TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS

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1 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS PETER ALLEN*, JULIA BÖTTCHER*, YOSHIHARU KOHAYAKAWA, AND YURY PERSON Abstract. We give an agorithmic proof for the existence of tight Hamiton cyces in a random r-uniform hypergraph with edge probabiity p = n 1+ε for every ε > 0. This party answers a question of Dudek and Frieze [Random Structures Agorithms], who used a second moment method to show that tight Hamiton cyces exist even for p = ω(n)/n (r 3) where ω(n) arbitrary sowy, and for p = (e+o(1))/n (r 4). The method we deveop for proving our resut appies to reated probems as we. 1. Introduction The question of when the random graph G(n, p) becomes hamitonian is we understood. Pósa [19] and Korshunov [15, 16] proved that the hamitonicity threshod is og n/n, Komós and Szemerédi [14] determined an exact formua for the probabiity of the existence of a Hamiton cyce, and Boobás [4] estabished an even more powerfu hitting time resut. The first poynomia time randomised agorithms for finding Hamiton cyces in G(n, p) were deveoped by Anguin and Vaiant [1] and Shamir [21]. Finay, Boobás, Fenner and Frieze [3] gave a deterministic poynomia time agorithm whose success probabiity matches the probabiities estabished by Komós and Szemerédi. Date: May 22, * Department of Mathematics, London Schoo of Economics, Houghton Street, London WC2A 2AE, U.K. E-mai: p.d.aen j.boettcher@se.ac.uk. Instituto de Matemática e Estatística, Universidade de São Pauo, Rua do Matão 1010, São Pauo, Brazi. E-mai: yoshi@ime.usp.br. Institut für Mathematik, Freie Universität Berin, Arnimaee 3-5, D Berin, Germany. E-mai: person@math.fu-berin.de. PA was partiay supported by FAPESP (Proc. 2010/ ). JB was partiay supported by FAPESP (Proc. 2009/ ). YK was partiay supported by CNPq (308509/2007-2, /2012-4), CAPES/DAAD (415/pppprobra/po/D08/11629, 333/09) and NUMEC (Project MaCLinC/USP). YP was partiay supported by GIF grant no. I /2005. The cooperation of the authors was supported by a joint CAPES-DAAD project (415/pppprobra/po/D08/11629, Proj. no. 333/09). The authors are gratefu to NU- MEC/USP, Núceo de Modeagem Estocástica e Compexidade of the University of São Pauo, for supporting this research. 1

2 2 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON For random hypergraphs much ess is known. The random r-uniform hypergraph G (r) (n,p) on vertex set [n] is generated by incuding each hyperedge from ( ) [n] r independenty with probabiity p = p(n). First, Frieze [9] considered oose Hamiton cyces in random 3-uniform hypergraphs. The oose r-uniform cyce on vertex set [n] has edges {i+ 1,...,i + r} for exacty a i = k(r 1) with k N and (r 1) n, where we cacuate moduo n. Frieze showed that the threshod for a oose Hamiton cyce in G (3) (n,p) is Θ(ogn/n 2 ). Dudek and Frieze [6] extended this to r-uniform hypergraphs with r 4, where the threshod is Θ(ogn/n r 1 ). Both resuts require that n is divisibe by 2(r 1) (which was recenty removed by Dudek, Frieze, Loh and Speiss [7]) and rey on the deep Johansson-Kahn-Vu theorem [13], which makes their proofs non-constructive. Tight Hamiton cyces, on the other hand, were first considered in connection with packings. The tight r-uniform cyce on vertex set [n] has edges {i+1,...,i+r} for a i cacuated moduo n. Frieze, Kriveevich and Loh [11] proved that if p (og 21 n/n) 1/16 and 4 divides n then most edges of G (3) (n,p) can be covered by edge disjoint tight Hamiton cyces. Further packing resuts were obtained by Frieze and Kriveevich [10] and by Ba and Frieze [2], but the probabiity range is far from best possibe. Subsequenty, Dudek and Frieze [5] used a second moment argument to show that the threshod for a tight Hamiton cyce in G (r) (n,p) is sharp and equas e/n for each r 4 and for r = 3 they showed that G (3) (n,p) contains a tight Hamiton cyce when p = ω(n)/n for any ω(n) that goes to infinity. Since their method is non-constructive they asked for an agorithm to find a tight Hamiton cyce in a random hypergraph. In this paper we present a randomised agorithm for this probem if p is sighty bigger than in their resut. Theorem 1. For each integer r 3 and 0 < ε < 1/(4r) there is a randomised poynomia time agorithm which for any n 1+ε < p 1 a.a.s. finds a tight Hamiton cyce in the random r-uniform hypergraph G (r) (n,p). The probabiity referred to in Theorem 1 is with respect to the random bits used by the agorithm as we as by G (r) (n,p). The running time of the agorithm in the above theorem is poynomia in n, where the degree of the poynomia depends on ε. Organisation. We first provide some notation and a brief sketch of our proof, formuate the main emmas and prove Theorem 1 in Section 2. In Sections 3 and 4 we prove the main emmas, and in Section 5 we end with some remarks and open probems. 2. Lemmas and proof of Theorem Notation. An s-tupe (u 1,...,u s ) of vertices is an ordered set of vertices. We often denote tupes by bod symbos, and occasionay

3 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 3 aso omit the brackets and write u = u 1,...,u s. Additionay, we may aso use a tupe as a set and write for exampe, if S is a set, S u := S {u i : i [s]}. The reverse of the s-tupe u is the s-tupe (u s,...,u 1 ). In an r-uniform hypergraph G the tupe P = (u 1,...,u ) forms a tight path if {u i+1,...,u i+r } is an edge for every 0 i r. For any s [] we say that P starts with the s-tupe (u 1,...,u s ) =: v and ends with the s-tupe (u (s 1),...,u ) =: w. We aso ca v the start s-tupe of P, w the end s-tupe of P, and P a v w path. The interior of P is formed by a its vertices but its start and end (r 1)-tupes. Note that the interior of P is not empty if and ony if > 2(r 1). For a hypergraph H we define the 1-density of H to be d (1) (H) := e(h)/ ( v(h) 1 ) if v(h) > 1, and d (1) (H) := 0 if v(h) = 1. We set m (1) (H) := max{d (1) (H ): H H}. We denote the r-uniform tight cyce on vertices by C (r). Observe that m (1) (C (r) ) = /( 1) Outine of the proof. A simpe greedy strategy shows that for p = n ε 1 it is easy to find a tight path (and simiary a tight cyce) in G (r) (n,p) which covers a but at most n ε of its vertices. Incorporating these few remaining vertices is where the difficuty ies. To overcome this difficuty we appy the foowing strategy, which we ca the reservoir method. We first construct a tight path P of a inear ength in n which contains a vertex set W, caed the reservoir, such that for any W W there is a tight path on V(P) \ W whose end (r 1)-tupes are the same as that of P. In a second step we use the mentioned greedy strategy to extend P to an amost spanning tight path P, with a eftover set L. The advantage we have gained now is that we are permitted to reuse the vertices in W : we wi show that, by using a subset W of vertices from W to incorporate the vertices from L, we can extend the amost spanning tight path to a spanning tight cyce C. More precisey, we sha deete W from P (observe that, by construction of P, the hypergraph induced on V(P)\W contains a tight path with the same ends) and use precisey a vertices of W to connect the vertices of L to construct C. We remark that our method has simiarities, in spirit, with the absorbing method for proving extrema resuts for arge structures in dense hypergraphs (see, e.g., Röd, Ruciński and Szemerédi [20]). The techniques to dea with muti-round exposure in our agorithm is simiar to those used by Frieze in [8]. Moreover, a method very simiar to ours was used independenty by Kühn and Osthus [17] to find bounds on the threshod for the appearance of the square of a Hamiton cyce in a random graph.

