Rainbow Matchings in Properly Colored Bipartite Graphs

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.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    3 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us