Discrete Mathematics

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1 Dscrete Mathematcs 311 (2011) Contents lsts avalable at ScenceDrect Dscrete Mathematcs ournal homepage: Loose Hamlton cycles n hypergraphs Peter Keevash a, Danela Kühn b, Rchard Mycroft a,, Deryk Osthus b a School of Mathematcal Scences, Queen Mary, Unversty of London, Mle End Road, London, E1 4NS, Unted Kngdom b School of Mathematcs, Unversty of Brmngham, Brmngham, B15 2TT, Unted Kngdom a r t c l e n f o a b s t r a c t Artcle hstory: Receved 30 October 2008 Receved n revsed form 25 November 2010 Accepted 30 November 2010 Keywords: Hamlton cycle Hypergraph Hypergraph regularty Blow-up lemma We prove that any k-unform hypergraph on n vertces wth mnmum degree at least n + o(n) contans a loose Hamlton cycle. The proof strategy s smlar to that used 2() by Kühn and Osthus for the 3-unform case. Though some addtonal dffcultes arse n the k-unform case, our argument here s consderably smplfed by applyng the recent hypergraph blow-up lemma of Keevash Elsever B.V. All rghts reserved. 1. Introducton A fundamental theorem of Drac [3] states that any graph on n vertces wth mnmum degree at least n/2 contans a Hamlton cycle. A natural queston s whether ths theorem can be extended to hypergraphs. For ths, we frst need to extend the notons of mnmum degree and of Hamlton cycles to hypergraphs. A k-unform hypergraph or k-graph H conssts of a vertex set V and a set of edges each consstng of k vertces. We wll often dentfy H wth ts edge set and wrte e H f e s an edge of H. Gven a k-graph H, we say that a set of k 1 vertces T has neghbourhood N H (T) = {x V : {x} T H}. The degree of T s d (T) = N H (T). The mnmum degree of H s the mnmum sze of such a neghbourhood, that s, δ (H) = mn d (T) : T. We say that a k-graph C s a cycle of order n f ts vertces can be gven a cyclc orderng v 1,..., v n so that every consecutve par v, v +1 les n an edge of C and every edge of C conssts of k consecutve vertces. A cycle of order n s tght f every set of k consecutve vertces forms an edge; t s loose f every par of adacent edges ntersects n a sngle vertex, wth the possble excepton of one par of edges, whch may ntersect n more than one vertex. Ths fnal condton allows us to consder loose cycles whose order s not a multple of k 1. Fg. 1 shows the structure of each of these cycle types. A Hamlton cycle n a k-graph H s a sub-k-graph of H whch s a cycle contanng every vertex of H. Rödl et al. [10,11] showed that for any η > 0 there s an n 0 so that f n > n 0 then any k-graph H on n vertces wth mnmum degree δ (H) n/2 + ηn contans a tght Hamlton cycle (ths mproved an earler bound by Katona and Kerstead [6]). They gave a constructon whch shows that ths result s best possble up to the error term ηn. In ths paper, we prove the analogous result for loose Hamlton cycles. V V Correspondng author. E-mal addresses: p.keevash@qmul.ac.uk (P. Keevash), kuehn@maths.bham.ac.uk (D. Kühn), r.mycroft@qmul.ac.uk (R. Mycroft), osthus@maths.bham.ac.uk (D. Osthus) X/$ see front matter 2010 Elsever B.V. All rghts reserved. do: /.dsc

2 P. Keevash et al. / Dscrete Mathematcs 311 (2011) Fg. 1. Segments of a tght cycle (top), a generc cycle (mddle) and a loose cycle (bottom). Theorem 1.1. For all k 3 and any η > 0 there exsts n 0 so that f n > n 0 then any k-graph H on n vertces wth 1 δ (H) > ( + η)n contans a loose Hamlton cycle. 2() The case when k = 3 was proved by Kühn and Osthus [9]. We wll use a smlar method of proof for general k-graphs, but ths wll be greatly smplfed by the use of the recent blow-up lemma of Keevash [7]. Proposton 2.1 shows that Theorem 1.1 s best possble up to the error term ηn. In fact, Proposton 2.1 actually tells us more than ths, namely that up to the error term, ths mnmum degree condton s best possble to ensure the exstence of any (not necessarly loose) Hamlton cycle n H. Ths means that the mnmum degree needed to ensure the exstence of a n Hamlton cycle n a k-graph of order n s 2() + o(n). Whlst fnalzng ths paper we learnt that Hàn and Schacht [5] ndependently and smultaneously proved Theorem 1.1, usng a dfferent approach. The result n [5] also covers the noton of a k-unform l-cycle for l < k/2 (here one requres consecutve edges to ntersect n precsely l vertces). More recently Kühn et al. [8] further developed the method of Hàn and Schacht to nclude all l such that k l k (the remanng values of l are covered by the results of Rödl et al. [10,11]). There s also the noton of a Berge-cycle, whch conssts of a sequence of vertces where each par of consecutve vertces s contaned n a common edge. Ths s less restrctve than the cycles consdered n ths paper. Hamltonan Berge-cycles were studed n [2]. 2. Extremal example and outlne of the proof The next proposton shows that Theorem 1.1 s best possble, up to the error term ηn. Proposton 2.1. For all ntegers k 3 and n 2k 1, there exsts a k-graph H on n vertces such that δ (H) n 2k 2 1 but H does not contan a Hamlton cycle. Proof. Let V 1 and V 2 be dsont sets of sze n 1 and n n + 1 respectvely. Let H be the k-graph on the vertex 2k 2 2k 2 set V = V 1 V 2, wth e an edge f and only f e V 1, that s, f e contans at least one vertex from V 1. Then H has V k mnmum degree δ (H) = n 1. However, any cyclc orderng of the vertces of H must contan 2k 2 consecutve 2k 2 vertces v 1,..., v 2k 2 from V 2, but then v and v k cannot be contaned n a common edge consstng of k consecutve vertces, and so H cannot contan a Hamlton cycle. In our proof of Theorem 1.1 we construct the loose Hamlton cycle by fndng several paths and onng them nto a spannng cycle. Here a k-graph P s a path f ts vertces can be gven a lnear orderng such that every edge of P conssts of k consecutve vertces, and so that every par of consecutve vertces of P le n an edge of P. Smlarly as for cycles, we say that a path P s loose f edges of P ntersect n at most one vertex. The orderng of the vertces of P naturally gves an orderng of the edges of P. We say that any vertex of P whch les n the ntal edge of P, but not the second edge of P, s an ntal vertex. Smlarly, any vertex of P whch les n the