The coupling method for inhomogeneous random intersection graphs

We present new results concerning threshold functions for a wide family of random intersection graphs. To this end we apply the coupling method used for establishing threshold functions for homogeneous random intersection graphs introduced by Karo\'nski, Scheinerman, and Singer--Cohen. In the case of inhomogeneous random intersection graphs the method has to be considerably modified and extended. By means of the altered method we are able to establish threshold functions for a general random intersection graph for such properties as $k$-connectivity, matching containment or hamiltonicity. Moreover using the new approach we manage to sharpen the best known results concerning homogeneous random intersection graph.


Introduction
The first random intersection graph model was introduced by Karoński, Scheinerman, and Singer-Cohen [11]. Since than it has been attracting attention mainly because of its wide applications, for example: "gate matrix layout" for VLSI design (see e.g. [11]), cluster analysis and classification (see e.g. [9]), analysis of complex networks (see e.g. [7, ?]), secure wireless networks (see e.g. [2]) and epidemics ( [6]). Several generalisations of the model has been proposed, mainly in order to adapt it to use in some particular purpose. In this paper we consider the G (n, m, p) model studied for example in [1,5,12]. Alternative ways of generalizing the model defined in [11] are given for example in [7] and [9].
In a random intersection graph G (n, m, p) there is a set of n vertices V = {v 1 , . . . , v n }, an auxiliary set of m = m(n) features W = {w 1 , . . . , w m(n) }, and a vector p(n) = (p 1 , . . . , p m(n) ) such that p i ∈ (0; 1), for each 1 ≤ i ≤ m. Each vertex v ∈ V adds a feature w i ∈ W to its feature set W (v) with probability p i independently of all other properties and features. Any two vertices v, v ′ ∈ V are connected by an edge in G (n, m, p) if W (v) and W (v ′ ) intersect. If p(n) = (p, . . . , p) for some p ∈ (0; 1) then G (n, m, p) is a random intersection graph defined in [11]. We denote it by G (n, m, p).
The random intersection graph model is very flexible and its properties change a lot if we alter the parameters. For example G (n, m, p) for some ranges of parameters behaves We consider monotone graph properties of random graphs. For the family G of all graphs with the vertex set V, we call A ⊆ G a property if it is closed under isomorphism. Moreover A is increasing if G ∈ A implies G ′ ∈ A for all G ′ ∈ G such that E(G) ⊆ E(G ′ ) and decreasing if G\A is increasing. Increasing properties are for example: k-connectivity, containing a perfect matching and containing a Hamilton cycle.

Remark 3.
Theorem is also valid for S 1 = Ω(n 2 ) but witĥ , for S 3 ≫ √ S 1 and ω ≪ S 3 / √ S 1 ; for S 3 = O( √ S 1 ); Denote by C k , PM and HC the following graph properties: a graph is k-connected, has a perfect matching and has a Hamilton cycle, respectively. We will use Theorem 1 to establish threshold functions for C k , PM and HC in G (n, m, p). By C k we denote here vertex connectivity. From the proof it follows that the threshold function for edge connectivity is the same as this for C k .
For any sequence c n with limit we write Theorem 2. Let max 1≤i≤m p i = o((ln n) −1 ) and S 1 and S 1,2 be given by (1).
where f (c n ) is given by (5).
(ii) Let k be a positive integer and a n = S 1,2 S 1 . If S 1 = n(ln n + (k − 1) ln ln n + c n ), then lim n→∞ Pr {G (n, m, p) ∈ C k } = 0 for c n → −∞ and a n → a ∈ (0; 1]; 1 for c n → ∞. Assumption max 1≤i≤m p i = o((ln n) −1 ) is necessary to avoid awkward cases. The problem is explained in more detail in Section 4. A straightforward corollary of the above theorem is that for S 1 = n(ln n + c n ), c n → −∞ and any k = 1, 2, . . . , n. Theorem 3. Let max 1≤i≤m p i = o((ln n) −1 ) and S 1 be given by (1). If S 1 = n(ln n + c n ) then lim where f (·) is given by (5).
