Loading in 5 sec....

Generating random regular graphsPowerPoint Presentation

Generating random regular graphs

- 359 Views
- Updated On :

Generating random regular graphs. J.H. Kim (Microsoft) V. Vu (UCSD).

Related searches for Generating random regular graphs

Download Presentation
## PowerPoint Slideshow about 'Generating random regular graphs' - PamelaLan

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Let 0 <d <n be two positive integers and S(n,d) be the set of all simple d-regular graphs on n vertices. A random regular graph G(n,d) is obtained by sampling from S(n,d) with respect to the uniform distribution. Random regular graphs are important objects in both theoretical computer science and graph theory.

Question: How to generate a random regular graph G(n,d) ?

- Practical: Fast, simple algorithm.
- Theoretical: Useful for proving theorems.

Configuration method (Bender-Canfield, Bollobas 80’s): Consider n sets of size d. Draw a random prefect matching on nd points. Shrink each set into a vertex.

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).

Markov chain method (Jerrum-Sinclair 90) Create a master graph whose perfect matchings correspond to simple regular graphs. Use a Markov chain to produce a random perfect matching.

- Practical: Work for all d in polynomial time. However the running time is n10.
- Theoretical:
McKay-Wormald’s algorithm (90): nd3 time, but hard to implement.

Steger-Wormald’s algorithm (97): Use the idea of the configuration method, but now we build the perfect matching gradually, one edge per step, avoiding loops and parallel edges.

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).

Trivially the output is a simple d-regular graph, but it is unclear what the distribution is.

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.

Steger and Wormald (1996) proved that the distribution in question is asymptotically uniform, for d= o(n1/28). They conjecture that it remains uniform as far as

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.

A reverse philosophy: question is asymptotically uniform, for

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.

McKay and Wormald shown that for question is asymptotically uniform, for d= o(n1/3) the number of d-regular graphs is:

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

Consider an ordering question is asymptotically uniform, for M =x1, …, xnd/2 where xm are the edges of M. Assume that the first m edges of M are obtained and let Gm(M) be the graph formed by the projection of these edges. Thus Gm(M) is a subgraph of G. To count the number of suitable edges, notice that there are ways to form an edge.

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).

Let suitable edges is δm be the expectation of ∆m It is easy to find a formula for δm. Moreover, replacing ∆m by δm we have

so all we need to do is to justify this replacing, namely to show that

The main point here now is to show that for a typical suitable edges is m, the ratio

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.

d suitable edges is u

dv

u

e

v

f

g

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

We want to show that suitable edges is

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.

(2) suitable edges is Concentration using average Lipschitz coefficient (Kim-Vu 98):

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.

What happen beyond suitable edges is n1/3 ?

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.

Download Presentation

Connecting to Server..