HAMILTON CYCLES IN 3-UNIFORM HYPERGRAPHS
ENNO BUSS, HI ˆE. P HAN, AND MATHIAS SCHACHT`
Abstract. We investigate minimum vertex degree conditions for 3-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 3-uniformn-vertex (neven) hypergraphHwith min- imum vertex degreeδ1(H)≥ 167 +o(1) n
2
contains a loose Hamilton cycle.
This bound is asymptotically best possible.
1. Introduction
We consider k-uniform hypergraphs H, that are pairs H= (V, E) with vertex sets V =V(H) and edge setsE =E(H)⊆ Vk
, where Vk
denotes the family of allk-element subsets of the setV. We often identify a hypergraphHwith its edge set, i.e.,H ⊆ Vk
, and for an edge{v1, . . . , vk} ∈ Hwe often suppress the enclosing braces and writev1. . . vk∈ Hinstead.
Given ak-uniform hypergraphH= (V, E) and a setS ={v1, . . . , vs} ∈ Vs let deg(S) = deg(v1, . . . , vs) denote the number of edges ofHcontaining the setS and let N(S) = N(v1, . . . , vs) denote the set of those (k−s)-element sets T ∈ k−sV such thatS∪T forms an edge inH. We denote by δs(H) the minimum s-degree of H, i.e., the minimum of deg(S) over all s-element sets S ⊆ V. For s = 1 the corresponding minimum degreeδ1(H) is referred to as minimum vertex degree whereas for s = k−1 we call the corresponding minimum degree δk−1(H) the minimum collective degree ofH.
We study sufficient minimum degree conditions which enforce the existence of spanning, so-called Hamilton cycles. A k-uniform hypergraph C is called an `- cycle if there is a cyclic ordering of the vertices of Csuch that every edge consists ofkconsecutive 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`vertices. For`= 1 we call the cycleloose whereas the cycle is calledtightif`=k−1. Naturally, we say that ak-uniform,n-vertex hypergraphH contains a Hamilton`-cycle if there is a subhypergraph ofHwhich forms an`-cycle and which covers all vertices of H. Note that a Hamilton `-cycle contains exactly n/(k−`) edges, implying that the number of vertices of H must be divisible by (k−`) which we indicate byn∈(k−`)N.
The second author was supported by GIF grant no. I-889-182.6/2005.
The third author was supported through the Heisenberg-Programme of the Deutsche Forschungsgemeinschaft (DFG Grant SCHA 1263/4-1).
The collaboration of the authors was supported by joint grant of the Deutschen Akademischen Austausch Dienst (DAAD) and the Funda¸c˜ao Coordena¸c˜ao de Aperfei¸coamento de Pessoal de N´ıvel (CAPES).
1
Minimum collective degree conditions which ensure the existence of tight Hamil- ton cycles were first studied in [6] and in [12,13]. In particular, in [12, 13] R¨odl, Ruci´nski, and Szemer´edi found asymptotically sharp bounds for this problem.
Theorem 1. For every k ≥ 3 and γ > 0 there exists an n0 such that every k- uniform hypergraphH= (V, E)on|V|=n≥n0vertices withδk−1(H)≥(1/2+γ)n
contains a tight Hamilton cycle.
The corresponding question for loose cycles was first studied by K¨uhn and Os- thus. In [10] they proved an asymptotically sharp bound on the minimum collective degree which ensures the existence of loose Hamilton cycles in 3-uniform hyper- graphs. This result was generalised to higher uniformity by the last two authors [4]
and independently by Keevash, K¨uhn, Osthus and Mycroft in [7].
Theorem 2. For all integers k≥3 and everyγ >0 there exists an n0 such that everyk-uniform hypergraphH= (V, E)on|V|=n≥n0vertices withn∈(k−1)N andδk−1(H)≥(2(k−1)1 +γ)ncontains a loose Hamilton cycle.
Indeed, in [4] asymptotically sharp bounds for Hamilton`-cycles for all ` < k/2 were obtained. Later this result was generalised to all 0< ` < kby K¨uhn, Mycroft, and Osthus [9]. These results are asymptotically best possible for all k and 0<
` < k. Hence, asymptotically, the problem of finding Hamilton`-cycles in uniform hypergraphs with large minimumcollective degree is solved.
The focus of this paper are conditions on the minimum vertex degree which ensure the existence of Hamilton cycles. For δ1(H) very few results on spanning subhypergraphs are known (see e.g. [3,11]).
In this paper we give an asymptotically sharp bound on the minimum vertex degree in 3-uniform hypergraphs which enforces the existence of loose Hamilton cycles.
Theorem 3(Main result). For all γ >0there exists ann0 such that the following holds. SupposeHis a3-uniform hypergraph onn > n0 with n∈2Nand
δ1(H)>
7
16+γ n 2
. ThenHcontains a loose Hamilton cycle.
In the proof we apply the so-called absorbing technique. In [12] R¨odl, Ruci´nski, and Szemer´edi 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 anearly spanning loose path and indeed, finding such a path will be the main obstacle to Theorem3.
As mentioned above, Theorem3 is best possible up to the error constantγ as seen by the following construction from [10].
Fact 4. For every n ∈ 2N there exists a 3-uniform hypergraph H3 = (V, E) on
|V| = n vertices with δ1(H3) ≥ 167 n2
−O(n), which does not contain a loose Hamilton cycle.
Proof. Consider the following 3-uniform hypergraphH3 = (V, E). Let A∪B˙ =V be a partition ofV with|A|=bn4c −1 and letE be the set of all triplets fromV with at least one vertex inA. Clearly,δ1(H3) = |A|2
+|A|(|B| −1) = 167 n2
−O(n).
