Minimum degree tresolds for bipartite grap tiling Albert Bus Yi Zao Department of Matematics and Statistics Georgia State University January 17, 011 Abstract Given a bipartite grap H and a positive integer n suc tat v(h) divides n, we define te minimum degree tresold for bipartite H-tiling, δ (n, H), as te smallest integer k suc tat every bipartite grap G wit n vertices in eac partition and minimum degree δ(g) k contains a spanning subgrap consisting of vertex-disjoint copies of H. Zao, Hladký-Scact, Czygrinow-DeBiasio determined δ (n, K s,t ) exactly for all s t and sufficiently large n. In tis paper we determine δ (n, H), up to an additive constant, for all bipartite H and sufficiently large n. Additionally, we give a corresponding minimum degree tresold to guarantee tat G as an H-tiling missing only a constant number of vertices. Our δ (n, H) depends on eiter te cromatic number χ(h) or te critical cromatic number χ cr (H) wile te tresold for te almost perfect tiling only depends on χ cr (H). Tese results can be viewed as bipartite analogs to te results of Kun and Ostus [Combinatorica 9 (009), 65-107] and of Sokoufande and Zao [Rand. Struc. Alg. 3 (003), 180-05]. 1 Introduction Let G be a grap on n vertices and H be a grap on vertices. Te tiling (sometimes called packing) problem in extremal grap teory is to find in G as many vertex-disjoint copies of H as possible. Researcers are interested in finding a tigt minimum degree condition for G to contain an H-factor a subgrap wic consists of n/ copies of H. Tis is also sometimes called a perfect H-tiling or H-packing. Dirac s teorem on Hamilton cycles [4] is one of te earliest tiling results. It implies tat every n-vertex grap G wit minimum degree δ(g) n/ contains a perfect matcing (K -factor). Te seminal result of Hajnal and Szemerédi [6] determines te minimum degree tresold for a K r -factor for all integers r. By applying Szemerédi s Regularity Lemma [19], Alon and Yuster [1] found minimum degree conditions tat guarantee an H-factor for an arbitrary H. Komlós, Sarközy, and Szemerédi [11] improved Alon-Yuster s result, giving a tigt minimum degree for H wit equal-sized color classes. Instead of using te cromatic number χ(h) as in [1, 11], Komlós [9] introduced te critical cromatic number χ cr (H) and sowed tat it played a critical role in grap tiling (is result was improved by Sokoufande and Zao [18]). Kün and Ostus [13] finally determined exactly wen te critical cromatic number or te cromatic number was te appropriate parameter. In order to accurately state teir result, we need te following definitions. Current address: Scool of Matematics, Georgia Institute of Tecnology, Atlanta, GA 3033. Researc supported in part by NSA grants H9830-07-1-0019 and H9830-10-1-0165. 1
Let H be a grap on vertices wit χ(h) = l. Te critical cromatic number χ cr (H) is defined as (l 1) σ(h), were σ(h) is te size of te smallest color class over all proper l-colorings of H. It is easy to see tat l 1 < χ cr (H) l wit equality if and only if all l-colorings of H are balanced, namely, all color classes ave te same size. Suppose H as connected components C 1,..., C kc. We define cf c (H) as cf( C 1,..., C kc ), te igest common factor of integers C 1,..., C kc. Given an l-coloring C of H wit x 1 x... x l as te sizes of te color classes, let D(C) = {x i+1 x i i = 1,..., l 1}. Let D(H) = D(C) were te union ranges over all l-colorings of H. Define cf χ (H) as te igest common factor of D(H). In particular, we set cf χ (H) = if D(H) = {0}. Lastly, we say tat H is in Class 1 if { cfχ (H) = 1 wen χ(h) cf χ (H) and cf c (H) = 1 wen χ(h) =, oterwise H is in Class. (Te autors of [13] used cf(h) = 1 to denote te case wen H is in Class 1.) Teorem 1.1 ([13]). For every grap H on vertices, tere exist integers C and m 0 suc tat for all integers m m 0, if G is a grap on n = m vertices ten te following olds. If { (1 1/χcr (H))n + C if H is in Class 1 δ(g) (1 1/χ(H))n + C if H is in Class, ten G contains an H-factor. It was also sown in [13] tat Teorem 1.1 is best possible up to te constant C. Oter results and metods for tiling problems can be found in a recent survey of Kün and Ostus [14]. Rater tan working wit an arbitrary grap G, one may restrict G to be r-partite and tile it wit some r-partite grap H. Altoug it sounds like a special case, multipartite tiling is stronger tan general tiling in te following sense. First, a result on multipartite tiling does not follow from te corresponding general result. For example, an arbitrary grap G of order n contains a perfect matcing if δ(g) n/ (Dirac [4]), wile a bipartite grap B wit two partition sets of size n/ contains a perfect matcing if δ(b) n/4 (König-Hall [7]). Second, a result on multipartite tiling often implies one for general tiling. For example, suppose we know tat every bipartite grap wit two partition sets of size n/ and minimum degree at least n/4 contains a perfect matcing (assumed tat n is even). Let G be an arbitrary grap G wit δ(g) n/ + ɛn for some ɛ > 0. By taking a random balanced bipartition of G, we get a bipartite spanning subgrap B wit δ(b) δ(g) o(n) n/4 (as n ). Ten B contains a perfect matcing, wic is also a perfect matcing of G. In tis paper we consider tiling in a balanced bipartite grap, were an r-partite grap is balanced if all partition sets ave te same size. Zao [0] determined te minimum degree tresold for a K s,s -factor in a balanced bipartite grap for all s. Hladký and Scact [8] and Czygrinow and DeBasio [3] later determined te minimum degree tresold for a K s,t -factor for s < t. Given any bipartite H of order, since K, contains an H-factor, te result in [0] gives a sufficient condition for an H-factor. Teorem 1. ([0]). Let H be a bipartite grap of order. Suppose tat n is sufficiently large and divisible by. If G is a balanced bipartite grap on n vertices suc tat δ(g) n + 3, ten G contains an H-factor.
