MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES IN 3UNIFORM HYPERGRAPHS., where


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1 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES IN UNIFORM HYPERGRAPHS ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT Abstract. We investigate minimum vertex degree conditions for uniform hypergraphs which ensure the existence of loose Hamilton cycles. A loose Hamilton cycle is a spanning cycle in which only consecutive edges intersect and these intersections consist of precisely one vertex. We prove that every uniform nvertex (n even hypergraph H with minimum vertex degree δ (H ( o( ( n contains a loose Hamilton cycle. This bound is asymptotically best possible.. Introduction We consider kuniform hypergraphs H, that are pairs H = (V, E with vertex sets V = V (H and edge sets E = E(H ( ( V k, where V k denotes the family of all kelement subsets of the set V. We often identify a hypergraph H with its edge set, i.e., H ( V k, and for an edge {v,..., v k } H we often suppress the enclosing braces and write v... v k H instead. Given a kuniform hypergraph H = (V, E and a set S = {v,..., v s } ( V s let deg(s = deg(v,..., v s denote the number of edges of H containing the set S and let N(S = N(v,..., v s denote the set of those (k selement sets T ( V k s such that S T forms an edge in H. We denote by δ s (H the minimum sdegree of H, i.e., the minimum of deg(s over all selement sets S V. For s = the corresponding minimum degree δ (H is referred to as minimum vertex degree whereas for s = k we call the corresponding minimum degree δ k (H the minimum collective degree of H. We study sufficient minimum degree conditions which enforce the existence of spanning, socalled Hamilton cycles. A kuniform hypergraph C is called an l cycle if there is a cyclic ordering of the vertices of C such that every edge consists of k consecutive vertices, every vertex is contained in an edge and two consecutive edges (where the ordering of the edges is inherited by the ordering of the vertices intersect in exactly l vertices. For l = we call the cycle loose whereas the cycle is called tight if l = k. Naturally, we say that a kuniform, nvertex hypergraph H contains a Hamilton lcycle if there is a subhypergraph of H which forms an lcycle and which covers all vertices of H. Note that a Hamilton lcycle contains exactly n/(k l edges, implying that the number of vertices of H must be divisible by (k l which we indicate by n (k ln. The second author was supported by GIF grant no. I /005. The third author was supported through the HeisenbergProgramme of the Deutsche Forschungsgemeinschaft (DFG Grant SCHA 6/4. The collaboration of the authors was supported by joint grant of the Deutschen Akademischen Austausch Dienst (DAAD and the Fundação Coordenação de Aperfeiçoamento de Pessoal de Nível (CAPES.
2 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT Minimum collective degree conditions which ensure the existence of tight Hamilton cycles were first studied in [6] and in [, ]. In particular, in [, ] Rödl, Ruciński, and Szemerédi found asymptotically sharp bounds for this problem. Theorem. For every k and γ > 0 there exists an n 0 such that every k uniform hypergraph H = (V, E on V = n n 0 vertices with δ k (H (/+γn contains a tight Hamilton cycle. The corresponding question for loose cycles was first studied by Kühn and Osthus. In [0] they proved an asymptotically sharp bound on the minimum collective degree which ensures the existence of loose Hamilton cycles in uniform hypergraphs. This result was generalised to higher uniformity by the last two authors [4] and independently by Keevash, Kühn, Osthus and Mycroft in [7]. Theorem. For all integers k and every γ > 0 there exists an n 0 such that every kuniform hypergraph H = (V, E on V = n n 0 vertices with n (k N and δ k (H ( (k + γn contains a loose Hamilton cycle. Indeed, in [4] asymptotically sharp bounds for Hamilton lcycles for all l < k/ were obtained. Later this result was generalised to all 0 < l < k by Kühn, Mycroft, and Osthus [9]. These results are asymptotically best possible for all k and 0 < l < k. Hence, asymptotically, the problem of finding Hamilton lcycles in uniform hypergraphs with large minimum collective degree is solved. The focus of this paper are conditions on the minimum vertex degree which ensure the existence of Hamilton cycles. For δ (H very few results on spanning subhypergraphs are known (see e.g. [, ]. In this paper we give an asymptotically sharp bound on the minimum vertex degree in uniform hypergraphs which enforces the existence of loose Hamilton cycles. Theorem (Main result. For all γ > 0 there exists an n 0 such that the following holds. Suppose H is a uniform hypergraph on n > n 0 with n N and ( ( 7 n δ (H > 6 + γ. Then H contains a loose Hamilton cycle. In the proof we apply the socalled absorbing technique. In [] Rödl, Ruciński, and Szemerédi applied this elegant approach to tackle minimum degree problems for spanning graphs and hypergraphs. In our case it reduces the problem of finding a loose Hamilton cycle to the problem of finding a nearly spanning loose path and indeed, finding such a path will be the main obstacle to Theorem. As mentioned above, Theorem is best possible up to the error constant γ as seen by the following construction from [0]. Fact 4. For every n N there exists a uniform hypergraph H = (V, E on V = n vertices with δ (H 7 6( n O(n, which does not contain a loose Hamilton cycle. Proof. Consider the following uniform hypergraph H = (V, E. Let A B = V be a partition of V with A = n 4 and let E be the set of all triplets from V with at least one vertex in A. Clearly, δ (H = ( A + A ( B = 7 n 6( O(n. Now consider an arbitrary cycle in H. Note that every vertex, in particular every
3 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES vertex from A, is contained in at most two edges of this cycle. Moreover, every edge of the cycle must intersect A. Consequently, the cycle contains at most A < n/ edges and, hence, cannot be a Hamilton cycle. We note that the construction H in Fact 4 satisfies δ (H n/4 and indeed, the same construction proves that the minimum collective degree condition given in Theorem is asymptotically best possible for the case k =. This leads to the following conjecture for minimum vertex degree conditions enforcing loose Hamilton cycles in kuniform hypergraphs. Let k and let H k = (V, E be the kuniform, nvertex hypergraph on V = A B with A = n (k. Let E consist of all ksets intersecting A in at least one vertex. Then H k does not contain a loose Hamilton cycle and we believe that any kuniform, nvertex hypergraph H which has minimum vertex degree δ (H δ (H k + o(n contains a loose Hamilton cycle. Indeed, Theorem verifies this for the case k =.. Proof of the main result The proof of Theorem will be given in Section.. It uses several auxiliary lemmas which we introduce in Section.. We start with an outline of the proof... Outline of the proof. We will build a loose Hamilton cycle by connecting loose paths. Formally, a uniform hypergraph P is a loose path if there is an ordering (v,..., v t of its vertices such that every edge consists of three consecutive vertices, every vertex is contained in an edge and two consecutive edges intersect in exactly one vertex. The elements v and v t are called the ends of P. The Absorbing Lemma (Lemma 7 asserts that every uniform hypergraph H = (V, E with sufficiently large minimum vertex degree contains a socalled absorbing loose path P, which has the following property: For every set U V \ V (P with U N and U βn (for some appropriate 0 < β < γ there exists a loose path Q with the same ends as P, which covers precisely the vertices V (P U. The Absorbing Lemma reduces the problem of finding a loose Hamilton cycle to the simpler problem of finding an almost spanning loose cycle, which contains the absorbing path P and covers at least ( βn of the vertices. We approach this simpler problem as follows. Let H be the induced subhypergraph H, which we obtain after removing the vertices of the absorbing path P guaranteed by the Absorbing Lemma. We remove from H a small set R of vertices, called the reservoir (see Lemma 6, which has the property that many loose paths can be connected to one loose cycle by using the vertices of R only. Let H be the remaining hypergraph after removing the vertices from R. We will choose P and R small enough, so that δ (H ( o( ( V (H. The third auxiliary lemma, the Pathtiling Lemma (Lemma 0, asserts that all but o(n vertices of H can be covered by a family of pairwise disjoint loose paths and, moreover, the number of those paths will be constant (independent of n. Consequently, we can connect those paths and P to form a loose cycle by using exclusively vertices from R. This way we obtain a loose cycle in H, which covers all but the o(n leftover vertices from H and some leftover vertices from R. We will ensure that the number of those yet uncovered vertices will be smaller than βn and, hence, we can appeal to the absorption property of P and obtain a Hamilton cycle.