In particular we may state the following extension of the result from [17].
Corollary 2. Let b n be a sequence, β and γ be constants such that βγ(1 − e −γ ) = 1. If m = βn ln n and p = γ n 1 + b n ln n then where f (·) is given by (5).
Sometimes the method of the proof enable to improve the best known results concerning G (n, m, p) even more.  One of the question posed in [14] concerned the range of m = m(n) for which the threshold function for C k for G (n, m, p) coincides with this for δ(G (n, m, p)) ≥ 1. Moreover we may ask when threshold function for C k for G (n, m, p) is the same as this for C k for G 2 (n,p) withp = mp 2 . Theorem 6 gives a final answer to these questions.
for ln 2 n ≪ m ≪ n ln n ln ln n ln n+(k−1) ln ln n+cn m , for m = Ω(n ln n); The proof is divided as follows. Section 3 describes the coupling used to establish threshold functions and presents the proof of Theorem 1. In Section 4 we give minimum degree thresholds for G (n, m, p). Section 5 is dedicated to the properties of the auxiliary random graphs used in the coupling established in Theorem 1. The proofs of the remaining theorems are presented in Section 6.

Coupling
In this section we present a proof of Theorem 1. In the proof we use auxiliary random graph models G * i (n, M), i = 2, 3, . . . , n, in which M is a random variable with nonnegative integer values. For i = 2, . . . , n, G * i (n, M) is constructed on the basis of a random hypergraph H * i (n, M). H * i (n, M) is a random hypergraph with the vertex set V in which the hyperedge set is constructed by sampling M times with repetition elements from the set of all i-element subsets of V (all sets which are chosen several times are added only once to the hyperedge set). G * i (n, M) is a graph with the vertex set V in which v, v ′ ∈ V are connected by an edge if {v, v ′ } is contained in at least one of the hyperedges of H * i (n, M). If M equals a constant t with probability one or has the Poisson distribution, we write G * i (n, t) or G * i (n, Po (·)), respectively. Recall that similarly G i (n,p) is constructed on the basis of H i (n,p) -a hypergraph with independent hyperedges.
In this paper we treat random graphs as random variables. By a coupling (G 1 , G 2 ) of two random variables G 1 and G 2 we mean a choice of a probability space on which a random vector (G ′ 1 , G ′ 2 ) is defined and G ′ 1 and G ′ 2 have the same distributions as G 1 and G 2 , respectively. For simplicity of notation we will not differentiate between (G ′ 1 , G ′ 2 ) and (G 1 , G 2 ). For two graph valued random variables G 1 and G 2 we write if there exists a coupling (G 1 , G 2 ), such that in the probability space of the coupling G 1 is a subgraph of G 2 with probability 1 or 1 − o(1), respectively. Moreover, we write if G 1 and G 2 have the same probability distribution (equivalently there exists a coupling (G 1 , G 2 ) such that G 1 = G 2 with probability one).
Note that, for any λ, in H * i (n, Po (λ)) each edge appears independently with probability 1 − exp(−λ/ n i ) (see [8]). Thus We gather here a few useful facts concerning couplings of random graphs. For proofs see [13,14]. Fact 1. Let M n be a sequence of random variables and let t n be a sequence of numbers.
..,m be sequences of independent random graphs. If Fact 4. Let G 1 = G 1 (n) and G 2 = G 2 (n) be two random graphs, such that

Then for any increasing property
Proof. Define event E := {G 1 ⊆ G 2 } on the probability space of the coupling (G 1 , G 2 ) existing by (8). Then for any increasing property A The result follows by taking n → ∞ Proof of Theorem 1. We will show only (4) in the case S 3 ≫ √ S 1 . The remaining cases follow by similar arguments. Here we should note that S 2 = S 1 − S 3 and S 2 = Θ(S 1 ).