We first sow tat Teorem 1. is best possible (up to an additive constant) wen H is in Class. Proposition 1.3. Let H be a bipartite grap on vertices. We assume G to be a balanced bipartite grap on n = m vertices were m N. 1. If H is in Class, ten tere exists a G suc tat δ(g) = n 1 and G does not contain an H-factor.. If H is in Class 1, ten tere exists a G suc tat ( δ(g) = 1 and G does not contain an H-factor. 1 χ cr (H) ) n 1 Zao [0] asked about te minimum degree condition for H-factors in bipartite graps and suggested using eiter χ(h) = or χ cr (H) as in Teorem 1.1. Te main result of tis paper answers tis affirmatively; it can be viewed as a bipartite analog of Teorem 1.1. Teorem 1.4. Let H be a bipartite grap in Class 1 wit vertices. Tere exist positive integers m 0 and c 1 (H) 4 3 suc tat te following olds for all integers m m 0. If G is a balanced bipartite grap on n = m vertices suc tat ( ) 1 δ(g) 1 n + c 1 (H), χ cr (H) ten G contains an H-factor. Proposition 1.3, Part, sows tat Teorem 1.4 is best possible up to te value of c 1 (H). Our constant c 1 (H) is on te order of 3, and its exact value is specified in (8) of Teorem 4.9. It is muc smaller tan te constant C in Teorem 1.1, wic depends on te Regularity Lemma. Neverteless, we are unable to determine te best possible value of c 1 (H) as in [3, 8, 0]. In oter words, we determine te minimum degree tresold for bipartite tiling as follows. Given a bipartite grap H of order, let δ (n, H) denote te smallest integer k suc tat every balanced bipartite grap G of order n, wic is divisible by, wit δ(g) k contains an H-factor. Proposition 1.3 and Teorem 1.4 togeter imply tat { (1 1/χcr (H))n + O(1) if H is in Class 1 δ (n, H) = (1 1/χ(H))n + O(1) if H is in Class. Zao [0] also asked for te minimum degree tresold for an almost perfect H-tiling. Komlós [9] sowed tat for any grap H, every grap G wit n vertices and δ(g) (1 1/χ cr (H))n contains an H-tiling tat covers all but at most o(n) vertices. Sokoufande and Zao [18] improved o(n) to a constant, O( ), were is te order of H. In tis paper we prove a similar result for bipartite tiling. Teorem 1.5. Let H be a bipartite grap wit vertices. Tere exist integers n 0 and c (H) < 8 suc tat every bipartite grap G wit n n 0 vertices in eac partition set contains an H-tiling tat covers all but at most c (H) vertices if δ(g) (1 1/χ cr (H))n. 3
It is important to note tat Kün and Ostus [13] started teir proof of Teorem 1.1 wit te result of Komlós (alternatively tey could use te one of Sokoufande and Zao), wic gives an almost tiling of G, and ten modified it into a perfect tiling under te assumption tat H is in Class 1. Wile proving Teorem 1.4, since tere is no Komlós teorem available, we first find an almost-tiling (wic leaves o(n) vertices uncovered) by using te approac in [18]. If H is in Class 1, ten we modify it into a perfect H-tiling, oterwise we modify it into an H-tiling tat leaves only O( ) vertices uncovered. Te structure of te paper is as follows. We prove Proposition 1.3 in Section. In Section 3, we lay some groundwork for our proofs: we state bipartite versions of te Regularity Lemma and Blowup Lemma. Section 4 provides te proof of Teorem 1.4, wic is divided into te nonextremal case and te extremal case. Section 5 gives te proof of Teorem 1.5 based on te one of Teorem 1.4. In te last section we give concluding remarks, including a conjecture on r-partite tiling. Notation. Fix a grap. For two vertices x, y, we write x y if x is adjacent to y. Let Γ(x) denote te set of neigbors of x and deg(x) = Γ(x). For a vertex set S, let Γ(x, S) = Γ(x) S and deg(x, S) = Γ(x, S). A bipartite grap G[X, Y ] means a bipartite grap wit partition sets X and Y. Wen G is given and A, B V (G) are two disjoint sets, we use G[A, B] to denote te bipartite subgrap induced on A B and its size is denoted by e(a, B). Te density of A and B is te ratio d(a, B) = e(a, B)/( A B ). We will use te notation δ(x, Y ) to denote te minimum degree of a vertex in X into a set Y. In oter words, δ(x, Y ) = min x X deg(x, Y ). Note tat in general δ(x, Y ) δ(y, X). Trougout tis paper we assume tat H is a bipartite grap on vertices suc tat σ(h) = u and σ(h) = w. Let C 1,..., C kc be its connected components. Ten eac component C i as a unique -coloring {U i, W i } wit W i U i. Let c i = C i = W i + U i and d i = W i U i. Recall tat cf c (H) = cf(c 1,..., c kc ). We now define cf χ,c (H) as cf(d 1,..., d kc ). Given integers a 1,..., a k wit cf(a 1,..., a k ) = d, it is well known tat tere are integers b 1,..., b k suc tat a 1 b 1 + + a k b k = d. Tey are called te Bézout numbers of a 1,..., a k. We need te following fact on te Bézout numbers. Fact 1.6. For any k positive integers a 1,..., a k wit cf(a 1,..., a k ) = d, we may find te Bézout numbers b 1,... b k suc tat max 1 i k b i max 1 i k a i /d. Proof. It suffices to prove te fact for d = 1. In fact, if cf(a 1,..., a k ) = d > 1, ten cf(a 1,..., a k ) = 1, were a i = a i/d. If b 1,..., b k are te Bézout numbers of a 1,..., a k suc tat max 1 i k b i max 1 i k a i, ten tey are te Bézout numbers of a 1,..., a k wit max 1 i k b i max 1 i k a i /d. Suppose tat cf(a 1,..., a k ) = 1. We first sow tat tere exist te Bézout numbers b 1,..., b k suc tat max i k b i a 1. In fact, let b 1,..., b k be te Bézout numbers wit te minimum k i= b i. We claim tat b i a 1 for all i > 1. Suppose instead, tere exists i > 1 suc tat b i > a 1. If b i > 0, ten define b i = b i a 1 and b 1 = b 1 + a i ; oterwise let b i = b i + a 1 and b 1 = b 1 a i. Let b j = b j for j > 1 and j i. Ten a 1 b 1 + + a kb k = a 1b 1 + a b + + a k b k = 1. Since k i= b i = k i= b i a 1 < k i= b i, we obtain a contradiction. Next, among all te Bézout numbers b 1,..., b k satisfying max i k b i a 1, we assume tat b 1,..., b k ave te minimum b 1. We claim tat tis b 1 max i k a i and consequently b 1,..., b k are te desired Bézout numbers. Suppose instead, tat b 1 > max i k a i. Since a 1 b 1 + +a k b k = 1 and b 1 > 1, tere exists i > 1 suc tat b i as te opposite sign of b 1. Let b 1 = b 1 a i and b i = b i+a 1 if b 1 > 0 (tus b i < 0); oterwise let b 1 = b 1 + a i and b i = b i a 1. Let b j = b j for j > 1 and j i. Ten a 1 b 1 + + a kb k = a 1b 1 + a k b k = 1 and b 1,..., b k are te Bézout numbers wit b 1 < b 1 and b j a 1 for j > 1, a contradiction. We call te Bézout numbers wit minimum max 1 i k b i te smallest Bézout numbers. 4
