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In an $a\times b$ board, two players take turns putting a mark on an empty square. Whoever gets $c\leq \max(a,b)$ consecutive marks horizontally, vertically, or diagonally first wins. (Someone must win because we use only one mark type.) For each triple $(a,b,c)$, who has a winning strategy?

For $a=b=c=3$ (tic-tac-toe size), the first player can win by first going on the middle square and winning in the next turn. A generalization is that for $a,b,c$ are all odd, the first player can go on the middle square, then reflect the second player's move across the middle square. (He also needs to keep his eyes open in case the second player marks the $(c-1)$st square of a $c$-in-a-row, so that he can win immediately.)

In the one-dimensional case ($a=1$), this may well be a known game, but I also cannot find a reference.

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    $\begingroup$ Remarks: $c=1,2$ are trivial of course, and $c=3$ is already interesting enough to make a full question. The first nontrivial case, $(a,b,c)=(4,3,3)$, is a second-player win: the two maximal configurations without immediate winning moves are: $$\begin{pmatrix} \rm X & \cdot & \cdot & \cdot \\ \cdot & \cdot & \rm X & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} \cdot & \rm X & \cdot & \cdot \\ \rm X & \cdot & \cdot & \rm X \\ \cdot & \cdot & \rm X & \cdot \end{pmatrix}$$ $\endgroup$ Jul 24 '20 at 17:13
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    $\begingroup$ Have you solved the one-dimensional case yet (i.e. when $a = 1$)? $\endgroup$ Jul 27 '20 at 0:06
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    $\begingroup$ The $a=1$, $c=3$ case is (equivalent to) an octal game, and has signature 0.11337 I think. $\endgroup$ Jul 29 '20 at 9:09
  • $\begingroup$ Related question on MO: mathoverflow.net/questions/24693/neutral-tic-tac-toe $\endgroup$
    – acupoftea
    Aug 26 '20 at 18:45
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    $\begingroup$ Now posted to MO, mathoverflow.net/questions/386447/… $\endgroup$ Mar 14 at 21:53
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I dont have a general-case answer rather some insight regarding the c=3 case leading to a general conclusion: (In the c=3 case:)Given some tic already placed 'p', you lose if you place a tic 'q' in a place allowing the opponent to complete a 3-line with p & q. One notices quickly that all such tics lie within the Manhattan-distance circle of radius 2 centered around p. This means that if when you start playing one can place a tic everywhere, each tic placed limits the non-losing space with the area of a square (a Manhattan-distance circle is a square rotated). This means each player 'survives' their turn by placing another square inside the remaining area, so this is a classic Packing problem. Many variants exist such as circle packing (https://en.wikipedia.org/wiki/Circle_packing#:~:text=In%20geometry%2C%20circle%20packing%20is,enlarged%20without%20creating%20an%20overlap.) or square packing (https://en.wikipedia.org/wiki/Square_packing_in_a_square#:~:text=Square%20packing%20in%20a%20square%20is%20a%20packing%20problem%20where,wasted%20space%20for%20non%2Dinteger), both of which dont have a closed form for their packing capacity, this leads me to believe that, other than some specific cases, this generalization your searching for does not exist!

I think in cases c>3 one can express the area taken by a move similarly (much more complex but still an area taken-up) and thus some weird packing-problem. This leads me to believe no generalization exists at all for any c.

The specific cases are cool though. Cheers!

(Manhattan-distance AKA Taxicab distance : https://en.wikipedia.org/wiki/Taxicab_geometry)

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    $\begingroup$ Sorry I accidently downvoted, now I can't undo it. Nice answer. I think the shape is an eight-pointed star (plus and times together) instead of a Manhattan circle $\endgroup$
    – Empy2
    Aug 2 '20 at 14:36
  • $\begingroup$ Empy2 is right about the "forbidden" area not being a square. It is in fact all squares distance $\le 2$ away in $x$, and $\le 2$ away in $y$ EXCEPT a knight's move away. Anyway, even if the shape was nicer, packing and game playing are entirely different problems. In packing, the only "player" tries to pack as many as possible. In a game, the optimal moves by the two players might not lead to any sort of tight packing at all. E.g. in the 1-D version ($a=1$) the forbidden shape is very nice, and the packing problem is easily solved, but that doesn't help to solve the game at all. $\endgroup$
    – antkam
    Aug 2 '20 at 22:08

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