VERTEX DEGREE SUMS FOR PERFECT MATCHINGS IN 3-UNIFORM HYPERGRAPHS

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1 VERTEX DEGREE SUMS FOR PERFECT MATCHINGS IN 3-UNIFORM HYPERGRAPHS YI ZHANG, YI ZHAO, AND MEI LU Abstract. We determine the minimum degree sum of two adjacent vertices that ensures a perfect matching in a 3-graph without isolated vertex. More precisely, suppose that H is a 3-uniform hypergraph whose order n is sufficiently large and divisible by 3. If H contains no isolated vertex and deg(u+deg(v > 3 n 8 3 n+ for any two vertices u and v that are contained in some edge of H, then H contains a perfect matching. This bound is tight. 1. Introduction A k-uniform hypergraph (in short, k-graph H is a pair (V, E, where V := V (H is a finite set of vertices and E := E(H is a family of k-element subsets of V. A matching of size s in H is a family of s pairwise disjoint edges of H. If the matching covers all the vertices of H, then we call it a perfect matching. Given a set S V, the degree deg H (S of S is the number of the edges of H containing S. We simply write deg(s when H is obvious from the context. Given integers l < k n such that k divides n, we define the minimum l-degree threshold m l (k, n as the smallest integer m such that every k-graph H on n vertices with δ l (H m contains a perfect matching. In recent years the problem of determining m l (k, n has received much attention, see, e.g., [, 4, 5, 6, 7, 8, 9, 11, 1, 13, 15, 16, 17, 19, 0, 1]. In particular, Rödl, Ruciński, and Szemerédi [17] determined m k 1 (k, n for all k 3 and sufficiently large n. Treglown and Zhao [19, 0] determined m l (k, n for all l k/ and sufficiently large n. For more Dirac-type results on hypergraphs, we refer readers to surveys [14, 5]. In this paper we consider vertex degrees in 3-graphs. Hàn, Person and Schacht [4] showed that ( ( 5 n m 1 (3, n = 9 + o(1. (1 Kühn, Osthus and Treglown [10] and independently Khan [6] later proved that m 1 (3, n = ( ( n 1 n/3 + 1 for sufficiently large n. Motivated by the relation between Dirac s condition and Ore s condition for Hamilton cycles, Tang and Yan [18] studied the degree sum of two (k 1-sets that guarantees a tight Hamilton cycle in k-graphs. Zhang and Lu [] studied the degree sum of two (k 1-sets that guarantees a perfect matching in k- uniform hypergraphs. It is more natural to consider the degree sum of two vertices that guarantees a perfect matching in hypergraphs. For two distinct vertices u, v in a hypergraph, we call u, v adjacent if there exists an edge containing both of them. The following are three possible ways of defining the minimum degree sum of 3-graphs. Let σ (H = min{deg(u + deg(v : u, v V (H}, σ (H = min{deg(u + deg(v : u and v are adjacent} and σ (H = min{deg(u + deg(v : u and v are not adjacent}. The parameter σ is closely related to the Dirac threshold m 1 (3, n. We can prove that when n is divisible by 3 and sufficiently large, every 3-graph H on n vertices with σ (H ( ( ( n 1 n/3 + 1 contains a perfect matching. Indeed, such H contains at most one vertex u with deg(u ( ( n 1 n/3. If deg(u (5/9 ε ( n for some ε > 0, then we choose an edge containing u and find a perfect matching in Date: October 11, 017. Key words and phrases. Perfect matchings; Hypergraphs; Dirac s theorem; Ore s condition. Yi Zhang and Mei Lu are supported by the National Natural Science Foundation of China (Grants and Yi Zhao is partially supported by NSF grants DMS and DMS

