Disjoint Subgraphs in Sparse Graphs 1

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1 Disjoint Subgraphs in Sparse Graphs 1 Jacques Verstraëte Department of Pure Mathematics and Mathematical Statistics Centre for Mathematical Sciences Wilberforce Road Cambridge CB3 OWB, UK jbav2@dpmms.cam.ac.uk Abstract Häggkvist and Scott asked whether there exists a quadratic function q(k) such that if G is a graph of minimum degree at least q(k), then G contains vertex-disjoint cycles of k consecutive even lengths. In this paper, it is shown that if G is a graph of minimum degree at least 1 2 (k2 + 35k + 2) and sufficiently many vertices, then G contains vertex-disjoint cycles of k consecutive even lengths, answering the above question in the affirmative. The coefficient of k 2 cannot be decreased and, in this sense, this result is best possible. We also consider the problem of finding isomorphic disjoint topological subgraphs in a graph of appropriate average degree. 1. Introduction It is well-known that a graph of average degree at least two contains at least one cycle. From this it is easy to deduce that if k is a natural number, a graph of order n containing at least n + k 1 edges contains k distinct cycles. Corradi and Hajnal [4] considered the requirement that the k cycles be disjoint. They proved that a graph of minimum degree at least 2k and order at least 3k contains k disjoint cycles. Under the same minimum degree condition, Egawa [5] showed that a sufficiently large graph contains k disjoint cycles of the same length. In a recent paper by Bondy and Vince [3], it was shown that if G is a graph of 1 accepted for publication,

2 order at least three containing at most two vertices of degree at most two, then G contains a pair of cycles of consecutive lengths or consecutive even lengths. The following general theorem was established by the author (a special case of Theorem 1 in [16]): Theorem 1. Let k be a natural number and G a bipartite graph of average degree at least 4k. Then there exist cycles of k consecutive even lengths in G. Moreover, the shortest of these cycles has length at most twice the radius of G. Theorem 1 addresses the problem of finding cycles of consecutive even lengths in a graph of sufficient average degree. In this paper, we attack the associated problem of finding disjoint cycles of consecutive even lengths. All graphs considered in this paper are simple and finite. Throughout this paper, K s,t denotes the complete bipartite graph with s vertices in one colour class and t vertices in the other. Also, H denotes the order of a graph H and e(h) denotes the number of edges in H. The word disjoint will be taken to imply vertex-disjoint. 2. Disjoint Cycles of Consecutive Even Lengths. Häggkvist and Scott [8] asked whether there exists a quadratic function q(k) such that if G is a graph of minimum degree at least q(k), then G contains disjoint cycles of k consecutive even lengths. We answer this question in the affirmative, by proving the following theorem: Theorem 2. Let k be a natural number. Then there exists n k such that if G is a graph of order at least n k and minimum degree at least 1 2 (k2 + 35k + 2), then G contains disjoint cycles of k consecutive even lengths. Theorem 2 is essentially best possible since for s k = k and n 2s k 2, the complete bipartite graph K sk 1,n s k +1 contains no disjoint cycles of k consecutive even lengths and has minimum degree s k 1 = 1 2 (k2 + 3k). From the proof of Theorem 2, we may also deduce that a sufficiently large graph of average degree at least k k + 1 contains disjoint cycles of k consecutive even lengths this is also best possible in view of the example of the graph F above. It will follow from the proof of Theorem 2 that a sufficiently large graph, of minimum degree at least 16k + 3 and containing no cycle of length four, contains disjoint cycles of k consecutive even lengths. This type of difference between general graphs and graphs containing no cycle of length four was first observed by Häggkvist (see Thomassen [15], page 138). A similar phenomenon occurs for graphs of bounded maximum degree. In order to prove Theorem 2, we first need two lemmas. The first lemma is a consequence of a lemma due to Kostochka and Pyber (see Lemma 1.1 page 84 in [12]): 2

3 Lemma 3. Let r be a natural number, c a positive real number, and let G be a graph of order n and size at least cn 1+ 1 r. Then G contains a subgraph of average degree at least c and radius at most r. The second lemma is proved using a standard counting argument: Lemma 4. Let s and t be natural numbers and let G be a bipartite graph with colour classes A and B. If A > t B s+1 and e(g) s A, then G contains K s,t. ) Proof. If v A and the degree of v is d(v), then there exist subsets of B of size s in the neighbourhood of v. Since A > t B s+1, d(v) s v A 1 s s+1 v A d(v) s 1 s s+1 1 A s 1 1 s s+1 1 s s+1 ( d(v) s s d(v) v A 1 e(g)s A s 1 1 A s 1 ss A s A /s > t B s t B. s So there exists a subset of B of size s contained in the intersection of the neighbourhoods of at least t vertices of A. This implies that K s,t G. We will see that the following result implies Theorem 2: Theorem 5. Let k, m be positive integers and let G be a graph of order n and minimum degree at least m + 16k + 1. Then G contains at least (8k log 2 n) 4 n 1/(m+1) disjoint sets of k cycles of the same k consecutive even lengths or G contains K m,m. Proof. Let N = (8k log 2 n) 4 n 1/(m+1). Suppose G = G 0 contains cycles of k consecutive even lengths. Define r 0 = min{r : G contains cycles of lengths 2r, 2r + 2,..., 2r + 2k 2}. Let X 0 be a subgraph of G 0 comprising the union of cycles of lengths 2r 0, 2r 0 + 2,..., 2r 0 + 2k 2 in G 0. Let G 1 = G 0 V (X 0 ). If G 0, G 1,..., G i are defined, and G i contains k cycles of consecutive even lengths, define r i = min{r : G i contains cycles of lengths 2r, 2r + 2,..., 2r + 2k 2}. Let X i be a subgraph of G i comprising the union of cycles of lengths 2r i, 2r i + 2,..., 2r i + 2k 2 and let G i+1 = G i V (X 0 ). Note that the integers r 0, r 1, r 2,... form a non-decreasing sequence. If at least N of the integers r i are identical, say r i = r i+1 =... = r i+k 1 = t, then G = G 0 contains k disjoint cycles of lengths 2t, 2t + 2,..., 2t + 2k 2, contained in X i, X i+1,..., X i+k 1 respectively. Therefore suppose no N of the integers r i are identical. 3

4 We claim that G = G 0 contains cycles of k consecutive even lengths, the shortest of which has length at most 16k(log 2 n) 2. Let B be a spanning bipartite subgraph of G 0 containing at least half of the edges of G 0. Set r = 16k(log 2 n) 2. Then n 1/r 2 1/(16k) 1 + 1/(16k) and e(b) 1 (m + 16k + 2)n 4 1 (16k + 1)n 4 4kn 1+ 1 r. By Lemma 3, applied to B, there exists a subgraph H of B of average degree at least 4k and radius at most r 16k(log 2 n) 2. By Theorem 1, H (and therefore G 0 ) contains cycles of k consecutive even lengths, the shortest of which has length at most 32k(log 2 n) 2. This shows that r 0 16k(log 2 n) 2. Conversely, whenever G i does not contain cycles of k consecutive even lengths, the shortest having length at most 32k(log 2 n) 2, G i has average degree at most 16k + 1. As G is finite, there exists a non-negative integer j such that r j 16k(log 2 n) 2 and G j+1 does not contain cycles of k consecutive even lengths, the shortest having length at most 32k(log 2 n) 2. Let G = G j+1 and G = G V (G ). Noting that X i 2k(r i + k 2) and that no N of the integers r i are identical, j r G X i (N 1) 2k(i + k 2) i=0 i=0 (N 1)kr(r + 1) + 2k(k 1)(k 2)(r + 1) Nk 2 r 2 N (4k log 2 n) 4 < 1 2 n1/(m+1). As G contains no cycles of k consecutive even lengths with shortest cycle of length at most 32k(log 2 n) 2, G has average degree at most 16k + 1. Let B be the bipartite graph spanned by the edges of G with one end in G and the other in G. Then e(b ) (m + 16k + 1) G (16k + 1) G m G We also have m G m+1 2 m 1 mn n G = G. Applying Lemma 4 to B, we find a complete bipartite subgraph K m,m in B, as required. Proof of Theorem 2. When s k = k+2 2 1, the complete bipartite graph K sk,s k contains disjoint cycles of lengths 4, 6,..., 2k+2. We therefore assume that G does not contain K sk,s k. By Theorem 5, with m = s k, G contains at least N = (8k log 2 n) 4 n 1/(m+1) sets of k disjoint cycles of the k same consecutive even lengths. If N k, we select an appropriate cycle 4

