Edit: This problem is equivalent to the problem $20.\space b)$ from Завдання ХХIIІ ТЮМ (2020 р.) which talks about the divisors of $10^{2n}$ where legal moves are $\{\div 10, \times 2,\times 5\}$ to move from the previous divisor to the next one, and you can't revisit a divisor. But, it appears they did not publish the solutions. $-$ Thank you Witold for the reference.

Consider an odd sized chessboard $(2n+1)\times(2n+1)$ with WLOG bottom left corner square at $(0,0)$ and top right corner square at $(2n,2n)$. A piece whose allowed moves are $\{(-1,-1),(0,+1),(+1,0)\}$ is placed on one of the squares.

Two players then alternate moving the piece, such that the piece never stands on the same square twice. (The piece can't revisit the squares.) The player that can't move the piece, loses. (The player that last moved, wins.)

On which starting squares will the first player have a winning strategy?

I have brute-forced the game for boards of sizes $(2n+1)=3,5,7,9,11$. If the second player has a winning strategy, the square is colored blue. Otherwise, the first player has a winning strategy and the square is green.

Some special cases have simple strategies.

• For example, the square $(2n,2n)$ is clearly a win for the second player: The only move the first player can do, is the diagonal move $(-1,-1)$. Then, if the second player chooses either of the two non-diagonal moves, the first player again is forced to move diagonally. Repeat this until we reach the adjacent corner, where the first player can't move and loses.
• Another example is the $(0,0)$ square which is also clearly a win for the second player. Notice that the first player can't use the move $(-1,-1)$. Hence, which ever of the two other moves the first player uses, the second player can always return to the main diagonal, which eventually leads to the win for the second player.

But in general, I'm not sure what strategies can be used to prove the patterns for all squares. For example, it seems that if the piece starts on a square $(a,b)$ where $a,b$ are both even, then the second player can always force a win. Can we prove this?

Otherwise, if $a,b$ are not both even, then it seems the first/green player can force a win, unless the square is one of the "exceptions" (additional squares where the second/blue player can force a win). Since the squares are mirrored relative to the main diagonal $(0,0)-(2n,2n)$, WLOG we list only the exceptions below that diagonal:

• If $n\le2$ there are no exceptions.
• If $n=3$, the exceptions below the diagonal are $(4,3),(6,3)$.
• If $n=4$, the exceptions below the diagonal are $(8,3),(5,4),(8,5)$.
• If $n=5$, the exceptions below the diagonal are $(6,1),(7,4),(6,5),(10,5),(7,6)$.

But what will be the exceptions for $n\ge 6$? I do not see a pattern-

An equivalent question, but more generally for both even and odd boards $(n+1)\times(n+1)$, was asked on MSE almost a month ago (from the time of posting this question): Winning strategy in a game with the positive divisors of $10^n$. It did not receive any answers other than results on brute-forced boards (which I already presented here). Therefore, I decided to cross-post the odd case of boards here. $-$ P.S. If anyone manages to also solve the general linked problem on MSE, I'll pass on to them the bounty which will be at least 150.