2 the remaining 3-graph by (1 immediately. Otherwise, δ 1 (H (5/9 ε ( n. We can prove that H contains a perfect matching by following the same process as in [10]. 1 On the other hand, no condition on σ alone guarantees a perfect matching. In fact, let H be the 3-graph whose edge set consists of all triples that contain a fixed vertex. This H contains no two disjoint edges even though it satisfies all conditions on σ (because any two vertices of H are adjacent. Therefore we focus on σ. More precisely, we determine the largest σ (H among all 3-graphs H of order n without isolated vertex such that H contains no perfect matching. (Trivially H contains no perfect matching if it contains an isolated vertex. Let us define a 3-graph H, whose vertex set is partitioned into two vertex classes S and T of size n/3 + 1 and n/3 1, respectively, and whose edge set consists of all the triples containing at least two vertices of T. For any two vertices u T and v S, ( n/3 ( n ( ( deg(u = n n/3 1 3 > = deg(v. Hence σ (H = ( ( n/3 + (n/3 + 1(n/3 + n/3 1 = n /3 8n/3 +. Obviously, H contains no perfect matching. The following is our main result. Theorem 1. There exists n 0 N such that the following holds for all integers n n 0 that are divisible by 3. Let H be a 3-graph of order n n 0 without isolated vertex. If σ (H > σ (H = 3 n 8 3n +, then H contains a perfect matching. Theorem 1 actually follows from the following stability result. Theorem. There exist ε > 0 and n 0 N such that the following holds for all integers n n 0 that are divisible by 3. Suppose that H is a 3-graph of order n n 0 without isolated vertex and σ (H > n /3 εn, then H H or H contains a perfect matching. Indeed, if σ (H > n /3 8n/3 +, then H H and by Theorem, H contains a perfect matching. Furthermore, Theorem implies that H is the unique extremal 3-graph for Theorem 1 because all proper subgraphs H of H satisfy σ (H < σ (H. This paper is organized as follows. In Section, we provide preliminary results and an outline of our proof. We prove an important lemma in Section 3 and we complete the proof of Theorem in Section 4. Section 5 contains concluding remarks and open problems. Notation: Given vertices v 1,..., v t, we often write v 1 v t for {v 1,..., v t }. The neighborhood N(u, v is the set of the vertices w such that uvw E(H. Let V 1, V, V 3 be three vertex subsets of V (H, we say that an edge e E(H is of type V 1 V V 3 if e = {v 1, v, v 3 } such that v 1 V 1, v V and v 3 V 3. Given a vertex v V (H and a set A V (H, we define the link L v (A to be the set of all pairs uw such that u, w A and uvw E(H. When A and B are two disjoint sets of V (H, we define L v (A, B as the set of all pairs uw such that u A, w B and uvw E(H. We write 0 < a 1 a a 3 if we can choose the constants a 1, a, a 3 from right to left. More precisely there are increasing functions f and g such that given a 3, whenever we choose some a f(a 3 and a 1 g(a, all calculations needed in our proof are valid. We will need small constants. Preliminaries and proof outline 0 < ε η γ γ ρ τ 1. Suppose H is a 3-graph such that σ (H > n /3 εn. Let W = {v V (H : deg(v n /3 εn /}, U = V \ W. If W =, then (1 implies that H contains a perfect matching. We thus assume that W 1. Any two vertices of W are not adjacent otherwise σ (H n /3 εn, a contradiction. If W n/3 + 1, then H H and we are done. We thus assume W n/3 for the rest of the proof. Our proof will use the following claim. 1 In fact, due to the absorbing method, we only need to verify the extremal case.