5 from each of these sets of k cycles to obtain disjoint cycles of k consecutive even lengths. The inequality N k is satisfied, for example, when n (16s k ) 16s k. We can also give an average degree version of Theorem 2: a sufficiently large graph of average degree at least k k + 2 contains disjoint cycles of k consecutive even lengths. This average degree version is essentially best possible as seen by the graph K k 2 +3k 1 and also by the example presented after the statement of Theorem 2. In view of this example, we pose the following conjecture: Conjecture. Let k be a positive integer and s k = k Then any graph of minimum degree at least s k and order at least 2s k contains disjoint cycles of k consecutive even lengths. If G is a graph of order 2s k and minimum degree at least s k then this conjecture says that G contains disjoint cycles of lengths 4, 6,..., 2k + 2: this is a special case of a conjecture of El-Zahar [6], which we return to in section 4. Erdős and Burr [7] conjectured that if k + lz contains an even number, then there exists c k such that every graph of average degree at least c k contains a cycle of length l modulo k. This conjecture was proved by a number of authors (see [16] for references). Their proofs are all strengthened by Theorem 1, which guarantees that c k can be taken to be linear in k. We consider the natural associated problem of finding vertex disjoint cycles of length l modulo k in a sufficient large graph of average degree at least d k. The existence of d k follows easily from Theorem 5, as we shall see in the proof of Theorem 6 below. In the statement of the next result, π(k) = 1 if k is odd and π(k) = 0 if k is even: Theorem 6. Let k 2 and r be positive integers and let t 1, t 2,..., t r Z k. Suppose t i +kz contains an even integer for i = 1, 2,..., r. Define r m = (t i + kπ(t i )π(k)) + kπ(k) {t i < 4 : π(t i ) = 0} i=1 Then every sufficiently large graph of average degree at least 1 2m + 16k + 1 contains disjoint cycles of lengths t 1, t 2,..., t r modulo k. Proof. It is not hard to check that if t i + kz contains an even integer, then m is even. If G contains K m/2,m/2, then G contains the required r disjoint cycles: this has to be verified by cases (for example, if t i < 4 is even and k is odd, then a shortest cycle of length t i (mod k) in a complete bipartite graph must have length 2k + t i ). By Theorem 5, provided (8k log 2 n) 4 n 1/(m+1) r, G contains the required disjoint cycles. There certainly exists n k,r such that if n n k,r, the inequality is satisfied. 3. Disjoint Topological Subgraphs 5

6 Given a graph H, a topological H is a graph obtained by arbitrarily subdividing the edges of H. We now turn to the problem of finding disjoint isomorphic topological H in a graph G. This is a natural generalization of the problem of finding disjoint cycles of the same length in a graph G: in this case, H = K 3. In this direction, Egawa [5] showed that a sufficiently large graph of minimum degree at least 2k contains k disjoint cycles of the same length. A short proof of this fact was found by the author [17]. Mader [14] was the first to establish the existence of a constant c t such that every graph of average degree at least c t contains a topological complete graph of order t. Mader s showed that c t can be taken to be at most 2 O(t2). Since then, c t has been reduced to a quadratic polynomial in t by deep theorems of Bollobás,Thomason [2] and Komlós, Szemerédi [11], which is best possible. Kostochka and Pyber [12] used Mader s result to show that dense graphs contain small topological complete graphs: Proposition 7. Let H be a graph of order h and size e(h) and let G be a graph of order n and size at least 16 e(h) n 1+ε, where ε > 0. Then G contains a topological H of order at most 16e(H)(log 2 h)/ε. The theorems of Bollobás and Thomason and Komlós and Szemerédi do not give an effective upper bound on the size of the topological complete graph. Given a graph H, we now intend to apply the method of the proof of Theorem 2 to find many disjoint isomorphic topological H subgraphs in a graph of high enough average degree. Our result will give an essentially best possible minimum degree (or average degree) condition for find k disjoint topological H in a graph G. We write e for the base of the natural logarithm and emphasize that we will not optimize constants in what follows. Theorem 8. Let k be a positive integer, and let H be a graph of order h and independence number α. Then there exist constants C and K, depending only on H, such that if G is a graph of minimum degree at least k(h α) + K k 1 1/(h α)2 and order n > Ck, then G contains k disjoint isomorphic topological H. Proof. If H is empty, then we may take C = h and K to be arbitrary. If h α = 1 (so e(h) = 1), then we may take C = h + 1. This follows since a graph of order at least 3k 2 contains k disjoint edges and we require k(h 2) other vertices in G to find k disjoint isomorphic topological H (in this case, we may even take K = 0). For the rest of the proof, we assume h α > 1 (and therefore e(h) > 1). We prove the Theorem with C = (64e(H) log 2 h) 2e(H) and K = 16 e(h)+1. Note that for k = 1, G already contains a topological K h and therefore a topological H. For the remainder of the proof k 2. The number of isomorphic topological H of order m is, very roughly, at most the number of representations of m h as an ordered sum of e(h) non-negative integers. This number is precisely m h + e(h) 1. e(h) 1 6