Now consider an arbitrary cycle inH3. Note that every vertex, in particular every
vertex fromA, is contained in at most two edges of this cycle. Moreover, every edge of the cycle must intersectA. Consequently, the cycle contains at most 2|A|< n/2
edges and, hence, cannot be a Hamilton cycle.
We note that the constructionH3in Fact4satisfiesδ2(H3)≥n/4−1 and indeed, the same construction proves that the minimum collective degree condition given in Theorem2is asymptotically best possible for the casek= 3.
This leads to the following conjecture for minimum vertex degree conditions enforcing loose Hamilton cycles ink-uniform hypergraphs. Letk≥3 and letHk= (V, E) be thek-uniform,n-vertex hypergraph on V =A∪B˙ with|A|=2(k−1)n −1.
Let E consist of all k-sets intersecting A in at least one vertex. Then Hk does not contain a loose Hamilton cycle and we believe that any k-uniform, n-vertex hypergraphHwhich has minimum vertex degreeδ1(H)≥δ1(Hk) +o(n2) contains a loose Hamilton cycle. Indeed, Theorem3 verifies this for the casek= 3.
2. Proof of the main result
The proof of Theorem 3 will be given in Section 2.3. It uses several auxiliary lemmas which we introduce in Section2.2. We start with an outline of the proof.
2.1. Outline of the proof. We will build a loose Hamilton cycle by connecting loose paths. Formally, a 3-uniform hypergraph P is a loose path if there is an ordering (v1, . . . , vt) 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 elementsv1 andvt are called theends ofP.
The Absorbing Lemma (Lemma7) asserts that every 3-uniform hypergraphH= (V, E) with sufficiently large minimum vertex degree contains a so-calledabsorbing loose pathP, which has the following property: For every setU ⊂V \V(P) with
|U| ∈2Nand|U| ≤βn(for some appropriate 0< β < γ) there exists a loose path Qwith the same ends asP, which covers precisely the verticesV(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 (1−β)n of the vertices. We approach this simpler problem as follows. Let H0 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 H0 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 ofR only.
Let H00 be the remaining hypergraph after removing the vertices from R. We will choose P and R small enough, so that δ1(H00) ≥ (167 +o(1))| V(H200)|
. The third auxiliary lemma, the Path-tiling Lemma (Lemma 10), asserts that all but o(n) vertices of H00 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 fromR. This way we obtain a loose cycle inH, which covers all but theo(n) left-over vertices fromH00and some left-over vertices from R. We will ensure that the number of those yet uncovered vertices will be smaller thanβn and, hence, we can appeal to the absorption property ofP and obtain a Hamilton cycle.
As indicated earlier, among the auxiliary lemmas mentioned above the Path- tiling Lemma is the only one for which the full strength of the condition (167 + o(1)) n2
is required and indeed, we consider Lemma 10 to be the main obstacle to proving Theorem 3. For the other lemmas we do not attempt to optimise the constants.
2.2. 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 3-uniform hypergraph and (ai, bi)i∈[k] a set consisting of k mutually disjoint pairs of vertices. We say that a set of triples (xi, yi, zi)i∈[k] connects (ai, bi)i∈[k] if
•
S
i∈[k]{ai, bi, xi, yi, zi}
= 5k, i.e. the pairs and triples are all disjoint,
• for alli∈[k] we have{ai, xi, yi},{yi, zi, bi} ∈ H.
Suppose thataandbare 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 ≥ 1 be an integer, and let H= (V, E)be a 3-uniform hypergraph onn vertices withδ1(H)≥ 14+γ n
2
and n≥γm/12.
For every set (ai, bi)i∈[m] of mutually disjoint pairs of distinct vertices, there exists a set of triples(xi, yi, zi)i∈[m] connecting (ai, bi)i∈[m].
Proof. We will find the triples (xi, yi, zi) to connectaiwithbifori∈[m] inductively as follows. Suppose, for somej < kthe triples (xi, yi, zi) withi < jare constructed so far and for (a, b) = (aj, bj) we want to find a triple (x, y, z) to connectaandb.
Let
U =V \ m
[
i=1
{ai, bi} ∪
j−1
[
i=1
{xi, yi, zi}
and for a vertexu∈V letLu= (V \ {u}, Eu) be thelink graph ofv defined by Eu={vw: uvw∈E(H)}.
We considerLa[U] andLb[U], the subgraphs ofLa and Lb induced onU. Owing to the minimum degree condition ofHand to the assumptionm≤γn/12, we have
e(La[U])≥ 1
4 +γ n 2
−5m(n−1)≥ 1
4+γ 6
n 2
(1) and the same lower bound also holds for e(Lb[U])). Note that any pair of edges xy∈La[U] andyz∈Lb[U] withx6=zleads to a connecting triple (x, y, z) for (a, b).
Thus, if no connecting triple exists, then for every vertexu∈U one of the following must hold: either u is isolated in La[U] or Lb[U] or it is adjacent to exactly one vertexw in both graphsLa[U] and Lb[U]. In other words, any vertex not isolated inLa[U] has at most one neighbour inLb[U]. LetIa be the set of isolated vertices
inLa[U]. Sincee(La[U])> 14 n2
we have |Ia|< n/2. Consequently, e(Lb[U])≤
|Ia| 2
+|{uw∈E(Lb[U]) : u∈U\Ia}|
≤ |Ia|
2
+ (|U| − |Ia|)<
bn/2c 2
+n .
Usingγ≤3/4 andn≥γ/12 we see that this upper bound violates the lower bound
one(Lb[U]) from (1).
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 < 1/4 there exists an n0 such that for every 3-uniform hypergraph H = (V, E) on n > n0 vertices with minimum vertex degree δ1(H) ≥ 14+γ n
2
there is a set R of size at most γn with the following property: For every system(ai, bi)i∈[k] consisting ofk≤γ3n/12mutually disjoint pairs of vertices from V there is a triple system connecting (ai, bi)i∈[k]
which, moreover, contains vertices fromR only.
