,S) is a semi-regular representation of G with two orbits. In this case P (G,S), or
more precisely L : G→ Aut(P (G,S)), is called a Cayley representation of G.
Note that if a finite group is represented semi-regularly with two orbits by a poset P ,
then P has height equal to 1, as points in the same orbit cannot be comparable (x < gx
would lead to an infinite chain). Let G be a non-necessarily finite group and let P be a
semi-regular representation with two orbits and of height 1. Then the orbits are the set
of minimal points and the set of maximal points. Thus, the set of minimal points can be
identified with G, the set of maximal points with G′, and the action of G on P can be
assumed to be the left regular action on each copy of G. If we denote by S the set of
elements h ∈ G such that e < h′, then S ̸= ∅ and P = P (G,S). Indeed, given g, h ∈ G,
we have that g < h′ if and only if e < (g−1h)′, which is equivalent to g−1h being in S,
i.e. h ∈ gS.
In conclusion the semi-regular representations with two orbits and of height 1 of a
group G are the Cayley representations of G, and for G finite, every semi-regular repre-
sentation with two orbits is of this form. Our problem, in the finite case, is then to classify
those finite groups G for which there is a subset S with Aut(P (G,S)) being equal to the
automorphisms induced by L. An infinite group G, however, could admit a semi-regular
representation with two orbits of a different kind. Let P be constructed from two copies
Z,Z′ of Z in which i, i′ are smaller than j every time i < j, where < is the usual order of
4 Ars Math. Contemp. 25 (2025) #P2.06
the integers. This is a semi-regular representation of Z with two orbits. The difference be-
tween Cayley representations and general semi-regular representations with two orbits, and
the relevance of the first when studying regular representations will be clear in Section 6.
Let G be a non-necessarily finite group and ∅ ≠ S ⊆ G. Then P (G,S) is a Cayley
representation of G if and only if every automorphism fixing the minimal point e ∈ G ⊆
P (G,S) is the identity. Indeed if ϕ ∈ Aut(P (G,S)), then ϕ(e) must be a minimal point,
say g ∈ G ∈ P (G,S). Let L(g) be the corresponding automorphism induced by L : G →
P (G,S). Thus, L(g)−1ϕ fixes e, and if this implies that L(g)−1ϕ = 1P (G,S), then ϕ =
L(g).
The following result is easy to prove and details are left to the reader.
Proposition 2.1. Let G be a group, ∅ ≠ S ⊊ G. Then
(i) For every h ∈ G,P (G,S) is a Cayley representation of G if and only if P (G,Sh) is.
(ii) For every automorphism ψ of G, P (G,S) is a Cayley representation of G if and only
if P (G,ψ(S)) is.
(iii) P (G,S) is a Cayley representation of G if and only if the complementP (G,G∖S) is.
Remark 2.2. If P is a semi-regular poset representation of a group G different from the
trivial group 1 and Z2, thenP is connected. Indeed ifP is not connected, by semi-regularity
each component must have trivial automorphism group. If the components are pairwise
non-isomorphic, then Aut(P ) is trivial, a contradiction. Let C,C ′ be two isomorphic
components of P . If there is no other component, then Aut(P ) = Z2, a contradiction. If
there is another component, there is a non-trivial automorphism of P fixing every point in
that component, contradicting semi-regularity again.
In the case of Cayley posets P (G,S), the fact that P (G,S) is connected is equivalent
to SS−1 = {st−1| s, t ∈ S} being a generating set of G, which in particular implies that
S generates G.
A Haar graphical representation of a group G is a semi-regular representation by a
bipartite graph with the orbits being the parts of the bipartition. In other words, it is a
bipartite graph B with parts V1, V2 together with an isomorphism between G and Aut(B),
such that the action of G on the vertex set of B is semi-regular with the two orbits being V1
and V2. Every representation of this kind is isomorphic to a graph B(G,S) for some S ⊆
G. This is the bipartite graph with parts G,G′ and edges (g, (gs)′) for g ∈ G, s ∈ S. The
characterization of graphs admitting a Haar graphical representation is an open problem
[12].
The comparability graph C(P (G,S)) of P (G,S) (in this case this is the underlying
undirected graph of the Hasse diagram) is a bipartite graph, and every automorphism of
the poset induces an automorphism of the graph. Thus, if G admits a Haar graphical rep-
resentation, it admits a Cayley representation. The converse is not true as no abelian group
admits a Haar graphical representation [12], but they can have a Cayley representation (see
Example 3.1 below).
Recall that a digraphical regular representation of a group G is a digraph Γ such that
Aut(Γ) ≃ G and the action of Aut(Γ) on the vertex set of Γ is regular. Babai proved the
following characterization [4, 5].
J. A. Barmak: Regular and semi-regular representations of groups by posets 5
Theorem 2.3 (Babai). A non-necessarily finite group G admits a digraphical regular rep-
resentation if and only if G is different from Z22,Z
3
2,Z
4
2,Z
2
3 and the quaternion group Q8.
Babai observes that every digraphical regular representation induces a semi-regular
poset representation with three orbits [4, Proposition 7.3]. This representation is con-
structed by taking three copies G,G′, G′′ of G and setting g < g′ < g′′ for every g ∈ G
and g < h′′ if (g, h) is an edge in the digraphical regular representation. For Q8 Babai
provides a separate construction of a semi-regular poset representation with three orbits.
Thus, every group different from Z22,Z
3
2,Z
4
2,Z
2
3 admits a semi-regular poset representation
with three orbits. The converse of this statement is not analyzed in [4, 5].
Proposition 2.4. The group Z22 does not admit a semi-regular poset representation with
three orbits.
We give a proof by case analysis which is included at the end of this article. This is also
easy to check with a computer. We do not inspect here the cases Z32,Z
4
2 and Z
2
3.
The idea for proving that the five groups in Theorem 2.3 have no digraphical regular
representation is to show that in each of those cases for every subset S ⊆ G there exists
an automorphism ϕ ̸= 1G of G such that ϕ(S) = S. Instead of applying this idea to our
problem directly, we observe the following.
Remark 2.5. If a group G admits a Cayley representation P (G,S), then it admits a di-
graphical regular representation. Define the digraph Γ with vertex set G and edges (g, gs)
for every g ∈ G, s ∈ S. Let ϕ ∈ Aut(Γ). Define ϕ : P (G,S) → P (G,S) by ϕ(g) = ϕ(g)
and ϕ(g′) = ϕ(g)′ for every g ∈ G. It is easy to see that ϕ is an automorphism of P (G,S),
so by assumption it coincides with L(g) for some g ∈ G. This implies that ϕ corresponds
to the left multiplication by g. Thus, Γ is a digraphical regular representation of G.
The converse of this statement does not hold. Moreover, we have the following result.
Proposition 2.6. The groups G = Z3,Z4,Z5,Z6,Z7,Z
2
2,Z
3
2,Z
4
2,Z
2
3, S3 and Q8 do not
admit a semi-regular poset representation with two orbits.
Proof. The cases Z3,Z4,Z5 and Z7 are covered by Corollary 12 of [8]. For Z
2
2,Z
3
2,Z
4
2,Z
2
3,
Q8 the result follows from Theorem 2.3 and Remark 2.5. Suppose then that G = Z6 =
{0, 1, 2, 3, 4, 5}. Assume there exists ∅ ≠ S ⊆ Z6 such that Aut(P (G,S)) is a Cayley
representation of Z6. Then S ̸= Z6 and by Proposition 2.1, we may assume that |S| ≤ 3
and that 0 ∈ S. By Remark 2.2, P (G,S) is connected. If |S| = 2, P (G,S) is a crown, and
there is an involution fixing 0. Therefore we may assume |S| = 3, and by Proposition 2.1
there are only two representatives to analyze: S = {5, 0, 1}, {0, 1, 3}. In the first case the
involution k → −k, k′ → (−k)′ fixes 0. In the second case there is an involution which
transposes 2 with 5 and 0′ with 3′, and fixes any other point.
Finally, suppose ∅ ̸= S ⊊ S3 is such that P (S3, S) is a Cayley representation of
S3 = {e, (12), (13), (23), (123), (132)}. Again by Proposition 2.1 we may assume that
|S| ≤ 3 and that e ∈ S. Since S3 is not cyclic, by Remark 2.2, |S| = 3. By applying an au-
tomorphism of S3 there are only two cases to analyze S = {e, (12), (13)}, {e, (12), (123)}.
In the first case there is an involution which transposes (12) with (13), (123) with (132),
(12)′ with (13)′ and (123)′ with (132)′. In the second, there is an involution transposing
(12) with (23), (123) with (132), e′ with (123)′ and (13)′ with (23)′.
6 Ars Math. Contemp. 25 (2025) #P2.06
3 Girth of Cayley graphs and representations with two orbits
Let G be a group and let ∅ ≠ S ⊆ G be a subset. For g ∈ G we define the S-neighborhood
of g by NS(g) = gSS
−1 = {gst−1| s, t ∈ S} ⊆ G. Note that two minimal elements
g, h ∈ G of the Cayley poset P = P (G,S) have a common upper bound if and only if
h ∈ NS(g). Thus, if ϕ ∈ Aut(P ), ϕ(NS(g)) = NS(ϕ(g)) for every g ∈ G. Given
g, h ∈ G, their affinity is α(g, h) = #(NS(g) ∩ NS(h)) (this could be infinite if S is).
Then α(ϕ(g), ϕ(h)) = α(g, h) for every automorphism ϕ.
Example 3.1. We claim that if n ≥ 9 and S = {0, 1, 3} ⊆ Zn, then P = P (Zn, S) is a
Cayley representation of Zn. Let ϕ ∈ Aut(P ) be an automorphim which fixes 0 ∈ P . We
only need to prove that ϕ is the identity 1P .
It is easy to see that for i ∈ Zn, NS(i) = {i−3, i−2, i−1, i, i+1, i+2, i+3}. Since
n ≥ 7, #NS(i) = 7 for every i. Since n ≥ 8, α(i, i + 1) = 6 for every i. Since n ≥ 9,
α(i, j) = 6 only when j ∈ {i − 1, i + 1}. Since ϕ preserves affinity, and 0 is fixed, then
ϕ(1) ∈ {−1, 1}. However, if ϕ(1) = −1, by induction ϕ(i) = −i for every i. This is a
contradiction as {0, 2, 3} has an upper bound, while {0,−2,−3} does not. Thus ϕ(1) = 1
and by induction ϕ(i) = i for every i. Each maximal element i′ is uniquely determined by
its smaller elements i, i− 1, i− 3. Thus ϕ(i′) = i′ for every i.
Note that 9 is three times the diameter of S = {0, 1, 3} in the Cayley graph Γ(Zn, {1}).
Also note that the argument in Example 3.1 can be used with no changes to prove that Z
admits a Cayley representation.
In [8] it is proved that Z8 also admits a Cayley representation. A similar argument is
used with S = {0, 1, 2, 4} and a different version of the notion of S-neighborhood. In our
next result both choices of S will be used simultaneously to represent groups generated by
two elements.
Suppose {x, y} is a set of generators of a group G, and let Γ = Γ(G, {x, y}) be the
corresponding Cayley graph, i.e. the undirected graph with vertex set G and an edge with
vertices g, gx and another with vertices g, gy for each g ∈ G (in particular Γ has loops if
any generator is the identity and it has multiple edges if one generator has order two or if
it equals the other or its inverse). Once again, recall that the girth of a graph is the smallest
length of a cycle (infinite if the graph is a forest), so the girth g(Γ) of Γ is greater than or
equal to r if and only if for every non-trivial word w of length smaller than r in the free
group F2 = F (X,Y ) of rank 2, w(x, y) ̸= e ∈ G. The girth of Γ measures how similar
the Cayley graphs Γ and Γ(F2, {X,Y }) are locally. Namely, if g(Γ) ≥ 2r + 2, then the
balls Br(Γ), Br(Γ(F2, {X,Y })) of radius r with center in e in these two graphs (i.e. the
subgraphs induced by the vertices whose distance to e is at most r) are isomorphic.
Proof of Theorem 1.1. Let S = {e, x, x2, x4, y, y3} and let P = P (G,S). Let ϕ be an au-
tomorphism of P fixing the minimal point e ∈ P . We want to prove that ϕ = 1P . Note that
the set NS(e) = SS
−1 consists of the elements x−4, x−3, x−2, x−1, e, x, x2, x3, x4, y−3,
y−2, y−1, y, y2, y3, xy−1, xy−3, x2y−1, x2y−3, x4y−1, x4y−3, yx−1, yx−2, yx−4, y3x−1,
y3x−2, y3x−4. These 27 elements are different since the girth g(Γ) of Γ = Γ(G, {x, y})
is greater than 14. The ball B7(Γ) with center e and radius 7 in Γ is isomorphic to the ball
B7(Γ(F2, {X,Y })) since g(Γ) ≥ 16. This ball is then a tree. In Figure 1 we have depicted
the smallest connected subgraph ofB7(Γ) which contains all the 27 vertices corresponding
to elements in NS(e).
J. A. Barmak: Regular and semi-regular representations of groups by posets 7
12 111211
9
8
7
9
8
7
6
6
6
6
66
66
e x x2 x3x -1x -2x -3x -4
6
6
x4
88 77
y
y 2
y 3y 3x -1y 3x -4
y x -4 yx -1
y 3x -2
yx -2
y -1
y -2
y -3
x
x
y -1
y -3
x2
x2
y -1 x4y -1
x4y -3y -3
6
6
Figure 1: The smallest connected subgraph of B7(Γ) containing the vertices of NS(e).
The 27 vertices corresponding to the points in NS(e) are represented with big dots,
while there are 5 vertices in the picture which are not in NS(e), represented with small
dots. In each vertex e ̸= g ∈ NS(e) we have indicated the affinity α(e, g) = #(NS(e) ∩
NS(g)). This can be computed in each of the 26 cases algebraically, using that no non-
trivial word in F2 of length smaller than or equal to 21 represents the trivial element ofG, or
graphically, using that g(Γ) > 21. For the first alternative, note that an element h ∈ NS(e)
is represented by a word w ∈ F2 of length smaller than or equal to 7, while an element
h′ ∈ NS(g) is represented by a wordw′ with length at most 7+7 = 14. Thus h = h′ ∈ G if
and only ifw = w′ ∈ F2. For the second, graphical, alternative, note that if g is represented
by a word u ∈ F2 of length at most 7, then the union of B7(Γ(F2, {X,Y })) and the ball
uB7(Γ(F2, {X,Y })) of center u is a subgraph of Γ(F2, {X,Y }) of diameter at most 21.
Thus, this graph is isomorphic to the (non-necessarily induced) subgraph B7(Γ)∪ gB7(Γ)
of Γ, so α(e, g) can be computed in Γ(F2, {X,Y }).
Since ϕ(e) = e, ϕ(NS(e)) = NS(e). Among the points in NS(e), there are only two,
x−1, x whose affinity with e is α(e, x−1) = α(e, x) = 12. Thus ϕ(x) ∈ {x−1, x}. If
ϕ(x) = x−1, then ϕ(NS(x)) = NS(x−1) and α(ϕ(x2), x−1) = α(x2, x) = 12. This
implies that ϕ(x2) ∈ {x−2, e}. Since ϕ is injective, ϕ(x2) = x−2. By induction ϕ(xn) =
x−n for every n ≥ 0. The set {e, x2, x3, x4} of minimal points of P has an upper bound.
However, ϕ({e, x2, x3, x4}) = {e, x−2, x−3, x−4} does not. Indeed, if g′ ∈ P was an
upper bound, then {e, x−2, x−3, x−4} ⊆ P≤g′ = {g, gx−1, gx−2, gx−4, gy−1, gy−3}. In
particular e must be contained in the later, so g ∈ {e, x, x2, x4, y, y3} = S. But for any of
these six cases we see that either x−2 or x−3 is not contained in P≤g′ , using that g(Γ) ≥ 10.
We conclude then that ϕ(x) = x, and by an inductive argument, that ϕ(xn) = xn for every
n ≥ 0. Similarly, ϕ(x−n) = x−n for every n ≥ 0 (when G is infinite, and moreover, x is
not of finite order, this does not follow from the previous claim). Moreover, this argument
shows that if g ∈ P is fixed by ϕ for some g ∈ G, then gxn ∈ P is fixed for every n ∈ Z.
Among the points inNS(e) (we only care about points inNS(e) which are not powers of x,
in fact), only y−1 and y have affinity equal to 9 with e. Thus ϕ(y) ∈ {y−1, y}. As above,
if ϕ(y) = y−1, then ϕ(yn) = y−n for every n ≥ 0, and this yields a contradiction since
{e, y2, y3} ⊆ P has an upper bound, while {e, y−2, y−3} does not. Indeed, either y−2 or
8 Ars Math. Contemp. 25 (2025) #P2.06
y−3 is not contained in P≤g′ for each g ∈ S, as g(Γ) ≥ 10. Therefore, ϕ(y) = y, and by
induction ϕ(yn) = yn for every n ≥ 0. The same holds for n ≤ 0. Moreover, for every
g ∈ G, if ϕ(g) = g, then ϕ(gyn) = gyn for every n ∈ Z. Now, since {x, y} generates G,
all the minimal points of P are fixed by ϕ. Finally, every maximal point of P is determined
by the set of points it covers. Concretely, e′ is the unique upper bound of {e, y−1} ⊆ P ,
since for any e ̸= g ∈ S, y−1 does not belong to P≤g′ , using that g(Γ) ≥ 9. Moreover, for
each g ∈ G, g′ is the unique upper bound of {g, gy−1}. Since the later is invariant, g′ is
also fixed.
Remark 3.2. Note that in the proof of Theorem 1.1 we do not really need all the non-
trivial words w(X,Y ) of length at most 21 to represent non-trivial elements of G, but just
a concrete list with much fewer words. This could be used to obtain representations of
examples not covered by the theorem. Also, even if some of the words in this list are trivial
in G, a similar proof could work as long as we have some control on the number of words
which are indeed trivial. For example, the fact that Z2 has a Cayley representation follows
from a variant of this type.
Generalizations of Theorem 1.1 for groups with d ≥ 3 generators can be obtained with
similar methods by combining d generator sets S, S′, . . . , S(d) of Z as we did for d = 2.
In this article we restrict ourselves to the case of two generators.
As a first example, we can apply Theorem 1.1 to G = F2.
The construction of regular graphs, and specifically Cayley graphs, with large girth,
has a long history, with application to low density codes, among others. A simple counting
argument due to Moore implies that a regular graph Γ of degree 4 and girth g(Γ) > 21 has
at least 311 − 1 vertices. Thus, Theorem 1.1 can only be applied to groups of order at least
that number.
In [29] Margulis proves that if p is a prime and G = SL2(Zp), then x =
(
1 2
0 1
)
, y =(
1 0
2 1
)
generate G and g(Γ(G, {x, y})) ≥ 2 log1+√2(p2 ) − 1. Thus, for p > 2(1 +
√
2)11,
SL2(Zp) admits a Cayley representation. For PGL2(Zp) and PSL2(Zp) there are similar
results by Lubotzky, Phillips and Sarnak [27], but their Cayley graphs are constructed with
q+1
2 generators, where q is a prime congruent to 1 modulo 4.
4 Representability of finite simple groups
In [41] Weigel proved the following result, answering a question originally raised by Mag-
nus:
Theorem 4.1 (Weigel). If C is an infinite family of isomorphic types of non-abelian finite
simple groups, then F2 = F (X,Y ) is residually C. That is, the intersection of all the
normal subgroups N ⊴ F2 such that F2/N ∈ C, is trivial.
In fact, this result holds for Fd, d ≥ 2. A much stronger result is proved by Dixon,
Pyber, Seress and Shalev in [11] using probabilistic methods. Although our result in this
section (Corollary 4.4) follows from [11, Theorem 3] (and Theorem 1.1), we choose an
approach using only Weigel’s result. As explined in [11], Weigel’s Theorem implies
Theorem 4.2. If S is an infinite family of non-isomorphic non-abelian finite simple groups
and w ∈ F2 = F (X,Y ) is a non-trivial word, then there exists S ∈ S and x, y ∈ S such
that x, y generate S and w(x, y) ̸= e ∈ S.
J. A. Barmak: Regular and semi-regular representations of groups by posets 9
Indeed, since w is non-trivial, by Theorem 4.1 there exists N ⊴ F2 such that w /∈
N and F2/N is isomorphic to some S ∈ S . Let ϕ : F2 → S be an epimorphism with
ker(ϕ) = N . Then x = ϕ(X), y = ϕ(Y ) satisfy the required property.
If P1, P2, . . . , Pk are non-trivial polynomials over some coefficient ring, then their
product P has the property that every root of some Pi is also a root of P . The following is
a similar construction for equations over groups. Recall that an equation in d variables is
just a word w ∈ Fd = F (X1, X2, . . . , Xd), and a solution of w on a group G is a d-tuple
(x1, x2, . . . , xd) ∈ Gd such that w(x1, x2, . . . , xd) = e ∈ G.
Lemma 4.3. Letw1, w2, . . . , wk ∈ Fd be non-trivial words. Then there exists a non-trivial
word w ∈ Fd with the following property. For every group G and for every 1 ≤ i ≤ k,
every solution of the equation wi on G, is also a solution of w.
Proof. By induction it suffices to prove the result for k = 2. Let F be the subgroup of Fd
generated by w1, w2. If rk(F ) = 1, F is generated by some non-trivial word u ∈ Fd and
there exist l,m ∈ Z with w1 = ul, w2 = um. By assumption l,m ̸= 0. Let w = ulm.
Then w satisfies the required property. If rk(F ) = 2, then {w1, w2} is a basis of F and, in
particular, w1 and w2 do not commute. Take w = [w1, w2] ∈ Fd. Then w is non-trivial,
and every solution of w1 and of w2 on a group G is a solution of w.
Corollary 4.4. There are only finitely many finite simple groups which do not admit a
Cayley representation.
Proof. By Example 3.1 there are only finitely many finite abelian simple groups which do
not admit such representation. Suppose there exists an infinite family S of non-isomorphic
non-abelian finite simple groups which do not admit such representation. Let w1, w2, . . . ,
wk be all the non-trivial words in F2 of length smaller than or equal to 21. Let w ∈ F2 be
a non-trivial word as in the statement of Lemma 4.3. By Theorem 4.2 there exists S ∈ S
and x, y ∈ S which generate S and such that w(x, y) ̸= e ∈ S. Then wi(x, y) ̸= e ∈ S for
every 1 ≤ i ≤ k. This is a contradiction by Theorem 1.1.
5 Dehn presentations and random groups
In this section we will find a large class of examples satisfying the hypothesis of Theo-
rem 1.1, and we will see that, in some sense, almost all finitely presented groups admit a
Cayley representation.
We briefly recall some basic concepts from small cancellation theory. Standard refer-
ence for this is [28, 39]. Let P = ⟨x1, x2, . . . , xd|R⟩ be a finite presentation of a group G.
That is, R is a finite set of words in the letters x±1i , and G ≃ F (x1, x2, . . . , xd)/N(R),
where N(R) denotes the normal closure of R in the free group. A word (in the x±1i ) is
reduced if no letter equals the inverse of the following, and it is cyclically reduced if it is
reduced and the last letter is not the inverse of the first. We will assume R consists just of
cyclically reduced words. The set R of relators is called symmetrized if it is closed under
taking cyclic permutations and inverses. The symmetrized closure R of R consists of all
cyclic permutations of both elements of R and of their inverses. Replacing R by R we
can turn any presentation into a presentation of the same group with symmetrized relator
set. If a word w is a concatenation uv of two words, u is said to be an initial subword of
w. A word u is called a piece with respect to P if there are two different elements of R
which contain u as an initial subword. Given 0 < λ < 1, we say that P satisfies the small
10 Ars Math. Contemp. 25 (2025) #P2.06
cancellation condition C ′(λ) if for every piece u which is a subword of an element r ∈ R
we have |u| < λ|r|. Here |u| denotes the length of u.
If w is a reduced word which contains a subword u that is an initial subword of an
element r = uv ∈ R and moreover, |u| > |r|2 , then we can replace u by v−1 in w and then
reduce, to obtain a new reduced word w′ which represents the same element as w in G and
it is shorter than w. We repeat this process until we obtain a word which cannot be replaced
by a shorter one in this sense. This is the Dehn algorithm. If the algorithm finishes with the
empty word, then w = e ∈ G. We say that Dehn’s algorithm solves the word problem for
P , or that the presentation is Dehn’s, if for every reduced word which is trivial in G, the
algorithm finishes with the empty word.
Note that if P is Dehn’s and w is a reduced nonempty word which is trivial in G, then
in the last step of the algorithm we have a word which lies in R. In particular, there is
a relator of P which is shorter than or equal in length to w. If P = ⟨x1, x2, . . . , xn|R⟩
satisfies the C ′( 16 ) condition, then by Greendlinger’s lemma [18], it is a Dehn presentation.
Thus, if the length of every relator in P is greater than or equal to some number l ≥ 1, then
g(Γ(G, {x1, x2, . . . , xn})) ≥ l. From Theorem 1.1 we deduce the following
Corollary 5.1. Let P = ⟨x, y|r1, r2, . . . , rm⟩ be a presentation which satisfies C ′( 16 ) and
such that |rj | ≥ 22 for every 1 ≤ j ≤ m. Then the presented group GP admits a Cayley
representation.
For 1-relator groups with torsion there is a stronger result than Corollary 5.1 in virtue
of the Newman Spelling Theorem ([28, Theorem 5.5]), which implies that when the relator
is a proper power, the presentation is Dehn.
Corollary 5.2. Let P = ⟨x, y|rm⟩ be a presentation, where r is cyclically reduced and
m ≥ 2. If |rm| ≥ 22, GP admits a Cayley representation.
When studying random groups, there are two models which have received more at-
tention than any other: Gromov’s density model introduced in [19, §9] and Arzhantseva
and Ol’shanskii’s few relators model [2]. We recall basic definitions. Standard refer-
ence on this is [14, 26, 37]. In the density model we fix a number n ≥ 2 of generators
x1, x2, . . . , xn and a density parameter 0 ≤ d ≤ 1. The number of cyclically reduced
words of length l is asymptotically (2n − 1)l. Given l ≥ 1, we choose (2n − 1)dl cycli-
cally reduced words of length l uniformly randomly and independently to get a presentation
P = ⟨x1, x2, . . . , xn|R⟩ whose relators are those words. We say that a random presentation
(or a random group) satisfies a certain property, if the probability that P (or GP ) satisfies
the property tends to 1 as l → ∞. We have the following result by Gromov and Ollivier
[37].
Theorem 5.3 (Gromov, Ollivier). In the density model, if d < 12 , a random group is infinite
hyperbolic, while for d > 12 , a random group is Z or Z2.
Although hyperbolic groups admit a Dehn presentation, this new presentation could
have relators of different lengths. However the following result holds.
Theorem 5.4 ([19, Gromov]). In the density model, if d < 112 , then a random presentation
satisfies C ′( 16 ), so the Dehn algorithm solves the word problem. If d >
1
12 , a random
presentation does not satisfy C ′( 16 ).
J. A. Barmak: Regular and semi-regular representations of groups by posets 11
The fact that a presentation does not satisfy C ′( 16 ) does not imply it is not a Dehn
presentation. Moreover, Ollivier proved in [38] the following result.
Theorem 5.5 (Ollivier). In the density model, if d < 15 , a random presentation is Dehn’s.
If d > 15 , then it is not.
This, together with Theorem 1.1 imply the following
Corollary 5.6. In the density model, if d < 15 , a random group with n = 2 generators
admits a Cayley representation.
Similar results for every fixed number n ≥ 3 of generators should be true, and a gener-
alization of Theorem 1.1 could be used in the proof.
We turn now to the few relators model by Arzhantseva and Ol’shanskii. In this model
both the number n of generators and number m of relators are fixed. For each l ≥ 1,
m cyclically reduced words of length at most l are chosen at random independently and
uniformly to form the set R of relators. We say that a random n-generator, m-relator
presentation (or group) satisfies certain property if it does with probability → 1 when
l → ∞.
Theorem 5.7 (Arzhantseva, Ol’shanskii [2]). For every n ≥ 2, m ≥ 1, λ > 0, a random
n-generator m-relator presentation satisfies C ′(λ).
Corollary 5.8. For every m ≥ 1, a random 2-generator m-relator group admits a Cayley
representation.
Proof. This follows from Corollary 5.1, by applying Theorem 5.7 for n = 2, λ = 16 and
this assertion: in a random 2-generator m-relator presentation all the relators have length
at least 22. This follows from a simple counting argument (cf. [1, page 3208]): the number
cl of cyclically reduced words in x
±1, y±1 of length l is smaller than or equal to 4.3l−1.
Thus, the number c≤l of cyclically reduced words of length ≤ l is at most 2.(3l−1) < 2.3l.
The number of m-tuples of cyclically reduced words of length at most l which contain at
least one word of length ≤ 21 is at most m.2.321(2.3l)m−1 = m2m3213lm−l. On the
other hand c≤l ≥ cl ≥ 4.3l−2.2 = 8.3l−2 for each l ≥ 2. Therefore, the number of m-
tuples of cyclically reduced words of length ≤ l is at least (8.3l−2)m = 8m3lm−2m. Since
m2m3213lm−l/8m3lm−2m → 0 as l → ∞, the assertion is proved.
6 Extensions of the integers by semi-regularly representented groups
Recall that a group G is said to be indicable if there exists an epimorphism G → Z. In
other words, they are the extensions of Z or, equivalently, the semidirect products N ⋊ψ Z.
Theorem 6.1. Let G ̸= Z2 be a group which admits a Cayley representation, and let
ψ : Z → Aut(G) be a group homomorphism. Then G ⋊ψ Z admits a regular poset repre-
sentation.
Proof. Let P (G,S) be a Cayley representation of G. Define P to be the poset obtained
from countable many copies P (G,S) × {n} (n ∈ Z) of P (G,S) with the identifications
(g′, n) ∼ (ψ(1)(g), n + 1) for every g ∈ G,n ∈ Z. That is, the order on P is the
transitive closure of the union of the orders in each copy. Let H = (Gop ⋊ψ Z)
op. In
other words, the underlying set of H is the cartesian product G × Z and the operation is
12 Ars Math. Contemp. 25 (2025) #P2.06
defined by (g1, n1)(g2, n2) = (ψ(n2)(g1)g2, n1 + n2). The groups G ⋊ψ Z and H are
isomorphic via the map (g, n) 7→ (ψ(−n)(g),−n). Note that H has a well-defined left
action on P given by (g1, n1) · (g2, n2) = (ψ(n2)(g1)g2, n1+n2) and (g1, n1) · (g′2, n2) =
((ψ(n2)(g1)g2)
′, n1 + n2). The action is clearly transitive and free. In order to show that
H → Aut(P ) is surjective it suffices to prove that any automorphism of P fixing (e, 0) is
the identity.
Suppose ϕ ∈ Aut(P ) fixes (e, 0). Since P (G,S) is connected by Remark 2.2, ϕ
restricts to an automorphism of P (G,S) × {0}. Concretely, the set D1 of points covering
(e, 0) is ϕ-invariant, the set D2 of points covered by those in D1 is then also is ϕ-invariant,
then the set D3 of points covering those in D2, and so on. Thus P (G,S) × {0} ⊆ P is
ϕ-invariant. Since it is also ϕ−1-invariant, ϕ(P (G,S)× {0}) = P (G,S)× {0}.