4 4 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT As indicated earlier, among the auxiliary lemmas mentioned above the Pathtiling Lemma is the only one for which the full strength of the condition ( o( ( n is required and indeed, we consider Lemma 0 to be the main obstacle to proving Theorem. For the other lemmas we do not attempt to optimise the constants... Auxiliary lemmas. In this section we introduce the technical lemmas needed for the proof of the main theorem. We start with the connecting lemma which is used to connect several short loose paths to a long one. Let H be a uniform hypergraph and (a i, b i i [k] a set consisting of k mutually disjoint pairs of vertices. We say that a set of triples (x i, y i, z i i [k] connects (a i, b i i [k] if i [k] {a i, b i, x i, y i, z i } = 5k, i.e. the pairs and triples are all disjoint, for all i [k] we have {a i, x i, y i }, {y i, z i, b i } H. Suppose that a and b are ends of two disjoint loose paths not intersecting {x, y, z} and suppose that (x, y, z connects (a, b. Then this connection would join the two paths to one loose path. The following lemma states that several paths can be connected, provided the minimum vertex degree is sufficiently large. Lemma 5 (Connecting lemma. Let γ > 0, let m be an integer, and let H = (V, E be a uniform hypergraph on n vertices with δ (H ( 4 + γ ( n and n γm/. For every set (a i, b i i [m] of mutually disjoint pairs of distinct vertices, there exists a set of triples (x i, y i, z i i [m] connecting (a i, b i i [m]. Proof. We will find the triples (x i, y i, z i to connect a i with b i for i [m] inductively as follows. Suppose, for some j < k the triples (x i, y i, z i with i < j are constructed so far and for (a, b = (a j, b j we want to find a triple (x, y, z to connect a and b. Let ( m U = V \ i= j {a i, b i } i= {x i, y i, z i } and for a vertex u V let L u = (V \ {u}, E u be the link graph of v defined by E u = {vw : uvw E(H}. We consider L a [U] and L b [U], the subgraphs of L a and L b induced on U. Owing to the minimum degree condition of H and to the assumption m γn/, we have ( ( ( n e(l a [U] 4 + γ 5m(n 4 + γ ( n ( 6 and the same lower bound also holds for e(l b [U]. Note that any pair of edges xy L a [U] and yz L b [U] with x z leads to a connecting triple (x, y, z for (a, b. Thus, if no connecting triple exists, then for every vertex u U one of the following must hold: either u is isolated in L a [U] or L b [U] or it is adjacent to exactly one vertex w in both graphs L a [U] and L b [U]. In other words, any vertex not isolated in L a [U] has at most one neighbour in L b [U]. Let I a be the set of isolated vertices
5 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES 5 in L a [U]. Since e(l a [U] > 4( n we have Ia < n/. Consequently, ( Ia e(l b [U] + {uw E(L b [U]: u U \ I a } ( ( Ia n/ + ( U I a < + n. Using γ /4 and n γ/ we see that this upper bound violates the lower bound on e(l b [U] from (. When connecting several paths to a long one we want to make sure that the vertices used for the connection all come from a small set, called reservoir, which is disjoint to the paths. The existence of such a set is guaranteed by the following. Lemma 6 (Reservoir lemma. For all 0 < γ < /4 there exists an n 0 such that for every uniform hypergraph H = (V, E on n > n 0 vertices with minimum vertex degree δ (H ( 4 + γ ( n there is a set R of size at most γn with the following property: For every system (a i, b i i [k] consisting of k γ n/ mutually disjoint pairs of vertices from V there is a triple system connecting (a i, b i i [k] which, moreover, contains vertices from R only. Proof. We shall show that a random set R has the required properties with positive probability. For this proof we use some exponential tail estimates. Here we will follow a basic technique described in [5, Section.6]. Alternatively Lemma 6 could be deduced more directly from Janson s inequality. For given 0 < γ < /4 let n 0 be sufficiently large. Let H be as stated in the lemma and v V (H. Let L(v be the link graph defined on the vertex set V (H \ {v}, having the edges e E(L(v if e {v} H. Note that L(v contains deg H (v edges. Since the edge set of the omplete graph K n can be decomposed into n edge disjoint matchings, we can decompose the edge set of L into i 0 = i 0 (v < n pairwise edge disjoint matchings. We denote these matchings by M (v,..., M i0 (v. We randomly choose a vertex set V p from V by including each vertex u V into V p with probability p = γ γ independently. For every i [i 0 ] let X i (v = M i (v denote the number of edges e M i (v contained in V p. This way X i (v is a binomially distributed random variable with parameters M i (v and p. Using the following Chernoff bounds for t > 0 (see, e.g., [5, Theorem.] ( Vp we see that P [Bin(m, ζ mζ + t] < e t /(ζm+t/ ( P [Bin(m, ζ mζ t] < e t /(ζm ( γ n V p pn + (n ln 0 / γn k (4 with probability at least 9/0. Further, using ( and M i (v n/ we see that with probability at most n there exists an index i [i 0 ] such that X i (v M i (v p (n ln n /. Using i [i 0] M i(v = deg H (v and recalling that deg Vp (v denotes the degree of v in
