# Does Black have a winning strategy in Gomoku(freestyle)?

Gomoku is actually a finite two-person game of perfect information. Moreover, if we consider draw as victory of White, then by Zermelo's theorem, exactly one of the two has a winning strategy, either Black or White. In other words, either Black is destined to win, if he does not make any error, or White can at least make a draw.

So my question is which one? the Black or the White?

I have asked a similar question for Go, however, the terminal answer for board 19$\times$19 is still unknown despite of Black having more or less some advantages.

However, for Gomoku there is another story. A programmer asserted that Black has a winning strategy in Gomoku(freestyle). Moreover, (s)he announced (s)he had found this winning strategy and written a program named Gomoku Terminator which "completely terminated the gomoku game". Furthermore, (s)he claimed that the one who first beat the program can earn a bonus $￥920000$ (about $€92000$). But no one has taken this bonus since 2006. So there seems to be sufficient reasons to believe (s)he is right. But I still have a doubt: Do PCs nowadays have enough capability to calculate the whole game tree? Note that the game-tree complexity of Gomoku is PSPACE-complete.

So another question arose: Does Gomoku Terminator(v1.22) really have the winning algorithm for Black?

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The Wikipedia article you linked to states that L. Victor Allis showed in $1994$ that black wins on a $15\times15$ board. The "Gomoku Terminator" site you link to has an image of the upper portion of a board with $15$ columns. Thus it seems that this program merely does what was known to be possible in $1994$, and this has no bearing on the open question of the $19\times19$ board. Allis' thesis states that Gomoku used to be played on $19\times19$ boards because that's the size of Go boards, but that the $15\times15$ board has now become the standard.