Let w i ∈ W. Denote by V i the set of vertices which have chosen feature where I A is an indicator random variable of the event A. Note that X i , 1 ≤ i ≤ m, are independent random variables with binomial distributions Bin (n, be a graph with the vertex set V and the edge set containing those edges from G (n, m, p) which have both ends in V ′ i (i.e. its edges form a clique with the vertex set vertices chosen uniformly at random from the remaining ones. This coupling implies independent. Therefore by Fact 2 and the definition of G (n, m, p), we have By definition Therefore by Chebyshev's inequality, for any Thus with probability 1 − o(1) Therefore by Fact 1 We may assume that in the above coupling G * 2 n, The main reason for this is the fact that even though M 2 and M 3 are dependent (i.e. also G * 2 (n, M 2 ) and G * 3 (n, M 3 ) are dependent), the choice of a hyperedge of H * i (n, ·) in a given draw in the construction of G * 2 (n, M 2 ) and G * 3 (n, M 3 ) is independent from choices in other draws. Moreover note that in the coupling, in order to get G * 2 (n, ) additional draws and add hyperedges to H * 2 (n, resp. ) hypredges attributed to the last draws to get exactly m i , i = 2, 3, draws.
Let M ′ 2 and M ′ 3 be random variables with the Poisson distribution Then by sharp concentration of the Poisson distribution Therefore by Fact 1, Fact 3 and (7) , ω = 5ω ′ , andp 2 andp 3 as in (2) Therefore using standard couplings of G 2 (n, ·) and H 3 (n, ·) finally we get Therefore the result follows by Fact 4.

Vertex degrees in G (n, m, p)
For any graph G denote by δ(G) the minimum vertex degree in G.
where f (·) is given by (5). for some b n , c n = o(ln n). In both cases but the expected number of vertices of degree 0 in G (n, m, p) is respectively.
(i) If Here and in the proof we assume that 2 t=3 S 1,t = 0. For G (n, m, p) the result of Lemma 2 may be improved. Proof of Lemma 1. In the proof we assume that . Therefore by monotonicity of the considered property, analysis of the case c n = o(ln n) is enough to prove the general result stated above.
Consider a coupon collector process in which in each draw one choose one coupon uniformly at random from V. In order to determine the minimum degree we establish a coupling of the coupon collector process on V and the construction of G (n, m, p). Define V ′ i and Y i as in (9) and (10). We consider the process in which we collect coupons from V and at the same time we construct a family of sets {V ′ i : i = 1, . . . , m} (i.e. equivalently we construct an instance of G (n, m, p)). Assume we have a given vector chosen so that Y i are independent random variables with distribution of the random variables defined in (10). We divide the process of collecting coupons into m phases. In the i-th phase, 1 ≤ i ≤ m, we draw independently one by one vertices uniformly at random from V until within the phase we get y i distinct vertices. Let V ′ i be the set of vertices chosen in the i-th phase. We construct an instance of G (n, m, p) by connecting by edges all pairs of vertices within V ′ i , for all 1 ≤ i ≤ m. Obviously after the m-th phase 1≤i≤m V ′ i is the set of non-isolated vertices in G (n, m, p). Denote by T i the number of vertices drawn in the i-th phase. If Y i = y i = 0 then T i = 0. If Y i = y i ≥ 2 and in the i-th phase we have already collected j < y i vertices then the number of draws to collect the j + 1-st vertex has geometric distribution with parameter n−j n . Therefore n for 2 ≤ y i < n/2; n ln n for y i ≥ n/2.

Note that
Thus where S 1 is defined as in (1). Moreover by Markov's inequality for any ω Therefore with high probability .
In the probability space of the coupling described above define events: A − -all coupons are collected in at most T − draws; A + -all coupons are collected in at most T + draws; A -δ(G (n, m, p)) ≥ 1; B -the construction of G (n, m, p) is finished between T − -th and T + -th draw;
Therefore by the classical results on the coupon collector problem and by (12) and (13) Pr Thus the lemma follows by substituting the above values to (14).
