Double-ended colourful Nim
Some partial results about the Grundy values for double-ended colourful Nim, as discussed at the end of [[Winning colourful Nim with the eeny-meeny rule]]B. At the moment, apart from these basic cases I’ve got no idea in order to how efficiently compute the values for an arbitrary position. The hope would be for a convenient rule like the eeny-meeny rule.
No colour changes
\[\mathcal G(x) \mapsto x\]One colour change
\[\mathcal G(x, y) \mapsto x \oplus y\]This is intuitive, you’re effectively playing on two separate Nim piles.
Two colour changes
\[\mathcal G(1, y, 1) = \begin{cases} 1 &\text{if }y =1 \\ 0 &\text{otherwise} \end{cases}\] \[\mathcal G(1, y, 2) = \begin{cases} 2 &\text{if }y=1 \\ 1 &\text{if }y=2 \text{ or }y \ge 4 \\ 3 &\text{if } y = 3 \end{cases}\] \[\mathcal G(1, y, 3) = \begin{cases} 3 &\text{if } y = 1 \\ 2 &\text{if } y = 2 \text{ or } y\ge 4 \\ 1 &\text{if } y = 3 \end{cases}\] \[\mathcal G(1, y, 4) = \begin{cases} 4 &\text{if }y =1,2,3,\text{or }7 \\ 3 &\text{otherwise} \end{cases}\] \[\mathcal G(1, y, 5) = \begin{cases} 5 &\text{if }y=1,2,3, \text{or }5 \\ 4 &\text{if }y=4,6,\text{or }y \ge 8 \\ 3 &\text{if }y = 7 \end{cases}\] \[\mathcal G(1, y, 6) \stackrel?= \begin{cases} 8 &\text{if }y=1,2,3,4,5 \\ 7 &\text{otherwise } \end{cases}\](haven’t proven the last one)
Other rules
- $\mathcal G(x _ 1, \ldots, x _ n) = \mathcal G(x _ n, \ldots, x _ 1)$ (because the game is symmetric)
- Guess:
- $\mathcal G(x, y, z) = 0$ iff $x = z$ and $y \ne x, z$
- If $x = y = z$, then $\mathcal G(x, y, z) = x$.
- (probably wouldn’t be hard by induction)
Table of values
| (n1, n2, n3) | G(n1, n2, n3) | | ———— | ————- | | $(1,1,1)$ | $1$ | | $(2,1,1)$ | $2$ | | $(3,1,1)$ | $3$ | | $(1,2,1)$ | $0$ | | $(2,2,1)$ | $1$ | | $(3,2,1)$ | $2$ | | $(1,3,1)$ | $0$ | | $(2,3,1)$ | $3$ | | $(3,3,1)$ | $1$ | | $(1,1,2)$ | $2$ | | $(2,1,2)$ | $0$ | | $(3,1,2)$ | $1$ | | $(1,2,2)$ | $1$ | | $(2,2,2)$ | $2$ | | $(3,2,2)$ | $3$ | | $(1,3,2)$ | $3$ | | $(2,3,2)$ | $0$ | | $(3,3,2)$ | $2$ | | $(1,1,3)$ | $3$ | | $(2,1,3)$ | $1$ | | $(3,1,3)$ | $0$ | | $(1,2,3)$ | $2$ | | $(2,2,3)$ | $3$ | | $(3,2,3)$ | $0$ | | $(1,3,3)$ | $1$ | | $(2,3,3)$ | $2$ | | $(3,3,3)$ | $3$ | Here’s a longer list of calculations for bigger positions.