Upper bounds on the minimum size of Hamilton saturated hypergraphs

For 1 6 ` < k, an `-overlapping k-cycle is a k-uniform hypergraph in which, for some cyclic vertex ordering, every edge consists of k consecutive vertices and every two consecutive edges share exactly ` vertices. A k-uniform hypergraph H is `-Hamiltonian saturated if H does not contain an `-overlapping Hamiltonian k-cycle but every hypergraph obtained from H by adding one edge does contain such a cycle. Let sat(n, k, `) be the smallest number of edges in an `-Hamiltonian saturated k-uniform hypergraph on n vertices. In the case of graphs Clark and Entringer showed in 1983 that sat(n, 2, 1) = d 2 e. The present authors proved that for k > 3 and ` = 1, as well as for all 0.8k 6 ` 6 k− 1, sat(n, k, `) = Θ(n`). In this paper we prove two upper bounds which cover the remaining range of `. The first, quite technical one, restricted to ` > k+1 2 , implies in particular that for ` = 23k and ` = 3 4k we have sat(n, k, `) = O(n `+1). Our main result provides an upper bound sat(n, k, `) = O(n(k+`)/2) valid for all k and `. In the smallest open case we improve it further to sat(n, 4, 2) = O(n14/5).


Introduction
Given integers 1 ≤ < k, we define an -overlapping k-cycle as a k-uniform hypergraph (kgraph for short) in which, for some cyclic ordering of its vertices, every edge consists of k consecutive vertices, and every two consecutive edges (in the natural ordering of the edges induced by the ordering of the vertices) share exactly vertices.The notion of an -overlapping kpath is defined similarly, that is, with vertices ordered v 1 , . . ., v s , the edges of the path are {v 1 , . . ., v k }, {v k− +1 , . . ., v k+ }, . . ., {v s−k+1 , . . ., v s }, Note that the number of edges of anoverlapping k-cycle with s vertices is s/(k − ) (and thus, s is divisible by k − ).Similarly, it can be easily seen that the number of vertices s of an -overlapping k-path equals modulo k − .

2
for n ≥ 52.For k-graphs with k ≥ 3 the problem was first mentioned in [3,4].It seems to be quite hard to obtain such precise results as for graphs.Therefore, the emphasis has been put on the order of magnitude of sat(n, k, ).The present authors proved in [5] that for k ≥ 3 and = 1, as well as for all 0.8k ≤ ≤ k − 1, sat(n, k, ) = Θ(n ), (2) see also [6] for the case = k − 1.
The facts that (2) holds for very small and very large (with respect to k) values of and that no better lower bound is known suggest, as conjectured already in [5], that (2) holds for all 1 ≤ ≤ k − 1 and k ≥ 2.
Our first result provides an upper bound on sat(n, k, ) higher than the conjectured O(n ), but for a broader range of than in [5].
Theorem 1 For all k ≥ 3 and ≥ k+1 Of course, this bound is good only when g is small, and when g = 0 it is only by a factor of n worse than the conjectured optimum.All cases of Theorem 1 which are not covered by the result from [5], but for which g = 0, are given in the following corollary.
Corollary 2 For every k divisible by three and = 2  3 k, as well as for every k divisible by four and = 3  4 k, we have sat(n, k, ) = O(n +1 ).
In the remaining range of , that is, for 2 ≤ ≤ k/2, nothing else than the trivial upper bound sat(n, k, ) = O(n k ) and the easy lower bound ([5, Prop.2.1]) sat(n, k, ) = Ω n have been known.Our main result in this paper provides a first, non-trivial, general upper bound on sat(n, k, ).
Theorem 3 For all k ≥ 3 and .
One consequence of Theorem 3, combined with the case = k − 1 of (2), is that for all and k we have sat(n, k, ) = O n k−1 .
Theorems 1 and 3 are proved, respectively, in Sections 3 and 4.