fnal edge of P but not the penultmate edge s a fnal vertex. Also, we refer to vertces of P whch le n more than one edge of P as lnk vertces. Thus, for example, a loose path P has k 1 ntal vertces, k 1 fnal vertces, and one lnk vertex n each par of consecutve edges. In Secton 3, we shall ntroduce varous deas we wll need n the proof of Theorem 1.1. In partcular, we wll state a verson of the hypergraph regularty lemma due to Rödl and Schacht [12] and Theorem 3.3 due to Keevash [7]. The latter provdes a useful way of applyng the hypergraph blow-up lemma. In Secton 4, we shall prove varous auxlary results, ncludng a result on fndng loose paths n complete k-partte k-graphs, and an approxmate mnmum degree condton to guarantee a near-perfect packng of H wth a partcular k-graph A k. Fnally, n Secton 5 we shall prove Theorem 1.1 as follows Imposng structure on H In Secton 5.1 we use the hypergraph regularty lemma to splt H nto k-partte k-graphs H on dsont vertex sets X. These k-graphs H wll be sutable for embeddng almost spannng loose paths, and all the vertces of H not contaned n any of the X wll be ncluded n an exceptonal loose path L e (actually, f V(H) s not dvsble by k 1, then L e wll contan two consecutve edges whch ntersect n more than one vertex). The requrement that H contans an almost spannng loose

3 546 P. Keevash et al. / Dscrete Mathematcs 311 (2011) path means that the vertex classes of the H must have sutable sze. We acheve ths by frst defnng a sutable reduced k-graph R of H. Then we cover almost all vertces of R by copes of a sutable auxlary k-graph A k. For each copy of A k, the correspondng sub-k-graph of H s then splt nto the same number of dsont H The lnkng strategy In Secton 5.3 we shall use the structure mposed on H to fnd a Hamlton cycle n H by the followng process. (a) The k-graphs H are connected by means of a walk W = e 1,..., e l n the supplementary graph. Ths graph (whch we wll defne n Secton 5.2) has vertces 1,..., t correspondng to the k-graphs H. (b) Usng Lemma 5.2, each edge e of W s used to create a short connectng loose path L n H onng two dfferent H s. (c) L e and the paths L are extended to prepaths (these can be thought of as a path mnus an ntal vertex and a fnal vertex) L = e I 0L e F 0 and L = I L F, where I 0, F 0 and all I, F are sets of sze k 2. These prepaths have the property that there are large sets I and F such that L can be extended to a loose path by addng any vertex of I as an ntal vertex and any vertex of F as a fnal vertex. Smlarly there are large sets I l+1 and F 0 so that L e can be extended to a path by addng any vertex of I l+1 as an ntal vertex and any vertex of F 0 as a fnal vertex. I +1 and F both le n the same H (for all = 0,..., l). (d) For each H and for all those pars I, +1 F whch le n H, we choose a loose path L +1 nsde H from F to I +1. For each, we wll use the hypergraph blow-up lemma (n the form of Theorem 3.3) to ensure that together all those L whch le n H use all the remanng vertces of H. (e) The loose Hamlton cycle s then the concatenaton L e L 1 L 1 L l L l L l Controllng dvsblty Note that the number of vertces of a loose path s 1 modulo k 1. So n order to apply Theorem 3.3 to obtan spannng loose paths n a subgraph of H, we need ths subgraph to satsfy ths condton. So we choose our paths sequentally to satsfy the followng congruences modulo k 1. (a) L e s chosen wth V(H) \ V(L e ) 1. (b) Let X ( 1) be the subset of X obtaned by removng V(L 1 ),..., V(L 1 ). (All the X wll be dsont from V(L e ).) Let d be the number of tmes that W vsts H. When choosng L, for every X t traverses (except the fnal one) we arrange to ntersect X ( 1) n a set of sze t () X ( 1) + d (the sze modulo k 1 of the ntersecton of L wth the fnal X t traverses s then determned by the szes of the other ntersectons). The choce of L e n (a) ensures that after all L have been pcked, the remanng part X (l) of X has sze d. (c) Each L s extended to a prepath L by addng I and F. Smlarly, L e s extended nto a prepath L e by addng I 0 and F 0. Now the remanng part of X has sze d. (d) It remans to select d paths L wthn each X : each uses 1 vertces, so the dvsblty condtons are satsfed. 3. Regularty and the blow-up lemma 3.1. Graphs and complexes A We begn wth some notaton. By [r] we denote the set of ntegers from 1 to r. For a set A, we use to denote the k collecton of subsets of A of sze k, and smlarly to denote the collecton of non-empty subsets of A of sze at most k. A k We wrte x = y ± z to mean that y z x y + z. We shall omt floors and celngs throughout ths paper whenever they do not affect the argument. A hypergraph H conssts of a vertex set V(H) and an edge set, such that each edge e of the hypergraph satsfes e V(H). So a k-graph as defned n Secton 1 s a hypergraph n whch all the edges are of sze k. We say that a hypergraph H s a k-complex f every edge has sze at most k and H forms a smplcal complex, that s, f e 1 H and e 2 e 1 then e 2 H. As for k-graphs we dentfy a hypergraph H wth the set of ts edges. So H s the number of edges n H, and f G and H are hypergraphs then G \ H s formed by removng from G any edge whch also les n H. If H s a hypergraph wth vertex set V then for any V V the restrcton H[V ] of H to V s defned to have vertex set V and all edges of H whch are contaned n V as edges. Also, for any hypergraphs G and H we defne G H to be the hypergraph G[V(G) \ V(H)]. We say that a hypergraph H s r-partte f ts vertex set X s dvded nto r parwse-dsont parts X 1,..., X r, n such a way that for any edge e H, e X 1 for each. We call the X the vertex classes of H and say that the partton X 1,..., X r of X s equtable f all the X have the same sze. We say that a set A X s r-partte f A X 1 for each. So every edge of an r-partte hypergraph s r-partte. In the same way we may also speak of r-partte k-graphs and r-partte k-complexes. Gven a k-graph H, we defne a k-complex H = {e 1 : e 1 e 2 and e 2 H} and a (k 1)-complex H < = {e 1 : e 1 e 2 and e 2 H}. Conversely, for a k-complex H we defne the k-graph H = to be the top level of H,.e. H = = {e H: e = k}. (Here V(H) = V(H ) = V(H < ) = V(H = ).)