3 Claim 3. If W n/4, then every vertex of U is adjacent to some vertex of W. Proof. To the contrary, assume that some vertex u 0 U is not adjacent to any vertex in W. Then we have deg(u 0 ( ( U 1 = n W 1. Since W n/4 and n is sufficiently large, ( n n/4 1 deg(u 0 = 9 3 n 9 8 n + 1 < n 3 ε n, which contradicts the definition of U. By Claim 3, when W n 4, we have deg(u (n /3 εn ( n W for every u U. This is stronger than the bound given by the definition of U because ( ( ( ( n W n n/4 3 n εn 3 n εn > n 3 ε n. Our proof consists of two steps. Step 1. We prove that H contains a matching that covers all the vertices of W. Lemma 4. There exist ε > 0 and n 0 N such that the following holds. Suppose that H is a 3-graph of order n n 0 without isolated vertex and σ (H > n /3 εn. Let W = {v V (H : deg(v n /3 εn /}. If W n/3, then H contains a matching that covers every vertex of W. Our approach towards Lemma 4 begins by considering a largest matching M such that every edge of M contains one vertex from W and suppose M < W. If W (1/3 γn, then we choose two adjacent vertices, one from W and the other from V \ W to derive a contradiction with σ (H. If n/3 W > (1/3 γn, we use three unmatched vertices, one from W and two from V \ W to derive a contradiction. Step. We show that H contains a perfect matching. Because of Lemma 4, we begin by considering a largest matching M such that M covers every vertex of W and suppose that M < n/3. After choosing three vertices from V \ V (M, we distinguish the cases when M n/3 ηn and when M > n/3 ηn and derive a contradiction by comparing upper and lower bounds for the degree sum of these three vertices. When M > n/3 ηn, we need to apply (1. In Step we need three simple extremal results. The first lemma is Observation 1.8 of Aharoni and Howard [1]. A k-graph H is called k-partite if V (H can be partitioned into V 1,, V k, such that each edge of H meets every V i in precisely one vertex. If all parts are of the same size n, we call H n-balanced. Lemma 5. [1] Let F be the edge set of an n-balanced k-partite k-graph. If F does not contain s disjoint edges, then F (s 1n k 1. The bound in the following lemma is tight because we may let G 1 be the empty graph and G = G 3 = K n. Lemma 6. Given two sets A V such that A = 3 and V = n 4, let G 1, G, G 3 be three graphs on V such that no edge of G 1 is disjoint from an edge from G or G 3. Then 3 v A deg G i (v 6(n 1. Proof. Assume A = {u 1, u, u 3 } and let b = n 3 1. We need to show that 3 3 (u j 6b+1. Let l i denote the number of the vertices in A of degree at least 3 in G i. We distinguish the following two cases: Case 1: l 1 1. If l 1, say, deg G1 (u j 3 for j = 1,, then E(G i {u 1 u } for i =, 3 otherwise we can find two disjoint edges, one from G 1 and the other from G or G 3. Therefore, 3 (u j for i =, 3. Moreover, 3 (u j 3b + 6. We have 3 3 (u j 3b + 10 < 6b + 1. If l 1 = 1, say, deg G1 (u 1 3, then G i is a star centered at u 1 for i =, 3 otherwise one edge of G 1 must be disjoint from one edge of G or G 3. In this case we have 3 (u j b and 3 (u j b + 4 for i =, 3. Therefore, 3 3 (u j 3b + 14 < 6b + 1 as b 1. Case : l 1 = 0. If l i = 3 for some i {, 3}, then E(G 1 =. In this case 3 3 (u j (3b + 6 6b

