Discrete Mathematics 257 (2002) 357–369 www.elsevier.com/locate/disc 2-factors in dense bipartite graphs A. Czygrinow∗, H.A. Kierstead∗ Department of Mathematics, Arizona State University, Tempe, AZ 85287, USA Received 15 November 1999; accepted 20 October 2000 Abstract An n-ladder is a balanced bipartite graph with vertex sets A = {a;:::;an} and B = {b1;:::;bn} such that ai ∼ bj i2 |i − j| 6 1. We use techniques developed recently by KomlÃos et al. (1997) to show that if G =(U; V; E) is a bipartite graph with |U| = n = |V |, with n su6ciently large, and the minimum degree of G is at least n=2 + 1, then G contains an n-ladder. This answers a question of Wang. c 2002 Elsevier Science B.V. All rights reserved. Keywords: Bipartite graphs; Blow-up lemma; Cycles 1. Introduction A bipartite graph is said to balanced if it has the same number of vertices in each part. Let G =(U; V; E) be a balanced bipartite graph with |U|=n=|V |. For the purposes of this article, we shall call G universal if every balanced bipartite graph with n vertices in each part and maximum degree at most 2 can be embedded into G. Wang [8] conjectured that G is universal if the minimum degree (G)ofG is at least n=2+1. As Wang observed, the conjecture is best possible: The graph H obtained by adding a perfect matching between the larger parts of two copies of the complete bipartite graph Km+1;m satisÿes (H)=m +1=n=2, but does not contain the union C2m + C2m+2 of a(2m)-cycle with an (2m + 2)-cycle. Aigner and Brandt [1] proved the nonbipartite version of this conjecture: If H is a graph on n vertices with minimum degree (H)¿(2n − 1)=3, then H contains every graph on n vertices with maximum degree at most 2. Again this is best possible, since ∗ Corresponding author. E-mail addresses: [email protected] (A. Czygrinow), [email protected] (H.A. Kierstead). 0012-365X/02/$ - see front matter c 2002 Elsevier Science B.V. All rights reserved. PII: S 0012-365X(02)00435-1 358 A. Czygrinow, H.A. Kierstead / Discrete Mathematics 257 (2002) 357–369 m triangles cannot be embedded in the complete tripartite graph Km−1;m+1;m+1. Fan and Kierstead [3] proved something stronger. In an attempt at proving PÃosa’s conjecture 2 that every graph on n vertices with minimum degree at least 3 n contains the square of a hamiltonian cycle, they showed that every graph H with (H)¿(2n − 1)=3 contains the square Q of a hamiltonian path. Every graph on n vertices with maximum degree at most 2 can be embedded into Q. Then KomlÃos et al. [4] proved PÃosa’s conjecture using a wonderful new technique involving SzemerÃedi’s regularity lemma [7] and their own blow-up lemma [5]. Deÿne an n-ladder to be the balanced bipartite graph Ln with vertex sets A= {a;:::;an} and B={b1;:::;bn} such that ai ∼ bj i2 |i − j|61. So Ln consists of two vertex disjoint n-paths a1b2a3b4 ::: and b1a2b3a4;::: together with rungs formed by the matching {a1b1;:::;anbn}. (Perhaps, twisted ladder would be a more descriptive term.) The set of ends of Ln is the set {a1;b1;an;bn}. It is easy to see that the graph Ln is universal. For example, an even cycle x1y1;:::;xk yk can be embedded into the beginning of Ln by x1y1x2 ···xk yk → a1b2a3 :::sk tk ;sk−1;tk−1;:::;a2;b1; where (sk ;tk )=(ak ;bk ), if k is odd and (sk ;tk )=(bk ;ak ) otherwise. We shall apply the methods of KomlÃos et al. to prove the following strengthening of Wang’s conjecture for su6ciently large n. Theorem 1. For su3ciently large n, every balanced bipartite graph G =(U; V; E) with |U|=n=|V | and (G)¿n=2+1 contains a spanning ladder. The rest of this article is devoted to a proof of Theorem 1. This involves two parts. First, in Section 2 we identify a particular extremal conÿguration and use standard graph theoretical methods to show that if G contains this conÿguration, then G contains a spanning ladder. Then in Section 4, we use the regularity lemma and the blow-up lemma to show that if G does not contain the extremal