Sign in to follow this  
spraff

dynamic fuzzy graph partitions

Recommended Posts

Consider a graph of the following symmetric relations: A~B B~C C~A X~Y Y~Z Z~X. Clearly this graph can be partitioned into {A,B,C}, {X,Y,Z} and a simple algorithm to detect this would be a sort of flood fill. Now suppose we add the edge A~X. A fuzzy algorithm could still "partition" the graph into {A,B,C}, {X,Y,Z} and this is the algorithm I'm looking for. Ideally the algorithm would allow arbitrary measurements of how strongly an edge joins sets of nodes, and, here's the trick part, be incrementally adaptable so edges could be added and removed at any time, perhaps in batch. Is this a well-understood problem? Does the solution have a common name?

Share this post


Link to post
Share on other sites
Do the edges have associated weights? If not, then you don't have any way to determine whether your groups are {A,B,C} & {X,Y,Z} or {A,B,C,X} & {Y,Z} or some other combination. If you do have weights, and the weight of the edge between A and X is lower than the typical weights between {A,B,C} and {X,Y,Z}, you could use a simple clustering algorithm.

-Kirk

Share this post


Link to post
Share on other sites
Quote:
Is this a well-understood problem?
Yes.

You want to partition your graph into two sets of vertices. Such a partition is called a cut of a graph.

You want this partition to cut the minimum number of edges (or the minimum weight if your edges are weighted): you want a minimal cut. Your graphs are undirected. This is known as the minimum cut problem for undirected graphs.

There are very fast and straightforward algorithms published fairly recently for solving exactly this problem: google for "david karger minimum cut" for papers; they're available on the ACM, but it's not free so maybe google will turn up free copies.

I think Karger's algorithms lend themselves to online update fairly naturally, so you're probably in luck. :)


[Edited by - Anonymous P on November 22, 2007 11:43:47 PM]

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this