4 4 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON 2.3. Lemmas. We sha rey on the foowing emmas. We state these emmas together with an outine of how they are used, and then give the detais of the proof of Theorem 1. Our first emma asserts that there are hypergraphs H with density arbitrariy cose to 1 which have a spanning tight path and a vertex w such that deeting w from H eaves a spanning tight path with the same start and end (r 1)-tupes. Lemma 2 (Reservoir emma). For a r 2 and 0 < ε < 1/(6r), there exist an r-uniform hypergraph H = H (r,ε) on ess than 16/ε 2 vertices, a vertex w, and two disjoint (r 1)-tupes u = (u 1,...,u r 1 ) and v = (v 1,...,v r 1 ) such that (i) m (1) (H ) 1+ε, (ii) H has a tight Hamiton u v path, and (iii) H w has a tight Hamiton u v path. We provide a proof of Lemma 2 in Section 2. We aso ca the graph H asserted by this emma the reservoir graph and the vertex w the reservoir vertex, since they wi provide us as foows with the reservoir mentioned in the outine. If we can find many disjoint copies of H in G (r) (n,p), and if we can connect these copies of H to form a tight path, then the set W of reservoir vertices w from these H -copies forms such a reservoir. In order to find many disjoint H -copies, we use the foowing standard theorem. Theorem 3 (see, e.g., [12, Theorem 4.9]). For every r-uniform hypergraph H there are constants ν > 0 and C N such that if p Cn 1/m(1) (H), then G (r) (n,p) a.a.s. contains νn vertex disjoint copies of H. For connecting the H -copies into a ong tight path P we use the next emma. Lemma 4 (Connection emma). Given r 3, 0 < ε < 1/(4r) and δ > 0, there exists η > 0 such that there is a (deterministic) poynomia time agorithm A which on inputs G = G (r) (n,p) with p = n 1+ε a.a.s. does the foowing. Let 1 k ηn, et X be any subset of [n] of size at east δn. Let u (1),...,u (k), v (1),...,v (k) be any 2k pairwise disjoint (r 1)-tupes in [n]. Then A finds in G a coection of vertex disjoint tight paths P i, 1 i k, of ength at most := (r 1)/ε + 2, such that P i is a u (i) v (i) path a of whose interior vertices are in X. We prove this emma in Section 3. In fact, we wi aso make use of this emma after extending P to a maxima tight path P in order to extend P (reusing vertices of the reservoir W ) to cover the eftover vertices L. It is for this reason that we require the emma to work with a set X which can be quite sma.

5 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS Proof of the main theorem. Our goa is to describe an agorithm which a.a.s. constructs a tight Hamiton cyce in the r-uniform random hypergraph G (r) (n,q) in five steps. For convenience we repace G (r) (n,p) in Theorem 1 by G (r) (n,q). We make use of a 5-round exposure of the random hypergraph, that is, each of the five agorithm steps wi individuay a.a.s. succeed on an r-uniform random hypergraph with edge probabiity somewhat smaer than q. Observe, however, that for the agorithm the input graph is given at once, and not as the union of five graphs. Therefore, in a preprocessing step the agorithm wi first spit the (random) input hypergraph into five (independent random) hypergraphs. The ony probabiistic component of our agorithm is in the preprocessing step. Our five agorithm steps wi then be as foows. Firsty, we appy Theorem 3 in order to find cn vertex disjoint copies of the reservoir graph H from Lemma 2. Secondy, we use the connection emma, Lemma 4, to connect the H copies to a tight path P of ength c n which contains a set W of ineary many reservoir vertices. Thirdy, we greediy extend P unti we get a tight path P on n n 1 (ε /2) vertices. In the fourth and fifth step we use W and Lemma 4 to connect the remaining vertices to the path constructed so far and to cose the path into a cyce. For technica reasons it wi be convenient to assume that the edge probabiity in each of the ast four steps is exacty q = n 1+ε for some ε. We therefore spit our random input hypergraph into five independent random hypergraphs, of which the first has edge probabiity q q and the remaining four have edge probabiity q. Proof of Theorem 1. Constants: Given r 3 and 0 < ε < 1/(4r), set ε := ε/2. Suppose in the foowing that n is sufficienty arge and define q = n 1+ε. Now et q > n 1+ε be given and observe that q 5q 1 (1 q ) 5. Finay, et q (0,1] be such that (1) 1 q = (1 q )(1 q ) 4 and note that since q 1 (1 q ) 5, we have q q. Let η 1 > 0 be the constant given by Lemma 4 with input r, ε and δ = 1/2. Let H = H (r,ε /2) be the r-uniform reservoir hypergraph given by Lemma 2 and n := v(h ). Let ν > 0 be the constant given by Theorem 3 with input H. We set ( 1 (2) c := min ν ) 2n, n,η 1. Finay, et η 2 > 0 and 2 be the constants given by Lemma 4 with input r, ε and δ = c/2. Preprocessing: We sha use a randomised procedure to spit the input graph G which is distributed according to G (r) (n,q) into five hypergraphsg 1,...,G 5, suchthatg 1 isdistributedaccordingtog (r) (n,q )