Proof. We shall show that a random setRhas 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 2.6]. Alternatively Lemma6could be deduced more directly from Janson’s inequality.
For given 0< γ <1/4 letn0be sufficiently large. LetHbe 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 degH(v) edges. Since the edge set of the omplete graph Kn can be decomposed inton−1 edge disjoint matchings, we can decompose the edge set ofL into i0 =i0(v)< n pairwise edge disjoint matchings. We denote these matchings byM1(v), . . . , Mi0(v).
We randomly choose a vertex set Vp from V by including each vertexu ∈ V intoVp with probabilityp=γ−γ3 independently. For everyi∈[i0] let
Xi(v) =
Mi(v)∩ Vp
2
denote the number of edges e ∈ Mi(v) contained in Vp. This way Xi(v) is a binomially distributed random variable with parameters|Mi(v)|andp2. Using the following Chernoff bounds fort >0 (see, e.g., [5, Theorem 2.1])
P[Bin(m, ζ)≥mζ+t]< e−t2/(2ζm+t/3) (2) P[Bin(m, ζ)≤mζ−t]< e−t2/(2ζm) (3) we see that
γn
2 ≤ |Vp| ≤pn+ (3nln 20)1/2≤γn−2k (4) with probability at least 9/10.
Further, using (3) and|Mi(v)| ≤n/2 we see that with probability at most n−2 there exists an index i ∈ [i0] such that Xi(v) ≤ |Mi(v)|p2−(3nlnn)1/2. Using P
i∈[i0]|Mi(v)| = degH(v) and recalling that degV
p(v) denotes the degree of v in
H[Vp∪ {v}] we obtain that degV
p(v) = X
i∈[i0]
Xi(v)≥p2degH(v)−n(3nlnn)1/2, (5) holds with probability at least 1−n−2.
Repeating the same argument for every vertex v ∈V we infer from the union bound that (5) holds for all verticesv∈V simultaneously with probability at least 1−1/n. Hence, with positive probability we obtain a setRsatisfying (4) and (5) for allv∈V.
Let (ai, bi)i∈[k] be given and letS =S
i∈[k]{ai, bi}. Then we have |R∪S| ≤γn and
degR∪S(v)≥degR(v)≥ 1
4+γ2 γn 2
≥ 1
4 +γ2 |R∪S|
2
for allv∈V. Thus, we can appeal to the Connecting Lemma (Lemma5) to obtain a triple system which connects (ai, bi)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 582
<167.
Lemma 7(Absorbing lemma). For allγ >0there existβ >0andn0such that the following holds. LetH= (V, E)be a3-uniform hypergraph onn > n0vertices which satisfies δ1(H) ≥ 58 +γ2 n
2
. Then there is a loose path P with |V(P)| ≤ γ7n such that for all subsetsU ⊂V\V(P)of size at mostβnand|U| ∈2Nthere 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 R¨odl, Ruci´nski, and Szemer´edi. They introduced the concept ofabsorption, which, roughly speaking, stands for a local extension of a given structure, which preserves the global struc- ture. In our context of loose cycle we say that a 7-tuple (v1, . . . , v7) absorbs the two verticesx, y∈V if
• v1v2v3, v3v4v5, v5v6v7∈ Hand
• v2xv4, v4yv6∈ H
are guaranteed. In particular, (v1, . . . , v7) and (v1, v3, v2, x, v4, y, v6, v5, v7) 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 7-tuples absorbing this pair, provided the minimum vertex degree ofHis sufficiently large.
Proposition 8. For all 0 < γ < 3/8 there exists an n0 such that the following holds. Suppose H = (V, E) is a 3-uniform hypergraph on n > n0 vertices which satisfiesδ1(H)≥ 58+γ2 n
2
, then for every pair of verticesx, y∈V the number of 7-tuples absorbing xandy is at least(γn)7/8.
Proof. For givenγ >0 we choosen0= 168/γ7. First we show the following.
Claim 9. For every pairx, y∈V(H)of vertices there exists a setD=D(x, y)⊂V of size |D|=γnsuch that one of the following holds:
• deg(x, d)≥γnanddeg(y, d)≥38nfor all d∈D or
• deg(y, d)≥γnanddeg(x, d)≥38nfor all d∈D.
Proof of Claim9. By assuming the contrary there exists a pairx, y such that no setD=D(x, y) fulfills Claim9.
Let A(z) = {d ∈ V: deg(z, d) < γn} and leta =|A(x)|/n and b =|A(y)|/n.
Without loss of generality we assume that a ≤ b. Note that there are at most (a+γ)nverticesv∈V satisfying deg(y, v)≥38n. LetB(y) ={v∈V: deg(y, v)≥ 3n/8} and note that|B(y)|<(a+γ)n. Hence, the number of ordered pairs (u, v) such thatu∈B(y) and{u, v, y} ∈ H is at most
|B(y)|(n− |A(y)|) +|A(y)|γn≤(a+γ)(1−b)n2+bγn2.
Consequently, with 2 deg(y) being the number of ordered pairs (u, v) such that {u, v, y} ∈ Hwe have
5 8
2
+9γ 8
!
n2≤2 deg(y)≤(1−b−a−γ)3
8n2+ (a+γ)(1−b)n2+ 2bγn2
≤ n2
8 (5a−3b−8ab) +(3 + 8γ)n2
8 ,
Hence, we obtain 5
8 2
+9γ 8
!
n2≤2 deg(y)≤ 3n2
8 (1−b) + (a+γ) 5
8 −b
n2+ 2bγn2
≤ n2
8 (5a−3b−8ab) +(3 + 8γ)n2
8 ,
where in the last inequality we use the fact that b ≤3/8 which is a direct conse- quence of the condition onδ1(H). It is easily seen that this maximum is attained bya=b= 1/8, for which we would obtain
deg(y)≤ 5
8 2
+γ
! n2,
a contradiction.