7 Let G = G 0 and define G 1 = G 0 V (H 0 ) where H 0 is a smallest topological H appearing in G. By Proposition 7, with ε = (2 log k n) 1, we can guarantee that H 0 32e(H) log 2 h log k n, since G 0 has size at least e(H)+1 k 3/4 n 16 e(h) kn 16 e(h) n 1+ε. In general, if G i 1 is defined, let G i = G i 1 V (H i 1 ) where H i 1 is the smallest topological H appearing in G i 1. We continue this procedure until we reach a stage j where G j does not contain a topological H of order at most 32e(H) log 2 h log k n. Let G = G j and G = G V (G j ). If k of the H i are isomorphic in G then the requirements of the theorem are met, so we suppose this is not the case. For convenience, let m = 32e(H) log 2 h log k n. Then, m h + e(h) 1 G (k 1)m e(h) 1 2m km e(h) 1 km(2m) e(h) 1 k(2m) e(h) k(64e(h) log 2 h) e(h) (log k n) e(h) k C(log k n) e(h). It is clear that G has average degree at most 2 16 e(h) k 1 2 K k1 1/(h α)2 otherwise, by Proposition 7, G = G j contains a small topological H. This implies that at most half the vertices of G have at least 1 2 K k1 1/(h α)2 neighbours in G. As G has minimum degree at least k(h α) + K k 1 1/(h α)2, this implies that a set S of at least half the vertices of G have at least k(h α) K k1 1/(h α)2 neighbours in G. We now consider the bipartite graph B spanned by edges between S and G. If B contains a complete bipartite graph K h α,e(h), then B contains a topological H in which every edge not incident with an independent set is subdivided. The graph B contains such a complete bipartite graph whenever ( d(v) G ) > (e(h) 1). h α h α v S If we find k disjoint complete bipartite graphs K (h α),e(h) in B, then Theorem 8 is proved. By successively removing K h α,e(h) from B, and applying the above condition in the remaining graph, we see that a sufficient condition for B to contain k disjoint complete bipartite graphs K h α,e(h) is d(v) (k 1)(h α) h α v S ( G ) > (e(h) 1) h α 7 ( )