6 6 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON and G 2,...,G 5 are distributed according to G (r) (n,q ), where the choice of parameters is possibe by (1). Moreover these five random hypergraphs are mutuay independent. Our randomised procedure takes a copy G of G (r) (n,q) and coours its edges as foows. It coours each edge e of G independenty with a non-empty subset c of [5] such that { q c (1 q ) 4 c (1 q )/q if 1 / c Pr(e receives coour c) = q c 1 (1 q ) 5 c q /q if 1 c. Then we et G i be the hypergraph with those edges whose coour contains i for each i [5]. For justifying that this randomised procedure has the desired effect, et us consider the foowing second random experiment. We take five independent random hypergraphs, G 1 = G (r) (n,q ) and four copies G 2,...,G 5 of G (r) (n,q ), and form an r-uniform hypergraph on n vertices, whose edges are the union of G 1,...,G 5, each receiving a coour which is a subset of [5] identifying the subset of G 1,...,G 5 containing that edge. Observe that we simpy obtain G (r) (n,q), when we ignore the coours in this union. It is straightforward to check that the two experiments yied identica probabiity measures on the space of n-vertex cooured hypergraphs. It foows that any agorithm which with some probabiity finds a tight Hamiton cyce when presented with the five hypergraphs G i of the first experiment succeeds with the same probabiity when presented with five hypergraphs obtained from the second experiment. Step 1: The first main step of our agorithm finds cn vertex disjoint copies of the reservoir graph H in G 1. To this end we woud ike to appy Theorem 3, hence we need to check its preconditions. We require that q Cn 1/m(1) (H ) for some arge C. By Lemma 2 we have m (1) (H ) ε, and 1/( ε ) > 1 ε. It foows that for a sufficienty arge n we have q = n 1+ε Cn 1/m(1) (H ), and so the same hods for q since q q. By Theorem 3 and (2), a.a.s. G 1 contains at east νn n cn vertex disjoint copies of H. Hence we can agorithmicay find a subset of at east cn of them as foows. We search the vertex subsets of size n of G 1. Whenever we find a subset that induces H and does not share vertices with a previousy chosen H -copy, then we choose it. Ceary, we can do this unti we chose cn vertex disjoint copies H 1,...,H cn of H. This requires running time O ( n n ), where n 16ε 2 does not depend on n. Step2: ThesecondstepconsistsofusingG 2 andlemma4withinput r,ε and δ = 1/2 to connect the cn vertex disjoint reservoir graphs into one tight path. Let W consist of the cn reservoir vertices, one in each of H 1,...,H cn. By (2), H 1,...,H cn cover at most n/2 vertices. By

7 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 7 Lemma 4 appied with X = [n]\ ( i [cn] V(H i) ) there is a poynomia time agorithm which a.a.s. for each 1 i cn 1 finds a tight path in G 2 connecting the end (r 1)-tupe of H i with the start (r 1)-tupe of H i+1, where these tight paths are disjoint and have their interior in X. This yieds a tight path P in G 1 G 2 containing a of the H i with the foowing property. For any W W, if we remove W from P, then we obtain (using the additiona edges of the H i ) a tight path P(W) whose start and end (r 1)-tupes are the same as those of P (see Lemma 2(i)). Step 3: In the third step we use G 3 to greediy extend P to a tight path P covering a but at most n ε vertices. Let P 0 = P and do the foowing for each i 0. Let e i be the end (r 1)-tupe of P i if there is an edge e i v i in G 3 for some v i [n] \ V(P i ) then append v i to P i to obtain the tight path P i+1. If no such edge exists, then hat. Observe that in step i of this procedure, it suffices to revea the edges e i w with w [n]\p i. Hence, by the method of deferred decision, the probabiity that v i does not exist is at most (1 q ) n P i. So, as ong as P i n n ε this probabiity is at most exp( q n ε ) exp( n 1 2 ε ). We take the union bound over a (at most n) i to infer that this procedure a.a.s. indeed terminates with a tight path P with P n n ε which contains P. Step 4: Now et L be the set of vertices not covered by P. Let L be obtained from L by adding at most r 2 vertices of W, such that L is divisibe by r 1. Let Y 1,...,Y t be a partition of L into L /(r 1) tupes of size r 1. Let Y 0 be the reverse of the start (r 1)-tupe of P, and Y t+1 be the reverse of its end (r 1)-tupe. In the fourth step, we use G 4 and Lemma 4 with input r, ε and δ = 1c to find for each 0 i 1t a tight path between Y 2 2 2i and Y 2i+1 of ength at most 2 using ony vertices in W \L, such that these paths are pairwise disjoint. This is possibe since W \L 1 cn and since 2 t L n ε +r 2 impies t +1 2 n1 1 3 ε η 2 n for n sufficienty arge. LetW bethesetofateastcn ( t+1) 2 2 cn n ε 2 2cn 3 vertices in W not used in this step. Step 5: Simiary, in the fifth step, we use G 5 and Lemma 4, with input r, ε and δ = c/2, to find for each 0 i 1 (t 1) a tight path 2 between Y 2i+1 and Y 2i+2 of ength at most 2 using ony vertices in W \L, such that these paths are pairwise disjoint. Again, W \L 1 cn and t +1 η 2 2 2n for n sufficienty arge. Thus Lemma 4 guarantees that this step a.a.s. succeeds aso and the tight paths can be found in poynomia time. But now we are done: Let W be the vertices of W used in steps 4 and 5. By definition of W we can deete the vertices of W from P and obtain a tight path P (W) through the remaining vertices of P

8 8 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON (using additiona edges of the reservoir graphs) and with the same start and end (r 1)-tupes. Then P (W) together with the connections constructed in steps 4 and 5 (which incorporated a vertices of L) form a Hamiton cyce in G. Remark 5. We note that the ony non-deterministic part of the agorithm presented in the above proof concerns the partition of the edges of the input graph into five random subsets at the beginning. The agorithm in the connection emma (Lemma 4) is poynomia time, where the power of the poynomia is independent of ε. The same is (obviousy) true for the greedy procedure of step 3. Finding many vertex disjoint reservoir graphs in step 1 however, we can ony do in time n 16ε Proof of the connection emma Preiminaries. For a binomiay distributed random variabe X and a constant γ with 0 < γ 3/2 we wi use the foowing Chernoff bound, which can be found, e.g., in [12, Coroary 2.3]: (3) P ( X EX γex ) 2exp( γ 2 EX/3). In addition we appy the foowing consequence of Janson s inequaity (see for exampe [12], Theorem 2.18): Let E be a finite set and P be a famiy of non-empty subsets of E. Now consider the random experiment where each e E is chosen independenty with probabiity p and define for each P P the indicator variabe I P that each eement of P gets chosen. Set X = P P I P and = 1 2 P P,P P E(I PI P ). Then (4) P(X = 0) exp( EX). For e ( n r) we say that we expose the r-set e in G (r) (n,p), if we perform (ony) the random experiment of incuding e in G (r) with probabiity p (reca that p := n 1+ε ). If this random experiment incudes e then we say that e appears. Ceary, we can iterativey generate (a subgraph of) G (r) (n,p) by exposing r-sets, as ong as we do not expose any r-set twice. For a tupe u of at most r 1 vertices in [n] we say that we expose the r-sets at u, if we expose a r-sets e ( n r) with u e. Simiary, we expose H ( n r) if we expose a r-sets e H. In our agorithm we use the foowing structure. A fan F(u) in an r-uniform hypergraph H is a set {P 1,...,P t } of tight paths in H which a have ength either or 1, start in the same (r 1)- tupe u, and satisfy the foowing condition. For any set S of at east r/2 vertices, et {P j } j I be the coection of tight paths in which the set S appears as a consecutive interva. Then the paths {P j } j I aso coincide between u and the interva S. The tupe u is aso caed the root of F(u). Moreover, is the ength of F(u), and t its width. The set of eaves L ( F(u) ) of F(u) is the set of (r 1)-tupes u such that