Definition 1.7. Let H be a bipartite grap wit connected components C 1,..., C kc. Suppose tat C i = C i [U i, W i ] wit W i U i. Let c i = W i + U i and d i = W i U i. Recall tat cf c (H) = cf(c 1,..., c kc ) and cf χ,c (H) = cf(d 1,..., d kc ). 1. We define ζ(h) = max 1 i k c ζ i, were ζ 1,..., ζ kc are te smallest Bézout numbers of c 1,..., c kc.. We define β(h) = max 1 i k c β i, were β 1,..., β kc are te smallest Bézout numbers of d 1,..., d kc. Given H as in Definition 1.7, Fact 1.6 implies tat Proof of Proposition 1.3 ζ(h) max 1 i k c c i and β(h) max 1 i k c d i w u. (1) We first observe connections among cf c (H), cf χ (H) and cf χ,c (H). Lemma.1. Let H be any bipartite grap. 1. Ten cf χ,c (H) cf χ (H) cf χ,c (H).. If cf χ,c (H) =, ten cf c (H). 3. Suppose cf c (H) = 1. Ten cf χ (H) if and only if cf χ,c (H) = 1. Proof. Suppose tat H as k c connected components C 1 [U 1, W 1 ],..., C kc [U kc, W kc ]. Let c i = C i and d i = W i U i. Part 1. We ave cf χ (H) = cf(a), were A = { k c i=1 e id i : e i { 1, 1}} is te set of all combinations of adding and subtracting d 1,..., d kc. Terefore it suffices to sow tat cf(d 1,..., d kc ) cf(a) cf(d 1,..., d kc ). In fact, letting d = cf(d 1,..., d kc ) and q = cf(a), we ave d q because d divides every element of A. On te oter and, for any i, q divides d 1 +... + d kc and d 1 + + d i 1 d i + d i+1 + + d kc and tus q divides d i. Terefore q cf(d 1,..., d kc ) = d. Part. Suppose tat cf χ,c (H) =. Ten for eac component C i of H, d i is even. Tis means U i and W i ave te same parity and c i is even for all i. Tis implies tat cf c (H). Part 3. If cf χ (H), ten by Part 1, cf χ,c (H). If cf χ,c (H) =, ten by Part, cf c (H) contradicting our assumption. Terefore cf χ,c (H) = 1. On te oter and, if cf χ,c (H) = 1, ten cf χ (H) directly follows from Part 1. We now prove Proposition 1.3 by using Lemma.1 and four constructions. Proof of Proposition 1.3. Te proof consists of four (mutually disjoint) cases. Te first tree cases togeter prove te existence of a grap G wit δ(g) = n 1 but containing no H-factor wen H is in Class. Te last case provides a grap G wit δ(g) = [1 (1/χ cr (H))]n 1 but containing no H-factor wen H is in Class 1. Case 1: cf c (H) 3. Let G = K n, n +1 K n, n 1. Since cf c (H) 3, and any component of H must fit entirely into one of te two connected components of G, we can deduce te following. Te size of te components of G differ by ; but te size of te components of H differ by multiples of cf c (H) wic is at least 3. Tus, tere is no way to arrange te components nor te copies of H to even out te sizes of te components of G. So G contains no H-factor. 5
Case : cf c (H) =. Ten eac component of H as an even size. If n is odd, let G = K n, n K n, n. If n is even, let G = K n, n +1 K n, n 1. In eiter case, since every component of G as an odd size, G does not contain an H-factor. Case 3: cf c (H) = 1 and cf χ (H) 3. Let G = K n +1, n 1 K n 1, n +1. It is an immediate consequence of Lemma.1 tat if cf χ (H) 3 and cf c (H) = 1, ten cf χ,c (H) 3. (Note tat tis does not imply cf χ,c (H) cf χ (H).) Now, te sizes of te color classes of te connected components of G differ by 1 or. Since cf χ,c (H) 3, we can only adjust te relative sizes of te color classes of te connected components of G by multiples of cf χ,c (H); so we can never get an H-factor. Case 4: cf c (H) = 1 and cf χ (H), namely, H is in Class 1. Recall tat H =, u = σ(h), w = σ(h), and 1 1/χ cr (H) = u. Let G = K nu 1, nw +1 K nw nu +1, 1. Ten δ(g) = [1 (1/χ cr (H))]n 1. Let C 1, C,..., C kc be te components of H. By contradiction, suppose G as an H-factor. Te color class of G wit size σ(g) tus contains one color class from eac of te mk c packed components of H. Tus k c σ(g) m σ(c i ) = mu. i=1 However, it is easy to see tat σ(g) = mu by simply placing te components of size nu 1 = 1 in te same color class. Tis is a contradiction. So G contains no H-factor. mu 3 Regularity Lemma and Oter Tools Te Regularity Lemma [19] and te Blow-up Lemma [10] are te backbone of our proof. Tey allow us to gain convenient structural properties from an arbitrary grap G. Before stating te lemmas, we define ɛ-regularity, and (ɛ, δ)-super-regularity. Definition 3.1. Let ɛ, δ > 0. Let G be a grap wit disjoint vertex sets X and Y. (1) We say te pair (X, Y ) is ɛ-regular if for every A X and B Y satisfying A > ɛ X, B > ɛ Y we ave d(a, B) d(x, Y ) < ɛ. () Te pair (X, Y ) is (ɛ, δ)-super-regular if (X, Y ) is ɛ-regular and deg(x, Y ) > δ Y for every x X and deg(y, X) > δ X for every y Y. Te next two lemmas follow from te definition of ɛ-regularity easily; teir proofs can be found in te survey [1]. Lemma 3. (Slicing Lemma). Let ɛ, d > 0 be constants. Let (X, Y ) be an ɛ-regular pair wit density d. For any γ > ɛ, if X X, Y Y and X γ X, Y γ Y, ten (X, Y ) is an ɛ -regular pair wit density d were d d < ɛ and ɛ = max{ɛ, ɛ γ }. Lemma 3.3 (Embedding Lemma). Let d ɛ > 0. If (X, Y ) is an ɛ-regular pair wit density d, ten for any positive integers a, b, tere exists an n 0 suc tat if X, Y n 0, ten K a,b (X, Y ). Now we are ready to state te bipartite form of Szemerédi s Regularity Lemma (see [1] for various forms of te Regularity Lemma). Lemma 3.4 (Regularity Lemma - Bipartite form). For every ɛ > 0, tere exists an M R + suc tat if G = (X, Y ; E) is any bipartite grap wit X = Y = n, and d [0, 1] is any real number, ten tere is a partition of X into clusters X 0, X 1,..., X k, a partition of Y into Y 0, Y 1,..., Y k, and a spanning subgrap G = (X, Y ; E ) wit te following properties: 6
k M X 0 = Y 0 ɛn X i = Y j ɛn for all 1 i, j k deg G (v) > deg G (v) (d + ɛ)n for all v / X 0 Y 0 All pairs (X i, Y j ), 1 i, j k, are ɛ-regular in G, eac wit density eiter 0 or greater tan d. Te Blow-up Lemma is very useful for grap tiling, especially wen combined wit te Regularity Lemma as it essentially says tat, wen embedding a grap of bounded maximum degree, an (ɛ, δ)- super-regular pair beaves like a complete bipartite grap. We only need te bipartite form of tis lemma. Lemma 3.5 (Blow-up Lemma - Bipartite form). For every δ, > 0, tere exists an ɛ > 0 suc tat te following olds. Let (X, Y ) be an (ɛ, δ)-super-regular pair. If a bipartite grap H wit (H) can be embedded in K X, Y, ten H can be embedded in (X, Y ). We now give a sufficient condition for a complete bipartite grap to contain an H-factor. Lemma 3.6. Let H be a bipartite grap on vertices suc tat cf χ,c (H) = 1. Suppose tat β = β(h), u = σ(h), and w = u. Let G = K mu+t,mw t suc tat t = q(w u) + r and r < w u for non-negative integers m, t, q, r. If m rβ + q and q rβ, ten G contains an H-factor. Proof. K mu,mw as a natural H-factor wit all copies of H aving teir smallest color classes on one side and te largest color classes on te oter side. We