My question is whether or not a cycle can occur in the game of Savage Go. That is, you kill some of mine, I kill some of yours, you kill some of mine... Endless cycle of turns. Game never finishes.

No level of player skill is assumed. In fact, assume the players are cooperating in trying to create a cycle.

If you can't find a cycle, a proof that cycles can't occur would be most helpful. Otherwise, please just describe your experience with this. It's helpful to know if people made a serious effort but couldn't find a cycle.

To summarize, I need proof that cycles can occur or proof that cycles can't occur.

Savage Go

Savage Go


2 Answers 2


I don't have a rigorous proof, but here's my take on it:

  1. Every turn, the board ends up with one more stone on it unless a player is unable to play every stone.

  2. If each player can play all of their stones every turn, the board will eventually be full. If that happens, then the player who plays last will win, because playing a stone that completely fills the board will necessarily surround all the opponent's groups, removing them from the board.

  3. The only way to prevent (2) is to repeatedly have a scenario where a player cannot play all of their stones on their turn.

  4. The only things that can prevent a player from having to play all their stones on a turn are:

    • There are no empty spaces available on the board => this results in a win for the current player, per (2), and an end to the game.
    • The only spaces available to play would result in a stone being self-bounded without allowing for removal of any of the stones bounding it.
  5. From (3) and (4), the only way to possibly have a cycle is for each player to be continually prevented from playing all their pieces because of a self-bounding situation.

  6. If there is a configuration of stones that prevents Black from playing a piece because it will be self-bounded, then that space is a valid move for White (since it is surrounded by white pieces or the board edges). Thus White will eventually need to play a stone there (since the number of pieces on the board will always increase, barring illegal moves), at which point it will no longer be an issue for Black.

  7. From (5), (6), and (3), since no board space can be an illegal move (self-bounded) for both Black and White, and thus will eventually be filled, the board will eventually be completely filled, always resulting in a win for one player or the other.

This seems intuitive to me, but it's possible I missed something. If so, I'm sure someone will point it out.

  • $\begingroup$ Ok, thanks. I will study this, and let's wait and see if anyone else has anything to say. $\endgroup$ Sep 21, 2023 at 21:23
  • $\begingroup$ I don't think 1. is strictly true. I've played a test game even where at move 13, there's a ko shape in Go, there's 13 stones on the board. White captures the ko stone, still 13 stones on the board at move 14. Black plays two moves and captures two stones, still 13 stones at move 15. I believe that the general tendency will be that more pieces will be added as time goes on though. $\endgroup$
    – snulty
    Sep 21, 2023 at 23:04
  • $\begingroup$ This was the test game: online-go.com/review/1142848 $\endgroup$ Sep 22, 2023 at 1:10
  • $\begingroup$ Any ideas @drxorile? $\endgroup$ Sep 22, 2023 at 2:43
  • $\begingroup$ Early in the game, when both players can play all of their stones, the only way for the total number of stones not to increase in consecutive turns is for more stones to capture more stones to capture more stones... Obviously that can't go on indefinitely. Cycles would have to happen later in the game and involve players not being able to place all their stones. $\endgroup$ Sep 22, 2023 at 3:45

A cycle on a degenerate case is possible:

On 1x2 board.

White is X, Black is O.
1. ..
2. X.
3. XO -> .O Black captures 1 White.
4. XO -> X. White needs to place 2, but can only place 1. Captures 1 Black.
5. XO -> .O Black needs to place 2, but can only place 1. Captures 1 White.
6. Back to step 4.

Similar idea can probably be done on larger boards, but need to show the build up until the cycle steady state.

  • $\begingroup$ No. It's game over at step 3. The goal is annihilation. O wins. $\endgroup$ Sep 21, 2023 at 13:43
  • 1
    $\begingroup$ Ah, noted! This answer is incorrect then. But will keep it here as a warning for others. $\endgroup$
    – justhalf
    Sep 21, 2023 at 13:47
  • $\begingroup$ Ok :D Good idea. $\endgroup$ Sep 21, 2023 at 13:49

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