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 $\{(-1,-1),(+1,0),(0,+1)\}$ 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 are easy. 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.

But in general, I'm not sure how to prove these observed patterns. 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 player (green player) can force a win, unless the square is one of the "exceptions" (additional squares where the second player can force a win). Since the squares are mirrored relative to the $(0,0)-(2n,2n)$ diagonal, 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.