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Rainbow Matchings in Properly Colored Bipartite Graphs

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Rainbow Matchings in Properly Colored Bipartite Graphs Open Journal of Discrete Mathematics, 2012, 2, 62-64 http://dx.doi.org/10.4236/ojdm.2012.22011 Published Online April 2012 (http://www.SciRP.org/journal/ojdm) Rainbow Matchings in Properly Colored Bipartite Graphs Guanghui Wang, Guizhen Liu School of Mathematics, Shandong University, Jinan, China Email: {ghwang, gzliu}@sdu.edu.cn Received January 19, 2012; revised February 16, 2012; accepted March 25, 2012 ABSTRACT Let G be a properly colored bipartite graph. A rainbow matching of G is such a matching in which no two edges have the same color. Let G be a properly colored bipartite graph with bipartition ( X , Y ) and Gk= 3. We show that if 7k 3k maxXY , , then G has a rainbow coloring of size at least . 4 4 Keywords: Rainbow Matching; Bipartite Graphs 1. Introduction and Notation of applications in design theory. A matching M is subor- thogonal if there is at most one edge from each k-factor. We use [1] for terminology and notations not defined Alspach [7] posed the above problem in the case k =2. here and consider simple undirected graphs only. Let Stong [8] proved that if nm32, then there is a such GVE=, be a graph. A proper edge-coloring of G is a orthogonal matching. For k =3, the answer is yes, see function c: E ( is the set of non-negative in- [9]. In the same paper, Anstee and Caccetta proved the tegers) such that any two adjacent edges have distinct following theorem when k =1. colors. If G is assigned such a coloring c, then we say Theorem 3. [9] Let G be an m-regular graph on n that G is a properly edge-colored graph, or simply a vertices. Then for any decomposition of EG into m properly colored graph. Let ce denote the color of 1-factors F ,,,FF , there is a matching M of p edges, the edge e E. For a subgraph H of G, let 12 m at most one edge from each 1-factor, with cH=: ce e EH . 2 2 nn333 A subgraph H of G is called rainbow if its edges have pm>min ,3 . m 222 2 distinct colors. Recently rainbow subgraphs have re- ceived much attention, see the survey paper [2]. Here we In any decomposition of EG into m k-factors, we are interested in rainbow matchings. The study of rain- can construct an edge colored graph by giving each k- bow matchings began with the following conjectures. factor a color. Then a rainbow matching of G corre- Conjecture 1. (Ryser [3,4]) Every Latin square of odd sponds to a suborthogonal matching of G. In particular, order has a Latin transversal. when k =1, the edge colored graph obtained above is Conjecture 2. (Stein [5]) Every Latin square of order properly colored. So we can pose a more general prob- n has a partial Latin transversal of size at least n 1. lem: Let G be a properly colored graph of minimum de- An equivalent statement is that every proper n-edge- gree G . Is there a rainbow matching of size G ? coloring of the complete bipartite graph K contains a nn, Unfortunately, the answer is negative, see [10]. More- rainbow matching of size n 1; Moreover, if n is odd, over, if G is a properly colored complete graph, then G there exists a rainbow perfect matching. Hatami and Shor [6] proved that there is always a partial Latin transversal G 2 has no rainbow matching of size more than . In (rainbow matching) of size at least nO log n. 2 Another topic related to rainbow matchings is or- addition, the following theorem was shown in [11]. thogonal matchings of graphs. Let G be a graph on n Theorem 4. [8] Let G be a properly colored graph, vertices which is an edge disjoint union of m k-factors GK , and VG G 2 . Then G contains a (i.e. k regular spanning subgraphs). We ask if there is a 4 G matching M of m edges with exactly one edge from each rainbow matching of size . k-factor? Such a matching is called orthogonal because 2 Copyright © 2012 SciRes. OJDM G. H. WANG ET AL. 63 However, we believe that if the order of a properly 3k Then t 1. Let X VM= X1 and YVMY =.1 colored graph G is much larger than its minimum degree 4 G , there should be a rainbow matching of size Put YY 12= Y and X XX12= . Let X11 denote the G . In [10], we propose the following problem. vertices in X which are incident with Y by three edges Problem 5. [10] Is there a function f n such that 2 with new colors. Clearly, X11 X 1 . Otherwise, we can for each properly colored graph G with get a rainbow matching of size at least t 1, which is a VGfG , G must contain a rainbow matching contradiction. Let Y denote the vertices which are of size G ? 