4 P. Keevash et al. / Dscrete Mathematcs 311 (2011) Gven a k-graph G and a set W of vertces of G, we denote by G[W] the sub-k-graph of G obtaned by removng all vertces and edges not contaned n W (n ths case, we say G s restrcted to W ). For a k-graph G and a sub-k-graph H G wrte G H for G[V(G) \ V(H)]. Let X 1,..., X r be parwse-dsont sets of vertces, and let X = X 1 X r. Gven A [r], we wrte K k A (X) for the complete A -partte A -graph whose vertex classes are all the X wth A. The ndex of an r-partte subset S of X s (S) = { [r] : S X }. Furthermore, gven any set B (S), we wrte S B = S B X. Smlarly, gven A and an r-partte k-graph or k-complex H on the vertex set X we wrte H A for the collecton of edges n H of ndex A and let H = { }. In partcular, f H s a k-complex then H {} s the set of all those vertces n X whch le n an edge of H (and thus form a (sngleton) edge of H). In general, we wll often vew H A as an r-partte A -graph wth vertex set X. Also, gven a k-complex H we smlarly wrte H A = B A H B and H A < = B A H B. We wrte H A for the A -graph whose edges are those r-partte sets S X of ndex A for whch all proper subsets of S belong to H. (In other words, a set S wth ndex A satsfes S H A f and only f for all < A the edges of H whch have sze and are subsets of S form a complete -graph on S vertces.) Then the relatve densty of H at ndex A s d A (H) = H A / H A. The absolute densty of H A s d(h A ) = H A / K A (X). (Note that K A (X) = A X.) If H s a k-partte k-complex we may smply wrte d(h) for d(h [k] ). Smlarly, the densty of a k-partte k-graph H on X = X 1 X k s d(h) = H / K [k] (X). Fnally, for any vertex v of a hypergraph H, we defne the vertex degree d(v) of v to be the number of edges of H whch contan v. Note that ths s not the same as the degree defned earler, whch was for sets of k 1 vertces. The maxmum vertex degree of H s then the maxmum of d(v) taken over all vertces v V(H). The vertex neghbourhood VN(v) of v s the set of all vertces u V(H) for whch there s an edge of H contanng both u and v. For a k-partte k-complex H on the vertex set X 1 X k we also defne the neghbourhood complex H(v) of a vertex v X for some to be the (k 1)-partte (k 1)-complex wth vertex set X and edge set {e H : e {x} H}. [r] k 3.2. Regular complexes In ths subsecton we shall defne the concept of regular complexes (whch was frst ntroduced n the k-unform case by Rödl and Skokan [14]) n the form used by Rödl and Schacht [12,13]. Ths s a generalzaton of the standard concept of regularty n graphs, where we say that a bpartte graph B on vertex classes U and V forms an ϵ-regular par f for any U U and V V wth U > ϵ U and V > ϵ V we have d(b[u V ]) = d(b) ± ϵ. In the same way, we say that a k-complex G s regular f the restrcton of G to any large subcomplex of lower rank has smlar denstes to G. More precsely, let G be an r-partte k-complex on the vertex set X = X 1 X r. For any A we say that G A s ϵ-regular f for any H G A < wth H A ϵ G A we have G A H A H A = d A (G) ± ϵ. We say G s ϵ-regular f G A s ϵ-regular for every A [r] k [r]. Note that f G s a graph wthout solated vertces, then the k defnton n the prevous paragraph s equvalent to the 2-complex G beng ϵ-regular. To llustrate the defnton for k = 3, suppose that A = [3]. Then for nstance the top level of G [2] s the bpartte subgraph of G nduced by X 1 and X 2 and G A s the set of (graph) trangles n G. So roughly speakng, the regularty condton states that f we consder a subgraph of G [2] G {1,3} G {2,3} whch spans a large number of trangles, then the proporton of these whch also form an edge of G A s close to d A (G),.e. close to the proporton of (graph) trangles n G between X 1, X 2 and X 3 whch form an edge of G. Roughly speakng, the hypergraph regularty lemma states that an arbtrary k-graph can be splt nto peces, each of whch forms a regular k-complex. The verson of the regularty lemma we shall use also nvolves the noton of a partton complex, whch s a certan partton of the edges of a complete k-complex. As before, let X = X 1 X r be an r-partte vertex set. A partton k-system P on X conssts of a partton P A of the edges of K A (X) for each A [r] k,. We refer to the partton classes of P A as cells. So every edge of K A (X) s contaned n precsely one cell of P A. P s a partton k-complex on X f t also has the property that whenever S, S K A (X) le n the same cell of P A, we have that S B and S B le n the same cell of P B for any B A. Ths property of S, S forms an equvalence relaton on the edges of K A (X), whch we refer to as strong equvalence. To llustrate ths, agan suppose that k = 3 and A = [3]. Then f P s a partton k-complex, P {1}, P {2} and P {3} together yeld a vertex partton Q 1 refnng X 1, X 2, X 3. Q 1 naturally nduces a partton Q 2 of the 3 complete bpartte graphs nduced by the pars X, X. P {1,2}, P {2,3} and P {1,3} also yeld a partton Q 2 of these complete bpartte graphs. The requrement of strong equvalence now mples that Q 2 s a refnement of Q 2. At the next level, Q 2 naturally nduces a partton Q 3 of the set of trples nduced by X 1, X 2 and X 3. As before, strong equvalence mples that the partton P {1,2,3} of these trples s a refnement of Q 3. Let P be a partton k-complex on X = X 1 X r. For [k], the cells of P {} are called clusters (so each cluster s a subset of some X ). We say that P s vertex-equtable f all clusters have the same sze. P s a-bounded f P A a for every A (.e. f K A (X) s dvded nto at most a cells by the partton P A ). Also, for any r-partte set Q, we wrte C Q for the set X k

5 548 P. Keevash et al. / Dscrete Mathematcs 311 (2011) of all edges lyng n the same cell of P as Q, and wrte C Q for the r-partte k-complex whose vertex set s X and whose edge set s Q Q C Q. (Snce P s a partton k-complex, C Q s ndeed a complex.) The partton k-complex P s ϵ-regular f C Q X s ϵ-regular for every r-partte Q. k Gven a partton (k 1)-complex P on X and A [r], we can defne an equvalence relaton on the edges of K k A (X), namely that S, S K A (X) are equvalent f and only f S B and S B le n the same cell of P for any strct subset B A. We refer to ths as weak equvalence. Note that f the partton complex P s a-bounded, then K A (X) s dvded nto at most a k classes by weak equvalence. If we let G be an r-partte k-graph on X, then we can use weak equvalence to refne the partton {G A, K A (X) \ G A } of K A (X) (.e. two edges of G A are n the same cell f they are weakly equvalent and smlarly for the edges not n G A ). Together wth P, ths yelds a partton k-complex whch we denote by G[P]. If G[P] s ϵ-regular then we say that G s perfectly ϵ-regular wth respect to P. Note that f G[P] s ϵ-regular then P must be ϵ-regular too. Fnally, we say that r-partte k-graphs G and H on X are ν-close f G A H A < ν K A (X) for every A [r], that s, f there k are few edges contaned n G but not n H and vce versa. We can now present the verson of the regularty lemma we shall use to splt our k-graph H nto regular k-complexes. It actually states that there s some k-graph G whch s close to H and whch s regular wth respect to some partton complex. Ths wll be suffcent for our purposes, as we shall avod the use of any edges n G \ H, so every edge used wll le n both G and H. There are varous other forms of the regularty lemma for k-graphs whch gve nformaton on H tself (the frst of these were proved n [14,4]) but these do not have the herarchy of denstes necessary for the applcaton of the blow-up lemma (see [7] for a fuller dscusson of ths pont). The verson below s due to Rödl and Schacht [12] (actually t s a very slght restatement of ther result). Theorem 3.1 (Theorem 14, [12]). Suppose ntegers n, a, r, k and reals ϵ, ν satsfy 1/n ϵ 1/a ν, 1/r, 1/k and where a!r dvdes n. Suppose also that H s an