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As an example, take the position (x1, x2, x3) = (13,12,8). Is this a P-position? If not,what is a winning move? We compute the nim-sum of 13, 12 and 8:13 =1101212 =110028 =10002nim-sum =10012= 9Since the nim-sum is not zero, this is an N-position according to Theorem 1. Can you finda winning move? You must find a move to a P-position, that is, to a position with an evennumber of 1’s in each column. One such move is to take away 9 chips from the pile of 13,leaving 4 there. The resulting position has nim-sum zero:4 =100212 =110028 =10002nim-sum =00002= 0Another winning move is to subtract 7 chips from the pile of 12, leaving 5. Check it out.There is also a third winning move. Can you find it?2.3 Nim with a Larger Number of Piles.We saw that 1-pile nim is trivial, andthat 2-pile nim is easy. Since 3-pile nim is much more complex, we might expect 4-pilenim to be much harder still. But that is not the case. Theorem 1 also holds for a largernumber of piles! A nim position with four piles, (x1, x2, x3, x4), is a P-position if and onlyifx1⊕x2⊕x3⊕x4= 0. The proof below works for an arbitrary finite number of piles.2.4 Proof of Bouton’s Theorem.LetPdenote the set of Nim positions with nim-sum zero, and letNdenote the complement set, the set of positions of positive nim-sum.We check the three conditions of the definition in Section 1.3.I – 10Ifi844弦໔门o
(1)All terminal positions are inP. That’s easy. The only terminal position is theposition with no chips in any pile, and 0⊕0⊕· · ·= 0.(2)From each position inN, there is a move to a position inP.Here’s how weconstruct such a move. Form the nim-sum as a column addition, and look at the leftmost(most significant) column with an odd number of 1’s. Change any of the numbers thathave a 1 in that column to a number such that there are an even number of 1’s in eachcolumn. This makes that number smaller because the 1 in the most significant positionchanges to a 0. Thus this is a legal move to a position inP.(3)Every move from a position inPis to a position inN.If (x1, x2, . . .) is inPandx1is changed tox′1< x1, then we cannot havex1⊕x2⊕· · ·= 0 =x′1⊕x2⊕· · ·,because the cancellation law would imply thatx1=x′1. Sox′1⊕x2⊕· · ·̸= 0, implyingthat (x′1, x2, . . .) is inN.These three properties show thatPis the set of P-positions.It is interesting to note from (2) that in the game of nim the number of winningmoves from an N-position is equal to the number of 1’s in the leftmost column with anodd number of 1’s. In particular, there is always an odd number of winning moves.2.5 Mis`ere Nim.What happens when we play nim under the mis`ere play rule? Canwe still find who wins from an arbitrary position, and give a simple winning strategy? Thisis one of those questions that at first looks hard, but after a little thought turns out to beeasy.