Are there combinatorial games $G$ such that $G+G$ is fuzzy?
In giving an incomplete answer to this question, on the existence of nonzero combinatorial games $G$ such that $nG=0$ and $mG≠0$ for $m < n$, I noted a hypothesis that would immediately solve the problem (in the negative):
\[G+G\text{ is non-fuzzy for all }G.\]I've verified this hypothesis for all the combinatorial games of depth 2. It is always true for symmetric games (in which $G+G = 0$). It is also true whenever Left's best move in the game $G$ is better for Left than Right's best move is for Right (because then $G+G$ is a win for Left), and mutatis mutandis when we switch Right and Left.
Is it true, though? Can we prove it or give a counterexample for it?
-
The reason this hypothesis would solve the original problem is as follows. If $G+G$ is always non-fuzzy, then $nG=0$ only if $G+G=0$; otherwise $G+G$ would have to be either positive or negative, and in either case $nG$ could not equal 0. ↩︎
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| clemens |
Thread: Works for me Thanks. More generally, I see that any game of the form ${X | * - X}$ will work the same way. |
May 7, 2026 at 18:07 |
Yes (such games exist). A simple example is \(G = \{1\mathop|* \mathop- 1\}\).
If Left goes first in \(G + G\), their only move is to \(G + 1\), after which Right's only move is to \(* \mathop- 1 + 1 = *\), after which Left moves to 0 and wins.
If Right goes first in \(G + G\), their only move is to \(G + * - 1\). Left has two options: play in \(G\) or in \(*\).
- If Left moves to \(1 + * - 1 = *\), Right moves to 0 and wins.
- If Left moves to \(G - 1\), Right may move to \(* - 2\), after which Left must move to \(-2\), and Right wins.
\(G + G\) is thus a win for the first player, and therefore a fuzzy game.
0 comment threads
0 comment threads