Forbidding Hamilton cycles in uniform hypergraphs

Journal of Combinatorial Theory, Series A 143 (2016) 107–115

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Journal of Combinatorial Theory, Series A www.elsevier.com/locate/jcta

Forbidding Hamilton cycles in uniform hypergraphs ✩ Jie Han a , Yi Zhao b a

Instituto de Matemática e Estatística, Universidade de São Paulo, Rua do Matão 1010, 05508-090, São Paulo, Brazil b Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30303, USA

a r t i c l e

i n f o

Article history: Received 17 August 2015 Available online 21 June 2016 Keywords: Hamilton cycles Hypergraphs Perfect matchings

a b s t r a c t For 1 ≤ d ≤  < k, we give a new lower bound for the minimum d-degree threshold that guarantees a Hamilton -cycle in k-uniform hypergraphs. When k ≥ 4 and d <  = k − 1, this bound is larger than the conjectured minimum d-degree threshold for perfect matchings and thus disproves a well-known conjecture of Rödl and Ruciński. Our (simple) construction generalizes a construction of Katona and Kierstead and the space barrier for Hamilton cycles. © 2016 Elsevier Inc. All rights reserved.

1. Introduction The study of Hamilton cycles is an important topic in graph theory. A classical result of Dirac [4] states that every graph on n ≥ 3 vertices with minimum degree n/2 contains a Hamilton cycle. In recent years, researchers have worked on extending this theorem to hypergraphs – see recent surveys [15,18,26]. ✩ The first author is supported by FAPESP (Proc. 2014/18641-5, 2015/07869-8). The second author is partially supported by NSF grant DMS-1400073. E-mail addresses: [email protected] (J. Han), [email protected] (Y. Zhao).

http://dx.doi.org/10.1016/j.jcta.2016.05.005 0097-3165/© 2016 Elsevier Inc. All rights reserved.

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To define Hamilton cycles in hypergraphs, we need the following definitions. Given k ≥ 2, a k-uniform hypergraph (in short, k-graph) consists of a vertex set V and an edge   set E ⊆ Vk , where every edge is a k-element subset of V . Given a k-graph H with a set S of d vertices (where 1 ≤ d ≤ k − 1) we define degH (S) to be the number of edges containing S (the subscript H is omitted if it is clear from the context). The minimum d-degree δd (H) of H is the minimum of degH (S) over all d-vertex sets S in H. For 1 ≤  ≤ k − 1, a k-graph is a called an -cycle if its vertices can be ordered cyclically such that each of its edges consists of k consecutive vertices and every two consecutive edges (in the natural order of the edges) share exactly  vertices. In k-graphs, a (k − 1)-cycle is often called a tight cycle. We say that a k-graph contains a Hamilton -cycle if it contains an -cycle as a spanning subhypergraph. Note that a Hamilton -cycle of a k-graph on n vertices contains exactly n/(k − ) edges, implying that k −  divides n. Let 1 ≤ d,  ≤ k − 1. For n ∈ (k − )N, we define hd (k, n) to be the smallest integer h such that every n-vertex k-graph H satisfying δd (H) ≥ h contains a Hamilton -cycle. Note that whenever we write hd (k, n), we always assume that 1 ≤ d ≤ k − 1. Moreover, we often write hd (k, n) instead of hk−1 (k, n) for simplicity. Similarly, for n ∈ kN, we d define md (k, n) to be the smallest integer m such that every n-vertex k-graph H satisfying δd (H) ≥ m contains a perfect matching. The problem of determining md (k, n) has attracted much attention recently and the asymptotic value of md (k, n) is conjectured as follows. Note that the o(1) term refers to a function that tends to 0 as n → ∞ throughout the paper. Conjecture 1.1. [6,14] For 1 ≤ d ≤ k − 1 and k | n,  md (k, n) =



