I read the book 'Tic-Tac-Toe Theory (author : Jozsef Beck)', and saw the author's mention like this.

"we know only two explicit winning strategies in the whole class of $n^d$ Tic-Tac-Toe games: the $3^3$ version, which has an easy winning strategy, and the $4^3$ version, which has an extremely complicated winning strategy."

I think author maybe assume $n \ge 4$, because I think that it is easy to make winning strategy for first player in any $3^d$ Tic-Tac-Toe. But I want to confirm definitely, so post this question. Is it right that author assume $n \ge 4$ without comment?

And I have one more question. Was following proposition proved?

"if first player have winning strategy in $n^d$ Tic-Tac-Tow, then first player also have winning strategy in $n^{d+1}$ Tic-Tac-Toe."

  • 1
    $\begingroup$ $3^2$ doesn't have a winning strategy, only a forced draw (or better if opponent makes bad decision). $\endgroup$ – Justin Benfield Jun 26 '16 at 11:59
  • $\begingroup$ Are you saying that a winning strategy for $3^d$ (I assume you mean for $d\gt3$) is easy because you think it can be derived from the winning strategy for $3^3$? If so, that's at least not entirely obvious. Can you provide an argument for that claim? $\endgroup$ – joriki Jun 26 '16 at 12:22
  • $\begingroup$ Sorry. I want to fix miss expression. I meant non-losing strategy. $\endgroup$ – PPPiRi Jun 26 '16 at 14:45
  • $\begingroup$ And I realized that I thought wrong way... I confused winning strategy with non-losing strategy. thank you for answer! :) $\endgroup$ – PPPiRi Jun 26 '16 at 14:48
  • $\begingroup$ Perhaps you can write up an answer to your own question and accept it so the question doesn't remain unanswered? $\endgroup$ – joriki Jun 26 '16 at 17:47

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