In the smallest open case, k = 4, = 2, we improve Theorem 3 a bit by showing the following result in Section 5.
Our proofs expand and refine a general approach to this type of problems first developed in [6] and modified in [5].In short, we begin with constructing two k-graphs, H and H , such that H is not -Hamiltonian, while H ⊃ H contains some "troublemaking" edges.Then we define H as a maximal non--Hamiltonian k-graph satisfying H ⊆ H ⊆ H .It then remains to show that for every e ∈ H, H + e is -Hamiltonian, but, what is crucial, in doing so we may restrict ourselves to e ∈ H .
In [6] the constructions of H and H were based on a special partition of the vertex set, while in [5] we used blow-ups of sparse Hamiltonian saturated graphs.In this paper we return to both these ideas: we use the approach from [5] in the proof of Theorem 1, and the approach from [6] in the proofs of Theorems 3 and 4.

Preliminaries
Our proofs utilize the following special construction of a k-graph.Given a partition of the vertex set For further use, note that For i = 1, . . ., h, let Proof.By the definition of H k, , for every e ∈ C i and f ∈ C j , where i < j, we have |e∩U i | ≥ k− +1 and f ∩ U i = ∅, and so |e ∩ f | < .Moreover, for every e ∈ C i there is an switch one vertex without violating the membership in C i ), so that C i satisfies the minimality condition in the definition of an -component. 2 Since every -overlapping k-path in a k-graph H must be entirely contained in one thecomponents of H, we have the following corollary of Proposition 5.

Corollary 6
For every -overlapping k-path P in H k, there is an i ∈ {1, . . ., h} such that P ⊆ C i , or equivalently, for every edge e of P , we have min(e) = i.
We now investigate the maximum length of an -overlapping k-path in C i , i < h, which traverses through exactly x vertices of U i .Our next, purely combinatorial, result provides an easy upper bound, independent of .Given a positive integer x, let A and B be two disjoint sets, with |A| = x and |B| = ∞.Let ν(x) = max P |V (P )|, where the maximum is taken over all -overlapping paths P with A ⊂ V (P ) ⊂ A ∪ B and |e ∩ A| ≥ k − + 1 for all e ∈ P .
Proof.Suppose there is a path P with A ⊂ V (P ) ⊂ A ∪ B, |e ∩ A| ≥ k − + 1 for all e ∈ P , and |V (P )| ≥ kx + 1.Let us view V (P ) as a binary sequence, where each vertex of A is replaced by symbol a and each vertex of V (P ) ∩ B is replaced by symbol b.If there is a pair of consecutive symbols a in the sequence then, by averaging, there is a run (=a sequence of consecutive symbols) of at least that is, of at least k symbols b.But then there is an edge of P with at most k − vertices of Aa contradiction.If, on the other hand, there are no consecutive symbols a in the sequence then, again by averaging, there is a run of at least that is, of at least k − 1 symbols b (here we use the assumption x ≥ k − 2).Thus, there is a segment b • • • bab where the run of b s is of length k − 1.The first (from the left) edge of P whose leftmost end is in this run may have at most k − symbols a -a contradiction, again. 2 We also have the following lower bound on ν(x).
Proof.Let a sequence Q begin with a vertex in B and then traverse, alternately, groups of k − 1 vertices of A followed by one vertex of B until fewer than k −1 vertices of A are left.The remaining vertices of A are placed all at one end of Q.Clearly, every k-tuple of consecutive vertices of To turn Q into an -overlapping path, the number of vertices of Q must equal modulo k − .Therefore, we may be forced to drop up to k − −1 ≤ k −2 vertices of B from Q.This is possible as by our assumption on x.The obtained path has the required properties and the claimed number of vertices.2 Note that ν(x) is a nondecreasing function of x (just replace any vertex of B with a new vertex of A).Our next observation shows that it cannot increase too fast.