We continue the proof of Proposition8. For a given pairx, y∈V we will select the tuplev1, . . . , v7 such that the edges
• v1v2v3, v3v4v5, v5v6v7∈ Hand
• v2xv4, v4yv6∈ H
are guaranteed. Note that (v1, . . . , v7) forms a loose path with the ends v1 and v7 and (v1, v3, v2, x, v4, y, v6, v5, v7) also forms a loose path with the same ends, showing that (v1, . . . , v7) is indeed an absorbing tuple for the paira, b. Moreover, we will show that the number of choices for eachvi will give rise to the number of absorbing tuples stated in the proposition.
First, we want to choosev4 and letD(x, y) be a set with the properties stated in Claim9. Without loss of generality we may assume that|N(y, d)| ≥ 38nfor all d∈D(x, y). Fixing somev4∈D(x, y) We choosev2∈N(x, v4) for which there are γnchoices. This gives rise to to hyperegde v2xv4 ∈ H. and applying Claim9 to v2 and v4 we obtain a set D(v2, v4) with the properties stated in Claim9 and we choose v3 ∈D(v2, v4). We choose v1 ∈ N(v2, v3) to obtain the edgev1v2v3 ∈ H.
Note that |N(v2, v3)| ≥ γn. Next, we choose v5 from N(v3, v4) which has size
|N(v3, v4)| ≥ γn. This gives rise to the edge v3v4v5 ∈ H. We choose v6 from
N(y, v4) with the additional property that deg(v5, v6)≥ γn/2. Hence, we obtain v4yv6 ∈ H and we claim that there are at least γn/2 such choices. Otherwise at least (|N(y, v4)| −γn/2) verticesv∈V satisfy deg(v5, v)< γn/2, hence
deg(v5)< 3γ 16n2+
(58+γ2)n 2
< δ(H),
which is a contradiction. Lastly we choose v7 ∈ N(v5, v6) to obtain the edge v5v6v7∈ Hwhich completes the absorbing tuple (v1, . . . , v7).
The number of choices for v1, . . . , v7 is at least (γn)7/4 and there are at most
7 2
n6 choices such thatvi =vj for somei6=j. Hence, we obtain at least (γn)7/8
absorbing 7-tuples for the pairx, y.
With Proposition8 and the connecting lemma (Lemma5) at hand the proof of the absorbing lemma follows a scheme which can be found in [12, 4]. We choose a familyF of 7-tuples by selecting each 7-tuples with probability p=γ7n−6/448 independently. Then, it is easily shown that with non-zero probability the family F satisfies
• |F | ≤γ7n/12,
• for all pairsx, y∈V there are at leastpγ7n7/16 tuples inFwhich absorbs x, y
• the number of intersecting pairs of 7-tuples inF is at mostpγ7n7/32 We eliminate intersecting pairs of 7-tuples by deleting one tuple for each such pair.
By definition each for the remaining 7-tuples (v1i, . . . , v7i)i∈[k]withk≤γ7n/12 forms a loose path with endsvi1 andvi7 and appealing to Lemma5 we can connect them to one loose path which can absorb anypγ7n7/32 =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 Theorem3. It asserts that the vertex set of a 3-uniform hypergraphHwith minimum vertex degreeδ1(H)≥
7
16+o(1) n 2
can be almost perfectly covered by a constant number of vertex disjoint loose paths.
Lemma 10(Path-tiling lemma). For allγ >0andα >0there exist integerspand n0 such that for n > n0 the following holds. SupposeHis a3-uniform hypergraph onn vertices with minimum vertex degree
δ1(H)≥ 7
16+γ n 2
.
Then there is a family of p disjoint loose paths inH which covers all but at most αnvertices of H.
The proof of Lemma10uses the weak regularity lemma for hypergraphs and will be given in Section3.
2.3. Proof of the main theorem. In this section we give the proof of the main result, Theorem3. The proof is based on the three auxiliary lemmas introduced in Section2.2and follows the outline given in Section 2.1.
Proof of Theorem 3. For given γ >0 we apply the Absorbing Lemma (Lemma 7) withγ/8 to obtainβ >0 andn7. Next we apply the Reservoir Lemma (Lemma6) forγ0= min{β/3, γ/8}to obtainn6 which isn0of Lemma6. Finally, we apply the
Path-tiling Lemma (Lemma10) withγ/2 andα=β/3 to obtainpandn10. Forn0
of the theorem we choosen0= max{n7,2n6,2n10,24(p+ 1)(γ0)−3}.
Now let n≥ n0, n∈ 2Nand let H = (V, E) be a 3-uniform hypergraph on n vertices with
δ1(H)≥ 7
16+γ n 2
.
LetP0⊂ Hbe the absorbing path guaranteed by Lemma 7. Leta0 andb0 be the ends ofP0and note that
|V(P0)| ≤γ0n < γn/8.
Moreover, the pathP0has the absorption property, i.e., for allU ⊂V\V(P0) with
|U| ≤βn and|U| ∈2Nthere exists
a loose pathQ ⊂ Hs.t.V(Q) =V(P0)∪U andQhas the endsa0 andb0. (6) Let V0 = (V \V(P0))∪ {a0, b0} and let H0 =H[V0] = (V0, E(H)∩ V30
) be the induced subhypergraph ofHonV0. Note that δ1(H0)≥(167 +34γ) n2
.
Due to Lemma6 we can choose a setR⊂V0 of size at mostγ0|V0| ≤γ0nsuch that for every system consisting of at most (γ0)3|V0|/12 mutually disjoint pairs of vertices fromV can be connected using vertices fromR only.
