/^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 319-327 https://doi.org/10.26493/1855-3974.1349.b67 (Also available at http://amc-journal.eu) The 4-girth-thickness of the complete graph Christian Rubio-Montiel UMILAFMIA 3175 CNRS at CINVESTAV-IPN, 07300, Mexico City, Mexico División de Matemáticas e Ingeniería, FES Ac-UNAM, 53150, Naulcalpan, Mexico Received 10 March 2017, accepted 21 June 2017, published online 19 September 2017 Abstract In this paper, we define the 4-girth-thickness 6(4, G) of a graph G as the minimum number of planar subgraphs of girth at least 4 whose union is G. We prove that the 4-girth-thickness of an arbitrary complete graph Kn, 6(4, Kn), is [np] for n = 6,10 and 6(4, K6) = 3. Keywords: Thickness, planar decomposition, girth, complete graph. Math. Subj. Class.: 05C10 1 Introduction A finite graph G is planar if it can be embedded in the plane without any two of its edges crossing. A planar graph of order n and girth g has size at most g—■ (n - 2) (see [6]), and an acyclic graph of order n has size at most n - 1, in this case, we define its girth as to. The thickness 6(G) of a graph G is the minimum number of planar subgraphs whose union is G; i.e. the minimum number of planar subgraphs into which the edges of G can be partitioned. The thickness was introduced by Tutte [11] in 1963. Since then, exact results have been obtained when G is a complete graph [1, 3, 4], a complete multipartite graph [5, 12, 13] or a hypercube [9]. Also, some generalizations of the thickness for the complete graph Kn have been studied such that the outerthickness 6o, defined similarly but with outerplanar instead of planar [8], and the S-thickness 6S, considering the thickness on a surfaces S instead of the plane [2]. See also the survey [10]. We define the g-girth-thickness 6(g, G) of a graph G as the minimum number of planar subgraphs of girth at least g whose union is G. Note that the 3-girth-thickness 6(3, G) is the usual thickness and the TO-girth-thickness 6(to, G) is the arboricity number, i.e. the minimum number of acyclic subgraphs into which E(G) can be partitioned. In this paper, we obtain the 4-girth-thickness of an arbitrary complete graph of order n = 10. E-mail address: christian@cs.cinvestav.mx (Christian Rubio-Montiel) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 320 Ars Math. Contemp. 14 (2018) 267-284 2 The exact value of 6(4, K„) for n = 10 Since the complete graph Kn has size and a planar graph of order n and girth at least 4 has size at most 2(n — 2) for n > 3 and n — 1 for n G {1,2} then the 4-girth-thickness of Kn is at least n(n — 1) 2(2n — 4) n + 1 1 n + 2 2n — 4 4 + for n > 3 and also |"n+2"| for n G {1,2}, we have the following theorem. Theorem 2.1. The 4-girth-thickness 0(4, Kn) of Kn equals for n = 6,10 and 0(4, Ke) =3. Proof. Figure 1 displays equality for n < 5. i 2 , 2 4 2 Figure 1: 0(4, Kn) = f^] for n =1,2, 3,4,5. To prove that 0(4, Ke) = 3 > = 2, suppose that 0(4, Ke) = 2. This partition define an edge coloring of Ke with two colors. By Ramsey's Theorem, some part contains a triangle obtaining a contradiction for the girth 4. Figure 2 shows a partition of Ke into tree planar subgraphs of girth at least 4. Figure 2: 0(4, Ke) = 3. For the remainder of this proof, we need to distinguish four