Proof.Consider a longest path P of length ν(x) and remove its first (from the left) s vertices, where ≤ s ≤ k and s = ν(x) mod k − .As there must be a vertex of A among the first vertices of any edge, the remaining path P satisfies x := |V (P ) ∩ A| ≤ x − 1 and, by the monotonicity of ν(x) we have Returning to the hypergraph H k, , Propositions 7-9 imply the following corollary.
Then the length of a longest path P in C i such that A ⊂ V (P ) ⊂ A ∪ B equals ν(x).Moreover, we have In addition to the basic construction H k, , the proof of Theorem 1 relies on the notion of a (hypergraph) blow-up of a graph which will be defined soon.First, however, we recall a simple fact about graphs proved in [5,Fact 2.2].For a graph G, let c(G) denote the number of components of G. Given a subset T ⊆ V (G), let G[T ] be the subgraph of G induced by T .Fact 11 ( [5]) Let k, , and ∆ be constants, and for n = 1, 2, . . ., let G n be a graph with n vertices and Given a graph G and an integer sequence a = (a 1 , . . ., a n ), the a-blow-up of G is the k-graph where K (k) (U ) is the complete k-graph on U and the sets U i are pairwise disjoint.For a subset S ⊂ V (H), let tr(S) = {i ∈ V (G) : The following immediate corollary of Fact 11 has been already noted in [5,Cor. 2.3].
Corollary 12 ( [5]) Let a 1 , . . ., a n , k, , and 3 Proof of Theorem 1 In this section we prove Theorem 1, where the construction of an -Hamiltonian saturated k-graph is based on a blow-up of a suitably chosen Hamiltonial saturated graph.
Our proof is a substantial modification of the proof of Theorem 1.1 in [5].Specifically, we have made the range of in ( 7) broader (it used to be 2k − + 1 ≤ a i ≤ 4 − 2k + 1) and, at the same time, we altered the definition of H 2 (by introducing the cores U i ).In what follows, we assume that since otherwise + 2g + 1 ≥ (k + )/2 and Theorem 1 follows from Theorem 3. We begin with a technical inequality.
Proof.The inequality in question is equivalent to To prove (5), note that, by the assumptions on , there exists some integer a ≥ 1 such that Then, by the lower bound on , Hence, by the upper bound on , we finally have It follows from Proposition 13, as in [5], that every sufficiently large integer N can be expressed as a sum for some n, where 2k (This is because the range of a i in (7) has at least two consecutive values.)Fix a large integer N which is divisible by (k − ) and let a = (a 1 , . . ., a n ), where the a i 's and n are as in (7).Note that N = Θ(n).Let G n be an n-vertex Hamiltonian saturated graph with ∆(G n ) = O(1), and let be the a-blow-up k-graph of G n (see the definition in Section 2) with Thus, by (6), It is easy to check that (4) implies that

, n, and let
Since H 2 ⊆ H k, , by (3), for every e ∈ H 2 we have, in fact, (Note that (4) implies that, indeed, g ≤ − 2, which guarantees that H 2 is nonempty.)We have the following immediate consequence of the definition of H 2 and Corollary 6.
Corollary 14 If P is a path in H 2 , then there is i ∈ {1, . . ., n} such that for every e ∈ P we have Observe also that for each e ∈ H 1 , the set tr(e) is either a vertex or an edge of G. Consequently, c(e) = 1 and the k-graphs H 1 and Lemma 15 H is not -Hamiltonian.
