mmammel,

The proof for game with the restriction is likely harder than the proof for game without, but the same principals should apply.

And yes, we need to prove it for every possible move for player 2, and we only need to show that there is at least one 'correct' move for player 1 at each turn.

The solution to the problem you mention is to create a new game, let's call it pente-X. The only rule change is that player 2's first move must be N squares from center AND that he doesn't have to declare exactly where that position is until he chooses. Clearly, pente-X includes all the possible pente games in which player 2's first move is N moves from center. Thus, once we prove that pente-X can always be won by player 1, we will also prove that the same holds true for this subset of games of pente.

You wouldn’t have to do an exhaustive search if you could show that player 1 can always win AND keep the game within the N-size circle so that player 2’s first stone never affects play, no matter where outside the circle it was actually placed.

Again, this is still a difficult proof, but easier than proving the game always winnable by player 1 directly.

Thad