max

  k−d   1 1 n−d ,1 − 1 − + o(1) . 2 k k−d

Conjecture 1.1 has been confirmed [1,17] for min{k − 4, k/2} ≤ d ≤ k − 1 (the exact values of md (k, n) are also known in some cases, e.g., [23,25]). On the other hand, hd (k, n) has also been extensively studied [2,3,5,7–13,16,19,20,22,24]. In particular, Rödl, Ruciński and Szemerédi [20,22] showed that hk−1 (k, n) = (1/2 + o(1))n. The same authors proved in [21] that mk−1 (k, n) = (1/2 + o(1))n (later they determined mk−1 (k, n) exactly [23]). This suggests that the values of hd (k, n) and md (k, n) are closely related and inspires Rödl and Ruciński to make the following conjecture. Conjecture 1.2. [18, Conjecture 2.18] Let k ≥ 3 and 1 ≤ d ≤ k − 2. Then hd (k, n) = md (k, n) + o(nk−d ). By using the value of md (k, n) from Conjecture 1.1, Kühn and Osthus stated this conjecture explicitly for the case d = 1.

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Conjecture 1.3. [15, Conjecture 5.3] Let k ≥ 3. Then  h1 (k, n) =



1 1− 1− k

k−1

  n−1 + o(1) . k−1

In this note we provide new lower bounds for hd (k, n) when d ≤ . Theorem 1.4. Let 1 ≤ d ≤ k − 1 and t = k − d, then  hd (k, n) ≥

   t

t/2t/2 (t/2 + 1)t/2 n 1− + o(1) . (t + 1)t t/2 t 

Theorem 1.5. Let 1 ≤ d ≤  ≤ k − 1 and t = k − d. Then     n hd (k, n) ≥ 1 − bt,k− 2−t + o(1) , t where bt,k− equals to the largest sum of k −  consecutive binomial coefficients taken   from { 0t , . . . , tt }. Theorem 1.4 disproves both Conjectures 1.2 and 1.3. Corollary 1.6. For all k,       n n 5 5 + o(1) , hk−3 (k, n) ≥ + o(1) , hk−2 (k, n) ≥ 9 2 8 3    n 409 hk−4 (k, n) ≥ + o(1) 625 4 and in general, for any 1 ≤ d ≤ k − 1,  hd (k, n) >

1

1−

3(k − d)/2 + 1



 n . k−d

(1.1)

These bounds imply that Conjecture 1.2 is false when k ≥ 4 and min{k − 4, k/2} ≤ d ≤ k − 2, and Conjecture 1.3 is false whenever k ≥ 4. We will prove Theorem 1.4, Theorem 1.5, and Corollary 1.6 in the next section. We believe that Conjecture 1.2 is false whenever k ≥ 4 but due to our limited knowledge on md (k, n), we can only disprove Conjecture 1.2 for the cases when md (k, n) is known.   This bound hk−2 (k, n) ≥ ( 59 + o(1)) n2 coincides with the value of m1 (3, n) – it   was shown in [6] that m1 (3, n) = (5/9 + o(1)) n2 , and it was widely believed that n h1 (3, n) = (5/9 + o(1)) 2 , e.g., see [19]. On the other hand, it is known [17] that

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    m2 (4, n) = ( 12 + o(1)) n2 , which is smaller than 59 n2 . Therefore k = 4 and d = 2 is the smallest case when Theorem 1.4 disproves Conjecture 1.2. More importantly, (1.1)  n  shows that hd (k, n)/ k−d tends to one as k − d tends to ∞. For example, as k becomes   sufficiently large, hk−ln k (k, n) is close to n−d k−d , the trivial upper bound. In contrast, Conjecture 1.1 suggests that there exists c > 0 independent of k and d (c = 1/e, where   e = 2.718 . . ., if Conjecture 1.1 is true) such that md (k, n) ≤ (1 − c) n−d . k−d √ n  Similarly, by Theorem 1.5, if k −  = o( t), hd (k, n)/ t tends to one as t tends to ∞ because 1 − bt,k− 2