Proof.Suppose that H contains an -Hamiltonian k-cycle C H = (e 1 , . . ., e m ).Unlike in [5], the proof breaks only into two cases: We omit the proof in this case, as it is identical to Case 1 of the proof of Lemma 4.1 in [5] (Indeed that proof relied only on the assumption that Let Z be the set of vertices that lie between e m and e s on C H . Formally, Then e 1 ⊆ e m ∪ Z ∪ e s and, consequently, (11) This further implies that e m and e s are disjoint, but more importantly, that e 1 and e s are disjoint too (since e m and e s cannot be consecutive disjoint edges).Thus, s ≥ 3 and by ( 9).Note, however, that due to the structure of -overlapping k-paths, Therefore, by ( 13), ( 12) and ( 11 Recall that We will build an -overlapping Hamiltonian cycle C H in H + e using the Hamiltonian saturation of G n .Let (u 1 , . . ., u N ) be the vertices of V in the order as they will appear on the C H under construction.Our goal is to define this ordering so that each segment of k consecutive vertices which begins at u i , where i ≡ 1 ( mod k − ), is an edge of H + e.We will denote by e 1 the edge beginning at u 1 , by e 2 -the edge beginning at u 1+k− and so on, until the last edge e m of C H which begins at u N −k+ +1 , where m = N k− .To achieve our goal, we will first construct an -overlapping path P ⊆ H 2 + e, extending e in both directions, and using only the vertices of U x and U y , one type at each end of e.Then, we will connect the endsets of P by an -overlapping path P ⊆ H 1 , covering all the remaining vertices and, thus, creating, together with P , an -overlapping Hamiltonian cycle in H + e.The construction of P will be facilitated by tracing a Hamiltonian path in G connecting x and y.
To construct P , let e 1 := e and order the vertices of e 1 = (u 1 , . . ., u k ) so that the first r x vertices belong to U x , the last r y vertices belong to U y , and the − r y vertices immediately preceding the r y vertices of U y ∩ e 1 all belong to sets U j with j > y. (We know from (15) that there are more than enough such vertices in e 1 .)In other words, we request that The remaining vertices of e 1 are labeled arbitrarily by u rx+1 , . . ., u k− .Our plan is to extend e 1 in either direction, but only for as long as the new edges still intersect e 1 .This means that we will have in P precisely κ := l k − new edges, and thus, precisely κ(k − ) = g + new vertices on each side of e 1 , where the last equality follows from (1).Formally, we set where, recall, the edge e j begins at the vertex u 1+(j−1)(k− ) .
We request that all vertices of P to the left of e 1 belong to U x and all vertices to the right of e 1 belong to U y , that is, This is possible, since, by ( 16) and (7).

min (|U
We also request that This can be easily accommodated, as each of these sets contains precisely k− vertices from outside of e 1 .Note that P is, trivially, an -overlapping path in the complete k-graph on V .We will show that, in fact, P ⊆ H 2 + e. Suppose first that m + 1 − κ ≤ j ≤ m.Then, by the definition of x, min(e j ) = x.By our construction (see ( 17), (20), and (21)), |e j ∩ U x | ≥ k − + 1 and e j ⊇ U x .The same is true for e j with j = 2, . . ., κ + 1, if we replace x by y (see ( 18), ( 19),(20), and ( 21)).
To conclude that P ⊆ H 2 + e, it remains to show that c(e j ) ≥ g + 2 for each e j , j = 1.As, clearly, |e j \ e 1 | ≤ + g, we also have Thus e j ∈ H 2 for each e j ∈ P , j = 1.Now we will build the rest of C H using only the edges of H 1 .Recall that x and y belong to different components of tr(e) and, hence, xy ∈ G. Therefore, by the Hamiltonian saturation of G, there is a Hamiltonian path Q = (v 1 = y, v 2 , . . ., v n−1 , v n = x) from y to x in G.We connect the two -element endsets of P by an -overlapping path P = (e κ+2 , . . ., e m−κ ) in H 1 ⊆ H which, by tracing Q, "swallows" all the remaining N − |V (P )| vertices of V . Set Observe that Let us order the elements R so that all elements of U vi precede all elements of U vi+1 , for i = 1, . . ., n − 1, and denote this ordering by (u k+g+ +1 , . . ., u N −g− ).The vertex set of P is then defined as Note that for v ∈ {x, y}, by ( 7) and ( 16), Hence, every edge of P stretches over at most two sets U v and each such two sets are always indexed by adjacent vertices of G.This implies that P ⊆ H 1 .2 4 Proof of Theorem 3 In this section we prove Theorem 3, where the construction of an -Hamiltonian saturated k-graph is based on a special partition of the vertex set into q + 1 sets U 1 , . . ., U q+1 (q to be chosen), and the associated with it notion of the hypergraph H k, (U 1 , . . ., U q+1 ), introduced at the beginning of Section 2.