SetV00 =V \(V(P0)∪R) and let H00=H[V00] be the induced subhypergraph ofHonV00. Clearly,
δ(H00)≥ 7
16+γ 2
n 2
Consequently, Lemma10applied toH00(withγ10andα) yields a loose path tiling of H00which covers all but at mostα|V00| ≤αnvertices fromV00and which consists of at mostppaths. We denote the set of the uncovered vertices inV00 byT. Further, letP1,P2. . . ,Pq withq≤pdenote the paths of the tiling. By applying the reservoir lemma appropriately we connect the loose pathsP0,P1, . . . ,Pq 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+γ6)n≤βn. Moreover, sinceC is a loose cycle and n∈2Nwe have|U| ∈2N. Thus, using the absorption property ofP0 (see (6)) we can replace the subpathP0 inCby a pathQ(sinceP0 andQhave the same ends) and sinceV(Q) =V(P0)∪U the resulting cycle is a loose Hamilton cycle ofH.
3. Proof of the Path-tiling Lemma
In this section we give the proof of the Path-tiling Lemma, Lemma10. Lemma10 will be derived from the following lemma. Let M be the 3-uniform hypergraph defined on the vertex set {1, . . . ,8} with the edges 123,345,456,678 ∈ M. We will show that the conditionδ1(H)≥ 167 +o(1) n
2
will ensure an almost perfect M-tiling ofH, i.e., a family of vertex disjoint copies ofM, which covers almost all vertices.
Lemma 11. For allγ >0 andα >0there existsn0 such that the following holds.
Suppose H is a 3-uniform hypergraph on n > n0 vertices with minimum vertex degree
δ1(H)≥ 7
16+γ n 2
.
Then there is an M-tiling ofHwhich covers all but at mostαnvertices of H.
The proof of Lemma 11 requires the regularity lemma which we introduce in Section3.1. Sections3.2and3.3are devoted to the proof of Lemma11and finally, in Section3.4, we deduce Lemma10from Lemma11by making use of the regularity lemma.
3.1. The weak regularity lemma and the cluster hypergraph. In this section we introduce theweak hypergraph regularity lemma, a straightforward extension of Szemer´edi’s regularity lemma for graphs [15]. Since we only apply the lemma to 3-uniform hypergraphs we will restrict the introduction to this case.
LetH= (V, E) be a 3-uniform hypergraph and letA1, A2, A3 be mutually dis- joint non-empty subsets ofV. We define e(A1, A2, A3) to be the number of edges with one vertex in eachAi,i∈[3], and the density ofHwith respect to (A1, A2, A3) as
d(A1, A2, A3) = eH(A1, A2, A3)
|A1||A2||A3| .
We say the triple (V1, V2, V3) of mutually disjoint subsets V1, V2, V3 ⊆V is (ε, d)- regular, for constantsε >0 andd≥0, if
|d(A1, A2, A3)−d| ≤ε
for all triple of subsetsAi ⊂Vi,i∈[3], satisfying|Ai| ≥ε|Vi|. The triple (V1, V2, V3) is called ε-regular if it is (ε, d)-regular for some d≥ 0. It is immediate from the definition that an (ε, d)-regular triple (V1, V2, V3) is (ε0, d)-regular for allε0 > εand ifVi0⊂Vi has size|Vi0| ≥c|Vi|, then (V10, V20, V30) is (ε/c, d)-regular.
Next we show that regular triples can be almost perfectly covered by copies of Mprovided 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 (v1, . . . , vt) 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 12. SupposeHis a 3-uniform hypergraph onmvertices with at least dm3edges. Then there is a tight path inHwhich covers at least2(dm+ 1)vertices.
In particular, if H is 3-partite with the partition classes V1, V2, V3 and 2dm >10 then for each i∈[3] there is a copy ofM in Hwhich intersectsVi in exactly two vertices and the other partition classes in three vertices.
Proof. Starting fromHwe remove all edges containingu, v for each pair u, v∈V of vertices such that 0<deg(u, v)<2dm. We keep doing this until every pairu, v satisfies deg(u, v) = 0 or deg(u, v)≥2dmin the current hypergraphH0. Since less than m2
·2dm < dm3≤e(H) edges were removed during the process we know that H0is not empty. Hence we can pick a maximal non-empty tight path (v1, v2, . . . , vt) in H0. Since the pair v1, v2 is contained in an edge inH0 it is contained in 2dm 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 2(dm+ 1) 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 3-partite 3-uniform hypergraph once the two starting vertices are fixed. SinceM 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.
Proposition 13. Suppose the triple (V1, V2, V3) is(ε, d)-regular with d ≥2ε and suppose the sizes of the partition classes satisfy
m=|V1| ≥ |V2| ≥ |V3| with5|V1| ≤3(|V2|+|V3|) (7) and 2ε2m > 7. Then there is an M-tiling of (V1, V2, V3) leaving at most 3εm vertices uncovered.
Proof. Note that if we take a copy of M intersecting Vi, i ∈ [3] in exactly two vertices then this copy intersects the other partition classes in exactly three vertices.
We define ti= (1−ε)1
8(3|Vj|+ 3|Vk| −5|Vi|) where i, j, k ∈[3] are distinct.
Due to our assumption allti are non-negative and we chooseti copies ofMinter- sectingVi in exactly two vertices. This would leave|Vi| −(2ti+ 3tj+ 3tk) =ε|Vi| vertices inVi uncovered, hence at most 3εmin 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 classVi
take a subsetUi,i∈[3] of sizeε|Vi|. Due to the regularity of the triple (V1, V2, V3) we havee(U1, U2, U3)≥(d−ε)(εm)3. Hence, by Proposition 12there is a copy of
M(of each type) in (U1, U2, U3).
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., [1,2,14]).
