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How can be it proved that tic-tac-toe on an infinite grid (winning with $12$ in a row, a column or a diagonal) can always end in a tie (with optimal strategies of both players)?

There is a hint: to use a "magic square of $4\times4$" and a "tessellation".

Peter Woolfitt
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John Smith
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  • Tic-tae-toe on an infinite grid can never end in a tie. Presumably you mean that neither player has a winning strategy. – Peter Taylor Apr 15 '14 at 10:39
  • @PeterTaylor I define tie as "not making 12 in a row, a column or a diagonal", so the infinite play is a tie. You are right about the inappropriateness of the verb "end", though. – John Smith Apr 15 '14 at 18:42
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    There's a proof in one of the later volumes of Winning Ways that the 9-in-a-row game is a draw on an infinite board. A fortiori the 12-in-a-row game is too. – MJD Aug 26 '14 at 14:10
  • @PeterTaylor According to the rules, the game ends after $\omega$ moves if nobody has made $12$ in a row. One could consider a variant, where play continues into the transfinite as long as there are any unoccupied points, but this is not so popular. – bof Apr 19 '15 at 12:25
  • @MJD Winning Ways (in the edition that I have) has only two volumes. A drawing strategy for $9$-in-a-row is shown in vol. 2, figure 12 on p. 677. By the way, your a fortiori would be hard to justify. Fortunately, the same strategy works for $n$-in-a-row for all $n\ge9.$ That's because it's purely defensive; it blocks the opponent from making $9$-in-a-row, it does not depend on counterattacking by threatening to make one's own $9$-in-a-row. – bof Apr 19 '15 at 12:37
  • @bof Mine is in two volumes also, but there is a new(er) edition by A.K. Peters in which the four parts (“Spade work”, “A change of heart”, etc.) are in four separate volumes. – MJD Apr 19 '15 at 12:41
  • @bof: Can you explain what you mean by "hard to justify"? MJD's argument seems straightforward: the second player has a drawing strategy in 9-in-a-row, so if the second player plays that strategy in the 12-in-a-row game, then the first player never gets a row of length 9, and so never gets a row of length 12 either. – Gareth Rees Apr 28 '15 at 09:46
  • @GarethRees Consider ordinary tic-tac-toe on $3\times3$ board. Note that the second player can not block the first player making a row of $3$: 1. a1 b2 2. c3 b1 3. a3 b3 4. a2 and player one has $3$ counters on the $a$-file. Of course player two got $3$ counters on the $b$-file. Player can force a draw in ordinary tic-tac-toe, but only because he can counterattack by threatening to make his own $3$-in-a-row. If the drawing strategy for $9$-in-a-row relied on counterthreats, it would not be clear that it works for $10$-in-a-row. – bof Apr 28 '15 at 10:48
  • @GarethRees See this answer and note the Open Problems, esp. 5.3. – bof Apr 28 '15 at 10:49
  • @bof: I understand your point now. But the proof from Winning Ways that MJD referred to does not rely on counterattacks: it's a Hales-Jewett pairing strategy. – Gareth Rees Apr 28 '15 at 10:54
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    @GarethRees Yes, I know. I remarked on that fact in my comment to MJD, and again in my reply to your comment, by bolding "If". The only reason for my comment to MJD was that he did not mention the pairing strategy but only reported that "the 9-in-a-row game is a draw", which I thought could be misleading. – bof Apr 28 '15 at 11:10

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