6 6 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT H[V p {v}] we obtain that deg Vp (v = i [i 0] X i (v p deg H (v n(n ln n /, (5 holds with probability at least n. Repeating the same argument for every vertex v V we infer from the union bound that (5 holds for all vertices v V simultaneously with probability at least /n. Hence, with positive probability we obtain a set R satisfying (4 and (5 for all v V. Let (a i, b i i [k] be given and let S = i [k] {a i, b i }. Then we have R S γn and ( ( ( ( γn R S deg R S (v deg R (v 4 + γ 4 + γ for all v V. Thus, we can appeal to the Connecting Lemma (Lemma 5 to obtain a triple system which connects (a i, b i i [k] and which consists of vertices from R only. Next, we introduce the Absorbing Lemma which asserts the existence of a short but powerful loose path P which can absorb any small set U V \ V (P. In the following note that ( 5 8 < 7 6. Lemma 7 (Absorbing lemma. For all γ > 0 there exist β > 0 and n 0 such that the following holds. Let H = (V, E be a uniform hypergraph on n > n 0 vertices which satisfies δ (H ( γ ( n. Then there is a loose path P with V (P γ 7 n such that for all subsets U V \ V (P of size at most βn and U N there exists a loose path Q H with V (Q = V (P U and P and Q have exactly the same ends. The principle used in the proof of Lemma 7 goes back to Rödl, Ruciński, and Szemerédi. They introduced the concept of absorption, which, roughly speaking, stands for a local extension of a given structure, which preserves the global structure. In our context of loose cycle we say that a 7tuple (v,..., v 7 absorbs the two vertices x, y V if v v v, v v 4 v 5, v 5 v 6 v 7 H and v xv 4, v 4 yv 6 H are guaranteed. In particular, (v,..., v 7 and (v, v, v, x, v 4, y, v 6, v 5, v 7 both form loose paths which, moreover, have the same ends. The proof of Lemma 7 relies on the following result which states that for each pair of vertices there are many 7tuples absorbing this pair, provided the minimum vertex degree of H is sufficiently large. Proposition 8. For all 0 < γ < /8 there exists an n 0 such that the following holds. Suppose H = (V, E is a uniform hypergraph on n > n 0 vertices which satisfies δ (H ( γ ( n, then for every pair of vertices x, y V the number of 7tuples absorbing x and y is at least (γn 7 /8. Proof. For given γ > 0 we choose n 0 = 68/γ 7. First we show the following. Claim 9. For every pair x, y V (H of vertices there exists a set D = D(x, y V of size D = γn such that one of the following holds: deg(x, d γn and deg(y, d 8n for all d D or
7 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES 7 deg(y, d γn and deg(x, d 8n for all d D. Proof of Claim 9. By assuming the contrary there exists a pair x, y such that no set D = D(x, y fulfills Claim 9. Let A(z = {d V : deg(z, d < γn} and let a = A(x /n and b = A(y /n. Without loss of generality we assume that a b. Note that there are at most (a + γn vertices v V satisfying deg(y, v 8n. Let B(y = {v V : deg(y, v n/8} and note that B(y < (a + γn. Hence, the number of ordered pairs (u, v such that u B(y and {u, v, y} H is at most B(y (n A(y + A(y γn (a + γ( bn + bγn. Consequently, with deg(y being the number of ordered pairs (u, v such that {u, v, y} H we have ( (5 + 9γ n deg(y ( b a γ n + (a + γ( bn + bγn Hence, we obtain ( (5 + 9γ 8 8 n 8 ( + 8γn (5a b 8ab +, 8 ( n deg(y n 5 ( b + (a + γ 8 8 b n + bγn n ( + 8γn (5a b 8ab +, 8 8 where in the last inequality we use the fact that b /8 which is a direct consequence of the condition on δ (H. It is easily seen that this maximum is attained by a = b = /8, for which we would obtain ( (5 deg(y + γ n, 8 a contradiction. We continue the proof of Proposition 8. For a given pair x, y V we will select the tuple v,..., v 7 such that the edges v v v, v v 4 v 5, v 5 v 6 v 7 H and v xv 4, v 4 yv 6 H are guaranteed. Note that (v,..., v 7 forms a loose path with the ends v and v 7 and (v, v, v, x, v 4, y, v 6, v 5, v 7 also forms a loose path with the same ends, showing that (v,..., v 7 is indeed an absorbing tuple for the pair a, b. Moreover, we will show that the number of choices for each v i will give rise to the number of absorbing tuples stated in the proposition. First, we want to choose v 4 and let D(x, y be a set with the properties stated in Claim 9. Without loss of generality we may assume that N(y, d 8n for all d D(x, y. Fixing some v 4 D(x, y We choose v N(x, v 4 for which there are γn choices. This gives rise to to hyperegde v xv 4 H. and applying Claim 9 to v and v 4 we obtain a set D(v, v 4 with the properties stated in Claim 9 and we choose v D(v, v 4. We choose v N(v, v to obtain the edge v v v H. Note that N(v, v γn. Next, we choose v 5 from N(v, v 4 which has size N(v, v 4 γn. This gives rise to the edge v v 4 v 5 H. We choose v 6 from
8 8 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT N(y, v 4 with the additional property that deg(v 5, v 6 γn/. Hence, we obtain v 4 yv 6 H and we claim that there are at least γn/ such choices. Otherwise at least ( N(y, v 4 γn/ vertices v V satisfy deg(v 5, v < γn/, hence deg(v 5 < γ 6 n + ( ( γ n < δ(h, which is a contradiction. Lastly we choose v 7 N(v 5, v 6 to obtain the edge v 5 v 6 v 7 H which completes the absorbing tuple (v,..., v 7. The number of choices for v,..., v 7 is at least (γn 7 /4 and there are at most n 6 choices such that v i = v j for some i j. Hence, we obtain at least (γn 7 /8 ( 7 absorbing 7tuples for the pair x, y. With Proposition 8 and the connecting lemma (Lemma 5 at hand the proof of the absorbing lemma follows a scheme which can be found in [, 4]. We choose a family F of 7tuples by selecting each 7tuples with probability p = γ 7 n 6 /448 independently. Then, it is easily shown that with nonzero probability the family F satisfies F γ 7 n/, for all pairs x, y V there are at least pγ 7 n 7 /6 tuples in F which absorbs x, y the number of intersecting pairs of 7tuples in F is at most pγ 7 n 7 / We eliminate intersecting pairs of 7tuples by deleting one tuple for each such pair. By definition each for the remaining 7tuples (v, i..., v7 i i [k] with k γ 7 n/ forms a loose path with ends v i and v7 i and appealing to Lemma 5 we can connect them to one loose path which can absorb any pγ 7 n 7 / = β pairs of vertices, proving the lemma. To avoid unnecessary calculations we omit the details here. The next lemma is the main obstacle when proving Theorem. It asserts that the ( vertex set of a uniform hypergraph H with minimum vertex degree δ (H o( ( n can be almost perfectly covered by a constant number of vertex disjoint loose paths. Lemma 0 (Pathtiling lemma. For all γ > 0 and α > 0 there exist integers p and n 0 such that for n > n 0 the following holds. Suppose H is a uniform hypergraph on n vertices with minimum vertex degree δ (H ( γ ( n. Then there is a family of p disjoint loose paths in H which covers all but at most αn vertices of H. The proof of Lemma 0 uses the weak regularity lemma for hypergraphs and will be given in Section... Proof of the main theorem. In this section we give the proof of the main result, Theorem. The proof is based on the three auxiliary lemmas introduced in Section. and follows the outline given in Section.. Proof of Theorem. For given γ > 0 we apply the Absorbing Lemma (Lemma 7 with γ/8 to obtain β > 0 and n 7. Next we apply the Reservoir Lemma (Lemma 6 for γ = min{β/, γ/8} to obtain n 6 which is n 0 of Lemma 6. Finally, we apply the
