On the connected components of a random permutation graph with a given number of edges

Abstract

A permutation σ of [n] induces a graph Gσ on [n] -its edges are inversion pairs in σ. The graph Gσ is connected if and only if σ is indecomposable. Let σ(n, m) denote a permutation chosen uniformly at random among all permutations of [n] with m inversions. Let p(n, m) be the common value for the probabilities P(σ(n, m) is indecomposable) and P(G σ(n,m) is connected). We prove that p(n, m) is non-decreasing with m by constructing a Markov process {σ(n, m)} in which σ(n, m + 1) is obtained by increasing one of the components of the inversion sequence of σ(n, m) by one. We show that, with probability approaching 1, G σ(n,m) becomes connected for m asymptotic to mn = (6/π 2 )n log n. We also find the asymptotic sizes of the largest and smallest components when the number of edges is moderately below the threshold mn.1991 Mathematics Subject Classification. 05A05; 60C05.

Degree sequence of random permutation graphs

“…To get the density of the limiting random variable, let Z : 1]. For z ≤ 1/2, conditioning on U the distribution function of Z can be calculated as…”

“…The probability p(n, m) that σ(n, m) is indecomposable, is same as the probability that the random permutation graph G σ(n,m) is connected. Acan and Pittel [1] showed that p(n, m) has a phase transition from 0 to 1 at m n := (6/π 2 )n log n. They also studied the behavior of G σ(n,m) at the threshold. One of the most important graph statistic is its degree sequence.…”

Regenerative random permutations of integers

“…These components I j form a partition of I, which is coarser than the partition by cycles of Π. For example, the permutation π = (1)(2, 4)(3) ∈ S 4 induces the partition by components [1] [2,3,4]. So Π acts on each of its components I j as an indecomposable permutation of I j , meaning that Π does not act as a permutation on any proper subinterval of I j .…”

Formation of a giant component in the intersection graph of a random chord diagram

“…In the proof of Lemma 3.4 in [2] it was shown that, whp, the maximum M (x) does not exceed (1 + ε)(m/n) log n when the sequence x is chosen uniformly at random from IN V. Using this fact and (3.13), we get I n,m C n−M0 T n,m (3.14) for M 0 = ⌈((1 + ε)m/n) log n⌉. Also, by the Stirling's formula for the factorials, Remark.…”