r-partte k-graph whose vertex classes X 1,..., X r form an equtable partton of ts vertex set X, where X = n. Then there s an a-bounded ϵ-regular vertex-equtable partton (k 1)-complex P on X and an r-partte k-graph G on X that s ν-close to H and perfectly ϵ-regular wth respect to P. Here (and later on) we wrte 0 < a 1 a 2 a 3 a 4 1 to mean that we can choose the constants a 1,..., a 4 from rght to left. More precsely, there are ncreasng functons f 1, f 2, f 3 such that, gven a 4, whenever we choose some a 3 f 3 (a 4 ), a 2 f 2 (a 3 ) and a 1 f 1 (a 2 ), all calculatons needed n the proof of the subsequent statement are vald. Herarches wth more constants are defned smlarly. One mportant property of regular complexes s that they reman regular when restrcted to a large subset of ther vertex set. For regular k-partte k-complexes ths property s formalsed by the followng lemma, a specal case of Lemma 6.18 n [7]. Lemma 3.2 (Restrcton of regular complexes). Suppose ϵ ϵ d c 1/k, and that G s an ϵ-regular k-partte k-complex on the vertex set X = X 1 X k such that G {} = X for each and d(g) > d. Let W be a subset of X such that W X c X for each. Then the restrcton G[W] of G to W s ϵ -regular, wth d(g[w]) > d(g)/2 and d [k] (G[W]) > d [k] (G)/ Robustly unversal complexes Apart from Theorem 3.1, the other man tool we shall use n the proof of Theorem 1.1 s the recent hypergraph blowup lemma of Keevash. Ths result nvolves not only a k-complex G, but also a k-graph M of marked edges on the same vertex set. If the par (G, M) s super-regular, then ths blow-up lemma can be appled to embed any spannng boundeddegree k-complex n G \ M, that s, wthn G but avodng any marked edges. We wll apply ths wth M = G \ H where G s the k-graph gven by Theorem 3.1. Super-regularty s a stronger noton than regularty. A result n [7] states that every ϵ-regular k-complex can be made super-regular by deletng a few of ts vertces. Unfortunately, the noton of hypergraph super-regularty s very techncal, but the followng defnton from [7] avods many of these techncaltes. Let J be a k- partte k-complex. Roughly speakng, we say that J s robustly D-unversal f the followng holds: even after the deleton of many vertces of J, the resultng complex J has the property that one can fnd n J a copy of any k-partte k-complex L whch has vertex degree at most D and whose vertex classes are the same as those of J. Condton () puts a natural restrcton on the number of vertces we are allowed to delete from the neghbourhood complex of a vertex of J and condton () states that for a few vertces u of L we can even prescrbe a target set n V(J) nto whch u wll be embedded. Defnton (Robustly unversal complexes). Suppose that J s a k-partte k-complex on V = V 1 V k wth J {} = V for each [k]. We say that J s (c, c 0 )-robustly D-unversal f whenever () V V are sets wth V c V for all [k], such that wrtng V = [k] V and J = J [V] we have J(v) = c J (v) = for any [k] and v V, () L s a k-partte k-complex of maxmum vertex degree at most D on some vertex set U = U 1 U k wth U = V for all [k],

6 P. Keevash et al. / Dscrete Mathematcs 311 (2011) () U U satsfes U U c 0 U for every [k], and sets Z u V (u) satsfy Z u c V (u) for each u U, where for each u we let (u) be such that u U (u), then J contans a copy of L, n whch for each [k] the vertces of U correspond to the vertces of V, and u corresponds to a vertex of Z u for every u U. So our use of the blow-up lemma wll be hdden through ths defnton. Of course, we shall also need to obtan robustly unversal complexes. Ths s the purpose of the next theorem, whch states that gven a regular k-partte k-complex G wth suffcent densty, and a k-partte k-graph M on the same vertex set whch s small relatve to G, we can delete a small number of vertces from ther common vertex set so that G \ M s robustly unversal. It s a specal case of Theorem 6.32 n [7]. Theorem 3.3. Suppose that 1/n ϵ c 0 d d a θ d, c, 1/k, 1/D, 1/C, G s a k-partte k-complex on V = V 1 V k wth n G {} = V Cn for every [k], G s ϵ-regular wth d [k] (G) d and d(g [k] ) d a, and M G = wth M θ G =. Then we can delete at most 2θ 1/3 V vertces from each V to obtan V = V 1 V k, G = G[V ] and M = M[V ] such that () d(g ) > d and G (v) = > d G = / V for every v V, and () G \ M s (c, c 0 )-robustly D-unversal. 4. Prelmnary results In ths secton we wll collect the prelmnary results we need to prove Theorem 1.1. In order to apply Theorem 3.3, we need to know under what condtons we can fnd partcular loose paths n complete k-partte k-graphs, whch s the topc of the next subsecton Loose paths n complete graphs The problem of when we can fnd partcular loose paths n a complete k-partte k-graph can be reformulated n terms of the queston of whch strngs satsfyng certan adacency condtons can be produced from a fxed character set; the followng lemma s the result we wll need. Lemma 4.1. Let l and a 1,..., a k be ntegers such that 0 a < l/2 for all, and l = k =1 a. Then for any s, t [k] there exsts a strng of length l on alphabet x 1,..., x k such that the followng propertes hold: (1) no two consecutve characters are equal, (2) the frst character s not x s and the fnal character s not x t, (3) the number of occurrences of character x s a. Proof. Note that the condtons on l and the a mply that l 3. We wll construct the requred strng by startng wth an empty strng of l blank postons, and for each nsertng precsely a copes of character x. Ths ensures that condton (3) wll be satsfed. We shall fll the empty postons n the followng order: frst the frst poston, then the thrd, and so on through the odd-numbered postons, untl we reach ether poston l or poston l 1 (dependent on whether l s odd or even). We then fll the second poston, then the fourth, and so on untl all postons are flled. Note that f we proceed by nsertng all copes of one character, then all the copes of another character, and so forth, then condton (1) must be satsfed. Ths s because to get two consecutve copes of x, we must have nserted a copy of x at some odd poston p, then p + 2, p + 4, and so on untl reachng l or l 1, and then flled even postons 2, 4, 6,..., p 1. However, ths would mply that we had nserted at least l/2 copes of character x, contradctng the fact that a < l/2. We therefore only need to determne an order to nsert the dfferent characters so as to satsfy (2). We frst consder the case s t, say s = 1 and t = 2. In ths case we nsert x 2 frst, x 1 last, and the remanng character blocks n any order n between. Clearly ths prevents the frst character from beng x 1 and the last from beng x 2, and so (2) s satsfed. Now we may assume s = t, say s = t = 1. Then f l s odd, we nsert the characters n the followng order: x 2, x 3,..., x k, x 1. Then all the copes of x 1 must be n even postons (snce a 1 < l/2), and so (2) s satsfed. Alternatvely, f l s even, we nsert frst x for some 1 wth a > 0, then x 1, and then the remanng blocks of characters n any order. (Note that these nclude at least one character other than x 1 and x snce l 3 and a < l/2 mply that at least three have a 1.) So nether the frst nor last character can be x 1, and so (2) s agan satsfed. The next lemma s the result we were amng for n ths secton, gvng nformaton about whch loose paths can be found n complete k-partte k-graphs. Note that the maxmum vertex degree of a loose path s two, and so ths lemma wll tell us when we can fnd a loose path n a robustly unversal k-complex. Lemma 4.2. Let G be a complete k-partte k-graph on the vertex set V 1 V k. Let b 1,..., b k be ntegers wth 0 b V for each. Suppose that n := 1 (( k =1 b ) 1) s an nteger, and n b n for all. Then for any s, t [k], there exsts a loose path n G wth an ntal vertex n V s, a fnal vertex n V t, and contanng b vertces from V for each [k].

