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I would like to know if a cycle of moves is possible in the Go variant, Savage Go. That is, you capture my stones, I capture your stones, you capture my stones... The game never ends. A position is eventually repeated, completing a cycle.

Savage Go

Savage Go

Cyclicity could be proved simply by finding a cycle. To disprove cyclicity would require a logical proof. Either way, it won't be easy.

Don't assume perfect or even reasonably intelligent play. I need to know if a cycle can possibly happen, even if it takes player cooperation to make it happen.

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  • $\begingroup$ No. Play on a 2x1 grid and White wins on his first move. Black is annihilated and it's game over. $\endgroup$
    – Mark Steere
    Commented Sep 26, 2023 at 7:22
  • $\begingroup$ The rules are already clear as day. First of all, Black plays first, just like in Go. As clearly stated in the Object Of The Game section, "Savage Go is a game of annihilation. You win by capturing all of the enemy stones on the board." Black places a stone and White kills it, thereby winning the game. $\endgroup$
    – Mark Steere
    Commented Sep 26, 2023 at 7:47
  • $\begingroup$ I am taking a break, since I've just constructively shown the existence of closed (Euclidean) knight's tours on any $\{2 \times 2 \times \cdots \times 2\} \subseteq \mathbb{Z}^k$ chessboard, as long as $k \geq 7$ is given (see arxiv.org/abs/2309.09639, Theorem 4.1 for the proof). Anyway, in this case, I suspect that the N+1 rule could prevent the existence of any cycle, unless we force a stalemate situation with the only exception of an even number of free squares. This is just a personal note in order to think about it in the next few days. $\endgroup$
    – Marco Ripà
    Commented Sep 26, 2023 at 9:13
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    $\begingroup$ Wow, very interesting. Congratulations on that. I will study your proof. $\endgroup$
    – Mark Steere
    Commented Sep 26, 2023 at 9:28
  • $\begingroup$ Trying by hand-drawn, I haven't been able to detect any cycle on grids up to 9 vertices. It seems that my first guess could be the correct one... the N+1 rule covers any hole that we can put on the board, reaching the final step where W or B has enough stones to take all the opponents stones, ending the game. I cannot find a way to circumvent this recurring configuration. $\endgroup$
    – Marco Ripà
    Commented Sep 26, 2023 at 9:29

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