By hypothesis ϕ|P (G,S)×{0} is the identity. In particular (e, 1) = (e′, 0) is fixed. By an
inductive argument ϕ|P (G,S)×{n} is the identity for every n ≥ 0. Also, since (e′,−1) =
(e, 0) is fixed, then P (G,S) × {−1} is invariant, and ϕ|P (G,S)×{−1} is the identity. By
induction we deduce that ϕ is the identity of P .
Example 6.2. Let G be the fundamental group of the Klein bottle, that is the group pre-
sented by ⟨x, y| xyx−1y⟩. Then G admits a regular poset representation. Indeed, there is a
split short exact sequence
1 → Z → G→ Z → 1,
where the map Z → G takes a generator of Z to y, and G → Z maps a word w to its total
x-exponent. ThusG = Z⋊ψZ for some homomorphism ψ : Z → Aut(Z). Since Z admits
a Cayley representation P (Z, S) for S = {0, 1, 3} (see the comments after Example 3.1),
Theorem 6.1 applies.
In view of the results in previous sections, many groups with torsion (non-trivial ele-
ments of finite order) admit a regular poset representation. On the other hand if G ̸= 1,Z2
is a torsion group, then it admits no regular representation. If P is a non-discrete regular
representation of a torsion group G, there are comparable elements x < y, so there exists
g ∈ G with gx = y. This implies that gnx < gn+1x for every n ≥ 0, and the fact that g
has finite order yields a contradiction. If P is a discrete regular representation of G, then G
is trivial or of order 2. For example, the Burnside groups B(d, n) of exponent n and d ≥ 2
generators are finitely generated groups which do not admit a regular representation, and
they are infinite for n large enough [25, 33].
The posetsP constructed in Theorem 6.1 are regular representations of indicable groups
which are graded in the sense that there exists a (rank) function ρ : P → Z such that
1. x < y implies ρ(x) < ρ(y) and 2. whenever y covers x, we have ρ(y) = ρ(x) + 1.
Not every regular representation of a group is graded. Let P be the poset with underlying
set Z and the order ◁ given by a ◁ b if a = b or b − a ≥ 2. Then it is clear that P is
a regular representation of G = Z. Indeed, if an automorphism ϕ of P fixes 0, the set
{−1, 1} of points not comparable with 0 must be ϕ-invariant. Since −1 ◁ 1, ϕ fixes both
−1 and 1, and by an inductive argument ϕ = 1P . On the other hand P is not graded as
0 ◁ 2 ◁ 4 ◁ 6 and 0 ◁ 3 ◁ 6 are two maximal chains from 0 to 6 of different length.
Proposition 6.3. Let G ̸= 1,Z2 be a group which admits a graded regular poset represen-
tation. Then G is indicable.
Proof. Let P be a graded regular representation of G with rank function ρ : P → Z. Since
G ̸= 1,Z2, by Remark 2.2, P must be connected. Fix x ∈ P . By adding a constant to ρ
J. A. Barmak: Regular and semi-regular representations of groups by posets 13
we may assume ρ(x) = 0. Define f : G→ Z by f(g) = ρ(gx). We prove that f is a group
homomorphism. Let g, h ∈ G. Condition 1 in the definition of rank function guarantees
that there are no infinite chains between two fixed elements of P . By connectivity there is
a sequence x = x0, x1, . . . , xn = gx in which for every 0 ≤ i < n either xi+1 covers xi or
xi covers xi+1. Then, the number of i for which the first happens minus the number of i for
which the latter happens is ρ(gx). Similarly, there is a sequence x = y0, y1, . . . , ym = hx
in which each yi is covered by or covers yi+1, and the corresponding difference is ρ(hx).
Thus,
x0, x1, . . . , xn, gy1, gy2, . . . ghx
is a sequence of the same kind, and the corresponding difference is ρ(gx) + ρ(hx) =
ρ(ghx). Thus f(gh) = f(g) + f(h). The fact that f is an epimorphism follows directly
from the fact that P is non-discrete.
Theorem 6.4. Let G be a group different from Z22,Z
3
2,Z
4
2,Z
2
3 and let ψ : Z → Aut(G) be
a group homomorphism. ThenG⋊ψZ admits a semi-regular poset representation with two
orbits.
Proof. The proof is very similar to that of Theorem 6.1. If G is the trivial group, the exis-
tence of a non-Cayley and a Cayley representation was already discussed in the beginning
of Section 2 and right after Example 3.1. Assume G is non-trivial. The representation we
construct will not be Cayley. If G ̸= Q8, let B be a semi-regular representation of G with
three orbits, as constructed by Babai in [4, 5] and recalled after Theorem 2.3 above. The
underlying set of B is G ∪ G′ ∪ G′′. The set of minimal points of B is G, and the set of
maximal points is G′′. A point g′ ∈ G′ covers just one element g ∈ G and it is covered
only by g′′ ∈ G′′. If G ̸= Z2, B is necessarily connected, and if G = Z2, B can and will
be assumed to be connected. In particular, since G is non-trivial, every minimal point g is
covered by at least one maximal point h′′ (and also by g′).
Consider countable many copiesB×{n} (n ∈ Z) ofB. Identify (g′′, n) with (ψ(1)(g),
n+1) for every g ∈ G,n ∈ Z, to obtain a posetP . Note that the setB′ of points (g′, n) ∈ P
for g ∈ G,n ∈ Z, is invariant by any automorphism of P (and so is its complement), since
those are the points covered by just one element.
LetH be the group isomorphic toG⋊ψZ defined in the proof of Theorem 6.1. There is
a left action λ of H on P given by (g1, n1) · (g2, n2) = (ψ(n2)(g1)g2, n1 + n2), (g1, n1) ·
(g′2, n2) = ((ψ(n2)(g1)g2)
′, n1 + n2), (g1, n1) · (g′′2 , n2) = ((ψ(n2)(g1)g2)′′, n1 + n2).
This action is free with two orbits: B′ and its complement. Since these are invariant by
any automorphism, to show that λ is the required representation we may prove that every
automorphism of P fixing (e, 0) ∈ G× {0} is the identity.
Let ϕ ∈ Aut(P ) be such that ϕ(e, 0) = (e, 0). Suppose that D is a subset of B × {0}
which is invariant in this strong sense: D∩(G×{0}),D∩(G′×{0}) andD∩(G′′×{0}) are
ϕ-invariant. We claim that the setD of points inB×{0} which are greater than some point
in D also satisfies this property. Indeed, if (g′, 0) covers some element in D ∩ (G× {0}),
then so does ϕ(g′, 0). This implies that ϕ(g′, 0) ∈ B × {0}, and since B′ is invariant,
ϕ(g′, 0) ∈ G′ × {0}. Now, if (h′′, 0) covers an element in D ∩ (G′ × {0}), or if it covers
an element in D ∩ (G × {0}), or if it covers an element (g′, 0) which covers a point in
D∩ (G×{0}) (as before), then so does ϕ(h′′, 0). In particular ϕ(h′′, 0) ∈ B×{0}. Since
the complement of B′ is ϕ-invariant, ϕ(h′′, 0) ∈ G′′ × {0}. Similarly, the set D of points
in B × {0} which are smaller than some point in D is also invariant in the strong sense.
Let D0 = {(e, 0)}. For k ≥ 1 odd define Dk = Dk−1, while for k ≥ 2 even, define Dk =
14 Ars Math. Contemp. 25 (2025) #P2.06
Dk−1. Then Dk is ϕ-invariant for every k ≥ 0. Since B is connected, B × {0} = ∪Dk
is ϕ-invariant. Since it is also ϕ−1-invariant, ϕ restricts to an automorphism of B × {0}
fixing a point. Therefore this is the identity. By induction ϕ restricts to the identity in each
copy of B, as we wanted to prove.
The case G = Q8 is very similar to the previous. In this case, the semi-regular repre-
sentation B of G with three orbits constructed by Babai in [5] (proof of Corollary 4.3) also
has G ∪ G′ ∪ G′′ as underlying set, with the set of minimal points being G and maximal
points being G′′. Every point in G is covered by three points: two in G′ and one in G′′,
while each point inG′ is covered by two points. With this, essentially the same proof works
as the set B′ of points (g′, n) ∈ P is invariant by any automorphism.
Since for each exception G = Z22,Z
3
2,Z
4
2,Z
2
3 there are only finitely many homomor-
phisms ψ : Z → Aut(G), we deduce the following
Corollary 6.5. With finitely many exceptions, every indicable group admits a semi-regular
representation with two orbits.
In particular every torsion-free indicable group admits a semi-regular representation
with two orbits, and thus every locally indicable group (that is, each non-trivial finitely
generated subgroup is indicable) admits such a representation. This includes knot groups
[21], torsion free one-relator groups [10, 21], amenable left-orderable groups [31].
We finish with the postponed proof that Z22 does not admit a semi-regular representation
with three orbits.
Proof of Proposition 2.4. Assume P is a semi-regular poset representation of G = Z22 =
{0, a, b, a + b} with three orbits. We can identify those orbits with three copies G,G′, G′′
of G, and assume the action of G is the regular left action on each orbit. Since points in the
same orbit are not comparable, the height h(P ) of P is 1 or 2.
Case 2: h(P ) = 2. We may assume by a relabeling that 0 < 0′ < 0′′. If 0′ is
only comparable with 0 and 0′′, then P is the poset associated to a digraphical regular
representation of G (see the proof of Theorem 7.3 in [4]), which is absurd by Theorem 2.3.
Since the opposite P op of P is also a semi-regular representation, we may assume 0′ < a′′.
We call an edge (x, y) in the Hasse diagram H(P ) of P long if x ∈ G and y ∈ G′′. If
there are two long edges starting at 0, they must be (0, b′′) and (0, (a+b)′′). In this case we
claim that the poset P̃ obtained from P by removing from the relation all the pairs (x, y)
such that (x, y) is a long edge, has the same automorphism group as P . Indeed, x ∈ G and
y ∈ G′′ are not comparable in P̃ if and only if (x, y) is a long edge of H(P ). Thus we may
assume there is at most one long edge starting at 0.
Case 2.0: There is no long edge starting at 0. In other words, there are no long edges at
all in H(P ). If 0′ is only covered by 0′′ and a′′ or if it is covered by all the elements in G′′,
the transposition which permutes 0′′ and a′′ shows that the action of G on P is not semi-
regular. We may assume that 0′ is covered by 0′′, a′′, b′′ and not by (a + b)′′. In this case
Aut(P ) is isomorphic to the automorphism group of the subposet induced by G ∪ G′, so
G admits a semi-regular representation with two orbits, a contradiction by Proposition 2.6.
Case 2.1: There is one long edge starting in 0, which may be assumed to be (0, b′′).
Since b′′ covers 0, 0′ ≮ b′′ and 0 ≮ b′, so b = b · 0 ≮ b · b′ = 0′. Since 0′ < a′′,
(a + b)′ = (a + b) · 0′ < (a + b) · a′′ = b′′, and since b′′ covers 0, 0 ≮ (a + b)′. Then
J. A. Barmak: Regular and semi-regular representations of groups by posets 15
a+ b ≮ 0′. Thus 0′ covers 0 and possibly a, but nothing else. And 0′ is covered by 0′′ and
a′′, and possibly (a+ b)′′, but not b′′. Thus, there are only four cases to analyze. However,
we cannot have a < 0′ and 0′ < (a + b)′′ simultaneously as (a, (a + b)′′) is a long edge.
The three possible cases appear in Figure 2.
a
a'
a''
b'
b
b''
a+b
(a+b)'
(a+b)''
0
0'
0''
a
a'
a''
b'
b
b''
a+b
(a+b)'
(a+b)''
0
0'
0''
a
a'
a''
b'
b
b''
a+b
(a+b)'
(a+b)''
0
0'
0''
Figure 2: Three candidates for Case 2.1.
In the first and second cases we have the involution which permutes 0′′ with a′′, b′
with (a + b)′ and b with a + b. In the third case there is an involution permuting a′′ with
(a+ b)′′, a′ with (a+ b)′, and a with a+ b. These are non-identity automorphisms fixing
0, a contradiction.
Case 1: h(P ) = 1. Since P is connected by Remark 2.2, by a relabeling we may
assume that 0′ > 0 < 0′′. The dual 0′ < 0 > 0′′ can be ignored as P op is another
semi-regular representation.
Since h(P ) = 1, 0′ covers only points from the orbit G. If 0′ covers four points, the
transposition which permutes 0′ and a′ shows that the action of G on P is not free. If 0′
covers one point or three, Aut(P ) is isomorphic to the automorphism group of the subposet
induced by G ∪G′′. Thus, G would admit a semi-regular representation with two orbits, a
contradiction. We may assume that 0′ covers two elements, 0 and a. Then the transposition
which permutes 0′ and a′ is a non-trivial automorphism of P which fixes 0.
ORCID iDs
Jonathan A. Barmak https://orcid.org/0000-0001-8583-2073
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.07
https://doi.org/10.26493/1855-3974.3015.f4a
(Also available at http://amc-journal.eu)
Perfect matching cuts partitioning a graph into
complementary subgraphs*
Diane Castonguay , Erika M. M. Coelho , Hebert Coelho ,
Julliano R. Nascimento †
Instituto de Informática, Universidade Federal de Goiás, Goiânia, Brazil
Uéverton S. Souza
Instituto de Computação, Universidade Federal Fluminense, Niterói, Brazil and
Institute of Informatics, University of Warsaw, Warsaw, Poland
Received 28 November 2022, accepted 10 May 2024, published online 20 March 2025
Abstract
In PARTITION INTO COMPLEMENTARY SUBGRAPHS (COMP-SUB) we are given a
graph G = (V,E), and an edge set property Π, and asked whether G can be decomposed
into two graphs, H and its complement H , for some graph H , in such a way that the
edge cut [V (H), V (H)] satisfies the property Π. Motivated by previous work, we consider
COMP-SUB(Π) when the property Π = PM specifies that the edge cut of the decomposi-
tion is a perfect matching. We prove that COMP-SUB(PM) is GI-hard when the graph G is
C5-free or G is {Ck≥7, Ck≥7}-free. On the other hand, we show that COMP-SUB(PM) is
polynomial-time solvable on hole-free graphs and on P5-free graphs. Furthermore, we
present characterizations of COMP-SUB(PM) on chordal, distance-hereditary, and ex-
tended P4-laden graphs.
Keywords: Graph partitioning, complementary subgraphs, perfect matching, matching cut, graph
isomorphism.
Math. Subj. Class. (2020): 05C70, 05C85, 05C76, 05C51, 05C69
*This research has received funding from Rio de Janeiro Research Support Foundation (FAPERJ) under grant
agreement E-26/201.344/2021, National Council for Scientific and Technological Development (CNPq) under
grant agreement 309832/2020-9, and the European Research Council (ERC) under the European Union’s Horizon
2020 research and innovation programme under grant agreement CUTACOMBS (No. 714704). An extended
abstract, containing some of the results of this paper, has appeared in the Proceedings of the 33rd International
Workshop on Combinatorial Algorithms (IWOCA 2022).
†Corresponding author.
E-mail addresses: diane@inf.ufg.br (Diane Castonguay), erikamorais@inf.ufg.br (Erika M. M. Coelho),
hebert@inf.ufg.br (Hebert Coelho), jullianonascimento@inf.ufg.br (Julliano R. Nascimento), ueverton@ic.uff.br
(Uéverton S. Souza)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.07
1 Introduction
Finding graph partitions with some special properties has been a topic of extensive re-
search. Several combinatorial problems can be viewed as partition problems, such as VER-
TEX COLORING and CLIQUE COVER. In addition, many graph classes, e.g. bipartite and
split graphs, can also be defined through a partition of its vertex set. In particular, the class
of complementary prisms [11] are defined over complementary parts. The complementary
prism GG of a graph G arises from the disjoint union of a labeled graph G and its com-
plement G by adding the edges of a perfect matching between vertices with same label in
G and G. Studies concerning the computational complexity of classical graph problems
restricted to the class of complementary prisms graphs can be found in [4, 10].
We say that a graph G = (V,E) is decomposed into two graphs G1 and G2 if V (G)
can be partitioned into V1 and V2, where G[V1] ≃ G1 and G[V2] ≃ G2. The edge cut
[V1, V2] is called the edge cut of this decomposition.
As a generalization of complementary prisms, Nascimento, Souza and Szwarcfiter [17]
introduced the problem defined as follows.
PARTITION INTO COMPLEMENTARY SUBGRAPHS (COMP-SUB)
Instance: A graph G = (V,E), and an edge set property Π.
Question: Can G be decomposed into two graphs, H and its complement H , for some
graph H , in such a way that the edge cut M of the decomposition satisfies the property
Π?
For short, we abbreviate PARTITION INTO COMPLEMENTARY SUBGRAPHS with the
edge set property Π as COMP-SUB(Π). We write G ∈ COMP-SUB(Π) to denote that G is
a yes-instance of COMP-SUB(Π) and we call (H,H) as a complementary decomposition
of G.
The COMP-SUB(Π) problem also finds motivation in parameterized complexity. Rec-
ognizing whether a graph has a complementary decomposition can be useful for solving
problems in FPT-time, as pointed out in [17]. Nascimento, Souza and Szwarcfiter [17]
considered the cases where the edge cut M is empty or induces a complete bipartite graph.
They also presented some remarks when Π is a general edge set property. In particular,
when M is empty, they make some links between COMP-SUB(Π) and the GRAPH ISO-
MORPHISM problem, from which they show that COMP-SUB(Π) is GI-hard.
It is known that the recognition of complementary prisms can be done in polynomial
time [5]. This implies that, when the property Π is a perfect matching M between corre-
sponding vertices in H and H , the COMP-SUB(Π) problem is polynomial-time solvable.
So, a natural question is the study of COMP-SUB(Π) when Π specifies that M is any
perfect matching. In this context, two related problems arise: MATCHING CUT [13, 18]
and PERFECT MATCHING CUT [12]. A (perfect) matching cut is a partition of vertices
of a graph into two parts such that the set of edges crossing between the parts forms a
(perfect) matching. Considering Π = PM as the property of being a perfect matching,
COMP-SUB(PM) can be seen as a variant of PERFECT MATCHING CUT with the ad-
ditional restriction that the two parts must induce complementary subgraphs. Note that
studies regarding matchings satisfying particular constraints have received wide attention
in the literature (c.f. [9, 14, 15, 20, 21]).
Motivated by Nascimento, Souza and Szwarcfiter [17], in this paper we deal with
COMP-SUB(Π), when Π = PM considers M as a perfect matching. We show that
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 3
COMP-SUB(PM) is GI-hard when the graph G is C5-free or G is {Ck≥7, Ck≥7}-free. On
the other hand, we present polynomial time algorithms able to solve COMP-SUB(PM)
when the input graph G is hole-free or P5-free. In addition, we characterize graphs
G ∈ COMP-SUB(PM) when G is chordal, distance-hereditary, or extended P4-laden.
Although extended P4-laden graphs generalize cographs, we also show a simpler charac-
terization for cographs.
The paper is organized as follows. Section 2 contains some fundamental concepts and
an auxiliary result. Sections 3 and 4 contains our results on some cycle-free graphs and
graphs with few P4’s, respectively. Further discussions are presented in Section 5.
2 Preliminaries
We consider only finite, simple, and undirected graphs, and we use standard terminology
and notation. See [1] for graph-theoretic terms not defined here.
Let G be a graph. For a vertex v ∈ V (G), we denote its open neighborhood by NG(v),
and its closed neighborhood, denoted by NG[v] := NG(v)∪ {v}. For a set U ⊆ V (G), let
NG(U) =
⋃
v∈U NG(v) \ U , and NG[U ] = NG(U) ∪ U .
The degree of a vertex v ∈ V (G) on a set U ⊆ V (G), denoted by dU (v), is dU (v) =
|NG(u) ∩ U |. If U = V (G), we simply write dG(u). We say that v ∈ V (G) is an isolated
(resp. universal) vertex if dG(v) = 0 (resp. dG(v) = |V (G)| − 1). We define the distance,
distG(u, v), between two vertices u and v of a graph G as the length of the shortest path
between u and v.
The subgraph of G induced by U , denoted by G[U ], is the graph whose vertex set is U
and whose edge set consists of all the edges in E(G) that have both endvertices in U .
Let G be a graph. A set U ⊆ V (G) is called a clique (resp. independent set) if the
vertices in U are pairwise adjacent (resp. nonadjacent). We denote by Kn a complete
graph, In an independent set, Pn a path graph, and Cn a cycle graph on n vertices. Let r
be a positive integer. An r-partite graph is one whose vertex set can be partitioned into r
subsets, in such a way that no edge has both ends in the same subset. An r-partite graph
is complete if any two vertices in different subsets are adjacent. When r is not specified,
we simply say (complete) multipartite. A split graph G is one whose vertex set admits a
partition V (G) = C ∪ I into a clique C and an independent set I . The complement G of a
graph G is the graph defined by V (G) = V (G) and uv ∈ E(G) if and only if uv /∈ E(G).
Let P = v1v2 . . . vn be a path. We call v2, . . . , vn−1 as inner vertices of P . Two or
more paths in a graph are independent if none of them contains an inner vertex of another.
A graph G is ℓ-connected if any two of its vertices can be joined by ℓ independent paths.
A 2-connected graph is called biconnected.
A vertex v in a graph G is a cutvertex or cutpoint, if G\{v} is disconnected. A maximal
connected subgraph without a cutpoint is a block. The block-cutpoint tree of a graph G is a
bipartite graph whose vertex set consists of the set of cutpoints of G and the set of blocks
of G. A cutpoint is adjacent to a block whenever the cutpoint belongs to the block in G.
Two graphs G = (V,E) and G′ = (V ′, E′) are isomorphic, denoted as G ≃ H , if and
only if there is a bijection, called isomorphism function, φ : V → V ′ such that uv ∈ E
if and only if φ(u)φ(v) ∈ E′, for every u, v ∈ V . A graph G is self-complementary if
G ≃ G. The GRAPH ISOMORPHISM problem receives as input two graphs G and G′ and
asks whether G ≃ G′. We denote by GI the class of problems that admit a polynomial-time
reduction to GRAPH ISOMORPHISM.
4 Ars Math. Contemp. 25 (2025) #P2.07
A problem Q is GI-complete if the two conditions are satisfied: (i) Q is a member of
GI; and (ii) Q is GI-hard, that is, for every problem Q′ ∈ GI, Q′ is polynomially reducible
to Q.
We denote the set of positive integers {1, . . . , k} by [k]. Let G and G1, . . . , Gk be
graphs. We say that G is {G1, . . . , Gk}-free if G does not contain Gi as an induced sub-
graph, for every i ∈ [k].
A module of a graph is a set X of vertices such that for each vertex x /∈ X , either every
member of X is adjacent to x, or no member of X is adjacent to x [16].
Let G and H be two graphs such that V (G) ∩ V (H) = ∅. The disjoint union of G and
H , denoted by G ∪H , is the graph with V (G ∪H) = V (G) ∪ V (H) and E(G ∪H) =
E(G) ∪ E(H). The join of G and H , denoted by G+H , is the graph with V (G+H) =
V (G) ∪ V (H) and E(G+H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G) and v ∈ V (H)}.
Let G be a graph and C a class of graphs. A set S ⊆ V (G) is a C-modulator if G \ S
belongs to C. We define the distance of G to class C as the size of a minimum S which is
a C-modulator.
Let G be a graph that has a complementary decomposition (G1, G2) with perfect
matching cut M = {u1v1, . . . , unvn}, where ui ∈ V (G1) and vi ∈ V (G2), i ∈ [n].
We say that ui (resp. vi) is the corresponding vertex of vi (resp. ui), for every i ∈ [n].
For X ⊆ V (G1), we call X
G2 = {vi ∈ V (G2) : ui ∈ X} as the corresponding set of
X over G2. Similarly, for X ⊆ V (G2), we call X
G1 = {ui ∈ V (G1) : vi ∈ X} as the
corresponding set of X over G1.
Next, we present an auxiliary result, defined for COMP-SUB(PM) with a restriction
on the graphs of the decomposition. A cograph is a P4-free graph.
Lemma 2.1. Let G be a graph. The problem of determining whether G can be decomposed
into two graphs, G1, and its complement G2, such that G1 is a cograph and the edge cut
of the decomposition is a perfect matching, can be solved in polynomial time.
Proof. Let Cbe the class of cographs and G a 2n-vertex graph. Suppose that G is decom-
posable into complementary subgraphs G1 and G2, such that G1 ∈ C and the edge cut M
of the decomposition is a perfect matching.
Since C is closed under complement, we have that G2 ∈ C. Given that GRAPH ISO-
MORPHISM is linear-time solvable on cographs [7], we perform a brute force algorithm to
check every relevant partition V (G1), V (G2) of V (G). For that, we propose Algorithm 1,
explained in sequel.
Algorithm 1: PARTITION-INTO-COMPLEMENTARY-COGRAPHS(G)
Input: A graph G.
Output: Whether G admits a complementary decomposition such that the edge cut of
the decomposition is a perfect matching.
1 forall x1, x2, y1, y2 ∈ V (G) do
2 V (G1) := NG[{x1, x2}] \ {y1, y2}
3 V (G2) := V (G) \ V (G1)
4 M := {xy ∈ E(G) : x ∈ V (G1), y ∈ V (G2)}
5 if M is a perfect matching and G1 is a cograph and G2 is a cograph and G1 ≃ G2
then
6 return yes
7 return no
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 5
We know that a cograph is connected if and only if its complement is disconnected [8].
Consequently, if a complementary decomposition (G1, G2) exists, then either G1 or G2
is disconnected, say G2. Then G1 can be obtained by a join between the corresponding
connected components of G2. Thus, there exist two adjacent vertices x1, x2 ∈ V (G1),
such that NG1 [{x1, x2}] = V (G1). Furthermore, the edge set M of the decomposition
implies that there exist y1, y2 ∈ V (G2) such that x1y1, x2y2 ∈ M .
By the above arguments, it is possible to find V (G1) by means of NG[{x1, x2}] except
for two vertices y1, y2 ∈ NG[{x1, x2}] that must belong to V (G2). This way, V (G2) is
obtained by {y1, y2} ∪ {v ∈ V (G) : v /∈ NG[{x1, x2}]}. Once found V (G1), V (G2), and
M , we test whether M is a perfect matching and whether G1 and G2 are cographs. If so,
we compute G2 and then we check isomorphism between G1 and G2.
The correctness of the algorithm follows from the fact that all the possible relevant
partitions (for the emergence of the cographs, if any) are considered.
Now, we show that Algorithm 1 runs in polynomial time.
For enumerating every 4-tuple of vertices x1, x2, y1, y2 ∈ V (G) it is required O(n
4)
time. After, in O(n + m) time we can check whether M is a perfect matching, as well
as checking whether G1 and G2 are cographs. Finally, for computing G2 and checking
isomorphism between G1 and G2 is also required O(n+m) time. Therefore, the running
time of Algorithm 1 takes O(n5 + n4m) time.
3 Results on some Ck-free graphs
We begin by showing two hardness results, in Theorems 3.1 and 3.2.
Theorem 3.1. COMP-SUB(PM) is GI-hard on {Ck≥7, Ck≥7}-free graphs.
Proof. Given that GRAPH ISOMORPHISM is GI-hard on split graphs [6], we show a poly-
nomial-time reduction from such a problem to COMP-SUB(PM).
Note that a split graph is connected if and only if it does not contain isolated vertices.
Therefore, we may assume that the instances of GRAPH ISOMORPHISM on split graphs are
pairs of connected split graphs.
Let A and B be connected split graphs such that |V (A)| = |V (B)| = n, for some
n ≥ 3. From an instance (A,B) of GRAPH ISOMORPHISM, we construct an instance G of
COMP-SUB(PM).
Let G arise from the disjoint union between A, B, Kn, and In. Denote Kn by K and In
by I . We make every vertex in V (A) adjacent to every vertex in V (K). Furthermore, we
add an arbitrary perfect matching between V (A) and V (I) and between V (K) and V (B).
An example of graph G follows in Figure 1. Aditionally, let H1 = G[V (A) ∪ V (K)] and
H2 = G[V (B) ∪ V (I)]. Clearly, the construction can be done in polynomial time.
We first show that G is {Ck≥7, Ck≥7}-free.
Claim 1. Let G be the graph obtained from the construction. It holds that G is a
{Ck≥7, Ck≥7}-free graph.
Proof of Claim 1. We prove that (I) G is Ck≥7-free, and (II) G is Ck≥7-free.
(I) Suppose by contradiction that G contains a Ck≥7, denoted as C, as induced sub-
graph. We may assume that k is minimum.
By construction, H1 and H2 are split graphs and it is clear that H1 and H2 are Cℓ+4-
free, for every ℓ ≥ 0. Then V (C) ̸⊆ V (H1) and V (C) ̸⊆ V (H2). So, we assume that
6 Ars Math. Contemp. 25 (2025) #P2.07
V (C) ∩ V (H1) ̸= ∅ and V (C) ∩ V (H2) ̸= ∅. Since I is a set of vertices with degree one
in G, we have that V (C) ∩ I = ∅. So, we may suppose that V (C) ∩ V (B) ̸= ∅ and since
C is a cycle, |V (C) ∩ V (B)| ≥ 2. Since B is split, we have that |V (C) ∩ V (B)| ≤ 4.
Since C is a cycle and K is a complete graph, C must contain exactly two vertices from
K and no vertex of A. Then, |V (C)| ≥ 7 implies that |V (C)∩V (B)| ≥ 5, a contradiction.
(II) Suppose by contradiction that G contains a Ck≥7, denoted as D, as induced sub-
graph. Let V (D) = {d1, . . . , dℓ}, for some ℓ ≥ 7, and E(D) = {didj : 1 ≤ i < j ≤
ℓ} \ ({didi+1 : 1 ≤ i ≤ ℓ− 1} ∪ {dℓd1}).
By definition of D, {d1, d2, d4, d5} induces a C4. Then, since H1 and H2 are split
graphs, V (D) ̸⊆ V (H1) and V (D) ̸⊆ V (H2). So, we assume that V (D) ∩ V (H1) ̸= ∅
and V (D) ∩ V (H2) ̸= ∅. Then, there exists i, j ∈ [ℓ] such that di ∈ V (H1), dj ∈ V (H2)
and didj ∈ E(D).
Without loss of generality, suppose that i = 1. Since {d1, d3, d5} induces a K3, we may
assume that {d1, d3, d5} ⊆ V (H1). Thus, d1dj ∈ E(D), for some j ∈ {4, 6, . . . , ℓ − 1}.
Notice that, if j = 4 (resp. j ≥ 6), then {d1, d4, d6} (resp. {d1, d3, dj}) induces a K3
which intersects both V (H1) and V (H2), a contradiction. Therefore G is Ck≥7-free.
In what follows, we prove that (A,B) is a yes-instance of GRAPH ISOMORPHISM if
and only if G is a yes-instance of COMP-SUB(PM).
Suppose that A ≃ B. Since In = Kn, B ≃ A, and there is no edge between a vertex
in I and a vertex in V (B), it is easy to see that H1 and H2 are isomorphic. Therefore, G is
a yes-instance of COMP-SUB(PM).
Let (V ′, V ′′) be a partition of V (G) into complementary parts such that [V ′, V ′′] is a
perfect matching. Since I is a set of vertices with degree one in G and A is connected, it
holds that either (I ⊂ V ′ and V (A) ⊂ V ′′) or (V (A) ⊂ V ′ and I ⊂ V ′′). Suppose that
V (A) ⊂ V ′. This implies that V ′ = V (A) ∪ K and V ′′ = V (B) ∪ I . Since G[V ′] and
G[V ′′] are complementary, we have that G[V ′] ≃ G[V ′′]. Hence, due to the automorphism
of universal vertices, it holds that A ≃ B.