4 Suppose l, l 3 and l = or l 3 =. Without loss of generality, assume l = and deg G (u j 3 for j = 1,. Then E(G 1 {u 1 u }. In this case 3 (u j and 3 (u j b for i =, 3. Hence 3 3 (u j 4b b + 1 as b 1. Assume l, l 3 1 and l = 1 or l 3 = 1. Without loss of generality, assume l = 1 and deg G (u 1 3. Then G 1 is a star centered at u 1. We have 3 (u j 4 and 3 (u j b for i =, 3. So 3 3 (u j b b + 1 as b 1. Suppose l, l 3 = 0. In this case 3 (u j 6 for i = 1,, 3. Therefore, 3 3 (u j 18 6b + 1 as b 1. The bound in the following lemma is tight because we may let G 1 = G = G 3 be a star of order n centered at a vertex of A. Lemma 7. Given two sets A V such that A = 3 and V = n 5, let G 1, G, G 3 be three graphs on V such that no edge of G i is disjoint from an edge from G j for any i j. Then 3 v A deg G i (v 3(n+1. Proof. Assume A = {u 1, u, u 3 } and let b = n 3. We need to show that 3 3 (u j 3b+1. Let l i denote the number of the vertices in A of degree at least 3 in G i. We distinguish the following two cases: Case 1: l i 1 for some i [3]. Without loss of generality, l 1 1 and deg G1 (u 1 3. If deg G1 (u 3 or deg G1 (u 3 3, say, deg G1 (u 3, then E(G i {u 1 u } for i =, 3 otherwise we can find two disjoint edges e 1 and e from two distinct graphs of G 1, G, G 3. In this case 3 (u j 3b + 6 and 3 (u j for i =, 3, which implies that 3 3 (u j 3b Assume deg G1 (u j for j =, 3. We know that G i, i =, 3 is a star centered at u 1 otherwise one edge of G 1 must be disjoint from one edge of G i, i {, 3}. If deg G (u 1 3 or deg G3 (u 1 3, then G 1 is also a star centered at u 1. In this case 3 (u j b+4 for i [3], so 3 3 (u j 3b+1. Otherwise deg Gi (u 1 for i =, 3, hence 3 (u j 4 for i =, 3. Since 3 (u j b + 6, we have 3 3 (u j b b + 1. Case : l i = 0 for i [3]. In this case 3 (u j 6 for i = 1,, 3. Hence 3 3 (u j 18 3b + 1 as b. 3. Proof of Lemma 4 Choose a largest matching of H, denoted by M, such that every edge of M is of type UUW. To the contrary, assume that M W 1. Let U 1 = V (M U, U = U \U 1, W 1 = V (M W, and W = W \W 1. Then U 1 = M, and U = n W M. We distinguish the following two cases. Case 1: 0 < W ( 1 3 γn. We further distinguish the following two sub-cases: Case 1.1: A vertex v 0 W is adjacent to a vertex u 0 U. Let M = {e M : u e, N(v 0, u U 3}. Assume {u 1, u, v 1 } M such that u 1, u U 1, v 1 W 1, and N(v 0, u 1 U 3. We claim that N(u 0, v 1 (U {u } =. ( Indeed, if {u 0, v 1, u 3 } E(H for some u 3 U, then we can find u 4 U \ {u 0, u 3 } such that {v 0, u 1, u 4 } E(H. Replacing {u 1, u, v 1 } by {u 0, v 1, u 3 } and {v 0, u 1, u 4 } gives a larger matching than M, a contradiction. The case when {u 0, v 1, u } E(H is similar. By the definition of M, there are at most ( U 1 M edges containing v 0 with one vertex in U 1 \V (M and one vertex in U. This implies that ( U1 deg(v 0 + M U + ( U 1 M = 4 ( U1 + U 1 + M ( U 4.

5 By (, there are at most U 1 W 1 M edges consisting of u 0, one vertex in U 1, and one vertex in W 1, and at most ( U 1( W 1 M edges consisting of u 0, one vertex in U, and one vertex in W 1. Therefore, ( U 1 deg(u 0 + U 1 W + U 1 W 1 M + ( U 1( W 1 M ( U 1 = + U 1 W + ( U 1 W 1 U M, and consequently, ( ( U1 U 1 deg(v 0 + deg(u 0 + U U 1 W + ( U 1 W 1 + M ( U 4. Since W ( 1 3 γn, we have U > 3γn > 4. As M M = W 1 = U1, it follows that ( ( U1 U 1 deg(v 0 + deg(u 0 + U U 1 W + ( U 1 U 1 + U 1 ( U 4 (( ( ( ( U U U 1 = + + W 1 U 1 ( = ( U 1 U + ( W 1 M. Since M W 1 and U n 3 W +, we derive that ( n 3 W + deg(v 0 + deg(u 0 (n W 1 + ( W 1( W 1 = 3 n 7 3 n ( n W. Since W ( 1 3 γn, 0 < ε γ and n is sufficiently large, we have deg(v 0 + deg(u 0 3 n 7 3 n ( γn + 7 < 6 3 n εn. This contradicts our assumption on σ (H because v 0 and u 0 are adjacent. Case 1.: No vertex in W is adjacent to any vertex in U. Fix v 0 W. Since v 0 is not adjacent to any vertex in U, we have deg(v 0 ( U 1 ( = M. Since v0 is not an isolated vertex, there exists a vertex u 1 U 1 that is adjacent to v 0. By the assumption, H contains no edge containing u 1 with one vertex in U, one vertex in W. Thus deg(u 1 ( U 1 +( U 1 W U W. Since M W 1 and U = n W, it follows that ( ( ( W 1 U 1 deg(v 0 + deg(u ( U 1 W (n 3 W + = 3 ( W n 5 n Furthermore, since W ( 1 3 γn and 0 < ε γ, we derive that deg(v 0 + deg(u 1 3 ( n 3 γn n 5 n + 13 ( 8 = 3 γ + 3 γ < 3 n εn, contradicting our assumption on σ (H. Case : W > ( 1 3 γn. Claim 8. M n/3 γ n. 5 n (3 3 γ n +