cases, namely, when n = 4k — 1, n = 4k, n = 4k + 1 and n = 4k + 2 for k > 2. Note that in each case, the lower bound of the 4-girth thickness require at least k + 1 elements. To prove our theorem, we exhibit a decomposition of K4k into k +1 planar graphs of girth at least 4. The other three C. Rubio-Montiel: The 4-girth-thickness of the complete graph 321 cases are based in this decomposition. The case of n = 4k - 1 follows because K4fc_i is a subgraph of K4k. For the case of n = 4k + 2, we add two vertices and some edges to the decomposition obtained in the case of n = 4k. The last case follows because K4fc+1 is a subgraph of K4fc+2. In the proof, all sums are taken modulo 2k. 1. Case n = 4k. It is well-known that a complete graph of even order contains a cyclic factorization of Hamiltonian paths, see [7]. Let G be a subgraph of K4k isomorphic to K2k. Label its vertex set V(G) as {vi; v2,..., v2k}. Let Fi be the Hamiltonian path with edges V1V2, V2V2k, V2kV3, V3V2fc_l, . . . , V2 + fcVi+fc. Let Fj be the Hamiltonian path with edges vivi+i,vi+ivi-i,vi-ivi+2, Vj+2Vj-2, . . . , v¿+fc+ivi+fc, where i e {2,3,..., k}. Such factorization of G is the partition {E(Fi), E(F2),..., E(Fk)}. We remark that the center of F¿ has the edge e = vj+ pk-j vj+ p 3k-j, see Figure 3. a) i + 1 M f1 +r t- i-i v,+ r 32k v,+r k i-i +r 21 v,+ r k Vi+k+1 vi+k-1 b) v, ^%i+1 vi+r3í i+1 v,+r k i-1 +r kk i Mt i vi+k-1 vi+k+1 Figure 3: The Hamiltonian path F^ Left a): The dashed edge e for k odd. Right b) The dashed edge e for k even. Now, consider the complete subgraph G' of K4k such that G' = K4k \ V(G). Label its vertex set V(G') as {vl, v2,..., v2k} and consider the factorization, similarly as before, {E(F'), E(F2),..., E(Fk)} where F/ is the Hamiltonian path with edges viv'+l, vi+lvi-l, vi-lv'+2, v'+2v'-2, . . . , vi+fc+lvi+fc, where i G {1,2,..., k}. Next, we construct the planar subgraphs Gl, G2,...,Gk-l and Gk of girth 4, order 4k and size 8k - 4 (observe that 2(4k - 2) = 8k - 4), and also the matching Gk+l, as follows. Let Gj be a spanning subgraph of K4k with edges E(Fj) U E(F/) and jv+i, vi+ivi_i, vi+ivi-i, vi-ivi+2, vi-ivi+2,...,vi+fc+ivi+fc, vj+fc+ivi+fc v v 322 Ars Math. Contemp. 14 (2018) 267-284 where i e {1,2,..., k}; and let Gk+1 be a perfect matching with edges vjvj for j e {1, 2,..., 2k}. Figure 4 shows Gj is a planar graph of girth at least 4. b) vi V2 G k+1 V2fc Figure 4: Left a): The graph Gj for any i e {1, 2,..., k}. Right b) The graph Gk+1. v v k + 1 To verify that K4k = |j G^ 1) If the edge v^ vi2 of G belongs to the factor Fj i=1 then vjj vi2 belongs to Gj. If the edge is primed, belongs to Gj. 2) The edge v^ v-2 belongs to Gk+1 if and only if i1 = i2, otherwise it belongs to the same graph Gj as vj1 vj2. Similarly in the case of vvj2 and the result follows. 2. Case n = 4k - 1. Since K4k-1 c K4k, we have k +1 < 0(4, K4fc_1) < 0(4, K4fc) < k + 1. 