See in Figure 1 an example of the construction presented in Theorem 3.1.
A
BKn
In
H1 H2G
Figure 1: Graph G constructed for Theorem 3.1.
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 7
Our next result shows that COMP-SUB(PM) is also GI-hard on C5-free graphs. No-
tice that the construction of G presented in Theorem 3.1 allows the existence of C5 as an
induced subgraph. For instance, if B contains an induced P3 = v1v2v3, the corresponding
vertices of v1 and v3 (in K) union with v1, v2, v3 induce a C5 in G. Hence, we adapt that
construction to ensure a C5-free graph.
Theorem 3.2. COMP-SUB(PM) is GI-hard on C5-free graphs.
Proof. As in the proof of Theorem 3.1, we reduce GRAPH ISOMORPHISM on split graphs
to COMP-SUB(PM).
Let A and B be split graphs. We may assume that A and B does not contain isolated
vertices, A and B are not complete graphs, and |V (A)| = |V (B)| = n, for some n ≥
3. From an instance (A,B) of GRAPH ISOMORPHISM, we construct an instance G of
COMP-SUB(PM) as follows.
• First, let H be a graph arising from the disjoint union of the graphs X1 = A, X2 =
B, Y1 = B, and Y2 = A.
• Add to E(H) all the edges between a vertex in T and a vertex in R, for {T,R} ∈
{{X1, Y1}, {Y1, Y2}, {Y2, X2}}. Figure 2 contains an example of graph H .
• Create a graph H ′ by a copy of H . To distinguish vertices from H and H ′ we let
v′ ∈ V (H ′) be the corresponding vertex of v ∈ V (H).
• Let the graph G arise from the disjoint union of H and H ′ by the addition of the
edges vv′ ∈ E(G), for every v ∈ V (H).
Clearly, the construction can be done in polynomial time. Next, we show that G is
indeed a C5-free graph.
Claim 1. Let G be the graph obtained from the construction. It holds that G is a C5-free
graph.
Proof of Claim 1. By contradiction, suppose that G has a C5, denoted as C, as an induced
subgraph. If C has edges crossing from H to H ′ then they should be exactly two edges,
because C is 2-connected and has size five. However, for every u, v ∈ V (H), we have
uv ∈ E(H) if and only if u′v′ ∈ E(H ′), which implies that C cannot have edges crossing
from H to H ′ due to the order of C. Thus, either V (C) ⊆ V (H) or V (C) ⊆ V (H ′).
Since H ≃ H ′, we assume that V (C) ⊆ V (H).
Recall that A and B are split graphs, which implies that Xi, Yi, for every i ∈ [2] are also
split graphs. Hence, Xi and Yi, for every i ∈ [2], are C5-free graphs. Consequently, there
exists T,R such that {T,R} ∈ {{X1, Y1}, {Y1, Y2}, {Y2, X2}}, with the property that
V (C) ∩ V (T ) ̸= ∅ and V (C) ∩ V (R) ̸= ∅. Thus, for such a pair (R, T ), the join R + T
implies that |V (C) ∩ V (R)|, |V (C) ∩ V (T )| ≤ 2 and |V (C) ∩ (V (R) ∪ V (T ))| = 3.
Since |V (C)∩(V (R)∪V (T ))| = 3 and |V (C)| = 5, there exists Z∈{X1, X2, Y1, Y2}\
{R, T} such that either the join Z + R or Z + T contains a subgraph of C, for which
V (C)∩V (Z) ̸= ∅, say Z+R. Since dC(v) = 2, for every v ∈ V (C), we have that |V (C)∩
V (R)| = 2, hence, |V (C) ∩ V (Z)| = 1 and V (C) ∩ ({V (A), V (A), V (B), V (B)} \
{R, T, Z}) = ∅. This implies that |V (C) ∩ V (H)| ≤ 4, a contradiction.
8 Ars Math. Contemp. 25 (2025) #P2.07
Now, assume that A ≃ B. We show that HH ′ is a partition of G such that H ′ ≃ H
and [V (H), V (H ′)] is a perfect matching cut of G, implying that G is a yes-instance of
COMP-SUB(PM). The construction of G implies that [V (H), V (H ′)] is a perfect match-
ing cut of G. Thus, it remains to show that H ′ ≃ H . To this end, recall that P4 is
self-complementary, and the graph obtained by contracting into a single vertex each of Xi,
Yi, i ∈ [2], is a P4.
Note that H has the shape
X1–Y1–Y2–X2
and its complement H must have shape
Y 1–X2–X1–Y 2,
where R–T means a join between the vertices of R and T , and R, T are induced subgraphs.
Also, if A ≃ B then X1 ≃ X2 and Y1 ≃ Y2, consequently X1 ≃ X2 ≃ Y1 ≃ Y2 and
Y 1 ≃ Y 2 ≃ X1 ≃ X2. Therefore,
H ′ ≃ X1–Y1–Y2–X2 ≃ Y 1–X2–X1–Y 2 ≃ H.
For the converse, we suppose that G is a yes-instance of COMP-SUB(PM).
Let (V ∗, V ∗∗) be a partition of V (G) into complementary parts such that [V ∗, V ∗∗] is a
perfect matching. By definition of perfect matching cut, it is clear that [V ∗, V ∗∗] ⊆ {wz ∈
E(G) : NG(w) ∩ NG(z) = ∅}. Also, due to the joins of the construction, it holds that
{wz ∈ E(G) : NG(w) ∩ NG(z) = ∅} = [V (H), V (H
′)]. Since |V ∗| = |V ∗∗| it follows
that (V ∗, V ∗∗) = (V (H), V (H ′)) and H ≃ H .
Let g : V (H) → V (H) be an isomorphism function between H and H . By per-
forming a modular decomposition we find the maximal modules X1, X2, Y1, Y2 in H , and
X1, X2, Y 1, Y 2 in H .
By construction, for every x ∈ X1 ∪ X2, 1 ≤ dH(x) ≤ 2n − 1, and, for every
y ∈ Y1 ∪ Y2, 2n ≤ dH(y) ≤ 3n − 2. Those degrees imply that g maps V (X1) ∪ V (X2)
to V (Y 1) ∪ V (Y 2), and V (Y1) ∪ V (Y2) to V (X1) ∪ V (X2). By the uniqueness of the
decomposition into maximal modules [16], we obtain that either V (X1) 7→ V (Y 1) or
V (X1) 7→ V (Y 2).
If V (X1) 7→ V (Y 1), then
V (Y1) 7→ V (X2),
V (Y2) 7→ V (X1), and
V (X2) 7→ V (Y 2).
Since H ≃ H , the mapping V (X1) 7→ V (Y 1) implies that A ≃ X1 ≃ Y 1 ≃ B ≃ B.
Otherwise, V (X1) 7→ V (Y 2) implies that
V (Y1) 7→ V (X1),
V (Y2) 7→ V (X2), and
V (X2) 7→ V (Y 1).
Then, the mapping V (Y2) 7→ V (X2) implies that A ≃ Y2 ≃ X2 ≃ B. Therefore,
A ≃ B.
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 9
X1
Y2
H
X2
Y1
Figure 2: Graph H constructed for Theorem 3.2.
See in Figure 2 an example of graph H constructed for Theorem 3.2, where A = P3
and B = K2 ∪K1.
Despite the hardness results presented in Theorems 3.1, and 3.2, next we show that
COMP-SUB(PM) can be solved in polynomial time on hole-free graphs. Recall that a
hole is a cycle on 5 or more vertices.
Theorem 3.3. COMP-SUB(PM) is polynomial-time solvable on hole-free graphs.
Proof. Let G be a hole-free graph having 2n-vertices. We assume that n is at least 5;
otherwise, the problem can be solved in O(1) time.
Suppose that G ∈ COMP-SUB(PM), then G is decomposable into complementary
subgraphs G1 and G2, such that the edge cut M of the decomposition is a perfect matching.
Recall that G1 or G2 is a connected graph. Thus, we assume that G1 is connected.
We go through the proof by analysing the structure of the graphs of the decomposition
by means of their connectivity (Claims 1 and 2), and we conclude by showing how to find
that decomposition when it exists.
Claim 1. Let G1 be a connected graph with at least five vertices and F ⊆ V (G1). If G[F ]
is biconnected, then G[FG2 ] is a cluster graph.
Proof of Claim 1. Suppose, by contradiction, that G[FG2 ] is not a cluster graph and let
v1v2v3 be a P3 in G[F
G2 ]. Since G[F ] is 2-connected, there exist two independent paths
between any two vertices in F . Consider u1, u2, u3 ∈ F as the corresponding vertices
of v1, v2, v3, respectively. Let P and P
′ be two independent paths between u1 and u3
in F . Since P and P ′ are independent, u2 does not belong to P or P
′, say P . Then
P ∪ {v1, v2, v3} induces a hole in G, a contradiction.
Next, we see more on the structure of G1 and G2.
Claim 2. Let G1 be a connected graph having at least five vertices. If G1 is non-biconnected,
then either there is S ⊂ V (G1) with |S| ≤ 2 such that G1 \ S is biconnected; or there is
S′ ⊂ V (G2) with |S
′| ≤ 2 such that G2 \ S
′ is biconnected.
10 Ars Math. Contemp. 25 (2025) #P2.07
Proof of Claim 2. Suppose that G1 is non-biconnected and let T be a block-cut-point tree
of G1. Let B = {B1, . . . , Bs} and C = {c1, . . . , ct} be the sets of blocks and cutpoints in
G1, respectively. The proof is divided in two cases: (I) |B| ≥ 2, |C | = 1; and (II) |B| ≥ 2,
|C | ≥ 2.
Recall that if |B| = 1, then |C | = 0 and G1 is biconnected.
(I) Suppose that |B| ≥ 2 and |C | = 1. Let C = {c}. We have that G1 \ {c} is
the disjoint union (B1 \ {c}) ∪ · · · ∪ (Bs \ {c}). This implies that G1 \ {c} is the join
(B1 \ {c}) + · · ·+ (Bs \ {c}).
• If s ≥ 3 then G2 \ {c} = (B1 \ {c}) + · · ·+ (Bs \ {c}) is biconnected.
• If s = 2, |B1 \{c}| ≥ 2, and |B2 \{c}| ≥ 2 then G2 \{c} = (B1 \{c})+(B2 \{c})
is also biconnected.
• If s = 2 and |B1 \ {c}| = 1 then |B2 \ {c}| ≥ 2. Otherwise, G1 (and G2) has only
three vertices. Thus, B2 is a block of G1 with size |V (G1)| − 1, and S = B1 \ {c}
is as required.
(II) Now, consider that |B| ≥ 2 and |C | ≥ 2. Let B,B′ ∈ B be two distinct leaves in
T and c, c′ ∈ C be two distinct cutpoints such that Bc,B′c′ ∈ E(T ).
Let D = V (G1) \ (B ∪B
′). Since B (resp. B′) is a leaf in T , we have that V (B) \ {c}
(resp. V (B′) \ {c′}) is not adjacent to B′ ∪D (resp. B ∪D). This implies that G1 \ {c, c
′}
is the join (B \ {c}) + (B
′
\ {c′}) +D.
• If D ̸= ∅, we have that (B \{c})+(B
′
\{c′})+D is biconnected. Thus, G2 \{c, c
′}
is biconnected as required.
• If D = ∅, |B\{c}| ≥ 2, and |B′\{c′}| ≥ 2, then G2\{c, c
′} = (B\{c})+(B
′
\{c′})
is also biconnected.
• If D = ∅ and |B \ {c}| = 1, then |B′ \ {c′}| ≥ 2. Otherwise, G1 (and G2) has only
four vertices. Thus, G1 \B is biconnected (notice that |B| = 2).
This completes the proof of Claim 2.
By Claim 1, if G1 is biconnected, then G2 is a cluster graph. Since G1 ≃ G2, we
have that G1 is a complete multipartite graph. Hence, G1 and G2 are cographs and, by
Lemma 2.1, we can find the complementary partition of G in polynomial time.
Now, if G1 is non-biconnected, recall that by Claim 2, either there is S ⊂ V (G1) with
|S| ≤ 2 such that G1 \ S is biconnected; or there is S
′ ⊂ V (G2) with |S
′| ≤ 2 such that
G2 \ S
′ is biconnected.
Thus, there is a fixed number of vertices (at most 2) such that removing from G1 or G2
leaves a biconnected graph. We deal with the case that there exist c, c′ ∈ V (G2) such that
G2 \ {c, c
′} is biconnected. The approach for the other case is similar.
If there exist c, c′ ∈ V (G2) such that G2 \ {c, c
′} is 2-connected, by Claim 1 (dual),
we have that the graph induced by (V (G2) \ {c, c
′})G1 is a cluster graph. Then G1 and G2
have distance to cluster equal to 2. We proceed by Algorithm 2.
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 11
Algorithm 2: PARTITION-INTO-COMPLEMENTARY-SUBGRAPHS(G)
Input: A graph G.
Output: Whether G is partitionable into two complementary graphs G1 and G2 such
that G1 and G2 have distance to cluster equal to 2 and the edge cut of the
decomposition is a perfect matching.
1 forall x1, . . . , x4, y1, . . . , y4 ∈ V (G) do
2 V (G2) := NG[{y1, . . . , y4}] \ {x1, . . . , x4}
3 V (G1) := V (G) \ V (G2)
4 M := {xy ∈ E(G) : x ∈ V (G1), y ∈ V (G2)}
5 if M is a perfect matching then
6 forall cluster-modulator S1 of G1, such that |S1| ≤ 2 do
7 forall cluster-modulator S2 of G2, such that |S2| = |S1| do
8 forall mapping f : S1 7→ S2 do
9 if f can be extended to an isomorphism from G1 to G2 then
10 return yes
11 return no
Since G2 has distance to complete multipartite equal to 2, there exist four vertices
y1, . . . , y4 ∈ V (G2) such that NG2 [{y1, . . . , y4}] = V (G2). Then, if a complementary
decomposition (G1, G2) exists, we have that |NG[{y1, . . . , y4}]| = n + 4. Thence, it is
possible to find V (G2) which is NG[{y1, . . . , y4}] except for four vertices x1, . . . , x4 ∈
NG[{y1, . . . , y4}]. We put x1, . . . , x4 in V (G1) as well as the remaining vertices {v ∈
V (G) : v /∈ NG[{y1, . . . , y4}]}. Given V (G1), V (G2), and M , we check whether M is
a perfect matching. If so, we compute G2 and we proceed to the step of finding cluster-
modulators S1 for G1 and S2 for G2, that are done by Lines 6–7. In a naive manner, all
the possible pair of modulators can be found in O(n4), but we show how to find them in a
more efficient way.
We first find a P3 = w1w2w3 in G1. We know that at least one vertex in {w1, w2, w3}
must be included in a cluster-modulator for G1. Then, for every w ∈ {w1, w2, w3} we put
w ∈ S1 and we branch by searching (if any) for a P3 = w
′
1w
′
2w
′
3 in G1 \ {w}. Again,
given that at least one vertex in {w′1, w
′
2, w
′
3} must be included in a cluster-modulator for
G1, for every w
′ ∈ {w1, w2, w3} we put w
′ ∈ S1. If G1 \S1 is a cluster graph, we proceed
to finding, in the same manner, a cluster-modulator S2 for G2. Note that this is basically a
bounded search tree algorithm for finding cluster vertex deletion sets.
Given a pair of modulators S1 and S2 such that |S1| = |S2|, and a mapping from S1
to S2, the final task is checking if such a mapping can be extended to an isomorphism
between G1 and G2. Note that, by the bounded search tree technique, the number of pairs
of modulators and mappings that must be considered is bounded by a constant.
Recall that G1 (resp. G2) is a disjoint union of complete graphs H1 ∪ · · · ∪Hp (resp.
H ′1 ∪ · · · ∪ H
′
p), for some p ≥ 2, with the addition of two vertices w,w
′ (resp. z, z′)
arbitrarily adjacent to H1 ∪ · · · ∪ Hp (resp. H
′
1 ∪ · · · ∪ H
′
p). With this structure, an
isomorphism from G1 to G2 can be determined as follows.
12 Ars Math. Contemp. 25 (2025) #P2.07
For a mapping w 7→ z, w′ 7→ z′, we can map Hi to H
′
j , i, j ∈ [p], if and only if
|V (Hi)| = |V (H
′
j)| and
|NHi(w) \NHi(w
′)| = |NH′
j
(z) \NH′
j
(z′)| and
|NHi(w
′) \NHi(w)| = |NH′j (z
′) \NH′
j
(z)| and
|NHi(w) ∩NHj (w
′)| = |NH′
i
(z) ∩NH′
j
(z′)|.
Therefore, each mapping w 7→ z, w′ 7→ z′ defines “types” of cliques, from which the
mapping can be extended to an isomorphism from G1 to G2 if and only if G1 and G2 have
the same number of cliques per type.
Next, we analyse the running time of Algorithm 2.
First, in Line 1, we check every 8-tuple of vertices in V (G) to separate those
x1, . . . , x4 ∈ V (G1) and y1, . . . , y4 ∈ V (G2), which requires O(n
8) time. Lines 2–4
define V (G1), V (G2), and M , which run in O(n + m) time. Checking whether M is a
perfect matching (Line 5) can be done in O(n+m) time.
Recall that a P3 in G can be found in O(n + m) time. By the method previously de-
scribed, Line 6 can be done by finding a P3 = w1w2w3 in G1; for every w ∈ {w1, w2, w3}
finding a P3 = w
′
1w
′
2w
′
3 in G1 \ {w} in G1; and finally, for every w
′ ∈ {w1, w2, w3},
checking whether G1 \ S1 = {w,w
′} is a cluster graph. This produces a ternary search
tree with height equal to 2. Hence with 9 leaf nodes, that are at most 9 possible cluster-
modulators {w,w′} for G1. This requires a running time of O(m + n). For every of
those possible cluster-modulators for G1 we proceed to finding every cluster-modulator for
G2 (Line 7) by the same method. This gives an amount of at most 81 possible 4-tuples
w,w′, z, z′ that must be checked, hence Lines 6–8 run in O(n+m) time.
Finally, for Line 9, checking whether an isomorphism from G1 to G2 can be extended
from f can be done by checking sizes of cliques and neighborhoods, which can be done in
O(n+m) time.
Therefore, the overall running time of Algorithm 2 is of order O(n8(n+m)) = O(n9+
n8m).
Next, it follows a characterization for COMP-SUB(PM) in the class of distance hered-
itary graphs, which is a subclass of hole-free graphs. A distance-hereditary graph is a
{domino, house, gem, hole}-free graph. A domino, a house, and a gem are depicted in
Figure 3. For the next result, let ϱ be the graph in Figure 3.
house domino gem
ϱ
Figure 3: Some small subgraphs.
Proposition 3.4. Let G be a distance-hereditary graph of order 2n. It holds that G ∈
COMP-SUB(PM) if and only if G ∈ {KnKn,ϱ}.
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 13
Proof. Let G be a distance-hereditary graph. Clearly, KnKn ∈ COMP-SUB(PM). Let
ϱ be the graph with vertex set V (ϱ) = {u1, u2, u3, v1, v2, v3} denoted as in Figure 4.
Let V1 = {u1, u2, u3}, and V2 = {v1, v2, v3}. Clearly G[V1] ≃ G[V2], then ϱ ∈
COMP-SUB(PM).
v1
v2
v3
u1
u2
u3
Figure 4: Graph ϱ.
Next, suppose that G ∈ COMP-SUB(PM). By definition, there exist a complementary
decomposition (G1, G2) of G such that the edge cut M of the decomposition is a perfect
matching. Let M = {u1v1, . . . , unvn} where ui ∈ V (G1) and vi ∈ V (G2), for every
i ∈ [n]. We may assume that G1 is connected. We begin by showing that G2 is a cluster
graph with no induced K3.
Claim 1. If G1 is connected, then G2 is a {P3,K3}-free graph.
Proof of Claim 1. Suppose, by contradiction, that G2 contains (I) an induced P3 or
(II) an induced K3.
(I) Let i, j, k ∈ [n] be pairwise distinct, such that vivj , vjvk ∈ E(G2) and vivk /∈
E(G2). Since G1 is connected, there exists a (up, uq)-path in G1, for every p, q ∈ {i, j, k}.
Since G is hole-free, it follows that distG1(ui, uj) ≤ 1. Since G1 is connected, ui and uj
lie in the same connected component of G1, then uiuj ∈ E(G1). With a similar argument,
we obtain that ujuk ∈ E(G1). Again, since G is hole-free, uiuk /∈ E(G1). However, the
set {ui, uj , uk, vi, vj , vk} induces a domino in G, a contradiction.
(II) Let i, j, k ∈ [n] be pairwise distinct, such that {vi, vj , vk} induces a K3 in G2. In
this case, we have that each of {ui, uj , uk} must lie in distinct connected components of
G1, otherwise {ui, uj , uk, vi, vj , vk} induces a house in G. Thus, we have a contradiction
to the connectedness of G1.
By Claim 1, G2 is a {P3,K3}-free graph, that is, G2 ≃ pK1 ∪ qK2, for some p, q ≥ 0.
Since G1 ≃ G2, we have that G1 = A+ B, where A is a complete graph Kp = pK1 and
B is a complete multipartite graph qK2. For the rest of the proof, we consider the possible
cases for q. We deal with q ≥ 2 in Claim 2. In the sequence, we address the cases q = 1
and q = 0.
Claim 2. If q ≥ 2, then for every edge e ∈ E(G1), e belongs to a K3 or a C4 in G1.
Proof of Claim 2. Suppose that q ≥ 2. Recall that G1 = A + B where A = Kp and
B = qK2. Let B
′ be a C4 subgraph of B with V (B
′) = {u1, . . . , u4} and E(B
′) =
{u1u2, u2u3, u3u4, u4u1}. Let e = xy ∈ E(G1). If x, y ∈ V (B
′), the conclusion that e
belongs to a C4 is immediate.
14 Ars Math. Contemp. 25 (2025) #P2.07
If x, y ∈ V (G1) \ V (B
′), then e = xy ∈ E(A). The join A + B implies that ui is
a common neighbor of x and y, for every i ∈ [4]. Then e belongs to a K3 induced by
{x, y, ui} in G1.
Otherwise, we may assume that x ∈ V (G1)\V (B
′) and y ∈ V (B′). Let w ∈ NB′(y).
Since x is adjacent to every vertex in B′, we have that e = xy belongs to a K3 induced by
{x, y, w} in G1.
Let q ≥ 2. By Claim 2, every edge e ∈ E(G1) belongs to a K3 or a C4 in G1. Recall
that |E(G2)| = q ≥ 2, then let i, j ∈ [n] such that vivj ∈ E(G2). Given that G1 is
connected and G is hole-free, we conclude that uiuj ∈ E(G1). By Claim 2, we have
that uiuj belongs to a K3 or a C4 in G1. In the former, G contains an induced house, a
contradiction. In the latter, G contains an induced domino, also a contradiction.
Next, let q = 1. Since G1 is connected, we have that p ≥ 1. If p = 1, then G1 ≃ P3
and G = ϱ. If p ≥ 2, then G1 is isomorphic to a Kp+2 minus one edge. It is easy to see
that every e ∈ E(G1) belongs to a K3 in G1. Since |E(G2)| = q = 1, let i, j ∈ [n] such
that vivj ∈ E(G2). Again, given that G1 is connected and G is hole-free, we conclude
that uiuj ∈ E(G1). But uiuj belongs to a K3 in G1, consequently G contains an induced
house, a contradiction.
Finally, let q = 0. Clearly, p = n, G1 = A = Kn, and G = KnKn.
We close this section with a characterization of COMP-SUB(PM) on chordal graphs.
Recall that a chordal graph is a Ck≥4-free graph.
Proposition 3.5. Let G be a chordal graph of order 2n. It holds that G ∈
COMP-SUB(PM) if and only if G = KnKn.
Proof. Let G be a chordal graph. Clearly, KnKn ∈ COMP-SUB(PM).
Consider that G ∈ COMP-SUB(PM). There exist a complementary decomposition
(G1, G2) of G such that the edge cut M of the decomposition is a perfect matching. Let
M = {u1v1, . . . , unvn} where ui ∈ V (G1) and vi ∈ V (G2), for every i ∈ [n].
Suppose, by contradiction, that G ̸= KnKn. We may assume that there exists i, j, k, l ∈
[n] such that uiuj ∈ E(G1) and vkvl ∈ E(G2). Since M is a perfect matching and
G is chordal, we have that vivj /∈ E(G2) and ukul /∈ E(G1). Then, we obtain that
upvq ∈ E(G1) if and only if vpvq /∈ E(G2), for every distinct p, q ∈ [n]. Furthermore,
the chordality of G implies that there exists no path between vi, vj in G2, and no path
between uk, ul in G1. Consequently both G1 and its complement G2 are disconnected, a
contradiction.
4 Results on some Pk-free graphs
In this section, we still consider Π = PM as the property that considers M as a perfect
matching. We begin by showing how to solve COMP-SUB(PM) in polynomial time when
the input graph G is P5-free.
Theorem 4.1. COMP-SUB(PM) is polynomial-time solvable on P5-free graphs.
Proof. Let G be a 2n-vertex P5-free graph. Recall that if G ∈ COMP-SUB(PM), then G
is decomposable into complementary subgraphs G1 and G2, such that the edge cut M of
the decomposition is a perfect matching. Since G is P5-free, the existence of M implies
that G1 and G2 are P4-free, that is, G1 and G2 are cographs. Then, the conclusion follows
by applying Lemma 2.1.
D. Castonguay et al.: Perfect matching cuts partitioning a graph into complementary . . . 15
A graph is extended P4-laden if every induced subgraph with at most six vertices that
contains more than two induced P4’s is {2K2, C4}-free. Extended P4-laden graphs gener-
alize cographs, P4-sparse, P4-lite, P4-laden and P4-tidy graphs, and they were considered
under the perspective of partitioning. For instance, Bravo et al. [3] show that partition-
ing an extended P4-laden graph into at most k independent sets and at most ℓ cliques is
linear-time solvable for k, ℓ ≥ 1, and Bravo et al. [2] show a linear time algorithm for rec-
ognizing graphs that can be partitionable into a clique and a forest. In addition, Pedrotti and
De Mello [19] describe a linear-time algorithm that lists the minimal separators of extended
P4-laden graphs.
Another related result to partitioning is implied by considering that extended P4-laden
graphs are P6-free. The result on 3-colorability by Randerath and Schiermeyer [22] implies
that the problem of partitioning a graph into 3 independent sets is polynomial-time solvable
on extended P4-laden graphs.
We present in Proposition 4.2 a characterization concerned to COMP-SUB(PM) on
extended P4-laden graphs.
Proposition 4.2. Let G be an extended P4-laden graph of order 2n. It holds that G ∈
COMP-SUB(PM) if and only if G = KnKn.
Proof. Let G = KnKn. We analyse the subgraphs of G with at most 6 vertices to show
that G is an extended P4-laden graph. Let G
′ be a subgraph of G such that |V (G′)| ≤ 6.
If G′ is a subgraph of Kn or Kn, it is clear that G
′ does not have induced P4’s. Then, we
suppose that V (G′) intersects both V (Kn) and V (Kn). Notice that two induced P4’s arise
in G′ only if |V (G′) ∩ V (Kn)| ≥ 3 and |V (G
′) ∩ V (Kn)| ≥ 3. Since G is a split graph,
G′ is also a split graph. This implies that G′ is {2K2, C4}-free and hence, G is extended
P4-laden.
Now, we show that G ∈ COMP-SUB(PM) implies that G = KnKn. Suppose that G ∈
COMP-SUB(PM), and, by contradiction, that G ̸= KnKn. Since G ∈ COMP-SUB(PM)
there exists a complementary decomposition (G1, G2) of G, such that the edge cut M of
the decomposition is a perfect matching. Let M = {u1v1, . . . , unvn} where ui ∈ V (G1)
and vi ∈ V (G2), for every i ∈ [n]. We assume that G1 is connected.
Given that G ̸= KnKn, let u1u2u3 be an induced P3 in G1 and G
′ = G[{ui, vi : i ∈
[3]}].
Since uivi ∈ E(G
′), for every i ∈ [3], we have that {u1, v1, u3, v3} induces a 2K2 in
G′. Then, we may suppose that v1v3 ∈ E(G
′). Notice that {u2, v2, v1, v3} induces a 2K2
in G′, then we consider that v1v2 ∈ E(G
′) or v2v3 ∈ E(G
′). In both possibilities we have
an induced C4 in G
′, by {u1, v1, u2, v2} in the first, and by {u2, v2, u3, v3} in the latter, a
contradiction.
Our last result characterizes cographs yes-instances of COMP-SUB(PM). Recall that
a cograph is a P4-free graph.
Proposition 4.3. Let G be a cograph of order 2n. Then, G ∈ COMP-SUB(PM) if and
only if G = K2.
Proof. Let G be a cograph. Trivially K2 ∈ COMP-SUB(PM).
Suppose that G ∈ COMP-SUB(PM), and, by contradiction, that G ̸= K2. Let
(G1, G2) be a complementary decomposition of G, in which the edge cut M of the de-
composition is a perfect matching. Let M = {u1v1, . . . , unvn} where ui ∈ V (G1) and
vi ∈ V (G2), for every i ∈ [n].
16 Ars Math. Contemp. 25 (2025) #P2.07
Let i, j ∈ [n] such that uiuj ∈ E(G1). We know that uivj /∈ E(G) whenever i ̸= j.
Since G is P4-free, we obtain that uiuj ∈ E(G1) if and only if vivj ∈ E(G2), for ev-
ery distinct i, j ∈ [n]. This implies that G1 ≃ G2, then G1 ≃ G1, i.e., G1 is self-
complementary. Since a cograph is connected if and only if its complement is discon-
nected [8], we conclude that G1 cannot be self-complementary, a contradiction.
5 Concluding remarks
We have considered COMP-SUB(PM) problem when PM states the edge cut of the de-
composition as a perfect matching. We have presented polynomial-time algorithms for
solving COMP-SUB(PM) when the input graph G is hole-free or P5-free and we have
shown characterizations on chordal, distance-hereditary, and extended P4-laden graphs.
Concerning complexity results, despite its resemblance with the NP-complete problem
PERFECT MATCHING CUT, we show that COMP-SUB(PM) is GI-hard when the given
input graph G is C5-free or {Ck≥7, Ck≥7}-free.
We remark that our results by Theorem 3.1, Theorem 3.3, and Proposition 3.5 address
the cases when G is a Ck≥ℓ-free graph, for every ℓ ≥ 4, except for ℓ = 6. Then, we leave
the following open question.
Question 5.1. Can COMP-SUB(PM) on Ck≥6-free graphs be solved in polynomial time?
We also leave the complexity of COMP-SUB(PM) on P6-free graphs open. Further-
more, we still do not know whether COMP-SUB(PM) is GI-complete.