Recall that the function ν(x) has been defined in Section 2. Given a large integer n divisible by k − , choose integers α = Θ n 1/2 , β = Θ n 1/2 , p = Θ n 1/2 , and where g = g(k, ) is given by ( 1) and ν := ν(α), such that β ≥ kα, To see that such a choice is feasible, one may set, for instance, α = 2k 2 √ n .Recall that, by Proposition 7, α ≤ ν ≤ kα.Next, choose p = n/ν − k − 1.Then, first of all, (24) holds.Furthermore, using (23) and the estimates g ≤ k, 2p ≥ k − 3, and 4kp ≤ α among others, we can sandwich the quantity as follows: Thus, there exists an integer β, kα ≤ β ≤ (k + 3)α, which satisfies (25).Note that, in particular, by (23) and Proposition 8, where and all sets U i , i = 1, . . ., q + 1, are pairwise disjoint.We begin our construction of the required -Hamiltonian saturated k-graph H, by letting Recall from Section 2 that H 1 breaks naturally into q+1 -components, that is, Thus, every path in H 1 is entirely contained in some C i , and, by Corollary 10, for all i ≤ q − 1 such paths are no longer than kν ≤ k 2 α.On the other hand, by the definition of C i , the vertex set of every path contained in C q ∪ C q+1 must be a subset of U q ∪ U q+1 .Therefore, in view of our assumptions on β, p and α, we have the following conclusion.
Corollary 17 The length of a longest path in H 1 is O( √ n).In particular, H 1 is not -Hamiltonian.

2
Following the outline described in the Introduction, we build a k-graph H by slightly enriching H 1 , but so that it still remains non--Hamiltonian.Let and Lemma 18 H is not -Hamiltonian.
Proof.Suppose that C is an -overlapping Hamiltonian cycle in H . Let M be a maximal set of disjoint edges in C ∩ H 2 .By Corollary 17, M = ∅.Set t := |M |.Since we have t ≤ p.
From C we now extract t vertex disjoint paths, all contained in H 1 , as follows.For every e ∈ M , denote by N (e) the union of the set of vertices of e, the set of g consecutive vertices lying just before e, and the set of g consecutive vertices lying just after e (here, 'before' and 'after' refer to an arbitrarily fixed direction of traversing C).Let W = e∈M N (e).Then C[V \ W ] consists of at most t paths (we treat a nonempty set of fewer than k consecutive isolated vertices as a single trivial path).Observe that Since each obtained path P is contained in H 1 , either min(V (P )) ≤ q − 1 or V (P ) ⊆ U q ∪ U q+1 .If all t paths are of the former kind, then their total number of vertices is at most tν, and otherwise, it is at most (t − 1)ν Finally, by ( 23), (28), and (29), and using t ≤ p, we get , and where i = min(e) and t = |e ∩ U min(e) |. 2 To complete the proof of Theorem 3, it remains to show the following lemma.Since e ∈ H 1 ∪ H 3 , we have |U x ∩ e| ≤ k − and x < y ≤ q − 2k − 1.
Our ultimate goal is to construct in H an -overlapping Hamiltonian cycle C.Recalling (26), let J = {j 1 , . . ., j p−2 } be the set of the p − 2 smallest indices in the set {1, . . ., q − 2k − 1} \ {x, y}.Further, let Since e ∈ H 2 , we have Furthermore, let us also put aside a set Q = A q ∪ A q of 2(g + 1) elements of U q \ e, where The vertices in B and Q will be used later in our construction.
