Shifted Nim-sum conjecture for chocolate-bar games with k congruent to 1 modulo 4

Counterexamples to the shifted chocolate-bar characterization for every nontrivial parameter

Ryohei Miyadera, Shunsuke Nakamura, and Hikaru Manabe, Previous Player's Positions of Impartial Three-Dimensional Chocolate-Bar Games, Thai Journal of Mathematics 21 (2023), 717--732, Conjecture 2; see also arXiv:2205.11884.

The conjecture is false for every parameter $k=4m+1$ with $m\geq1$. In each case, its proposed characterization declares two positions connected by a legal move to be previous-player winning positions, which is impossible. The smallest instance also admits a complete direct winning-move certificate.

The exact game and conjecture

Fix $k=4m+1$, and put

$$f(x,z)=\left\lfloor\frac{x+z}{k}\right\rfloor. \tag{1}$$

The positions are triples $(x,y,z)$ of nonnegative integers satisfying $y\leq f(x,z)$. By Definition 1.10 of the published paper, their legal followers are

$$\begin{aligned} \bigl(u,\min\{y,f(u,z)\},z\bigr) &\qquad(0\leq u<x),\\ (x,v,z) &\qquad(0\leq v<y),\\ \bigl(x,\min\{y,f(x,w)\},w\bigr) &\qquad(0\leq w<z). \end{aligned} \tag{2}$$

Writing $\oplus$ for bitwise exclusive-or, Conjecture 2 asserts

$$(x,y,z)\text{ is a }\mathcal P\text{-position} \quad\Longleftrightarrow\quad (x+1)\oplus y\oplus(z+1)=0. \tag{3}$$

A fundamental property of every finite impartial normal-play game is that no legal move joins two $\mathcal P$-positions.

An infinite family of adjacent conjectured winning positions

For any integer $m\geq1$, set $k=4m+1$ and consider

$$X_k=(k-1,2,k+1), \qquad Y_k=(k-1,1,k-2). \tag{4}$$

Both positions are admissible, since

$$f(k-1,k+1)=2, \qquad f(k-1,k-2) =\left\lfloor\frac{2k-3}{k}\right\rfloor =1. \tag{5}$$

Reducing the third coordinate of $X_k$ from $k+1$ to $k-2$ is legal, and the prescribed truncation changes its middle coordinate from $2$ to $1$. Thus

$$X_k\longrightarrow Y_k \tag{6}$$

is a legal move in precisely the published game.

Since $k=4m+1$, its two least significant binary digits are $01$. Consequently,

$$k\oplus(k+2)=2, \qquad k\oplus(k-1)=1. \tag{7}$$

Therefore both positions satisfy the conjectured zero condition:

$$\begin{aligned} (k-1+1)\oplus2\oplus(k+1+1) &=k\oplus2\oplus(k+2)=0,\\ (k-1+1)\oplus1\oplus(k-2+1) &=k\oplus1\oplus(k-1)=0. \end{aligned} \tag{8}$$

If (3) were correct, both ends of the legal move (6) would be $\mathcal P$-positions. This contradicts the defining property of $\mathcal P$-positions. Hence Conjecture 2 fails for every $m\geq1$.

The smallest explicit position and its exact outcome

At $m=1$, we have $k=5$, and (4) becomes

$$(4,2,6)\longrightarrow(4,1,3). \tag{9}$$

Both shifted nim-sums vanish:

$$5\oplus2\oplus7=0, \qquad 5\oplus1\oplus4=0. \tag{10}$$

We can identify the actual outcomes without assuming any unproved characterization. Whenever $y=0$, every follower still has middle coordinate zero, and (2) reduces to ordinary two-heap Nim on $(x,z)$. Hence

$$(t,0,t)\text{ is a }\mathcal P\text{-position} \qquad(t\geq0). \tag{11}$$

The position $(4,1,3)$ has exactly the following eight followers. Each follower has the indicated legal move to one of the established $\mathcal P$-positions (11):

$$\begin{array}{c|c} \text{Follower of }(4,1,3)&\text{Legal reply to a }\mathcal P\text{-position}\\ \hline (0,0,3)&(0,0,0)\\ (1,0,3)&(1,0,1)\\ (2,1,3)&(2,0,2)\\ (3,1,3)&(3,0,3)\\ (4,0,3)&(3,0,3)\\ (4,0,0)&(0,0,0)\\ (4,1,1)&(1,0,1)\\ (4,1,2)&(2,0,2) \end{array} \tag{12}$$

Every follower in the left column is therefore an $\mathcal N$-position. It follows that

$$(4,1,3)\text{ is a }\mathcal P\text{-position}, \qquad (4,2,6)\text{ is an }\mathcal N\text{-position}. \tag{13}$$

In particular, $(4,2,6)$ is an explicit position whose shifted nim-sum vanishes but which is not a $\mathcal P$-position. The argument for (4)--(8) proves failure simultaneously for every nontrivial parameter in the conjectured family; it does not address the remaining case $k=1$.