Once Nimber arithmetic is explained with 3 stacks, there is no need to go any further. The value of four or more stacks can be determined as combinations of 1, 2 and/or 3 stacks.

The starting point here is to use 3 stacks containing 1, 2 and 3 counters. These 3 stacks (1* + 2* + 3*) have the combined value of 0. If the first person removes one of the stacks, the second person can then equalize the remaining stacks and leave the Game with a value of 0 again. Alternatively if the first person equalizes two of the three stacks, then the second person can remove the un-equal third stack and leave the game with 2 equal stacks again.

So the first set of equivalences in Nimber Addition are;

1* + 2* + 3* = 0

1* + 2* = 3*

1* + 3* = 2*

2* + 3* = 1*

The next triplet of Nimbers in Winning Ways, Vol. 1 that equal 0 can be played out in much the same manner. In addition to the moves mentioned before (first player: removes a stack, second player: equalizes the remaining stacks and first player: equalizes 2 stacks, second player: removes the un-equal stack), the second player also has the following strategy. If the first player reduces one of the bigger stacks to either 2* or 3*, the second player can remove counters from the other large stack to give it a value of 3* or 2* respectively. The game now has the value of 1* + 2* + 3* or, as previously determined, 0.

Evaluating any random set of 3 stacks is not terribly difficult. One just has to find the Nimber value of two smaller stacks. If this value equals the third stack, the total value of the triplet is 0 and the first person to play loses. If the combined Nimber value of the smaller stack does NOT equal the third stack, all the first player needs to do to win is to reduce the third stack to the value calculated and leave the Game with the value of 0.

Further examples of Nimber addition can be found in Berkamp, Conway and Guy's "Winning Ways, Vol. 1", pages 42 and 59 - 59.

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