# Strategy for Whim

Whim is played like nim, except one time, a player can use their whim to decide whether they are playing regular or misère nim.

Many positions have an easy analysis. To play misère nim, just play normal nim, except when it would leave only heaps of 0 and 1, and the number of 1 heaps is even. Instead, make the number odd. So from now on assume no one has used their whim.

Call a position firm if it has at least one heap of 2 or more, otherwise fickle. If you're in a fickle position, and the number of 1 heaps is even or odd, use your whim to play normal of misère nim, respectively. If you're in a firm position with value Grundy value 0, use your whim as a pass move.

The only difficult case is when you're in a firm position with nonzero Grundy value. This probably turns into she-loves-me, she-loves-me-not because there are only three ways to "exit" from this scenario:

1. Moving to a position of Grundy value 0
3. Moving to a fickle position

And in each case, the person who exits loses.

Is there a simple strategy to play in this scenario?

• If you're in a firm position with nonzero Grundy value, why don't you move to another firm position with nonzero Grundy value? Jul 30 '20 at 21:35
• @saulspatz Eventually someone won't be able to do that, and you want to force your opponent to be that person. Jul 31 '20 at 0:21
• Yes, of course, but the way the question is stated, you seem to discount the possibility of making such a move. Jul 31 '20 at 0:44
• I just thought of something else. Suppose there's only one pile left, and no one has yet exercised the whim. (I don't know if this situation cab arise in rational play.) At that point, if I pick up all the counters, I assume it's a draw, since we don't know whether the game is ordinary or misère. Do you agree? Jul 31 '20 at 1:01
• Can't the next player then just use their whim to call normal play and win then? Jul 31 '20 at 13:32

This game is described on page $$500$$ in volume $$2$$ of "Winning Ways". They use the following terminology.

• $$0$$-nim is ordinary nim;
• $$1$$-nim is misère nim;
• $$2$$-nim is whim.

Generalizing, they consider $$d$$-nim.

This is played with an extra pile of $$d$$ counters (the "quiddity heap") which indicates what kind of nim is being played. At each turn, the player may either

• move as in nim, or
• reduce $$d$$,

but not both.

Of course, once $$d$$ has been reduced to $$1$$ or $$0$$, it cannot be reduced any further.

An analysis is not given, but they make the following statement, which I have not yet verified.

Then each of these games becomes like Nim with an extra heap. If $$2^k\leq d<2^{k+1}$$, the quiddity heap behaves like a heap of size $$d$$ when all other heaps are of size less than $$2^{k+1}$$ and like a heap of size $$d-1$$ when they're not.

When $$d=0$$ or $$d=1$$ the game is just ordinary of misère nim, and we can forget about the quiddity heap. When $$d=2$$ we have $$k=1$$ and the above says that if all the other heaps are of size less than $$4$$, the quiddity heap behaves like a heap of size $$2$$, otherwise like a heap of size $$1$$.