9 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES 9 Pathtiling Lemma (Lemma 0 with γ/ and α = β/ to obtain p and n 0. For n 0 of the theorem we choose n 0 = max{n 7, n 6, n 0, 4(p + (γ }. Now let n n 0, n N and let H = (V, E be a uniform hypergraph on n vertices with ( ( 7 n δ (H 6 + γ. Let P 0 H be the absorbing path guaranteed by Lemma 7. Let a 0 and b 0 be the ends of P 0 and note that V (P 0 γ n < γn/8. Moreover, the path P 0 has the absorption property, i.e., for all U V \ V (P 0 with U βn and U N there exists a loose path Q H s.t. V (Q = V (P 0 U and Q has the ends a 0 and b 0. (6 Let V = (V \ V (P 0 {a 0, b 0 } and let H = H[V ] = (V, E(H ( V be the induced subhypergraph of H on V. Note that δ (H ( γ( n. Due to Lemma 6 we can choose a set R V of size at most γ V γ n such that for every system consisting of at most (γ V / mutually disjoint pairs of vertices from V can be connected using vertices from R only. Set V = V \ (V (P 0 R and let H = H[V ] be the induced subhypergraph of H on V. Clearly, ( 7 δ(h 6 + γ ( n Consequently, Lemma 0 applied to H (with γ 0 and α yields a loose path tiling of H which covers all but at most α V αn vertices from V and which consists of at most p paths. We denote the set of the uncovered vertices in V by T. Further, let P, P..., P q with q p denote the paths of the tiling. By applying the reservoir lemma appropriately we connect the loose paths P 0, P,..., P q to one loose cycle C H. Let U = V \ V (C be the set of vertices not covered by the cycle C. Since U R T we have U (α + γ 6 n βn. Moreover, since C is a loose cycle and n N we have U N. Thus, using the absorption property of P 0 (see (6 we can replace the subpath P 0 in C by a path Q (since P 0 and Q have the same ends and since V (Q = V (P 0 U the resulting cycle is a loose Hamilton cycle of H.. Proof of the Pathtiling Lemma In this section we give the proof of the Pathtiling Lemma, Lemma 0. Lemma 0 will be derived from the following lemma. Let M be the uniform hypergraph defined on the vertex set {,..., 8} with the edges, 45, 456, 678 M. We will show that the condition δ (H ( o( ( n will ensure an almost perfect Mtiling of H, i.e., a family of vertex disjoint copies of M, which covers almost all vertices. Lemma. For all γ > 0 and α > 0 there exists n 0 such that the following holds. Suppose H is a uniform hypergraph on n > n 0 vertices with minimum vertex degree ( ( 7 n δ (H 6 + γ. Then there is an Mtiling of H which covers all but at most αn vertices of H.
10 0 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT The proof of Lemma requires the regularity lemma which we introduce in Section.. Sections. and. are devoted to the proof of Lemma and finally, in Section.4, we deduce Lemma 0 from Lemma by making use of the regularity lemma... The weak regularity lemma and the cluster hypergraph. In this section we introduce the weak hypergraph regularity lemma, a straightforward extension of Szemerédi s regularity lemma for graphs [5]. Since we only apply the lemma to uniform hypergraphs we will restrict the introduction to this case. Let H = (V, E be a uniform hypergraph and let A, A, A be mutually disjoint nonempty subsets of V. We define e(a, A, A to be the number of edges with one vertex in each A i, i [], and the density of H with respect to (A, A, A as d(a, A, A = e H(A, A, A. A A A We say the triple (V, V, V of mutually disjoint subsets V, V, V V is (ε, d regular, for constants ε > 0 and d 0, if d(a, A, A d ε for all triple of subsets A i V i, i [], satisfying A i ε V i. The triple (V, V, V is called εregular if it is (ε, dregular for some d 0. It is immediate from the definition that an (ε, dregular triple (V, V, V is (ε, dregular for all ε > ε and if V i V i has size V i c V i, then (V, V, V is (ε/c, dregular. Next we show that regular triples can be almost perfectly covered by copies of M provided the sizes of the partition classes obey certain restrictions. First note that M is a subhypergraph of a tight path. The latter is defined similarly to a loose path, i.e. there is an ordering (v,..., v t of the vertices such that every edge consists of three consecutive vertices, every vertex is contained in an edge and two consecutive edges intersect in exactly two vertices. Proposition. Suppose H is a uniform hypergraph on m vertices with at least dm edges. Then there is a tight path in H which covers at least (dm + vertices. In particular, if H is partite with the partition classes V, V, V and dm > 0 then for each i [] there is a copy of M in H which intersects V i in exactly two vertices and the other partition classes in three vertices. Proof. Starting from H we remove all edges containing u, v for each pair u, v V of vertices such that 0 < deg(u, v < dm. We keep doing this until every pair u, v satisfies deg(u, v = 0 or deg(u, v dm in the current hypergraph H. Since less than ( m dm < dm e(h edges were removed during the process we know that H is not empty. Hence we can pick a maximal nonempty tight path (v, v,..., v t in H. Since the pair v, v is contained in an edge in H it is contained in dm edges and since the path was chosen to be maximal all these vertices must lie in the path. Hence, the chosen tight path contains at least (dm + vertices. This completes the first part of the proof. For the second part, note that there is only one way to embed a tight path into a partite uniform hypergraph once the two starting vertices are fixed. Since M is a subhypergraph of the tight path on eight vertices we obtain the second part of the statement by possibly deleting up to two starting vertices.