will sow ow to transform tis into an H-factor of G. First, since m q we can take q copies of H and swap teir sides (ere swapping means switcing te sides of te color classes). Tis now results in a spanning subgrap of K mu+t r,mw t+r. Let us call te part of tis tiling tat was not swapped as G 1 and te part tat was swapped as G. Since cf χ,c (H) = 1, tere exist integers β 1,..., β kc as in Definition 1.7. Let us say tat β 1,..., β i are nonnegative and β i+1,..., β kc are all negative. Now, in G 1 swap rβ j copies of C j for all j = 1,..., i. Note tat since m q rβ, we ave enoug copies of eac component to perform te aforementioned swaps. In G, swap rβ j copies of C j for all j = i + 1,..., k c. We can perform tis swap because q rβ. So, te left side gains r = rβ 1 d 1 +... + rβ i d i + rβ i+1 d i+1 +... + rβ kc d kc vertices. Similarly, te rigt side loses r vertices, and we now ave a spanning subgrap of K mu+t,mw t = G. We will use te following corollary of Lemma 3.6 in Section 4.1, wic is sligtly stronger tan te bipartite version of Lemma 1 in [13]. Corollary 3.7. Let H be a bipartite grap in Class 1 wit vertices. Let u = σ(h), w = σ(h), and 0 < γ < w u u. Suppose tat G[X, Y ] is a complete bipartite grap on m vertices for some sufficiently large integer m suc tat (1 + γ) u w X Y 1. Ten G contains an H-factor. 7
Proof. We will prove tat G satisfies te conditions of Lemma 3.6 in order to get an H-factor. First, since X + Y is divisible by, we may write G = K mu+t,mw t were m = ( X + Y )/ and t is some integer. Furter write t = q(w u) + r for some integers q, r suc tat 0 r < w u. Let β = β(h) and m (w u) ( + uγ)β. () uwγ We must prove m rβ + q and q rβ. Since q = w u t m rβ + and (ii) rβ. Since t w u w u we ave tat t + tuγ mwuγ or t X Y = mu + t mw t (1 + γ) u w, uw +uγ mγ. Now, by (), we ave uw t t w u, it is sufficient to prove tat (i) +uγ mγ (w u) β, wic implies tat t (w u) β > (w u)rβ tus proving (ii). On te oter and, X Y 1 implies tat mu + t mw t, or t m(w u). Since t (w u)rβ, we ave m(w u) (w u)rβ + t, wic gives (i). 4 Proof of Teorem 1.4 Let H be a bipartite grap on vertices wit positive integers u = σ(h) and w = u. We assume tat u < w oterwise χ cr (H) = and Teorem 1. gives te proof. We tus ave w, and 3. Te proof of our main teorem consists of two parts: te nonextremal case and te extremal case. Rougly speaking, a balanced bipartite grap wit n = m vertices is in te extremal case if it is relatively similar to K nu 1, nw +1 K nw nu +1, 1, te construction we gave in Case 4 of te proof of Proposition 1.3. 4.1 Nonextremal Case In tis subsection we prove te following teorem, wic covers te nonextremal case. Teorem 4.1. Let H be a bipartite grap on vertices suc tat H is in Class 1. Let u = σ(h) and w = σ(h). For every α > 0 tere exist γ > 0 and a positive integer m 0 suc tat if m m 0 and G[X, Y ] is a balanced bipartite grap on n = m vertices wit δ(g) ( 1 1 χ cr (H) γ ten G eiter contains an H-factor or tere exist sets A X, B Y suc tat A = B = and d(a, B) α. ) n wn We say tat a bipartite grap G[X, Y ] is in te extremal case wit parameter α if tere exist sets A X, B Y suc tat A = B = wn and d(a, B) α. Te proof of Teorem 4.1 is divided into two lemmas. Te first lemma puts most vertices of G into super-regular pairs suc tat te ratio of te sizes between te pairs is sligtly larger tan u/w. Having a ratio sligtly larger tan u/w allows us to remove a small amount of vertices from te super-regular pair yet its remaining vertices can be tiled by H perfectly by applying Corollary 3.7 and Lemma 3.5. We make tis precise by te following definition. 8
Definition 4.. Given 0 < ɛ < d < 1 and positive integers p, q, N, let G[X, Y ] be a balanced bipartite grap. A partition of V (G) = X 0 Y 0 P 1 Q 1... P k Q k is called an almost (ɛ, d, p, q, N)-cover of G if X 0 X, Y 0 Y, and X 0, Y 0 ɛn For all i, P i /p = Q i /q N and eiter P i X and Q i Y, or P i Y and Q i X For all i, (P i, Q i ) is (ɛ, d)-super-regular. Lemma 4.3. Let w > u be positive integers and = w + u. For every α > 0 and integer N, tere exists a positive integer n 0 and constants 0 < ɛ d γ α suc tat if G[X, Y ] is a balanced bipartite grap on n vertices wit n n 0, and δ(g) ( u γ) n, ten eiter G is in te extremal case wit parameter α or G contains an almost (ɛ, d, p, q, N)-cover, were p = w + u γ and q = w γ are integers. Tere are two reasons wy we cannot immediately apply Corollary 3.7 to eac (P i, Q i ) in te cover. First, we need to get rid of te exceptional sets X 0 and Y 0. Second, we may not ave P i + Q i divisible by. Acieving tese two additional properties is te content of Lemma 4.4, in wic we also assume H is in Class 1. By definition, if H is in Class 1, ten cf c (H) = 1 and cf χ (H). By Part 3 of Lemma.1, tis implies tat cf c (H) = 1 and cf χ,c (H) = 1. Te condition of cf c (H) = 1 is used for acieving te divisibility of P i + Q i. Te condition of cf χ,c (H) = 1 is needed for Corollary 3.7. Lemma 4.4. Let H be a bipartite grap wit cf c (H) = 1 and cf χ,c (H) = 1. Let u = σ(h) and w = u. Let G be a balanced bipartite grap on n = m vertices suc tat δ(g) (1 1/χ cr (H) γ) n. Suppose tat G contains an almost (ɛ, d, p, q, N)-cover for some positive ɛ d γ 1, integers p, q satisfying p/q = (1 + γ)u/w, and sufficiently large N. Ten G contains an H-factor. Proof of Lemma 4.3. In te proof we will omit te floor function wen it does not affect our calculations. Assume n is large. Witout loss of generality, assume α 1. We coose parameters ɛ 0, d 0, γ so tat tey satisfy te following relations ɛ 0 d 0 γ = 1 z α (3) for some integer z. Let p = uz + w and q = wz be two integers. Ten p and q ave te following property: u w < p q = u + γ 1. (4) w We apply te Regularity Lemma (Lemma 3.4) wit parameters ɛ 0 and d 0 to G. We obtain an integer k 0 M(ɛ 0 ) and a spanning subgrap G consisting of clusters X 1, Y 1,... X k0, Y k0 of size N 0 ɛ 0 n and exceptional sets X 0 and Y 0 of size at most ɛ 0 n. Every pair of clusters (X i, Y j ) is ɛ 0 -regular, wit density eiter 0 or greater tan d 0. Te degrees of te vertices in G are very close to teir degrees in G: ( u ) deg G (v) > deg G (v) (d 0 + ɛ 0 )n = γ d 0 ɛ 0 n Let R be te reduced grap of G were eac vertex corresponds to a cluster in G (X 0 Y 0 ), and we say tere is an edge between X i and Y j if te density d(x i, Y j ) > d 0, written as X i Y j. 9
