Hex variant: randomly removed cells Hex, Havannah

8 replies. Last post: 2018-01-30

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Hex variant: randomly removed cells
  • apetresc ★ at 2018-01-23

    One simple variant that I’ve been pondering is just normal 13x13 or 19x19 hex, but with some small number (say, 1-3) cells removed from the board; they’re just not valid plays for either player. These would just be randomly generated at the very beginning, and I trust the swap rule would still keep everything balanced.

    The main problem, of course, is that once you remove cells from the board, you can now have draws again. My question is, do any of the stronger players here have any sort of intuition about how often such draws would occur in high-level play? For, say, a single removed cell? Two? Three? I’ve played a few test games myself and it’s added some interesting dynamism, but I’m a lowly ~1500 rated player. Hoping to get some insight from someone with real intuition.


  • ypercube at 2018-01-24

    Interesting variant. I think I’ve read the idea somewhere in the past but never actually played it.

  • HappyHippo at 2018-01-24

    That’s an interesting idea.

    Without having tried it, I do think that draws would be common. The empty hexes would serve as natural “blocks” that could be used against the other player. While in standard hex, blocking an opponent necessarily results in making your own connection, in this variant one could block the opponent without making a connection of your own. But without trying it I can’t be sure.

  • Arek Kulczycki at 2018-01-24

    It seems that playing for a draw would be relatively easy. Example, say e7 is a missing cell (instead of white like in the picture)

    The reasoning:

    Assume the missing cell is around center like above. Assume white uses this cell like he would use a white stone. 1 additional stone in center is a huge advantage hence it’s a relatively easy for white to connect both sides through it, therefore forcing a draw.

    I don’t know what can happen if both players play for a win, but maybe eventually the same.

  • David J Bush ★ at 2018-01-25

    If the removed cells are on the perimeter, and there are still four corner cells which are part of both borders, a draw could be avoided. Maybe one player removes cells and places the first stone & the other chooses to swap or not.

  • Tasmanian Devil at 2018-01-26

    I think in a game between two players of uneven strength, the weaker player will remove the centre cell and try to force a draw if he gets to make the first move, while the stronger player will remove perhaps an acute corner cell. This suggests that the removed cell should be fixed, not selected by either player. 

  • apetresc ★ at 2018-01-30

    Thanks for the feedback, all. Arek’s point is well-taken; either player forming a path through the cell would likely result in a draw, and thinking of it as a handicap stone makes it clear just how easy that should be. So this idea probably wouldn’t work in practice.

    So how about this: instead of random cells being removed from the board, instead they are made to count for both players. This doesn’t create drawing opportunities as far as I can tell, and maybe creates some cool racing situations that don’t exist in regular Hex, although I haven’t tried it. Does this give player 1 too large an advantage, so that no swap move could possibly outweigh it? Unlike my previous idea, I haven’t actually experimented with this one at all.

  • Arek Kulczycki at 2018-01-30

    This seems to be a valid variant to me, however I’m not too fond of playing variants in general. But IMO your game should be playable equally to normal Hex.

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