First, however, we construct p vertex disjoint paths P j1 , . . ., P jp−2 , P xy and P q .Together, these p paths will contain all elements of V , except for some k − + g + 1 vertices of U x , the same number of vertices of U y , twice as many vertices of each U j , j ∈ J, and except for the vertices in B ∪ Q.Using these exceptional vertices, the paths will be connected by p 'bridges', made mostly of the edges of H 2 , to form an -overlapping Hamiltonian cycle C in H.
Construction of P xy .Order the vertices of e so that the set e ∩ U x constitutes the leftmost segment of e, while the rightmost vertex of e belongs to U y .Next, we will extend e in both directions (see Fig. 1).Let A x be a set of arbitrary k − + g vertices of U x \ e and A y be a set of arbitrary k − + g vertices of U y \ e (the reader should not worry, we will later construct sets A x and A y too).Let Further, for each z ∈ {x, y}, let P z ⊆ C z be a path containing precisely vertices of U z \ (e ∪ A x ∪ A y ) and ν(α z ) − α z vertices of R, where V (P x ) ∩ V (P y ) = ∅.Since, by Proposition 7, each of P x and P y requires no more than (k − 1)α vertices of R, while |R| ≥ 2kα − k, we will not run out of the vertices of R.
To finish the construction of P xy , we extend e • to the left, by adding the set A x , followed by P x , and • to the right, by adding the set A y , followed by P y . Thus, Set and observe that Fact 21 Proof.The path P xy consists, besides the edges of P x , P y , and e itself, also of a set on each side of e.These are precisely those edges of P xy which intersect the set A x ∪ A y .Thus, to prove that P xy ⊆ H 1 + e, it remains to show that each edge from A belongs to H 1 .
Let us consider an edge e intersecting A x .Obviously, min(e ) = x.Also, |e ∩ A x | ≥ k − , and so |e ∩ U x | ≥ k − .Furthermore, if |e ∩ A x | = k − then either e contains also the leftmost vertex of e (which belongs to U x ), or |e ∩ V (P x )| = .In the latter case, recall that each edge of Construction of P q .Let P q be a longest path with V (P q ) ⊂ U q \ (e ∪ Q).Clearly, at most k − − 1 vertices of U q will be left out, that is, Trivially, P q ⊂ H 1 .
Construction of P j , j ∈ J. Set and, for each j ∈ J, let P j ⊆ C j ⊆ H 1 be a path with V (P j ) ⊆ U j ∪ W which uses precisely vertices of U j \e and as many as possible vertices from W (we maintain that all paths P j , j ∈ J, are pairwise vertex-disjoint).Since i > j for every i ∈ [q + 1] \ (J ∪ {x, y}), we do have min(V (P j )) = j.Also, Split arbitrarily the set U j \ (V (P j ) ∪ e) into two sets A q and A q of equal size Next, we perform crucial calculations showing that we have, indeed, used all the vertices of W , that is, there are no vertices outside the constructed paths except for those listed in (32,34) and those put aside in B ∪ Q.
Proof.We have, by the definition of P xy , and by (31) and (33), Recall that each path P j , j ∈ J, may have the maximum length ν(α j ), and thus cover up to ν(α j ) − α j vertices of W .Therefore, to complete the proof it suffices to show that Note that for each j ∈ J ∪ {x, y} Hence, by the monotonicity of the function ν(•) and by Proposition 9, we have and it remains to show that To this end, (Since there is some margin in the above estimates, it means that not all the paths P j , j ∈ J, are of maximum length.) 2 Now comes the final stage of our construction, where we glue together the paths P j1 , . . ., P jp−2 , P q , and P xy , in this order, to form a Hamiltonian cycle C.We do it as indicated in Fig. 4, with the set A x placed at the left end of P xy , that is, next to the end of the path P x (see Fig. 4).