Note tat we use te same notation for a cluster in G and a vertex in R; we clearly say weter it is a cluster of G or a vertex of R wen tis is not clear from te context. In order to bound δ(r), we consider an arbitrary X i and an arbitrary vertex x X i. We ave ( u γ d 0 ɛ 0 ) n deg G (x) Y 0 Y j X i Y j = deg R (X i )N 0. (5) Using (3) and k 0 N 0 n, we derive tat deg R (X i ) ( u γ)k 0. Te same olds for any cluster in Y. Tus we ave δ(r) ( u γ)k 0. (6) We need a simple fact on te size of a maximum matcing in bipartite graps; for completeness, we include a proof. Fact 4.5. If G[X, Y ] is a bipartite grap wit minimum degree δ suc tat X Y, ten G as a matcing of size at least min{δ, X }. Proof. Let M = {x 1 y 1,..., x t y t } be a maximum matcing in G. Assume t < X. Ten, tere exists a vertex x X {x 1,..., x t }. Since Y X, tere also exists y Y {y 1,..., y t }. Let I = {1 i t : y i Γ(x)} and J = {1 j t : x j Γ(y)}. Ten I, J δ. Since M is a maximum matcing, we ave I J = and I, J δ (oterwise we may extend te matcing). Tis implies tat t I + J δ. Let M be a maximum matcing in te reduced grap R. Since u < w + u =, by Fact 4.5, we ave M δ(r) ( u γ)k 0. Denote by U 1 and U te set of unmatced clusters from X and Y respectively. Ten U 1, U ( w u + γ)k 0. Te next part of te proof will be decomposing clusters to get pairs of ratio p q. We first prove tat we can find two disjoint subgraps P 1 and P of R tat satisfy te following properties. Te subgrap P 1 will ave vertex sets U 1 and Γ(U 1 ) := Xi U 1 Γ(X i ). Moreover, for any vertex X i U 1, deg P1 (X i ) = p, and for any vertex Y j Γ(U 1 ), deg P1 (Y j ) q p. Te subgrap P will ave vertex sets U and Γ(U ); for any Y j U, deg P (Y j ) = p, and for any X i Γ(U ), deg P (X i ) q p. Note tat since M is maximal, Γ(U 1 ), Γ(U ) V (M) and no edge of M as one end in Γ(U 1 ) and te oter end in Γ(U ). W 1 P 1 Γ(U 1 ) Γ(U ) W U 1 U P Figure 1: Finding P 1 and P Let α = α/1. We prove te following claim: Claim 4.6. (a) If U 1, U ( w u α )k 0, ten tere exist two disjoint subgraps P 1 and P wit te above properties. (b) If U 1, U > ( w u α )k 0, ten G is in te extremal case wit parameter α. 10
Proof. We first prove (a). We will only prove tat we can find P 1 because te proof for P is te same. We will find P 1 by te greedy algoritm. Arbitrarily order te vertices in U 1. For eac vertex in U 1, we find p neigbors in Γ(U 1 ) wit te restriction tat we cannot coose any vertex in Γ(U 1 ) more tan q p times. Wen considering te it vertex in U 1, suppose tat tere are t vertices in Γ(U 1 ) tat ave been cosen q p times. Since t (i 1)p/(q p) < U 1 p/(q p), it suffices to sow tat δ(r) p + U 1 p/(q p). Using (6) and U 1 ( w u α )k 0, we ave δ(r) p ( u ) q p U 1 γ k 0 p q p ( ) w u α k 0. From te Regularity Lemma, we know tat k 0 1 ɛ 0. Tus, it suffices to sow tat ( u ) ( ) w u p φ := γ α q p ɛ 0p. In fact, te definition of p, q and te assumption z w w u, wic follows from γ 1, give tat p q p u w u = uz + w (w u)z w u w u = w ((w u)z w)(w u) w (w u) z. By using (3), we obtain tat ( u ) ( ) ( w u u φ γ α w u + w ) (w u) z > γ w γ (w u) + Tus, te greedy algoritm is sufficient to find te subgraps P 1 and P. u w u α ɛ 0 p. Now, we prove (b). We assume U 1, U > ( w u α )k 0. Let W i be te neigbors of Γ(U i ) in M for i = 1,. It is easy to see tat te following four quantities must all be equal to 0 or we can extend te matcing in G: e(u 1, U ) = e(u 1, W ) = e(u, W 1 ) = e(w 1, W ) = 0. For example, if tere exists an edge X i Y j between W 1 and W, ten we can extend te matcing as follows. Let Y i denote te matced neigbor of X i, X j denote te matced neigbor of Y j, X i denote a vertex in U 1 adjacent to Y i, and Y j denote a vertex in U adjacent to X j. Ten we can enlarge te matcing by replacing X i Y i, X j Y j by X i Y i, X i Y j, and X j Y j. Now, letting A = U 1 W 1, and B = U W, ten e R (A, B) = 0. Moreover, ( ) w u ( u ) ( w ) A = U 1 + W 1 U 1 + δ(r) α k 0 + γ k 0 = α γ k 0. Let A and B be te sets of vertices of G in all te clusters of A and of B respectively. Since k 0 N 0 (1 ɛ 0 )n and ɛ 0 γ α, we derive tat A ( w α )n. Te same olds for B. Since e G (A, B ) = 0, for any subset S A, we ave e G (S, B ) e G (A, B ) + S (d 0 + ɛ 0 )n d 0 n S. Now, by adding at most α n vertices to A and B, we get two sets A, B of size exactly wn ; wen A or B is greater tan wn wn, we simply take a subset of size. Since eac of te new vertices in A (or B) migt be adjacent to all te vertices in B (or A), we ave d(a, B) e G(A A, B ) + α n B + α n A A B So, we are in te extremal case wit parameter α. 11 = d 0n + 4α n B 1α = α
We assume tat G is not in te extremal case wit parameter α, and tus Claim 4.6 (a) olds. Now we use te structures of P 1 and P to guide us to break up clusters. In order to evenly divide a cluster into small pieces, we ensure te size of all clusters is divisible by pq(q p ) by moving at most pq(q p ) 1 vertices from eac cluster to te exceptional set. Tis increases X 0 and Y 0 by a constant, less tan pq(q p )k 0. For simplicity we still use N 0 for te size of te clusters. Now we only give te details on ow to andle te clusters in U 1 Γ(U 1 ). We evenly decompose every cluster X i U 1 into p subclusters and adjoin eac subcluster to a unique neigbor of X i in P 1. Since deg P1 (X i ) = p for eac X i U 1, tis is possible. However, we do not adjoin eac subcluster of X i to te entire cluster. Instead, we adjoin it to a subcluster of size N 0 q. Tus, te ratio between two adjoining subclusters is p q. Size N 0 p Size N 0 q X i U 1 p neigbors of X i Figure : Decomposing one cluster in U 1 Let Y j Y be a cluster covered by te matcing M. We know tat Y j as degree i q p in P 1 (i = 0 wen Y j Γ(U 1 )). In total, in 0 q vertices of Y j are already used. We matc up te remaining N 0 in 0 q vertices in Y j wit its neigbor X j in M forming at most 3 cluster pairs of ratio p q as follows. First take in 0 q p vertices from X j and matc tem wit ipn 0 q(q p) vertices from Y j. Tis makes a cluster pair wit ratio p q. Now, te number of remaining vertices in X j is N 0 in 0 q p, wile te number of remaining vertices in Y j is N 0 in 0 q ipn 0 q(q p), also equal to N 0 in 0 q p. Finally, we make two more cluster pairs wit ratio p q by pairing togeter (N 0 in 0 q p )( p q+p cluster wit (N 0 in 0 ) from te oter. q p )( q q+p ) vertices from one In summary, we broke all te clusters into subclusters and group tem into pairs wit sizes { N0 p, N } { } { 0 in0, q q p, ipn 0 q p i q, q(q p) q p p + q N 0, q p i } p q p p + q N 0, (7) were 0 i q p. Let γ = min{ 1 q, p }. (ten γ > d > ɛ q p 0 by (3)). Te size of any subcluster is at least γ N 0, wic is larger tan te given integer N because N 0 (1 ɛ 0 ) n k 0 is sufficiently large. Let (P 1, Q 1 ),..., (P k, Q k ) denote tese cluster pairs. After relabeling, we may assume tat te first k 1 of tem ave P i in X and Q i in Y (see Figure 3). We ave k pk 0 because eac cluster in U 1 U generates at most p pairs, wile eac cluster covered by M generates at most 3 pairs, and p 3. Te ɛ 0 -regularity between te original clusters implies tat all (P i, Q i ) ave density witin ɛ 0 of d 0. Lemma 3. furter guarantees tat all (P i, Q i ) are ɛ 1 -regular wit ɛ 1 = ɛ 0 /γ. 1