−t

√   k− t o( t) ≥1− . ≈1−

2t t/2 πt/2

Theorem 1.5 also implies the following special case: suppose k is odd and  = d = 1  n k − 2. Then t = 2 and bt,k− = b2,2 = 3, and consequently hk−2 k−2 (k, n) ≥ 4 + o(1) 2 . n k 2 Previously it was only known that hk−2 (k, n) ≥ (1 − ( ) + o(1)) (2.1) with by k−2 2 k+1 a = k/(k − ) = (k + 1)/2 from Section 2. When k is large, the bound provided by Theorem 1.5 is much better. Finally, we do not know if Theorems 1.4 and 1.5 are best possible. Glebov, Person, and Weps [5] gave a general upper bound (far away from our lower bounds)  hd (k, n)

≤

1−



1 ck3k−3

 n−d , k−d

where c is a constant independent of d, , k, n. 2. The proofs Before proving our results, it is instructive to recall the so-called space barrier. ˙ for X ∪ Y when sets X, Y are disjoint. Throughout the paper, we write X ∪Y ˙ and Proposition 2.1. [16] Let H = (V, E) be an n-vertex k-graph such that V = X ∪Y V  1 E = {e ∈ k : e ∩ X = ∅}. Suppose |X| < a(k−) n, where a := k/(k − ), then H does not contain a Hamilton -cycle. A proof of Proposition 2.1 can be found in [16, Proposition 2.2] and is actually included in our proof of Proposition 2.2 below. It is not hard to see that Proposition 2.1 shows that  hd (k, n)

≥



1 1− 1− a(k − )

k−d

 + o(1)

 n−d . k−d

(2.1)

Now we state our construction for Hamilton cycles – it generalizes the one given by Katona and Kierstead [11, Theorem 3] (where j = k/2 ) and the space barrier (where

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j =  + 1 − k) simultaneously. The special case with k = 3,  = 2, j = 1, and |X| = n/3 appears in [19, Construction 2]. Proposition 2.2. Given an integer j such that  + 1 − k ≤ j ≤ k, let H = (V, E) be an   ˙ and E = {e ∈ Vk : |e ∩ X| ∈ n-vertex k-graph such that V = X ∪Y / {j, j + 1, . . . , j + k − j−1 j+k−   − 1}. Suppose a (k−) n < |X| < a(k−) n, where a := k/(k − ) and a := k/(k − ), then H does not contain a Hamilton -cycle. Proof. Suppose instead, that H contains a Hamilton -cycle C. Then all edges e of C / {j, j + 1, . . . , j + k −  − 1}. We claim that either all edges e of C satisfy satisfy |e ∩ X| ∈ |e ∩ X| ≤ j − 1 or all edges e of C satisfy |e ∩ X| ≥ j + k − . Otherwise, there must be two consecutive edges e1 , e2 in C such that |e1 ∩ X| ≤ j − 1 and |e2 ∩ X| ≥ j + k − . However, since |e1 ∩ e2 | = , we have ||e1 ∩ X| − |e2 ∩ X|| ≤ k − , a contradiction. Observe that every vertex of H is contained in either a or a edges of C and C contains n k− edges. This implies that a |X| ≤



|e ∩ X| ≤ a|X|.

e∈C

n n On the other hand, we have e∈C |e ∩ X| < (j − 1) k− or e∈C |e ∩ X| > (j + k − ) k− . j−1 j+k− In either case, we get a contradiction with the assumption a (k−) n < |X| < a(k−) n. 2 Note that by reducing the lower and upper bounds for |X| by small constants, we can conclude that H actually contains no Hamilton -path.   In the proofs of Theorems 1.4 and 1.5, we will consider binomial coefficients pq with   q < 0 – in this case pq = 0. We will conveniently write |X| = xn, where 0 < x < 1, instead of |X| = xn – this does not affect our calculations as n is sufficiently large. In the proofs of Theorem 1.4 we apply Proposition 2.2 with j and x such that j−1 k < j+1 j−1 j+1 x < k . Note that (when j ≤ k, 1 ≤ d ≤ k − 1, and t = k − d) k < x < k implies j+1 that j−d t+1 < x < t+1 , in particular, j − d ≤ x(t + 1) ≤ j. Proof of Theorem 1.4. Let x = t/2/(t + 1).1 Since t/2 t+1