Clearly, every edge of p−2 i=1 P ji ∪ P xy ∪ P q belongs to H + e.As the last ingredient of our proof of Theorem 3, we now show that every other edge of C belongs to Note that each edge of C \ p−2 i=1 P ji ∪ P xy ∪ P q intersects some set A ∈ A. recall that between any two disjoint edges of C there are exactly g + t(k − ) vertices on C, for some t ≥ 0. In that case we say that the edge to the right (in some fixed ordering of C) t-follows the other edge.Let f 1 , be the edge of C which 1-follows the rightmost edge of P xy .Similarly, for i = 1, . . ., p − 2, let f i+1 be the edge of C which 1-follows the rightmost edge of P ji .Finally, let f p be the edge of C which 1-follows the rightmost edge of P q , see Fig. 4. Note that for each i = 1, . . ., p, we have B i ⊂ f i , and thus f i ∈ H 2 .Furthermore, these are the only edges of C which intersect more than one set from A.
Consider now some Recall that each edge of P ji contains at least k − + 1 vertices of U ji , and consequently there is always a vertex of U ji among any vertices of such an edge.This implies that |f ∩ U ji | ≥ k − + 1 and so, f ∈ H 1 .The same argument works for any f ∈ C intersecting some set A ∈ A.
2 Thus, we have constructed an -overlapping Hamiltonian cycle C in H + e, which completes the proof of Lemma 20, which, in turn, together with Fact 19, implies Theorem 3.
5 The smallest open case: k = 4 and = 2 In this section we prove Theorem 4. Our ultimate goal is, given large even integer n, to construct a maximally non-2-Hamiltonian 4-graph H.In doing so we refine the technique used in the proof of Theorem 3.
because p >> α.Recalling that q > 4(α−1) 3α (p − 1) and using the above claim as well as (40,43), we finally argue that 2 To complete the proof of Theorem 4, it remains to show the following lemma.
Lemma 27 For every e ∈ V 4 \ H the 4-graph H + e is 2-Hamiltonian.
Set I = [q − 1] \ {x, y}, note that p − 2 is (much) smaller than q − 3, and let J = {j 1 , . . ., j p−2 } be the set of the p − 2 smallest indices in I.We will construct p paths P j1 , . . ., P jp−2 , P xy , and P q+1 , such that for each j ∈ J, we have and V (P q+1 ) ⊂ U q+1 .Together, these paths will contain all vertices in V except some 2p vertices of U q+1 .Using these exceptional vertices, the paths will be connected by p 'bridges' made of the edges of H 2 , to form a 2-Hamiltonian cycle in H.
For the ease of notation assume that x = q − 2 and y = q − 1.Then J = [p − 2].To display the structure of each path we will use a shorthand notation j for any element of U j , j = 1, . . ., p − 2, x, y, q, q + 1.Finally, we designate by * each of the two unknown elements of e = {u 1 , u 2 , u 3 , u 4 } (other than x and y); recall that u Construction of P xy .We consider five cases with respect to the multiplicities of the vertices of V x and V y in e. Case 1.In the case when u 1 ∈ U x , u 2 ∈ U y and none of u 3 , u 4 belongs to U y , the path P xy is constructed as follows: To describe the remaining paths, let symbol w represent any element of the set Construction of P j , j = 1, . . ., p − 2. For j = 1, . . ., p − 2, we build path P j by splitting α − 4 vertices of U j into (α − 4)/3 blocks of length 3, separating them by arbitrary vertices from W and putting the remaining 4 vertices of U j at the end.In a diagram form P j = jj|jw|jj|jw| . . .|jj|jw|jj|jj.
On the other hand, the difference between the L-H-S and R-H-S of (44) is less than 4 α 3 << p, so that the surplus w-spots can be filled with some elements of U q+1 .Construction of P q+1 .The last path, P q+1 , consists of all the remaining vertices of U q+1 whose number is even, because n is even and every so far built path, as well as the set B, consists of an even number of vertices.
