On matchings in stochastic Kronecker graphs

The stochastic Kronecker graph is a random structure whose vertex set is a hypercube and the probability of an edge depends on the structure of its ends. We prove that when a.a.s. the stochastic Kronecker graph becomes connected it a.a.s. contains a perfect matching. Let n ∈ N, N = 2, and let 0 6 α, β, γ 6 1, where γ 6 α be some constants. Denote by P a symmetric matrix P = ( 1 0 1 α β 0 β γ ) , where 0’s and 1’s are labels of rows and columns of P. A stochastic Kronecker graph K(n,P) is a random graph with vertex set V = {0, 1}, the set of all binary sequences of length n, where the probability that two vertices u = (u1, . . . , un), v = (v1, . . . , vn) ∈ V are adjacent is given by pu,v = n ∏

Let n ∈ N, N = 2 n , and let 0 α, β, γ 1, where γ α be some constants.Denote by P a symmetric matrix where 0's and 1's are labels of rows and columns of P. A stochastic Kronecker graph K(n, P) is a random graph with vertex set V = {0, 1} n , the set of all binary sequences of length n, where the probability that two vertices u = (u 1 , . . ., u n ), v = (v 1 , . . ., v n ) ∈ V are adjacent is given by This model was introduced by Leskovec, Chakrabarti, Kleinberg and Faloutsos in [4].They showed empirically that the desired graph properties of real world networks hold in this model (see [5], [6]).
In [2] Mahdian and Xu showed that the threshold for connectivity of stochastic Kronecker graph is β + γ = 1.They also proved that if α, β, γ are such that K(n, P) is asymptotically almost surely (a.a.s.) connected, and moreover γ β α, then the graph has a.a.s. a constant diameter.Horn and Radcliffe [7] studied the emergence of the giant component and verified that a.a.s. it appears in K(n, P) as soon as (α + β)(β + γ) > 1. Kang, Karoński, Koch and Makai [8] showed that the degree distribution of K(n, P) does not obey a power-law and determined the threshold for the existence of some small subgraphs.
In this paper we denote by d(v, u) the Hamming distance between two vertices v and u and by w(v) the weight of a vertex v = (v 1 , . . ., v n ) the number of 1's in its label Note that the expected degree of a vertex v with weight w = w(v) is We start with stating the connectivity result.As we have already mentioned Mahdian and Xu [2] showed that if β + γ > 1, then a.a.s.K(n, P) is connected while for β + γ < 1 a.a.s. it contains isolated vertices.Later their result was supplemented (in much larger generality by Radcliffe and Young [9]).From their result we can derive the following observation concerning the connectivity of K(n, P) at the threshold, i.e. when β + γ = 1.Theorem 1.
We shall show that the threshold for the emergence of a perfect matching in K(n, P) is basically the same as the connectivity threshold.Our main result can be stated as follows.
Proof.Let β + γ 1 and β = 1.In the proof of Theorem 1, Radcliffe and Young have shown that a.a.s.K(n, P) contains an isolated vertex and so a.a.s. it does not contain a perfect matching. Let Thus, with probability 1, K(n, P) contains a perfect matching.Now let us consider the most interesting case, β + γ > 1.The main idea of our argument is the following.We shall choose a dense bipartite subgraph H of K(n, P) and show that it contains a perfect matching by verifying Hall's condition.
To this end for a given odd number t let H = H(n, t) denote a graph with a vertex set {0, 1} n , and edges between the pairs of vertices of Hamming distance d(u, v) = t.Denote by V 1 and V 2 the subsets of vertices of H of odd and even weights respectively.Since t is odd, all the edges in H must have one end in V 1 and the other in V 2 , i.e.H is bipartite.Now let H = H(t, n, P) be a subgraph of K(n, P), which contains only the edges between the vertices of Hamming distance t, i.e.H contains only those edges of K(n, P) which belong to H. Clearly, H is bipartite.We shall show that for which is basically the value of t which maximizes the expected number of edges in H, the random bipartite graph H a.a.s.fulfills Hall's condition, and so a.a.s. it contains a perfect matching.
In order to do that we show first that the underlying bipartite graph H has good expanding properties.Let us first introduce some notation.For two subsets W and U of the vertex set of H, let e H (W, U ) denote the number of edges with one end in W and another in U .Let W denote the complement of W in the vertex set of H.By Vol(W ) we denote the sum of vertex degrees of W .
Let us recall that a graph G is edge-transitive, if for any two edges e 1 , e 2 ∈ E(G) there exists a graph automorphism F : V (G) → V (G), which transforms e 1 into e 2 .The following result of Chung [1] (Theorem 7.1) is crucial for our argument.
Theorem 3. Let G be an edge-transitive graph with diameter D. Then for every .
In order to apply this result we need to check if H is edge transitive and has small diameter.
Lemma 4. H n, 2 β 2(β+γ) n + 1 is edge-transitive and its diameter can be bounded from above by a constant D which depends only on constants β and γ but not on n.