7 550 P. Keevash et al. / Dscrete Mathematcs 311 (2011) Proof. Note frst that n s the number of edges such a path must contan. Let a = n b for each, so that 0 a < (n 1)/2. By Lemma 4.1 we can fnd a strng S of length n 1 on the alphabet V 1, V 2,..., V k such that V appears a tmes, no two consecutve characters are dentcal, the frst character s not V s and the fnal character s not V t. Let S be the th character of S. To construct a loose path P n G, frst choose any vertex from V s to be the ntal vertex of P, and any vertex from V t to be the fnal vertex of P. We also use S to choose the lnk vertces of P: choose the th lnk vertex (.e. the vertex lyng n the ntersecton of the th and ( + 1)th edges of P) to be any member of S not yet chosen. We have now assgned two vertces to each edge of P. Fnally, we complete P by assgnng to each edge one as yet unchosen vertex from each of the k 2 classes not yet represented n that edge. Ths s possble snce precsely a lnk vertces are from the class V and so the total number of vertces used from V s n a = b. Snce G s complete we know that each edge of P s an edge of G, and so P s a loose path satsfyng all the condtons of the lemma Walks and connectedness n k-graphs A walk W n a hypergraph H conssts of a sequence of edges e 1,..., e l of H and a sequence x 0,..., x l of (not necessarly dstnct) vertces of H, satsfyng x 1 x for all [l], and also x 0 e 1, x l e l and x e e +1 for all [l 1]. The length of W s the number of ts edges. We say that x 0 s the ntal vertex of W, x l s the fnal vertex of W, and that x 1,..., x l 1 are the lnk vertces of W. By a walk from x to y we mean a walk wth ntal vertex x and fnal vertex y. Note that the vertces of a hypergraph H can be parttoned usng the equvalence relaton, where x y f and only f ether x = y or there exsts a walk from x to y. We call the equvalence classes of ths relaton components of H. We say that H s connected f t has precsely one component. Observe that all vertces of an edge of H must le n the same component. Fnally, note that f H s a connected hypergraph of order n, then for any two vertces x, y of H we can fnd a walk from x to y of length at most n n H Random splttng In ths secton we shall obtan, wth hgh probablty, a lower bound on the densty of a subgraph of a k-partte k-graph chosen unformly at random. We wll use Azuma s nequalty on the devaton of a martngale from ts mean. Lemma 4.3 (Azuma [1]). Suppose Z 0,..., Z m s a martngale,.e. a sequence of random varables satsfyng E(Z +1 Z 0,..., Z ) = Z, and that Z Z 1 c for some constants c and all [m]. Then for any t 0, P( Z m Z 0 t) 2 exp t2 m 2 c 2 =1. Lemma 4.4. Suppose 1/n c, β, 1/k, 1/b < 1, and that H s a k-partte k-graph on the vertex set X = X 1 X k, where n X bn for each [k]. Suppose also that H has densty d(h) c and that for each we have β X t X. If we choose a subset W X wth W = t unformly at random and ndependently for each, and let W = W 1 W k, then the probablty that H[W] has densty d(h[w]) > c/2 s at least 1 1/n 2. Moreover, the same holds f we choose W by ncludng each vertex of X ndependently wth probablty t / X. Proof. Let m = X. To prove the frst asserton, we obtan our subsets W X through the followng two-stage random process, ndependently for each. Frst we assgn the vertces of each X nto sets X 1 and X 2 ndependently at random, wth each vertex beng assgned to X 1 wth probablty t / X, and assgned to X 2 otherwse. Then, n the (hghly probable) event that we have X 1 t we shall select unformly at random a set of vertces to transfer between X 1 and X 2 to obtan from X 1 the set W wth W = t. For each, no subset W X of sze t s more lkely to result from ths process than any other, so we have chosen each W unformly at random. It remans to show that H[W] s lkely to have hgh densty. We do ths by notng that H[X 1 ] s lkely to have hgh densty (where X 1 = X 1 1 X 1 k ) and that wth hgh probablty we wll only need to transfer a small number of vertces to form W = W 1 W k, whch can have only a lmted effect on the densty. More precsely, let x 1,..., x m be an orderng of the vertces of X, and for each [m] let the random varable Y take the value 1 f x X 1, and 0 otherwse. Recall that we wrte H to denote the number of edges of a k-graph H. For all = 0,..., m we now defne random varables Z by Z = E( H[X 1 ] Y 1,..., Y ). Then the sequence Z 0,..., Z m s a martngale, Z m = H[X 1 ], and as we formed each X 1 by assgnng vertces of X ndependently at random nto X 1 and X 2, we have Z 0 = E( H[X 1 ] ) k c =1 t. Also, for any vertex x, let f () be such that x X f () (.e. f () s the ndex of x ). Then Z Z 1 f () X (bn) for all [m]. Thus we can apply Lemma 4.3 to obtan k c t c 2 k t 2 P Z =1 m Z exp =1 32mb 2k 2 n 2k 2 1 n. 3