P 1 Q 1 P k1 Q k1 Cluster-Pairs of Ratio p q Q k1 +1 P k1 +1 Q k P k X 0 Exceptional Sets Y 0 Figure 3: Grap G after decomposition In order to obtain super-regularity for eac (P i, Q i ), we now remove vertices wit small degree into te opposite cluster to te exceptional sets X 0, Y 0. Suppose tat, for example, P i X and Q i Y. We move any vertex x P i suc tat deg(x, Q i ) < (d(p i, Q i ) ɛ 1 ) Q i to X 0, and any vertex y Q i suc tat deg(y, P i ) < (d(p i, Q i ) ɛ 1 ) P i to Y 0. Te ɛ 1 -regularity between P i and Q i guarantees tat we move at most ɛ 1 C vertices from eac C {P i, Q i }. In order to maintain te ratio to be exactly p q, we may ave to move more vertices from P i to X 0 and from Q i to Y i suc tat, in total, P i loses at most p ɛ 1 P i /p ɛ 1 P i + p ɛ 1 P i vertices wile Q i loses at most q ɛ 1 Q i /q ɛ 1 Q i + q ɛ 1 Q i vertices. We still denote te resulting clusters by P i and Q i. Since te original P i as at least γ N 0 vertices, te modified P i as at least (1 ɛ 1 )γ N 0 vertices. By Lemma 3., te modified (P i, Q i ) is ɛ 1 -regular. Since te density between te original P i and Q i is at least d 0 ɛ 0, te modified (P i, Q i ) satisfies deg(x, Q i ) (d 0 ɛ 0 ɛ 1 ) Q i for any vertex x P i, and deg(y, P i ) (d 0 ɛ 0 ɛ 1 ) P i for any vertex y Q i. Let ɛ = ɛ 1 and d = d 0 ɛ 0 ɛ 1. Ten all (current) (P i, Q i ) are (ɛ, d)- super-regular. In total, we moved at most C (ɛ 1 C + q) ɛ 1 n + kq vertices to X 0 were te sum ranges over all current clusters contained in X. As a result, X 0 ɛ 0 n + pq(q p )k 0 + ɛ 1 n + kq ɛn. Te same olds for Y 0. Proof of Lemma 4.4. Let X 0, Y 0, P 1, Q 1,..., P k, Q k be te given almost (ɛ, d, p, q, N)-cover of G. As before, we call X 0, Y 0 exceptional sets, and P i, Q i, i = 1,..., k clusters. We know tat X 0, Y 0 ɛn, all pairs (P i, Q i ) are (ɛ, d)-super-regular wit P i / Q i = p q = u w + γ. Our first goal will be to take vertices in X 0 Y 0 and find disjoint copies of K u,w (a supergrap of H) for eac of tem. Claim 4.7. We may remove X 0 Y 0 disjoint copies of K u,w from G, eac of wic contains exactly one vertex from X 0 Y 0, suc tat eac cluster C {P i, Q i } loses at most d 3 C vertices. Proof. We say tat a vertex v is adjacent to a cluster C (written as v C) if deg(v, C) d C. Following an arbitrary order of X 0 and Y 0, we associate eac vertex x X 0 Y 0 to a cluster C tat x is adjacent to. We also say tat x is associated wit te cluster pair (P i, Q i ) if C {P i, Q i }. First assume tat C = P i. By Lemma 3., (Γ(x, P i ), Q i ) is ε/d-regular and by Lemma 3.3, (Γ(x, P i ), Q i ) contains a copy of K u,w 1 wit w 1 vertices in Q i. We ten remove tis copy of K u,w 1 togeter 13
wit x (tey form a copy of K u,w ). Wen C = Q i, we remove a copy of K u 1,w from (P i, Γ(x, Q i )) wit u 1 vertices in P i. Togeter wit x, te removed vertices form a copy of K u,w. To ensure tat eac cluster C loses at most d d 3 C vertices, we associate at most 3w Q i vertices of X 0 Y 0 to any pair (P i, Q i ). Ten Q i loses at most d 3 Q i vertices because eac associated vertex of X 0 Y 0 makes Q i lose at most w vertices. On te oter and, P i loses at most u vertices for eac associated vertex. Since Q i /w P i /u, P i loses at most u d 3w Q i d 3 P i vertices. We need to prove tat under tis restriction, tere are enoug clusters for all te vertices in te exceptional sets. First we give a lower bound for x C C for all x X 0 Y 0. Fix x X 0 (te case wen x Y 0 is similar). By te minimum degree condition and te definition of x C, ( u γ)n d G(x) Y 0 + C + d C ɛn + dn + C x C C Y :x C x C wic implies tat x C C ( u γ)n by using ɛ d γ. For a cluster C {P i, Q i } wit x C, if we ave associated d 3w Q i d 3w C exceptional vertices wit (P i, Q i ), ten we can not associate x wit C. If all te clusters C adjacent to x can not be used, ten te number of exceptional vertices tat ave been considered is at least d 3w C d 3w (u γ)n > ɛn, a contradiction. x C Oter tan a small number of copies of K u,w, te grap G now consists of cluster pairs (P i, Q i ) wit ratio near p q. In order to apply Corollary 3.7 to tese (P i, Q i ), we want P i + Q i to be divisible by. We use te fact tat cf c (H) = 1 and let ζ = ζ(h). Claim 4.8. We may remove at most ζk disjoint copies of H suc tat eac cluster C {P i, Q i } loses at most ζ vertices, and all P i + Q i are divisible by. Proof. Recall tat k c i=1 ζ ic i = 1 and ζ = max 1 i kc ζ i, were c 1,..., c kc are te sizes of te components of H. In order to ensure tat te size of eac cluster pair is divisible by, we sow ow to increase or decrease te size of a cluster pair by 1 modulo. Let G 1 and G denote te subgraps induced by two cluster pairs (P i, Q i ) and (P j, Q j ) respectively. We will decrease te order of G 1 by 1 modulo and increase te order of G by 1 modulo. To do tis, we remove ζ copies of H by selectively coosing were te components of H come from. Since te cluster pairs are regular, we can find tese copies of H by Lemma 3.3. From G 1 we remove ζ ζ i copies of C i for 1 i k c. Totally G 1 loses k c i=1 (ζ ζ i)c i = ζ 1 vertices. From G we remove ζ +ζ i copies of C i for 1 i k c. Ten G loses k c i=1 (ζ +ζ i)c i = ζ+1 vertices. Since it is impossible tat all te removed ζ + 1 vertices come from one of P j and Q j, eac of P j, Q j loses at most ζ vertices. Let r i be te remainder of P i + Q i mod for i = 1,..., k. Suppose tat r i is te smallest nonzero remainder and r j is te largest remainder. By applying te procedure above at most min{r i, r j } times, we eiter reduce r i to 0 or enlarge r j to. Repeat tis process at most k 1 times and obtain r i 0 mod for all i = 1,..., k (note tat r i 0 mod all te time). Te total number of te removed copies of H is at most ζ( 1)(k 1) < ζk, and eac cluster loses at most ζ( 1) < ζ vertices. Pairing (P i, Q i ) and (P j, Q j ) togeter and performing tis process until eiter r i 0 mod or r j 0 mod, it is easy to see tat one may apply tis procedure totally at most ( 1) k i=1 r i times to ensure tat P i + Q i is divisible by for all i = 1,..., k. 14