ORCID iDs
Diane Castonguay https://orcid.org/0000-0002-6640-0213
Erika M. M. Coelho https://orcid.org/0000-0002-3234-5789
Hebert Coelho https://orcid.org/0009-0003-9727-0006
Julliano R. Nascimento https://orcid.org/0000-0003-3002-5172
Uéverton S. Souza https://orcid.org/0000-0002-5320-9209
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.08
https://doi.org/10.26493/1855-3974.3231.f5e
(Also available at http://amc-journal.eu)
Adjacent vertex distinguishing total coloring of
corona product of graphs*
Hanna Furmańczyk †
Institute of Informatics, Faculty of Mathematics, Physics and Informatics,
University of Gdańsk, Wita Stwosza 57, 80-309 Gdańsk, Poland
Rita Zuazua ‡
Department of Mathematics, Faculty of Sciences, National Autonomous University of
Mexico, Ciudad Universitaria, Coyoacan, 04510 Mexico, DF, Mexico
Received 21 September 2023, accepted 12 May 2024, published online 21 March 2025
Abstract
An adjacent vertex distinguishing total k-coloring f of a graph G is a proper total k-
coloring of G such that no pair of adjacent vertices has the same color sets, where the
color set at a vertex v, CGf (v), is {f(v)} ∪ {f(vu)|u ∈ V (G), vu ∈ E(G)}. In 2005
Zhang et al. posted the conjecture (AVDTCC) that every simple graph G has adjacent
vertex distinguishing total (∆(G) + 3)-coloring. In this paper we confirm the conjecture
for many types of coronas, in particular for generalized, simple and l-coronas of graphs,
not relating the results to particular graph classes of the factors.
Keywords: Corona graph, l-corona, generalized corona graph, adjacent vertex distinguishing total
coloring, AVDTC Conjecture.
Math. Subj. Class. (2020): 05C15, 05C76, 68R10
1 Introduction
The processes occurring in the world around us can very often be modeled by the language
of graph theory. The graph coloring problems, vertex, edge as well as total version, are
ones of the best known problems of graph theory. In this paper we consider one of graph
*The short version of the paper was accepted to be published in proceedings of LAGOS2023 [12].
†Corresponding author.
‡This study was funded by UNAM (Grant PAPIIT-UNAM-IN117219).
E-mail addresses: hanna.furmanczyk@ug.edu.pl (Hanna Furmańczyk), ritazuazua@ciencias.unam.mx
(Rita Zuazua)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.08
coloring models connected with distinguishing elements in a graph. The authors of the first
paper concerning such a topic - Alberson and Collins in 1996 [1] - provided the following
motivational example based on a puzzle formulated in 1970 by Frank Rubin: ”Professor X ,
who is blind, keeps keys on a circular key ring. Suppose there are a variety of handle shapes
available that can be distinguished by touch. Assume that all keys are symmetrical so that
a rotation of the key ring about an axis in its plane is undetectable from an examination
of a single key. How many shapes does Professor X need to use in order to keep n keys
on the ring and still be able to select the proper key by feel?” It turned out that we can
solve the puzzle with particular distinguishing coloring of vertices in the relevant cycle.
Since that time the topic of distinguishing colorings has been developing a lot and now
many researchers considered colorings (proper, total or from lists) such that vertices (all or
adjacent) are distinguished either by sets or multisets or sums. In this paper we investigate
the problem of proper total distinguishing adjacent vertices by sets.
Let G = (V,E) be a simple graph with maximum degree ∆(G). A subset of vertices
in a graph G is an independent set if no pair of its vertices is adjacent. The independence
number of G, denoted by α(G), is the maximum cardinality of an independent set in G.
Let [k] denote the set {1, . . . , k} for any positive integer k. Suppose that f : V ∪ E → [k]
is a proper total coloring of G, i.e. no two adjacent edges, no two adjacent vertices, and
no edge and its endvertices are assigned the same color. The smallest number k admitting
such a proper total k-coloring is named total chromatic number and is denoted by χ′′(G).
Clearly, χ′′(G) ≥ ∆(G) + 1. Vizing [23], and independly Behzad et al. [2], posted the
following conjecture.
Conjecture 1.1 (TCC [2, 23]). For any graph G, χ′′(G) ≤ ∆(G) + 2.
For a given proper total k-coloring of G and for a vertex v ∈ V (G), let Cf (v) denote
the color set of v with respect to f , i.e. the set {f(v)} ∪ {f(vu)|u ∈ V (G), vu ∈ E(G)}.
This means that the color set Cf (v) consists of colors which are assigned to a vertex v and
all the edges incident to it. By Cf (v) we denote the set of colors that are not used to color
either vertex v or any edge incident to it, i.e. Cf (v) = [k] \ Cf (v). Sometimes we will
consider the color set restricted to some subgraph of G. Let H be a subgraph of G and let
v ∈ V (H). Then, CHf (v) denotes the set {f(v)} ∪ {f(vu)|u ∈ V (H), vu ∈ E(H)}. If
the total coloring is clear, we can use the notation of C(v) and CH(v), respectively.
In this paper, we are interested in the smallest number k of colors such that there is a
proper total k-coloring of G with the adjacent vertices being distinguished by their color
sets. Such a model was introduced by Zhang et al. [28] in 2005. The topic of graph coloring
models connected with distinguishing elements of a graph is very widely discussed in the
literature (cf. [9, 19, 20]). In the paper we consider one of them. More formally, in adjacent
vertex distinguishing total k-coloring f (avd total k-coloring, for short), we have Cf (u) ̸=
Cf (v) for every pair of vertices u, v such that uv ∈ E(G). The smallest k admitting
such coloring is called the adjacent vertex distinguishing total chromatic number (avd total
chromatic number, for short) and is denoted by χ′′a(G). Of course, χ
′′
a(G) ≥ χ
′′(G). It
turns out that there are a lot of examples of graphs for which this inequality is strict, e.g.
2l + 1 = χ′′(K2l+1) < χ
′′
a(K2l+1) = 2l + 3 [16].
As a direct consequence of the definition, we have the following relation between avd
total chromatic number and chromatic number, χ(G), and chromatic index of a graph G,
χ′(G).
Proposition 1.2. For any graph G, χ′′a(G) ≤ χ(G) + χ
′(G).
H. Furmańczyk and R. Zuazua: Adjacent vertex distinguishing total coloring of corona . . . 3
Indeed, in any proper total coloring f of the vertices and edges of a graph G, which
uses two different sets of colors to color the vertices and edges of the graph respectively,
every two adjacent vertices u and v (uv ∈ E(G)) are colored with different colors that
we do not use to color the edges of this graph. So the color sets for these two vertices are
different: c(u) ∈ Cf (u)\Cf (v) and c(v) ∈ Cf (v)\Cf (u). Hence, any such total coloring
is also avd total coloring. Taking into account for example Vizing and Brook’s theorems
and applying them to Proposition 1.2 we get the following.
Proposition 1.3. Let G ̸= Kn and G ̸= C2k+1. Then χ
′′
a(G) ≤ 2∆(G) + 1.
Huang et al. [15] proved that the bound from the last proposition can be improved to
2∆(G). Whereas, for planar graphs Proposition 1.2 implies
Proposition 1.4. For any planar graph G we have χ′′a(G) ≤ ∆(G) + 5.
Zhang et al. in [28] determined χ′′a(G) for many basic families of graphs, including
cycles, complete graphs, fans, wheels or trees. Additionally, the following bound on χ′′a(G)
in terms of the maximum degree of a graph ∆(G) was conjectured.
Conjecture 1.5 (AVDTCC [28]). For any simple graph G, χ′′a(G) ≤ ∆(G) + 3.
The conjecture has been attracting the attention of many graph theorists since 2005. It
has been proved for some families of graphs, including planar [3, 7, 13, 14], outerplanar
[26], subcubic [4, 16, 25], bipartite [4], and 4-regular graphs [18]. The last result is proved
by giving a relevant algorithm for avd total 7-coloring of 4-regular graphs. Coker and
Johannson [8] used probabilistic methods to prove that there exists a constant c such that
χ′′a(G) ≤ ∆(G) + c.
Zhang et al. [28] proved also
Lemma 1.6 ([28]). If G has two vertices of maximum degree which are adjacent, then
χ′′a(G) ≥ ∆(G) + 2.
Lemma 1.7 ([28]). If G has m components Gi, i ∈ [m], and |V (Gi)| ≥ 2, i ∈ [m], then
χ′′a(G) = max{χ
′′
a(G1), χ
′′
a(G2), . . . , χ
′′
a(Gm)}.
That is why we assume that all graphs considered in this paper are connected.
In this paper we put our attention to graph products. They are interesting and useful
in many situations. The complexity of many problems that deal with very large and com-
plicated graphs is reduced greatly if one is able to fully characterize the properties of less
complicated prime factors. In the literature we have some results concerning adjacent ver-
tex distinguishing total coloring for join graphs of paths with cycles and fans [17, 27], and
some Cartesian products of simple graphs [5, 6, 21, 24]. We consider the objective problem
for corona product of graphs - generalized, simple and l-coronas. They are often close to
the boundary between easy and hard coloring problems [11].
Definition 1.8. For a given simple graph G with V (G) = {v1, . . . , vnG}, and graphs
H1, . . . , HnG , the generalized corona, denoted by G◦̃Λ
nG
i=1Hi or by G◦̃(H1, . . . , HnG), is
the graph obtained by taking one copy of graphs G, H1, ..., HnG and joining the vi vertex
of G to every vertex of Hi (cf. Figure 1).
4 Ars Math. Contemp. 25 (2025) #P2.08
3 4
4 35
6
4
65 7
3 2
4
5
6
7
1
4
4 1
3
3
5 6 7 8
2
1
2
1
3
1
2 3
4
3
4
1
2 3 2 4
Figure 1: An example of a generalized corona - graph C4◦̃(C3, P3, C4, P2) and its exem-
plary avd total coloring.
In the cases when all graphs Hi are isomorphic, i.e. H1 ≃ H2 ≃ · · · ≃ HnG ≃ H , the
generalized corona is reduced to simple corona G ◦H . Graph G is called the center graph,
while graph H is named the outer graph. Such type of graph product was introduced by
Frucht and Harary [10].
Let v ∈ V (G), by Fv we denote the set of edges of G ◦H linking v with the relevant
copy of H , and we name it by fan in v.
Definition 1.9. For any integer l ≥ 2, the graph G ◦ℓ H is defined as G ◦ℓ H = (G ◦ℓ−1
H) ◦H , where G ◦1 H = G ◦H . Graph G ◦ℓ H is also named as l-corona product of G
and H .
In this paper we confirm Conjecture 1.5 for many coronas, generalized, simple, or l-
coronas under the assumption that the factors fulfill the conjecture (or even sometimes
without this assumption), not relating the results to particular graph classes. We conclude
the paper with some open questions.
2 Main results
We start with some basic observations.
Remark 2.1. Let G be a simple graph. If there is avd total k-coloring of G then there is
also its avd total (k + 1)-coloring.
Remark 2.2. In any avd total k-coloring f of a graph G, if the degree of a vertex u is
different of the degree of v, u, v ∈ V (G), then Cf (u) ̸= Cf (v).
Note that every bipartite graph G has (∆(G) + 2)-adjacent-vertex-distinguishing total
coloring such that ∆(G) colors are used to color edges of G, while the remaining two
H. Furmańczyk and R. Zuazua: Adjacent vertex distinguishing total coloring of corona . . . 5
colors are used to color vertices. Moreover, we may extend this coloring into adjacent-
vertex-distinguishing total coloring that uses more than ∆(G) + 2 colors by introducing
new colors only for vertices.
Theorem 2.3 ([4]). If G is a bipartite graph, then χ′′a(G) ≤ ∆(G) + 2.
We start our main results from the theorem concerning generalized coronas G◦̃Λni=1Hi
where the maximum degree of each graph Hi does not exceed the maximum degree of G. It
is worth emphasizing that in the assumption of the theorem we use only avd total coloring
of graph G while graphs Hi are colored totally in any proper way.
Theorem 2.4. Let G and H1, H2, . . . HnG be connected simple graphs, each one on at
least two vertices, such that χ′′a(G) ≤ ∆(G) + t, t ≥ 2, and χ
′′(Hi) ≤ ∆(Hi) + ti,
1 ≤ ti ≤ t, for each i ∈ [nG]. In addition, let ∆(G) ≥ ∆(Hi) for all 1 ≤ i ≤ nG. Then,
χ′′a(G◦̃Λ
n
i=1Hi) ≤ ∆(G◦̃Λ
n
i=1Hi) + t.
Proof. Without loss of generality we can assume that ∆(H1) ≥ ∆(H2) ≥ · · · ≥ ∆(HnG).
Let |V (G)| = nG and |V (Hi)| = nHi , i ∈ [nG]. From the assumption, we have nG ≥ 2
and nHi ≥ 2 for every i ∈ [nG]. It is clear that δ(G) + mini nHi ≤ ∆(G◦̃Λ
n
i=1Hi) ≤
∆(G)+maxi nHi and the maximum degree of the generalized corona can be realized only
by vertices of G.
In order to obtain adjacent vertex distinguishing total (∆(G◦̃Λni=1Hi) + t)-coloring f
we start from any avd total (∆(G) + t)-coloring of G. We will extend the coloring for
all graphs Hi, i ∈ [nG], and the relevant fans. We consider vi ∈ V (G) with the relevant
graph Hi, for consecutive i ∈ [nG]. We apply one of the following cases, depending on the
relation between degrees of G and Hi.
Case 1: ∆(G) > ∆(Hi) or ∆(G) = ∆(Hi), but ti < t (cf. the cases of H1, H2, and H4
in the generalized corona from Figure 1).
We color vertices and edges of Hi in a proper way with ∆(Hi) + ti colors from the set
[∆(Hi) + ti + 1], but we do not use the color assigned to vi in f |G. Since ∆(Hi) + ti <
∆(G) + t, it is doable. Finally, we use colors ∆(G) + t+ 1, . . . ,∆(G) + nHi + t to color
nHi edges in the fan Fvi . Note, that these nHi colors are not used in either G or Hi.
Case 2: ∆(G) = ∆(Hi) and ti = t (cf. the case of H3 in the generalized corona from
Figure 1).
If we are able to color graph Hi with ∆(Hi) + t colors in a proper total way such
that color f(vi) is not used to color vertices in Hi, then we do it. Next, we assign colors
∆(G) + t + 1, . . . ,∆(G) + nHi + t to color nHi edges in the fan Fvi . Otherwise, we
consider any total (∆(Hi) + t)-coloring of Hi. Note that taking exactly such a coloring
of Hi to G◦̃Λ
n
i=1Hi results in improper partial total coloring. The vertices in Hi that are
assigned color f(vi) need to be recolored into a new color ∆(G)+t+1 and such a modified
total coloring of Hi is taken to G◦̃Λ
n
i=1Hi. Because nHi ≥ 2 and Hi is connected, then
we are able to choose one vertex in Hi not colored with ∆(G) + t + 1, let us say vertex
u′, and the edge viu
′ in the fan Fvi can be assigned color ∆(G) + t+ 1. The rest of edges
in the fan Fvi are colored with ∆(G) + t+ 2, . . . ,∆(G) + nH + t colors, not used earlier
either in G or in Hi.
Finally, after coloring all outer graphs and the relevant fans, the total coloring of the
generalized corona is proper. We claim that it is adjacent vertex distinguishing. In order to
justify this we consider the following cases.
6 Ars Math. Contemp. 25 (2025) #P2.08
• Let a and b be two adjacent vertices in Hi, i.e. ab ∈ E(Hi) for any i ∈ [nG]. Note
that CHi(a), as well as CHi(b), is completed only by one color in the whole avd
total coloring of G◦̃Λni=1Hi. In Case 1 the colors used to color edges in Fvi were
different from those in CHi(a) ∪ CHi(b) and they have not been used in H before.
In Case 2 , at least one of these two vertices a and b is completed by a new color not
used earlier. Thus, CG◦̃Λ
n
i=1
Hi(a) ̸= CG◦̃Λ
n
i=1
Hi(b).
• Let a and b be any two adjacent vertices in G. We colored them taking into account
only graph G, i.e. CG(a) ̸= CG(b), next their color sets were completed by the set
of new colors, not used earlier in G. Thus, CG◦̃Λ
n
i=1
Hi(a) ̸= CG◦̃Λ
n
i=1
Hi(b).
• Let a be a vertex of Hi and let b be a vertex of G. Since in the corona G◦̃Λ
n
i=1Hi,
deg(a) ≤ nHi while deg(b) ≥ nHi +1, then deg(a) ̸= deg(b) and C
G◦̃Λn
i=1
Hi(a) ̸=
CG◦̃Λ
n
i=1
Hi(b).
As a consequence of the previous theorem, for the case when Hi ≃ Hj for any i, j ∈
[nG], we get the following corollary.
Corollary 2.5. Let G and H be connected simple graphs on at least two vertices, for which
χ′′a(G) ≤ ∆(G)+t with t ≥ 2 and χ
′′(H) ≤ ∆(H)+t′ with 1 ≤ t′ ≤ t. If ∆(G) ≥ ∆(H),
then
χ′′a(G ◦H) ≤ ∆(G ◦H) + t.
When t ∈ {2, 3} in the assumption of Corollary 2.5 we can conclude what follows.
Corollary 2.6. Let G and H be connected simple graphs on at least two vertices, for which
∆(G) ≥ ∆(H).
(1) If Conjecture 1.5 holds for G and Conjecture 1.1 holds for H , then
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
(2) If Conjecture 1.5 holds for G and H , then
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
(3) If G and H are bipartite graphs. Then,
χ′′a(G ◦H) ≤ ∆(G ◦H) + 2.
Observe that ∆(G ◦ℓ H) = ∆(G) + l · nH , for any l ≥ 1. If ∆(H) ≤ ∆(G) then we
immediately have ∆(H) ≤ ∆(G ◦ℓ−1 H), for any l ≥ 2. Hence we have the following
generalization of Corollary 2.5.
Corollary 2.7. Let G and H be connected simple graphs on at least two vertices, for which
∆(G) ≥ ∆(H).
(1) If Conjecture 1.5 holds for G and Conjecture 1.1 holds for H , then
χ′′a(G ◦
ℓ H) ≤ ∆(G ◦ℓ H) + 3,
for any integer l ≥ 2.
H. Furmańczyk and R. Zuazua: Adjacent vertex distinguishing total coloring of corona . . . 7
(2) If Conjecture 1.5 holds for G and H , then
χ′′a(G ◦
ℓ H) ≤ ∆(G ◦ℓ H) + 3,
for any integer l ≥ 2.
(3) If G and H are bipartite graphs, then
χ′′a(G ◦
ℓ H) ≤ ∆(G ◦ℓ H) + 2,
for any integer l ≥ 2.
One can ask what about the Conjecture 1.5 for coronas G ◦H where ∆(H) > ∆(G).
We also ask how big the difference between maximum degrees can be to be sure that
AVDTC Conjecture holds. We partially answer these questions.
Theorem 2.8. Let G and H be connected simple graphs on at least two vertices, for which
Conjecture 1.5 holds. Let ∆(H) = ∆(G) + 1. Then,
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
Proof. If H is a bipartite graph, then, due to Theorem 2.3, we may start with any avd total
(∆(H) + 2)-coloring of each copy of H . In this case we need the same number of colors
for avd total (∆(G) + 3)-coloring of G. Hence, this case can be seen as equivalent to the
one given in Case 2 in the proof of Theorem 2.4. So, let us assume H is not bipartite. Then
an avd total (∆(G ◦H) + 3)-coloring f of G ◦H can be obtained as follows.
1. Color vertices and edges of graph G with ∆(G) + 3 colors in adjacent vertex distin-
guishing way. We will refer to this part of the coloring as to f |G.
2. We will extend our avd total coloring f |G into avd total coloring of each copy of H
in G◦H . Let v ∈ V (G), c = f(v), and we consider the relevant copy of H . Let f |H
denote an avd total (∆(H) + 3)-coloring of H such that there is a vertex u ∈ V (H)
for which f(u) ̸= c and the color ∆(H) + 3 does not belong to color set of u in H ,
i.e. ∆(H) + 3 ̸∈ CH
f |H
(u).
Note that such a coloring of H always exists due to the fact that we have at least
two missing colors in a color set of every vertex in H . So we color the chosen H ,
H ⊂ G ◦H , in the desirable way.
Since ∆(H) + 3 = ∆(G) + 4, the color ∆(H) + 3 does not belong to CH
f |H
(u), and
∆(G) + 4 is not used in f |G, we can color an edge uv in the fan Fv with ∆(G) + 4.
Note that after this step the partial total coloring f of G ◦H may not be proper. We
need to fix it, if it is the case. We do it in the following way. If the coloring is im-
proper, i.e. there are vertices in H colored with c, we recolor them to ∆(G)+5. Note
that after this recoloring the new coloring, limited to graph H , is certainly proper avd
total coloring, while the partial total coloring of the whole corona, received so far, is
proper.
3. Next, we will complete the coloring of the fan Fv . If the coloring was initially not
proper, we choose one vertex w ∈ V (H) such that w ̸= u and w is not colored with
c. Such a vertex certainly exists, because H is not bipartite. Note that vw can be
8 Ars Math. Contemp. 25 (2025) #P2.08
colored with ∆(G)+5. We do it. The rest uncolored nH −2 edges in Fv are colored
with new colors: ∆(G)+6, . . . ,∆(G)+nH +3. Otherwise, we color all uncolored
nH − 1 edges in Fv with ∆(G) + 5, . . . ,∆(G) + nH + 3.
We repeat Step 2 and Step 3 for every v ∈ V (G) with the relevant copy of H .
Observe that for any vertex v ∈ V (G), CG◦H(v) = CG(v) ∪ {∆(G) + 4, ...,∆(G) +
nH + 3}. Since the coloring f |G was adjacent vertex distinguishing, then also f , in ac-
cordance to vertices of G, is avd. In addition, for every uw ∈ E(H), there exists at least
one color t ∈ {1, 2, ..,∆(G) + 3} such that t /∈ CH(u) ∩ CH(w). Hence, the obtained
(∆(G◦H)+3)-coloring of the whole corona G◦H is proper adjacent vertex distinguishing
total coloring.
Theorem 2.9. Let G be a connected simple graph on at least two vertices, for which
Conjecture 1.5 holds, and let H = K∆(G)+3. Then
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
Proof. We start from any adjacent vertex distinguishing total (∆(G) + 3)-coloring of G.
We will refer to this part of the coloring as to f |G. An extension of f |G into the whole
(∆(G ◦H) + 3)-coloring f of G ◦H is obtained as follows.
Let v be any vertex in G and let f |G(v) = c, where c ∈ [∆(G) + 3]. Since ∆(H) =
∆(G) + 2, we have that c ∈ [∆(H) + 1]. We consider the copy of H joint to the vertex v.
Let f |H be an adjacent vertex distinguishing total (∆(H)+3)-coloring of H , such that
for all u ∈ V (H), f |H(u) /∈ {c,∆(G) + 5}. It is possible because H = K∆(G)+3 and we
use exactly ∆(G) + 3 colors to color vertices.
Observe that for any vertex u ∈ V (H), |CH(u)| = 2, and for any two vertices u,w in
H , the following holds 0 ≤ |CH(u) ∩ CH(w)| ≤ 1. We claim that there are two vertices
u,w ∈ V (H) such that CH(u) ∩ CH(w) = ∅. Indeed, let us assume that every pair of
vertices in H has one missing color in common, i.e. for any x1, x2 ∈ V (H) we have
|CH(x1) ∩ CH(x2)| = 1. We can assume that CH(x1) = {a, d}, CH(x2) = {b, d}
with a ̸= d ̸= b. Now, the color d need to belong to a color set of at least one vertex in H ,
otherwise we do not use this color in the coloring of H . Suppose d ∈ CH(x3), x3 ∈ V (H).
By the definition, d ̸∈ CH(x3). If the missing colors are not a, i.e. CH(x3) ∩ {a} = ∅
then vertices x1 and x3 are those whose sets of missing colors are disjoint. So, let assume
that a ∈ CH(x3). We can repeat the same reasoning for color b. So, we can assume now
that CH(x3) = {a, b}, a ̸= d ̸= b.
As the order of H is at least 4, nH ≥ 4, there is x4 ∈ V (H) such that, by hypothesis,
|CH(x4) ∩ CH(xi)| = 1, for i = 1, 2, 3.. If d ∈ CH(x4) then the second missing color
must be different than a and b and then vertices x3 and x4 have disjoint sets of missing
colors. If d ̸∈ CH(x4) then, similarly as we justified for x3, CH(x4) = {a, b}, but then
the coloring is not distinguishing, contradiction.
Thus, we always have a pair of vertices in H with disjoint sets of missing colors. Let
{u,w} ⊂ V (H) such that CH(u) ∩ CH(w) = ∅ with CH(u) = {a, b}, CH(w) = {d, g},
where a, b, d, g are four different colors. It is clear that at least one of the missing colors in
these two sets is different than c. W.l.o.g. we can assume that b ̸= c and g ̸= c. We want
to lead to the situation when CH(u) = {a,∆(G) + 4}, CH(w) = {d,∆(G) + 5}. So,
if it is necessary, we can swap colors b and ∆(G) + 4, as well as g and ∆(G) + 5. Next
H. Furmańczyk and R. Zuazua: Adjacent vertex distinguishing total coloring of corona . . . 9
we color the edges f(uv) = ∆(G) + 4, f(wv) = ∆(G) + 5 and we use the new colors
{∆(G) + 6,∆(G) + 7, ...,∆(G) + nH + 3} to the edges xv, where x ∈ V (H)\{u,w}.
We repeat the same actions and arguments for subsequent copies of H , finally obtaining
the coloring of the entire graph G ◦H (cf. Figure 2).
Observe that the obtained total (∆(G◦H)+3)-coloring of ∆(G◦H) is adjacent vertex
distinguishing. Indeed, due to the fact that deg(x) ̸= deg(y) for any pair of vertices such
that x ∈ V (G) and y ∈ V (H), we need to consider only the case of different color sets
for two adjacent vertices within graph G or H . If x and y are two adjacent vertices in G,
since CG(x) ̸= CG(y) and next their color sets were completed by the set of new colors
not used earlier in G, then CG◦H(x) ̸= CG◦H(y). On the other hand, if x and y are two
adjacent vertices in H , then since their color sets from f |H were completed by one color
and in the case where u ̸= x ̸= w and u ̸= y ̸= w it was a new color not used earlier in
H , then we get CG◦H(x) ̸= CG◦H(y). The only doubt could appear for vertices u,w in
H , chosen as above. But since CH(u) ∩ CH(w) = ∅, then CG◦H(u) ∩ [∆(G) + 5] = a
and CG◦H(w) ∩ [∆(G) + 5] = d, a ̸= d. Thus CG◦H(u) ̸= CG◦H(w) and the proof is
complete.
3
2
6
1 1
3
4
1
6
2 2
3
4
7 6 5 8 6 7 8 5
1 2
2 3
45 5
3
4
1
Figure 2: An example of a corona K2 ◦K4 and its avd total coloring illustrating the proof
of Theorem 2.9.
Lemma 2.10. Let H be a connected simple graph with ∆(H) ≥ 3 for which Conjecture 1.5
holds. Let α(H) ≥ 2 and let u1 and u2 be any two non-adjacent vertices in H . Then for
any color c ∈ [∆(H) + 1] there is an avd total (∆(H) + 3)-coloring f of H such that all
four conditions hold:
(1) ∆(H) + 2 ∈ CHf (u1),
(2) ∆(H) + 3 ∈ CHf (u2),
(3) f(u1) ̸= c,
(4) f(u2) ̸= c.
Proof. We can easily start from any avd total (∆(H) + 3)-coloring f of H such that
f(u1) = c1 ̸= c and f(u2) = c2 ̸= c and c1, c2 ∈ [∆(H) + 1]. It may happen that
c1 = c2.
10 Ars Math. Contemp. 25 (2025) #P2.08
Now, let us assume that CHf (u1) does not contain ∆(H) + 2. Since each vertex has
at least two missing colors, let us say {a1, a2} ⊂ CHf (u1), then we can exchange one
of missing colors, let us say a1, with ∆(H) + 2, and vice versa, in the entire graph H .
Similarly, if CHf (u2) does not contain ∆(H) + 3, let {b1, b2} ⊂ C
H
f (u2). We can choose
one of missing colors, different than a1, and exchange it into ∆(H) + 3, and vice versa,
also in the whole graph H .
Finally, we received an avd total (∆(H) + 3)-coloring f of H fulfilling the desirable
conditions.
Theorem 2.11. Let G an H be connected simple graphs on at least two vertices, for which
Conjecture 1.5 holds. Let ∆(H) = ∆(G) + 2 and α(H) ≥ 2. Then
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
Proof. We start from any avd total (∆(G) + 3)-coloring of G. We will refer to this part of
the coloring as to f |G. Further, we extend this coloring into the entire corona depending
on the form of H .
Let v ∈ V (G) and f |G(v) = c, where c ∈ [∆(G) + 3], or in other words c ∈ [∆(H) +
1]. We consider the relevant copy of H . If H is bipartite, H = (V1 ∪ V2, E), then by
Theorem 2.3, χ′′a(H) ≤ ∆(H) + 2. By König’s theorem, we color the edges of H with
∆(H) colors: color c and ∆(H) − 1 other colors are used from the set [∆(H) + 1]. One
color among {1, . . . ,∆(H) + 1} is not used to color edges in H . We assign it to color all
vertices in V1, while ∆(G) + 4 is used to color all vertices in V2. Next, we choose one
vertex x ∈ V1 and assign ∆(G) + 4 to vx. Further, we complete the coloring of edges in
the fan Fv by coloring the residual uncolored nH − 1 edges with different colors from the
set {∆(G) + 5, . . . ,∆(G) + nH + 3}. We repeat the procedure for all copies of H . It is
easy to see that the coloring f of G ◦H is adjacent vertex distinguishing total coloring.
Otherwise, i.e. if H is not bipartite, an extension of f |G into the whole (∆(G◦H)+3)-
coloring f of G ◦ H is obtained as follows. Let I be any independent vertex set in H of
size at least 2 and let u1, u2 be any two vertices from I .
1. Color vertices and edges of H in avd total way with ∆(H) + 3 = ∆(G) + 5 colors
in such a way that color ∆(G) + 4 is missing in the color set of u1, i.e. ∆(G) + 4 ∈
CH(u1), and ∆(G) + 5 ∈ CH(u2), and f(u1) ̸= c, and f(u2) ̸= c. This is possible
due to Lemma 2.10.
It may happen that the partial total coloring of G ◦H is improper at this stage. We
will fix it later.
2. Assign color ∆(G) + 4 to an edge vu1 and color ∆(G) + 5 to vu2.
3. If the partial total coloring of G ◦ H is improper then we recolor vertices colored
initially with c into ∆(G) + 6.
4. Choose any vertex x ∈ V \{u1, u2} not colored with ∆(G)+6 and assign ∆(G)+6
to vx.
Note that such a vertex always exists because H is not bipartite and there exists at
least one edge with both endvertices in V \H . At least one of them is not colored
with ∆(G) + 6.
H. Furmańczyk and R. Zuazua: Adjacent vertex distinguishing total coloring of corona . . . 11
5. Complete the coloring of edges in the fan Fv by coloring the remaining uncolored
nH − 3 edges with new colors ∆(G) + 7, . . . ,∆(G) + nH + 3.
Note, that the only cases for which we really need to check the color sets are adjacent
vertices of x which are colored with ∆(G) + 6. But edges in the fan Fv joining such
a vertex are colored with completely new colors not used any more in Fv and in
the copy of H under consideration. So, even for such adjacent vertices, their color
sets are different. Thus, the partial coloring of G ◦H is proper and adjacent vertex
distinguishing.