8 P. Keevash et al. / Dscrete Mathematcs 311 (2011) Fg. 2. The 3-graph A 3 (only edges nvolvng U 1 are shown). Therefore the event that d(h[x 1 ]) > 3c/4 has probablty at least 1 1/n 3. Also, by a standard Chernoff bound, for each [k] the event that X 1 = t ± X 2/3 has probablty at least 1 1/n 3. Thus wth probablty at least 1 1/n 2 all of these events wll happen. Now, f X 1 > t, we choose a set of X 1 t vertces of X 1 unformly at random and move these vertces from X Smlarly, f X vertces of X unformly at random and move these to X 2 < t, then we choose a set of t X 1 vertces to X 1. In ether case, for any ths acton can decrease d(h[x 1 ]) by at most X 1 t / X 1 c. Thus f we let W be the set obtaned from X 1 n ths way, we have d(h[w]) > c/2, provng the frst part of the lemma. The proof of the moreover part s the same except that we can omt the transfer step at the end of the proof Decomposton of G nto copes of A k Let A k denote the k-graph whose vertex set V(A k ) s the unon of 2k 2 dsont sets U 0, U 1, U 2,..., U 2k 3 of sze k 1 and whose edges consst of all k-tuples of the form U {x}, wth > 0 and x U 0 (see Fg. 2). So V(A k ) = 2(k 1) 2. An A k -packng n a k-graph G s a collecton of parwse vertex-dsont copes of A k n G. most θm sets S [m] 1 2() Lemma 4.5. Suppose 1/m θ ψ 1/k, and that G s a k-graph on [m] such that N G (S) > (. Then G has an A k -packng whch covers more than (1 ψ)m vertces of G. + θ)m for all but at Proof. Let A 1,..., A t be an A k -packng of G of maxmum sze, so t m/(2(k 1) 2 ). Let X be the setof uncovered X vertces, and suppose that X > ψm. Let b = θ X. Our frst am s to choose dsont sets S 1,..., S b n so that N G (S ) > (1/(2())+θ)m and N G (S ) X < θm/2 for all [b]. Note that θ ψ X 2b() mples that θm. So we can greedly choose dsont S 1,..., S 2b such that N G (S ) > (1/(2(k 1)) + θ)m for all [2b]. Let X T = { [2b] : N G (S ) X θm/2}. We clam that T b. Otherwse, consder the bpartte graph B wth vertex classes T and X, where we on T to x X f S {x} s an edge of G. Note that B cannot contan a complete bpartte graph wth 2k 3 vertces n T and k 1 vertces n X, as ths would correspond to a copy of A k contaned n X, whch s mpossble as A 1,..., A t s a maxmum sze A k -packng. However, by defnton of T we have d B () θm/2 for every T, and double-countng pars (, P) wth T and P θm/2 T k 1 X #{(, P)} < (2k 3), k 1 N B () gves a contradcton. Ths proves the clam, and by relabellng the S we can assume that N G (S ) > (1/(2(k 1)) + θ)m and N G (S ) X < θm/2 for all [b]. Now we show how to enlarge the A k -packng A 1,..., A t. For [b] let F = { [t] : N G (S ) V(A ) k}. Snce V(A ) = 2(k 1) 2 for each [b] we have 1 2(k 1) + θ t m < N G (S ) \ X = N G (S ) V(A ) 2 =1 F 2(k 1) 2 + (t F ) (k 1) < 2(k 1) 2 F + (k 1)m 2(k 1) 2,

9 552 P. Keevash et al. / Dscrete Mathematcs 311 (2011) and so F > θm/(4(k 1) 2 F ). We now double-count pars (, Q ) wth [b] and Q. The number of such pars s b F θm > θψm 4() 2 > t m. k 1 k 1 k 1 =1 So we can fnd some Q and R [b] wth R > m such that Q for every r R. For each r R and [t] each q Q fx some k-set K r,q N G (S r ) V(A q ) (whch s possble by defnton of F r ). Then we can choose R R wth R = k(2k 3) so that K r,q = K r,q for all r, r R and every q Q. For each q Q we wrte K q for K r,q wth r R. We wll now use the K q to fnd k new copes of A k that only ntersect k 1 of the copes n our packng. We arbtrarly dvde R nto k sets R 1,..., R k of sze 2k 3 and label V(K q ) = {v q,1,..., v q,k } for all q Q. The new copes A 1,..., A k of A k are obtaned for each [k] by dentfyng U 1,..., U 2k 3 wth {S r : r R } and U 0 wth {v q, } q Q. Replacng the copes {A q : q Q } by A 1,..., A k we obtan a larger A k-packng. Ths contradcton completes the proof. Corollary 4.6. Lemma 4.5 stll holds f we nsst that the sub-k-graph of G nduced by the vertces covered by the A k -packng must be connected. Proof. Apply Lemma 4.5 to obtan an A k -packng A 1,..., A l n G wth m 0 := l =1 V(A ) > (1 ψ/2)m, and let A be l the sub-k-graph of G nduced by =1 V(A ). By hypothess at most θm sets S [m] have fewer than m/(2(k 1)) neghbours n G and so at most θm sets T have no neghbours n V(A). By the defnton of a component, no V(A) edges of A contan vertces from dfferent components of A. Therefore the largest component C of A must contan at least (1 m ψ)m vertces. Indeed, f not then there are at least 0 k 2 (ψm/2)/(k 1) θm V(A) sets T whch meet at least two components of A and thus have no neghbours n A, a contradcton (we can obtan such a set T by choosng k 2 vertces arbtrarly n V(A) and then choosng the fnal vertex n a dfferent component of A than the frst vertex). Thus we may take the A k -packng consstng of all those copes A of A k wth V(A ) V(C). 5. Proof of Theorem 1.1 In our proof we wll use constants that satsfy the herarchy 1 n ϵ d d a 1 a ν, 1 r θ d c φ δ η 1 k. Furthermore, for any of these constants α, we use α α α and assume that the above herarchy also extends to the addtonal constants, e.g. d c c φ Imposng structure on H F r Step 1. Applyng the regularty lemma Let H 1 be the sub-k-graph obtaned from H by removng up to a!r vertces so that V(H 1 ) s dvsble by a!r. Let T = T 1 T r be an equtable r-partton of the vertces of H 1, and let H 2 consst of all those edges of H 1 that are r- partte sets n T. Then H 2 s an r-partte k-graph wth order dvsble by a!r, and so we may apply the regularty lemma (Theorem 3.1), whch yelds an a-bounded ϵ-regular vertex-equtable partton (k 1)-complex P on T and an r-partte k-graph G on T that s ν-close to H 2 and perfectly ϵ-regular wth respect to P. Let M = G \ H 2. So any edge of G \ M s also an edge of H. Let V 1,..., V m be the clusters of P. So T = V 1 V m and G s m-partte wth vertex classes V 1 V m. Note that m ar snce P s a-bounded. Moreover, snce P s vertex-equtable, each V has the same sze. So let n 1 = V = T /m. As s usual n regularty arguments, we shall consder a reduced k-graph, whose vertces correspond to the clusters V, and whose edges ndcate that wthn the cells of P correspondng to the edge we can fnd a subcomplex to whch we can apply Theorem 3.3. For ths we would lke G to have hgh densty n these cells, and M to have low densty. Thus we defne the reduced k-graph R on [m] as follows: a k-tuple S of vertces of R corresponds to the k-partte unon S = S V of clusters. The edges of R are precsely those S [m] k for whch G[S ] has densty at least c (.e. G[S ] > c K S (S ) ) and for whch M[S ] has densty at most ν 1/2 (.e. M[S ] < ν 1/2 K S (S ) ). Now, the edges n the reduced graph are useful n the followng way. Gven an edge S R, let S = S V agan. Usng weak equvalence (defned n Secton 3.2), the cells of P nduce a partton C S,1,..., C S,m S of the edges of K S (S ). Recall that m S a k. Therefore at most c K S (S ) /3 edges of K S (S ) can le n sets C S, wth C S, c K S (S ) /(3a k ). Furthermore, M[S ] < ν 1/2 K S (S ) (as S R) and so at most ν 1/4 K S (S ) edges of K S (S ) can le n sets C S, wth M C S, ν 1/4 C S,. Together wth the fact that G[S ] > c K S (S ) ths now mples that more than c K S (S ) /2 edges of G[S ] le n sets

10 P. Keevash et al. / Dscrete Mathematcs 311 (2011) C S, wth C S, > c K S (S ) /(3a k ) and M C S, < ν 1/4 C S,. Thus there must exst such a set C S, that also satsfes G C S, > c C S, /2. Fx such a choce of C S, and denote t by C S. Let G S be the k-partte k-complex on the vertex set S consstng of G C S and the cells of P that underle C S,.e. for any edge Q G C S we have G S = (G C S ) C Q. (1) Q Q (Recall that C Q was defned n Secton 3.2.) We also defne the k-partte k-graph M S = G S M on the vertex set S. Then the followng propertes hold: (A1) G S s ϵ-regular. (A2) G S has k-th level relatve densty d [k] (G S ) d. (A3) G S has absolute densty d(g S ) d a. (A4) M S satsfes M S < 2ν 1/4 (G S ) = /c. (A5) (G S ) {} = V for any S. Indeed, (A1) follows from (1) snce G s perfectly ϵ-regular wth respect to P. To see (A2), note that (G S [k] ) = C S and so d [k] (G S ) = G S [k] / (GS [k] ) = G S C S / C S > c /2 by our choce of C S. Smlarly, (A3) follows from our choce of C S snce d(g S ) = GS [k] K S (S ) = G S C S C S ) 2 C S K S (S ) > (c > d 6a. k a (A4) holds snce (G S ) = = G C S > c C S /2 and M S M C S < ν 1/4 C S. Fnally, (A5) follows from (1) and the fact that C {v} = V for all v V Step 2. Choosng an A k -packng of R The next step n our proof s to use Corollary 4.6 to fnd an A k -packng n the reduced k-graph R. For ths we shall need an approxmate mnmum degree condton for R. Let [m] J = I : N R (I) k 1 1 2(k 1) + φ m. We shall show that J s small, that s, that almost all (k 1)-tuples of vertces of R have degree at least (1/(2(k 1)) + φ)m n R. Consder how many edges of H do not belong to G[S ] for some edge S R. (Recall that S = S V.) There are three possble reasons why an edge e H does not belong to such a restrcton: () e s not an edge of G. Ths could be because e les n H but not H 1, n H 1 but not H 2, or n H 2 but not G. There are at most a!rn edges of the frst type, at most n k /r of the second type, and at most νn k of the thrd type. () e G contans vertces from V 1,..., V k such that the restrcton of M to S = S V satsfes M[S ] ν 1/2 K S [S ], where S = { 1,..., k }. (Note that snce G and thus M s m-partte, 1,..., k are all dstnct.) Snce G and H 2 are ν-close and thus M νn k there are at most ν 1/2 n k edges of ths type. () e G contans vertces from V 1,..., V k such that the restrcton of G to S V has densty less than c. There are at most c n k edges of ths type. Therefore there are fewer than 2c n k edges of H that do not belong to the restrcton of G to S for some S R, and so we have J n 1 1 2(k 1) + η n < N H ({x : I}) I J x V, I < 2c kn k + N R (I) n k 1 2c kn k 1 + J 2(k 1) + φ mn k. 1 I J Snce n a!r mn 1 n we deduce that J n 1 (η φ)n < 2c kn k < 3c k(mn 1 ) n, and so J < φm (snce c φ η). Ths allows us to apply Corollary 4.6 (wth G = R) to obtan an A k -packng A 1,..., A t n R wth t =1 A > (1 t δ)m, such that the sub-k-graph of R nduced by =1 V(A ) s connected. For each [t], let the vertex set of A be U 0 U 1 U 2k 3, wth each U of sze k 1, so that the edge set s {U {x} : [2k 3], x U 0 } Step 3. Formng the exceptonal path Gven a sub-k-graph R of R and a cluster V, we say that V belongs to R f V(R ). Let V 0 contan the at most a!r vertces of H we removed at the start of the proof, and also the vertces n all those clusters not belongng to some copy of A k n our packng (there are at most δn of the latter). We wll ncorporate these vertces nto a path L e whch wll later form part of our loose Hamlton cycle. We also nclude n V 0 an arbtrary choce of δn 1 vertces from each V y for whch y U for some [2k 3] and some [t] (we do not modfy any of the V y for whch y U 0 ). We add up to k 3 more vertces from U 1 1 (say) to V 0 so that V 0 0 mod k 2. We delete all these vertces from the clusters they belonged to and stll wrte V y for the subcluster of a cluster V y obtaned n ths way. Ths gves V 0 5δn/2.