Fix i = 1,..., k. Let P i, Q i denote te clusters obtained from P i, Q i after applying Claim 4.7 and Claim 4.8. We observe tat P i, Q i are large and (1 + γ ) u w P i Q i 1. In fact, by Claims 4.7 and 4.8, eac cluster C loses at most d C /3+ζ d C / vertices, and consequently C (1 d ) C. Since d γ 1, we derive tat ( 1 + γ ) ( u w 1 d ) ( u ) w + γ = (1 d ) P i Q i P i Q i P i u (1 d ) Q i = w + γ 1 d < 1 By Corollary 3.7, te complete bipartite grap K P i, Q i contains an H-factor. If we can sow tat (P i, Q i ) is super-regular, ten te Blow-up Lemma implies tat G[P i, Q i ] also contains an H-factor. In fact, since (P i, Q i ) is (ɛ, d)-super-regular, we ave Γ(x, Q i ) d Q i d Q i / d Q i / for all x P i and similarly Γ(y, P i ) d P i / for all y Q i. By te Slicing Lemma, (P i, Q i ) is (ɛ, d/)-super-regular. Note tat V (G) \ k i=1 (P i Q i ) consists of disjoint copies of H. We tus obtain te desired H-factor of G. 4. Te Extremal Case We now prove tat we can tile G in te extremal case. teorem: More precisely, we prove te following Teorem 4.9. Let H be a bipartite grap wit H is in Class 1, u = σ(h), w = σ(h), ζ = ζ(h), and β = β(h). Let c 1 (H) := ζ + β(w u) + (w u) + w. (8) Ten, tere exist α > 0 and an integer m 0 suc tat for any m m 0, if G[X, Y ] is a balanced, bipartite grap on n = m vertices suc tat (i) G as minimum degree ( u δ(g) n + c 1 (H), ) and (ii) tere are subsets A X, B Y, were A = B = wn wit d(a, B) α, ten G contains an H-factor. By (1) and (8), we derive tat c 1 (H) 4 3 and tus complete te proof Teorem 1.4. To prove Teorem 4.9, let us start wit a simple corollary of te Blow-up Lemma. Recall tat δ(x, Y ) denotes min x X deg(x, Y ). Lemma 4.10. For every positive integer, tere exists a positive number ρ = ρ( ) < 1 suc tat te following olds. For any bipartite grap F, if (F ) and F can be embedded into K X, Y, ten F can be embedded into every bipartite grap G[X, Y ] wit δ(x, Y ) (1 ρ) Y, δ(y, X) (1 ρ) X. (9) Proof. We first prove tat for any 0 < ρ < 1, every bipartite grap G[X, Y ] satisfying (9) is ρ- regular. In fact, consider subsets A X, B Y wit A = γ 1 X and B = γ Y for some γ 1, γ > ρ. By (9), we ave δ(a, Y ) Y ρ Y and consequently δ(a, B) B ρ Y = (γ ρ) Y. Te density between A and B satisfies d(a, B) δ(a, B) A A B (γ ρ) Y B = γ ρ γ > 1 ρ ρ = 1 ρ. 15
Since 1 ρ < d(a, B) 1 and in particular, 1 ρ < d(x, Y ) 1, we ave d(a, B) d(x, Y ) < ρ. Now assume tat K X, Y contains a copy of F and let ɛ be given by te Blow-up Lemma (Lemma 3.5) wit δ = 1/ and (F ) =. Let ρ = min{ɛ, 1/} and G[X, Y ] be a bipartite grap satisfying (9). Ten G is (ɛ, 1/)-super-regular and tus contains a copy of F. Proof of Teorem 4.9. Recall tat A X and B Y are sets of size wn wit d(a, B) α. Let A c = X A and B c = Y B. Ten A c = B c = un. We define te following subsets: A 1 = {x X : deg(x, B) < α 1 3 B }, A = {x X : deg(x, B) > (1 α 1 3 ) B }, B1 = {y Y : deg(y, A) < α 1 3 A } A 0 = X A 1 A, B 0 = Y B 1 B. B = {y Y : deg(y, A) > (1 α 1 3 ) A } Clearly A 1 A A 0 is a partition of X and B 1 B B 0 is a partition of Y. We claim tat A 1, B 1, A, B are very close to A, B, A c, B c respectively (so A 0 and B 0 are fairly small) and subgraps G[A 1, B ] and G[A, B 1 ] are almost complete. Claim 4.11. Assume tat α 1 3 < 1 and δ(g) u n (so c 1(H) is unnecessary ere). 1. (1 α 3 ) A { A 1, B 1 } (1 + α 3 ) A and A c α 3 A { A, B } A c + α 3 A.. δ(b, A 1 ) (1 α 1 3 ) A 1, δ(a, B 1 ) (1 α 1 3 ) B 1 and δ(a 1, B ) (1 α 1 3 w u ) B, δ(b 1, A ) (1 α 1 3 w u ) A. 3. (B 1, A 1 ), (A 1, B 1 ) A (α 3 + α 1 3 ). 4. A 0, B 0 α 3 A and δ(a 0, B 1 ), δ(b 0, A 1 ) (α 1 3 α 3 ) A. Proof. Part 1. We only prove bounds for A 1 and A ; te calculations for B 1 and B are exactly te same. By definition of A 1, e(a A 1, B) δ(a A 1, B) A A 1 α 1 3 B A A1. On te oter and, e(a A 1, B) e(a, B) α A B. Togeter tey imply tat A A 1 α 1 3 B α A B or A A 1 α 3 A. Since A A 1 A A 1, it follows tat A 1 (1 α 3 ) A. In order to derive an upper bound for A c A, we need te minimum degree condition δ(g) u n. Since δ(g) is an integer, we actually ave δ(g) u n. Ten e(b, A c ) = e(b, X) e(a, B) u n B α A B. Let ē(b, A c ) denote te size of te bipartite complement of G on [B, A c ]. Since w n + u n = n, we ave ē(b, A c ) = B A c e(b, A c ) B (n w n ) ( u n B α A B ) = α A B. 16
By definition of A, Terefore, e(a c A, B) (1 α 1 3 ) B A c A. ē(a c A, B) A c A B (1 α 1 3 ) B A c A = α 1 3 B A c A. Te upper and lower bounds for ē(a c A, B) togeter imply tat α 1 3 B A c A α A B, wic gives A c A α 3 A. We tus deduce tat A A c α 3 A. Since A 0 + A 1 + A = n = A + A c, we furter ave A 0 + A 1 A +α 3 A. Togeter wit A 1 (1 α 3 ) A, it yields tat A α 3 A A 1 A + α 3 A. Te lower bound for A 1 also implies tat A A c + α 3 A. Togeter wit A A c α 3 A, we tus obtain desired bounds for A. Te proof above actually gives tat A A 1, B B 1, A c A, B c B α 3 A. Part. Let us consider te minimum degree between A 1 and B ere; te same olds for te degree between B 1 and A. First δ(b, A 1 ) δ(b, A) A A 1 (1 α 1 3 α 3 ) A. By using δ(g) un = Bc, we derive tat δ(a 1, B ) δ(a 1, B c ) B c B δ(g) α 1 3 B B c B B c (α 1 3 + α 3 ) B. We now prove tat δ(b, A 1 )/ A 1 1 α 1 3. By Part 1, A 1 (1 + α 3 ) A. Ten δ(b, A 1 ) A 1 (1 α 1 3 α 3 ) A (1 + α 3 ) A 1 α 1 3 because α 1 3 > α 3. Similarly we can prove δ(a 1, B )/ B 1 α 1 3 w u toug we also need B w n w u Bc : δ(a 1, B ) B Bc (α 1 3 + α 3 ) B B c + α 3 B Bc (α 1 3 + α 3 ) B c w u B c + α 3 B c w u. By using α 1 3 > α 3 again, we derive tat δ(a 1, B )/ B 1 α 1 3 w u. Part 3. By using A 1 A A c A α 3 A, we obtain (B 1, A 1 ) (B 1, A) + A 1 A (α 1 3 + α 3 ) A. Te same olds for (A 1, B 1 ). Part 4. Part 1 immediately implies tat A 0, B 0 α 3 A. By definition of A 1, we ave δ(a 0, B 1 ) α 1 3 B B B 1 (α 1 3 α 3 ) B. Te same olds for δ(b 0, A 1 ). Recall tat n = m. We now separate te proof into two parts, wen m is even and wen m is odd. We give all details in Part 1, including te exact values of α and n, and wile reducing Part to Part 1, we only justify te value of c 1 (H). Part I: m is even. Let ρ = ρ(w) be given as in Lemma 4.10. We define α > 0 suc tat { α 1 1 3 = min 5, ρ }, (10) 17