11 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES Proposition. Suppose the triple (V, V, V is (ε, dregular with d ε and suppose the sizes of the partition classes satisfy m = V V V with 5 V ( V + V (7 and ε m > 7. Then there is an Mtiling of (V, V, V leaving at most εm vertices uncovered. Proof. Note that if we take a copy of M intersecting V i, i [] in exactly two vertices then this copy intersects the other partition classes in exactly three vertices. We define t i = ( ε 8 ( V j + V k 5 V i where i, j, k [] are distinct. Due to our assumption all t i are nonnegative and we choose t i copies of M intersecting V i in exactly two vertices. This would leave V i (t i + t j + t k = ε V i vertices in V i uncovered, hence at most εm in total. To complete the proof we exhibit a copy of M in all three possible types in the remaining hypergraph, hence showing that the choices of the copies above are indeed possible. To this end, from the remaining vertices of each partition class V i take a subset U i, i [] of size ε V i. Due to the regularity of the triple (V, V, V we have e(u, U, U (d ε(εm. Hence, by Proposition there is a copy of M (of each type in (U, U, U. The connection of regular partitions and dense hypergraphs is established by regularity lemmas. The version introduced here is a straightforward generalisation of the original regularity lemma to hypergraphs (see, e.g., [,, 4]. Theorem 4. For all t 0 0 and ε > 0, there exist T 0 = T 0 (t 0, ε and n 0 = n 0 (t 0, ε so that for every uniform hypergraph H = (V, E on n n 0 vertices, there exists a partition V = V 0 V... V t such that (i t 0 t T 0, (ii V = V = = V t and V 0 εn, (iii for all but at most ε ( t sets {i,..., i } ( [t], the triple (Vi, V i, V i is εregular. A partition as given in Theorem 4 is called an (ε, tregular partition of H. For an (ε, tregular partition of H and d 0 we refer to Q = (V i i [t] as the family of clusters (note that the exceptional vertex set V 0 is excluded and define the cluster hypergraph K = K(ε, d, Q with vertex set [t] and {i, i, i } ( [t] being an edge if and only if (V i, V i, V i is εregular and d(v i, V i, V i d. In the following we show that the cluster hypergraph almost inherits the minimum vertex degree of the original hypergraph. The proof which we give for completeness is standard and can be found e.g. in [8] for the case of graphs. Proposition 5. For all γ > d > ε > 0 and all t 0 there exist T 0 and n 0 such that the following holds. Suppose H is a uniform hypergraph on n > n 0 vertices which has vertex minimum degree δ (H ( γ ( n. Then there exists an (ε, tregular partition Q with t 0 < t < T 0 such that the cluster hypergraph K = K(ε, d, Q has minimum vertex degree δ (K ( γ ε d ( t. Proof. Let γ > d > ε and t 0 be given. We apply the regularity lemma with ε = ε /44 and t 0 = max{t 0, 0/ε} to obtain T 0 and n 0. We set T 0 = T 0 and
12 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT n 0 = n 0. Let H be a uniform hypergraph on n > n 0 vertices which satisfies δ(h (7/6 + γ ( n. By applying the regularity lemma we obtain an (ε, t  regular partition V 0 V... V t of V and let m = V = ( ε n/t denote the size of the partition classes. Let I = {i [t ]: V i is contained in more than ε ( t /8 non ε regular triples} and observe that I < 8ε t /ε due to the property (iii of Theorem 4. Set V 0 = V 0 i I V i and let J = [t ]\I and t = J. We now claim that V 0 and Q = (V j j J is the desired partition. Indeed, we have T 0 > t t > t ( 8ε /ε t 0 and V 0 < ε n + 8ε n/ε εn/6. The property (iii follows directly from Theorem 4. For a contradiction, assume now that deg K (V j < ( γ ε d( t for some j J. Let x j denote the number of edges which intersect V j in exactly one vertex and each other V i, i J, in at most one vertex. Then, the assumption yields [( ( 7 t x j V j 6 + γ ε d m + ε ( t 8 ( V j n γ ε m + ε 6 n + d On the other hand, from the minimum degree of H we obtain ( ( ( ( 7 n x j V j 6 + γ Vj Vj n ( ( n 7 V j 6 + γ 4 t a contradiction. ( ] t m.. Fractional hom(mtiling. To obtain a large Mtiling in the hypergraph H, we consider weighted homomorphisms from M into the cluster hypergraph K. To this purpose, we define the following. Definition 6. Let L be a uniform hypergraph. A function h: V (L E(L [0, ] is called a fractional hom(mtiling of L if ( h(v, e 0 v e, ( h(v = e E(L h(v, e, ( for every e E(L there exists a labeling of the vertices of e = uvw such that h(u, e = h(v, e h(w, e h(u, e By h min we denote the smallest nonzero value of h(v, e (and we set h min = if h 0 and the sum over all values is the weight w(h of h w(h = h(v, e. (v,e V (L E(L The allowed values of h are based on the homomorphisms from M to a single edge, hence the term hom(mtiling. Given one such homomorphism, assign each vertex in the image the number of vertices from M mapped to it. In fact, for any such homomorphism the preimage of one vertex has size two, while the preimages of the other two vertices has size three. Consequently, for any family of homomorphisms of M into a single edge the smallest and the largest class of preimages can differ by a factor of / at most and this observation is the reason for condition ( in Definition 6. We also note the following.