Proof. Clearly, for i ∈ [n] the function τ
) is an automorphism of H.We show that the group generated by all automorphisms of the above two kinds acts transitively on edges of H.
Although it is a rather easy observation let us prove it more formally.Let e 1 = {u 1 , v 1 }, e 2 = {u 2 , v 2 } be two edges of H.For i ∈ {1, 2}, there exist precisely t positions j such that u i j = v i j .Let I i ⊆ [n] be the set of those positions (for i ∈ {1, 2}).Let φ be a permutation of [n] such that φ(I 1 ) = I 2 and φ * = Aut(φ) be the automorphism of H induced by φ.This automorphism is uniquely defined for fixed permutation φ.Note that the pairs {φ * (u 1 ), φ * (v 1 )} and {u 2 , v 2 } differ on the same positions, i.e. φ * (u 1 ) j = φ * (v 1 ) j if and only if u 2 j = v 2 j .Define ψ : {0, 1} n → {0, 1} n by putting Clearly ψ (φ * (u 1 )) = u 2 .Moreover ψ(x) j = x j iff φ * (u 1 ) j = u 2 j and it happens iff φ * (v 1 ) j = v 2 j .Thus ψ (φ * (v 1 )) = v 2 so ψ • φ * is an automorphism of H which maps e 1 into e 2 .Hence H is edge-transitive.It remains to find an upper bound for the diameter of H. Let v, v be two vertices of H such that d(v, v ) is even.We show that they are connected by a short path.We split our argument into several cases.
In this case there exists a vertex v which is adjacent to both v and v .Indeed, to find v it is enough to change v on d(v, v )/2 positions on which v and v differ and t−d(v, v )/2 positions on which they coincide.
For each pair of such vertices v and v there exists a vertex v adjacent to v such that d(v , v ) = d(v, v ) − t.To get v we only need to change v on t positions on which v and v differ.Applying this observation 2r times, where Notice that in this case 2t < n, so 2t < 2n−2t and thus d(v 2r , v ) min{2t, 2n − 2t}.As d(v 2r , v ) is even, one can connect vertices v 2r and v by a path of length two using the argument from Case 1.
For each such v and v there exist a path vv To obtain v 1 from v, we need to change all n−d(v, v ) positions on which v, v do not differ and t−n+d(v, v ) among other positions.Then, To obtain v 2 from v 1 , we need to change all n − d(v, v ) positions on which v, v do not differ, all n − t positions on which v and v 1 are the same and 2t Arguing in the same way we find a path ) is even, we can apply Case 1 to connect v 2s and v by a path of length two.
Consequently, we have shown that the diameter D of H is bounded from above by As a direct consequence of Theorem 3 and Lemma 4 we get the following result on expansion properties of H. Lemma 5. Let W be a subset of the vertex set of H such that Then there exists a constant c = c(β, γ) > 0 such that Proof.Let W , |W | 2 n−1 , be a set of vertices of H. Since H is an n t -regular graph, Since H is edge transitive, by Theorem 3 we get where D is the diameter of H.By Lemma 4, D is bounded above by a constant, so for some positive constant c we have Let us return to the random graph H. Recall that H is a bipartite graph with a bipartition (V 1 , V 2 ), where |V 1 | = |V 2 | = 2 n−1 .We will use the Hall's condition, which states that a bipartite graph G(V, U ), |V | = |U | does not have a perfect matching iff there exists a set R ⊆ where N G (R) is the set of all vertices adjacent in G to the vertices from R. Suppose G does not have a perfect matching.Let S be the smallest set S ⊆ V or S ⊆ U which satisfies Therefore if H does not have a perfect matching, there exists a set Let A be the event that such a subset S exists in H. Let A 1 be the event that such a subset S ⊆ V 1 exists.Let A 2 be the event that such a subset S ⊆ V 2 exists.Then P(A 1 ) = P(A 2 ), hence P(H does not contain a perfect matching) = P(A) 2P(A 1 ).
For two fixed sets S ⊆ V 1 , |S| 2 n−2 = N/4 and T ⊆ V 2 , |T | = |S| − 1, let A S,T denote the event that T is the neighbourhood of S in the random graph H.In order to estimate the probability of A S,T we apply Lemma 5 to the set W = S ∪ T .Clearly For deterministic H we have while from the regularity of H we get e H (S, S) = e H (S, and Adding ( 3) and (4) and subtracting (5), we obtain that in H, e H (S, V 2 \ T ) 1 2 for some constant c > 0. Thus if A S,T occurs, c n t |S| fixed pairs of vertices which are adjacent in H are not adjacent in H. Observe that for each pair u, v of vertices with Hamming distance t, the probability that there exists an edge {u, v} is at least β t γ n−t .Thus, the probability of A that Hall's condition fails for some set S, |S| 2 n−2 = N/4, is bounded from above by the electronic journal of combinatorics 23(4) (2016), #P4.6 Since and t was chosen to correspond the largest term in the sum, so for n large enough and since the term in brackets is exponentially small, it is maximal when s = 1 that is Consequently, a.a.s.H, and thus also K(n, P), contains a perfect matching.
In the proof we have found a perfect matching in a bipartite subgraph H of K(n, P), containing only the edges joining vertices which are at Hamming distance respectively, we can mimic our argument to construct k edge-disjoint perfect matchings.Thus, let k-PM denote the property, that a graph contains k edge-disjoint perfect matchings.Theorem 6.Let k ∈ N, k 2 be a constant.
In particular Note the difference between the cases k = 1 and k 2 for β = 1 and γ = 0 when, as we have already observed, a.a.s. the minimum degree of K(n, P) is one.