9 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 9 some path in F ends in u. For intuition, observe that in the graph case r = 2, a fan is simpy a rooted tree a of whose eaves are at distance either or 1 from the root. For r 3, a fan is a more compicated structure. Idea. We sha consecutivey buid the u (i) v (i) paths P i in the setx, startingwithp 1. TheconstructionofthepathP i wecaphase i, and the strategy in this phase is as foows. We sha first expose a the hyperedges at u (i), excuding a set of used vertices U (ike those not in X, or in any u (i ) or v (i ) ). The edges {u (i),c} appearing in this process form possibe starting edges for a path connecting u (i) and v (i). For each such (one edge) path P we next consider the (r 1)-endtupe of P and expose a edges at this tupe, excuding edges that were exposed earier and used vertices (where now we count vertices in P as used). And so on. In this way we obtain a (consecutivey growing) fan F(u (i) ) with root u (i). Whie growing this fan we sha aso insist that no j- tupe of vertices with j < r is used too often. We stop when the fan has width n 1 ε/2. We wi show that with high probabiity the fan then has ony constant depth. Then we simiary construct a fan F(v (i) ) of width n 1 ε/2 with root v (i) (again avoiding used vertices and exposed edges). In a ast step, for each eaf ũ (i) of F(u (i) ) and each eaf ṽ (i) of F(v (i) ) we expose a ũ (i) ṽ (i) pathsofength 2(r 1), avoiding exposed edges. We sha show that with high probabiity at east one of these paths appears (and the fans F(u (i) ) and F(v (i) ) can be constructed), and hence we have successfuy constructed P i. We sha aso show that, in phase i we ony exposed much ess than a 1/n fraction of the r-sets in X. Hence it is pausibe that we can avoid these exposed r-sets in future phases. We note that this ast statement makes use of the fact r 3: our connection agorithm does not work for 2-graphs. Proof of Lemma 4. Setup: Given r 3, δ > 0 and 0 < ε < 1/(4r), we set (5) ξ := δ/(48r 2 ), ξ := (ξ ) r /(r 2 (r 1)!) and η = δ/(16r). Without oss of generaity we wi assume X = δn: this simpifies our cacuations. In the agorithm described beow, we maintain various auxiiary sets. We have a set U of used vertices, which contains a vertices in the sets u (i) andv (i), andinpreviousyconstructedconnectingpaths. Inphasei we maintain additionay a (non-uniform) mutihypergraph U i of used sets, which keeps track of the number of times we have so far used a vertex, or pair of vertices, et cetera, consecutivey in some path of the fan currenty under construction. Actuay, it wi greaty simpify the anaysis if any such used set can ony appear in a unique order on these paths. Hence we choose the

10 10 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON foowing setup. We arbitrariy fix an equipartition X = Y 1 Y 2r Y 1 Y 2r, and set Y := Y 1 Y 2r and Y := Y 1 Y 2r. We sha construct the fan F(u (i) ) with root u (i) in Y 1,...,Y 2r, taking successive eves of the fan from successive sets (in cycic order), and simiary F(v (i) ) in Y 1,...,Y 2r. Further, we maintain an r-uniform exposé hypergraph H, which keeps trackofther-setswhichwehaveexposed. WeetH i bethehypergraph with the edges of H at the beginning of phase i. We define hypergraphs D (1) i,...,d (r 1) i of dangerous sets for phase i as foows: (6a) (6b) D (r 1) i := D (j) i := { x ( } X r 1) : deghi (x) ξn, and { x ( } X j) : degd (j+1)(x) ξn, r 2 j 1. i We wi not use any set in any D (j) i consecutivey in a path in the fans constructed in phase i. Given two vertex-disjoint (r 1)-sets u and v, we say that the path (u,v) of ength 2r 2 is bocked by the exposé hypergraph H if any r consecutive vertices of the (2r 2)-set {u,v} is in H. When constructing the fan F ( v (i)) with root v (i), we need to ensure that not too many of its eaves are bocked by H together with too many eaves of the previousy constructed fan F ( u (i)). For this purpose we define D (j) i hypergraphs of temporariy dangerous sets in phase i as foows. We ca an (r 1)-set y in Y temporariy dangerous if there are at east ξ L ( F(u (i) ) ) eaves x of F(u (i) ) such that {x,y} is bocked by H i. We define (7a) (7b) D (r 1) i := D (j) i := { y ( ) } Y r 1 : y is temporariy dangerous, and { y ( ) } Y j : deg (y) ξ n, for r 2 j 1. D(j+1) i Summarising, we do not want to append a vertex c X \U to the end (r 1)-tupe a of a path in one of our fans, if for a or for any end (j 1)-tupe a j 1 of a with j [r 2] we have (i) {a,c} is in H, (ii) {a j 1,c} is an edge of D (j) (j) i or of D i, or (iii) {a j 1,c} has mutipicity greater than ξ r j n (r 1)/2 j(1 ε) in U i. Hence we define the set B(a) of bad vertices for a to be the set of vertices in X \U for which at east one of these conditions appies. Agorithm: The desired paths P i wi be constructed using Agorithm 1. This agorithm constructs for each i [k] two fans F(u (i) ) and F(v (i) ), using Agorithm 2 as a subroutine.

11 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 11 Agorithm 1: Connect each pair u (i), v (i) with a path P i U := i [k] {u(i),v (i) } ; H := ; foreach i [k] do 1 construct the fan F(u (i) ) in Y 1... Y 2r ; 2 construct the fan F(v (i) ) in Y 1... Y 2r ; et L := L(u (i) ) be the eaves of F(u (i) ) ; et L := L(v (i) ) be the eaves of F(v (i) ) reversed ; P := a L L -paths of ength 2r 2 not bocked by H ; 3 expose a edges which are in some P P ; if one of these paths ũ (i),ṽ (i) appears then P(u (i) ) := the path in F(u (i) ) ending with ũ (i) ; P(v (i) ) := reversa of the path in F(v (i) ) ending with ṽ (i) ; P i := P(u (i) ),ũ (i),ṽ (i),p(v (i) ) ; 4 ese hat with faiure; 5 U := U V(P i ) ; 6 foreach x L(u (i) ),y L(v (i) ) do H := H ( ) x y r ; end Agorithm 2: Construct the fan F(u (i) ) F(u (i) ) := {u (i) } ; U i := ; t := 1 ; repeat forever P := F(u (i) ) ; 7 foreach path P P do et a be the end (r 1)-tupe of P ; 8 expose a edges {a,c} with c C := Y t \(V(P) U B(a)) ; 9 C := {c: {a,c} appears in previous step} ; 10 if not δn ε /(16r) C δn ε /(2r) then hat with faiure; 11 F(u (i) ) := ( F(u (i) )\{P} ) { (P,c): c C } ; 12 a j := ast j vertices of P for j [r 2] ; 13 U i := U i C c C { {aj,c}: j [r 2] } ; 14 H := H { (a,c): c C } ; 15 if F(u (i) ) n (r 1)/2 ε/2 then return ; end t := (t mod 2r)+1 ; end It is cear that the running time (whether the agorithm succeeds or fais) is poynomia: Steps 10 and 15 guarantee that in one ca,