13 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES Fact 7. There is a fractional hom(mtiling h of the hypergraph M which has h min / and weight w(h = 8. Proof. Let x, x, w, y, y, w, z, and z be the vertices of M and let x x w, w y y, y y w, and w z z be the edges of M. On the edges x x w and z z w we assign the vertex weights (,, /, where the weight / is assigned to w and w. The vertex weights for edges y y w and y y w are (/, /, /, where w and w get the weight /. It is easy to see that those vertex weights give rise to a hom(mtiling h on M with h min = / and w(h = 8. The notion hom(mtiling is also motivated by the following proposition which shows that such a fractional hom(mtiling in a cluster hypergraph can be converted to an integer Mtiling in the original hypergraph. Proposition 8. Let Q be an (ε, tregular partition of a uniform, nvertex hypergraph H with n > ε and let K = K(ε, 6ε, Q be the corresponding cluster hypergraph. Furthermore, let h: V (K E(K [0, ] be a fractional hom(m tiling of K with h min /. Then there exists an Mtiling of H which covers all but at most (w(h 7tε V vertices. Proof. We restrict our consideration to the subhypergraph K K consisting of the hyperedges with positive weight, i.e., e = abc K with h(a, h(b, h(c h min. For each a V (K let V a be the corresponding partition class in Q. Due to the property ( of Definition 6 we can subdivide V a (arbitrarily into a collection of pairwise disjoint sets (Ua e a e K of size Ua e = h(a, e V a. Note that every edge e = abc K corresponds to the (ε, 6εregular triplet (V a, V b, V c. Hence we obtain from the definition of regularity and h min / that the triplet (Ua, e Ub e, U c e is (ε, 6ε regular. From the property ( in Definition 6 and Proposition we obtain an Mtiling of (Ua, e Ub e, U c e incorporating at least ( h(a, e + h(b, e + h(c, e 9ε V a vertices. Applying this to all hyperedges of K we obtain an Mtiling incorporating at least ( h(a, e + h(b, e + h(c, e 9ε V a ( w(h 9 K ε V a abc=e K vertices. Noting that K t (because of h min / and V a V we obtain the proposition. Owing to Proposition 8, we are given a connection between fractional hom(m tilings of the cluster hypergraph K of H and Mtilings in H. A vertex i V (K corresponds to a class of vertices V i in the regular partition of H. The total vertex weight h(i essentially translates to the proportion of vertices of V i which can be covered by the corresponding Mtilings in H. Consequently, w(h essentially translates to the proportion of vertices covered by the corresponding Mtiling in H. This reduces our task to finding a fractional hom(mtiling with weight greater than the number of vertices previously covered in K. The following lemma (Lemma 9, which is the main tool for the proof of Lemma, follows the idea discussed above. In the proof of Lemma we fix a maximal Mtiling in the cluster hypergraph K of the given hypergraph H. Owing to the minimum degree condition of H and Proposition 5, a typical vertex in the cluster hypergraph K will be contained in at least (7/6 + o( ( V (K hyperedges of K. We will show that a typical vertex u of K which is not covered by the
14 4 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT maximal Mtiling of K, has the property that (7/6 + o( 64 > 8 of the edges incident to u intersect some pair of copies of M from the Mtiling of K. Lemma 9 asserts that two such vertices and the pair of copies of M can be used to obtain a fractional hom(mtiling with a weight significantly larger than 6, the number of vertices of the two copies of M. This lemma will come in handy in the proof of Lemma, where it is used to show that one can cover a higher proportion of the vertices of H than the proportion of vertices covered by the largest Mtiling in K. We consider a set of hypergraphs L 9 definied as follows: Every L L 9 consists of two (vertex disjoint copies of M, say M and M, and two additional vertices u and v such that all edges incident to u or v contain precisely one vertex from V (M and one vertex from V (M. Moreover, L satisfies the following properties for every a V (M and b V (M we have uab E(L iff vab E(L deg(u = deg(v 9. Lemma 9. For every L L 9 there exists a fractional hom(mtiling h with h min / and w(h 6 +. The following proof of Lemma 9 is based on straightforward, but somewhat tedious case distinction. Proof. For the proof we fix the following labeling of the vertices of the two disjoint copies of M. Let V (M = {x, x, w, y, y, w, z, z } be the vertices and x x w, w y y, y y w, w z z be the edges of the first copy of M. Analogously, let V (M = {x, x, w, y, y, w, z, z } be the vertices and x x w, w y y, y y w, w z z be the edges of the other copy of M (see Figure.a. Moreover, we denote by X = {x, x }, Y = {y, y }, and Z = {z, z } and, let X, Y, and Z be defined analogously for M. Figure. Labels and case: a b, a b L with {b, b } {X, Y, Z } X x x X a x x a w w Y y y y y Y b b u v w w Z z z Z z z.a: Vertex labels of M and M in L.b: All edges are (aedges The proof of Lemma 9 proceeds in two steps. First, we show that in any possible counterexample L, the edges incident to u and v which do not contain any vertex
