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Random graph process and connectivity hitting time

PDF p. 38: Section 7, second paragraph

Definition

Assume countable choice. Let n≥1 be an integer, let V={1,…,n}, and put E={{i,j}:i,j∈V, i<j} and N=∣E∣=n(n−1)/2. A graph on V has an edge set contained in E; it is connected if every pair of vertices is joined by a finite path of edges. An edge in E that is not in the graph is called missing.

Choose an ordering (e1,…,eN) uniformly from the N! orderings of E. The random graph process {G(n,m)}m=0N has vertex set V and edge set {e1,…,em} at time m, with the empty set at m=0. Thus, conditional on any revealed ordering of the first m<N edges, the next edge is uniformly chosen from the missing edges. Its connectivity hitting time is τ=min⁡{m∈{0,…,N}:G(n,m) is connected}.

Facts & Assumptions

[F1]

The finite edge-order construction gives each ordering probability 1/N! and, conditional on a specified initial ordered segment of length m<N, gives each remaining edge probability 1/(N−m) of being next (Uniform edge orderings, graph-process times and independent labels).

[F2]

A finite undirected graph is represented by its vertex and edge sets, with the stated simple graph convention as a special case (Adjacency, degrees, graph Laplacian, and spectral notation).

[F3]

Countable choice is assumed in the edge-order supplier's construction (The Axiom of Countable Choice (AC_ω), Uniform edge orderings, graph-process times and independent labels).

Justification

Given: An integer n≥1 and the finite uniform edge ordering under countable choice.

Goal: Verify that the process and its connectivity hitting time are well defined and have the stated transition rule.

1.1 The set E has one element for each pair i<j, so it has N elements. Each prefix of an ordering is a subset of E and therefore defines a graph on V; the prefix of length zero is empty. By the uniform-order construction, conditional on any specified prefix of length m<N, each of the N−m missing edges is next with probability 1/(N−m). Thus the stated transition rule holds. [F1, F2, F3, given]

2.1 At time N every pair of distinct vertices is an edge, so the graph is connected: distinct vertices are joined by a one-edge path, and the one-vertex case is vacuous. Consequently the set of connected times is a nonempty subset of the finite set {0,…,N} and has a least element. This establishes τ, including τ=0 when n=1. [given, step 1.1] ∎

Source notes

The process and hitting-time notation are attributed to Expander graphs are globally synchronizing, Section 7, second paragraph. The article is attribution for this formulation, not a premise of the justification.

Depends on

Used by · 1

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