Generating random regular graphs. J.H. Kim (Microsoft) V. Vu (UCSD).
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J.H. Kim (Microsoft)
V. Vu (UCSD)
Question: How to generate a random regular graph G(n,d) ?
Trouble: With very high probability, the resulting graph is not simple. In fact, the probability that it is simple is only exp (- d2 / 4).
(1) Practical: Reasonable only if d very small (less than 10, say).
(2) Theoretical: Can be used to prove theorems for small d, typically d= O(ln n).
McKay-Wormald’s algorithm (90): nd3 time, but hard to implement.
Start with n sets of size d. Assume that m pairs have been chosen. Delete all pairs that are parallel with these edges and also all pairs that form loops. Among the available pairs, choose one uniformly randomly.
Start with n points. Assume that m edges have been drawn. Draw a new edge uv with probability proportional to (d-du)(d-dv).
This algorithm is fast (nd2) and very easy to implement. Recently Kim-Vu, using a coupling argument, show that the graph created by this algorithm behaves essentially like an Erdos-Renyi’s random graph with the same density. So if we can prove that the output is indeed G(n,d), we can derive lots of results for random regular graphs.
d =o( n1/3).
Main theorem (Kim-V. 02) The distribution is asymptotic uniform for d = o(n1/3).
Corollary: all properties of Erdos-Renyi graphs (hamiltonianity, colorability etc) hold for random regular graphs G(n,d) in this range.
In a typical situation, one uses Markov chains to generate a uniform random sample and use this to compute the number of objects.
In the current problem, we know the number of random regular graphs (via combinatorial arguments) and use this information to prove that the graph generated by the algorithm is indeed uniformly random.
It is well-known (and easy to check) that for each d-regular graph G, there are exactly (d!)n different simple matchings which give rise to G. So to prove our theorem, it suffices to show that if d=o(n1/3), then for any simple perfect matching M
However, an edge is suitable if and only if it does not join two vertices from the same group or two vertices come from two already adjacent groups. The number of edges of the first type is
and the number of unsuitable edges of the second type is
where du is the degree of u in Gm (M).
It follows that
Now it suffices to prove that
so all we need to do is to justify this replacing, namely to show that
is very very close to 1.
For a moment let us assume that the main part of ∆m (M) is
We want to show that this random variable is very close to its expectation.
We can think of the projected graph Gm (M) as a random subgraph of G where each edge is chosen with probability p= m/(nd/2); te is the indicator of the event that the edge e is chosen. So can be written as
is very strongly concentrated
(1) Azuma’s inequality (Steger-Wormald):d =o(n 1/28 ).
Azuma’s inequality tells that a random variable with small Lipschitz coefficient is strongly concentrated around its mean.
The trouble here is that the Lipschitz coefficient is rather large, order d2.
Where E is the expectation of Y, E’ is the average Lipschitz coefficient, and k is the degree of Y (k=3). This gives d=o(n1/10).
(3)For d= o(n1/3) we need a stronger version (V.00)
and many extra combinatorial arguments.
We conjecture that the probability is still asymptotically uniform for most d-regular graph G. Maybe some graph properties of G need to be exploited.