Theorem 14. For allt0≥0andε >0, there existT0=T0(t0, ε)andn0=n0(t0, ε) so that for every3-uniform hypergraph H= (V, E)on n≥n0 vertices, there exists a partitionV =V0∪V˙ 1∪˙ . . .∪V˙ t such that
(i) t0≤t≤T0,
(ii) |V1|=|V2|=· · ·=|Vt|and|V0| ≤εn, (iii) for all but at most ε 3t
sets {i1, . . . , i3} ∈ [t]3
, the triple (Vi1, Vi2, Vi3) is
ε-regular.
A partition as given in Theorem14is called an (ε, t)-regular partition ofH. For an (ε, t)-regular partition ofHand d≥0 we refer toQ= (Vi)i∈[t] as the family of clusters (note that the exceptional vertex setV0is excluded) and define the cluster hypergraphK=K(ε, d,Q) with vertex set [t] and{i1, i2, i3} ∈ [t]3
being an edge if and only if (Vi1, Vi2, Vi3) isε-regular andd(Vi1, Vi2, Vi3)≥d.
In the following we show that the cluster hypergraph almost inherits the mini- mum vertex degree of the original hypergraph. The proof which we give for com- pleteness is standard and can be found e.g. in [8] for the case of graphs.
Proposition 15. For all γ > d > ε >0 and allt0 there existT0 andn0 such that the following holds. SupposeHis a3-uniform hypergraph onn > n0 vertices which has vertex minimum degreeδ1(H)≥ 167 +γ n
2
.Then there exists an(ε, t)-regular partition Q with t0 < t < T0 such that the cluster hypergraph K =K(ε, d,Q) has minimum vertex degreeδ1(K)≥ 167 +γ−ε−d t
2
.
Proof. Let γ > d > ε and t0 be given. We apply the regularity lemma with ε0 = ε2/144 andt00 = max{2t0,10/ε} to obtain T00 and n00. We set T0 =T00 and
n0 = n00. Let H be a 3-uniform hypergraph on n > n0 vertices which satisfies δ(H) ≥ (7/16 +γ) n2
. By applying the regularity lemma we obtain an (ε0, t0)- regular partitionV00∪V˙ 1∪˙. . .∪V˙ t0 of V and let m=|V1|= (1−ε0)n/t0 denote the size of the partition classes.
Let I = {i ∈ [t0] : Vi is contained in more than ε t20
/8 nonε0-regular triples}
and observe that|I| <8ε0t0/ε due to the property (iii) of Theorem 14. Set V0= V00∪S
i∈IViand letJ = [t0]\Iandt=|J|. We now claim thatV0andQ= (Vj)j∈J
is the desired partition. Indeed, we have T0 > t0 ≥ t > t0(1−8ε0/ε) ≥ t0 and
|V0|< ε0n+ 8ε0n/ε≤εn/16. The property (iii) follows directly from Theorem14.
For a contradiction, assume now that degK(Vj) < (167 +γ−ε−d) t2
for some j∈J. Letxj denote the number of edges which intersectVj in exactly one vertex and each otherVi, i∈J, in at most one vertex. Then, the assumption yields
xj≤ |Vj| 7
16+γ−ε−d t 2
m2+ε 8
t0 2
m2+ ε 16n2+d
t 2
m2
≤ |Vj|n2 2
7
16+γ−ε 2
On the other hand, from the minimum degree ofHwe obtain xj ≥ |Vj|
7
16+γ n 2
−2 |Vj|
2
n−3 |Vj|
3
≥ |Vj| n
2 7
16+γ− 4 t0
a contradiction.
3.2. Fractionalhom(M)-tiling. To obtain a largeM-tiling in the hypergraphH, we consider weighted homomorphisms fromMinto the cluster hypergraph K. To this purpose, we define the following.
Definition 16. Let L be a3-uniform hypergraph. A functionh: V(L)×E(L)→ [0,1]is called a fractional hom(M)-tiling of Lif
(1) h(v, e)6= 0⇒v∈e, (2) h(v) =P
e∈E(L)h(v, e)≤1,
(3) 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)≥2 3h(u, e)
Byhmin we denote the smallest non-zero value of h(v, e)(and we sethmin=∞ if h≡0) and the sum over all values is the weight w(h)of h
w(h) = X
(v,e)∈V(L)×E(L)
h(v, e).
The allowed values ofh are based on the homomorphisms fromM to a single edge, hence the term hom(M)-tiling. Given one such homomorphism, assign each vertex in the image the number of vertices fromM 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 homomor- phisms ofMinto a single edge the smallest and the largest class of preimages can differ by a factor of 2/3 at most and this observation is the reason for condition (3) in Definition16. We also note the following.
Fact 17. There is a fractional hom(M)-tilingh of the hypergraph M which has hmin≥1/3 and weightw(h) = 8.
Proof. Letx1,x2,w1, y1,y2,w2,z1, andz2 be the vertices ofMand letx1x2w1, w1y1y2, y1y2w2, andw2z1z2 be the edges of M. On the edgesx1x2w1 andz1z2w2
we assign the vertex weights (1,1,2/3), where the weight 2/3 is assigned to w1
andw2. The vertex weights for edgesy1y2w1andy1y2w2are (1/2,1/2,1/3), where w1 andw2 get the weight 1/3. It is easy to see that those vertex weights give rise to a hom(M)-tilinghonMwithhmin= 1/3 andw(h) = 8.
The notion hom(M)-tiling is also motivated by the following proposition which shows that such a fractional hom(M)-tiling in a cluster hypergraph can be “con- verted” to an integerM-tiling in the original hypergraph.