conÿguration, then G contains a spanning ladder. In Section 3 we review the regularity and blow-up lemmas. In the remainder of this section, we review our notation and work out some easy details of the theory of dense bipartite graphs. For a positive integer n, let [n] denote the set {1; 2;:::;n}. Let deg(v) denote the degree of the vertex v and let deg(v; X ) denote the number of neighbors of v in the set X . For two disjoint, nonempty sets of vertices U and W , let e(U; W ) denote the number of edges with end points in both U and W . We will need Lemma 3 for our argument and the rest of this section serves as a warm-up. The next lemma is due to Bondy and ChvÃatal [2]. Lemma 2. Let P =x1y1 :::xsys be an even path of a bipartite graph H = (X; Y; E) with s¿2. If deg (x1;P) + deg (ys;P)¿s +1 then G has a cycle C with V (C)=V (P). Proof. By the pigeon hole principle, there exists i∈[s] such that x1 ∼ yi and ys ∼ xi. Then x1y1 :::xiysxsys−1xs−1 :::yi is the desired cycle. A. Czygrinow, H.A. Kierstead / Discrete Mathematics 257 (2002) 357–369 359 Lemma 3. Let H =(X; Y; E) be a connected bipartite graph with |X |=|Y |=n. If (H)¿k then H contains a path on min{4k − 1; 2n} vertices. Proof. Let P be the longest path in H and assume that the length of P is less than min{4k − 1; 2n}. Case 1: P =x1y1 :::xsys where xi ∈X; yi ∈Y . Then d(x1;P)+d(ys;P)¿2k¿s. Thus, by Lemma 2, there is a cycle C with V (C)=V (P). Since s¡n there is a vertex v not on C. Since H is connected one can obtain a longer path starting from v and using all of C. Case 2: P =x1y1 :::xsysxs+1. First we claim that H contains a cycle on 2s ver- tices. Indeed, since P is maximal, x1 and xs+1 have all their neighbors on P. Since deg(x1)+deg(xs+1)¿2k¿s+1, there exists i∈[s−1] such that x1 ∼ yi+1 and xs+1 ∼ yi. This gives a cycle x1y1 :::xiyixs+1ysxs :::yi+1. Assume now that C =x1y1 :::xsys is a cycle. Suppose that v is a vertex not on C. Then v is only adjacent to vertices on C, since otherwise there is a path longer than P. Since s¡n, there exist x∈X \V (C) and y∈Y \V (C). Since deg(x) + deg(y)¿2k¿s, there exists i∈[s] such that x ∼ yi and y ∼ xi. This yields a path on 2(s + 1) vertices and a contradiction. Note that it now follows that a balanced bipartite graph with n vertices in each part and minimum degree greater than n=2 is hamiltonian: By the degree condition, it is connected. By Lemma 3 it has a hamiltonian path. So by the degree condition and Lemma 2, it has a hamiltonian cycle. 2. The extremal case We say that G =(U; V; E)is$-splittable if there exist X ⊂U and Y ⊂V such that 1 2 |X |=(2 − $)n=|Y | and e(X; Y )6$n : In this case we say that X and Y are an $-splitting of G. 1 Lemma 4. For $¿0 su3ciently small, in particular $6 640;000 , if G =(U; V; E) is a balanced bipartite graph with n vertices in each part that is $-splittable and (G)¿n=2+1, then G contains a spanning ladder. Proof. Let X and Y bean$-splitting of G. Our ÿrst goal is to partition the vertices of G into two almost complete, bipartite, spanning subgraphs with about n=√2 vertices in each part. Let XM =U\X and YM =V \Y . Let S ={x∈X : deg(x; YM)¡( 1 − $)n}. For √ √ 2 each x∈S, deg(x; Y )¿ $n. Thus |S|6 $n, since √ |S| $n6 deg(x; Y )6e(X; Y )6$n2: x∈S 360 A. Czygrinow, H.A. Kierstead / Discrete Mathematics 257 (2002) 357–369 Counting the edges of the complement GM of G from X to YM we get: M M eGM (X; Y )=|X ||Y |− deg(x) − e(X; Y ) x∈X 1 1 1 n2 6 − $ + $ n2 − − $ + $n2 2 2 2 2 3 2 6 2 $n : √ √ M 1 Let T ={y∈Y : deg(y; X )¡( 2 − $ − $)n}. For each y∈T, degGM (y; X )¿ $n. Thus 3 √ |T|6 2 $n, since √ 3 |T| $n6 deg (y; X )6e (X; YM)6 $n2: GM GM 2 y∈T M Thus, we can choose X1 ⊂X \S and Y1 ⊂Y \T such that 1 √ (1) |X1|; |Y1|=(2 − 2 $)n and 1 √ (2) deg(x1;Y1); deg(y1;X1)¿( 2 − 4 $)n, for all x1 ∈X1, y1 ∈Y1. M Similarly, we can choose X2 ⊂X and Y2 ⊂Y such that 1 √ (1) |X2|; |Y2|=(2 − 2 $)n and 1 √ (2) deg(x; Y2); deg(y; X2)¿( 2 − 4 $)n, for all x∈X2 and y∈Y2.