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I am considering generalized domineering. By that I mean domineering played as usual, except that both left and right can either place one or two dominoes on the board during their turn.

I think that the position in the picture (which can arise by playing $6\times 4$ rectangle of generalized domineering) has value $\{4|-1\}$ and therefore temperature of $\frac{5}{2}$. Is that correct?

position in generalized domineering

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The rules of generalized Domineering make the canonical form of this position annoying to find; separate pieces are not always disjoint sums because of the new move rule. With cgsuite's help, I found that the canonical form is not $\left\{4\mid-1\right\}$, but rather $\left\{\left.4,\left\{4\mid\frac32\right\}\right|-1,\left\{\frac32\mid-1\right\}\right\}$. However, the temperature is still $\frac52$.

I checked the canonical form two ways. First, I basically up the game tree by hand in cgsuite, with code seen below. But that was too complicated for me to trust that I hadn't made a mistake, so I asked Mathematica to naively calculate the entire gametree. Cgsuite gave the same canonical form in either case.

twobytwo:={1,0|-1,0};
longL:={-1|0,1};
twobythree:={twobytwo,1,2|longL,1,0};
twobytwoandone:={twobytwo,2,1|0,1};
twobytwoandminus1:= 0 - twobytwoandone;
twobytwotwice:={2,twobytwo,twobytwoandone|-2,twobytwo,twobytwoandminus1};
longLandone:={longL,-1,0|1,2};
twobythreeandone:={twobythree,twobytwo,2,twobytwoandone,3|longLandone,1,2};
longLandminus1:={-2|longL,-1,0,1};
twobythreeandminus1:={twobytwoandminus1,0,1|longLandminus1,0,-1,twobythree};
twobytwoand2:={3,2,twobytwo,twobytwoandone|1,2};
longLandtwobytwo:={twobytwoandminus1,longLandone,longL,0|twobytwo,twobytwoandone,longLandminus1,longL,-1,0};
twobythreeandtwobytwo:={twobytwotwice,twobytwoand2,twobythreeandone,twobythree,twobytwoandone,3|twobythreeandminus1,twobythree,longLandtwobytwo,twobytwoandone,twobytwo,longLandminus1};
longLand2:={1,longLandone,0|2,3};
longLtwice:={longLandminus1,-2|0,1,2,longLandone,longL};
twobythreeandlongL:={longLandtwobytwo,longLand2,twobythreeandminus1,twobytwoandminus1,1,longLandone|twobythree,twobythreeandone,longLtwice,longLandone,longL,longLandone,longL};
twobythreeand2:={twobythreeandone,twobythree,twobytwoand2,4,3,twobytwoandone,3|longLand2,3,2};
twobythreetwice:={twobythreeandtwobytwo,twobythreeand2,twobythreeandone,twobytwotwice,twobytwoand2,4|twobythreeandlongL,longLtwice,longLandone,longL};
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    $\begingroup$ Thanks. I also implemented this in cgsuite (after asking this question) and determined that while the canonical form is not {4|-1}, the reduced canonical form indeed is {4|-1}. $\endgroup$ – Adam Apr 26 '17 at 18:40
  • $\begingroup$ @Adam, good idea. That makes good sense. $\endgroup$ – Mark S. Apr 26 '17 at 18:52

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