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In the general case finding a Maximum Independent Subset of a Graph is NP-Hard.

However consider the following subset of graphs:

  • Create an $N \times N$ grid of unit square cells.
  • Build a graph $G$ by creating a vertex corresponding to every cell. Notice that there are $N^2$ vertices.
  • Create an edge between two vertices if their cells share a side. Notice there are $2N(N-1)$ edges.

A Maximum Independent Subset of $G$ is obviously a checker pattern. A cell at the $R$th row and $C$th column is part of it if $R+C$ is odd.

Now we create a graph $G'$ by copying $G$ and removing some vertices and edges. (If you remove a vertex also remove all edges it ended of course. Also note you can remove an edge without removing one of the vertices it ends.)

By what algorithm can we find a Maximum Independent Subset of $G'$?

Gilles 'SO- stop being evil'
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Andrew Tomazos
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    Hint: Grid graphs are bipartite. – JeffE Aug 03 '12 at 16:47
  • @JeffE: There is one way to partition a grid graph into two sets of verticies such that they have no internal edge (by R+C parity). Once you start removing verticies and edges doesn't the number of ways to partition it grow exponentially. (maybe I have it. a single connected component maybe only still has one way to partition it) – Andrew Tomazos Aug 03 '12 at 16:56
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    Hint 2: Matching. – JeffE Aug 03 '12 at 16:56
  • Yeah I think I see it. Sort it into connected components. Split each component into the unique two partition. Select the larger partition from each component. – Andrew Tomazos Aug 03 '12 at 16:58
  • Nope. The maximum independent set is not necessarily one side of the bipartition. – JeffE Aug 03 '12 at 16:59
  • Yeah you're right. Example: G' = (A-B-C, B-E, D-E-F). Bipartition (A,C,E | D,F,B). Maximum Indep Set (A,C,D,F). – Andrew Tomazos Aug 03 '12 at 17:05
  • @JeffE: The Hopcroft–Karp algorithm finds a maximum set of edges such that none of them share an endpoint. Maybe I can use that somehow. – Andrew Tomazos Aug 03 '12 at 17:09
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    @JeffE: I still haven't figured out how to get from a bipartite graph to a maximum independent set. See my post here. – Andrew Tomazos Aug 03 '12 at 19:22

1 Answers1

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As JeffE mentions in a comment, a grid graph is an example of a bipartite graph. If you take a bipartite graph, and remove some of the vertices, it will still be a bipartite graph.

Königs Theorem states, that for bipartite graphs, the number of vertices in a minimum covering, equals the number of edges in a maximum matching.

As mentioned in the answer for Maximum Independent Set of a Bipartite Graph:

The complement of a maximum independent set is a minimum vertex cover

What you are asking, is how to find a maximum independent set in a grid graph with some vertices removed, and the way to do that, is to use some of the algorithms for finding a maximum matching in a bipartite graph, from that construct a minimum vertex cover, and then invert it to get a maximum independent set.

The figure below shows a bipartite graph with a maximum matching (in blue), a minimum vertex cover (in red) and consequently also a maximum independent set (in white):

from Wikipedias article on Königs Theorem

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