The proofs of Lemmas 2 and 3 rely on a technique of dividing G (n, m, p) into subgraphs. We gather here some simple facts concerning this division. For each 2 ≤ t ≤ k, let G t (n, m, p) be a random graph with a vertex set V and an edge set consisting of those edges from G (n, m, p), which are contained in at least one of the sets {V i : |V i | = t}. Moreover let G k+1 (n, m, p) be a subgraph of G (n, m, p) containing only those edges which are subsets of at least one of the sets {V i : Note that, for all t = 2, . . . , k, we have G t (n, m, p) = G * t (n, M t ), where G * t (n, ·) is defined as in Section 3. Moreover where S 1 and S 1,t are defined as in (1). For all 1 ≤ t ≤ k, M t is a sum of independent Bernoulli random variables and S 1,t ≤ S 1 . Therefore by Chebyshev's inequality, Markov's inequality, and by (12) for any ω → ∞ with high probability max 0, for all t = 2, 3, . . . , k for all t = 2, 3, . . . , k.
Moreover, for any t = 2, 3, . . . , k, if S 1,t → ∞ then with high probability Set ω a function tending slowly to infinity. In the proofs we will assume that ω is small enough to get the needed bounds.
For t = 2, . . . , k let , otherwise. For all t = 2, . . . , k G t (n, m, p) = G * t (n, M 2 ), thus by (15) and (18), using the same methods as in the proof of Theorem 1 we can show the following fact.
As in the proof of Lemma 1 we will use the coupling of the coupon collector process on V and the construction of G (n, m, p). However here, for some t = 2, . . . , k, in the process we omit rounds in which Y i = t. Therefore we construct j=2,...,k+1;j =t G t (n, m, p) instead of G (n, m, p). Reasoning in the same way as in the proof of Lemma 1, by the definition of M t , (13), and (17) we get that with high probability the construction of j=2,...,k+1;j =t G t (n, m, p) is finished before the T t+ -th draw. Similarly, if in the process we omit rounds with Y i ≤ k, then with high probability we construct G k+1 (n, m, p) in at least T − draws. Therefore analogous reasoning as this used in the proof of Lemma 1 gives the following facts.  Proof of Lemma 2. In the proof we use notation introduced in (19). Moreover let a n = S 1,2 /S 1 . (i) We restrict our attention to the case c n = o(ln n). In the latter cases the result follows by Lemma 1.
First consider the case a n = S 1,2 /S 1 ≫ 1/ ln n. Then S 1,2 → ∞. Take any probability space on which we define the coupon collector process on V and G 2 (n,p 2+ ), in such a manner that they are independent. Let X + be a random variable counting vertices which have not been collected during the coupon collector process in T 2+ draws and have degree at most k − 1 in G 2 (n,p 2+ ). If S 1 = n(ln n + (k − 1) ln(a n ln n) − c n ) then for ω tending to infinity slowly enough andp 2+ ∼ a n ln n n . Therefore (a n ln n) k−1 exp(− ln n − (k − 1) ln(a n ln n) − c n ) ∼ e −cn (k − 1)! and EX + (X + − 1) ∼n 2 Thus by the second moment method with high probability X + > 0 as c n → −∞. Note that if there is a vertex which is isolated in k t=3 G t (n, m, p) and has degree at most k − 1 in G 2 (n, m, p), then δ(G (n, m, p)) ≤ k − 1. By Facts 6 and 5 there is a probability space such that with high probability G 2 (n, m, p) ⊆ G 2 (n,p 2+ ), the number of isolated vertices in k t=3 G t (n, m, p) is at least the number of non-collected coupons after T 2+ draws, and G 2 (n,p 2+ ) and the coupon collector process are independent. Thus X + > 0 imply that with high probability δ(G (n, m, p)) ≤ k − 1.
(ii) In the proof we restrict our attention to the case c n = O(ln n). In the latter case the result follows after combining (3) with known results on G 2 (n,p).
Let a n = S 1,2 /S 1 ≫ 1/ ln n (thus S 1,2 → ∞ and a n → ∞). Consider any probability space on which we define the coupon collector process on V and G 2 (n,p 2− ), in such a manner that they are independent. Let X − be a random variable defined on this probability space and counting vertices which have not been collected during the coupon collector process with T − draws and have degree at most k − 1 in G 2 (n,p 2− ). If S 1 − k t=3 S 1,t = n(ln n + (k − 1) ln(a n ln n) + c n ) then for ω tending to infinity slowly enough = ln n + (k − 1) ln(a n ln n) + c n + o(1). Thus (a n ln n) k−1 exp(− ln n − (k − 1) ln(a n ln n) − c n ) = O(1) e −cn (k − 1)! Therefore with high probability X − = 0, i.e. with high probability each vertex is collected in T − draws or has degree at least k in G 2 (n,p 2− ). Note that if each vertex is non-isolated in G k+1 (n, m, p) or has degree at least k in G 2 (n, m, p), then δ(G (n, m, p)) ≥ k. Therefore Facts 5 and 7 imply that with high probability δ(G (n, m, p)) ≥ k. Now let S 1,2 /S 1 = O((ln n) −1 ), i.e. S 1,2 = O(n). Then S 1 − k−1 t=2 S 1,t = n(ln n + c n + O(1)) and T − = n(ln n + c n + O(1)).
Thus by Fact 7 with high probability there is no isolated vertex in G k+1 (n, m, p), i.e. with high probability δ (G (n, m, p)) ≥ k.
Proof of Lemma 3. We use notation from (19). Moreover let Note that where S 1 and S 1,t are defined as in (1). Therefore as far as np 2 = o(1) If S 1,t = O(S 1 / ln n) for all t then a n = O(1) and . Therefore the lemma follows by Lemma 2. It remains to consider the case: S 1,2 ≫ S 1 / ln n or S 1,k ≫ S 1 / ln n. Note that in this case if c n = O(ln n) then np 2 = o(1) and a n → ∞.
Let now c n → −∞. As in the proof of Lemma 2 we may restrict our attention to c n = o(ln n). If e −np ln n/(1 − e −np ) = Ω(1), then the lemma follows by Lemma 2, therefore assume that (20) e −np ln n 1 − e −np = o(1) i.e. a n ∼ (np) k−1 e −np ln n .
By (20) S 1,k → ∞, thus we may apply the second part of Fact 5. Take any probability space on which we define independent coupon collector process on V and G k (n,p k+ ). Let X + be a random variable counting vertices which have not been collected during the coupon collector process with T k+ draws and have degree at most k − 1 in G k (n,p k+ ). For ω tending to infinity slowly enough S 1 ω ln n = ln n + ln a n + c n + o(1), Thus with high probability X + > 0 as c n → −∞ (i.e. with high probability there is a vertex, which has degree at most k − 1 in G k (n,p k+ ) or has not been collected in T k+ draws). Thus by Facts 5 and 6 with high probability δ(G (n, m, p)) ≤ k − 1. Now assume that c n → ∞. As it is explained in the proof of Lemma 2 we may restrict our considerations to the case c n = O(ln n). Recall that we assume that S 1,2 ≫ S 1 / ln n or S 1,t ≫ S 1 / ln n, i.e. a n → ∞. Note that by definition Take any probability space on which we may define independent coupon collector process and k t=2 G t (n,q t ). Let X − be a random variable counting vertices which have not been collected during the coupon collector process in T − draws and have degree at most k − 1 in k t=2 G t (q t ). Note that if v has degree at most k − 1 in k t=2 G t (q t ), then for some 0 ≤ k 0 ≤ k − 1 and a sequence of integers r 2 , . . . , r t such that k t=2 (t − 1)r t = k 0 (i) there is a set V ′ ⊆ V of k 0 vertices such that for each t there are r t hyperedges in H t (q t ) (generating G t (q t )) contained in V ′ ∪ {v}.
(ii) and all hyperedges in k t=2 H t (q t ) containing v are subsets of V ′ ∪ {v}. Let r = k t=2 r t , then for c n = O(ln n) event (i) occurs with probability S 1,t + O ω S 1 + S 2 1 n −2 = ln n + ln a n + c n + o(1). Thus Therefore with high probability X − = 0. By Facts 5 and 7 with high probability δ(G (n, m, p)) ≥ k.
