We prove that any $r$-coloring of the edges of $K_m$ contains a monochromatic even cycle, where $m = 3r + 1$ if $r$ is odd and $m =3r$ if $r$ is even. We also prove that $K_{m−1}$ has an $r$-coloring without monochromatic even cycles

Gyárfás, András

Pálvölgyi, Dömötör

English

Contributions to Discrete Mathematics (E-Journal, University of Calgary)

Volume 7, Number 1, Pages 48–50ISSN 1715-0868MONOCHROMATIC EVEN CYCLESANDRA´S GYA´RFA´S AND DO¨MO¨TO¨R PA´LVO¨LGYIAbstract. We prove that any r-coloring of the edges of Km containsa monochromatic even cycle, where m = 3r + 1 if r is odd and m =3r if r is even. We also prove that Km−1 has an r-coloring withoutmonochromatic even cycles.An easy exercise, perhaps folkloristic, says that in any r-coloring of theedges of K2r+1 there is a monochromatic odd cycle (and this is not true forK2r).This note explores what happens if we ask the same question for even cy-cles. Let f(r) denote the smallest integer m for which there is a monochro-matic even cycle in every edge coloring of Km.Theorem 1. For odd r, f(r) = 3r + 1 and for even r, f(r) = 3r.Every graph with n vertices and with more than m = b3(n− 1)/2c edgescontains a Θ-graph, i.e. three internally vertex disjoint paths connecting thesame pair of vertices (see [1], Exercise 10.1). Since a Θ-graph obviouslycontains an even cycle, any graph with n vertices and more than m edgescontains an even cycle. This easily implies that the stated values are upperbounds of f(r) in Theorem 1. Indeed, considering the majority color, onecan easily check that⌈(3r+12)r⌉>⌊3(3r)2⌋if r is oddand ⌈(3r2)r⌉>⌊3(3r − 1)2⌋if r is even.Therefore to prove Theorem 1 we need a construction, a partition of theedge set of K3r (K3r−1) into r graphs, each without even cycles. Let H1 be atriangle with vertices v1, v2, v3. For odd r > 1 let Hr be the graph formed bythree vertex disjoint copies of (r−1)/2 triangles sharing one common vertexvi, i = 1, 2, 3 and the triangle v1, v2, v3 which is called the central triangle ofHr. Note that each block (maximal biconnected subgraph or cut-edge) ofReceived by the editors May 23, 2011.2000 Mathematics Subject Classification. 05C15: Coloring of graphs and hypergraphs.The second author acknowledges support from OTKA CNK 77780.c©2012 University of Calgary48MONOCHROMATIC EVEN CYCLES 49Hr is a triangle, so it has no even cycles. Thus for odd r Theorem 1 followsfrom the next proposition.Proposition 2. For odd r, K3r can be partitioned into r copies of Hr.Proof. The proof is based on a well-known construction of Steiner triplesystems on 6t + 3 vertices (see [2], Theorem 9.1). Set r = 2t + 1, then3r = 6t + 3. The vertex set of K = K3r is partitioned into {ai, bi, ci}, fori = 1, 2, . . . , 2t+ 1. For r = 1, {ai, bi, ci} is an H1, for r > 1 consider a nearfactorization of a complete graph S2t+1 with vertex set {1, 2, . . . , 2t+1} intofactors Fi, where Fi avoids vertex i. To each factor Fi we define a copy ofH ir as follows. Place the edges of the following triangles to Hir:(1) {biakal, cibkbl, aickcl : kl ∈ Fi}, {aibici}.One can easily see that H ir is isomorphic to Hr and for i = 1, . . . , 2t + 1they give a partition on the edge set of K (in fact their blocks are trianglesforming a Steiner triple system on K). For r = 2 note that K5 can be partitioned into two pentagons. However,K5 can be also partitioned into two “bulls”, which is a triangle with twopendant edges (see Figure 1). This latter works well to reduce the even caseto the odd one in Proposition 3.For even r define the graph Ar from Hr−1 by removing the edges of itscentral triangle v1, v2, v3 and adding two new vertices u,w together withthe five edges v1v2, uvi, wv2 (see Figure 2). Let Br be the graph with r − 1triangles sharing a common vertex x plus r pendant edges, one from x andone from each triangle (from a vertex different from x). Note that Ar, BrFigure 1. A bull with its complementary bull dotted, drawnas later used.w u w uwuFigure 2. The Hr−1, Ar and Br monochromatic subgraphsfor r = 6.50 ANDRA´S GYA´RFA´S AND DO¨MO¨TO¨R PA´LVO¨LGYIhave 3r − 1 vertices and their blocks are cut-edges and triangles so they donot have even cycles. The graphs A2, B2 are both bulls.Proposition 3. For even r, K3r−1 can be partitioned into r − 1 copies ofAr and one copy of Br.Proof. Let r be even and consider the construction of Proposition 2 for r−1colors. This gives a partition of K3r−3 into r−1 copies of Hr−1. Notice thatthe central triangles Ti = {ai, bi, ci} of the i-th copies are vertex disjoint(i = 1, 2, . . . , r − 1). Adding two new vertices d, e to V (K3r−3) transformthe i-th copy of Hr−1 as follows: remove the edges aici, bici from Ti and adddai, dbi, dci, ebi. This gives r − 1 copies of Ar for (i = 1, 2, . . . , r − 1). The“missing edges”, de, eai, eci, aici, bici for i = 1, 2, . . . , r − 1 define one copyof Br. Proposition 3 shows that for even r, f(r) ≥ 3r, thus completing the proofof Theorem 1.References1. L. Lova´sz, Combinatorial problems and exercises, 2nd ed., North-Holland, 1993.2. W. D. Wallis, One-Factorizations, Kluwer Academic Publishers, 1997.Alfre´d Re´nyi Institute, Hungarian Academy of Sciences,Budapest, P.O. Box 127, Budapest, Hungary, H-1364.E-mail address: gyarfas@renyi.huComputer Science Department, Institute of Mathematics,Eo¨tvo¨s Lora´nd University, Pa´zma´ny Pe´ter se´ta´ny 1/c,Budapest, Hungary, H-1117.E-mail address: dom@cs.elte.hu

Monochromatic even cycles

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Monochromatic even cycles

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Contributions to Discrete Mathematics (E-Journal, University of Calgary)