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

Scattercut rules

Scattercut rules 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.

Solution by Christopher Field:


"If red plays on the left side and blue plays on the right side and they never play in each other's spots, then it's a coop cycle."

Now adding rule... "If you have a choice of placements, you must choose one that results in the maximum number of your stones on the board."

Same question. Are cycles possible, and if not, can it be proved?

  • $\begingroup$ If placing your stone in a crosscut position forms a path, do you instantly win? Or must you have that connection after the crosscut stone has been removed? $\endgroup$
    – Dr Xorile
    Aug 22, 2023 at 0:24
  • 1
    $\begingroup$ Very good question. Not something I had considered or included in the rule sheet. I think it should be that at the conclusion of your turn there must be a connecting path in order for you to win. I've updated the rule sheet to reflect that. Thanks for pointing this out. $\endgroup$ Aug 22, 2023 at 2:05
  • $\begingroup$ In the little playing around I've done with this, I suspect that's the better answer. $\endgroup$
    – Dr Xorile
    Aug 22, 2023 at 2:25
  • $\begingroup$ Christopher Field's solution is essentially the same as my solution, although the bigger board makes it possible to have neither side winning. I think it would be better to make a separate question rather than edit this question if you change the rules. $\endgroup$
    – Dr Xorile
    Aug 22, 2023 at 4:38

1 Answer 1


When you fill up all the squares one side or other will have won (I'm pretty sure it's similar to this).

But if we ignore that for now, I think you can get a cycle with cooperative play (but not in practice because one side will have won already).

Suppose you start in this position with red to play (for clarity I'm only showing the colors that matter. Grey squares can be red or blue but must be filled.):

Position 1

Suppose red plays B4. Then they pick up C5, but cannot play it:

Position 2

Then suppose blue plays E3. Then they pick up D2, but cannot play it:

Position 3

Then red plays C5. They pick by B4, but cannot play it:

Position 4

And finally, blue plays D2. They pick up E3, but cannot play it and we're back to the first position.

  • $\begingroup$ Very interesting. You have found a cycle. But why must one of the players have won before the cycle has completed? And is that true for all cycles or just this cycle? $\endgroup$ Aug 22, 2023 at 2:07
  • $\begingroup$ It's probably true for all, but I don't know. In this case, it's just that if the board is this full someone will have won. I'd need to think how to prove that. $\endgroup$
    – Dr Xorile
    Aug 22, 2023 at 2:23
  • $\begingroup$ Ok, well great work so far. Thank you for looking into it. $\endgroup$ Aug 22, 2023 at 2:33
  • $\begingroup$ In a couple of test games, I suspect there's a winning strategy for the first play. Have you looked at that? Like if they get onto a main diagonal and then keep working it, it seems hard to stop... $\endgroup$
    – Dr Xorile
    Aug 22, 2023 at 2:46
  • $\begingroup$ If you have all squares occupied, then if red hasn't won, then blue must have cut red off the whole width of the board and therefore blue must have won. So one or other will have won when the board is full (or both). But the crosscut may prevent the board getting full, of course. $\endgroup$
    – Dr Xorile
    Aug 22, 2023 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.