6 Proof. To the contrary, assume that M < n/3 γ n. Fix v 0 W. Then deg(v 0 ( ( U U because there is no edge of type U U W. Suppose u U is adjacent to v 0. Trivially deg(u ( U 1 + ( U 1 W. Thus ( U 1 deg(v 0 + deg(u + ( U 1 W + ( U ( ( U U = (n 1( U 1 Our assumptions imply that U n/3 + γn and U γ n. As a result, ( ( γ deg(v 0 + deg(u (n 1 3 n + γn 1 n < 3 n εn, because ε γ γ and n is sufficiently large. This contradicts our assumption on σ (H. Fix u 1 u U and v 0 W. Trivially deg(w ( ( U for any vertex w W and deg(u U 1 + W ( U 1 for any vertex u U. Furthermore, for any two distinct edges e 1, e M, we observe that at least one triple of type UUW with one vertex from each of e 1 and e and one vertex from {u 1, u, v 0 } is not an edge otherwise there is a matching M 3 of size three on e 1 e {u 1, u, v 0 } and M 3 M \ {e 1, e } is thus a matching larger than M. By Claim 8, M n/3 γ n. Thus, (( U 1 deg(u 1 + deg(u + deg(v 0 + W (U 1 + ( U ( n/3 γ n On the other hand, since W > ( 1 3 γn n/4, Claim 3 implies that u i is adjacent to some vertex in W for i = 1,. We know that v 0 is adjacent to some vertex in U. Therefore, deg(u i > ( n /3 εn ( U for i = 1,, and deg(v 0 > ( n /3 εn ( ( U 1 + W ( U 1. It follows that ( ( ( n U U 1 deg(u 1 + deg(u + deg(v 0 > 3 3 εn W ( U 1. The upper and lower bounds for deg(u 1 + deg(u + deg(v 0 together imply that (( ( ( U 1 U n/3 γ ( n n 3 + W ( U 1 + > 3 3 εn, or ( U 1(n 1 1 ( n/3 γ n > n 3 3 εn, which is impossible because U n/3+γn, 0 < ε γ γ 1 and n is sufficiently large. This completes the proof of Lemma Proof of Theorem Choose a matching M such that (i M covers all the vertices of W ; (ii subject to (i, M is the largest. Lemma 4 implies that such a matching exists. Let M 1 = {e M : e W }, M = M \ M 1, and U 3 = V (H \ V (M. We have M 1 = W, M = M W, U 3 = n 3 M. Suppose to the contrary, that M n/3 1. Fix three vertices u 1, u, u 3 of U 3. We distinguish the following two cases. Case 1: M n/3 ηn. Trivially there are at most 3 M edges in H containing u i and two vertices from the same edge of M for i = 1,, 3. For any distinct e 1, e from M, we claim that L ui (e 1, e 18. Indeed, let H 1 be the 3-partite subgraph of H induced on three parts e 1, e, and {u 1, u, u 3 }. We observe that H 1 does not contain a perfect matching otherwise, letting M 1 be a perfect matching of H 1, (M \ {e 1, e } M 1 is a larger matching than M, a contradiction. Apply Lemma 5 with n = k = s = 3, we obtain that E(H Therefore 3 L u i (e 1, e

7 For any e M 1, we claim that L ui (e, U 3 6( U 3 1. Indeed, assume e = {v 1, v, v 3 } M 1 with v 1 W. Apply Lemma 6 with A = {u 1, u, u 3 }, V = U 3, and G i = (U 3, L vi (U 3 for i = 1,, 3. Since M n/3 4, we have B = U 3 3. By the maximality of M, no edge of G 1 is disjoint from an edge of G or G 3. By Lemma 6, 3 3 (u j 6( U 3 1. Hence 3 L u i (e, U 3 = 3 3 (u j 6( U 3 1. Similarly, for any e M, we can apply Lemma 7 to obtain that L ui (e, U 3 3( U Putting these bounds together gives ( M deg(u i 18 ( M M + L ui (V (M 1, U 3 + L ui (V (M, U M + 6 M 1 ( U M ( U Since M 1 = W, M = M W, U 3 = n 3 M, we derive that ( M deg(u i M + 6 W (n 3 M 1 + 3( M W (n 3 M + 1 = (3n 9 