(iii) B 3,k -for all S ⊆ V if 1 ≤ |S| ≤ n γ and all vertices in S have degree at least 4k + 15 in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) we have (iv) B 4,C -any two vertices of degree at most C are at distance at least 6.
(v) B 5 -contains a path of length at least 1 − 8 ln 2 ln n .
First consider the case n γ ≤ |S| ≤ 4n ln ln n/ ln n. Let s = |S|. Then |N 2 (S)| has the binomial distribution Bin (n − s, 1 − (1 −p 2 ) s ) and E|N 2 (S)| ≫ 2|S|. Denote by X the number of sets S of cardinality n γ ≤ |S| ≤ 4n ln ln n/ ln n such that |N 2 (S)| ≤ 2|S|. Using Chernoff's inequality. Therefore with high probability for all sets S ⊆ V such that n γ ≤ |S| ≤ 4n ln ln n/ ln n we have |N 2 (S)| ≥ 2|S|. Now consider the case |S| ≥ 4n ln ln n/ ln n. Let r = 4n ln ln n/ ln n. Let moreover K r and K r,r be the complements of the complete graph on r vertices and the complete bipartite graph with each set of bipartition of cardinality r. Denote by X r and X r,r the number of K r and K r,r in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ), respectively. Then Therefore with high probability X r = 0 which implies that with high probability α(G 2 (n,p 2 ) ∪ G 3 (n,p 3 )) ≤ r.
Moreover with high probability X r,r = 0 thus with high probability for any S ⊆ V such that r ≤ |S| we have N 3 (S) ≥ n − |S| − r.
(Otherwise G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) would contain K r,r .) Therefore with high probability for any S ⊆ V such that r ≤ |S| ≤ 2n/3 Moreover with high probability for any S ⊆ V such that r ≤ |S| ≤ n/4 we have This finishes the proof of (i) and (ii).
(iii) For any two disjoint sets S ⊆ V and S ′ ⊆ V we denote by e(S; S ′ ) the number of edges in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) with one end in S and one end in S ′ and by e(S) the number of edges in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) with both ends in S. We will bound numbers e(S) and e(S, S ′ ) for |S| ≤ n γ and . At least one of the two first events occur with probability at mostp 2 + (n − (2k + 1)s)p 3 = O(ln n/n) independently for all v ∈ S and v ′ ∈ S ′ . Moreover each hyperedge with all vertices in S ∪ S ′ appears independently with probabilityp 3 = O(ln n/n 2 ) and generates two edges between S and S ′ . Therefore Therefore with high probability for any set S (1 ≤ |S| ≤ n γ ) of vertices of degree at least 4k + 15 in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) we have Otherwise for S ′ = N 3 (S) there would be |S ′ | = |N 3 (S)| ≤ 2k and e(S, N 3 (S)) + 2e(S) would exceed (4(k + 1) + 10)|S|.
(iv) The probability that in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) there are two vertices of degree at most C at distance at most 5 is at most (v) Follows by Theorem 8.1 from [4].
Lemma 5. Let k be a positive integer p 2 + n 2p 3 = ln n + (k − 1) ln ln n + c n n .
Thus with high probability X t = 0 for all t ≤ k − 1.
In particular, if we substitute G(n) = G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) then by Lemmas 5 and 6 we obtain the following result.
