A game with graphs

A graph is a set of dots (vertices) connected by a number of line segments (edges). In figure 2.1 you see a finite directed graph with one top and one bottom vertex. In a directed graph each edge has a direction, designated by an arrow, indicating the direction in which to “walk” along the edge. The top vertex has only arrows starting from it, while the bottom vertex has only arrows ending in it. There are no cycles in our graph. We say that a vertex is higher than another one if going from the first to the second vertex one can follow the directions of the arrows. Then of course the second vertex is lower than the first.

The game is played by two players. In turns they put a coin on one of the vertices. By doing that they block all lower vertices, meaning that on the blocked vertices no further coins may be laid. For instance, if a coin is placed on vertex A, all vertices indicated by a hand are blocked. The player who is forced to place a coin on vertex P loses the game.

One can prove that the first player, supposing he plays perfectly, can always win this game. This is even true for an arbitrary finite directed graph with one top vertex and one bottom vertex. Before reading further, try to find the proof yourself.

Proof

Let's take an arbitrary directed graph G with one top vertex P and one bottom vertex Q. The graph leaving out the bottom vertex Q we call G′. Two players A and B play on graph G; A puts down the first coin. If A and B are playing on G′, there are two possibilities:

1. 1. There is a winning strategy for A.

2. 2. There is no winning strategy for A, meaning that B can play in such a way that A will lose.

If in G′ player A has a winning strategy, A follows that strategy in G. The bottom vertex is blocked after A's first move.

If in G′ player B would be the winner, A puts a coin on the bottom vertex Q as his first move. The graph he leaves is G′, but the roles have been switched: B has to start on G′ and consequently A can follow a winning strategy.