11 554 P. Keevash et al. / Dscrete Mathematcs 311 (2011) Fg. 3. Splttng up A n the case k = 3. Now, we shall construct a path L e n H, whch wll contan all the vertces n V 0 and avod all the clusters V y wth y U 0. Let V >0 = {V y : y U, [2k 3], [t]}. So we shall use only vertces from V 0 and V >0 n formng L e. Recall that f V(H) s not a multple of, then a loose Hamlton cycle contans a sngle par of edges whch ntersect n more than one vertex: we shall make allowance for ths here. Choose A, B V >0 satsfyng A = B =, A B 1 V(H) mod and 1 A B k 1. Now choose dstnct x 0, x 1 V >0 \ (A B) such that {x 0 } A H and {x 1 } B H (we shall see n a moment that such x 0, x 1 exst). These edges wll be the frst 2 edges of L e. To complete L e, let Z 1,..., Z s be any partton of the vertces of V 0 nto sets of sze k 2. We proceed greedly n formng L e: for each = 1,..., s choose any x +1 V >0 \ (A B) such that Z {x, x +1 } H (where the x are all chosen to be dstnct). Let us now check that there wll always be such a vertex avalable. Indeed, every set n V(H) has at least (1/(2(k 1)) + η)n neghbours and we can choose any such neghbour whch les n V >0 and has not already been used. But V(H) \ V >0 n/(2(k 1)) + V 0 and at most V 0 + 2k 3δn vertces have been used before. Thus (snce δ η) for each choce of an x we have at least ηn/2 vertces of V >0 to choose from. Moreover, these vertces must be contaned n at least ηn/(2n 1 ) dfferent V y such that y U ( > 0). Thus we can avod choosng a vertex from any sngle V y more than 6δn 1 /η δ n 1 /2 tmes. The path L e thus formed has edges {x 0 } A, B {x 1 } and {x, x +1 } Z for all [s]. So all the vertces of V 0 are ncluded n L e. For each cluster V y, we stll denote the subset of V y lyng n V(H L e ) by V y. Then each V y wth y U 0 for some stll satsfes V y = n 1, and each V y wth y U for some > 0 satsfes (1 δ )n 1 1 δ δ n 1 (k 3) V y (1 δ)n 1. (2) 2 In addton V(H) \ V(L e ) V(H) A B {x 0, x 1 } 1 mod k 1. (3) Note that L e need not be a loose path, but that even f t s not t may stll form part of a loose Hamlton cycle. Also observe that V(L e ) 6δn Step 4. Splttng our copes of A k. The next step of the proof wll be to splt the copes A 1,..., A t of A k (more precsely the clusters belongng to the A ) nto sub-k-complexes of G that we shall later use to embed spannng loose paths. Consder any A. For convenent notaton we dentfy each U n A wth [k 1] (but recall that they are dsont sets). For each y U 0 = [k 1] we have V y = n 1, and so we can partton V y unformly at random nto 2k 3 parwse dsont subsets S y,1,..., S y,2k 3, each of sze n 1 2k 3. Smlarly, gven z U = [k 1] wth [2k 3], (2) and the fact that δ η mply that we can partton V z unformly at random nto k 1 parwse dsont subsets T,z and {U,z,w } n w []\{z} so that 1 T (1 η)2n 1 2k 3,z and U 2k 3,z,w = (1 η)2n 1 for all 2k 3 w [k 1] \ {z}. Fg. 3 shows how we do ths n the case k = 3. We arrange these peces nto (k 1)(2k 3) collectons of k sets as follows: for each y U 0 and each [2k 3] we have a collecton consstng of S y,, T,y and {U },z,y z y. (3 of these collectons are llustrated n Fg. 3.) For convenent notaton we relabel these collectons as {X,1,..., X,k } wth 1 t = (k 1)(2k 3)t, where for all [t ] we have X,1 = n 1 2k 3, n 1 2k 3 X,2 (1 η)2n 1 2k 3 and X, = (1 η)2n 1 2k 3 for 3 k, (4)

12 P. Keevash et al. / Dscrete Mathematcs 311 (2011) and (1 δ )n 1 k X, (1 δ)n 1. =2 (5) ((5) follows from (2) usng the fact that all the U,z,w have equal sze.) Let X = [k] X,, so each X s a k-partte set, on whch we shall now fnd a sub-k-complex G of G that s sutable for applyng Theorem 3.3. Consder any copy A n our A k -packng. Note that for each of the (k 1)(2k 3) collectons {X,1,..., X,k } obtaned by splttng up the clusters belongng to A there s an edge S() A such that each X, les n a cluster belongng to S() (and these clusters are dstnct for each of X,1,..., X,k ). Recall that S () denotes the unon l S() V l of all the clusters belongng to S(). Let G denote the restrcton of the k-partte k-complex G S() (whch was defned n Secton 5.1.1) to X,.e. G = G S() [X ]. Let M = M G = M S() [X ]. We clam that we may choose the above collectons {X,1,..., X,k } such that d(h[x ]) c 4 for all [t ]. Indeed, snce S() R, G[S ()] has absolute densty at least c and M[S ()] has densty at most ν 1/2. Snce G \ M H and ν c ths shows that H[S ()] has densty at least c /2. Lemma 4.4 now mples that each H[X ] has densty at least c /4 wth probablty 1 1/n 2 1, and so wth non-zero probablty ths s true for all [t ]. Lemma 3.2 and propertes (A1) (A3) and (A5) mply that G s an ϵ -regular k-partte k-complex on the vertex set X, wth absolute densty d(g ) d(g S() )/2 d a, relatve densty d [k] (G ) d, and (G ) {} = X, for each. Moreover, usng ν θ c, property (A4) and the fact that d(g ) d(g S() )/2 we see that M M S() < 2ν1/4 (G S() ) = c θ (G ) =. So by Theorem 3.3 we can delete at most θ X, vertces from each X, so that f we let X, X, consst of the undeleted vertces, and let X := k =1 X,, G := G [X ] and M := M [X ], then G \ M s (c, ϵ )-robustly 2 k -unversal, d(g ) > d and G (v) = > d (G ) = / X, for every v X,. In partcular, the latter two condtons together mply that d(g (v) =) > (d ) 2 for every v X. Let X denote the set of vertces deleted from any X,, so X θ n. By deletng up to k 3 more vertces f necessary, we may assume that X s dvsble by k 2. The latter wll help us to extend L e nto a path whch contans all the vertces n X. (6) Step 5. Extendng the exceptonal path L e When extendng L e n order to ncorporate X, we shall have to remove some more vertces from some of the X,, and we wsh to do ths so that the remander satsfes () n the defnton of robust unversalty. For ths reason, we partton each X, nto two parts AX, and BX, as follows (where we wrte BX for [k] BX, ): (B1) For all, and every v X, we have (G (v)[bx ]) = 2c G (v) =. (B2) Every set of k 1 vertces of H has at least n/(4k) neghbours n, AX,. (Recall that for a (k 1)-complex F, F = denotes the (k 1)th level of F.) To see that such a partton exsts, consder a partton obtaned by assgnng each vertex to a part wth probablty 1/2 ndependently of all other vertces. (B2) s then satsfed wth hgh probablty by a standard Chernoff bound. Now consder (B1). The moreover part of Lemma 4.4 mples that wth hgh probablty we have for all, and for all v X, that d((g (v)[bx ]) =) d(g (v) =)/2. Also, a standard Chernoff bound mples that wth hgh probablty BX, X, /3 for all [k]. Thus (G (v)[bx ]) = = d((g (v)[bx ]) =) BX, d(g (v) =) 2 X, 3 2c G (v) =. Now, we shall extend our path L e to nclude the vertces n X, usng only vertces from, AX,. We proceed smlarly to when constructng L e. So we splt X nto sets Z 1,..., Z s of sze k 2 (so s θ n). Lettng x 0 be a fnal vertex of L e, for [s ], we successvely choose x to be a neghbour of the (k 1)-tuple Z {x 1 } contaned n some AX, and not already ncluded n L e, and extend L e by the edge Z {x 1, x }, contnung to denote the extended path by L e. Recall that L e orgnally contaned at most 6δn vertces. Snce X θ n, after each extenson of L e we shall have V(L e ) < ηn. So (B2) mples that for each choce of x we have at least n/(5k) sutable vertces and hence at least t /(5k) of the sets AX contan such a sutable vertex. Ths shows that we can choose the x n such a way that at most θ n 1 vertces are chosen from any sngle AX For each [t ] let X = X 1 X k be the vertces remanng after the removal from X of the at most θ n 1 vertces used n extendng L e, let G = G [X ], and let M = M [X ]. By (6) there are at least cn vertces v V(H) such that v les n some X for whch at least H[X ] /(2 X ) edges of H[X ] contan v. So we may add two further edges of H to L e (one at each end) so that the new path L e has an ntal vertex x e and a fnal vertex y e whch each le n at least H[X ] /(2 X ) edges of.