Here are some distressing facts about 4 pile Nim. This is a theorem (which I once knew how to prove). For piles (a,b,c,d), hold a and b constant and define f(c) as the (unique) value of d such that (a,b,c,d) is a P position. Then f(c)-c is eventually periodic. Some of you if you have a good program might find it amusing to look at these periods even for the special case where (2,b,c,d) as you increase b. There seems to be no discernable pattern. Adries Brouwer of Eindhoven Technical University ran a program and concluded that most conjectures one might make turn out to be (eventually) false. t 01:16 AM 1/21/04 -0800, you wrote:
Case III:
(0,1+b,1+c,1+d) is a P-position iff nimsum (b, c, d) = 0.
Case IV.1:
(1,b,c,d) is a P-position iff nimsum (b, c, d) = 0.
But case IV.2 (positions (2,b,c,d)) and the general case VI don't seem to have any such nice form. I've found P-positions (2,b,c,d) wiht 2 <= b <= 6.
- Scott
David Gale writes:
Before I start here's something more in the math-fun spirit. I'm offering tw o hundred bucks ($200) for a solution of the following Nim variant. There are 4 piles, ordinary Nim rules, the only difference being that the ga me ends when three of the four piles are empty.
OK---I'll bite, if no one else will.
Case I: (Single non-empty heap) Answer: *0 (the game is already over) everything is a P position, ie 2nd play er to move wins.
Case II: (Two nonempty heaps, sizes m and n). Everything is an N-position. A wiinning move to take everything in one of the heaps. For extra credit (but not for the $200), after a few minutes of noodling I got the nim heap equivalent is BitXor[m-1,n-1] +1.
Case III: (Three nonempty heaps, sizes r,s,t). Assume r <= s <= t. Every move involving taking a whole heap leads to a (Case II) N position. Wha t are the three heap P positions? 1,1,1 is a P position for example. Then 1,1,t for other t>1 is always an N po sition, since there's a move to 1,1,1. It looks like the P positions are in one-to-one correspondence with every pair (a,b) = (s-r, t-s) with a, b nonnegative. Ie, the P position (1,1,1) corresponds to the choice (a,b) = (0,0), and (for example), the P position (2,3,4) corresponds to the choice (1,1), (and for another example), (4,6,7) corresponds to (2,1), E T CETERA, where I don't really know what ET CETERA means. Assuming this correspondence is valid, how to reco ver r from a and b? This should be achievable---but not after two glasses of chianti. Is it hopeless to give a formula for the Case III nim equivalent?
Case IV: (Four nonempty heaps).
Hmmm....
Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org
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