Transfinite Nim | Joel David Hamkins

Shall we have a game of transfinite Nim? One of us sets up finitely many piles of wooden blocks, each pile having some ordinal height, possibly transfinite, and the other of us chooses who shall originate the first relocate. Taking turns, we each successively erase a top part of any one pile of our choosing, making it strictly lessen. Whoever consents the very last block triumphs. (It is fine to erase an entire pile on a turn or to erase blocks from a branch offent pile on a tardyr turn.)

In my contest problem last week, for example, I set up six piles with heights:
$$1qquad omega+3qquad omega^omega+5 qquad omega^{omega+3}+omega^omegacdot3+omegacdot 5+7qquad epsilon_0qquad omega_1$$Would you want to go first or second? What is the best relocate? In vague, we can commence with any finite number of piles of arbitrary ordinal heights — what is the vague triumphning strategy?

Before progressing with the transfinite case, however, let’s appraise the triumphning strategy in frequent finite Nim, which I elucidateed in my post last week troubleing my visit to the 7th/8th grade Math Team at my son’s school. To say it rapidly aget, a finite Nim position is equitable, if when you think about the binary recurrentations of the pile heights, there are an even number of ones in each binary place position. Another way to say this, and this is how I elucidateed it to the school kids, is that if you skinnyk of each pile height as a sum of exceptional powers of two, then any power of two that occurs in any pile does so an even number of times overall for all the piles. The mathematical facts to set up are that (1) any relocate on a equitable position will unstability it; and (2) any unequitable position acunderstandledges a balancing relocate. Since the triumphning relocate of taking the very last block is a balancing relocate, it adheres that the triumphning strategy is to stability wdisenjoyver position with which you are faced. At the commence, if the position is unequitable, then you should go first and stability it; if it is already equitable, then you should go second and adchoose the balancing strategy. It may be fascinating to remark that this triumphning strategy is exceptional in the sense that any relocate that does not stability the position is a losing relocate, since the opposing joiner can adchoose the balancing strategy from that point on. But of course there is normally a choice of balancing relocates.

Does this balancing strategy idea progress to utilize to transfinite Nim? Yes! All we necessitate to do is to enbig a little of the theory of transfinite binary recurrentation. Let me presume that you are all recognizable with the common ordinal arithmetic, for which $alpha+beta$ is the ordinal whose order type is isomorphic to a duplicate of $alpha$ adhereed by a duplicate of $beta$, and $alphacdotbeta$ is the ordinal whose order type is isomorphic to $beta$ many copies of $alpha$. Consider now ordinal exponentiation, which can be detaild recursively as adheres:
$$alpha^0=1$$ $$alpha^{beta+1}=alpha^betacdotalpha$$ $$alpha^lambda=sup_{beta

Theorem. Every ordinal $beta$ has a exceptional recurrentation as a decreasing finite sum of ordinal powers of two. $$beta=2^{beta_n}+cdots+2^{beta_0}, qquad beta_n>cdots>beta_0$$

The proof is effortless! We srecommend show it by transfinite induction on $beta$. If the theorem hgreaters below an ordinal $beta$, first let $2^alpha$ be the bigst power of two that is at most $beta$, so that $beta=2^alpha+gamma$ for some ordinal $gamma$. It adheres that $gamma<2^alpha$, for otheralerted we could have made $2^{alpha+1}leqbeta$. Thus, by induction, $gamma$ has a recurrentation with powers of two, and so we may srecommend comprise $2^alpha$ at the front to recurrent $beta$. To see that the recurrentations are exceptional, first set up that any power of two is equivalent to or more than the supremum of the finite decreasing sums of any strictly minusculeer powers of two. From this, it adheres that any recurrentation of $beta$ as above must have engaged $2^alpha$ equitable as we did for the first term, becaengage otheralerted it couldn’t be big enough, and then the recurrentation of the remaining part $gamma$ is exceptional by induction, and so we get exceptionalness for the recurrentation of $beta$. QED

Thus, the theorem shows that every ordinal has a exceptional binary recurrentation in the ordinals, with finitely many nonzero bits. Suppose that we are given a position in transfinite Nim with piles of ordinal heights $eta_0,ldots,eta_n$. We detail that such a position is equitable, if every power of two materializeing in the recurrentation of any of the piles materializes an even number of times overall for all the piles.

The mathematical facts to set up are (1) any relocate on a equitable position will unstability it; and (2) every unequitable position has a balancing relocate. These facts can be showd in the transfinite case in essentiassociate the same manner as the finite case. Namely, if a position is equitable, then any relocate impacts only one pile, changing the ordinal powers of two that materialize in it, and thereby raze the equitable parity of whichever powers of two are impacted. And if a position is unequitable, then see at the bigst unequitable ordinal power of two materializeing, and originate a relocate on any pile having such a power of two in its recurrentation, reducing it so as exactly to stability all the minusculeer powers of two materializeing in the position.

Finassociate, those two facts aget recommend that the balancing strategy is a triumphning strategy, since the triumphning relocate of taking the last block or blocks is a balancing relocate, down to the all-zero position, which is equitable.

In the case of my contest problem above, we may recurrent the ordinals in binary. We understand how to do that in the case of 1, 3, 5 and 7, and actuassociate those numbers are equitable. Here are some other advantageous binary recurrentations:

$omega+3=2^omega+2+1$

$omega^omega+5 = (2^omega)^omega+5=2^{omega^2}+4+1$

$omega^{omega+3}=(2^omega)^{omega+3}=2^{omega^2+omegacdot 3}$

$omega^omegacdot3=(2^omega)^omegacdot 3=2^{omega^2}cdot 2+2^{omega^2}=2^{omega^2+1}+2^{omega^2}$

$omegacdot 5+7 =2^{omega}cdot 2^2+2^omega+7=2^{omega+2}+2^omega+4+2+1$

$epsilon_0 = 2^{epsilon_0}$

$omega_1=2^{omega_1}$

I emphasize aget that this is ordinal exponentiation. The Nim position of the contest problem above is easily seen to be unequitable in cut offal ways. For example, the $omega_1$ term among others materializes only once. Thus, we definitely want to go first in this position. And since $omega_1$ is the bigst unequitable power of two and it materializes only once, we understand that we must join on the $omega_1$ pile. Once one recurrents all the ordinals in terms of their powers of two recurrentation, one sees that the exceptional triumphning relocate is to lessen the $omega_1$ pile to have ordinal height
$$epsilon_0+omega^{omega+3}+omega^omegacdot 2+omegacdot 4.$$This will exactly stability all the minusculeer powers of two in the other piles and therefore departs a equitable position overall. In vague, the triumphning strategy in transfinite Nim, equitable as for finite Nim, is always to depart a equitable position.

Special honors to Pedro Sánchez Terraf for being the only one to post the triumphning relocate in the comments on the other post!