12 12 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON Agorithm 2 runs at most n (r 1)/2 times through its repeat oop. Our anaysis wi show that a.a.s. the agorithm indeed succeeds. Before we proceed with the anaysis, et us remind the reader that H denotes the aready exposed hyperedges that appeared so far, H i consists of the hyperedges of H before the start of phase i, U is the set of aready used vertices and U i is the auxiiary mutihypergraph which is maintained through phase i and records those j-tupes (j [r 1]) that were used for constructing the fan F(u (i) ) (F(v (i) ) resp.). Anaysis: First, we caim that the agorithm is vaid in that it does not try to expose any r-set twice. To see this, we need to check that at steps 3 and 8, we do not attempt to re-expose an aready exposed r-set. Since we do not expose any r-set in H at either step (by the definition of B(a)), it is enough to check that after either step, a exposed r-sets are added to H before the next visit to either step. This takes pace in steps 6 and 14. In order to show that the agorithm succeeds, we need to show that the foowing hod with sufficienty high probabiity for each i [k]. (A1) Agorithm 2 successfuy buids the fans F(u (i) ) and F(v (i) ), that is, the condition in step 15 eventuay becomes true, and the condition in step 10 never becomes true. (A2) If this is the case, then Agorithm 1 successfuy constructs P i, that is, one of the paths exposed in step 3 appears. (A3) If this is the case, then P i is of ength at most s = r 1, that is, ε the fans F(u (i) ) and F(v (i) ) have ength at most s/2. It is straightforward to see that (A3) hods. Indeed, if Agorithm 2 succeeds in step i, then in the ast repetition of the for-oop creating F(u (i) ), the width of F(u (i) ) finay exceeds n (r 1)/2 ε/2. Since by step 10 at most C δn ε /(2r) < n (r 1)/2 ε/2 paths are added to F(u (i) ) in this ast for-oop (and the same hods for F(v (i) )), we obtain for the width of F(u (i) ) and F(v (i) ) (which equas the number of their eaves) that (8) n (r 1)/2 ε/2 L(u (i) ), L(v (i) ) 2n (r 1)/2 ε/2. Now observe that by step 10 the fan F(u (i) ) (and simiary F(v (i) )) has width at east ( δn ε /(16r) ) s i, where s i is the ength of F(u (i) ). For s i (r 1)/(2ε) this woud impy L(u (i) ) ( δn ε /(16r) ) (r 1)/(2ε) > n (r 1)/2 ε/2, contradicting (8). Hence we have (A3). For proving (A1) and (A2), we first show bounds on various quantities during the running of the agorithm. For a set a in the mutiset U i with i [k], we write mut Ui (a) for the mutipicity of a in U i. Caim 6. If phase i and a phases before succeed, then the foowing hod throughout phase i. (a) U k ( s+2(r 1) ) 2kr/ε.

13 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 13 (b) For each j [r 1] and each j-set a U i we have mut Ui (a) ξ r j n ((r 1)/2) j(1 ε) +1. (c) For each j [r 1] and each (j 1)-set a in [n], for a but ξn vertices c X we have mut Ui ({a,c}) ξ r j n ((r 1)/2) j(1 ε). (d) e(h i+1 ) 2 2r+1 (i+1)n r 1 ε/2. (e) At step 8 in Agorithm 2, we have Y t \ (V(P) U B(a)) δn/(8r). Observe that for j r/2 Caim 6(b) impies that we aways have mut Ui (a) 1 for any j-tupe used in any F(u (i) ) or F(v (i) ). This shows that F(u (i) ) and F(v (i) ) are indeed fans, as we caim. Proof of Caim 6. We first prove (a). The set U contains the 2k(r 1) vertices of the k pairs of (r 1)-tupes which we wish to connect, together with a the vertices of the paths thus far constructed. Since by (A3) these paths are of ength at most s, it foows that U 2k(r 1)+(i 1)s k ( s+2(r 1) ). To see that (b) hods, observe that j-sets are added to U i ony at step 13, and at this point the sets added are distinct: two sets either contain different members of Y t, or they are of different sizes. Moreover, they are added ony if their mutipicity in U i is at most ξ r j n (r 1)/2 j(1 ε) by (iii) in the definition of B(a). For (c) we proceed by induction on j. First consider the case j = 1. Observe that c X is added to U i in step 13 ony if it is added at the end of a path P. Since step 10 guarantees that each fan grows by a factor of at east 2 in each iteration, we have mut Ui (c) 2 ( L(u (i) ) +L(v (i) ) ) (8) 4n (r 1)/2 ε/2 < ξ r n (r 1)/2. c X We concude that there are at most ξ r n (r 1)/2 = ξ r 1 ξn1 ε n ((r 1)/2) 1+ε vertices c X with mut Ui (c) > ξ r 1 n ((r 1)/2) 1+ε. Now assume that (c) hods for j 1 and et a be a (j 1)-set in [n]. Simiary as before, for c X the set {a,c} is in U i with mutipicity equatothenumberoftimesthatahasappearedastheendofapathp in one of the two fans constructed in this phase and the path (P,c) was subsequenty added to the fan in step 11. Since we did not previousy hat in step 10, for any P there are at most δn ε /(2r) 1 2 nε vertices c X such that (P,c) is added in this way. Thus we have (9) c X mut Ui (a,c) mut Ui (a) 1 2 nε.

14 14 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON By (b) we know in addition that mut Ui (a) ξ r j+1 n ((r 1)/2) (j 1)(1 ε) +1. Note that if this bound is ess than 2 then (9) directy impies that there are at most ξn vertices c with mut Ui (a,c) 1 and we are done. Hence we may assume mut Ui (a) 2ξ r j+1 n ((r 1)/2) (j 1)(1 ε). This together with (9) aso impies that there are at most 2ξ r j+1 n ((r 1)/2) (j 1)(1 ε) 1 2 nε ξ r j n ((r 1)/2) j(1 ε) = ξn vertices c X with mut Ui (a,c) ξ r j n ((r 1)/2) j(1 ε), as desired. For the remaining parts of the caim, we proceed by induction on the phase i [k]. So assume that the caim hods at the end of the (i 1)st phase. Wenextprove(d). Attheendofphasei, thehypergraphh contains a the r-sets which it had at the end of phase i 1, together with a those added in phase i. Now consider the construction of one fan in phase i, say of F(u (i) ). Since we did not hat in step 10, the width of the fan grows exponentiay, more than doubing at each step. Thus we can bound the tota number of iterations of the for-oop by the number L(u (i) ) of eaves of this fan (cf. step 11). In each of these iterations, we exposed Y t \ (P U B(a)) < n of the r-sets and added them to H. Hence, whie constructing F(u (i) ) (and simiary for F(v (i) )), we added at most L(u (i) ) n new r-sets to H. The ony other step where we add r-tupes to H is step 6. In this step, for each pair of eaves of F(u (i) ) and F(v (i) ), we add ( ) 2r 2 r new r-sets to H. Using the induction hypothesis we thus concude that at the end of phase i we have e(h i+1 ) e(h i )+ ( L(u (i) ) + L(v (i) ) ) n+ ( 2r 2 r (8) 2 2r+1 i n r 1 ε/2 +4n (r+1)/2 ε/2 n+ ( 2r 2 r 2 2r+1 (i+1)n r 1 ε/2, ) L(u (i) ) L(v (i) ) ) 4n r 1 ε where for the fina inequaity we use the fact that (r + 1)/2 r 1, which hods since r 3. This is the ony step in the anaysis where we use r 3, but this anaysis is reasonaby tight: the agorithm does fai for r = 2. Last we prove (e), for which we additionay proceed by induction on the number f of iterations through the for-oop of Agorithm 2 done in the ith phase so far. So we assume that the caim hods at the end of the (i 1)st phase and after f 1 iterations. Let P be the path considered in iteration f of this for-oop, and a the (r 1)-tupe ending P. We woud ike to estimate the size of B(a) Y t. Keep in mind in the foowing anaysis that for j [r 1] the