W + 3 M + 3 W n 9 W. Furthermore, 3n 9 W + 3 > 0 and M n/3 ηn implies that If W n/4, from (3, we have ( n deg(u i (3n 9 W ηn + 3 W n 9 W = (9ηn 9 W + (1 3η n + (1 3η n. (3 deg(u i (9ηn 9 n 4 + (1 3η n + (1 3η n = (1 34 ( η n 3η + 5 n, 4 which contradicts the condition ( 3 deg(u n i 3 3 εn because u i U 3 for i [3] and ε η. If W > n/4, Claim 3 implies that u i is adjacent to one vertex of W, i = 1,, 3. Furthermore, deg(w for w W. So ( U ( n deg(u i > 3 3 εn ( ( U n = 3 3 εn The upper and lower bounds for 3 deg(u i together imply that ( n W (9ηn 9 W + (1 3η n + (1 3η n + 3 ( n W. ( n > 3 3 εn, which is a contradiction because W > n/4, 0 < ε η 1 and n is sufficiently large. Case : M > n/3 ηn. If M = n/3 1, then U 3 = 3 and we can not apply Lemmas 6 and 7. In fact, whenever M > n/3 ηn, Lemma 5 suffices for our proof. 7

8 Let W = {v W : deg(v (5/18+τn }. Let M be the sub-matching of M covering every vertex of W. If W ρn, we claim that deg H (u ( γ ( n for every vertex u V (H, where H := H[V \ V (M ]. Indeed, from the definition of W, deg H (u > (5/18 + τn for every vertex u V (H. Hence, ( 5 deg H (u deg H (u 3n W > 18 + τ n 3n W. Since W ρn, 0 < γ ρ τ 1 and n is sufficiently large, we have ( ( ( 5 5 n deg H (u > 18 + τ n 3ρn > 9 + γ. In addition, n is divisible by 3, so V (H is divisible by 3. (1 implies that H contains a perfect matching M. Now M M is a perfect matching of H. Therefore, we assume that W ρn in the rest of the proof. If one vertex of u 1, u, u 3, say, u 1, is adjacent to one vertex in W, the definition of W implies that deg(u 1 > n /3 εn ( τ n. Recall that deg(u i > n /3 εn / for i =, 3. Thus deg(u i > By Lemma 5, we have ( ( ( n εn 18 + τ n = 18 ε τ n. (4 ( M deg(u i 18 = M + 9 M (n 3 M 1 ( M 1 4 n n 9 4 n + 9 8, where 18 ( M accounts for edges between pairs of matching M, 9 M for edges with two vertices in the same matching edge from M, and 9 M (n 3 M 1 for edges with one vertex in V (M, one vertex in U 3. Since M > n/3 ηn, it follows that ( n deg(u i 18 3 ηn 1 4 n n 9 4 n = (1 + 3η 18η n + (9η 3n. However, (1 + 3η 18η n + (9η 3n < ( ε τ n because 0 < ε η τ 1 and n is sufficiently large. It contradicts (4. If none of these three vertices u 1, u, u 3 are adjacent to the vertices in W, we have ( M M deg(u i 18 ( M + 3 = 3 + 9( M M + 9( M M (n 3 M 1 + 3( M (n 3 M 1 ( M + 1 n 3 M 45 4 M + 9 n M 9 M n. Here 18 ( M M accounts for edges between pairs of matching M \ M, 9( M M for edges with two vertices in the same matching edge from M \ M, 9( M M (n 3 M 1 for edges with one vertex in V (M \M, one vertex in U 3, 3 ( M for edges with two vertices in V (M \W, and 3( M (n 3 M 1 for edges with one vertex in V (M \ W, one vertex in V (H \ V (M. Since n/ + 3 M / < 0 and 8

9 M = W ρn, then ( deg(u i 3 ρn + 1 n 3 M 45 4 M + 9 n M 9 M n ( = 18 M 1 4 n ρn + 1 ( ρ 3 4 ρ n 9 4 ρn 9 4 n Recall that 0 < ρ 1, so 1 4 n ρn 1 4 < n 3 ηn. Furthermore, M > n 3 ηn, hence we have ( n deg(u i 18 3 ηn 1 4 n 1 4 ρn + 1 ( ρ 3 4 ρ n 9 4 ρn 9 4 n = ( 1 3ρ 9ηρ + 3η 18η n + (9η 3n, which contradicts the condition 3 deg(u i 3 ( n /3 εn / because 0 < ε η ρ 1 and n is sufficiently large. This completes the proof of Theorem. 