15 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES 5 from {w, w, w, w } form a subgraph of K,, (see Claim 0. In the second step we show that every edge contained in this subgraph of K,, forbids too many other edges incident to u and v, which will yield a contradiction to the condition deg(u = deg(v 9 of L (see Claim. We introduce the following notation to simplify later arguments. For a given L L 9 with M and M being the two copies of M, let L be set of those pairs (a, b V (M V (M such that uab E(L. We split L into L L according to { L, if {a, b} {w, w, w (a, b, w } =, L, otherwise. It will be convenient to view L and L as bipartite graphs with vertex classes V (M and V (M. We split the proof of Lemma 9 into the following two claims. Claim 0. For all L L 9 without a fractional hom(mtiling with h min / and w(h 6 + /, we have L K,, where each of the sets X, Y, Z and X, Y, Z contains precisely one of the vertices of the K,. Claim 0 will be used in the proof of the next claim, which clearly implies Lemma 9. Claim. Let F = { a b V (M V (M : a {w, w } or b {w, w } } and for every edge ab L let F(a, b F be the set of those e F, whose appearance in L (i.e. e L implies the existence of a fractional hom(mtiling h with h min / and w(h 6 + /. Then there is an injection f : L F such that f(a, b F(a, b for every pair ab L. Clearly, F = 8 and L F. Hence, from L + f(l = L + L 9 we derive that L and f(l must intersect. By Claim this yields the desired fractional hom(mtiling and Lemma 9 follows. In the proofs of Claim 0 and Claim we will consider fractional hom(m tilings h which use vertex weights of special types. In fact, for an edge e = a a a, the weights h(a, e, h(a, e, and h(a, e will be of the following forms (a h(a, e = h(a, e = h(a, e = (a h(a, e = h(a, e = h(a, e = (a h(a, e = h(a, e = h(a, e = (b h(a, e = h(a, e = and h(a, e = (b h(a, e = h(a, e = and h(a, e = (b h(a, e = h(a, e = and h(a, e = An edge that satisfies (a is called an (aedge, etc. Note that all these types satisfy condition ( of Definition 6. Proof of Claim 0. Given L L 9 satisfying the assumptions of the claim and with the labeling from Figure.a. Observe that for any A {X, Y, Z}, the hypergraph M A contains two disjoint edges. Similarly, for every B {X, Y, Z }, M B contains two disjoint edges. First we exclude the case that there is a matching {a b, a b } of size two in L between some {a, a } = A {X, Y, Z} and some {b, b } = B {X, Y, Z }. In this case we can construct a fractional hom(mtiling h as follows: Choose two
16 6 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT Figure. Case: ab, ab L with {b, b } {X, Y, Z }.a: (aedges w y y, w z z, and w z z, (bedges ax u and ax v, (b edge w y y, and (aedge x x w..b: (aedges w y y, w z z, and w z z, (bedges ay u and ay v, (b edge x x w, and (aedge w y y. edges ua b, va b. Using these and the four disjoint edges in (M A (M B, we obtain six disjoint edges (see Figure.b. Letting all these six edges be (a edges, we obtain a fractional hom(mtiling h with h min = and w(h = 8. Next, we show that each vertex a A {X, Y, Z} has at most one neighbour in each B {X, Y, Z }. Assuming the contrary, let a A {X, Y, Z} and {b, b } = B {X, Y, Z }, such that ab, ab L. For symmetry reasons, we only need to consider the case B = X and B = Y. The case B = Z is symmetric to B = X. In those cases, we choose h as shown in Figure.a (B = X and Figure.b (B = Y and in either case we find a fractional hom(mtiling h satisfying h min = / and w(h = 6 + /. Note that the cases A = Y and A = Z can be treated in the same manner since the only condition needed to define h is that M A contains two disjoint edges. To show that L is indeed contained in a K, it remains to verify that every a b, a b with {a, a } = A {X, Y, Z}, b B {X, Y, Z }, and b B {X, Y, Z } \ B guarantees the existence of a fractional hom(mtiling h with h min / and w(h 6 + /. Again owing to the symmetry, the only cases we need to consider are B = X, B = Y (see Figure.a and B = X, B = Z (see Figure.b. In fact, the fractional hom(mtilings h given in Figure.a and Figure.b satisfy h min / and w(h = 7. Again the cases A = Y and A = Z can be treated in the same manner. This concludes the proof of Claim 0. To complete the proof of Lemma 9 it is left to prove Claim. Proof of Claim. Before defining the injection f : L F we collect some information about F(a, b with ab L. Owing to Claim 0, we may assume without loss of generality that x, y, z and x, y, z are the vertices which span all edges of L. First we consider e = y y. As shown in Figure 4.a the appearance
17 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES 7 Figure. Case: a b, a b L with {b, b } {X, Y, Z }.a: (aedges w y y, w z z, and w z z, (bedges a x u and a y v, and (bedges x x w and w y y..b: (aedges w y y and w z z, (b edges a x u and a z v, (bedges x x w and w z z, and (aedges w y y and y y w. of w y L would give rise to a fractional hom(mtiling h with h min / and w(h = 6.5. Consequently, we have w y F(y, y. For the case e = x x L, Figure 4.b, shows that x w F(x, x and by symmetry, it follows that {x w, w x } F(x, x. By applying appropriate automorphisms to M and M we immediately obtain information on F(x, z, F(z, x, and F(z, z. Indeed, we have {x w, w z } F(x, z, {w x, z w } F(z, x, {z w, w z } F(z, z. Next we consider e = y x. In this case Figure 5.a shows that y w F(y, x. Moreover, as shown in Figure 5.b we also have w x F(y, x and, consequently, we obtain {y w, w x } F(y, x. Again applying appropriate automorphisms to M and M we immediately obtain information on F(x, y, F(y, z, and F(z, y. Indeed one can show {w y, x w } F(x, y, {y w, w z } F(y, z, {w y, z w } F(z, y. Finally, we define an injection f : L F L such that f(a, b F(a, b for every pair ab L. Due to Claim 0 we have L {x, y, z } {x, y, z } and it
18 8 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT Figure 4. F(y, y and F(x, x 4.a: (aedge w z z, (aedge y y u, (bedges x x w, w z z, and x x w, and (bedges w y v, y y w, and w y y. 4.b: (aedges x x u, w y y, and w z z, (bedges x w v and w z z, and (bedges w y y and y y w. Figure 5. F(y, x 5.a: (aedges y x u, x x w, and w z z, (bedges y w v and w z z, and (bedges w y y and y y w. 5.b: (aedge w z z, (aedge y x u, (bedges x x w, w z z, and w y y, and (bedges w x v, y y w, and x x w. follows from the discussion above that we can define f as follows f(x, x = w x, f(x, y = x w, f(x, z = x w, f(y, x = y w, f(y, y = w y, f(y, z = w z, f(z, x = w x, f(z, y = w y, f(z, z = z w.