We repeat the same procedure for all copies of H . Finally, we get an avd total (∆(G ◦
H)+3)-coloring of G◦H . The justification for two adjacent vertices within G is the same
as in the proofs of the previous theorems. The proof is complete.
Up to now, since actually Theorem 2.9 concerns the case when α(H) = 1 and ∆(H) =
∆(G) + 2, we proved Conjecture 1.5 for all coronas G ◦ H of graphs G and H with
∆(H) − ∆(G) ≤ 2, under some additional assumptions for G and H . Now we present
a partial result concerning graphs with a greater difference, i.e. let ∆(H) = ∆(G) + k,
k ≥ 3. We start with H being bipartite and k = 3.
Theorem 2.12. Let G be connected simple graphs on at least two vertices, for which
Conjecture 1.5 holds. Let H = (V1 ∪V2, E) be a bipartite graph with ∆(H) = ∆(G)+3.
Then
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
Proof. By Theorem 2.3, χ′′a(H) ≤ ∆(H) + 2. We start from any (∆(G) + 3)-adjacent-
vertex-distinguishing total coloring of G. Let v ∈ V (G) and f |G(v) = c, where c ∈
[∆(G) + 3], or in other words c ∈ [∆(H) + 6]. We consider the relevant copy of H .
By König’s theorem, we color the edges in the relevant copy of H with ∆(H) colors,
∆(H) = ∆(G) + 3, among them the color c. We assign color ∆(G) + 4 to all vertices in
V1 while color ∆(G) + 5 is assigned to all vertices from V2. Now, let us choose a vertex
colored with ∆(G) + 4, let us name it x1, x1 ∈ V1, and a vertex colored with ∆(G) + 5,
let us name it by x2, x2 ∈ V2. At first, try to choose non-adjacent vertices or vertices of
different degrees. When it is not possible, choose any two vertices colored appropriately.
Let f(vx1) = ∆(G) + 5 and let f(vx2) = ∆(G) + 4. If C
G◦H(x1) = C
G◦H(x2) then
recolor x2 into ∆(G)+6 and choose any vertex in the same partition, let us say x3, x3 ∈ V2,
and color vx3 with ∆(G) + 6, otherwise we assign ∆(G) + 6 to any uncolored edge in
the Fv . Next, we complete the coloring of edges in the fan Fv by coloring the remaining
uncolored nH − 3 edges with new colors ∆(G) + 7, . . . ,∆(G) + nH + 3. Note that the
partial coloring of G ◦ H is proper and adjacent vertex distinguishing from the point of
view of vertices from the copy of H .
We repeat the same procedure for all vertices v ∈ V (G) and the relevant copies of H .
Since we completed CGf (v), for each v ∈ V (G), by the same set of colors {∆(G) +
4, . . . ,∆(G) + nH + 3}, and taking into account the previous reasoning for color sets of
any two adjacent vertices from H , the total coloring of the whole corona is adjacent vertex
distinguishing. The proof is complete.
Note that a similar idea applied to H being bipartite graph, in particular complete bi-
partite graph, with ∆(H) = ∆(G) + k for k ≥ 4 will not work. We mean assigning only
∆(H) colors to edges of H and assigning colors ∆(G) + 4, . . . ,∆(G) + nH + 3 to edges
12 Ars Math. Contemp. 25 (2025) #P2.08
of a fan Fv . Since in such solutions all ∆(H) colors used to color edges of H are present
in color sets of all vertices in H , then we can use none of these colors to color edges in
Fv and thus such an approach would involve more than ∆(G ◦H) + 3 colors, for k ≥ 4.
Hence, we certainly need to color edges of H with more colors than only ∆(H).
Now, let us consider more general graphs. We take an attempt of generalization of
the method given in the proof of Theorem 2.11. The basis of this method is such an avd
total (∆(H) + 3)-coloring f |H of H that there is a set of non-adjacent vertices U =
{u1, . . . , uk} for which the colors from [∆(H) + 3]\[∆(G) + 3] are missing colors for
vertices in U . For k = 2 such a coloring was easy to achieve and it was guaranteed by
Lemma 2.10. For a greater k we need additional conditions for degrees of u ∈ U .
Lemma 2.13. Let H be a connected simple graph with ∆(H) ≥ k + 1 for which Conjec-
ture 1.5 holds. Let α(H) ≥ k and let u1, . . . , uk be any k non-adjacent vertices in H such
that deg(u1) ≤ ∆(H), deg(u2) ≤ ∆(H), deg(ui) ≤ ∆(H)− i+2 for i ∈ {3, . . . , k−2}.
Then for any color c ∈ [∆(H)− k + 4] there is an avd total (∆(H) + 3)-coloring f of H
such that all the following conditions hold:
(1) ∆(H) + 1 + i ∈ CHf (ui), i ∈ [k],
(2) f(ui) ̸= c, i ∈ [k].
We leave the full proof to the reader, but it is easy to see that the conditions for degrees of
chosen vertices in an independent set of size at least k guarantee us that |
⋃
i∈[k] C
H
f (ui)| ≥
k. So we are able to exchange colors in H to achieve the desirable conditions. Of course,
the conditions for degrees in Lemma 2.13 are not the only one guaranteeing us ”good” avd
total coloring of H .
Theorem 2.14. Let G an H be connected simple graphs on at least two vertices, for which
Conjecture 1.5 holds. Let k ≥ 3 be an integer, ∆(H) = ∆(G)+k, and α(H) ≥ k. If there
exist k non-adjacent vertices in H u1, . . . , uk such that deg(u1) ≤ ∆(H), deg(u2) ≤
∆(H), deg(ui) ≤ ∆(H)− i+ 2 for i ∈ {3, . . . , k − 2} then
χ′′a(G ◦H) ≤ ∆(G ◦H) + 3.
Proof. We start from any adjacent vertex distinguishing total (∆(G) + 3)-coloring of G.
We will refer to this part of the coloring as to f |G. An extension of f |G into the whole
(∆(G ◦ H) + 3)-coloring f of G ◦ H is obtained as follows. Let u1, . . . , uk be any k
non-adjacent vertices fulfilling the assumption of the theorem.
1. Color vertices and edges of H in avd total way with ∆(H) + 3 = ∆(G) + k + 3
colors in such a way that color ∆(G) + 3 + i is missing in the color set of ui, i.e.
∆(G) + 3 + i ∈ CH(ui), i ∈ [k], and none of vertices ui, . . . , uk is colored with c.
This is possible due to Lemma 2.13.
It may happen that the partial total coloring of G ◦H is improper at this stage. We
will fix it later.
2. Assign color ∆(G) + 3 + i to an edge vui, for every i ∈ [k].
3. If the partial total coloring of G ◦ H is improper then we recolor vertices colored
initially with c into ∆(G) + k + 4.
H. Furmańczyk and R. Zuazua: Adjacent vertex distinguishing total coloring of corona . . . 13
4. Choose any vertex x ∈ V \{u1, . . . , uk} not colored with ∆(G) + k + 4 and assign
∆(G) + k + 4 to vx. Note that such a vertex always exists. Since ∆(H) ≥ k + 1
there must exist an edge whose endvertices are from the set {u1, . . . , uk}, then at
least one of the endvertices of such an edge is not colored with ∆(G) + k + 4.
5. Complete the coloring of edges in the fan Fv by coloring the remaining uncolored
nH −k−1 edges with new colors ∆(G)+k+5, . . . ,∆(G)+nH +3. Note, that the
only cases for which we really need to check the color sets are adjacent vertices of x
which are colored with ∆(G) + k + 4. But edges in the fan Fv joining such vertices
are colored with completely new colors not used any more in Fv and in the copy of H
under consideration. So, even for such adjacent vertices, their color sets are different.
Thus, the partial coloring of G ◦H is proper and adjacent vertex distinguishing from
the point of view of vertices from the copy of H .
We repeat the same procedure for all copies of H . Finally, we get a total (∆(G ◦H) + 3)-
coloring of G ◦ H . Since we completed CGf (v), for each v ∈ V (G), by the same set of
colors {∆(G)+4, . . . ,∆(G)+nH +3}, the total coloring of the whole corona is avd.
3 Conclusion
In the paper we considered adjacent vertex distinguishing total coloring of corona graphs in
the context of AVDTC Conjecture posted by Zhang in 2005. We confirmed this conjecture
for:
• generalized coronas G◦̃Λni=1Hi with ∆(G) ≥ ∆(Hi), under the assumption that
Conjecture 1.5 holds for G and Conjecture 1.1 holds for every Hi, i ∈ [nG] (this is
the extension of the result published recently for simple coronas [22]);
• all simple coronas G ◦H with ∆(H)−∆(G) ≤ 2, under the assumption that Con-
jecture 1.5 holds for G and H;
Actually, the assumption for graph H can be a little weaker. Our proofs show that it
is enough that Conjecture 1.1 holds for H .
• all simple coronas G ◦ H with ∆(H) = ∆(G) + 3, where H is bipartite and Con-
jecture 1.5 holds for G;
• some simple coronas G ◦H with ∆(H) = ∆(G) + k, k ≥ 3 under some additional
constraints - for details see Theorem 2.14.
It is worth to emphasize that our results do not concern coronas of particular graph
classes, but taking into account the results known from the literature and taking the results
from this work, we can replace our general graphs G and H with particular graph classes
fulfilling Conjecture 1.1 and 1.5. In Table 1 we present only exemplary results.
One can ask what about the remaining coronas not covered by the results of this paper.
We retain this for a further investigation and as an open problem for other graph theorists.
ORCID iDs
Hanna Furmańczyk https://orcid.org/0000-0001-8057-4108
Rita Zuazua https://orcid.org/0000-0002-5570-254X
14 Ars Math. Contemp. 25 (2025) #P2.08
❤
❤
❤
❤
❤
❤
❤
❤
❤
❤
❤
❤
❤
❤❤
G
H
path cycle 3-regular 4-regular
path ✓ ✓ ✓ ✓
cycle ✓ ✓ ✓ ✓
3-regular ✓ ✓ ✓ ✓
4-regular ✓ ✓ ✓ ✓
complete graph Kn, n ≥ 6 ✓ ✓ ✓ ✓
Table 1: An exemplary graph classes of G and H such that Conjecture 1.5 holds for G◦H .
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.09
https://doi.org/10.26493/1855-3974.3081.d6c
(Also available at http://amc-journal.eu)
On a conjecture of Erdős on size Ramsey
number of star forests
Akbar Davoodi *
The Czech Academy of Sciences, Institute of Computer Science, Pod Vodárenskou věžı́ 2,
182 07 Prague, Czech Republic
Ramin Javadi †
Department of Mathematical Sciences, Isfahan University of Technology,
84156-83111, Isfahan, Iran and
School of Mathematics, Institute for Research in Fundamental Sciences (IPM),
19395-5746, Tehran, Iran
Azam Kamranian
Department of Mathematical Sciences, Shahrekord University, 115, Shahrekord, Iran
Ghaffar Raeisi ‡
Department of Mathematical Sciences, Shahrekord University, 115, Shahrekord, Iran and
School of Mathematics, Institute for Research in Fundamental Sciences (IPM),
19395-5746, Tehran, Iran
Received 4 May 2023, accepted 10 May 2024, published online 1 April 2025
Abstract
Given two graphs F1 and F2, their size Ramsey number, denoted by r̂(F1, F2), is the
minimum number of edges of a graph G such that for any edge coloring of G by colors
red and blue, G contains either a red copy of F1 or a blue copy of F2. In this paper, we
deal with the size Ramsey number of star forests (disjoint union of stars) and following a
conjecture by Burr, Erdős, Faudree, Rousseau, and Schelp in 1978, we determine the exact
value of r̂(⊔si=1K1,ni ,⊔
t
i=1K1,mi) in several cases including when either mi’s and ni’s
are odd, or s = 1 or s = 2 and n1 = n2.
Keywords: Size Ramsey number, star forest, Ramsey minimal graph.
Math. Subj. Class. (2020): 05D10, 05C55
*The author was supported by the Czech Science Foundation, grant number GA19-08740S.
†The research of the author was in part supported by a grant from IPM No. 1400050420.
‡Corresponding author. The research of the author was in part supported by a grant from IPM No. 1403050321.
E-mail addresses: davoodi@cs.cas.cz (Akbar Davoodi), rjavadi@iut.ac.ir (Ramin Javadi),
azamkamranian@stu.sku.ac.ir (Azam Kamranian), g.raeisi@sku.ac.ir (Ghaffar Raeisi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.09
1 Introduction
In this paper, we are only concerned with undirected simple finite graphs without isolated
vertices and we follow [1] for terminology and notations not defined here. For a given
graph G, we denote its vertex set, edge set, maximum degree, minimum degree and edge
chromatic number (chromatic index) of G by V (G), E(G), ∆(G), δ(G) and χ′(G), re-
spectively. The order of a graph G is the cardinality of its vertex set, and the size of G,
e(G), is the cardinality of its edge set. For a vertex v ∈ V (G), we use deg (v) and N(v)
to denote the degree and the set of neighbors of v in G, respectively. Also, G − v stands
for the graph obtained from G by deleting v and all leave neighbors of v. The star graph
on n + 1 vertices is denoted by K1,n, consisting of one root vertex connected to n leaves.
Also, by a matching of size m, mK2, we mean m independent edges and by Pk we mean
a path on k vertices. Also, for given graphs G and H , we use G ⊔H to denote the disjoint
union of G and H and we use mG to denote the disjoint union of m copies of G.
A factor in a graph G is a spanning subgraph of G and a k-factor is a k-regular factor
in G. In particular, 1-factors are usually called perfect matchings and a 2-factor is the
disjoint union of cycles covering all vertices. A graph G is called k-factorable if the edges
of G can be decomposed into k-factors. It is well known that graphs of even degree can be
decomposed into edge-disjoint cycles [1]. For regular graphs of even degree, the edge-set
can be decomposed into 2-factors, as presented in the following theorem.
Theorem 1.1 (Petersen [7]). Every regular graph of even degree is 2-factorable.
For given graphs F1, F2 and G, we write G → (F1, F2) and we say G is Ramsey graph
for the pair (F1, F2) if in any red-blue coloring of the edges of G, there is either a red copy
of F1 or a blue copy of F2. A 2-edge coloring of G is called (F1, F2)-free coloring if G
avoids monochromatic Fi in the i-th color class, for each i ∈ {1, 2}. For given graphs
F1, F2, the size Ramsey number r̂(F1, F2) is defined as the minimum number of the edges
of a graph G such that G → (F1, F2). The investigation of size Ramsey numbers was first
initiated by Erdős et al. [3] in 1978. For results concerning size Ramsey numbers, we refer
the reader to [3, 4, 6, 8, 9] and references therein. In this paper, we study the size Ramsey
number of star forests (a star forest, sometimes also referred as a galaxy, is a forest whose
each component is a star) and the exact value of r̂(F1, F2) is determined for some pairs
(F1, F2) of star forests.
In 1978 [2], Burr et al. proposed the following conjecture, determining the exact value
of the 2-color size Ramsey number of star forests.
Conjecture 1.2 (Burr, Erdős, Faudree, Rousseau and Schelp [2]). For given positive in-
tegers n1 ≥ n2 ≥ · · · ≥ ns and m1 ≥ m2 ≥ · · · ≥ mt, let F1 = ⊔
s
i=1K1,ni and
F2 = ⊔
t
j=1K1,mj . Then, r̂(F1, F2) =
∑s+t
k=2 lk, where, lk = max{ni+mj−1 : i+j = k}.
This conjecture was verified in [2] for star forests containing stars of the same sizes.
Theorem 1.3 (Burr, Erdős, Faudree, Rousseau and Schelp [2]). For positive integers s, t,
m and n, r̂(sK1,n, tK1,m) = (s+ t− 1)(m+ n− 1). Moreover, if G → (sK1,n, tK1,m)
and has (s+ t− 1)(m+ n− 1) edges, then G = (s+ t− 1)Kn+m−1 or n = m = 2 and
G = lK3 ⊔ (s+ t− l − 1)K1,3, for some l, 1 ≤ l ≤ s+ t− 1.
Subsequently, the restriction that each star forest could only have stars of the same
size was relaxed and a more general condition was considered by Győri and Schelp in [5].
A. Davoodi et al.: On a conjecture of Erdős on size Ramsey number of star forests 3
Although they did not completely verify the conjecture for all star forests, they did so for a
large class of forests, which provides substantial support for Conjecture 1.2.
Theorem 1.4 (Győri and Schelp [5]). For positive integers n1 ≥ n2 ≥ · · · ≥ ns and
m1 ≥ m2 ≥ · · · ≥ mt, let F1 = ⊔
s
i=1K1,ni and F2 = ⊔
t
j=1K1,mj . For k = 2, . . . , s+ t,
set lk = max{ni +mj − 1 : i + j = k}. If
(
lk
2
)
>
∑s+t
i=k li, for all 2 ≤ k ≤ s + t, then
r̂(F1, F2) =
∑s+t
k=2 lk.
In Theorem 1.3, some cases are missed for Ramsey minimal graphs. So, in this paper,
we first present a much shorter proof for Theorem 1.3 including all equality cases (see The-
orem 2.2). In addition, we prove Conjecture 1.2 for many pairs (F1 = ⊔
s
i=1K1,ni , F2 =
⊔tj=1K1,mj ) of star forests, including the cases s = 1 (Theorem 2.3), s = 2 and n1 = n2
(Theorem 2.4), all ni’s and mi’s are odd (Theorem 2.5) and all ni’s are equal to an odd
number and m1 is also odd (Theorem 2.6). Moreover, for some pairs (F1, F2) of star
forests, Ramsey minimal graphs, i.e. graphs F with F → (F1, F2) and e(F ) = r̂(F1, F2)
will be classified exactly.
2 Main results
In this section, the main results of the paper will be presented. Note that by the notations
of Conjecture 1.2 and using the simple fact that K1,n+m−1 → (K1,n,K1,m), we deduce
that if F = ⊔s+tk=2K1,lk , then F → (F1, F2) and so, r̂(F1, F2) is upper bounded by e(F ) =
∑s+t
k=2 lk. This is certainly the case for all results of this section, therefore we shall always
prove just the claimed lower bound for the size Ramsey number of star forests. Thus, to
determine r̂(F1, F2) for a given pair (F1, F2) of star forests, it is sufficient to show that if
F is a graph such that F → (F1, F2), then e(F ) ≥
∑s+t
k=2 lk.
In what follows, we present a much shorter proof for Theorem 1.3. Beforehand, we
prove the following simple lemma that will be used in the proof of next results.
Lemma 2.1. Let n,m be positive integers and let G be a graph such that either ∆(G) ≤
m+n−3, or ∆(G) ≤ m+n−2 and m,n are both odd. Then, G has a (K1,n,K1,m)-free
coloring.
Proof. First, suppose that ∆(G) ≤ m + n − 3. Then, by Vizing’s Theorem, there is a
proper edge-coloring c for G with at most m+n− 2 colors. Partition the color classes in c
into two sets A1, A2 of sizes at most n− 1,m− 1, respectively. For each i ∈ {1, 2}, let Gi
be the subgraph of G induced by the edges of colors in Ai. Clearly, G1 and G2 decompose
the edge set of G and ∆(G1) ≤ n − 1 and ∆(G2) ≤ m − 1. Coloring all edges of Gi,
i ∈ {1, 2}, by the i-th color yields a (K1,n,K1,m)-free coloring for G, as desired.
Now, suppose that ∆(G) = m+n−2 and m,n are both odd. Then, G can be embedded
into a ∆(G)-regular graph H . By Theorem 1.1, H can be decomposed into (m+n− 2)/2
two-factors. Now, color all the edges of G which are in the first (n − 1)/2 two-factors of
H by red and other edges of G by blue. Clearly, the maximum degree of the red (resp.
blue) subgraph of G is at most n− 1 (resp. m− 1). Therefore, this is a (K1,n,K1,m)-free
coloring for G. This completes the proof.
Note that the lemma above cannot be extended to the case in which ∆(G) = m+n−2
and either n or m is even. For instance, when m = 2, an n-regular graph without a perfect
matching would be a counterexample. Now, we are ready to give an alternative proof for
Theorem 1.3 including missing extremal cases.
4 Ars Math. Contemp. 25 (2025) #P2.09
Theorem 2.2. For positive integers s, t, m and n, n ≥ m, we have r̂(sK1,n, tK1,m) =
(s+t−1)(m+n−1). Moreover, if G → (sK1,n, tK1,m) and e(G) = (s+t−1)(m+n−1),
then G = (s + t − 1)Kn+m−1 or n = m = 2 and G = lK3 ⊔ (s + t − l − 1)K1,3, or
s = m = 1 and n = 2 and G = lC4 ⊔ (t− 2l)K1,2, for some nonnegative integer l.
Proof. Since K1,m+n−1 → (K1,n,K1,m), we have (s+t−1)K1,m+n−1 → (sK1,n, tK1,m).
Thus, r̂(F1, F2) ≤ (s+ t− 1)(m+ n− 1). Now, we prove the lower bound by induction
on s + t. Let G be a graph such that G → (sK1,n, tK1,m). If ∆(G) ≤ m + n − 3,
then by Lemma 2.1, there is a (K1,n,K1,m)-free coloring of G, a contradiction. Thus,
∆(G) ≥ m+ n− 2. Let v be a vertex of maximum degree in G.
Claim. (G− v) → ((s− 1)K1,n, tK1,m).
To prove the claim, on the contrary, suppose that (G − v) ̸→ ((s − 1)K1,n, tK1,m).
Then, by the definition, there is a ((s−1)K1,n, tK1,m)-free coloring of G−v. This coloring
can be extended to a (sK1,n, tK1,m)-free coloring of G by coloring all edges incident with
v by the first color, contradicting that G → (sK1,n, tK1,m). This observation completes
the proof of the claim.
Thus, G − v → ((s − 1)K1,n, tK1,m) and by the induction hypothesis, e(G − v) ≥
(s+ t− 2)(m+n− 1). Now, if ∆(G) ≥ m+n− 1, then e(G) ≥ (s+ t− 1)(m+n− 1)
and we are done.
Thus, assume that ∆(G) = m+n−2 and there are s+t−1 vertices, say v1, . . . , vs+t−1,
in G such that dGi−1(vi) = m + n − 2, in which G0 = G and Gi = G − {v1, . . . , vi}.
Therefore, vi’s form an independent set in G.
Let W = {v1, . . . , vs+t−1} and let B be the bipartite graph induced by the edges
between W and V (G)\W in G. By Vizing’s theorem and the fact that bipartite graphs
are in class I , χ′(B) ≤ ∆(B) = ∆(G) = m + n − 2. Color all the first n − 1 color
classes by red and the m−1 remaining classes by blue. Clearly, this is a (K1,n,K1,m)-free
coloring of B. We extend this coloring to a coloring for G by giving a 2-coloring for (at
most) s+ t− 2 remaining edges. Obviously, these edges lie in V (G)\W . Select arbitrarily
s − 1 of them and color these edges by red. Finally, color the remaining edges by blue.
Since B has no red copy of K1,n and each red edge in V (G)\W participates in at most
one vertex disjoint red copy of K1,n, there is no red copy of sK1,n in G. Similarly, there
is no blue tK1,m in G. But this contradicts the assumption G → (sK1,n, tK1,m). Hence,
r̂(F1, F2) = (s+ t− 1)(m+ n− 1).
Now, we are going to characterize the extremal structures. For simplicity, define the
graph Hm,n to be either K1,3 or K3 if m = n = 2 and to be K1,m+n−1, otherwise. Now,
let G → (F1, F2) and e(G) = (s+t−1)(n+m−1). There are two possibilities for G. The
first is ∆(G) = m+n−1 and the second is ∆(G) = m+n−2 and e(Gs+t−1) = s+t−1.
For the first case, e(G−v) = (s+t−2)(m+n−1) and G−v → ((s−1)K1,n, tK1,m)
and G − v → (sK1,n, (t − 1)K1,m). Without loss of generality, we may assume that all
isolated vertices of G− v are removed.
First, suppose s = m = 1 and n = 2. By the induction hypothesis, G−v is the disjoint
union of some C4’s and K1,2’s. Note that ∆(G) = m+n− 1 = 2 and so v is not adjacent
to vertices of C4’s and the centers of K1,2’s. If N(v) is disjoint from V (G − v) or v is
adjacent to both leaves of one K1,2, then G is also union of some C4’s and K1,2’s and we
are done. Otherwise, G has a connected component isomorphic to either a path P5 or P7.
It is easy to see that P5 ̸→ (K1,2, 2K1,1) and P7 ̸→ (K1,2, 3K1,1) (for this, we can color
first and last edges of the path by red and others by blue). Hence, G ̸→ (K1,2, tK1,1).
A. Davoodi et al.: On a conjecture of Erdős on size Ramsey number of star forests 5
Now, suppose that either s ̸= 1, or m ̸= 1, or n ̸= 2. Then, by the induction hypothesis,
G − v is the disjoint union of s + t − 2 graphs Hm,n. Again, if N(v) is disjoint from
V (G − v), then G is the disjoint union of s + t − 1 graphs Hm,n and we are done. Now,
suppose that N(v) contains a vertex u in a copy of Hm,n, say H0, in G− v. If H0 is a star,
let w be the center of H0 and if H0 is a triangle then let w ̸= u be an arbitrary vertex of
H0.
Denote the edges incident with v by H1 and color edges of H0 and H1 such that in
each Hi, i = 1, 2, there are exactly n red edges and m − 1 blue edges, the edges wu and
uv are both red and there is no blue copy of K1,m in H0 ⊔H1 (this can be done because v
and w cannot be the center of a blue K1,m and in case m = 2, since n ≥ m, we can avoid
a blue K1,2). Also, since wu and uv are both red, there is no two vertex disjoint red copies
of K1,n in H0 ⊔H1. Now, for the remaining edges, color s− 2 copies of Hm,n by red and
t − 1 copies of Hm,n by blue. So, there is at most s − 1 disjoint red copies of K1,n and
t− 1 disjoint blue copies of K1,m. Thus, G ̸→ (sK1,n, tK1,m).
For the second case, i.e. when ∆(G) = m + n − 2 and e(Gs+t−1) = s + t − 1, if
m = 1, then ∆(G) = n− 1 and we color all the edges by red. Therefore, we may suppose
that m,n ≥ 2. If m = n = 2, then ∆(G) = 2 and G is disjoint union of some paths and
cycles. We color edges of each path and cycle alternatively by blue and red. Let λ be the
number of odd cycles in G. Clearly, the number of monochromatic copies of K1,2 is equal
to λ. Since e(G) = 3(s + t − 1), we have λ ≤ s + t − 1. If λ ≤ s + t − 2, then we can
color G such that there are at most s − 1 red copies of K1,2 and at most t − 1 blue copies
of K1,2, and thus, G ̸→ (F1, F2). If λ = s + t − 1, then G = (s + t − 1)K3 and we are
done.
Now, let m ≥ 2, n ≥ 3 and we show that G ̸→ (F1, F2). Recall that the edge set
of B can be colored by red and blue so that B ̸→ (K1,n,K1,m). Consider the induced
subgraph F on V (G)\W . By the assumption, F has exactly s + t − 1 edges. If two of
these edges, say e1 and e2, are adjacent, then we color e1, e2 and s − 2 more arbitrarily
selected edges of F by red and t − 1 remaining edges of F by blue. Clearly, there are at
most t − 1 disjoint blue copies of K1,m and at most (s − 1) disjoint red copies of K1,n
in G which is a contradiction. Hence, the edges e1, e2, . . . , es+t−1 form a matching in F .
Note that every red K1,n in F1 or blue K1,m in F2 should contain one of ei’s. Now, we
claim that for all i, ei has one endpoint, say ui, which is incident with n − 1 red edges in
B and another endpoint, say vi, which is incident with m− 1 blue edges in B. To see this,
note that if both endpoints of ei are incident with at most n − 2 red edges in B, then we
can color ei and s− 1 more edges in F by red and t− 1 remaining edges by blue and this
implies that G ̸→ (F1, F2). Thus, there is an endpoint of ei, say ui, with red degree equal
to n− 1 in B. Similarly, there is an endpoint of ei, say vi, with blue degree equal to m− 1
in B. Also, since ∆(B) = m+ n− 2, ui and vi are distinct. This proves the claim.
On the other hand, since B has exactly (s+t−1)(m+n−2) edges, we have degB(ui) =
n−1 and degB(vi) = m−1. Recall that we applied Vizing’s theorem to color the edges of
B. Now, we exchange color of the first and the last classes. In this way, every ui is incident
with n−2 red edges and one blue edge, and every vi is incident with m−2 blue edges and
one red edge. Now, we color all the s + t− 1 edges in F by red. Being n ≥ 3 guarantees
that neither a red K1,n nor a blue K1,m is seen. Hence, G ̸→ (sK1,n, tK1,m).
Now, for the next step, we are going to prove Conjecture 1.2 when s = 1. In other
words, we determine the size Ramsey number of a star versus an arbitrary star forest.
Moreover, Ramsey minimal graphs will be completely characterized.
6 Ars Math. Contemp. 25 (2025) #P2.09
Theorem 2.3. For given positive integers n and m1 ≥ m2 ≥ · · · ≥ mt ≥ 2, we have
r̂(K1,n,⊔
t
j=1K1,mj ) =
∑t
j=1(n +mj − 1). Moreover, if F → (K1,n,⊔
t
j=1K1,mj ) and
e(F ) =
∑t
j=1(n+mj−1), then F =
⊔t
j=1Gj , where for each j, either Gj = K1,n+mj−1,
or n = mj = 2 and Gj = K3.
Proof. We use induction on n to show that if F → (K1,n,⊔
t
j=1K1,mj ), then e(F ) ≥
∑t
j=1(n + mj − 1). If n = 1, then F → (K2,⊔
t
j=1K1,mj ) and so if all edges of F are
colored by the second color, we should have a monochromatic copy of ⊔tj=1K1,mj in F ,
which implies that e(F ) ≥
∑t
j=1 mj .
Now, let n ≥ 2 and F be a graph such that F → (K1,n,⊔
t
j=1K1,mj ). Also, let M be
the maximum matching in F . Clearly |M | ≥ t, because if all edges of F are colored by
blue, then ⊔tj=1K1,mj ⊆ F . Let F \M be the graph obtained from F by deleting all edges
of M .
Claim. (F \M) → (K1,n−1,⊔
t
j=1K1,mj ).
To prove the claim, consider a red/blue coloring of F \M and extend it to a coloring of
F by coloring the edges of M by red. Hence, there is either a red K1,n or a blue ⊔
t
j=1K1,mj
in F . In the latter case, F \M contains a blue ⊔tj=1K1,mj and in the earlier case, at most
one edge of the red K1,n is in M , so F \ M contains a red K1,n−1. This completes the
proof of the claim.
Applying the induction hypothesis and the claim, we have
e(F \M) ≥
t
∑
j=1
(mj + n− 2).
Therefore, e(F ) = e(M) + e(F \M) ≥
∑t
j=1(n+mj − 1) and we are done.
Now, we use induction on n to characterize the extremal structures. Suppose that
F → (K1,n,⊔
t
j=1K1,mj ) and e(F ) =
∑t
j=1 n + mj − 1. For n = 1, it is obvi-
ous that F =
⊔t
j=1 K1,mj . Now, let n ≥ 2 and M be the maximum matching of
F . By the above arguments, e(M) ≥ t and (F \ M) → (K1,n−1,⊔
t
j=1K1,mj ). Thus,
e(F \M) =
∑t
j=1(n+mj − 2) and e(M) = t.
