Converting a Gomoku winning strategy from a small board to a winning strategy on a larger board

Gomoku is the game where Black and White take turns placing stones of their own color, and the winner is the player who first gets five of their own stones in a row. Black moves first.

In Gomoku on an infinite big board, I made the following claim:

If Black can win on a $15\times 15$ board, he can certainly win on an infinite board, since he can pretend to be playing on a $15\times 15$ board, and if White plays outside the imaginary boundary, so much the better for Black.

I am no longer happy with this argument. Here is a counterargument:

Suppose Black pretends to be playing on a $15\times 15$ board, making all his moves inside the imaginary boundary of a $15\times 15$ region. If White has the opportunity to make any five moves outside the imaginary boundary before Black wins inside the boundary, White will win first. Thus if White can find a strategy $S$ for the $15\times 15$ board that delays Black's win long enough for her to pass five times, then she can win on the infinite board by playing $S$ against Black, and using her pass moves to play outside the imaginary boundary.

My idea is that Black's winning strategy might take a long time to execute, say 103 moves, and perhaps pass moves by White only allow Black to speed up his win by a moderate amount; say by passing at the correct times, White can still force Black to take 37 moves to win. Then despite Black's winning strategy for the $15\times 15$ board, White can win on the infinite board if Black does not defend outside the $15\times15$ region.

One of the two arguments must fail. Right now, I think it's the first one that fails. My questions are:

1. Is there some reason that I've overlooked that the notional winning strategy $S$ for White can't exist?

2. If there isn't, does such a strategy exist?

3. Is there any way in general to convert winning strategies on smaller boards to winning strategies on a larger boards?

• This is an interesting question! Is there also a possible counterstrategy where white combines some moves at the edge of the 15 by 15 region with some moves outside that region so as to win? – paw88789 Oct 28 '14 at 2:36
• @paw88789: only if black refuses to move outside the $15 \times 15$ Otherwise, the strategy stealing argument works just fine. – Ross Millikan Oct 28 '14 at 3:00

In József Beck's book Combinatorial Games: Tic-Tac-Toe Theory, he states the following open problems ("unrestricted $5$-in a row" is Gomoku on an infinite board):

Open Problem 4.1. Is it true that unrestricted $5$-in-a-row is a first player win?

Open Problem 4.2. Is it true that unrestricted $n$-in-a-row is a draw for every $n\ge6$?

There are some relevant counterexamples in the more general setting of "positional games". The following information is from pp. 78-84 of József Beck's book.

A positional game is defined by a hypergraph $(V,E)$ where $V$ is the vertex set (or board) and $E$ is the set of (hyper-)edges (or winning sets), a collection of nonempty finite subsets of $E$. Two players take turns picking (previously unpicked) vertices; the game is won by the first player to succeed in picking all the vertices of a winning set.

The following example, attributed to Fred Galvin, illustrates what Beck calls "the Induced Extra Set Paradox". The board is $V=\{1,2,3,4,5,6,7\}$; the winning sets are $\{1,2,3\},\{1,3,4\},\{1,4,5\},\{1,2,5\},\{4,5,6\},\{4,5,7\},\{6,7\}$. This game is a draw, but it becomes a first-player win if the board is restricted to the subset $\{1,2,3,4\}$ with winning sets $\{1,2,3\},\{1,3,4\},\{1,4,5\},\{1,2,5\}$.

In Galvin's example the hypergraph is non-uniform, meaning that the winning sets are not all the same size. Here is a slightly larger example of the same phenomenon, attributed to Sujith Vijay, where the winning sets are all triples: the board is $V=\{1,2,3,4,5,6,7,8,9\}$, the winning sets are $\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,3,4\},\{1,5,6\},\{3,5,7\},\{2,4,8\},\{2,6,9\}$. This game is a draw, but the restriction to the board $\{1,2,3,4,5,6,7\}$ is a first-player win.

Beck states the following open problems concerning generalized Tic-Tac-Toe, analogous to your question 3:

Open Problem 5.2 Is it true that, if the $n^d$ Tic-Tac-Toe is a first player win, then the $n^D$ game, where $D\gt d$, is also a win?

Open Problem 5.3. Is it true that, if the $n^d$ game is a draw, then the $(n+1)^d$ game is also a draw?

Yes, the strategy stealing argument works on all board sizes to prove white cannot win. I agree that it is possible that black must play outside the original $15 \times 15$ to block white. You are assuming that black will select at the start a $15 \times 15$ area to play in that will be a win, but that is not required. If the board is $19 \times 19$, black is allowed to play anywhere on the board. I can't see an easy proof that black winning on $15 \times 15$ proves he wins on all larger boards. It could be that the flow in the$15 \times 15$ case lets white make threats on a larger board that take more moves for black to counter than white took to set up. That would allow some board sizes greater than $15 \times 15$ to be a draw, but no way white wins.