Closing gaps in problems related to Hamilton cycles in random graphs and hypergraphs

We show how to adjust a very nice coupling argument due to McDiarmid in order to prove/reprove in a novel way results concerning Hamilton cycles in various models of random graph and hypergraphs. In particular, we firstly show that for $k\geq 3$, if $pn^{k-1}/\log n$ tends to infinity, then a random $k$-uniform hypergraph on $n$ vertices, with edge probability $p$, with high probability (w.h.p.) contains a loose Hamilton cycle, provided that $(k-1)|n$. This generalizes results of Frieze, Dudek and Frieze, and reproves a result of Dudek, Frieze, Loh and Speiss. Secondly, we show that there exists $K>0$ such for every $p\geq (K\log n)/n$ the following holds: Let $G_{n,p}$ be a random graph on $n$ vertices with edge probability $p$, and suppose that its edges are being colored with $n$ colors uniformly at random. Then, w.h.p\ the resulting graph contains a Hamilton cycle with for which all the colors appear (a rainbow Hamilton cycle). Lastly, we show that for $p=(1+o(1))(\log n)/n$, if we randomly color the edge set of a random directed graph $D_{n,p}$ with $(1+o(1))n$ colors, then w.h.p.\ one can find a rainbow Hamilton cycle where all the edges are directed in the same way.


Introduction
In this paper we show how to adjust a very nice coupling argument due to McDiarmid [7] in order to prove/reprove problems related to the existence of Hamilton cycles in various random grpahs/hypergraphs models. The first problem we consider is related to the existence of a loose Hamilton cycle in a random k-uniform Hypergraph.
A k-uniform hypergraph is a pair H = (V, E), where V is the set of vertices and E ⊆ n k is the set of edges. In the special case where k = 2 we simply refer to it as a graph and denote it by G = (V, E). The random k-uniform hypergraph H (k) n,p is defined by adding each possible edge with probability p independently at random, where for the case k = 2 we denote it by G n,p (the usual binomial random graph). We define a loose Hamilton cycle as a cyclic ordering of V for which the edges consist of k consecutive vertices, and for each two consecutive edges e i and e i+1 we have |e i ∩ e i+1 | = 1 (where we consider n + 1 = 1). It is easy to verify that if n is not divisible by k − 1 then such a cycle cannot exist.
Frieze [4] and Dudek and Frieze [2] showed that for p = ω (log n/n), the random k-uniform hypergraph H (k) n,p w.h.p. (with high probability) contains a loose Hamilton cycle in H (k) n,p whenever 2(k − 1)|n. Formally, they showed: Pr H (k) n,p contains a loose Hamilton cycle = 1.
The assumption 2(k − 1)|n is clearly artificial, and indeed, in [3] Dudek, Frieze, Loh and Speiss removed it and showed analog statement to 1.1 where there the only restriction on n is to be divisible by k − 1 (which is optimal).
As a first result in this paper, we give a very short proof for the result of Dudek, Frieze, Loh and Speiss while weakening (a) a bit. Formally, we prove the following theorem: (b) Suppose that k ≥ 4 and that pn k−1 / log n tends to infinity. Then Pr H (k) n,p contains a loose Hamilton cycle = 1.
Another problem we handle with is the problem of finding a rainbow Hamilton cycle in a randomly edge-colored random graph. For an integer c, let us denote by G c n,p the random graph G n,p , where each of its edges is being colored, uniformly at random with a color from [c]. A Hamilton cycle in G c n,p is called rainbow if all its edges receive distinct colors. Clearly, a rainbow Hamilton cycle can not exists whenever c < n. Bal and Frieze [1] showed that for some constant K > 0, if p ≥ (K log n)/n, the G n (n, p) w.h.p. contains a rainbow Hamilton cycle, provided that n is even. For the odd case, they proved similar statement but for p = ω((log n)/n). We overcome this and show the following:

contains a rainbow Hamilton cycle.
It is well known (see e.g. [6]) that a Hamilton cycle appear (w.h.p.) in G n,p for p ≈ (log n)/n. Therefore, one would expect to prove an analog for Theorem 1.3 in this range of p. However, it is easy to see that in this range, while randomly color the edges of G n,p with n colors, w.h.p. not all the colors appear. Frieze and Loh [5] proved that for p = (1 + ε)(log n)/n and for c = n + Θ(n/ log log n), a graph G c n,p w.h.p. contains a rainbow Hamilton cycle. It is thus natural to consider the same problem for a randomly edge-colored directed random graph, denoted by D c n,p (we allow edges to go in both directions). Note that in directed graphs we require to have a directed Hamilton cycle, which is a Hamilton cycle with all arcs pointing to the same direction.
The following theorem will follow quite immediately: Theorem 1.4. Let p = (1 + ε)(log n)/n and let c = n + Θ(n/ log log n). Then D c n,p w.h.p. contains a rainbow Hamilton cycle.
Our proof is based on a very nice coupling argument due to McDiarmid [7] and on Theorem 1.1.