By the induction hypothesis, F \ M contains exactly t components such that each
component is either a star or a triangle. We claim that there is no triangle in F \ M .
Otherwise, we have n = 3 and in this case, color all the triangles and two arbitrary edges
of each star in F \ M by red and all the remaining edges of F by blue. Clearly, there is
no red copy of K1,n and the number of blue edges is strictly less than
∑t
j=1 mj which is a
contradiction. Hence, F \M = ⊔tj=1K1,n+mj−2.
Now, we claim that for every edge e ∈ M , the endpoints of e cannot be in two different
components of F \ M . On the contrary, suppose that the endpoints of e are in two com-
ponents of F \ M say S1 and S2. Color n − 1 edges of each star in F \ M by red and
all other edges of F by blue such that S1 ∪ S2 ∪ e contains a blue copy of P4, a path of
length three. It is obvious that there is no red copy of K1,n and the number of blue edges
is exactly
∑t
j=1 mj . Existence of a blue P4 in F implies that F contains no blue copy of
F2. This observation proves the claim.
The maximality of M implies that each edge e of M is either incident with a center of
a star in F \M , or n = 3 and e is incident with the leaves of a K1,2 in F \M . Therefore,
A. Davoodi et al.: On a conjecture of Erdős on size Ramsey number of star forests 7
F =
⊔t
j=1Gj , where for each j, either Gj = K1,n+mj−1, or n = mj = 2 and Gj = K3.
This completes the proof.
Here, we determine the exact value of the size Ramsey number of two stars of the same
sizes versus an arbitrary star forest.
Theorem 2.4. For positive integers n and m1 ≥ m2 ≥ · · · ≥ mt ≥ 2, let F1 = 2K1,n
and F2 = ⊔
t
i=1K1,mi . Then r̂(F1, F2) = n+m1 − 1 +
∑t
i=1(n+mi − 1).
Proof. It is sufficient to show that if G is a graph such that G → (F1, F2), then e(G) ≥
n+m1 − 1 +
∑t
i=1(mi + n− 1).
Let G be a graph such that G → (F1, F2). If ∆(G) ≤ n+m1−3, then by Lemma 2.1,
there exists a (K1,n,K1,m1)-free coloring of G, contradicting G → (F1, F2). Now, let
∆(G) ≥ m1 + n− 1. Thus, there is a vertex v ∈ V (G) such that deg(v) ≥ m1 + n− 1.
With the same argument as in the proof of Theorem 2.2, G − v → (K1,n, F2). Thus,
by Theorem 2.3, e(G − v) ≥
∑t
i=1(mi + n − 1) and so, e(G) ≥ deg(v) + e(G − v) ≥
n+m1 − 1 +
∑t
i=1(mi + n− 1).
Hence, we may assume that ∆(G) = n + m1 − 2 and deg(v) = n + m1 − 2. If
e(G−v) >
∑t
i=1(mi+n−1), then e(G) > n+m1−2+
∑t
i=1(mi+n−1) and we are
done. Thus, we may assume that e(G−v) =
∑t
i=1(mi+n−1). As, G−v → (K1,n, F2),
by Theorem 2.3, G− v = ⊔ti=1Gi such that for every i, Gi = K1,n+mi−1 or Gi = K3.
If Gj = K1,n+m1−1 for some j, then ∆(G) ≥ n+m1 − 1 which contradicts ∆(G) =
n+m1 − 2. Therefore, G− v = tK3 and for all i, 1 ≤ i ≤ t, we have n = mi = 2. Since
∆(G) = 2, v has no neighbor in G−v and so G = K1,2∪ tK3. Now, we color an arbitrary
edge of each component of G by red and the rest by blue. Then, we exchange the color of a
blue edge in exactly one of the triangles by red. Clearly, there is only one red copy of K1,2
and t− 1 blue copies of K1,2 which contradicts F → (2K1,n,∪
t
i=1K1,mi).
Now, in the sequel, we determine the size Ramsey number of star forests containing
stars of odd sizes.
Theorem 2.5. Let n1 ≥ n2 ≥ · · · ≥ ns and m1 ≥ m2 ≥ · · · ≥ mt be odd positive
integers and F1 = ⊔
s
i=1K1,ni and F2 = ⊔
t
j=1K1,mj . Then, r̂(F1, F2) =
∑s+t
k=2 lk, where,
lk = max{ni +mj − 1 : i+ j = k}.
Proof. Suppose that G is a graph such that G → (F1, F2). We are going to prove that
e(G) ≥
∑s+t
k=2 lk. First, note that by Lemma 2.1, ∆(G) ≥ m1 + n1 − 1.
Now, let v1 be a vertex of degree at least l2 in G. Then, one may see that G − v1 →
(⊔si=2K1,ni , F2) and G − v1 → (F1,⊔
t
j=2K1,mj ). Thus, again by Lemma 2.1, we have
∆(G−v1) ≥ max{n2+m1−1, n1+m2−1} = l3. Now, choose a vertex v2 of degree at
least l3 in G− v1. By continuing this process, one we may find vertices v1, v2, . . . , vs+t−1
such that for all k ∈ {1, . . . s + t − 1}, the degree of vk in G \ {v1, . . . , vk−1} is at least
lk+1. This proves that e(G) ≥
∑s+t
k=2 lk. This completes the proof.
As the last result of this paper, we prove the following theorem which determines the
size Ramsey number of arbitrary star forests under a certain condition.
8 Ars Math. Contemp. 25 (2025) #P2.09
Theorem 2.6. Let s, n and m1 ≥ m2 ≥ · · · ≥ mt ≥ 2 be positive integers. If n and m1
are both odd, then
r̂(sK1,n,⊔
t
j=1K1,mj ) = (s− 1)(n+m1 − 1) +
t
∑
j=1
(n+mj − 1).
Moreover, if n and m1 are both odd, G → (sK1,n,⊔
t
j=1K1,mj ) and e(G) = (s− 1)(n+
m1 − 1) +
∑t
j=1(n+mj − 1), then G =
(
(s− 1)K1,n+m1−1
)
⊔
(
⊔tj=1 K1,n+mj−1
)
.
Proof. We use induction on s to prove the theorem. The base case s = 1, follows from
Theorem 2.3. Let s ≥ 2 and G be a graph such that G → (sK1,n,⊔
t
j=1K1,mj ).
First, by Lemma 2.1, we have ∆(G) ≥ n + m1 − 1. Now, let v be a vertex of max-
imum degree in G, i.e. deg(v) ≥ n + m1 − 1. By a similar argument, used in the proof
of Theorem 2.2, (G − v) → ((s − 1)K1,n,⊔
t
j=1K1,mj ). By the induction hypothesis,
e(G− v) ≥ (s− 2)(n+m1 − 1) +
∑t
j=1(n+mj − 1) and so
e(G) = deg(v) + e(G− v) ≥ (s− 1)(n+m1 − 1) +
t
∑
j=1
(n+mj − 1).
This observation shows that
r̂(sK1,n,⊔
t
j=1K1,mj ) = (s− 1)(n+m1 − 1) +
t
∑
j=1
(n+mj − 1).
Now, let G → (sK1,n,⊔
t
j=1K1,mj ) and e(G) = (s−1)(n+m1−1)+
∑t
j=1(n+mj−
1). Also, suppose that m1 = · · · = mℓ and mℓ+1 < mℓ, for some ℓ ≥ 1. By the above
argument, we have deg(v) = n+m1−1 and e(G−v) = (s−2)(n+m1−1)+
∑t
j=1(n+
mj−1). By the induction hypothesis, G−v =
(
(s−2)K1,n+m1−1
)
⊔
(
⊔tj=1K1,n+mj−1
)
.
We claim that N(v) ∩ V (G− v) = ∅.
On the contrary, let u ∈ N(v) ∩ V (G − v). Set A = (s − 1)K1,n+m1−1 and B =
⊔tj=2K1,n+mj−1. First, suppose that u ∈ N(v) ∩ S, for some component S of A. Note
that the maximum degree condition on G implies that u is not the root vertex of S. In this
case, color n edges of each component in A and also n− 1 edges of each component in B
by red and the rest by blue. Also, color n edges incident with v by red and the rest by blue
such that edges incident with u are red. It is obvious that the red subgraph contains at most
s − 1 disjoint copies of K1,n and the blue subgraph contains at most ℓ − 1 disjoint copies
of K1,m1 . Hence, we have a (sK1,n,⊔
t
j=1K1,mj )-free coloring of G, a contradiction.
Now, let u ∈ N(v) ∩ S for some component S of B. In this case, color all edges in
A and also n − 1 edges of each component in B by red and the rest by blue. Also, color
n− 1 edges incident with v by red and the rest by blue such that S ∪N(v) contains a blue
copy of P4. Clearly, there are at most (s−1) disjoint red copies of K1,n. Also, the number
of blue edges is exactly
∑t
j=1 mj . Existence of a blue P4 in the blue subgraph implies
that G contains no blue copy of ⊔tj=1K1,mj and so we have a (sK1,n,⊔
t
j=1K1,mj )-free
coloring of G, a contradiction. This contradiction shows that N(v) ∩ V (G − v) = ∅ and
so G =
(
(s− 1)K1,n+m1−1
)
⊔
(
⊔tj=1 K1,n+mj−1
)
. This completes the proof.
A. Davoodi et al.: On a conjecture of Erdős on size Ramsey number of star forests 9
3 Concluding remarks
We close the paper with some supplementary remarks. First, one may think of a generaliza-
tion of the results of the paper to the multicolor size Ramsey numbers. Let p1, p2, . . . , pq
and n1i ≥ n
2
i ≥ · · · ≥ n
pi
i , (1 ≤ i ≤ q), be positive integers. Also, let F1, F2, . . . , Fq
be star forests such that for i = 1, 2, . . . , q, Fi = ⊔
pi
j=1K1,nj
i
. Set p =
∑q
i=1 pi and for
k = q, q + 1, . . . , p, suppose that
lk = max{(n
j1
1 − 1) + (n
j2
2 − 1) + · · ·+ (n
jt
t − 1) + 1 : j1 + j2 + · · ·+ jt = k}.
Using the simple fact that K1,(m1−1)+(m2−1)+···+(mq−1)+1 → (K1,m1 ,K1,m2 , . . . ,
K1,mq ), we deduce that if F = ⊔
p
k=qK1,lk , then F → (F1, F2, . . . , Fq). Thus,
r̂(F1, F2, . . . , Fq) ≤
p
∑
k=q
lk. (3.1)
We believe that equality holds in (3.1) which is stated as the following conjecture and
can be considered as an extension of Conjecture 1.2 to the multicolor case.
Conjecture 3.1. Let p1, p2, . . . , pq and n
1
i ≥ n
2
i ≥ · · · ≥ n
pi
i , (1 ≤ i ≤ q), be positive
integers and p =
∑q
i=1 pi. If F1, F2, . . . , Fq are star forests such that Fi = ⊔
pi
j=1K1,nj
i
,
then
r̂(F1, F2, . . . , Fq) =
p
∑
k=q
lk,
where, lk = max{(n
j1
1 − 1) + (n
j2
2 − 1) + · · ·+ (n
jt
t − 1) + 1 : j1 + j2 + · · ·+ jt = k}.
Using the same methods as in the proofs, we can generalize the results of Theorems 2.2,
2.3, 2.5, 2.6. In fact, we have the exact values of the following parameters.
• r̂(s1K2, s2K2, . . . , sqK2,⊔
t
j=1K1,mj ),
• r̂(K1,n1 ,K1,n2 , . . . ,K1,nq ,⊔
t
j=1K1,mj ),
• r̂(F1, F2, . . . , Fq), when each component of Fi’s is odd,
• r̂(s1K1,n1 , s2K1,n2 , . . . , sqK1,nq ,⊔
t
j=1K1,mj ), when for each i, ni and m1 are odd.
ORCID iDs
Akbar Davoodi https://orcid.org/0000-0001-6403-5091
Ramin Javadi https://orcid.org/0000-0003-4401-2110
Ghaffar Raeisi https://orcid.org/0000-0003-0313-8059
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for multiple copies, Nederl. Akad. Wetensch. Indag. Math. 81 (1978), 187–195, doi:10.1016/
S1385-7258(78)80009-2, https://doi.org/10.1016/S1385-7258(78)80009-2.
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[3] P. Erdős, R. J. Faudree, C. C. Rousseau and R. H. Schelp, The size Ramsey number, Pe-
riod. Math. Hungar. 9 (1978), 145–161, doi:10.1007/BF02018930, https://doi.org/10.
1007/BF02018930.
[4] R. J. Faudree and R. H. Schelp, A survey of results on the size Ramsey number, in: Paul
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of Bolyai Soc. Math. Stud., pp. 291–309, 2002, https://digitalcommons.memphis.
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[5] E. Győri and R. H. Schelp, Two-edge colorings of graphs with bounded degree in both col-
ors, volume 249, pp. 105–110, 2002, doi:10.1016/S0012-365X(01)00238-2, https://doi.
org/10.1016/S0012-365X(01)00238-2.
[6] R. Javadi and G. Omidi, On a question of Erdős and Faudree on the size Ramsey numbers,
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BF02392606, https://doi.org/10.1007/BF02392606.
[8] O. Pikhurko, Size Ramsey numbers of stars versus 3-chromatic graphs, Combinator-
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 25 (2025) #P2.10
https://doi.org/10.26493/1855-3974.3047.c2d
(Also available at http://amc-journal.eu)
Arc-disjoint hamiltonian paths in Cartesian
products of directed cycles*
Iren Darijani †
Memorial University of Newfoundland St. John’s, NL A1C 5S7 P.O. Box 4200 Canada
Babak Miraftab ‡
Computational Geometry Lab, School of Computer Science, Carleton University Ottawa,
Ontario, K1S 5B6 Canada
Dave Witte Morris
Department of Mathematics and Computer Science, University of Lethbridge,
4401 University Drive, Lethbridge, Alberta, T1K 3M4, Canada
Received 25 January 2023, accepted 16 May 2024, published online 2 April 2025
Abstract
We show that if C1 and C2 are directed cycles (of length at least two), then the Cartesian
product C1 □ C2 has two arc-disjoint hamiltonian paths. (This answers a question asked
by J. A. Gallian in 1985.) The same conclusion also holds for the Cartesian product of any
four or more directed cycles (of length at least two), but some cases remain open for the
Cartesian product of three directed cycles. We also discuss the existence of arc-disjoint
hamiltonian paths in 2-generated Cayley digraphs on (finite or infinite) abelian groups.
Keywords: Abelian groups, Cayley digraphs, hamiltonian paths.
Math. Subj. Class. (2020): 05C25, 20K01, 20K25
*We thank two anonymous referees for helpful comments that corrected errors and improved the exposition.
†The author acknowledges research support from NSERC (grant number RGPIN- 2017-04905) while holding
a postdoctoral position at the department of Mathematics and Computer Science of the University of Lethbridge.
‡Corresponding author. The research was conducted while the author had a postdoctoral position at the de-
partment of Mathematics and Computer Science of the University of Lethbridge.
E-mail addresses: i.darijani@mun.ca (Iren Darijani), bobby.miraftab@gmail.com (Babak Miraftab),
dmorris@deductivepress.ca (Dave Witte Morris)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 25 (2025) #P2.10
1 Introduction
It is easy to see (and well known) that the Cartesian product of any two directed cycles
has a hamiltonian path. (See Definition 3.3 for the definition of the Cartesian product.)
In 1985, J. A. Gallian (personal communication) asked whether there are two arc-disjoint
hamiltonian paths. The main result of this paper establishes that the answer is “yes”:
Theorem 4.4. If C1 and C2 are directed cycles (of length ≥ 2), then the Cartesian product
C1 □ C2 has two arc-disjoint hamiltonian paths.
In fact, if the lengths of the directed cycles are large, then there are many pairs of
arc-disjoint hamiltonian paths:
Proposition 4.5. Let N(m,n) be the number of (unordered) pairs {P, P ′} of arc-disjoint
hamiltonian paths in the Cartesian product of a directed cycle of length m and a directed
cycle of length n. If m and n are sufficiently large, then
N(m,n) >
m2 n2
10
.
Although the theorem only considers the Cartesian product of precisely two directed
cycles, it implies that arc-disjoint hamiltonian paths exist in the Cartesian product of any
larger number of directed cycles, except three:
Corollary 5.1. If C1, C2, . . . , Cr are directed cycles (of length ≥ 2), and r ≥ 4, then the
Cartesian product C1 □ C2 □ · · ·□ Cr has two arc-disjoint hamiltonian paths.
Thus, r = 3 is the only open case of the following conjecture.
Conjecture 1.1. If C1, C2, . . . , Cr are directed cycles (of length ≥ 2), and r ≥ 2, then the
Cartesian product C1 □ C2 □ · · ·□ Cr has two arc-disjoint hamiltonian paths.
Although the conjecture has not been proved for all Cartesian products of three directed
cycles, we know that it is true in most of these cases:
Proposition 5.2. Assume C1, C2, and C3 are directed cycles (of length ≥ 2). If either
(1) the Cartesian product of two of the directed cycles has a hamiltonian cycle, or
(2) the lengths of the three directed cycles do not all have the same parity (i.e., if there
is a directed cycle of even length and a directed cycle of odd length), or
(3) at least one of the directed cycles has length 2,
then C1 □ C2 □ C3 has two arc-disjoint hamiltonian paths.
Question 1.2. Does there exist a function f(r) → ∞, such that every Cartesian product
of r directed cycles has at least f(r) pairwise arc-disjoint hamiltonian paths?
It is well known that any Cartesian product of two directed cycles is isomorphic to a
2-generated Cayley digraph on an abelian group. (See Definition 3.2 and Example 3.4.) It
is natural to expect that Theorem 4.4 can be extended to this setting:
Conjecture 1.3. If {a, b} is a 2-element generating set of a finite abelian group G, and a
and b are nontrivial, then C−→ay(G; a, b) has two arc-disjoint hamiltonian paths.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 3
We have some evidence that this is true:
Proposition 1.4. Conjecture 1.3 is true in all cases where either:
(1) |G : ⟨a− b⟩| is even (see Proposition 6.1), or
(2) |G : ⟨a⟩| > 600 (see Proposition 6.6).
Remark 1.5. Computer calculations verified Conjecture 1.3 for all cases where |G| ≤
104. (The sagemath source code for these computations is available online at https://
arxiv.org/src/2203.11017/anc.) Although the calculations are not guaranteed
to be error-free, they provide additional evidence for the truth of the conjecture.
We also solve the analogous problem for 2-generated Cayley digraphs on infinite abelian
groups. The natural analogue of a hamiltonian path in this setting is a one-way infinite
hamiltonian path, which means a list
v0, v1, v2, v3, . . . ,
of the vertices of the digraph, such that there is a directed edge from vi to vi+1 for every i.
However, it is known (and easy to see) that a 2-generated Cayley digraph on an infinite
abelian group never has a one-way infinite hamiltonian path (cf. [13, Theorem 5.1] and
[15, Theorem 3.1]), so it certainly does not have two of them (whether arc-disjoint or not).
On the other hand, the corresponding natural analogue of a hamiltonian cycle is a two-way
infinite hamiltonian path, which means a doubly-infinite list
. . . , v−2, v−1, v0, v1, v2, . . .
of the vertices of the digraph, such that there is a directed edge from vi to vi+1 for every i.
It is well known that these can exist, and we determine exactly when there are two of them
that are arc-disjoint:
Proposition 7.5. Assume G is an infinite abelian group.
(1) If there exist a, b ∈ G, such that C−→ay(G; a, b) has two arc-disjoint two-way infinite
hamiltonian paths, then G is isomorphic to either Z or Z× Zm, for some m ≥ 2.
(2) For a, b ∈ Z, the Cayley digraph C−→ay(Z; a, b) has two arc-disjoint two-way infinite
hamiltonian paths if and only if a and b are odd, and
either {a, b} = {1,−1} or a+ b = ±2.
(3) For a, b ∈ Z × Zm, with m ≥ 2, the Cayley digraph C
−→ay(Z × Zm; a, b) has two
arc-disjoint two-way infinite hamiltonian paths if and only if either
(a) {a, b} = {(1, x), (−1, y)}, for some x, y ∈ Zm, such that ⟨x+ y⟩ = Zm, or
(b) m = 2, a = (0, 1), and b ∈ {±1} × Z2, perhaps after interchanging a and b.
This paper is structured in the following manner. Section 2 sets the paper in context by
providing a brief review of related results. (None of these results are needed elsewhere in
the paper.) Section 3 sets some notation and recalls known results that will be used in later
sections. (These include basic properties of the “arc-forcing” subgroup and some work
of Curran-Witte [7].) Section 4 studies the Cartesian product of two directed cycles, and
Section 5 studies the Cartesian product of more than two directed cycles. Section 6 studies
2-generated Cayley digraphs on finite abelian groups, and Section 7 closes the paper with
a discussion of 2-generated Cayley digraphs on infinite abelian groups.
4 Ars Math. Contemp. 25 (2025) #P2.10
2 Brief review of related results
Several researchers have studied hamiltonian properties of the Cartesian product of two
directed cycles. For example, the following result presents three different necessary and
sufficient conditions for the existence of a hamiltonian cycle. (We use C⃗n to denote a
directed cycle of length n.)
Theorem 2.1. If m,n ≥ 2, then the following are equivalent:
(1) The Cartesian product C⃗m □ C⃗n has a hamiltonian cycle.
(2) (Rankin, cf. [22]) There exist a, b ≥ 1, such that
a+ b = gcd(m,n) and ⟨(a, b)⟩ = ⟨(1,−1)⟩,
where ⟨(x, y)⟩ denotes the subgroup generated by (x, y) in the abelian group Zm ×
Zn.
(3) (Trotter-Erdős [24]) There exist a, b ≥ 1, such that
a+ b = gcd(m,n) and gcd(a,m) = gcd(b, n) = 1.
(4) (Curran [25, Theorem 4.3]) There exist a, b ≥ 1, such that
gcd(a, b) = 1 and am+ bn = mn.
Remark 2.2. References to generalizations, such as determining the lengths of all of the
directed cycles in C⃗m □ C⃗n, can be found in the bibliography of [14].
The same methods also determine whether there are two arc-disjoint hamiltonian cy-
cles:
Theorem 2.3 (Keating [16, Corollary 2.4]). C⃗m □ C⃗n has two arc-disjoint hamiltonian
cycles if and only if there exist a, b ≥ 1, such that
a+ b = gcd(m,n) and gcd(ab,mn) = 1.
Remark 2.4. Theorem 2.3 settles the existence of two arc-disjoint hamiltonian cycles, and
the main result of this paper settles the existence of two arc-disjoint hamiltonian paths (see
Theorem 4.4).
Mixing the two, one might ask about the existence of a hamiltonian path that is arc-
disjoint from a hamiltonian cycle. However, this is not interesting, because the complement
of a hamiltonian cycle in C⃗m □ C⃗n must be a union of disjoint directed cycles. Therefore,
if the complement contains a hamiltonian path, then it is a hamiltonian cycle.
The undirected case is easier. The starting point here is the easy observation that the
Cartesian product of two (undirected) cycles is hamiltonian. In fact, it was proved by
A. Kotzig in 1973 that the Cartesian product of two cycles always has two edge-disjoint
hamiltonian cycles. In other words, the Cartesian product of two cycles can be decomposed
into hamiltonian cycles. This has been generalized to any number of cycles:
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 5
Theorem 2.5 (Aubert-Schneider, cf. [2, Theorem A]). If C1, C2, . . . , Cr are undirected cy-
cles (of length at least 3), then the Cartesian product C1□C2□· · ·□Cr can be decomposed
into edge-disjoint hamiltonian cycles.
It is conjectured that this extends to all Cayley graphs on abelian groups:
Conjecture 2.6 (Alspach, cf. [1, Unsolved Problem 4.5, page 454]). Let G be an abelian
group with a symmetric generating set S. If the Cayley graph Cay(G;S) has no loops, then
it can be decomposed into hamiltonian cycles (plus a 1-factor if the valency is odd).
This conjecture is known to be true for valency 4:
Theorem 2.7 (Bermond-Favaron-Maheo [3]). If S is a 4-element symmetric generating set
of a finite abelian group G (and 0 /∈ S), then Cay(G;S) has two edge-disjoint hamiltonian
cycles.
Remark 2.8.
(1) It is not known whether all connected Cayley graphs on nonabelian finite groups
have hamiltonian cycles (or hamiltonian paths), but it was proved by D. Bryant and
M. Dean [4] that not all of them can be decomposed into hamiltonian cycles.
(2) The most recent survey on hamiltonian paths and cycles in Cayley graphs and Cayley
digraphs seems to be [18].
3 Preliminaries on 2-generated Cayley digraphs
This section starts with a few basic definitions, and then recalls some essential results from
the 1980’s on hamiltonian paths in 2-generated Cayley digraphs on finite abelian groups.
We use standard terminology and notation from the theory of graphs and digraphs. (In
particular, “arc” is synonymous with “directed edge.”) All paths in a digraph are assumed
to be directed paths.
Notation 3.1. Throughout this section,
{a, b} is a 2-element generating set of a finite abelian group G.
We use o(g) to denote the order of an element g of G, and we denote the cardinality of a
set S by #S.
Definition 3.2 (cf. [10, page 504]). The Cayley digraph C−→ay(G; a, b) of G with respect to
the generators a and b is the digraph whose vertex set is G and which has a directed edge
from v to v + s for every v ∈ G and s ∈ {a, b}.
Definition 3.3. Recall that the Cartesian product X □ Y of two digraphs X and Y is
the digraph with vertex set V (X) × V (Y ) where the vertex (x1, y1) is joined to (x2, y2)
by a directed edge if and only if either x1 = x2 and (y1, y2) ∈ E(Y ) or y1 = y2 and
(x1, x2) ∈ E(X).
Example 3.4 (cf. [17, proof of Lemma 1]). The Cartesian product C⃗m□C⃗n of two directed
cycles can be realized as the Cayley digraph C−→ay(Zm × Zn; e1, e2), where {e1, e2} =
{(1, 0), (0, 1)} is the standard generating set of the abelian group Zm × Zn.
6 Ars Math. Contemp. 25 (2025) #P2.10
Definition 3.5.
(1) We use the usual terminology that, for s ∈ {a, b}, a directed edge of C−→ay(G; a, b) is
called an s-edge if it is of the form (v, v + s), for some v ∈ G. (So every directed
edge is either an a-edge or a b-edge, but not both.)
(2) Suppose P is a subdigraph of C−→ay(G; a, b), and let s ∈ {a, b}.
(a) We let δs(P ) be the number of s-edges in P .
(b) We say that a vertex v travels by s in P if v is a vertex of outdegree 1 in P , and
the out-edge of v is an s-edge [12, page 82].
(c) A set of vertices travels by s in P if every element of the set travels by s (cf. [12,
page 83]).
3.1 Elementary theory of the arc-forcing subgroup
Definition 3.6 ([9, Definition 2.6]). A spanning quasi-path in a digraph Γ is a spanning
subdigraph P , such that precisely one connected component of P is a directed path, and all
other components of P are directed cycles.
In particular, every hamiltonian path is a spanning quasi-path. Next we provide another
characterization.
Remark 3.7 (cf. [9, Definition 2.6]). A spanning subdigraph P of Γ is a spanning quasi-
path if and only if there exist vertices ι and τ , such that:
(1) the indegree of ι is 0 in P , but the indegree of all other vertices is 1, in P and,
(2) the outdegree of τ is 0 in P , but the outdegree of all other vertices is 1 in P .
We call ι the initial vertex of P , and call τ the terminal vertex of P . (This acknowledges
the fact that they are the initial vertex and terminal vertex of the path component of P .)
Definition 3.8. Recall that G = ⟨a, b⟩ is a finite abelian group.
(1) The subgroup ⟨a− b⟩ is called the arc-forcing subgroup [25, §2.3].
(2) Suppose τ is the terminal vertex of a spanning quasi-path P of C−→ay(G; a, b). Then
the coset τ + ⟨a − b⟩ is called the terminal coset [20, Definition 5(ii)] and all other
cosets are called non-terminal cosets. (In the older terminology of Housman [12],
τ + ⟨a− b⟩ is called the special coset and all other cosets are called regular cosets.)
Since no vertex of a spanning quasi-path has indegree 2, it is clear that if a vertex v
travels by a, then v + (a − b) cannot travel by b. (Similarly, if v travels by b, then v −
(a − b) cannot travel by a.) This observation (which is originally due to R. A. Rankin
[22, Lemma 1], in the setting of hamiltonian cycles, rather than hamiltonian paths) has the
following consequence:
Lemma 3.9 (Housman, cf. [12, pages 82–83]). If P is a spanning quasi-path P in
C−→ay(G; a, b), then:
(1) Each non-terminal coset either travels by a or travels by b .
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 7
(2) If d is the number of vertices in the terminal coset that travel by b, then:
(a) ι = τ + a+ d(a− b) ∈ τ + a+ ⟨a− b⟩, and
(b) if v = τ + i(a− b) is any vertex of the terminal coset (with 0 ≤ i < o(a− b)),
then:
• v travels by b iff 1 ≤ i ≤ d, and
• v travels by a iff i > d.
Remark 3.10 (Housman, cf. [12, proof of Theorem 1]). This implies that a spanning quasi-
path in C−→ay(G; a, b) is uniquely determined by specifying:
(1) the terminal vertex,
(2) how many vertices of the terminal coset travel by b, and
(3) the non-terminal cosets that travel by b.
Determining whether a spanning quasi-path is a hamiltonian path does not depend on
knowing which particular non-terminal cosets travel by b, or which particular vertices of
the terminal coset travel by b, but only on the number of each:
Proposition 3.11 (Housman, cf. [7, Proposition 6.7]). Suppose P and P ′ are spanning
quasi-paths in C−→ay(G; a, b), such that
• the number of non-terminal cosets that travel by b in P is equal to the number of
non-terminal cosets that travel by b in P ′, and
• the number of vertices in the terminal coset that travel by b in P is equal to number
of vertices in the terminal coset that travel by b in P ′.
If P is a hamiltonian path, then P ′ is also a hamiltonian path.
This can be restated in terms of the total number of vertices that travel by b in all of P :
Corollary 3.12. Assume that P and P ′ are spanning quasi-paths in C−→ay(G; a, b), such that
δb(P ) = δb(P
′). If P is a hamiltonian path, then P ′ is also a hamiltonian path.
Proof. Let t (resp. t′) be the number of non-terminal cosets that travel by b in P (resp. in P ′),
and let d (resp. d′) be the number of vertices of the terminal coset that travel by b in P
(resp. in P ′). Then 0 ≤ d, d′ < o(a− b), and we have
δb(P ) = t o(a− b) + d and δb(P
′) = t′ o(a− b) + d′.
Since δb(P ) = δb(P
′) by assumption, we conclude that t = t′ and d = d′. So Proposi-
tion 3.11 tells us that P ′ is a hamiltonian path.
The following construction provides a standard example of a spanning quasi-path in
which a particular number t of non-terminal cosets travel by b, and a particular number d
of vertices in the terminal coset travel by b. See Figure 1 for some examples.
