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Strategies for the Shannon Switching Game

We present a proof that the Shannon switching game on a graph with distinguished vertices A and B has a winning strategy for player Short iff the graph has a subgraph connecting A to B with two edge disjoint spanning trees. This theorem was first proved by Lehman in [11]. The difficulty is that his proof is phrased in terms of matroids and appears very difficult. However, I was able to cull a neat and short but elegant proof out of Lehman's work and that is what I want to present here. Using this proof, we will be able to give a strategy for player Cut for the graphs in which he has a win. This strategy is not quite as neat as Short's strategy, and does not appear to be computationally feasible, but at least it does not involve any look ahead analysis. We are given a multi-graph G with two distinguished vertices, A and B. The two players, Cut and Short play alternately. Short's move consists of marking an edge while Cut may remove any unmarked edge from the graph. If Short can succeed at marking an entire path from A to B, he wins. Otherwise Cut wins. As we shall see shortly, if G contains two trees connecting A to B having no edges in common but sharing the same vertex set, then Short has a winning strategy even if he goes second. Our main theorem is the converse, that if he has a winning strategy from the second position, then there must be two such trees. Brualdi has written an excellent expository article about the switching game. See [1]. This article has been essentially repeated as a section in his book [2, Section 11.6]. Let us say that a graph is positiue if it has an edge disjoint pair of spanning trees. Then what we wish to show is that G has a positive subgraph containing both A and B if and only if Short has a winning strategy even if he plays second. The reader may easily verify that the union of two positive graphs is either disconnected or positive. The reader may also easily verify that if there is a positive subgraph connecting A to B, then Short has a winning strategy. He proceeds as follows: If Cut removes an edge from one of the two trees, Short finds an edge in the other tree which reconnects the broken tree and marks it. He then has two spanning trees for the subgraph with only marked edges in common. Let us prove the converse. Suppose Short goes second and has a winning strategy. We must prove the existence of a positive subgraph. So Cut goes first and deletes an edge. Short then marks an edge, call it a. By induction, we may assume that there are two trees, S1 and T1 from A to B having a common vertex set but only the edge a in common. We now need a lemma. We shall state and use the lemma and only when the main proof is finished will we come back and prove it. The lemma is the following: