[Variant] No small chains Dots and Boxes

11 replies. Last post: 2015-05-23

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[Variant] No small chains
  • purgency at 2015-05-18

    Here is an idea for a variant where small chains count as long chains:

    Additional rule: drawing the third wall of a box is forbidden unless it’s a loony move.

    example
    - The position has 8 long chains
    - Forbidden moves: a6, b5, b9, f9, h9, h11, j1, k2, j7, j11, k10
    - final score is 13-10 (the two isolated 1-chains won’t be taken at all since it will never be allowed to sacrifice them)
    - interesting note: there’s no quad

    Would somebody be interested in trying this out in an unrated game? I don’t know if one can be good at this since controlling the number of really small regions like 1-chains seems kind of impossible, but I’d give it a try. Alternatively one could try a version where sacrificing 2-chains in the center is forbidden, but sacrificing 1-chains is allowed (that would make it easier to pursue an aim).

  • purgency at 2015-05-18

    Oups, I discovered a small flaw.

    The additional rule should be: drawing the third wall of a box is forbidden unless the move is loony or completes a box.
    Otherwise it wouldn’t be allowed to take the last 2 boxes of a chain which would be stupid.


    No wait, that’s still incorrect.

    The additional rule should be: drawing the third wall of a box is forbidden unless the move is loony, completes a box or completes a domino.
    Otherwise it wouldn’t be allowed to use doublecross-strategy since completing a domino would also draw the third wall of a box.


    I am sorry about that, the rule sounds kind of ugly now... I didn’t think this through and forgot there’s other ways of drawing the third wall of a box than by sacrificing.

  • William Fraser at 2015-05-19

    Sure, I’ll give it a try.

  • The_Shark_c at 2015-05-20

    For those who want to watch this gruesome affair, http://www.littlegolem.net/jsp/game/game.jsp?gid=1710952

  • ypercube at 2015-05-21

    Let me get some pop corn ;)

  • Carroll at 2015-05-21

    Please do not provide hints on where the chains will pop out.

  • William Fraser at 2015-05-22

    Thanks, purgency.
    I really had no idea what was going on until move 29, when I spent about 45 minutes working out the nim-values of all regions.
    From that point on, I knew I was winning the control battle, which meant that I couldn’t lose (as sacrifices don’t exist, so I couldn’t be forced to make one), but I couldn’t see any way to force a win, either.
    Does anyone else see a winning line for me?
    And does anyone see a way for purgency to avoid giving me control on move 28?

  • purgency at 2015-05-22

    Thanks for volunteering Bill and well done destroying me nim-wise! Also, I think you meant move 18 and 19?

    Interesting outcome for the first game. I wonder if this is the score that results from perfect play?I really enjoyed the endgame as a nice nimstring exercise and I think that’s not unusual for the variant which I find pretty cool.

    I didn’t really figure out what to do during the first half so I played pretty randomly for the first 16 moves too and after that I miscalculated but somehow managed to force a draw by counting which wasn’t planned out at all. One interesting thought I had later is that isolated one chains can be prevented completely just from playing one move in each corner which would also make a draw impossible, that’s why I don’t think the score with perfect play is a draw, but it’s a total mystery to me whoever has the advantage.

  • William Fraser at 2015-05-22

    The correct move number was 21.  (And thus my question was about moves 21 and 20).  I half-heartedly tried doing the calculation on move 19, but it seemed a bit much at that point.

    Somehow I don’t think it’s worth trying to reprogram the shark, although it would run much faster, as I wouldn’t need to consider the huge class of positions in which boxes are captured when none of their neighbors are captured.

  • al_ef13 at 2015-05-23

    Well, what I can see immediately for any rectangular board nxm is that (with optimal play):

    1) if n and m have the same parity, then P2 has at least a draw;

    2) if n and m have different parity, then P1 has at least a draw.

    In the case 1), P2 just plays central symmetry strategy until P1 makes a loony move, and then by stealing strategy argument he cannot lose with optimal play.

    In the case 2), P1 plays the central line on move 1, and then similarly applies the central symmetry strategy.

    Also, I am strongly convinced that if n and m are both odd and greater than 1 (e.g. n=m=5), then P2 actually wins. The only possible way for P1 to equalize in this case is to play at some point a loony move that opens a 2-chain or a 4-loop, but it seems unlikely that this can lead to a draw. However, I did not check the case 3x3, which seems to be not that hard.

  • purgency at 2015-05-23

    Oh yeah, you’re right indeed... RIP “no small chains”, i guess... Idk, i came up with the variant when i watched a numberphile video about removing the fifth axiom in geometry or something like that which apparently opened a new and super cool dimension of mathematics so I thought: "Hey, let’s just remove a random element from dots and boxes and see what happens."
    Removing small chains admittedly wasn’t quite one of the more ingenious ideas i’ve had in my life, but i kind of like the concept of simply removing (or possibly adding) something and experimenting with whatever is being created. Someone more versatile in the art of game designing may find some better variant and when symmetry strikes once more you can simply add the rule "turning a non-symmetric position symmetric is forbidden."
    As for this variant though, I didn’t really like the concept of removing sacrifices along with it anyways since it’s a rather nice element of dnb, so I’ll personally just forget about it. One may be interested in a variant where only 2 chains are long, possibly a variant where 3-chains are short or one where loops are chains and chains are loops, if that’s possible (I don’t think it is).

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