3. Case n = 4k + 2 (for k = 2). Let {G1,..., Gk+1} be the planar decomposition of K4k constructed in the Case 1. We will add the two new vertices x and y to every planar subgraph Gj, when 1 < i < k + 1, and we will add 4 edges to each Gj, when 1 < i < k, and 4k +1 edges to Gk+1 such that the resulting new subgraphs of K4k+2 will be planar. Note that (42kt + 4k + 4k + 1 = (4k+2t. To begin with, we define the graph ffk+1 adding the vertices x and y to the planar subgraph Gk+1 and the 4k +1 edges {xy,xv1,xv2,xv3,xv4,... ,xv2k_1,xv2k, yv'x,yv2,yv3,yv4,..., yv2k_1,yv2k}. The graph Hk+1 has girth 4, see Figure 5. In the following, for 1 < i < k, by adding vertices x and y to Gj and adding 4 edges to Gj, we will get a new planar graph Hj such that {H1,..., Hk+1} is a planar decomposition of K4k+2 such that the girth of every element is 4. To achieve it, the given edges to the graph Hj will be vjx, xvj_1,vjy, yvj_1, for some odd j e {1, 3,..., 2k - 1}. According to the parity of k, we have two cases: C. Rubio-Montiel: The 4-girth-thickness of the complete graph 323 Figure 5: The graph Hk+1. • Suppose k odd. For odd i G {1, 2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+ [f] -1, xvi+ [32k 1 , yvi+ \32k 1-1' yvi+ [21 } when is even, otherwise {y<+r f 1-1'yVi+[ 32k 1 'XM 32k 1-1'XVI+[ f 1}. Additionally, for even i G {1, 2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+r 11-1, xvi+r 11, yvi+ r t1-1, yvi+ r 11} when is even, otherwise {yvi+rk 1-1,yvi+r 11 ,xvi+r 11-1 ,xvi+r 11}. Note that the graph Hi has girth 4 for all i, see Figure 6. • Suppose k even. Similarly that the previous case, for odd i G {1,2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+r 32k 1+1, xvi+ r f 1, yvi+ r f 1+1, yv r 32k 1} when is even, otherwise {yvi+ r 32k 1+1, yvi+ r 3^ 1, xvi+r t 1+1, xvi+ r 32k 1}. On the other hand, for even i G {1, 2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+r 11, xvi+ r 11 -1, yvi+ r 11, yv*+r k 1 -1} when is even, otherwise {yvi+rk 1, yvI+r 11-1, xvi+r 11, xv r 11 -1}. Note that the graph Hi has girth 4 for all i, see Figure 7. 324 Ars Math. Contemp. 14 (2018) 267-284 a) Vk+i k+i Vk+i k+i vx Vi+\k 1-1 •Vi+r 11 v,-+r 3k W^Vi + k Vi+k+1 Vi+k-1 V. _«Vi + 1 Vi+r 11-1 •M 11 V,-+r 3k •r^^Vi+k Vi+k+1 Vi+k-1 V V Figure 6: The graph H when k is odd and its auxiliary graph F*. Above a) When i is odd. Botton b) When i is even. C. Rubio-Montiel: The 4-girth-thickness of the complete graph 325 a) Vk + i k + i b) Vk + i k+i vx »M k i+ k -i v,+\ -k l+i »M k i M * i vi+k-1 Vi+k+1 V, J'i + l »M k i+ k -i Vi+\3k l+i »M k i Vi+\ ¥ l Vi+k+l vi+k-1 V V Figure 7: The graph H when k is even and its auxiliary graph F*. Above a) When i is odd. Botton b) When i is even. 326 Ars Math. Contemp. 14 (2018) 267-284 In order to verify that each edge of the set [xv'l,xv2,xv'3,XV3,. .. ,xv2k_1;xv2k,yvi,yv'2,yv3,yv's,. .. ,yv2k-i,yv'^k}. is in exactly one subgraph Hi, for i e {1,..., k}, we obtain the unicyclic graph F* identifying vj and vj resulting in vj; identifying x and y resulting in a vertex which is contracted with one of its neighbours. The resulting edge, in dashed, is showed in Figures 6 and 7. The set of those edges are a perfect matching of K2k proving that the added two paths of length 2 in Gi have end vertices vj and vj_1, and the other vj and vj_1. The election of the label of the center vertex is such that one path is vevenxv odd and vevenyvodd and the result follows. 4. Case n = 4k + 1 (for k = 2). Since K4k+1 c K4k+2, we have k +1 < 0(4, K4k+i) < 0(4, K4k+2) < k + 1. For k = 2, Figure 8 displays a decomposition of three planar graphs of girth at least 4 proving that 0(4, Kg) = [^] = 3. 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