Proposition 18. Let Qbe an(ε, t)-regular partition of a3-uniform,n-vertex hy- pergraph H with n >21ε−2 and let K = K(ε,6ε,Q) be the corresponding cluster hypergraph. Furthermore, let h: V(K)×E(K)→ [0,1] be a fractional hom(M)- tiling of K with hmin ≥1/3. Then there exists an M-tiling ofH which covers all but at most(w(h)−27tε)|V1|vertices.
Proof. We restrict our consideration to the subhypergraph K0 ⊂ K consisting of the hyperedges with positive weight, i.e.,e=abc∈ Kwithh(a), h(b), h(c)≥hmin. For each a∈ V(K0) let Va be the corresponding partition class inQ. Due to the property (2) of Definition16we can subdivide Va (arbitrarily) into a collection of pairwise disjoint sets (Uae)a∈e∈Kof size|Uae|=h(a, e)|Va|. Note that every edgee= abc∈ Kcorresponds to the (ε,6ε)-regular triplet (Va, Vb, Vc). Hence we obtain from the definition of regularity andhmin≥1/3 that the triplet (Uae, Ube, Uce) is (3ε,6ε)- regular. From the property (3) in Definition 16 and Proposition 13we obtain an M-tiling of (Uae, Ube, Uce) incorporating at least h(a, e) +h(b, e) +h(c, e)−9ε
|Va| vertices. Applying this to all hyperedges ofK0we obtain anM-tiling incorporating at least
X
abc=e∈K0
h(a, e) +h(b, e) +h(c, e)−9ε
!
|Va| ≥ w(h)−9|K0|ε
|Va| vertices. Noting that |K0| ≤3t (because ofhmin≥1/3) and |Va| ≥ |V1|we obtain
the proposition.
Owing to Proposition18, we are given a connection between fractional hom(M)- tilings of the cluster hypergraph K of Hand M-tilings in H. A vertex i∈ V(K) corresponds to a class of verticesVi in the regular partition ofH. The total vertex weight h(i) essentially translates to the proportion of vertices of Vi which can be covered by the correspondingM-tilings inH. Consequently,w(h) essentially trans- lates to the proportion of vertices covered by the correspondingM-tiling inH. This reduces our task to finding a fractional hom(M)-tiling with weight greater than the number of vertices previously covered inK.
The following lemma (Lemma 19), which is the main tool for the proof of Lemma 11, follows the idea discussed above. In the proof of Lemma 11 we fix a maximalM-tiling in the cluster hypergraphKof the given hypergraphH. Ow- ing to the minimum degree condition ofHand Proposition15, a typical vertex in the cluster hypergraphK will be contained in at least (7/16 +o(1)) |V(K)|2
hyper- edges ofK. We will show that a typical vertexuofK which is not covered by the
maximalM-tiling ofK, has the property that (7/16 +o(1))·64>28 of the edges incident touintersect some pair of copies ofMfrom theM-tiling ofK. Lemma19 asserts that two such vertices and the pair of copies ofM can be used to obtain a fractional hom(M)-tiling with a weight significantly larger than 16, the number of vertices of the two copies ofM. This lemma will come in handy in the proof of Lemma11, where it is used to show that one can cover a higher proportion of the vertices ofHthan the proportion of vertices covered by the largestM-tiling inK.
We consider a set of hypergraphsL29definied as follows: EveryL ∈L29consists of two (vertex disjoint) copies ofM, sayM1andM2, and two additional verticesu andvsuch that all edges incident touorvcontain precisely one vertex fromV(M1) and one vertex fromV(M2). Moreover,Lsatisfies the following properties
• for everya∈V(M1) andb∈V(M2) we haveuab∈E(L) iffvab∈E(L)
• deg(u) = deg(v)≥29.
Lemma 19. For every L ∈ L29 there exists a fractional hom(M)-tiling h with hmin≥1/3 andw(h)≥16 +13.
The following proof of Lemma 19 is based on straightforward, but somewhat tedious case distinction.
Proof. For the proof we fix the following labeling of the vertices of the two dis- joint copies ofM. Let V(M1) = {x1, x2, w1, y1, y2, w2, z1, z2} be the vertices and x1x2w1,w1y1y2, y1y2w2,w2z1z2 be the edges of the first copy ofM. Analogously, let V(M2) = {x01, x02, w10, y01, y20, w20, z01, z20} be the vertices and x01x02w01, w01y10y20, y10y20w02, w20z10z02 be the edges of the other copy ofM (see Figure 1.a). Moreover, we denote byX ={x1, x2},Y ={y1, y2}, andZ ={z1, z2}and, letX0,Y0, andZ0 be defined analogously forM2.
Figure 1. Labels and case: a1b1, a2b2∈L1with{b1, b2} ∈ {X0, Y0, Z0} x1
x2
w1
y1
y2
w2
z1
z2
x′1 x′2 w′1 y1′ y2′ w′2 z1′ z2′ X
Y
Z
X′
Y′
Z′
1.a: Vertex labels ofM1 andM2 inL
a1
a2
b1
b2
u v
1.b: All edges are (a1)-edges
The proof of Lemma19proceeds in two steps. First, we show that in any possible counterexampleL, the edges incident touandv which do not contain any vertex
from{w1, w2, w10, w20}form a subgraph ofK2,3,3(see Claim20). In the second step we show that every edge contained in this subgraph of K2,3,3 forbids too many other edges incident to uand v, which will yield a contradiction to the condition deg(u) = deg(v)≥29 ofL(see Claim21).
We introduce the following notation to simplify later arguments. For a given L ∈L29with M1 andM2 being the two copies of M, letLbe set of those pairs (a, b)∈V(M1)×V(M2) such thatuab∈E(L). We splitLintoL1∪L˙ 2 according to
(a, b)∈
(L1, if{a, b} ∩ {w1, w2, w01, w02}=∅, L2, otherwise.
It will be convenient to view L1 and L2 as bipartite graphs with vertex classes V(M1) andV(M2).