8 Ars Math. Contemp. 25 (2025) #P2.10
H0(0) H0(1) H0(2)
H1(0) H1(1) H1(2)
Figure 1: The spanning quasi-path Ht(d) in the Cayley digraph C
−→ay
(
Z4×Z6; (1, 0), (0, 1)
)
for t ∈ {0, 1} and 0 ≤ d ≤ 2. (Vertices in the non-terminal coset are black, and vertices in
the terminal coset are white.) Of these six subdigraphs, only H1(0) is a hamiltonian path.
Notation 3.13 ([7, Definition 5.2]). For 0 ≤ t < |G : ⟨a − b⟩| and 0 ≤ d < o(a − b), we
let Ht(d) be the spanning subdigraph of C
−→ay(G; a, b), such that:
(1) for 0 ≤ i < o(a− b), the element −a− i(a− b) of the coset −a+ ⟨a− b⟩:
• travels by b if i < d,
• has no outedges if i = d, and
• travels by a if i > d.
(2) the cosets ⟨a− b⟩, a+ ⟨a− b⟩, . . . , (t− 1)a+ ⟨a− b⟩ travel by b, and
(3) all other cosets travel by a.
This is a spanning quasi-path whose initial vertex is 0, and whose terminal vertex is −a+
d(b− a). By construction, exactly t non-terminal cosets travel by b, and exactly d vertices
in the terminal coset travel by b.
Since t and d are arbitrary in Notation 3.13 (and any non-negative integer can be written
in the form t o(a− b)+d, with 0 ≤ d < o(a− b)), it follows that there is a spanning quasi-
path with any desired number of b-edges (in the allowable range):
Lemma 3.14. For 0 ≤ k < |G|, there is a spanning quasi-path P in C−→ay(G; a, b), such
that δb(P ) = k.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 9
3.2 Some results of Curran-Witte
We now state (without proof) two results from [7]. One is the following theorem. The
other is Theorem 3.21, which provides a way to determine whether the subdigraph Ht(d)
is a hamiltonian path in C−→ay(G; a, b). We will then use this latter result to derive some
rather technical information about hamiltonian paths in C−→ay(G; a, b).
Theorem 3.15 (Curran-Witte [7, Theorem 9.1]). If C1, C2, . . . , Cr are directed cycles (of
length ≥ 2), and r ≥ 3, then the Cartesian product C1 □C2 □ · · ·□Cr has a hamiltonian
cycle.
The other main result of [7] relies on some additional notation. Recall that a point
(x, y) in R2 is a lattice point if its coordinates are integers (i.e., if x, y ∈ Z), and that a
lattice point (x, y) is primitive if gcd(x, y) = 1 (cf. [6, Proposition 7.2, page 36]).
Notation 3.16. Recall that G = ⟨a, b⟩ is a finite abelian group. Let:
(1) m = o(a).
(2) n = |G : ⟨a⟩|, so |G| = mn.
(3) e ∈ Z, such that nb = ea and 0 ≤ e < m. (A reader who is interested only in
Cartesian products of directed cycles can always assume e = 0.)
(4) T (m,n, e) be the closed triangle with vertices (0, 0), (n, 0), and (e,m).
Notation 3.17 ([7, Notation 3.4]). For 0 ≤ t ≤ |G : ⟨a− b⟩|, let
Xt =
(
1−
t
I
)
(n, 0) +
t
I
(e,m) where I = |G : ⟨a− b⟩|.
Note 3.18 ([7, Remark 3.6]). The sequence (X0, X1, . . . , XI) is a list of all the lattice
points on the line segment joining (n, 0) and (e,m). This implies that
|G : ⟨a− b⟩| = gcd(n− e,m).
See Figures 2 and 3 for a concrete example of a triangle T (m,n, e), the associated
lattice points Xt, and the values defined in the following notation.
Notation 3.19 (cf. [7, pages 37–38]). For 0 ≤ t < |G : ⟨a− b⟩|:
(1) Let Tt be the closed triangle with vertices (0, 0), Xt, and Xt+1, so {Tt | 0 ≤ t <
|G : ⟨a− b⟩| } is a decomposition of the triangle T (m,n, e) into smaller triangles.
(2) Let At(1), At(2), . . . , At(ft) be a list of the primitive lattice points in Tt, in order of
increasing slope. (So ft is the number of primitive lattice points in Tt.)
(3) For 1 ≤ k ≤ ft, let ht(k) be the number of lattice points that
• are a positive scalar multiple of At(k), and
• are in the closed triangle Tt, but
• are not on the side of Tt that is opposite (0, 0), that is, not on the line segment
from Xt to Xt+1.
10 Ars Math. Contemp. 25 (2025) #P2.10
More concretely, we have
ht(k) =
⌊
mn− 1
mxk + (n− e) yk
⌋
where At(k) = (xk, yk).
(4) For 1 ≤ k < ft, let
ut(k) = ht(1) + 2
k
∑
j=2
ht(j).
This means that ut(1) = ht(1), and ut(k) = ut(k − 1) + 2ht(k) for 2 ≤ k < ft. Also
note that u0(1) = h0(1) = n− 1.
x
y
0 1 2 3 4 5 6 7 8
0
1
2
3
4
5
6
X0
X1
X2
X3
Figure 2: Primitive lattice points in the triangle T (6, 8, 5).
Remark 3.20. For each t, Curran and Witte [7, Notation 3.14] also defined a certain func-
tion
Bt : {0, 1, 2, . . . , o(a− b)− 1} → Z× Z,
such that
Bt(d) = (0, 0) ⇔ d = ut(k) for some k (with 1 ≤ k < ft). (∗)
We will not define the function Bt, because we have no need for it, other than its appearance
in the statement of the following theorem, which is the other main result of [7]. (Actually,
the statement of [7, Theorem 7.1] only includes ((1) ⇔ (2)) of the following result, but
((2) ⇔ (3)) is the fundamental property of Bt(d) that is stated in (∗) above. All we need is
the equivalence ((1) ⇔ (3)).)
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 11
t ft k At(k) ht(k) ut(k)
0 6 1 (1, 0) 7 7
2 (7, 1) 1 7 + 2× 1 = 9
3 (6, 1) 1 9 + 2× 1 = 11
4 (5, 1) 1 11 + 2× 1 = 13
5 (4, 1) 1 13 + 2× 1 = 15
6 (7, 2) 0
1 6 1 (7, 2) 0 0
2 (3, 1) 2 0 + 2× 2 = 4
3 (5, 2) 1 4 + 2× 1 = 6
4 (2, 1) 3 6 + 2× 3 = 12
5 (5, 3) 1 12 + 2× 1 = 14
6 (3, 2) 1
2 5 1 (3, 2) 1 1
2 (4, 3) 1 1 + 2× 1 = 3
3 (5, 4) 1 3 + 2× 1 = 5
4 (1, 1) 5 5 + 2× 5 = 15
5 (5, 6) 0
Figure 3: The functions At, ht, and ut in the case where (m,n, e) = (6, 8, 5). The primi-
tive lattice points At(k) are taken from Figure 2.
Theorem 3.21 (Curran-Witte [7, Theorem 7.1]). For 0 ≤ t < |G : ⟨a − b⟩| and 0 ≤ d <
o(a− b), the following are equivalent:
(1) Ht(d) is a hamiltonian path.
(2) Bt(d) = (0, 0).
(3) d = ut(k) for some k ∈ {1, 2, . . . , ft − 1}.
Corollary 3.22 (Curran-Witte [7, Remark 8.4]).
#{ (t, d) | Ht(d) is a hamiltonian path in C
−→ay(G; a, b) }
is exactly one less than the number of primitive lattice points in the closed triangle
T (m,n, e).
Lemma 3.23 ([7, Proposition 3.13]). For 0 ≤ t < |G : ⟨a− b⟩|, we have
ut(ft − 1) + ht(ft) = o(a− b)− 1.
This has the following immediate consequences:
Corollary 3.24. For 0 < t < |G : ⟨a− b⟩|, we have
ut−1(ft−1 − 1) = o(a− b)− 1− ut(1).
Proof. Note that At−1(ft−1) = At(1) (because this is the primitive lattice point on the
edge that Tt−1 and Tt have in common), so we have ht−1(ft−1) = ht(1). Since ht(1) =
ut(1) (by the definition of ut(k)), the desired conclusion is now immediate from
Lemma 3.23 (after replacing t with t− 1).
12 Ars Math. Contemp. 25 (2025) #P2.10
Corollary 3.25. Assume 0 ≤ t, t′ < |G : ⟨a− b⟩|, 1 ≤ k < ft, and 1 ≤ k
′ ≤ ft′ .
(1) If o(a− b) is odd, then ut(k) ≡ ut′(k
′) (mod 2).
(2) If o(a− b) is even, then ut(k)− ut′(k
′) ≡ t− t′ (mod 2).
Proof. It is immediate from Notation 3.19(4) that the parity of ut(k) depends only on t.
Then we see from Corollary 3.24 that ut−1(∗) has the same parity as ut(∗) if and only if
o(a− b) is odd.
In the remainder of this section, we will give two consequences of Theorem 3.21. They
are stated in [7] only for the special case where C−→ay(G; a, b) is the Cartesian product of
two directed cycles, but the same arguments apply to the general case. For completeness,
we provide proofs.
Proposition 3.26 (Curran-Witte). Assume that m and n are as in Notation 3.16. If m ≥ 2,
then there exists n′ ∈ {n− 1, n}, such that, for each i in
{n′, n′ + 2, n′ + 4, . . . , n′ + 2⌊(n− 1)/2⌋},
there is a hamiltonian path Pi in C
−→ay(G; a, b), such that δb(Pi) = i.
Proof (cf. [7, (9.2) on page 67, and Case 3 on page 69]). For convenience, we let T =
T (m,n, e). Let x1, x2 ∈ R, such that (x1, 1) and (x2, 2) are on the line segment from
(n, 0) to (e,m). (This line segment is the side of T that is opposite the vertex (0, 0).) Then
(x1, 1) is the midpoint of the line segment from (n, 0) to (x2, 2), so
x1 =
x2
2
+
n
2
.
We consider two cases, but the argument is almost the same for both.
Case I. Assume x1 /∈ Z. For convenience, let x
∗
1 = ⌊x1⌋, and let
I = {x∗1, x
∗
1 − 1, x
∗
1 − 2, . . . , x
∗
1 −
(
⌊n/2⌋ − 1
)
}.
For all x ∈ I , it is obvious that x ≤ x∗1 < x1. Also, we have
x ≥ x∗1 −
(
⌊n/2⌋ − 1
)
> (x1 − 1)−
(
(n/2)− 1
)
=
x2
2
.
So the set { (x, 1) | x ∈ I } is contained in the interior of T , and the slope of each point in
this set is strictly smaller than the slope of any nonzero lattice point in T that is not in this
set, other than the points (1, 0), (2, 0), . . . , (n, 0) that are on the x-axis (and therefore have
slope 0). This implies that (x, 1) is in T0 for all x ∈ I . More precisely, since it is obvious
that the lattice point (x, 1) is primitive, we have
{A0(k) | 2 ≤ k ≤ ⌊n/2⌋+ 1 } = { (x, 1) | x ∈ I }.
Since 2x > x2 for all x ∈ I , this implies 2A0(k) /∈ T , so
h0(k) = 1 for 2 ≤ k ≤ ⌊n/2⌋+ 1.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 13
By the definition of ut(k), we then have
u0(k)− u0(k − 1) = 2h0(k) = 2
for these values of k, so
{u0(k) | 1 ≤ k ≤ ⌊n/2⌋+ 1 }
= {n− 1, n+ 1, n+ 3 . . . , n− 1 + 2⌊n/2⌋}
Hence, Theorem 3.21 tells us that H0(d) is a hamiltonian path for all d in this set. Since
δb
(
H0(d)
)
= 0 · o(a− b) + d = d,
and 2⌊n/2⌋ ≥ 2⌊(n− 1)/2⌋, this implies that we may let n′ = n− 1 and Pi = H0(i).
Case II. Assume x1 ∈ Z. This means that X1 = (x1, 1). Let
I = {x1, x1 − 1, x1 − 2, . . . , x1 − ⌊(n− 1)/2⌋ }.
For all x ∈ I , we have
x ≥ x1 − ⌊(n− 1)/2⌋ > x1 − (n/2) =
x2
2
,
so, as in the previous case, the points in { (x, 1) | x ∈ I } are in T , and have strictly smaller
slope than any other lattice points in T , other than the lattice points on the x-axis. Since
(x1, 1) = X1, this implies (x, 1) is in T1. More precisely, we have
{A1(k) | 1 ≤ k ≤ ⌊(n− 1)/2⌋+ 1 } = { (x, 1) | x ∈ I }.
Since (x, 1) is primitive (and (x1, 1) is on the boundary of T , not in the interior), this
implies that
h0(k) =
{
0 if k = 1
1 otherwise
for 1 ≤ k ≤ ⌊(n− 1)/2⌋+ 1.
Therefore H1(d) is a hamiltonian path for all d in
{u1(k) | 1 ≤ k ≤ ⌊(n− 1)/2⌋+ 1 } = {0, 2, 4, . . . , 2⌊(n− 1)/2⌋}.
Note that, since the y-coordinate of X1 = (x1, 1) is 1, it follows from the definition in
Notation 3.17 that m = |G : ⟨a− b⟩|, so
o(a− b) =
|G|
|G : ⟨a− b⟩|
=
mn
m
= n.
Therefore
δb
(
H1(d)
)
= 1 · o(a− b) + d = 1 · n+ d = n+ d.
Therefore, we may let n′ = n and Pi = H1(i).
14 Ars Math. Contemp. 25 (2025) #P2.10
Proposition 3.27 (Curran-Witte, cf. [7, (9.5) on page 68]). Assume that m, n, and e are as
in Notation 3.16. Also assume m ≥ 2 and n ≥ e.
For every integer k, such that n− 1 ≤ k ≤ m(n− 1), there is a hamiltonian path P in
C−→ay(G; a, b), such that
k ≤ δb(P ) ≤ k + 2
⌊
mn− 1
m+ n− e
⌋
.
Therefore, if n ≥ m+ e, then
k ≤ δb(P ) ≤ k + 2
⌊
n− 1
2
⌋
.
Proof. First, note that if n ≥ m+ e, then
mn− 1
m+ n− e
≤
mn− 1
2m
<
mn
2m
=
n
2
,
so
⌊
mn− 1
m+ n− e
⌋
≤
⌊
n− 1
2
⌋
.
Therefore, the final sentence of the proposition follows from the one that precedes it.
For convenience, let N = o(a− b) and M = |G|/N − 1. In this notation, we have
{ δb(P ) |P is a hamiltonian path }
= {u0(1), u0(2), . . . , u0(f0 − 1),
N + u1(1), N + u1(2), . . . , N + u1(f1 − 1),
2N + u2(1), 2N + u2(2), . . . , 2N + u2(f2 − 1),
...
MN + uM (1),MN + uM (2), . . . ,MN + uM (fM − 1)}.
Since u0(1) = n− 1 and
MN + uM (fM − 1) =
(
|G| −N
)
+
(
N − 1− hM (f)
)
= mn− 1− hM (f) ≥ mn−m = m(n− 1),
the assertion we wish to prove is that the difference between successive terms in this list is
never larger than 2⌊(mn− 1)/(m+ n− e)⌋+ 1. It therefore suffices to show
ut(k)− ut(k − 1) ≤ 2
⌊
mn− 1
m+ n− e
⌋
+ 1 for 2 ≤ k < ft (3.1)
and
N + ut+1(1)− ut(ft − 1) ≤ 2
⌊
mn− 1
m+ n− e
⌋
+ 1 for 0 ≤ t < M. (3.2)
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 15
To establish (3.1), note that (much as in [7, Case 1 on page 67]) we have
ut(k)− ut(k − 1) = 2hk
= 2
⌊
mn− 1
mxk + (n− e)yk
⌋
(where xk, yk ∈ Z
+)
≤ 2
⌊
mn− 1
m+ n− e
⌋ (
n− e ≥ 0
and xk, yk ≥ 1
)
.
To establish (3.2), we make three observations. First, note that, by definition, we have
ut+1(1) = ht+1(1). Second, we know from Lemma 3.23 that
ut(ft − 1) = N − 1− ht(f).
Third, we have At(ft) = At+1(1). (This is the primitive lattice point on the boundary
between Tt and Tt+1.) So
ht+1(1) = ht(f) =
⌊
mn− 1
mxf + (n− e)yf
⌋
≤
⌊
mn− 1
m+ n− e
⌋
.
Therefore
N + ut+1(1)− ut(ft − 1)
= N + ht(f)−
(
N − 1− ht(f)
)
= 2ht(f) + 1
≤ 2
⌊
mn− 1
m+ n− e
⌋
+ 1.
To provide a convenient reference, we now restate Proposition 3.27 for the special case
where C−→ay(G; a, b) is the Cartesian product of two directed cycles:
Corollary 3.28. Let {e1, e2} be the standard generating set of Zm × Zn, and assume 2 ≤
m ≤ n. If 1 ≤ k ≤ m(n−1), then there is a hamiltonian path P in C−→ay(Zm×Zn; e1, e2),
such that
k ≤ δe2(P ) ≤ k + 2
⌊
n− 1
2
⌋
.
Proof. We have e = 0, so the assumption that m ≤ n implies n ≥ m + e. Therefore, if
k ≥ n− 1, then the desired conclusion is obtained from the final sentence in the statement
of Proposition 3.27.
So we may now assume k < n− 1. Then, since
n− 1− k ≤ n− 2 ≤ 2⌊(n− 1)/2⌋,
we may let P = H0(n−1). (Since u0(1) = n−1, we know that H0(n−1) is a hamiltonian
path. We also have δe2
(
H0(n− 1)
)
= n− 1.)
4 Cartesian product of two directed cycles
In this section, we show that the Cartesian product of two directed cycles has two arc-
disjoint hamiltonian paths. We start with the following lemma.
16 Ars Math. Contemp. 25 (2025) #P2.10
Lemma 4.1. Let {a, b} be a 2-element generating set of a finite abelian group G, and let
k and ℓ′ be integers such that
0 ≤ k, ℓ′ < |G| and |k − ℓ′| ≤ 1.
Then there exist two arc-disjoint spanning quasi-paths P and P ′ in C−→ay(G; a, b), such that
δb(P ) = k and δa(P
′) = ℓ′.
Proof. By Lemma 3.14, there is a spanning quasi-path P , such that δb(P ) = k. Let ι and τ
be the initial vertex and terminal vertex of P , respectively.
Let P be the complement of P in C−→ay(G; a, b). (Thus, a subdigraph of C−→ay(G; a, b) is
arc-disjoint from P if and only if it is contained in P .) The vertices that have some a-edge
as an out-edge in P are precisely the vertices that do not travel by a in P ; this consists of
the vertices that travel by b in P , plus the terminal vertex τ . Hence, we have
δa(P ) = δb(P ) + 1 = k + 1.
Note that P is not a spanning quasi-path, because the indegree of ι is 2, and the outde-
gree of τ is 2. (The indegree and outdegree of each of the other vertices is 1.)
Case I. Assume P does not have a directed edge from τ to ι. If we construct a subdigraph P ′
of P by removing either of the two in-edges of ι and either of the two out-edges of τ , then
P ′ will be a spanning quasi-path. (The initial vertex of the removed in-edge will have
outdegree 0, and the terminal vertex of the removed out-edge will have indegree 0.) We
can freely decide whether neither, precisely one, or both of the directed edges we remove
is an a-edge, so we can construct P ′ so that δa(P
′) is any desired element of
{δa(P ), δa(P )− 1, δa(P )− 2} = {k + 1, k, k − 1}.
Since |k− ℓ′| ≤ 1, this means we can construct P ′ so that δa(P
′) is equal to ℓ′, as desired.
Case II. Assume (τ, ι) is a directed edge of P . Assume, without loss of generality, that
(τ, ι) is a b-edge (by interchanging a and b if necessary).
If ℓ′ = k − 1, then ℓ′ = δa(P ) − 2, so we construct P
′ by removing two a-edges
from P : the out-edge of τ that is labelled a, and the in-edge of ι that is labelled a. (These
are two different directed edges, because we have assumed that the directed edge (τ, ι) is a
b-edge.)
We may now assume that
ℓ′ ∈ {k + 1, k} = {δa(P ), δa(P )− 1}.
Removing the b-edge (τ, ι) from P results in a spanning subdigraph P ∗ in which the inde-
gree and outdegree of every vertex is 1, so each connected component of P ∗ is a directed
cycle. Hence, removing any directed edge from P ∗ will result in a spanning quasi-path P ′.
Thus, if P ∗ has both an a-edge and a b-edge, then we can choose which type of directed
edge to remove, so we can arrange for δa(P
′) to be either element of
{δa(P ), δa(P )− 1},
so we can certainly arrange for δa(P
′) to be equal to ℓ′.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 17
Since P ∗ does have a-edges (such as the out-edge of τ ), we may now assume that every
directed edge of P ∗ is an a-edge. This means that δa(P
∗) = |G|, so
k = δa(P )− 1 = δa(P
∗)− 1 = |G| − 1.
Since ℓ′ ∈ {k+1, k} and ℓ′ < |G|, this implies that ℓ′ = k = |G|−1. So we can construct
the desired spanning quasi-path P ′ by removing one of the a-edges of P ∗.
Remark 4.2. It is easy to see that the converse of Lemma 4.1 is true. Namely, if P and P ′
are arc-disjoint spanning quasi-paths in C−→ay(G; a, b), then we have
|δb(P )− δa(P
′)| ≤ 1.
Proposition 4.3. Let {a, b} be a 2-element generating set of a finite abelian group G,
and assume that C−→ay(G; a, b) has hamiltonian paths P and P ′ that satisfy either of the
following two equivalent conditions:
(1) |δb(P )− δa(P
′)| ≤ 1, or
(2) |δb(P ) + δb(P
′)| ∈ {|G|, |G| − 1, |G| − 2}.
Then C−→ay(G; a, b) has two arc-disjoint hamiltonian paths.
Proof. First, note that the equivalence of the two conditions is immediate from the obser-
vation that δa(P
′) = |G| − 1− δb(P
′). Therefore, we may assume that (1) holds.
Lemma 4.1 tells us that there exist two arc-disjoint spanning quasi-paths P0 and P
′
0 in
C−→ay(G; a, b), such that δb(P0) = δb(P ) and δa(P
′
0) = δa(P
′). Since δb(P0) = δb(P ) and
δb(P
′
0) = |G| − 1− δa(P
′
0) = |G| − 1− δa(P
′) = δb(P
′),
we see from Corollary 3.12 that P0 and P
′
0 are hamiltonian paths. Since P0 and P
′
0 are
arc-disjoint, this completes the proof.
We are now ready to prove the main theorem of the paper.
Theorem 4.4. If C1 and C2 are directed cycles (of length ≥ 2), then the Cartesian product
C1 □ C2 has two arc-disjoint hamiltonian paths.
Proof. Let m and n be the lengths of C1 and C2, so
C1 □ C2 ∼= C
−→ay(Zm × Zn; e1, e2),
where {e1, e2} = {(1, 0), (0, 1)} is the standard generating set of the abelian group Zm ×
Zn. Assume without loss of generality that m ≤ n. Then Corollary 3.28 provides a
hamiltonian path P ′, such that
mn− n− 2
⌊
n− 1
2
⌋
≤ δe2(P
′) ≤ mn− n.
Since δe1(P
′) = mn− 1− δe2(P
′), we have
n− 1 ≤ δe1(P
′) ≤ n− 1 + 2
⌊
n− 1
2
⌋
.
Then it is immediate from Proposition 3.26 that there is a hamiltonian path Pi, such that we
have |δe2(Pi)−δe1(P
′)| ≤ 1. The desired conclusion now follows from Proposition 4.3(1).
18 Ars Math. Contemp. 25 (2025) #P2.10
The above result shows that there is always at least one pair of arc-disjoint hamiltonian
paths. In fact, the number of such pairs is large if both directed cycles in the product have
large length:
Proposition 4.5 (see Corollary 6.4 below). Let N(m,n) be the number of (unordered)
pairs P, P ′ of arc-disjoint hamiltonian paths in the Cartesian product of a directed cycle
of length m and a directed cycle of length n. If m and n are sufficiently large, then
N(m,n) >
m2 n2
10
.
If C−→ay(G; a, b) has two arc-disjoint hamiltonian cycles, then removing a directed edge
from each of these hamiltonian cycles yields two arc-disjoint hamiltonian paths P and P ′.
Note that the initial vertex ι′ of P ′ is completely arbitrary, even after the path P has been
chosen, because there is no restriction on the directed edge that is removed. (The terminal
vertex of P ′ is also arbitrary, but it is determined by the choice of ι′.) In all other situations,
the following observation shows that P ′ is almost completely determined by the choice
of P .
Proposition 4.6. Suppose P and P ′ are two arc-disjoint hamiltonian paths in C−→ay(G; a, b).
Let ι and ι′ be the initial vertices of these hamiltonian paths, and let τ and τ ′ be the terminal
vertices. Assume there is no directed edge from τ to ι (or, equivalently by Remark 2.4, that
there is no directed edge from τ ′ to ι′). Then:
(1) ι′ ∈ {τ + a, τ + b},
(2) τ ′ ∈ {ι− a, ι− b},
(3) ι and τ ′ are in the same coset of ⟨a− b⟩,
(4) ι′ and τ are in the same coset of ⟨a− b⟩, and
(5) there are at most two hamiltonian paths that are arc-disjoint from P .
Proof. Let P be the complement of P in C−→ay(G; a, b), as in the proof of Lemma 4.1. Note
that P ′ can be obtained from P by removing one of the two in-edges of ι and one of the two
out-edges of τ . (Since there is no directed edge from τ to ι, the in-edges of ι are distinct
from the out-edges of τ .) Thus, we have (1) ι′ ∈ {τ + a, τ + b} and (2) τ ′ ∈ {ι− a, ι− b}.
Combining these with Lemma 3.9(2)(a) yields (3) and (4).
(5) Since there are only two possible choices for the in-edge of ι to remove, and two
possible choices for the out-edge of τ to remove, it is obvious that no more than 4 hamilto-
nian paths are arc-disjoint from P .
To complete the proof, we show there cannot be more than one hamiltonian path that
starts at τ +a and is arc-disjoint from P . (By symmetry, there is also no more than one that
starts at τ + b). Suppose, for a contradiction, that the terminal vertex of P ′ is τ ′ = ι − a
and the terminal vertex of some other arc-disjoint hamiltonian path P ′′ is τ ′′ = ι− b. Then
τ ′ − τ ′′ = b− a. Thus, if d is the number of vertices in τ ′ + ⟨a− b⟩ that travel by b in P ′,
then the number of vertices in this coset that travel by b in P ′′ is d+ 1.
Note that if a non-terminal coset travels by a in P , then it must travel by b in both P ′
and P ′′ (since P ′ and P ′′ are arc-disjoint from P ). Conversely, if a non-terminal coset
travels by b in P ′ or P ′′, then it must travel by a in P . Therefore, a non-terminal coset
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 19
travels by b in P ′ if and only if it travels by b in P ′′. Hence, if we let t be the number of
non-terminal cosets that travel by b in P ′, then t is also the number of non-terminal cosets
that travel by b in P ′′.
Since P ′ and P ′′ are hamiltonian paths, we see from Proposition 3.11 that Ht(d) and
Ht(d + 1) are hamiltonian paths. So Theorem 3.21 tells us that d = ut(k
′) and d + 1 =
ut(k
′′), for some k′ and k′′. However, it is clear from Notation 3.19(4) that two values of
ut(∗) cannot differ by 1. This is a contradiction.
The following observation will be used in the proof of Proposition 5.2(2):
Corollary 4.7. Let {a, b} be a 2-element generating set of a finite abelian group G, such
that
• C−→ay(G; a, b) does not have a hamiltonian cycle,
• |G| is even, and
• |G : ⟨a− b⟩| is odd.
If P and P ′ are two arc-disjoint hamiltonian paths in C−→ay(G; a, b), with initial vertices ι
and ι′, and terminal vertices τ and τ ′, then
ι′ + τ ′ = ι+ τ.
Proof. Since C−→ay(G; a, b) is not hamiltonian, there is no directed edge from τ to ι, so we
see from Proposition 4.6 that ι′ ∈ {τ + a, τ + b} and τ ′ ∈ {ι− a, ι− b}, so
(ι′ + τ ′)− (ι+ τ) ∈ {0,±(a− b)}.
Therefore, since a − b /∈ 2G (because |G| is even and |G : ⟨a − b⟩| is odd), it will suffice
to show ι′ + τ ′ ≡ ι+ τ (mod 2G).
Write
δb(P ) = t o(a− b) + d and δb(P
′) = t′ o(a− b) + d′,
so we see from Theorem 3.21(3) (and Proposition 3.11) that d = ut(k) and d
′ = ut′(k
′)
for some k and k′.
Also recall that Lemma 3.9(2)(a) tells us τ ∈ ι − a + ⟨a − b⟩. Since (from above)
τ ′ ∈ {ι−a, ι−b}, this implies τ+⟨a−b⟩ = τ ′+⟨a−b⟩, which means that P and P ′ have
the same terminal coset. So they also have the same non-terminal cosets. Since P and P ′
are arc-disjoint, each non-terminal coset travels by b in either P or P ′, but not both. So
t+ t′ = |G : ⟨a− b⟩| − 1.
Since |G : ⟨a− b⟩| − 1 is even, we conclude that t ≡ t′ (mod 2), so Corollary 3.25(2)
tells us that d ≡ d′ (mod 2). We know from Lemma 3.9(2)(a) that
ι− τ = d(a− b) + a and ι′ − τ ′ = d′(a− b) + a,
so this implies
ι′ + τ ′ ≡ ι′ − τ ′ (mod 2G)
= d′(a− b) + a
≡ d(a− b) + a (mod 2G)
= ι− τ
≡ ι+ τ (mod 2G).
20 Ars Math. Contemp. 25 (2025) #P2.10
5 Cartesian product of more than two directed cycles
Corollary 5.1. If C1, C2, . . . , Cr are directed cycles (of length ≥ 2), and r ≥ 4, then the
Cartesian product C1 □ C2 □ · · ·□ Cr has two arc-disjoint hamiltonian paths.
Proof. This argument is very easy (and classical: see the proof of [7, Lemma 9.2], for
example). Theorem 3.15 provides a hamiltonian cycle C in the Cartesian product C1 □
C2 □ · · ·□Cr−1. Then C1 □C2 □ · · ·□Cr has a spanning subdigraph that is isomorphic
to C □ Cr, and Theorem 4.4 provides two arc-disjoint hamiltonian paths in this spanning
subdigraph. Hamiltonian paths in a spanning subdigraph are also hamiltonian paths in the
entire digraph.
We do not know whether the Cartesian product of three directed cycles always has two
arc-disjoint hamiltonian paths, but we can prove that the paths exist in most cases:
Proposition 5.2. Assume C1, C2, and C3 are directed cycles (of length ≥ 2). If either
(1) the Cartesian product of two of the directed cycles has a hamiltonian cycle, or
(2) the lengths of the three directed cycles do not all have the same parity (i.e., if there
is a directed cycle of even length and a directed cycle of odd length), or
(3) at least one of the directed cycles has length 2,
then C1 □ C2 □ C3 has two arc-disjoint hamiltonian paths.
Proof. (1) This uses the same simple argument as the proof of Corollary 5.1. Assume,
without loss of generality, that C1 □ C2 has a hamiltonian cycle C. Then C □ C3 is
isomorphic to a spanning subdigraph of C1 □ C2 □ C3, and Theorem 4.4 provides two
arc-disjoint hamiltonian paths in this spanning subdigraph.