(2.2)

k−1

j−1 j+1 j=1 ( k , k ) = (0, 1) and 1/3 ≤ j+1 such that j−1 < t/2 k t+1 < k . Let

≤ 1/2, there exists an integer j ∈ [k − 1] ˙ , |X| = xn and E = {e ∈ H = (V, E) be an n-vertex k-graph such that V = X ∪Y V  j−1 j+1 k : |e ∩ X| = j}. Since k n < |X| < k n, H contains no tight Hamilton cycle by Proposition 2.2 with  = k − 1. Now let us compute δd (H). For 0 ≤ i ≤ d, let Si be any d-vertex subset of V that contains exactly i vertices in X. By the definition of H, 1 It can be shown that among all 0 < x < 1 and 0 ≤ j ≤ k such that j−1
j+1 k ,

our choice of x and

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 degH (Si ) =

    n−d |X| − i |Y | − (d − i) − . t j−i t−j+i

Note that this holds for i > j or i < j − t trivially. So we have      n−d |X| − i |Y | − (d − i) δd (H) = min − 0≤i≤d t j−i t−j+i      |X| |Y | n + o(nt ). = − max  j−d≤i ≤j i t − i t Write |X| = xn and |Y | = yn. When 0 ≤ i ≤ t, we have 

        (xn)i (yn)t−i |X| |Y | t i t−i n t = x + o(n ) = y + o(nt ). i !(t − i )! i t − i t i

When i < 0 or i > t, we have δd (H) =

|X| |Y |  i

t−i

=0=

 t  i t−i n i x y t . In all cases, we have

      t i t−i n n x + o(nt ). − max y j−d≤i ≤j i t t

(2.3)

 Let ai := ti xi y t−i . Since x = t/2/(t + 1) and y = 1 − x, it is easy to see that i max0≤i≤t ai = at/2 (e.g., by observing aai+1 = xy · i+1 t−i for 0 ≤ i < t). Moreover, by (2.2), we have j − d ≤ t/2 ≤ j. Therefore,  max {ai } = at/2 =

j−d≤i≤j

 t

t/2t/2 (t/2 + 1)t/2 , (t + 1)t

t/2

and we complete the proof by substituting it into (2.3).

2

Now we turn to the proof of Theorem 1.5, in which we assume that |X| = n/2, though a further improvement of the lower bound may be possible by considering other values of |X|. Proof of Theorem 1.5. The proof is similar to the one of Theorem 1.4. Let H = (V, E)   ˙ , |X| = n/2 and E = {e ∈ Vk : |e ∩ X| ∈ / be an n-vertex k-graph such that V = X ∪Y { /2, . . . , /2 + k −  − 1}}. Note that   k  (k − ) ≥ k − (k −  − 1) =  + 1 > 2( /2 − 1), and a (k − ) = k−   k a(k − ) = (k − ) ≤ k + (k −  − 1) < 2(k − /2 ) = 2( /2 + k − ). k− So we have n

/2 + k − 

/2 − 1 n < |X| = < n. a (k − ) 2 a(k − )

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Thus, H contains no Hamilton -cycle by Proposition 2.2. Fix 1 ≤ d ≤ k − 1 and let t = k − d. Now we compute δd (H). For 0 ≤ i ≤ d, let Si be any d-vertex subset of V that contains exactly i vertices in X. It is easy to see that

degH (Si ) =

  i +k−−1  |X| |Y |  n + o(nt ), − p t − p t  p=i

where i = /2 − i. Using |X| = |Y | = n/2 and the similar calculations in the proof of Theorem 1.4, we get   i +k−−1   t  1 n n degH (Si ) = + o(nt ). − p 2t t t  p=i