19 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES 9 Consequently, L F L and Claim follows from L + L F 8... Proof of the Mtiling Lemma. In this section we prove Lemma. Let H be a uniform hypergraph on n vertices. We say H has a βdeficient Mtiling if there exists a family of pairwise disjoint copies of M in H leaving at most βn vertices uncovered. Proposition. For all / > d > 0 and all β, δ > 0 the following holds. Suppose there exists an n 0 such that every uniform hypergraph H on n > n 0 vertices with minimum vertex degree δ (H d ( n has a βdeficient Mtiling. Then every uniform hypergraph H on n > n 0 vertices with δ (H (d δ ( n has a (β + 5 δdeficient Mtiling. Proof. Given a uniform hypergraph H on n > n 0 vertices with minimum degree δ (H (d δ ( n. By adding a set A of δn new vertices to H and adding all triplets to H which intersect A we obtain a new hyperpgraph H on n = n + A vertices which satisfies δ (H d ( n. Consequently, H has a βdeficient Mtiling and by removing the Mcopies intersecting A, we obtain a (β + 5 δdeficient Mtiling of H. Proof of Lemma. Let γ > 0 be given and we assume that there is an α > 0 such that for all n 0 there is a uniform hypergraph H on n > n 0 vertices which satisfies δ (H ( γ( n but which does not contain an αdeficient Mtiling. Let α0 be the supremum of all such α and note that α 0 is bounded away from one due to Proposition. We choose ε = (γα 0 / 00. Then, by definition of α 0, there is an n 0 such that all uniform hypergraphs H on n > n 0 vertices satisfying δ (H ( γn have an (α 0 + εdeficient Mtiling. Hence, by Proposition all uniform hypergraphs H on n > n 0 vertices satisfying δ (H ( γ ε( n have an (α0 + ε + 5 ε deficient Mtiling. We will show that there exists an n (to be chosen such that all uniform hypergraphs H on n > n vertices satisfying δ (H ( γ( n have an (α 0 εdeficient Mtiling, contradicting the definition of α 0. To this end, we apply Proposition 5 with the constants γ, ε/, d = ε/ and t 5 = max{n 0, (ε/ } to obtain an n 5 and T 5. Let n max{n 5, n 0 } be sufficiently large and let H be an arbitrary uniform hypergraph on n > n vertices which satisfies δ (H ( γ( n but which does not contain an α0 deficient Mtiling. We apply Proposition 5 to H with the constants chosen above and obtain a cluster hypergraph K = K(ε/, ε/, Q on t > t 5 vertices which satisfies δ (K ( γ ε( t. Taking M to be the largest Mtiling in K we know by the definition of α 0 and by Proposition that M is an α deficient Mtiling of K, for some α α ε. We claim that M is not (α 0 /deficient and for a contradiction, assume the contrary. Then, from Fact 7, we know that for each M j M there is a fractional hom(mtiling h j of M j with h j min / and weight w(hj = 8. Hence, the union of all these fractional hom(mtiling gives rise to a fractional hom(mtiling h of K with h min / and weight w(h 8 M t( α 0 /. By applying Proposition 8 to the fractional hom(mtiling h (and recalling that the vertex classes V,..., V t of the regular partition has the same size, which was
20 0 ENNO BUSS, HIỆP HÀN, AND MATHIAS SCHACHT at least ( ε/n/t we obtain an Mtiling of H which covers at least ( ( w(h tε ε n t ( α 0 + εn vertices of H. This, however, yields a (α 0 ε deficient Mtiling of H contradicting the choice of H. Hence, M is not (α 0 /deficient from which we conclude that X, the set of vertices in K not covered by M, has size X α 0t. (8 For a pair M i M j ( M we say the edge e K is ijcrossing if e V (Mi = e V (M j =. Claim. Let C be the set of all triples xij such that x X, M i M j ( M and there are at least 9 ijcrossing edges containing x. Then we have C γ ( t X /7. Proof. Let A be the set of those hyperedges in K which are completely contained in X and let B be the set of all the edges with exactly two vertices in X. Then it is sufficient to show that A 7 6 ( X and B 7 ( X M. (9 Indeed, assuming (9 and C γ ( t X /7 we obtain the following contradiction ( ( ( M 8 deg(x A + B + 8 X C + 64 C + M X x X [ ( 7 X X + 7 ( M 6 X M ( ( ] t 8 7 γ + M [( 7 X 6 + γ ( ( ] t 8 + M < X δ (K where in the third inequality we used ( ( t = X ( + 8 X M + 8 M. Note that the first part of (9 trivially holds since in the opposite case, using the first part of Proposition we obtain a tight path in X of length at least eight. However, this path contains a copy of M as a subhypergraph which yields a contradiction to the maximality of M. To complete the proof let us assume B > 7 there is an M M such that V (M intersects at least 7 ( X M from which we deduce that ( X edges from B. From V (M we remove the vertices which are contained in less than X edges from B. Note that there are at least four vertices, say v,..., v 4, and at least ( + ε ( X edges from B left which intersect these vertices. Hence, there exists a pair x, x such that {x, x, v i } H for all i =,..., 4. Removing all edges intersecting x, x we still have at least ( + ε/ ( X edges intersecting v,..., v 4 and we can find another pair x, x 4 disjoint from x, x with {x, x 4, v i } H for i =,..., 4. For each v i we can find another edge from B containing v i, keeping them all mutually disjoint and also disjoint from {x, x } and {x, x 4 }. This is possible since each v i is contained in more than X edges from B. This, however, yields two copies of M which contradicts the fact that M was a largest possible Mtiling.
21 MINIMUM VERTEX DEGREE CONDITIONS FOR LOOSE HAMILTON CYCLES The set X will be used to show that there is an L L 9 such that K contains many copies of L. Claim 4. There is an element L L 9 and a family L of vertex disjoint copies of L in the cluster hypergraph K(ε/, ε/, Q such that L γα 0 t/ 75. Proof. We consider the uniform hypergraph C (as given from Claim on the vertex set X M. Note that for fixed ij a vertex x is contained in at most 64 ijcrossing edges, thus there are at most 64 different hypergraphs with the property that x is contained in at least 9 edges which are ijcrossing. We colour each edge xij by one of the 64 colours, depending on the partite hypergraph induced on x, M i, andm j. On the one hand, we observe that a monochromatic tight path consisting of the two edges xij, x ij C corresponds to a copy of L. On the other hand, Claim implies that there is a colour such that at least C 64 γ ( t X ( α 0γt 7 edges in C are coloured by it. Hence, by Proposition there is a tight path with α 0 γt/ 7 vertices using edges of this colour only. Note that in this tight path every three consecutive vertices contain one vertex from X and the other two vertices are from M. Thus, this path gives rise to at least α 0 γt/ 75 pairwise vertex disjoint tight paths on four vertices such that the ends are vertices from X. For any L i L we know from Lemma 9 that there is a fractional hom(m tiling h i of L i with h i min / and weight w(hi 6 + /. Furthermore, for every M j M which is not contained in any L i L we know from Fact 7 that there is a fractional hom(mtiling of h j of M j with h j min / and weight w(h j = 8. Hence, the union of all these fractional hom(mtiling gives rise to a fractional hom(mtiling h of K with h min / and weight ( w(h 6 + L + 8( M L = 8 M + L. By applying Proposition 8 to the fractional hom(mtiling h (and recalling that the vertex classes V,..., V t of the regular partition has the same size which was at least ( ε/n/t we obtain an Mtiling of H which covers at least ( ε ( w(h tε ( ε n t ( 8 M + L tε n t vertices of H. Since M was an (α 0 +6 εdeficient Mtiling of K, the tiling we obtained above is an (α 0 εdeficient Mtiling of H due to the choice of ε. This, however, is a contradiction to the fact that H does not permit an (α 0 εdeficient Mtiling..4. Proof of the pathtiling lemma. In this section we prove Lemma 0. The proof will use the following proposition which has been proven in [4] (see Lemma 0 in an even more general form, hence we omit the proof here. Proposition 5. For all d and β > 0 there exist ε > 0, integers p and m 0 such that for all m > m 0 the following holds. Suppose V = (V, V, V is an (ε, dregular triple with V i = m for i =, and V = m. Then there there is a loose path tiling of V which consists of at most p pairwise vertex disjoint paths and which covers all but at most βm vertices of V.
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