8 where S is any subset of S obtained by removing (k 1)e(H) vertices from S. We now show that this holds under the minimum degree condition for vertices in S and our choice of C = (64e(H) log 2 h) 2e(H). The first inequality we demonstrate is: ( (h α) G n G + ke(h) + e(h)k 1/(h α) ) h α. k Since G k C(log k n) e(h) and h α 2, it is sufficient to show that: n 2k C(log k n) e(h) + e(h) ( k (h α) C(log k n) e(h)) h α. If this inequality holds when n = Ck+ kc h, then it holds for all n Ck+ kc h. However, as e(h)(h α)(1 + log 2 C) e(h) C, we have: 2k C(log k Ck) e(h) + ( k e(h) (h α) C(log k Ck) e(h)) h α Ck + kc h. Therefore the lower bound for n holds. This implies that: The minimum degree condition gives: S e(h)k 1/(h α) [(h α) G /k] h α. d(v) (k 1)(h α) > k 1 1/(h α)2, so it follows that the summands in ( ) are each greater than: (h α) (h α) k h α 1/(h α), using the inequality ( a b ) (a/b) b. Our sufficient condition for k disjoint K h α,e(h) in B is now satisfied: the sum on the left in ( ) is at least: S (h α) (h α) k h α 1 (h α) ( G > e(h)k h α k > (e(h) 1) ( G h α ) h α So B contains k disjoint K h α,e(h) and therefore G contains k disjoint isomorphic topological H. The above theorem is best possible in the sense that if H has order h and independence number α, then the complete bipartite graph K k(h α) 1,n does not contain k disjoint isomorphic topological H, as the following simple lemma attests: Lemma 9. Let H be a graph of independence number α and order h. Let m be the minimum number of vertices in the smaller colour class in a bipartite topological H, taken over all ). 8

9 bipartite topological H. Then m = h α. Consequently, K k(h α) 1,n does not contain k vertex disjoint topological H. Proof. By subdividing all the edges of H not incident with a maximum independent set, we see that m h α. Let A be the minimum colour class over all bipartite topological H so A = m and let B be the larger colour class. If A < h α, then B contains a set B of more than α vertices of H. This set of vertices of H induces a graph with at least B α edges, by definition of α. Each of these edges is therefore subdivided and the new vertices form a set A A\V (H) of at least α B. This gives m A + A V (H) B α + h B V (H) = h α, as required. This implies that if k disjoint topological H fit into a complete bipartite graph, then the smaller colour class must have size at least k(h α). The important point is that C is a constant depending on H only and the deviation from k(h α) in the minimum degree condition is o(k). The proof of Theorem 8 can also be used to show that there exist constant C and K, depending only on H, such that if G is a graph of order n C k and average degree at least 2k(h α) + K k 1 1/(h α)2 contains k disjoint isomorphic topological H. 4. Some Questions Tiling and Disjoint Topological Subgraphs. Alon and Yuster [1] proved that if H is any graph of order h, then for all ε > 0 then there exists n 0 (ε, h) such that whenever n n 0 (ε, h) and h divides n, each graph of order n and minimum degree at least (1 1/χ(H) + ε)n contains n/h disjoint copies of H (an H-factor). Komlós, Sárközy and Szemerédi (see [10]) have recently generalized this, proving that there exists n 1 (H) and a constant K, depending only on H, such that if G is a graph of minimum degree at least (1 1/χ(H)) + K then G contains n/h disjoint copies of H (An H-tiling). In general, K cannot be taken to be zero: if H is an unbalanced complete bipartite graph, then K n,n, in general, does not contain 2n/h disjoint copies of H. A conjecture of Alon and Yuster remains: it may be true we can tile all but a constant number (depending on H) of vertices of a sufficiently large graph of order n and minimum degree at least (1 1/χ(H))n with copies of H. Theorem 8 implies that there exists k(ε, h) and n(h) such that if n n(h) kh and k k(ε, h), then any graph of order n and minimum degree at least (1 1/χ(H)+ε)kh contains k disjoint isomorphic topological H. By the result of Alon and Yuster, this is also true when n = kh. For 2kh < n < h 2, it is not necessarily true. This follows since (1 1/χ(H) + ε)kh < n/2 for ε < 1/χ(H) whereas k disjoint topological K h are all forced to be nonbipartite. If H is already bipartite, then we ask the following question: 9

