I'll pitch in one more detail, which is that this game is tame in misere play. This means that the misere nim strategy, which says follow the normal play strategy unless its recommended move leads to a position with heaps of size one only, carries over to this variant in misere play. Which gives me a lame excuse to mention an open algebraic problem in misere nim variants. Suppose misere nim is played with the restrictions that #1) 4k beans can never be removed from a heap, and #2) 4k+3 can be removed only if it is the whole heap Is the misere indistinguishability quotient of this game isomorphic to the following commutative monoid Q with 226 elements and the following presentation? Q = <a,b,c,d,e,f,g,h,i,j | a^2=1,b^3=b,b^2c=c,c^4=ac^3,b^2d=d,d^2=b^2,b^2e=e,c^3e=ac^2e, cde=bce,ce^2=ae^2,de^2=be^2,e^3=abe^2,b^2f=f,cf=c^2,ef=ac^2e,f^2=ac^3, b^2g=g,c^2g=ac^3d,eg=bc^2e,fg=c^3d,g^3=dg^2,b^2h=h,c^2h=be^2,eh=ae^2,fh=abe^2, g^2h=dgh,h^2=e^2,b^2i=i,c^3i=be^2,c^2ei=e^2,e^2i=abe^2,cgi=abch,hi=abfi,c^2i^2=abc^2i, ei^2=abei,fi^2=abfi,g^2i^2=e^2,i^3=abi^2,b^2j=j,cj=ci,ej=ei,fj=fi,g^2j=abgh,hj=abfi,ij=i^2,j^2=i^2>; See http://arxiv.org/abs/math.CO/0609825, section 3.6 for more Happy new year Thane On 1/3/07, Michael Kleber <michael.kleber@gmail.com> wrote:
Since RKG didn't say it, let me give the three-line explanation of why the option to split piles doesn't matter:
In regular nim, you may change a pile of size n to one of size m for any m<n. Here you may also split it into piles of size m1, m2 with m1+m2 < n (regular addition), but these two piles are equivalent (by induction) to one pile of size m1 (+) m2 (nim-addition), which is also <n. Every move in the new game is therefore equivalent to one that was already available; you can win by simply playing nim and ignore the fact that the other player may occasionally make a move you hadn't known existed.
--Michael Kleber
-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
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