By the definition of bt,k− , we have δd (H) = min degH (Si ) ≥ 0≤i≤d

    n n + o(nt ). − bt,k− 2−t t t

2

Corollary 1.6 follows from Theorem 1.4 via simple calculations. Proof of Corollary 1.6. Let t = k − d and  f (t) :=

 t

t/2t/2 (t/2 + 1)t/2 . (t + 1)t t/2

Theorem 1.4 states that hk−t (k, n) ≥ (1 − f (t) + o(1)) f (2) =

4 , 9

f (3) =

3 , 8

n t

for any 1 ≤ t ≤ k − 1. Since

and f (4) =

216 , 625

the bounds for hk−t (k, n), t = 2, 3, 4, are immediate. To see (1.1), it suffices to show that for t ≥ 1, 1 − f (t) > 1 −

t/2

(t/2+1) When t is odd, t/2 (t+1) t t ( t+1 ) . Thus, for all t, we have 2

t/2

1 3t/2 + 1

.

(2.4)

= 1/2t ; when t is even, t/2t/2 (t/2 + 1)t/2 < 

f (t) ≤

 t 1 , t/2 2t

  √ 2m where a strict inequality holds for all even t. Now we use the fact 2m / 3m + 1, m ≤2

which holds for all integers m ≥ 1. Thus, for all even t, we have f (t) ≤ 1/ 3t/2 + 1; for all odd t,

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   t+1 1 t 1 1 1 1 = ≤

<

. 2 t/2 + 1 2t t/2 2t 3(t + 1)/2 + 1 3t/2 + 1

 f (t) ≤

Hence f (t) ≤ 1/ 3t/2 + 1 for all t ≥ 1. Moreover, by the computation above, regardless of the parity of t, the strict inequality always holds and thus (2.4) is proved. We next show that whenever k ≥ 4 and 2 ≤ t ≤ k − 1,  1 − f (t) > max

 t  1 1 ,1 − 1 − . 2 k

This implies that Conjecture 1.3 fails for k ≥ 4, and Conjecture k ≥ 4 and  1.2 fails for  n  k−d  min{k − 4, k/2} ≤ d ≤ k − 2 (because md (k, n)/ k−d = max 12 , 1 − 1 − k1 + o(1) in this case). It suffices to show that for k ≥ 4 and 2 ≤ t ≤ k − 1,  1−

f (t) < 1/2 and f (t) <

1 k

t .

The first inequality immediately follows from (2.4) and 1/ 3t/2 + 1 ≤ 1/2. For the second inequality, note that f (t) <

1 3t/2 + 1

<

1 < e

 1−

1 k

k−1

 ≤

1−

1 k

t

 t for all t ≥ 5. For t = 2, 3 and all k ≥ 4, one can verify f (t) < (3/4)t ≤ 1 − k1 easily. Also, for t = 4 and all k ≥ 5, we have f (4) < (4/5)4 ≤ (1 − k1 )4 . 2 Acknowledgment We thank two referees for their valuable comments. References [1] N. Alon, P. Frankl, H. Huang, V. Rödl, A. Ruciński, B. Sudakov, Large matchings in uniform hypergraphs and the conjecture of Erdős and Samuels, J. Combin. Theory Ser. A 119 (6) (2012) 1200–1215. [2] E. Buß, H. Hàn, M. Schacht, Minimum vertex degree conditions for loose Hamilton cycles in 3-uniform hypergraphs, J. Combin. Theory Ser. B 103 (6) (2013) 658–678. [3] A. Czygrinow, T. Molla, Tight codegree condition for the existence of loose Hamilton cycles in 3-graphs, SIAM J. Discrete Math. 28 (1) (2014) 67–76. [4] G.A. Dirac, Some theorems on abstract graphs, Proc. Lond. Math. Soc. (3) 2 (1952) 69–81. [5] R. Glebov, Y. Person, W. Weps, On extremal hypergraphs for Hamiltonian cycles, European J. Combin. 33 (4) (2012) 544–555. [6] H. Hàn, Y. Person, M. Schacht, On perfect matchings in uniform hypergraphs with large minimum vertex degree, SIAM J. Discrete Math. 23 (2009) 732–748. [7] H. Hàn, M. Schacht, Dirac-type results for loose Hamilton cycles in uniform hypergraphs, J. Combin. Theory Ser. B 100 (2010) 332–346.

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