10 Question 1. Let H be a bipartite graph of order h and independence number α. Does there exists k 0 (H) and a constant K, depending only on H, such that if G is a graph of minimum degree at least k(h α)+k and order at least kh and k k 0 (H), then G contains k disjoint isomorphic topological H? The question has a positive answer when the order of G is exactly kh and H is balanced, by the result of Komlós, Sárközy and Szemerédi. Another difficult question is the optimal value of K when k is fixed: for example, if k = 2, can we take K to be a polynomial in h for all H with H = h? Cycles of the Same Length. Our second question concerns disjoint cycles of the same length. Komlós, Sárközy and Szemerédi (see [10]) have shown that every sufficiently large graph of minimum degree at least n/2 has a perfect tiling with quadrilaterals. More generally, a beautiful conjecture of El-Zahar [6] was recently proved by Abbasi (see [10]): if ri=1 n i = n and n i 3 for i = 1, 2,..., r, then every sufficiently large graph of minimum degree at least r i=1 n i /2 contains disjoint cycles of lengths n 1, n 2,..., n r. We ask the following question, which extends the result when all n i are equal and even: Question 2. Let k, g 2. Does every graph of minimum degree at least gk and order at least 2gk contain k equicardinal disjoint cycles of length at least 2g? Even in the case k = 2 and g = 2, the answer is not known. Wang [18] proposed the following conjecture for balanced bipartite graphs: does every sk sk bipartite graph of minimum degree greater than (s 1)k contain k disjoint K s,s? This conjecture was proved by Wang for k 4. Chromatic Number. It is difficult to obtain good upper bounds on the chromatic number of an arbitrary graph. We ask the following question: Question 3. Does there exists an absolute constant c such that if G is a graph of chromatic number at least c, then G contains a pair of disjoint odd cycles of the same length? It is easily seen that a graph of chromatic number at least 3k contains k disjoint odd cycles. As a graph of chromatic number at least c > 0 contains an odd cycle of length O(n 1/(c 2) ), we might try to answer the question in the affirmative by showing that a graph of sufficient chromatic number c contains at least, say, n ε disjoint subgraphs of chromatic number at least 2 + 1/ε. However, this cannot be done since Erdős proved that there exist graphs of chromatic number at least c in which every subgraph of order εn is 3-chromatic. Difficulties also arise if we try to remove disjoint short odd cycles since the chromatic number is severely affected by such action (i.e. we can t hope to do it more than c times if the chromatic number is 3c.) Question 4. Does there exist a constant c such that every graph of chromatic number at least c contains a pair of disjoint cycles of consecutive odd lengths? 10

11 The same question can be asked for cycles of consecutive lengths. The proof of Theorem 1 in [15] easily shows that a graph of chromatic number at least 2k + 2 contains cycles of k consecutive lengths. However, the same problems arise as for Question 3. Finally, we could ask some questions such as: does there exist c such that every graph of chromatic number at least c contains a pair of disjoint totally odd (i.e. every edge is subdivided an odd number of times) topological K 4, but even finding one such topological K 4 is very difficult. Asymptotic Tiling of Regular Graphs. We now briefly consider disjoint (but not necessarily isomorphic) topological subgraphs in regular graphs. It is easily seen that for r sufficiently large, at least one half of the vertices of an r-regular graph can be covered by disjoint topological H (see below). Jørgensen and Pyber [9] have discussed problems involving covering the edges of a graph with topological subgraphs. We consider an analogue for covering vertices. Let H be a fixed graph. We let f(r, H) denote the maximum proportion of vertices that can be covered with disjoint topological H in any r-regular graph. We define f (H) = lim inf f(r, H) r f + (H) = lim sup f(r, H). r We observe, as mentioned above, that 1/2 f (H) f + (H) 1 for every H. The lower bound follows by removing topological H from an r-regular graph G until we obtain a graph F containing no topological H. Then the average degree of F is bounded above by a constant c > 0 depending only on H, by the results of Mader [14]. As G is r-regular, the number of edges between F and the remainder of G is at most r( G F ), but at least (r c) F. Therefore F 1 2 (1 c r ) 1 G and hence f (H) 1 2. We say that regular graphs have an asymptotic topological H tiling if f (H) = 1. When H is a complete graph of order two or three, the problem above becomes the problem of tiling with paths or cycles respectively. Petersen s 2-factor theorem (see [13] page 54) states that for k 1, every 2k-regular graph contains a 2-factor. This shows that if H has order at most three, then f + (H) = 1. The author does not know of any graph H for which regular graphs do not have an asymptotic topological H tiling. This problem may prove to be interesting: particular cases of interest are when H is a cycle of fixed length, or H is a complete graph of order at least four. Acknowledgements I would like to thank Andrew Thomason and the referee for their invaluable suggestions and comments. References 11

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