(2) Write
C1 □ C2 □ C3 = C
−→ay(Zm × Zn × Zr; e1, e2, e3),
where {e1, e2, e3} is the standard generating set of Zm × Zn × Zr. For 0 ≤ i < r, let Xi
be the subdigraph that is induced by Zm × Zn × {i}, so Xi ∼= C
−→ay(Zm × Zn; e1, e2).
By Theorem 4.4, we may let P0 and P
′
0 be two arc-disjoint hamiltonian paths in X0. Let
ι0 and ι
′
0 be their initial vertices, and let τ0 and τ
′
0 be their terminal vertices. By assumption,
we may assume that m and n have opposite parity (by permuting the factors). We may also
assume X0 does not have a directed hamiltonian cycle, for otherwise (1) applies. Then we
see from Corollary 4.7 that
τ − ι′ = τ ′ − ι.
Therefore, if we let ∆ := τ − ι′, then
τ = ι′ +∆ and τ ′ = ι+∆.
For 0 ≤ i < r, let Pi and P
′
i be the translates of P0 and P
′
0 by (∆i, i), so Pi and P
′
i are
arc-disjoint hamiltonian paths in Xi.
Let (ιi, i) and (ι
′
i, i) be the initial vertices of Pi and P
′
i and let (τi, i) and (τ
′
i , i) be their
terminal vertices. Then
(τi, i) + e3 = (τ +∆i, i+ 1) =
(
(ι′ +∆) +∆i, i+ 1
)
)
=
(
ι′ +∆(i+ 1), i+ 1
)
= (ι′i+1, i+ 1)
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 21
is the initial vertex of P ′i+1. This means there is a directed e3-edge from the terminal vertex
of Pi to the initial vertex of P
′
i+1. Similarly, since τ
′ = ι+∆, we see that there is a directed
e3-edge from the terminal vertex of P
′
i to the initial vertex of Pi+1.
Therefore, if we start with the union of all of the paths Pi for i even and P
′
j for j odd,
then we can add certain e3-edges to construct a hamiltonian path P in C1□C2□C3. Sim-
ilarly, we can construct a hamiltonian path P ′ by adding appropriate e3-edges to the union
of all of the paths Pi for i even and P
′
j for j odd. (In the terminology of [7, Definition 9.4],
P and P ′ are “laminated” hamiltonian paths.) It is clear from the construction that these
hamiltonian paths are arc-disjoint.
(3) Assume, without loss of generality, that C1 has length 2. If C2 has even length, then
(since C1 has length 2) it is easy to see (and well known) that C1 □ C2 has a hamiltonian
cycle, so (1) applies. If C2 has odd length, then (2) applies.
6 2-generated Cayley digraphs on finite abelian groups
Proposition 6.1. Conjecture 1.3 is true in all cases where the arc-forcing subgroup ⟨a− b⟩
has even index in G.
Proof. Let t = |G : ⟨a−b⟩|/2, and let d = ut(1), so by Theorem 3.21 we know that Ht(d)
is a hamiltonian path. By Corollary 3.24, we have
ut−1(ft−1 − 1) = o(a− b)− 1− d,
so Ht−1
(
o(a− b)− 1− d
)
is also a hamiltonian path. Furthermore,
δb
(
Ht(d)
)
+ δb
(
Ht−1
(
o(a− b)− 1− d
))
=
[
t o(a− b) + d
]
+
[
(t− 1) o(a− b) +
(
o(a− b)− 1− d
)]
= 2t o(a− b)− 1
= |G| − 1.
Therefore, Proposition 4.3(2) provides two arc-disjoint hamiltonian paths.
The open cases of Conjecture 1.3 reduce to two types of examples (and G = Zn is
cyclic in both types):
Proposition 6.2. It suffices to prove Conjecture 1.3 for the following two families of Cayley
digraphs:
(1) C−→ay(Zk; a, a+ 1), where k, a ∈ Z
+ (and neither a nor a+ 1 is divisible by k), and
(2) C−→ay(Zk;−a, a + 1), where k, a ∈ Z
+, and k is divisible by 2a + 1 (and neither a
nor a+ 1 is divisible by k).
Proof. Assume that {a, b} is a 2-element generating set of a finite abelian group G. For
convenience, let F = ⟨a− b⟩. We may assume that |G : F| is odd, for otherwise Proposi-
tion 6.1 applies. Write |G : F| = 2t+ 1 and let c = ta+ tb.
Case I. Assume that c + a ̸= 0 and c + b ̸= 0. Let a′ = c + a and b′ = c + b. Since
a′, b′ ∈ ⟨a− b⟩, and a′ − b′ = a− b, we have
C−→ay(F ; a′, b′) ∼= C−→ay(Zk; ℓ, ℓ+ 1),
22 Ars Math. Contemp. 25 (2025) #P2.10
where k = o(a − b) and b′ = ℓ(a − b). This Cayley digraph is listed in (1), so we may
assume that it has two arc-disjoint hamiltonian paths. Thus (by Proposition 3.11), there
exist d and d′, such that H0(d) and H0(d
′) are hamiltonian paths in C−→ay(F ; c + a, c + b),
and (by Remark 4.2) we have d + d′ = o(a − b) − 1 + ϵ, where |ϵ| ≤ 1. Then Ht(d) and
Ht(d
′) are hamiltonian paths in C−→ay(G; a, b) (by the “skewed generators argument,” see the
proof of [19, Lemma 2.6]).
Since
δb
(
Ht(d)
)
+ δb
(
Ht(d
′)
)
=
(
t o(a− b) + d
)
+
(
t o(a− b) + d′
)
= 2t o(a− b) + (d+ d′)
=
(
|G : F| − 1
)
|F|+
(
o(a− b)− 1 + ϵ
)
= |G| − 1 + ϵ,
we conclude from Proposition 4.3(2) that C−→ay(G; a, b) has two arc-disjoint hamiltonian
paths.
Case II. Assume that either c+ a = 0 or c+ b = 0. Assume without loss of generality that
c+ a = 0. For convenience, let w = a− b and k = |G|. We have
0 = c+ a = (ta+ tb) + a = (t+ 1)a+ tb = (2t+ 1)b+ (t+ 1)w,
so (2t+ 1)b = −(t+ 1)w. Also note that
o(w) = o(a− b) =
k
2t+ 1
(so k is divisible by 2t+ 1). Hence, as an abelian group, G has the presentation
G =
〈
b, w
∣
∣
∣
(2t+ 1)b = −(t+ 1)w,
k
2t+ 1
w = 0
〉
,
so G, b, and w are uniquely determined (up to isomorphism) by k and t, and the assump-
tions that |G| = k, ta+ tb+ a = 0, and w = a− b has order k/(2t+1). Since a = b+w,
it is also uniquely determined.
On the other hand, it is clear that letting G = Zk, a = 1 + t, b = −t, and w = 2t + 1
provides an example (for any k and t, such that k is divisible by 2t+1). Hence, we conclude
that C−→ay(G; a, b) is isomorphic to the Cayley digraph that is listed in (2).
It is well known [11, Theorem 459, page 541] that the probability that two random
integers are relatively prime is 6/π2 = 0.6079 · · · . This has the following elementary
consequence:
Lemma 6.3 ([7, Theorem 8.5]). If N(m,n, e) is the number of primitive lattice points in
the interior of the triangle T (m,n, e), then
lim
m,n→∞
N(m,n, e)
Area
(
T (m,n, e)
) =
6
π2
.
Corollary 6.4. Let N(G; a, b) be the number of (unordered) pairs {P, P ′} of arc-disjoint
hamiltonian paths in C−→ay(G; a, b). If m = o(a) and n = |G|/o(a) are sufficiently large,
then
N(G; a, b) >
|G|2
10
.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 23
Proof. Let
R =
{
t o(a− b) + ut(k)
∣
∣
∣
0 ≤ t < |G : ⟨a− b⟩|, 1 ≤ k < ft,
At(k) is in the interior of T (m,n, e)
}
⊆ {1, 2, . . . , |G| − 2}.
Also let
R+ = R ∪ (R+ 1) and R− = R ∪ (R− 1).
We claim that |r − r′| ≥ 2 for all distinct r, r′ ∈ R. To see this, first recall that
ut(k)− ut(k − 1) = 2ht(k) ≥ 2.
Second, note that, by Corollary 3.24, we have
(
t o(a− b) + ut(1)
)
−
(
(t− 1) o(a− b) + ut−1(ft−1 − 1)
)
= 2ut(1) + 1.
If At(1) is in the interior of T (m,n, e), then ht(1) ≥ 1. Since ut(1) = ht(1), this implies
2ut(1) + 1 > 2, which completes the proof of the claim.
From the claim, we see that
#R+ = #R− = 2 ·#R ≈ 2 ·
6
π2
Area
(
T (m,n, e)
)
=
6
π2
|G|.
Therefore
#
(
R+ ∩ (|G| − 1−R−)
)
≥ #R+ +#R− − |G|
≈ 2 ·
6
π2
|G| − |G|
> 0.2|G|.
For each r ∈ R, we know from the definition of R that there is a hamiltonian path P , such
that δb(P ) = r. Hence, for each r in the above intersection, we have hamiltonian paths P
and P ′, such that
δb(P ) ∈ {r, r − 1}
and
δb(P
′) ∈ {|G| − 1− r, |G| − r}.
Therefore
δb(P ) + δb(P
′) ∈ {|G|, |G| − 1, |G| − 2}|,
so we see from (the proof of) Proposition 4.3(2) that these hamiltonian paths can be made
arc-disjoint.
This provides more than 0.2|G| ordered pairs of arc-disjoint hamiltonian paths. So the
number of unordered pairs is more than 0.1|G|. Furthermore, if {P1, P
′
1} and {P2, P
′
2} are
two such pairs, then we know from the construction that δb(P1) /∈ {δb(P2), δb(P
′
2)}, so
{P1, P
′
1} is not a translate of {P2, P
′
2}. Therefore, since each pair has |G| translates, the
total number of unordered pairs of arc-disjoint hamiltonian paths is more than 0.1|G|2.
24 Ars Math. Contemp. 25 (2025) #P2.10
(Actually, there is a slight technical issue that the translates of {P, P ′} might not all
be distinct. However, this can only happen if P ′ is a translate of P , which implies that
δb(P ) = δb(P
′). Since P and P ′ are arc-disjoint, this determines δb(P ) up to an error of at
most 1. So this problem is avoided by deleting a small number of values of r that are near
|G|/2.)
The following variant of Lemma 6.3 provides an explicit lower bound on m and n when
6/π2 is replaced with the smaller constant 1/2:
Lemma 6.5 (cf. [7, Theorem 8.5]). Let N(m,n, e) be the number of primitive lattice points
in the interior of the triangle T (m,n, e). If m,n ≥ 300, then
N(m,n, e) >
1
2
Area
(
T (m,n, e)
)
.
Sketch of proof (cf. [7, proof of Theorem 8.5]). For every triangle T , let N(T ) (resp. P (T ))
be the number of lattice points (resp. primitive lattice points) that are in the interior of T
and are not on the y-axis.
Case I. Assume −n ≤ e ≤ n. Then min
(
|x|, |y|
)
≤ min(m,n) for all (x, y) ∈
T (m,n, e), so we have N
(
1
k
T
)
= 0 for all k > min(m,n). Also, it is elementary to
see that
∣
∣
∣
∣
N
(
1
k
T
)
−Area
(
1
k
T
)∣
∣
∣
∣
≤
2 max(m,n)
k
for all k ∈ Z+. Therefore, letting A = Area(T ) and min = min(m,n), we have
P (T ) =
min
∑
k=1
µ(k)N
(
1
k
T
)
=
min
∑
k=1
µ(k)
(
A
k2
±
2 max(m,n)
k
)
= A
(
6
π2
−
∑
k>min
µ(k)
k2
)
± 2 max(m,n)(1 + logmin)
> A
(
6
π2
−
1
min
)
− 2max(m,n)(1 + logmin)
= A
(
6
π2
−
1
min
−
4(1 + logmin)
min
)
.
Since min ≥ 300 (in fact, min ≥ 252 would suffice), we have
1
min
+
4(1 + logmin)
min
< 0.1079,
so we conclude that P (T ) > 0.5A.
Case II. The general case. By applying an appropriate element
(
1 ∗
0 1
)
of SL(2,Z), we
may assume n − m ≤ e ≤ n. Since we may assume that Case I does not apply, we then
have n−m ≤ e < −n. The y-axis divides T (m,n, e) into two smaller triangles T1 and T
′
1.
Letting p = mn/(n− e) ≥ n, we have
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 25
• the vertices of T1 are (0, 0), (n, 0), and (0, p), and
• the vertices of T ′1 are (0, 0), (0, p), and (e,m).
Since p ≥ n, we see that Case I applies to the triangle T1, so
P (T1) >
1
2
Area(T1).
Now, rotating T ′1 by 90
◦ clockwise around the origin yields the triangle T ′′1 with vertices
(0, 0), (p, 0) and (m,−e). Since p > n and −e > n, this triangle satisfies the hypotheses
of the proposition. Therefore, either Case I applies to T ′′1 , or the argument in the preceding
paragraph divides T ′′1 into two triangles T2 and T
′
2, such that Case I applies to T2, and
T ′2 satisfies the hypotheses of the proposition (after applying an appropriate element of
SL(2,Z)). Continuing in this way, we see that T (m,n, e) can be decomposed into a finite
union
T = T1 ∪ T2 ∪ · · · ∪ Tℓ
such that Case I applies to each Ti (after applying an appropriate element of SL(2,Z)).
Therefore, we have P (Ti) >
1
2
Area(Ti) for each i, so
P (T ) ≥
∑
i
P (Ti) >
∑
i
1
2
Area(Ti) =
1
2
Area(T ).
Proposition 6.6. Conjecture 1.3 is true in all cases where we have either |G : ⟨a⟩| ≥ 600
or |G : ⟨b⟩| ≥ 600.
Proof. Assume, without loss of generality, that |G : ⟨a⟩| ≥ 600. We consider two cases.
Case I. Let m ≥ 300. Then one can see that by Lemma 6.5 we have 2N(m,n, e) > |G|/2,
so the argument in the proof of Corollary 6.4 establishes that the number of pairs of arc-
disjoint hamiltonian paths is greater than 0.
Case II. Assume m < 300. Since e < m and n = |G : ⟨a⟩| ≥ 600, this implies n > m+e.
Hence, we see from Proposition 3.27 that for all k, such that n− 1 ≤ k ≤ m(n− 1), there
is a hamiltonian path P in C−→ay(G; a, b), such that
k ≤ δb(P ) ≤ k + 2
⌊
n− 1
2
⌋
.
The desired conclusion can now be obtained by combining this with Proposition 3.26, as
in the proof of Theorem 4.4.
7 2-generated Cayley digraphs on infinite abelian groups
In this section, we study the natural analogue of Conjecture 1.3 for infinite digraphs. This
is an addition to the literature on two-way infnite hamiltonian paths in Cayley (di)graphs
on infinite abelian groups, which includes:
• Every connected Cayley graph on any countably infinite abelian group has a two-way
infinite hamiltonian path [21, Theorem 1].
26 Ars Math. Contemp. 25 (2025) #P2.10
• A connected, 4-valent, infinite, circulant graph Cay(Z;±a,±b) has two edge-disjoint
two-way infinite hamiltonian paths if and only if a+ b is even [5, Theorem 9].
• A connected, 4-valent Cayley graph on an infinite abelian group G has two edge-
disjoint two-way infinite hamiltonian paths if and only if, for every partition of G
into two infinite subsets X and Y , the number of edges from X to Y is either infinite
or even [8, Theorem 2].
• Every connected Cayley graph of infinite valency on a finitely generated abelian
group can be decomposed into edge-disjoint two-way infinite hamiltonian paths [5,
Theorem 8].
• Every finitely generated, infinite abelian group G has an irredundant generating set S,
such that C−→ay(G;S) has a two-way infinite hamiltonian path [15, Corollary 5.2].
• Assume G is an abelian group that has no cyclic subgroup of finite index, and X is
a Cayley digraph on G that has infinite valency. Then X can be decomposed into
arc-disjoint two-way infinite hamiltonian paths if and only if X is strongly connected
([16, Theorem 4.1] and [15, proofs of Theorems 5.3 and 6.3]).
The foundation of our results is the observation that the basic theory of the arc-forcing
subgroup easily generalizes to the infinite case:
Lemma 7.1 (cf. Lemma 3.9(1)). Let {a, b} be a 2-element subset of a group G (which
may be infinite). If P is a spanning subdigraph of C−→ay(G; a, b), such that the indegree and
outdegree of every vertex is 1, then every right coset of ⟨a− b⟩ either travels by a or travels
by b.
This has the following easy consequence, which is the infinite analogue of Theo-
rem 2.1(2):
Proposition 7.2. Let {a, b} be a 2-element generating set of an abelian group G (such that
a ̸= b), and let I = |G : ⟨a− b⟩|.
(1) If C−→ay(G; a, b) has a two-way infinite hamiltonian path, then I < ∞.
(2) Suppose P is a two-way infinite hamiltonian path in C−→ay(G; a, b). If k is the number
of cosets of ⟨a − b⟩ that travel by a, and ℓ is the number of cosets of ⟨a − b⟩ that
travel by b, then k + ℓ = I , and ⟨ka+ ℓb⟩ = ⟨a− b⟩.
(3) Conversely, suppose k + ℓ = I < ∞ and ⟨ka+ ℓb⟩ = ⟨a− b⟩. If P is any spanning
subdigraph of C−→ay(G; a, b), such that the outdegree of every vertex is 1, and exactly
k cosets of ⟨a− b⟩ travel by a and exactly ℓ cosets of ⟨a− b⟩ travel by b, then P is a
two-way infinite hamiltonian path in C−→ay(G; a, b).
Proof (cf. [22, Theorem 4]). (1) Let P = . . . , v−2, v−1, v0, v1, v2, . . . be a two-way infi-
nite hamiltonian path in C−→ay(G; a, b). Assume, without loss of generality, that v0 = 0.
Then, since a ≡ b (mod ⟨a − b⟩), we have vi ∈ ia + ⟨a − b⟩ for every i ∈ Z. If I = ∞,
this implies that each vi is in a different coset of ⟨a− b⟩. Since {vi}i∈Z is a list of all of the
elements of G, we conclude that vi is the only element of its coset, so each coset has only
one element. This means |⟨a− b⟩| = 1, which contradicts the fact that a ̸= b.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 27
(2) Since every coset of ⟨a−b⟩ travels by either a or b (but not both), we have k+ℓ = I .
Now, write P = . . . , v−2, v−1, v0, v1, v2, . . ., and assume, without loss of generality, that
v0 = 0. Then
vi ∈ ⟨a− b⟩ ⇐⇒ i is divisible by I
and
vjI = j(ka+ ℓb) for every j ∈ Z.
Since {vi}i∈Z is a list of all of the elements of G, this implies that {j(ka+ ℓb)} is a list of
all of the elements of ⟨a− b⟩, which means that ka+ ℓb generates ⟨a− b⟩.
(3) Let . . . , v−2, v−1, v0, v1, v2, . . . be the path component of P that contains 0 (with
v0 = 0). We wish to show that every vertex is in this component.
Suppose v ∈ G. Since
⟨a, ka+ ℓb⟩ = ⟨a, a− b⟩ = ⟨a, b⟩,
we have v ∈ ia+ ⟨ka+ ℓb⟩ for some i ∈ Z. Since vi ≡ ia (mod a− b), this implies there
exists j ∈ Z, such that
v = vi + j(ka+ ℓb) = vi+jI .
The following observation could easily be proved directly, but we present it as a simple
application of Proposition 7.2.
Corollary 7.3. If {e1, e2} is the standard generating set of Z×Zm (and m ≥ 2), then the
Cayley digraph C−→ay(Z×Zm; e1, e2) has a unique two-way infinite hamiltonian path, up to
translations.
Proof. (existence) Let k = 1 and ℓ = m− 1. Then
k + ℓ = m = |Z× Zm : ⟨e1 − e2⟩ |
and
ke1 + ℓe2 = e1 + (m− 1)e2 = e1 − e2,
so Proposition 7.2(3) tells us that C−→ay(Z× Zm; e1, e2) has a two-way infinite hamiltonian
path.
(uniqueness) Let k and ℓ be as in Proposition 7.2(2). Then
⟨(k, ℓ)⟩ = ⟨ke1 + ℓe2⟩ = ⟨e1 − e2⟩ = ⟨(1,−1)⟩,
so the projection of this subgroup to Z is surjective. We conclude that k = 1. This means
that precisely one coset of ⟨a−b⟩ travels by e1 (and all other cosets travel by e2). Therefore,
a two-way infinite hamiltonian path is determined by choosing which coset of ⟨e1 − e2⟩
travels by e1. Since cosets are translates of each other, this implies that all two-way infinite
hamiltonian paths are translates of each other.
Corollary 7.4. Let {a, b} be a 2-element generating set of an abelian group G (such that
a ̸= b), and let I = |G : ⟨a − b⟩|. The digraph C−→ay(G; a, b) has two arc-disjoint two-way
infinite hamiltonian paths if and only if I < ∞ and there exist k, ℓ ∈
{
0, 1, . . . , I
}
, such
that k + ℓ = I and
⟨a− b⟩ = ⟨ka+ ℓb⟩ = ⟨ℓa+ kb⟩.
28 Ars Math. Contemp. 25 (2025) #P2.10
Proof. (⇒) Let P be a two-way infinite hamiltonian path, and let k and ℓ be as in Proposi-
tion 7.2(2), so k + ℓ = I and ⟨ka+ ℓb⟩ = ⟨a− b⟩. If P ′ is a two-way infinite hamiltonian
path that is arc-disjoint from P , then the cosets that travel by a in P ′ are the cosets that
travel by b in P , so the number of cosets that travel by a in P ′ is ℓ, and the number of cosets
that travel by b is k. So Proposition 7.2(2) tells us that ⟨ℓa+ kb⟩ = ⟨a− b⟩.
(⇐) Choose k cosets of ⟨a − b⟩. Let P be the spanning subdigraph in which the
outdegree of every vertex is 1, and these particular k cosets of ⟨a− b⟩ travel by a, and the
other ℓ cosets travel by b. Then let P ′ be the spanning subdigraph in which the outdegree of
every vertex is 1, and the k chosen cosets of ⟨a− b⟩ travel by b, and the other ℓ cosets travel
by a. It is clear from the construction that P ′ is arc-disjoint from P (since a vertex travels
by a in P ′ if and only if it travels by b in P ). Furthermore, we see from Proposition 7.2(3)
that P and P ′ are two-way infinite hamiltonian paths.
This can be made much more explicit:
Proposition 7.5. Assume G is an infinite abelian group.
(1) There exist a, b ∈ G, such that C−→ay(G; a, b) has a two-way infinite hamiltonian path
if and only if G is isomorphic to either Z or Z× Zm, for some m ≥ 2.
(2) For a, b ∈ Z, the Cayley digraph C−→ay(Z; a, b) has two arc-disjoint two-way infinite
hamiltonian paths if and only if a and b are odd, and
either {a, b} = {1,−1} or a+ b = ±2.
(3) For a, b ∈ Z × Zm, with m ≥ 2, the Cayley digraph C
−→ay(Z × Zm; a, b) has two
arc-disjoint two-way infinite hamiltonian paths if and only if either
(a) {a, b} = {(1, x), (−1, y)}, for some x, y ∈ Zm, such that ⟨x+ y⟩ = Zm, or
(b) m = 2, a = (0, 1), and b ∈ {±1} × Z2, perhaps after interchanging a and b.
Proof. ((1) ⇐) It is obvious that C−→ay(Z; 1,−1) has a two-way infinite hamiltonian path.
The remaining case is immediate from Corollary 7.3.
((2) ⇐) If a, b ∈ {±1}, then C−→ay(Z; a) and C−→ay(Z; b) are two arc-disjoint two-way
infinite hamiltonian paths in C−→ay(Z; a, b).
We may now assume a + b = ±2, and a ̸= b. Since a + b is even, we may write
a − b = 2ℓ, for some nonzero ℓ ∈ Z. Assume without loss of generality that k > 0 (by
replacing a and b with their negatives if necessary). Letting k = ℓ, we have
k + ℓ = 2ℓ = a− b = |Z : ⟨a− b⟩|
and
⟨ka+ ℓb⟩ = ⟨ℓa+ kb⟩ = ⟨ℓ(a+ b)⟩ = ⟨2ℓ⟩ = ⟨a− b⟩,
so Corollary 7.4 tells us that C−→ay(G; a, b) has two arc-disjoint two-way infinite hamiltonian
paths.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 29
((3)(a) ⇐) Let : Z × Zm → Z be the natural projection, and assume, without loss
of generality, that a = (1, x) and b = (−1, y). Then a− b = 2, so |G : ⟨a − b⟩| = 2m.
Therefore, if we let k = m+ 1 and ℓ = m− 1, then
k + ℓ = 2m = |G : ⟨a− b⟩|
and
ka+ ℓb = ka+ ℓb = (m+ 1) · 1 + (m− 1) · (−1) = 2,
so
|G : ⟨ka+ ℓb⟩| = 2m = |G : ⟨a− b⟩|,
so
⟨ka+ ℓb⟩ = ⟨a− b⟩.
A similar calculation shows ⟨ℓa+kb⟩ = ⟨a−b⟩. So Corollary 7.4 tells us that C−→ay(G; a, b)
has two arc-disjoint two-way infinite hamiltonian paths.
((3)(b) ⇐) Since a − b is of the form (±1, ∗), we have I = 2. Let k = ℓ = 1. Then
k + ℓ = I , and, since a = −a, we have
⟨a− b⟩ = ⟨a+ b⟩ = ⟨ka+ ℓb⟩ = ⟨ℓa+ kb⟩.
So Corollary 7.4 tells us that C−→ay(G; a, b) has two arc-disjoint two-way infinite hamiltonian
paths.
((1) ⇒) The structure theorem for finitely generated abelian groups [23, 4.2.10] tells us
that
G ∼= Zr × Zn1 × Zn2 × · · · × Zns ,
where ni+1 is a multiple of ni for 1 ≤ i < s (and ni ≥ 2 for all i).
Since G is infinite, we have r ≥ 1. On the other hand, we know from Proposition 7.2(2)
that |G : ⟨a− b⟩| < ∞, so G has a cyclic subgroup of finite index, so r ≤ 1. We conclude
that r = 1.
The minimum cardinality of a generating set of G is r + s. Since G = ⟨a, b⟩, this
implies r + s ≤ 2. Since r = 1, we conclude that s ∈ {0, 1}. If s = 0, then G ∼= Z. If
s = 1, then we may let m = n1.
((2) ⇒, (3) ⇒) Let G = Z×Zm for (where m = 1 if the group is Z), let : Z×Zm → Z
be the natural projection, and let I be the absolute value of a− b, so ⟨a− b⟩ = ⟨I⟩.
If I = 0, then |G : ⟨a − b⟩| = ∞, which contradicts the conclusion of Proposi-
tion 7.2(1), so we must have a = b. Then G = ⟨a, b⟩ = ⟨a⟩ ∼= Z and a = b = ±1, so
a + b = ±2. This implies that a and b have the same parity. They cannot both be even
(since ⟨a, b⟩ = Z), so a and b are odd. Thus, the situation is described in part (2) of the
statement of the proposition.
We may now assume I > 0. Then we claim that |G : ⟨a− b⟩| = mI . We have
|G : ⟨a− b⟩| =
∣
∣
∣
∣
G
⟨a− b⟩
∣
∣
∣
∣
=
∣
∣
∣
∣
G
⟨a− b⟩+ Zm
∣
∣
∣
∣
·
∣
∣
∣
∣
⟨a− b⟩+ Zm
⟨a− b⟩
∣
∣
∣
∣
.
Also note that
G
⟨a− b⟩+ Zm
∼=
G/Zm
(
⟨a− b⟩+ Zm
)
/Zm
∼=
Z
⟨a− b⟩
=
Z
⟨I⟩
30 Ars Math. Contemp. 25 (2025) #P2.10
has order I , and
⟨a− b⟩+ Zm
⟨a− b⟩
∼=
Zm
⟨a− b⟩ ∩ Zm
=
Zm
{0}
∼= Zm
has order m. Therefore the claim is proved. By Corollary 7.4, there exist k, ℓ ≥ 0, such
that k + ℓ = mI , and
⟨ka+ ℓb⟩ = ⟨ℓa+ kb⟩ = ⟨a− b⟩.
Therefore
⟨ka+ ℓb⟩ = ⟨ℓa+ kb⟩ = ⟨a− b⟩ = ⟨I⟩,
so we may assume
ka+ ℓb = I and ℓa+ kb = ±I.
Adding these two equations, we conclude that
(k + ℓ)(a+ b) is either 0 or 2I .
Since k + ℓ = mI , this implies
m(a+ b) ∈ {0, 2}.
Case I. Assume m(a + b) = 0. This means a = −b. Since gcd(a, b) = 1, this implies
a = ±1 and b = ±1 (so I = 2). If G = Z, then the situation is described in part (2) of the
statement of the proposition.
Therefore, we may assume G ̸∼= Z, so m > 1. Write a = (1, x) and b = (−1, y)
(perhaps after interchanging a and b). Since
G = ⟨a, b⟩ = ⟨a, a+ b⟩ = ⟨(1, x), (0, x+ y)⟩,
it is clear that ⟨x + y⟩ = Zm. So we are in the situation specified by part (3)(a) of the
statement of the proposition.
Case II. Let m(a+ b) = 2. This immediately implies that m ∈ {1, 2}.
If m = 1 (and a+ b = 2), then G = Z, and we have a+ b = a+ b = 2. (This implies
that a and b have the same parity. They cannot both be even, since ⟨a, b⟩ = G, so a and b
must be odd.) So we are in a situation that is described in part (2) of the statement of the
proposition.
Assume now that m = 2 (and I = 1). Then a + b = 1. Note that ⟨2a⟩ ≠ Z and
⟨2b⟩ ≠ Z. Since ⟨ka + ℓb⟩ = Z and k + ℓ = mI = 2, we conclude that k = ℓ = 1. This
implies
a+ b = ka+ ℓb = I = 1.
Since we also have a− b = ±I = ±1, we conclude that {a, b} = {0, 1}. So we are in the
situation that is described in part (3)(b) of the statement of the proposition.
Corollary 7.6. C−→ay(Z × Zm; e1, e2) has two arc-disjoint two-way infinite hamiltonian
paths if and only if m = 2.
I. Darijani et al.: Arc-disjoint hamiltonian paths in Cartesian products of directed cycles 31
Proof. (⇐) Let k = ℓ = 1. Then
k + ℓ = 2 = |Z× Z2 : ⟨e1 − e2⟩|
and
⟨ke1 + ℓe2⟩ = ⟨ℓe1 + ke2⟩ = ⟨e1 + e2⟩ = ⟨e1 − e2⟩,
so Corollary 7.4 provides two arc-disjoint two-way infinite hamiltonian paths.
(⇒) Since e2 = (0, 1) is obviously not of the form (±1, ∗), it is clear that the generating
set {e1, e2} is not of the form specified in Proposition 7.5(3)(a), so it must be part (3)(b) of
the proposition that applies. So m = 2.
ORCID iDs
Iren Darijani https://orcid.org/0000-0003-2876-3782
Babak Miraftab https://orcid.org/0000-0002-8566-5364
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