# Tic-tac-toe with one mark type

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.

• 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: $$\begin{pmatrix} \rm X & \cdot & \cdot & \cdot \\ \cdot & \cdot & \rm X & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} \cdot & \rm X & \cdot & \cdot \\ \rm X & \cdot & \cdot & \rm X \\ \cdot & \cdot & \rm X & \cdot \end{pmatrix}$$ Jul 24 '20 at 17:13
• Have you solved the one-dimensional case yet (i.e. when $a = 1$)? Jul 27 '20 at 0:06
• The $a=1$, $c=3$ case is (equivalent to) an octal game, and has signature 0.11337 I think. Jul 29 '20 at 9:09
• Related question on MO: mathoverflow.net/questions/24693/neutral-tic-tac-toe Aug 26 '20 at 18:45
• Now posted to MO, mathoverflow.net/questions/386447/… Mar 14 at 21:53

• 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. Aug 2 '20 at 22:08