15 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 15 hypergraphd j i doesnotchangeduringphasei, bydefinition. Simiary, D j i does not change once the fan F(u i) is constructed. Now et us first assess the effect of (i) of the definition of B(a). Since a is the end of a path constructed by Agorithm 2, step 8 impies that the ast vertex b of a is not contained in B(b) where b is the end (r 1)-tupe of P b. From (ii) in the definition of B(b) we concude that a / D (r 1) i. Thus, by the definition of D (r 1) i in (6a), the number of edges in H i containing a is smaer than ξn. But how many edges {a,c} with c Y t did phase i add to H so far? By (b) the set a has mutipicity at most ξn (r 1)(2ε 1)/2 +1 < 2 in U i. It foows that since the start of phase i ony one edge containing a was added to H in step 14: the end r-tupe a r of P. However, since a r contains no vertices of Y t because the agorithm takes successive eves of the fan in successive Y t (or Y t ), we concude that the current phase did not add any additiona edges {a,c} to H with c Y t. Now et us estimate the number of vertices c Y t which (ii) of the definition of B(a) forbids. First, we need to consider the case j = 1, and show that the number of vertices in D (1) i is at most ξn. Suppose not, and observe that by definition (6b), each vertex in D (1) i extends to at east ξn pairs in D (2) i, and so on, where at the fina step each constructed member of D (r 1) i extends to at east ξn members of H i. We can construct any given member of H i in at most r! ways, so we concude that e(h i ) (ξn) r /r!, which (for sufficienty arge n) contradicts part (d). Next, again for the case j = 1, we need to show that further there are at most ξ (1) n vertices in D i. Again, suppose not: then as above this impiesthatthenumberofpairsof(r 1)-tupes(x,y)withx L ( u (i)) and y contained in Y 1... Y 2r is at east (10) ξ L ( u (i) ) (ξ n) r 1 /(r 1)! (5) r 2 ξn r 1 L ( u (i) ). However, by construction of F(u (i) ), for each j [r 1] and each x L ( u (i)), we have the property that the ast j vertices of x are not in D (j) i. We caim that this impies that the number of (r 1)-tupes y contained in Y 1... Y 2r such that (x,y) is bocked by H i, is at most (r 1) 2 ξn r 1, which is a contradiction to (10). To see this, consider the foowing property P of tupes y. For each r 1 j 1 and each 1 k r j, the tupe consisting of the ast j vertices of x foowed by the first k vertices of y is not in D (j+k) i (if j +k < r) and not in H i (if j +k = r). If y has property P, then ceary the pair (x,y) is not bocked by H i. On the other hand, if y does not have P, then there is a smaest k for which P fais. By definition of the sets D (j+k) i, for a fixed j given the first k 1 vertices of y there are at most ξn choices for the k-th vertex of y. Hence, in tota, given the first k 1 vertices of

16 16 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON y there are at most (r 1)ξn choices for the k-th vertex of y. Thus the number of (r 1)-tupes y which do not have P is at most (r 1) 2 ξn r 1 as desired. Now given 2 j r 2, et a j 1 be the set consisting of the ast j 1 vertices of a. By construction, a j 1 is in neither D (j 1) i nor in D (j 1) i. It foows from the definition of these sets in (6b) and (7b) that there are at most ξn vertices c such that {a j 1,c} D (j) i, and at most ξ (j) n such that {a j 1,c} D i. Together with the case j = 1, this gives at most (r 2)(ξ +ξ )n forbidden vertices c Y t. Finay, for (iii), observe that by part (b), ( for each j [r 2] there are at most ξn vertices c X with mut Ui {aj 1,c} ) > ξ r j n r 1 2 j(1 ε). Hence, in tota, B(a) Y t contains at most ξn+(r 2)(ξ +ξ )n+(r 2)ξn (5) δn 4r vertices. Moreover, it foows from (A3) that P r/ε, and from (a) that U 2kr/ε. Since we have Y t = δn/(2r), we concude that Y t \(P U B(a)) δn 2r r ε 2kr ε δn 4r δn 8r Now we can use a Chernoff bound to show that a.a.s. Agorithm 2 does not fai in step 10. Caim 7. At any given visit to step 10, Agorithm 2 hats with probabiity at most 2exp ( δn ε /(96r) ). Proof. By Caim 6(e), we have δn/(8r) Y t \(P U B(a)) Y t = δn/(4r). Since C is a p-random subset of Y t \(P U B(a)) with p = n 1+ε, we obtain δn ε /(8r) E C δn ε /(4r). Using the Chernoff bound (3) with γ = 1/2, we concude that δn ε /(16r) C δn ε /(2r) with probabiity at east 1 2exp ( δn ε /(96r) ). We woud ike to show that aso a.a.s. Agorithm 1 does not fai in step 4. Since the events considered in this step are not mutuay independent, we use Janson s inequaity for this purpose. Caim 8. Atanygivenvisittostep4,Agorithm1hatswithprobabiity at most exp( n (r 2)ε /4). Proof. Let E = P be the famiy of r-sets exposed in step 3 in this iteration of the foreach-oop. For each P P et I P be the indicator variabe for the event that the path P appears, which occurs with probabiity p = p r 1. Then X = P P I P is the random variabe counting the number of L L -paths appearing in this iteration. We

