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Nimber multiplication

Mathematics Asked by yberman on July 30, 2020

In the question on nimbers, the original poster asks for the meaning of Nimber multiplication in the context of impartial games.


Edit: As noted by Mark Fischler in the comments below, the following is wrong

My gut instinct is $*a times *b$ means that if $*a$ is a game equivalent to $a$ stones, and $*b$ is a game equivalent to $b$ stones, then if you replace every stone in the $*a$ game with a copy of the $*b$ game, you get a game with the Nimber $*a times *b$, but I haven’t been able prove it.

2 Answers

I don't think there's an intuitive way to understand nimber multiplication. Also note that it is distinct from repeated addition, so it isn't as simple as replacing stones with copies of piles. Multiplication is defined recursively as $$ab = text{mex}left({a'b+ab'+a'b': a'<a, b'<b}right)$$ where $text{mex}$ is the minimum excluded element.

Although it's hard to see what this means in terms of a concrete game, it's possible to understand why it's defined this way. The point is that we want to create an algebraic system without zero divisors, so $(a-a')(b-b') neq 0$ whenever $aneq a'$ and $bneq b'$. In other words, $$ab - ab' - a'b + a'b' neq 0,$$ so $$ab neq a'b + ab' - a'b'.$$ Since subtraction and addition are the same operation, we have $$ab neq a'b + ab' + a'b'.$$ The definition above takes the first nimber that meets this criterion.

Correct answer by Théophile on July 30, 2020

Winning Ways Volume 3, Chapter 14, considers coin-turning games. One such game is Turning Corners, where a move is to turn over the four corners of a rectangle with horizontal and vertical sides, but the North-East most coin needs to be turned from heads to tails. A sample move shows that it defines nimber multiplication:

a           X           H
.  
.  
.  
a'          X           X
.  
.  
.  
0  .  .  .  b' .  .  .  b

A move will turn the H to tails and Xs over, so from the position $(a,b)$, a move moves to the sum $(a',b)+(a,b')+(a',b')$, so the value of the position $(a,b)$ is the nimber product $ab$ inductively.

Answered by E. Z. L. on July 30, 2020

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