5. Concluding remarks In this paper we consider the minimum degree sum of two adjacent vertices that guarantees a perfect matching in 3-graphs. Given 3 k < n and s n/k, can we generalize this problem to k-graphs not containing a matching of size s? For 1 l k, let Hn,k,s l denote the k-graph whose vertex set is partitioned into two sets S and T of size n sl + 1 and sl 1, respectively, and whose edge set consists of all the k-sets with at least l vertices in T. Apparently Hn,k,s l contains no matching of size s. A well-known conjecture of Erdös [3] says that Hn,k,s 1 or Hk n,k,s is the densest k-graph on n vertices not containing a matching of size s. It is reasonable to speculate that the largest σ (H among all k-graphs H on n vertices not containing a matching of size s is also attained by Hn,k,s l. Note that Hk n,k,s is a complete k-graph of order sk 1 together with n sk + 1 isolated vertices and thus σ (H n,k,s k = ( sk k 1. When 1 l k, any two vertices of Hn,k,s l are adjacent and thus σ (Hn,k,s l = δ 1(Hn,k,s l. When l = k 1, it is easy to see that σ (H k 1 n,k,s = ( ( s(k 1 k 1 + (n s(k 1 + s(k 1 k. Assume s = n/k. Since Hn,k,n/k k contains isolated vertices and δ 1(Hn,k,n/k l δ 1(Hn,k,n/k 1 for 1 l k, we only need to compare σ (H n,k,n/k 1 and σ (H k 1 n,k,n/k. For sufficiently large n, it is easy to see that σ (H n,k,n/k 1 < σ (H k 1 n,k,n/k when k 6 and σ (Hn,k,n/k 1 > σ (H k 1 n,k,n/k when k 7. Problem 9. Does the following hold for any sufficiently large n that is divisible by k? Let H be a k-graph of order n without isolated vertex. If k 6 and σ (H > σ (H k 1 n,k,n/k or k 7 and σ (H > σ (H n,k,n/k 1, then H contains a perfect matching. Now assume k = 3 and s n/3. Note that ( (( ( 3s n 1 n s σ (H n,3,s 3 =, σ (Hn,3,s 1 =, and ( ( ( s s s 1 σ (H n,3,s = + (n s = (s (n 1. 1 It is easy to see that σ (H n,3,s > σ (H 1 n,3,s. Zhang and Lu [3] made the following conjecture. Conjecture 10. [3] There exists n 0 N such that the following holds. Suppose that H is a 3-graph of order n n 0 without isolated vertex. If σ (H > (( ( n 1 n s and n 3s, then H contains no matching of size s if and only if H is a subgraph of Hn,3,s. Zhang and Lu [3] showed that the conjecture holds when n 9s. Later the same authors [4] proved the conjecture for n 13s. If Conjecture 10 is true, then it implies the following theorem of Kühn, Osthus and Treglown [10]. 9

10 Theorem 11. [10] There exists n 0 N such that if H is a 3-graph of order n n 0 with δ 1 (H ( n and n 3s, then H contains a matching of size s. ( n s Our Theorem 1 suggests a weaker conjecture than Conjecture 10. Conjecture 1. There exists n 1 N such that the following holds. Suppose that H is a 3-graph of order n n 1 without isolated vertex. If σ (H > σ (H n,3,s and n 3s, then H contains a matching of size s. On the other hand, we may allow a 3-graph to contain isolated vertices. Note that σ (H n,3,s σ (H 3 n,3,s if and only if s (n + 4/9. We make the following conjecture. Conjecture 13. There exists n N such that the following holds. Suppose that H is a 3-graph of order n n and s n/3. If σ (H > σ (H n,3,s and s (n+4/9 or σ (H > σ (H 3 n,3,s and s > (n+4/9, then H contains a matching of size s. In fact, we can derive Conjecture 13 from Conjecture 1 as follows. Let n = max{ ( n 1, 3 n 1} and H be a 3-graph of order n n satisfying the assumption of Conjecture 13. If H contains no isolated vertex, then H contains a matching of size s by Conjecture 1. Otherwise, let W be the set of isolated vertices in H. Let H = H[V (H \ W ] and n = n W. Then H is a 3-graph without isolated vertex and σ (H = σ (H. When s (n + 4/9, we have σ (H > σ (H n,3,s > σ (H n,3,s. In addition, since n ( n 1 and ( n 1 σ (H > (s (n 1 (n 1, we have n n 1. When s > (n + 4/9, we have σ (H > σ (H n,3,s 3 > σ (H n,3,s > σ (H n,3,s. In addition, since n 3n 1 / and ( n ( ( 1 3s (n 1/3 σ (H > >, we have n n 1. In both cases, Conjecture 1 implies that H contains a matching of size s. References [1] R. Aharoni and D. Howard, A rainbow r-partite version of the Erdős-Ko-Rado theorem, Combin. Probab. Comput. 