17 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 17 woud ike to use Janson s inequaity (4) to show that X > 0 with high probabiity, in which case Agorithm 1 does not hat in step 4. To this end we first bound EX, for which we need to estimate P. Firsty, since F(u (i) ) and F(v (i) ) are disjoint fans, no vertex is in a eaf both of F(u (i) ) and of F(v (i) ), and in particuar L(u (i) ) and L(v (i) ) are disjoint. Now et ṽ be any (r 1)-tupe in L(v (i) ). By construction D (r 1) i ṽ is not in (see step 8 and the definition of B(a)). By (7a) the path (ũ,ṽ) therefore is bocked by H for at most ξ L(u (i) ) tupes ũ L(u (i) ). Thus we have which gives P L(v (i) ) (1 ξ ) L(u (i) ) (8) (1 ξ )n r 1 ε, (11) EX = P p (1 ξ )n r 1 ε n (ε 1)(r 1) = (1 ξ )n (r 2)ε. Next we woud ike to estimate E(I P I P ) for two distinct paths P = (ũ,ṽ) and P = (ũ,ṽ ) which share at east one edge. If P and P are distinct paths sharing at east one edge, then in particuar, either ũ and ũ have the same end r/2-tupe, or ṽ and ṽ have the same start r/2-tupe. Without oss of generaity assume the former and suppose that ṽ and ṽ match in the start j-tupe, but not in the (j+1)st vertex. Ceary 1 j, and since F(u (i) ) and F(v (i) ) are fans we have ũ = ũ and j < r/2. Hence P and P share precisey an interva of ength r 1+j, and thus j edges. Therefore E(I P I P ) p 2r 2 j. In addition, the discussion above shows that for a fixed path P = (ũ,ṽ), the number N P,j of paths P = (ũ,ṽ ) such that P and P share j edges, is at most the number of choices of a eaf ṽ L(v (i) ) such that ṽ ony has the end (r 1 j)-tupe v different from ṽ, pus the number ofchoicesofaeafũ L(u (i) )suchthatũ onyhasthestart(r 1 j)- tupe u different from ũ. By Caim 6(b) the start j-tupe of ṽ and the end j-tupe of ũ have mutipicity in U i at most n (r 1)/2 j(1 ε) +1. By step 13 this impies that there are at most n (r 1)/2 j(1 ε) choices for u and for v, and hence N P,j 2n (r 1)/2 j(1 ε). With this we are ready to estimate ( ) = E(I P I P ), P P,P P = E(I P I P ) = P 1 j<r/2 P P =j where P,P P. We have N P,j p 2r 2 j P 1 j<r/2 L(u (i) ) L(v (i) ) 2n (r 1)/2 j(1 ε) p 2r 2 j, 1 j<r/2

18 18 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON which, by (8), is at most n (r 1) ε 1 j<r/2 2n (r 1)/2 j(1 ε) n (ε 1)(2r 2 j), n (r 1) ε r n 3 2 (r 1)+2ε(r 1) < 1. Hence, inequaities (4) and (11) impy that P(X = 0) exp( EX) exp( n (r 2)ε /4), and thus Agorithm 1 fais with at most this probabiity in this visit to step 4 Since Agorithm 1 visits step 4 at most k n times, we can use Caim 8 and a union bound to infer that (A2) hods with probabiity at east 1 n exp ( n (r 2)ε /4 ) 1 1 exp( δn ε /(100r) ). Simiary, 2 step 10 of Agorithm 2 is caed at most once per eaf in any of the at most 2k constructed fans, which is at most 2k 2n (r 1)/2 ε/2 n r times by (8). It foows from Caim 7 that (A1) hods with probabiity at east 1 n r 2exp ( δn ε /(96r) ) exp( δnε /(100r) ). Summarising, we showed that Agorithm 1 constructs the k desired tight paths of ength at most with probabiity at east 1 exp( δn ε /(100r) ). 4. Proof of the reservoir emma In this section we prove Lemma 2. Proof of Lemma 2. Choose := 1/ ( 2(r 1)ε ) + 2. Our strategy wi be as foows. We wi start by defining an auxiiary r-uniform hypergraph D (r) with 2(r 1)(2 1)+1 vertices and as many edges, which impies (12) d (1) (D (r) 1 ) = 1+ 2(r 1)(2 1) 1+ε. After defining D (r) we sha construct a graph H which satisfies (ii) and (iii) and is such that D (r) H and D (r) has maximum 1-density among a subhypergraphs of H. The vertex set of D (r) is V(D (r) ) := U V i [ 1] A i i [ 2] where U := (u 1,...,u r 1,w,u r,...,u 2(r 1) ), V := (v 1,...,v 2(r 1) ), A i := (a (i) 1,...a (i) 2(r 1) ) for i [ 1], and B i := (b (i) 1,...b (i) 2(r 1) ) for i [ 2] are ordered sets of vertices. The edge set of D (r) contains exacty the edges of the tight paths determined by U, by V, by A i for B i.

19 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS 19 B 1 w U A 1 A 2 V Figure 1. H for r = 3 and = 3. The vertices of D (r) are drawn bigger than the vertices newy inserted in H. The continuous ine indicates the tight Hamiton path in H from (13), the dashed ine the tight Hamiton path in H w. each i [ 1], by B i for each i [ 2], as we as by the foowing vertex sequences: and Ũ A := (u 1,...,u r 1,a (1) r 1,...,a (1) 1 ), Ṽ A := (a ( 1) 2(r 1),...,a( 1) r,v r,...,v 2(r 1)), Ũ B := (u 2(r 1),...,u r,b (1) r 1,...,b (1) 1 ), Ṽ B := (b ( 2) 2(r 1),...,b( 2) r,v r 1,...,v 1), Ã i,i+1 := (a (i) 2(r 1),...,a(i) r,a (i+1) r 1,...,a (i+1) 1 ) for a i [ 2], B i,i+1 := (b (i) 2(r 1),...,b(i) r,b (i+1) r 1,...,b (i+1) 1 ) for a i [ 3]. It is not difficut to check that D (r) has exacty 2(r 1)(2 1) + 1 vertices and edges as caimed. In order to obtain H from D (r) we first et v i := v (r 1)+i for each i [r 1]. Then we insert k := 3(r 1) 2 (2 1) new vertices between each of the foowing pairs of vertex sets in D (r) : U and A 1, A 1 and V, A i and B i for each i [ 2], B i and A i+1 for each i [ 2]. We et I(X,Y) denote the ordered set of vertices inserted between the sets X and Y in this process (where we choose any ordering). In addition, we add to this graph the tight Hamiton path (13) U,I(U,A 1 ),A 1,I(A 1,B 1 ),B 1,I(B 1,A 2 ),A 2,......,B 2,I(B 2,A 1 ),A 1,I(A 1,V),V running from u to v (which uses some edges aready present in D (r) ). The resuting hypergraph is H (see aso Figure 1).