Lemma 7. Letp 2 andp 3 fulfil (21). Ifp 2 + n 2p 3 = (ln n + c n )/n then where f (·) is defined by (5). Ifp 2 + n 2p 3 = (ln n + (k − 1) ln ln n + c n )/n and c n → ∞ then Proof of Lemma 6. Denote by G(n) δ≥k a graph G(n) under condition δ(G(n)) ≥ k. If Pr {δ(G(n) ≥ k)} is bounded away from zero and G(n) has certain property with high probability, then also G(n) δ≥k has this property with high probability. From now on we assume that G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) and G(n) are defined on the same probability space existing by (22). We call a vertex v ∈ V small if its degree in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) is at most 4k + 14. Otherwise we call a vertex large. Let S ⊆ V and |S| ≤ n γ . Denote by S + and S − the subset of large and small vertices of S, respectively. Then by Lemma 4(iii) with high probability Moreover in G(n) δ≥k all vertices in S − have degree at least k and with high probability are at distance at least 6. Therefore with high probability in G(n) δ≥k no two vertices in S − are connected by an edge or have a common neighbour and at most |S + | of them have neighbours in N 3 (S + ) ∪ S + (otherwise they would be connected by a path of length at most 5). Thus with high probability If we combine this with Lemma 4(i) and (ii) we get that with high probability (25) N G(n) δ≥k (S) ≥ min{|S|, 4n ln ln n/ ln n} ≥ min{|S|, α(G(n) δ≥k )} for all S ⊆ V, 1 ≤ |S| ≤ 2n 3 ; (26) Finally (23) follows immediately by (25). Moreover if (26) is fulfilled then G(n) δ≥k has a perfect matching (for the proof see for example [3]). Therefore (24) follows. (27) will be used later to establish threshold function for a Hamilton cycle.
Let moreover G(n) be a random graph such that: (ii) with high probability δ(G(n)) ≥ 2; (iii) in a probability space existing by (i) all vertices of degree at most 22 in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) are at distance at least 6 in G(n).
Proof. We will follow the lines of the proof of Theorem 8.9 from [4]. Let t = 8n/ ln n and p 4,0 = 64 ln n/n 2 . Then tp 4,0 =p 4 . For any graph G let l(G) be the length of the longest path and l(G) = n if G has a Hamilton cycle. We say that G has property Q if G is connected and |N G (S)| ≥ 2|S|, for all S ⊆ V, |S| ≤ n/4.

Sharp thresholds
Proof of Theorems 2-4. First let S 1 = n(ln n + c n ) andp 2 andp 3 be given by (2). Note thatp 2 > 1+ε 2 ln n for some constant ε > 0. Then by Theorem 1 and Lemma 7 In the following proofs we will assume that c n = O(ln n). In the other cases theorems follow by Lemma 1 or (3) combined with known results concerning G 2 (n,p). v. Given v ∈ V and 1 ≤ i ≤ m ′ . Denote by A v,i,1 event that v is an isolated vertex in G * 2 (n, (Y i − 3Z i )/2) ∪ G * 3 (n, Z i ) and A v,i,2 event that v ∈ V ′ i . Then  Recall that random graph G * 2 (n, ·) ∪ G * 3 (n, ·) is constructed by making independent draws of edges and hyperedges in an auxiliary hypergraph. The number of draws is given by random variables. Couplings (i)-(iii) may relay on this construction. In the coupling (i)-(iii) in order to get from G * 2 n, Po n 2 ln(1 −p 2 ) −1 ∪ G * 3 n, Po n 3 ln(1 −p 3 ) −1 a graph G * 2 (n, M 2 ) ∪ G * 3 (n, M 3 ) with high probability we make some additional draws. Moreover by the sharp concentration of the Poisson distribution and (12) with high probability the number of additional draws is at most Kω √ n ln n for some constant K. Under condition that the number of additional draws is bounded at most Kω √ n ln n, probability that an edge (hyperedge) containing v is chosen in at least 2 additional draws is at most.
Moreover each draw (each chosen hyperedge) generates at most 2 edges incident to v. Concluding in the probability space of the coupling (28) with high probability the number of edges incident to v in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) is at least W ′ (v) − 2 − 2. Therefore if degree of v in G 2 (n,p 2 ) ∪ G 3 (n,p 3 ) is at most C then W ′ (v) ≤ C + 4.
Finally, probability that there are two vertices v, v ′ such that |W ′ (v)| ≤ C + 4 and |W ′ (v ′ )| ≤ C + 4 connected by a path of length at most 5 in G (n, m ′ , p) is at most Therefore by Lemma 8 with high probability G(n) ∪ G 2 n, 512 n ∈ HC.