The constructed paths P 1 , . . ., P p−2 , P xy , and P q+1 are now connected together, in arbitrary order, by the 2-element blocks B 1 , . . ., B p .Note that each B j makes edges of H 2 with arbitrary 2-element sets from some U i , i = 1, . . ., q.This completes the construction of a 2-Hamiltonian cycle in H + e.
2 The proof of Theorem 4 follows immediately from Lemma 27 and Fact 26.

{i} ⊆ tr(e 1 )
⊆ tr(e m ) ∪ tr(Z) ∪ tr(e s ).(10) What is more, e m ∩ U i = ∅ and e s ∩ U i = ∅.Since e m ∈ H 1 and e s ∈ H 1 , by the definition of H 1 , each of tr(e m ) and tr(e s ) is either the singleton {i} or an edge of G containing vertex i.Hence, by (10), c(e 1 ) ≤ 1 + |Z|, which combined with the bound g + 2 ≤ c(e 1 ) from the definition of H 2 , yields |Z| ≥ g + 1.

Lemma 20
For every e ∈ V k \ H the k-graph H + e is -Hamiltonian.Proof.Fix e ∈ V k \ H.If e ∈ H , then, by the definition of H, H + e is -Hamiltonian.Therefore, we may assume that e ∈ H .This implies that |tr(e)| ≥ 2, since otherwise e ∈ H 1 .Define x = min(e) and y = min 2 (e).

Figure 1 :
Figure 1: Construction of P xy

Figure 2 :
Figure 2: Construction of C
In view of Lemma 25, H does exist.By Corollary 12, graph of a Hamiltonian saturated n-vertex graph G n .It means that for all e ∈ H 1 we have c(e) = 1, while, by (8), for all e ∈ H 2 we have c(e) ≤ |tr(e)| ≤ .Thus, H = H 1 ∪ H 2 ⊆ H . Finally, let H be a maximal non--Hamiltonian k-graph on V such that H ⊆ H ⊆ H . Thus, to complete the proof of Theorem 1, it remains to show the following lemma.Lemma 16 For every e ∈ H c , H + e is -Hamiltonian.Proof.By the maximality of H, H + e is -Hamiltonian for each e ∈ H \ H. Hence, we may restrict ourselves only to e ∈ (H ) c , that is, such that c(e) ≥ + 2g + 2. Let us fix one such e.Let j 1 , j 2 , . . ., j +2g , y, and x = min(e) belong to + 2g + 2 different components of G[tr(e)] and satisfy min{j 1 , j 2 , . . ., j +2g } > y > x. (15) Let r x = |e ∩ U x | and r y = |e ∩ U y |.Note that, since |tr(e)| ≥ c(e) ≥ + 2g + 2, max{r x , r y } ≤ max 1≤i≤n Before we finalize our construction, we need one more piece of notation.For each e ∈ V Proof.By the definitions of H and H ,|H| ≤ |H | ≤ |H 1 | + |H 2 | + |H 3 |.Now, noticing that max 1≤i≤q+1 |U i | = β, we have which is a contradiction.Hence, there is no -overlapping Hamiltonian cycle in H .2 3 = e ∈ V k : |tr(e)| ≥ 2 and min 2 (e) ≥ q − 2k , H = H 1 ∪ H 2 ∪ H 3 ,and let H be a maximal non--Hamiltonian k-graph such that H ⊆ H ⊆ H .By Lemma 18, such a k-graph H exists.
which is a contradiction.Hence, no 2-overlapping Hamiltonian cycle exists in H 1 ∪ H 2 . 2 be the same as in the proof of Theorem 3. Finally, let H = H 1 ∪ H 2 ∪ H 3 and let H be a maximal non-2-Hamiltonian hypergraph such that H ⊆ H ⊆ H .By Lemma 25, such a 4-graph exists.
Proof.By the definitions of H and H ,