20 20 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON By construction v(h ) = 2(r 1)(2 1) k( 1) which by definition of k is smaer than 16(r 1) ε 2, and e(h ) = 2(r 1)(2 1)+1+2(k +r 1)( 1). Since > 1 this impies (14) d (1) (H 2(r 1)( 1)+1 ) = 1+ 2(r 1)(2 1)+2k( 1) 2(r 1)( 1)+1 1 < k( 1) 2(r 1)(2 1) (12) = d (1) (D (r) ). By (13) the hypergraph H satisfies (ii). We define Ĩ(Y,X) to be the reversa of I(X,Y). It can be checked that H aso contains the tight path Ũ A,Ĩ(A 1,U),ŨB,Ĩ(B 1,A 1 ),Ã1,2,Ĩ(A 2,B 1 ), B 1,2,Ĩ(B 2,A 2 ),Ã2,3,......,Ã 2, 1,Ĩ(A 1,B 2 ),ṼB,Ĩ(V,A 1),ṼA. This is a tight path from u to v running through a vertices of H but w, and so H aso satisfies (iii). It remains to show that D (r) has maxima 1-density among a subgraphs of H. Suppose that H is a subgraph of H with maxima 1-density. It foows that H is an induced subgraph of H, and that we have d (1) (H) d (1) (D (r) ) > 1. It foows that H cannot contain any vertex of degree one (otherwise we coud deete it and increase the 1-density). In particuar, this means that if I(X,Y) is any of the sets of k vertices which form a tight path in H and which are not present in D (r), then either every vertex of I(X,Y) is in H, or none are. Simiary, by the definition of k we have k d (1) (D (r) ) > k+(r 1) and so H cannot contain any k vertices meeting ony k +r 1 edges. Accordingy H cannot contain I(X,Y). It foows that H must be a subgraph of D (r). It is straightforward to check that if any of the vertices S := {u 2,...,u 2(r 1) 1,w,v 2,...,v 2(r 1) 1, a (i) 2,...,a (i) 2(r 1) 1,b(i) 2,...,b (i) 2(r 1) 1 } of D (r) is removed from D (r), then we obtain a graph which can be decomposed by successivey removing vertices of degree at most one (i.e. it is 1-degenerate) and which therefore has 1-density at most 1. It foows that S V(H). Now et x be any vertex of D (r) which is not in H. Since x S we have deg (r) D (x) = 2, and both edges containing x have a their remaining vertices in S. Thus we have d (1)( D (r) [ ]) ( V(H) {x} min d (1) (H),2 ) and since d (1) (D (r) ) < 2, we concude that d (1) (H) d (1) (D (r) ) as required.

21 TIGHT HAMILTON CYCLES IN RANDOM HYPERGRAPHS Concuding remarks Graphs. We remark that our approach does not work (as such) in the case r = 2, even for the sub-optima edge probabiity n ε 1. For this case, in the proof of the connection emma, Lemma 4, when growing a fan we woud have to revea in each iteration of the foreach-oop in Agorithm 2 more than n 1 ε edges at a vertex a. In the construction of one fan we woud have to repeat this operation at east n (1/2) 2ε times: ony then we coud hope for the fan to have n (1/2) ε eaves, which we need in order to get a connection between two such fans at east in expectation. But then we woud have reveaed at east n 1 ε n (1/2) 2ε = n (3/2) 3ε edges to obtain a singe connection. Hence, wecannotobtainainearnumberofconnectionsinthisway, asrequired by our strategy. Vertex disjoint cyces. It is easy to modify our approach to show the foowing theorem. Theorem 9. For every integer r 3 and for every ε,δ > 0 the foowing hods. Suppose that n 1,...,n are integers, each at east 2r/ε, whose sum is at most n, and n 1 δn. Then for any n 1+ε < p = p(n) 1, the random r-uniform hypergraph G (r) (n,p) contains a coection of vertex disjoint tight cyces of engths n 1,...,n with probabiity tending to one as n tends to infinity. A proof sketch is as foows. We refer to the steps used in the proof of Theorem 1. First, we woud run step 1 as before, except that we woud find reservoir graphs covering ony at most εδn/(8r) vertices. Step 2 remains unchanged. We woud then in an extra step (requiring an extra round of probabiity) to create greediy a coection of vertex disjoint tight paths of engths sighty shorter than n 2,...,n, and another extra step using Lemma 4 to connect these paths into tight cyces of engths n 2,...,n. Here we require that the connecting paths aways have a precisey specified ength. As written, Lemma 4 does not guarantee this (the output paths have engths differing by at most two, since the paths in each fan can differ in ength by one) but it is easy to modify the emma to obtain this (we woud simpy extend each of the shorter fan paths by one vertex whie avoiding dangerous sets). The remainder of the proof can remain amost unchanged. We extend the reservoir path greediy to cover most of the remaining vertices. Then we appy Lemma 4 twice to cover a the eftover vertices and compete a cyce. Then this cyce has ength n 1 as desired. (The ony difference is that some of our constants wi need to be adapted sighty.) Again, for fixed r, ε and δ we obtain a randomised poynomia time agorithm from this proof. Note that the condition that the cyces shoud not be too short cannot be competey removed: in order to have ineary many cyces of ength g with high probabiity, we require

22 22 P. ALLEN, J. BÖTTCHER, Y. KOHAYAKAWA, AND Y. PERSON that ineary many such cyces exist in expectation. This expectation is of the order n g p g, which is in o(n) if p = o ( n (g 1)/g). Derandomisation. Our approach to Theorem 1 yieds a randomised agorithm. However we ony actuay use the power of randomness in order to preprocess our input hypergraph and simuate muti-round exposure. This motivates the foowing question. Question 10. For a constructive proof which uses muti-round exposure, how can one obtain a deterministic agorithm? Repacing the randomised preprocessing step with a deterministic spitting of the edges of the compete r-uniform hypergraph into disjoint dense quasirandom subgraphs might be a promising strategy here. Muti-Round exposure is a very common technique in probabiistic combinatorics. Hence this question might be of interest for other probems as we. Resiience. A very active recent deveopment in the theory of random graphs is the concept of resiience: under which conditions can one transfer a cassica extrema theorem to the random graph setting? Lee and Sudakov [18], improving on previous work of Sudakov and Vu [22], showed that Dirac s theorem can be transferred to random graphs amost as sparse as at the threshod for hamitonicity. More precisey, they proved that for each ε > 0, if p Cogn/n for some constant C = C(ε), then amost surey the random graph G = G(n,p) has the foowing property. Every spanning subgraph of G which has minimum degree ( 1 +ε)pn, contains a Hamiton cyce. 2 It woud be interesting to prove a corresponding resut for tight Hamiton cyces in subgraphs of random hypergraphs. It is unikey that the Second Moment Method wi provide hep for this. Our methods, however, might be robust enough to provide some assistance. 6. Acknowedgements We woud ike to thank Kas Markström for suggesting Theorem 9. References [1] D. Anguin and L. G. Vaiant, Fast probabiistic agorithms for Hamitonian circuits and matchings, J. Comput. System Sci. 18 (1979), no. 2, [2] D. Ba and A. Frieze, Packing tight Hamiton cyces in uniform hypergraphs, SIAM Journa on Discrete Mathematics 26 (2012), no. 2, [3] B. Boobás, T. I. Fenner, and A. Frieze, An agorithm for finding Hamiton paths and cyces in random graphs, Combinatorica 7 (1987), no. 4, [4] B. Boobás, The evoution of sparse graphs, Graph theory and combinatorics (Cambridge, 1983), Academic Press, London, 1984, pp [5] A. Dudek and A. Frieze, Tight Hamiton cyces in random uniform hypergraphs, Random Structures Agorithms, to appear. [6], Loose Hamiton cyces in random uniform hypergraphs, Eectron. J. Combin. 18 (2011), no. 1, Paper 48, 14.