We split the proof of Lemma19into the following two claims.
Claim 20. For allL ∈L29 without a fractional hom(M)-tiling withhmin ≥1/3 andw(h)≥16 + 1/3, we haveL1⊆K3,3, where each of the setsX,Y,Z andX0, Y0,Z0 contains precisely one of the vertices of theK3,3.
Claim 20 will be used in the proof of the next claim, which clearly implies Lemma19.
Claim 21. Let F =
a0b0 ∈V(M1)×V(M2) : a0 ∈ {w1, w2} orb0 ∈ {w01, w02} and for every edge ab ∈ L1 let F(a, b) ⊆ F be the set of those e ∈ F, whose appearance inL (i.e.e∈L2) implies the existence of a fractionalhom(M)-tilingh withhmin≥1/3 andw(h)≥16 + 1/3. Then there is an injectionf: L1→F such that f(a, b)∈ F(a, b)for every pair ab∈L1.
Clearly, |F|= 28 and L2 ⊂F. Hence, from |L2|+|f(L1)|= |L2|+|L1| ≥ 29 we derive that L2 andf(L1) must intersect. By Claim 21 this yields the desired
fractional hom(M)-tiling and Lemma19follows.
In the proofs of Claim 20 and Claim 21 we will consider fractional hom(M)- tilingshwhich use vertex weights of special types. In fact, for an edgee=a1a2a3, the weightsh(a1, e),h(a2, e), andh(a3, e) will be of the following forms
(a1) h(a1, e) =h(a2, e) =h(a3, e) = 1 (a2) h(a1, e) =h(a2, e) =h(a3, e) = 12 (a3) h(a1, e) =h(a2, e) =h(a3, e) = 13 (b1) h(a1, e) =h(a2, e) = 1 andh(a3, e) =23 (b2) h(a1, e) =h(a2, e) =12 andh(a3, e) =13 (b3) h(a1, e) =h(a2, e) =23 andh(a3, e) =12
An edge that satisfies (a1) is called an (a1)-edge, etc. Note that all these types satisfy condition (3) of Definition16.
Proof of Claim20. GivenL ∈L29satisfying the assumptions of the claim and with the labeling from Figure1.a. Observe that for anyA∈ {X, Y, Z}, the hypergraph M1−Acontains two disjoint edges. Similarly, for everyB∈ {X0, Y0, Z0},M2−B contains two disjoint edges.
First we exclude the case that there is a matching {a1b1, a2b2} of size two in L1 between some{a1, a2}=A∈ {X, Y, Z} and some{b1, b2}=B∈ {X0, Y0, Z0}.
In this case we can construct a fractional hom(M)-tilinghas follows: Choose two
Figure 2. Case: ab1, ab2∈L1 with{b1, b2} ∈ {X0, Y0, Z0}
1 1
1 1 1 1
1 1 1
1 1
2 3 1 3 1 3 1 3 1
2
2 3
2 3 1
2 2
3
2 3
2.a: (a1)-edges w1y1y2, w2z1z2, and w02z01z20, (b3)-edges ax01u and ax02v, (b1)- edgew10y10y20, and (a3)-edgex01x02w1.
1 1
1 1 1 1
1 1 1
1 1
2 3 1 3 1 3
1 3 1
2
2 3
2 3 1
2
2 3
2 3
2.b: (a1)-edges w1y1y2, w2z1z2, and w02z01z20, (b3)-edges ay10u and ay02v, (b1)- edgex01x02w01, and (a3)-edgew01y01y02.
edgesua1b1, va2b2. Using these and the four disjoint edges in (M1−A) ˙∪(M2−B), we obtain six disjoint edges (see Figure 1.b). Letting all these six edges be (a1)- edges, we obtain a fractional hom(M)-tilinghwithhmin= 1 andw(h) = 18.
Next, we show that each vertexa ∈A ∈ {X, Y, Z} has at most one neighbour in each B ∈ {X0, Y0, Z0}. Assuming the contrary, let a ∈ A ∈ {X, Y, Z} and {b1, b2} = B ∈ {X0, Y0, Z0}, such that ab1, ab2 ∈ L1. For symmetry reasons, we only need to consider the case B = X0 and B = Y0. The case B = Z0 is symmetric toB=X0. In those cases, we choosehas shown in Figure2.a(B =X0) and Figure2.b (B =Y0) and in either case we find a fractional hom(M)-tilingh satisfyinghmin= 1/3 andw(h) = 16 + 1/3. Note that the casesA=Y andA=Z can be treated in the same manner since the only condition needed to define his thatM1−Acontains two disjoint edges.
To show thatL1 is indeed contained in a K3,3 it remains to verify that every a1b1, a2b2 with {a1, a2} =A∈ {X, Y, Z}, b1 ∈B1 ∈ {X0, Y0, Z0}, andb2 ∈B2 ∈ {X0, Y0, Z0} \B1 guarantees the existence of a fractional hom(M)-tiling h with hmin ≥1/3 and w(h) ≥ 16 + 1/3. Again owing to the symmetry, the only cases we need to consider are B1=X0, B2 =Y0 (see Figure 3.a) andB1=X0, B2=Z0 (see Figure3.b). In fact, the fractional hom(M)-tilings hgiven in Figure3.a and Figure 3.bsatisfyhmin≥1/3 and w(h) = 17. Again the casesA=Y andA=Z can be treated in the same manner. This concludes the proof of Claim20.
To complete the proof of Lemma19it is left to prove Claim21.
Proof of Claim21. Before defining the injection f: L1→F we collect some infor- mation about F(a, b) with ab∈L1. Owing to Claim 20, we may assume without loss of generality that x1, y1, z1 and x01, y01, z10 are the vertices which span all edges ofL1. First we consider e =y1y10. As shown in Figure 4.a the appearance