Auxiliary results
In this section we present some variants of a very nice argument by McDiarmid [7]. For the convenient of the reader we add a proof for one of them, and the rest will be left as easy exercises. Before stating our lemmas, let us define the directed random k-uniform hypergraph D (k) n,p in the following way. Each ordered k-tuple (x 1 , . . . , x k ) consisting of k distinct elements of [n] appears as an arc with probability p, independently at random. In the special case where k = 2 we simply write D n,p . A directed loose Hamilton cycle is a loose Hamilton cycle where consecutive vertices are now arcs of D (k) n,p and the last vertex of every are is the first of the consecutive one. In the following lemma we show that the probability for D  To this end, assume we exposed all arcs but those coming from e i . There are three possible scenarios: Note that in (a) and (b) there is nothing to prove. In case (c), the probability for Γ i−1 to have a directed loose Hamilton cycle is p, where the probability for Γ i to have such a cycle is at least p. This completes the proof of the lemma.
In the second lemma, we show that given an integer c, one can lower bound the probability of D c n,p to have a rainbow directed Hamilton cycle by the probability of G c n,p to have such a cycle. Note that by combining the result of Bal and Frieze [1] with Lemma 2.2 we immediately obtain the following corollary: There exists a constant K > 0 such that for every p ≥ (K log n)/n we have Pr D n n,p contains a rainbow Hamilton cycle = 1, provided that n is even.

Proofs of our main results
In this section we prove Theorems 1.2, 1.3 and 1.4. We start with proving Theorem 1.2.
Proof of Theorem 1.2: Suppose that (k − 1)|n and that 2(k − 1) does not divide n. Let f 2 (n) be a function that tends arbitrarily slowly to infinity and suppose that p = f 2 (n) log n n k−1 . Note that by deleting the orderings of a D (k) n,q , using a similar argument as a multi-round exposure (we refer the reader to [6] for more details), we obtain a H as an undirected hypergraph such that for every e ∈ n k there are k! independent trials to decide whether to add it). Now, let us choose q in such a way that (1−p/2)(1−q) k!f (n) = 1−p, and observe that q ≥ p 2k!f (n) = ω log n/n k−1 . We generate H (k) n,p in a multi-round exposure and present it as a union n,q (which, as stated above, is like H (k) n,s with (1 − q) k! = 1 − s) for each 1 ≤ i ≤ f (n) (of course, ignoring the orientations). In addition, all the H i 's are considered to be independent.
Our strategy goes as follows: First, take H 0 = H (k) n,p/2 and pick an arbitrary edge e * = {x 1 . . . , x k } (trivially, H 0 contains an edge w.h.p.). Now, fix an arbitrary ordering (x 1 , . . . , x k ) of e * and let V * = ([n] \ {x 1 , . . . , x k }) ∪ {e * } (that is, V * is obtained by deleting all the elements of e * and adding an auxiliary vertex e * ). For each i ≥ 1, whenever we expose H i we define an auxiliary k-uniform directed random hypergraph D i on a vertex set V * in the following way. Every arc e of H i is being added to D i if it satisfies one of the following: • e ∩ e * = ∅, or • e ∩ e * = {x 1 }, and x 1 is not the first vertex of the arc e, or • e ∩ e * = {x k } and x k is the first vertex of the arc e.
Note that indeed, by definition, every k-tuple of V * now appear with probability p, independently at random and that |V * | = n − (k − 1). Therefore, we clearly have that each of the D i 's is an independent D (k) n−(k−1),q . Moreover, note that 2(k − 1)|n and that each directed loose Hamilton cycle of D i with the special vertex e * as a starting/ending vertex of the edges touching it corresponds to a (undirected) loose Hamilton cycle of H (k) n,p . To see the latter, suppose that e * v 2 . . . v t e * is such a cycle in D i . Now, by definition we have that both x k v 2 , . . . v k and v t−k+2 . . . v t e * are arcs of H i , and therefore, by replacing e * with its entries x 1 . . . x k , one obtains a loose Hamilton cycle in H i .
Next, by combining Theorem 1.1 with Lemma 2.1, we observe that w.h.p. D i contains a directed loose Hamilton cycle. Note that by symmetry we have that the probability for e * to be an endpoint of an edge on the Hamilton cycle is 2/k. Therefore, after exposing all the D i 's, the probability to fail in finding such a cycle is (1 − 2/k) f (n) = o(1) as desired. This completes the proof.
Next we prove Theorem 1.3.
Proof of Theorem 1.3: Let us assume that n is odd (since otherwise there is nothing to prove) and that K > is a sufficiently large constant for our needs. Now, let q be such that (1−p/2)(1−q) 2 = 1−p, and present G n n,p as a union G 1 ∪ G 2 , where G 1 is G n n,p/2 and G 2 is D n n,q (as in the proof of Theorem 1.2, by ignoring orientations one can see D n n,q as G n n,s with s satisfying (1 − q) 2 = 1 − s). Next, let e * = (x, y) be an arbitrary edge of G 1 (trivially, w.h.p. there exists an edge), let c 1 denote its color, and define an auxiliary edge-colored random directed graph D as follows. The vertex set of D is V * = ([n] \ x, y) ∪ {e * } (that is, we delete x and y and add an auxiliary vertex e * ). The arc set of D consist of all arcs uv of G 2 with colors distinct than c 1 for which one of the following holds: • {u, v} ∩ {x, y} = ∅, or • v = x, or • u = y.
A moment's thought now reveals that D is D n−1 n−1,s , where s = (1 − 1/n)q, that n − 1 is even, and that a rainbow Hamilton cycle of D corresponds to a rainbow Hamilton cycle of G n n,p . Now, since s ≥ (K ′ log n)/n for some K ′ (we can take it to be arbitrary large), it follows from Corollary 2.3 that w.h.p. D contains a rainbow Hamilton cycle, and this completes the proof.

Proof of Theorem 1.4:
The proof is an immediate corollary of the result of Frieze and Loh [5] and Lemma 2.2.