11 incident with the vertices in X11 by the edges in M. We Since when n is even, an nn Latin square has no have the following claim. Latin transversal (perfect rainbow matching) (see [3]), if k the function f n exists, f n should be greater than Claim 1. X11 . 2n. Motivated by this problem, we prove the following 4 results in [10]. Proof. Let yYi 2 . If there is an edge xyi such that Theorem 6. [10] Let G be a properly colored graph cxy i cM , then x X1 . Otherwise, there is a rainbow matching M xy of size t 1 , which is a 8 G i and VG . Then G has a rainbow matching contradiction. Let E denote the edges which are inci- 5 1 dent with vertices in Y2 and have new colors. Since 3 G each vertex i n Y has degree at lea st k , of size at least . 2 5 k Theorem 7. [10] Let G be a properly colored triangle- EktY12 1 Y2. 4 free graph. Then G has a rainbow matching of size at On the other hand, EYX2 XX. So 2 G 1211111 least . we have the following equality 3 k In [12], Wang, Zhang and Liu proved that if 12YYX22111 XX 11. 4 nn2 14 1 fn , 4 Hence then G has a rainbow matching of size , which an- k 12YX21 swers the above question in the affirmative. Eiemunsch 4 X 13n 23 41 11 Y 2 et al. [13] improved this bound to 1. 2 228n kk 3 95n 12Y2 1 Later, this bound was improved to by Lo in [14]. 44 2 Y 2 In this paper, we consider the rainbow matching of the 2 properly colored bipartite graph, and prove the following k 122Yk2 2 result. 4 = . Theorem 8. Let G be a properly colored bipartite Y 2 graph with bipartition X ,Y and Gk= 3. If 2 7k maxXY , , then G has a rainbow coloring of 7k 73kk Since Y , Yk2 11, thus 4 4 44 3k size at least . k 4 X11 . 4 For more result about rainbow matchings under the Without loss of generality, we assume that color degree conditions, we refer to [15,16]. X11=,,,xx 1 2 xp , 2 Proof of Theorem 8 k Let GXY=,. Without loss of generality, we assume where p . Let yi denote the vertex which is 4 that maxX ,YY = . Suppose that our conclusion is incident with x by an edge in M. not true, we choose a maximum rainbow matching M. i Claim 2. Let x be any vertex in X and y de- Let tM= . Without loss of generality, we assume that i 11 i note the vertex which is incident with xi by an edge in cM=1,2,, t . M. If yi is incident with a vertex x X 2 , then Copyright © 2012 SciRes. OJDM 64 G. H. WANG ET AL. Applications,” Macmillan Press, New York, 1976. cxyi p1, , t. Proof. Suppose our conclusion does not hold. First, we [2] M. Kano and X. Li, “Monochromatic and Heterochro- matic Subgraphs in Edge-Colored Graphsa Survey,” know that cxyi can not be a new color. Otherwise, Graphs and Combinatorics, Vol. 24, No. 4, 2008, pp. since xi are incident with three edges having new col- 237-263. doi:10.1007/s00373-008-0789-5 ors, we can choose one edge (say e). Then we can get a [3] R. A. Brualdi and H. J. Ryser, “Combinatorial Matrix Theory,” Cambridge University Press, Cambridge, 1991. new rainbow matching of size t 1 by adding xyei , and deleting xiiy . Thus we will get a contradiction. So [4] H. J. Ryser, “Neuere Probleme der Kombinatorik,” in we conclude that cxy1,2,, p . Since G is prop- Vortrage uber Kombinatorik Ober-Wolfach, Mathema- i tisches Forschungsinstitut Oberwolfach, 1967, pp. 24-29. erly colored, cxy i. So without loss of generality, i [5] S. K. Stein, “Transversals of Latin Squares and Their we assume that cxy =1. Moreover, we assume that i Generalizations,” Pacific Journal of Mathematics, Vol. the edges with new colors are incident with x1 are e1, 59, No. 2, 1975, pp. 567-575. e2 and e3 . Now we can choose ejj =1,2, 3 such [6] P. Hatami and P. W. Shor, “A Lower Bound for the that cej ce and e j is not incident with y. Hence Length of a Partial Transversal in a Latin Square,” Jour- nal of Combinatorial Theory, Series A, Vol. 115, No. 7, we have a new rainbow matching by adding exyeji,, 2008, pp.
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