Let ζ = ζ(h). By coosing m 0 sufficiently large, we may assume m ζ /α 3 and consequently n = m/ satisfies nα 3 ζ 3. (11) Let G 1 = G[A 1, B B 0 ] and G = G[B 1, A A 0 ] denote te induced subgraps of G on A 1 B B 0 and B 1 A A 0, respectively. Our first step is to remove some copies of H so tat te orders of G 1 and G are divisible by. Suppose tat v(g 1 ) r (mod ) and accordingly v(g ) r (mod ) for some 0 r <. Claim 4.1. We may remove rζ copies of H from G were rζ + r vertices come from G 1 and rζ r vertices come from G. On te oter and, rζ vertices are from eac of X and Y. Proof. We first note tat since G 1 [A 1, B ] and G [A, B 1 ] are almost complete, we may find many disjoint copies of H in tem. In fact, since A 1 / B is about w/u, K A1, B contains an H- tiling tat covers most of its vertices. By Claim 4.11, δ(b, A 1 ) (1 α 1 3 ) A 1 and δ(a 1, B ) (1 α 1 3 w u ) B. By (10), α 1 3 w u ρ. Lemma 4.10 tus implies tat G 1[A 1, B ] contains an H-tiling tat covers most of its vertices. We remove rζ copies of H as follows: from G 1 [A 1, B ], remove r(ζ + ζ i ) copies of C i, and from G [A, B 1 ], remove r(ζ ζ i ) copies of C i for all i = 1,..., k c. Now fix an index i. Note tat r(ζ +ζ i ) and r(ζ ζ i ) ave te same parity. If tey are even, ten we remove r(ζ + ζ i )/ copies of C i from G 1 wit te larger side in X, and te oter r(ζ + ζ i )/ copies of C i from G wit te smaller side in X. Similarly, remove r(ζ ζ i )/ copies of C i from G wit te larger side in X, and te oter copies of H wit te smaller side in X. Clearly X and Y lose te same number of vertices for eac i. Since at te end X and Y togeter lose rζ vertices, eac of tem loses rζ vertices. If r(ζ + ζ i ) is odd, ten remove r(ζ + ζ i )/ copies of C i from G 1 wit te larger side in X and r(ζ + ζ i )/ copies of C i from G 1 wit te smaller side in X (terefore X loses w i u i more vertices tan Y ). On te oter and, we remove r(ζ ζ i )/ copies of C i from G wit te larger side in X and r(ζ ζ i )/ copies of C i from G wit te smaller side in X (tis makes Y lose w i u i more vertices tan X). Tus X and Y again lose te same number of vertices: eac loses rζ vertices at te end. Te total number of vertices tat G 1 loses is r(ζ + ζ 1 )c 1 +... + r(ζ + ζ kc )c kc = rζ(c 1 +... + c kc ) + r(ζ 1 c 1 +... + ζ kc c kc ) = rζ + r A similar calculation sows tat G loses rζ r vertices. Denote te sets of te remaining vertices in X, Y, A 1, A, B 1, B by X, Y, A 1, A, B 1, B, respectively. Te difference between A 1 and A 1 (similarly between B and B, etc.) is at most rζ. Our coice (11) of n is equivalent to w ζ α 3 w n. Since r 1 and A = w n, we derive tat wrζ α 3 A. (1) Let à = A A 0 and B = B B 0. Te current G 1, G are G 1 [A 1, B ] and G [B 1, Ã], respectively. By Claim 4.1, bot v(g 1 ) and v(g ) are divisible by. Let m 1 = v(g 1 )/, m = v(g )/, and write A 1 = m 1 w + s, B 1 = m w + t, à = m u t, B = m 1 u s for some integers s and t. Since X and Y ave equal number of vertices, we ave m 1 w+s+m u t = m w + t + m 1 u s, wic implies tat (m 1 m )(w u) = (t s). (13) 18
G 1 G A 1 B 1 rζ vertices à B rζ vertices Figure 4: Grap G wit sets A 1, Ã, B 1, B, and removed copies of H Witout loss of generality, assume tat m 1 m. Tis implies t s. By using δ(g) u n + c 1(H) and m m = n, we obtain a lower bound on δ(b 1, A 1 ): δ(b 1, A 1) δ(g) à rζ = u n + c 1(H) m u + t rζ c 1 (H) + t rζ. (14) Now we use te assumption tat m is even: m rζ = m 1 + m is even, tus m 1 m is even. Ten, by (13), we see tat w u divides t s. We now separate te cases wen t 0 and wen t < 0. Case 1: Assume t 0. We claim tat t is reasonably small. In fact, by Claim 4.1, v(g ) = A + A 0 + B 1 (rζ r). From Claim 4.11, we know tat A + B 1 n α 3 A and consequently m = v(g )/ (n α 3 A rζ)/. By definition, t = B 1 m w A + α 3 A w n + w α 3 A + wrζ = α 3 A + w α 3 A + wrζ By (1), we ave wrζ 1 α 3 A and tus t 3α 3 A. We want to move t vertices from A 1 to à and t vertices from B 1 to B. To move tese vertices, we will find t w-stars from B 1 to A 1 and t w-stars from A 1 to B 1 by te following lemma (Lemma 1 from [0]), and ten move te centers of tese stars. Lemma 4.13. ([0]) Let 1 k δ M be positive integers, and 0 < c < 1 6k+7. Let F [V 1, V ] be a bipartite grap suc tat V i M cm for i = 1,. If δ δ(v 1, V ) cm and (V, V 1 ) cm, ten F contains (δ k +1) vertex disjoint k-stars of wic δ k +1 are centered in V 1 and δ k +1 are centered in V. By (8), we ave c 1 (H) > rζ + w 1. Wit (14), it implies tat δ(b 1, A 1) w + 1 c 1 (H) + t rζ w + 1 > t. (15) On te oter and, δ(b 1, A 1 ), (A 1, B 1 ) (α 1 3 + α 3 ) A by Claim 4.11. From (10) and te fact tat w, we can derive tat α 1 3 15(w+1) < 1 6w+7. Tus, Lemma 4.13 provides t vertex disjoint w-stars centered in A 1 and t vertex disjoint w-stars centered in B 1. We now move te centers of tese stars from A 1 to à and from B 1 to B. Te resulting A 1, Ã, B 1, B satisfy A 1 = m 1 w + s t, B = m 1 u s + t; B 1 = m w, à = m u. Below we explain ow to find an H-factor in G 1 ; te same procedure works for G. Te resulting G 1 contains t 3α 3 A disjoint w-stars centered at B. By definition, B 0 B. We next find B 0 disjoint w-stars centered at B 0 from G 1 wic are also disjoint from te existing 19
w-stars. From Claim 4.11, we ave B 0 < α 3 A and δ(b 0, A 1 ) (α 1 3 α 3 ) A. Since A 1 A 1 rζ t and rζ α 3 A, we ave δ(b 0, A 1) δ(b 0, A 1 ) (t + rζ) (α 1 3 α 3 ) A 3α 3 A α 3 A = (α 1 3 5α 3 ) A. Since α 1 3 5α 3 5(w + 1)α 3 by (10), we derive tat δ(b 0, A 1) (α 1 3 5α 3 ) A 5wα 3 A w( B0 + t). We may terefore coose disjoint w-stars for te vertices of B 0 greedily. Now, we ave t + B 0 w-stars centered in B. For eac star, we will find a copy of K u,w (a supergrap of H), suc tat u 1 vertices come from B, and te rest are from te w-star. Recall tat B B rζ. Suppose tat a w-star as leaves v 1,..., v w in A 1. We claim tat w i=1 Γ(v i, B ) (u 1)( B 0 + t), tus we can greedily find a copy of K u,w for eac star suc tat it is vertex disjoint from te existing copies of K w,u. In fact, by Claim 4.11 and (1), w i=1γ(v i, B ) (1 w w u α 1 3 ) B rζ (1 w u α 1 3 )(1 α 3 ) B α 3 B (1 w u α 1 3 α 3 ) B. By (10), we ave 5uα 3 < α 1 3 and ( w u + 1)α 1 3 < α 1 3 < 1. Consequently w i=1γ(v i, B ) (u 1)( B 0 +t) (1 w u α 1 3 α 3 ) B (u 1)5α 3 B > (1 w u α 1 3 α 1 3 ) B > 0. We remove tese copies of K w,u, and let A 1 and B. We know tat A 1 A 1 and B B satisfy and B denote te set of remaining vertices in A 1 A 1 A 1 A 1 rζ t w( B 0 + t), B B B rζ (u 1)( B 0 + t). Furtermore, A 1 = m 1w + s t and B = m 1 u s + t for some large integer m 1. Since w u divides t s, we can apply Lemma 3.6 and obtain an H-factor of K A 1, B. It remains to sow tat G[A G[A 1, B 1, B ] satisfy te condition (9) of Lemma 4.10 (ten Lemma 4.10 provides an H-factor of ]). In fact, by Claim 4.11, δ(b, A 1) (1 α 1 3 ) A1 rζ t w( B 0 + t) (1 α 1 3 ) A1 α 3 A 3α 3 A w(5α 3 A ). By (10) and w + 1, we ave α 1 3 5(w + 1)α 3, wic implies tat, by Claim 4.11, α 1 3 A1 α 1 3 (1 α 3 ) A (4α 3 + 5wα 3 ) A. Consequently δ(b, A 1 ) (1 3α 1 3 ) A 1 (1 3α 1 3 ) A 1. On te oter and, δ(a 1, B ) (1 w u α 1 3 ) B rζ (u 1)( B 0 + t) (1 w u α 1 3 ) B α 3 B (u 1)5α 3 B. By (10), we ave α 1 3 u 5uα 3. Togeter wit B B c α 3 B ( u α 3 α 1 3 B α 1 3 ( u α 3 ) B α 3 B + 5(u 1)α 3 B. ) B, we ave 0