6 (017, [] N. Alon, P. Frankl, H. Huang, V. Rödl, A. Ruciński, and B. Sudakov. Large matchings in uniform hypergraphs and the conjectures of Erdős and Samuels, J. Combin. Theory Ser. A 119 (01, [3] P. Erdős, A problem on independent r-tuples, Ann. Univ. Sci. Budapest. Eőtvős Sect. Math. 8 (1965, [4] H. Hàn, Y. Person, and M. Schacht, On perfect matchings in uniform hypergraphs with large minimum vertex degree, SIAM J. Discrete Math. 3 (009, [5] J. Han, Perfect matchings in hypergraphs and the Erdős matching conjecture, SIAM J. Discrete Math. 30 (016, [6] I. Khan, Perfect matching in 3-uniform hypergraphs with large vertex degree, SIAM J. Discrete Math. 7 (013, [7] I. Khan, Perfect matchings in 4-uniform hypergraphs, J. Combin. Theory Ser. B 116 (016, [8] D. Kühn and D. Osthus, Matchings in hypergraphs of large minimum degree, J. Graph Theory 51 (006, [9] D. Kühn and D. Osthus, Embedding large subgraphs into dense graphs. In Surveys in combinatorics 009, volume 365 of London Math. Soc. Lecture Note Ser., pages Cambridge Univ. Press, Cambridge, 009. [10] D. Kühn, D. Osthus and A. Treglown, Matchings in 3-uniform hypergaphs, J. Combin. Theory Ser. B 103 (013, [11] D. Kühn, D. Osthus and T. Townsend, Fractional and integer matchings in uniform hypergraphs, European J. Combin. 38 (014, [1] K. Markström and A. Ruciński, Perfect matchings (and Hamilton cycles in hypergraphs with large degrees, European J. Combin. 3 (011, [13] O. Pikhurko, Perfect matchings and K4 3 -tilings in hypergraphs of large codegree, Graphs Combin. 4 (008, [14] V. Rödl and A. Ruciński, Dirac-type questions for hypergraphs a survey (or more problems for Endre to solve, An Irregular Mind, Bolyai Soc. Math. Studies 1 (010, [15] V. Rödl, A. Ruciński, and E. Szemerédi, Perfect matchings in uniform hypergraphs with large minimum degree, European J. Combin. 7 (006, [16] V. Rödl, A. Ruciński, and E. Szemerédi, An approximate Dirac-type theorem for k-uniform hypergraphs, Combinatorica, 8 (008, [17] V. Rödl, A. Ruciński, and E. Szemerédi, Perfect matchings in large uniform hypergraphs with large minimum collective degree, J. Combin. Theory Ser. A 116 (009,

11 [18] Y. Tang, and G. Yan, An approximate Ore-type result for tight hamilton cycles in uniform hypergraphs, Discrete Math. 340 (017, [19] A. Treglown and Y. Zhao, Exact minimum degree thresholds for perfect matchings in uniform hypergraphs, J. Combin. Theory Ser. A 119 (01, [0] A. Treglown and Y. Zhao, Exact minimum degree thresholds for perfect matchings in uniform hypergraphs II, J. Combin. Theory Ser. A 10 (013, [1] A. Treglown and Y. Zhao, A note on perfect matchings in uniform hypergraphs, Electron. J. Combin. 3 (016, [] Y. Zhang and M. Lu, Some Ore-type results for matching and perfect matching in k-uniform hypergraphs, submitted. [3] Y. Zhang and M. Lu, d-matching in 3-uniform hypergraphs, submitted. [4] Y. Zhang and M. Lu, Matching in 3-uniform hypergraphs, submitted. [5] Y. Zhao, Recent advances on dirac-type problems for hypergraphs. In Recent Trends in Combinatorics, volume 159 of the IMA Volumes in Mathematics and its Applications. Springer, New York, 016. Department of Mathematical Sciences, Tsinghua University, Beijing, address: shouwangmm@sina.com Department of Mathematics and Statistics, Georgia State University, Atlanta, GA address: yzhao6@gsu.edu Department of Mathematical Sciences, Tsinghua University, Beijing, address: mlu@math.tsinghua.edu.cn 11