The domination game conjecture for Cartesian products of paths

Counterexample: the smallest even-order grid on which the first player wins.

In the latest revision of Fiscus, Hurlbert, Myzelev, and Pence, arXiv:2504.03448v2, Conjecture 24 asserts

$$k,m>1,\quad km\equiv0\pmod2 \quad\Longrightarrow\quad \mathbb D(P_k\square P_m)\in\mathcal P.$$

An older author-hosted manuscript numbers the identical statement Conjecture 23.

The conjecture fails for

$$\boxed{\mathbb D(P_2\square P_7)\in\mathcal N.}$$

Although this grid has even order $14$, the first player has four winning opening moves:

$$(1,2),\quad(1,6),\quad(2,2),\quad(2,6),$$

using one-based row and column coordinates.

The source's game permits selection of any previously unchosen vertex, including a vertex already dominated. Play stops immediately when the chosen vertices form a dominating set, and the player making that move wins. The following complete finite backward-induction certificate uses precisely these rules.

Index $(r,c)$ by $i=7(r-1)+(c-1)$, put $F=2^{14}-1$, and encode the chosen vertex set by a mask $s$. Let

$$C_i=2^i+\sum_{j:v_j\sim v_i}2^j$$

be its closed-neighborhood masks. Compute

$$D(0)=0,\qquad D(s)=D(s-2^i)\mathbin{\mathrm{OR}}C_i,$$

where $i$ is the least occupied bit of $s$. Then $D(s)$ represents the dominated set.

Define the exact outcome indicator in decreasing mask order:

$$w(s)= \begin{cases} 0,&D(s)=F,\\ 1,&D(s)\ne F\text{ and } w(s\mathbin{\mathrm{OR}}2^i)=0 \text{ for some unchosen }i,\\ 0,&\text{otherwise}. \end{cases}$$

Every follower mask is strictly larger than $s$, so every required child has already been evaluated. Induction on the number of unchosen vertices proves

$$w(s)=1\iff s\text{ is a first-player-winning position}.$$

Exhaustive evaluation of all $2^{14}=16{,}384$ masks yields the following complete certificate:

Chosen vertices	Dominating terminal	Nondominating $\mathcal P$	Nondominating $\mathcal N$
0	0	0	1
1	0	4	10
2	0	13	78
3	0	56	308
4	2	268	731
5	86	320	1596
6	588	153	2262
7	1518	24	1890
8	2046	6	951
9	1706	0	296
10	949	0	52
11	360	0	4
12	91	0	0
13	14	0	0
14	1	0	0
Total	7361	844	8179

In particular, $w(0)=1$, and exactly four singleton masks have outcome $\mathcal P$, corresponding to the four winning openings above.

For the opening $(1,2)$, the following gives a winning response to every possible second-player move; each resulting three-vertex position has outcome $\mathcal P$:

$$\begin{array}{c|c} \text{Second-player move}&\text{First-player response}\\hline (1,1)&(1,6)\\ (1,3)&(1,7)\\ (1,4)&(1,5)\\ (1,5)&(1,4)\\ (1,6)&(1,1)\\ (1,7)&(1,3)\\ (2,1)&(1,6)\\ (2,2)&(1,5)\\ (2,3)&(2,7)\\ (2,4)&(1,7)\\ (2,5)&(1,4)\\ (2,6)&(1,6)\\ (2,7)&(1,5) \end{array}$$

The full outcome table, interpreted as $16{,}384$ consecutive indicator bytes, has SHA-256 digest

$$\texttt{a4b410df13688cfb66aecba5067a6fc6f1208d939647a6f98f1500f69065b7a9}.$$

The same exact recurrence gives outcome $\mathcal P$ for $P_2\square P_m$, $2\le m\le6$, and for $P_3\square P_4$. These exhaust all nontrivial rectangular grids of even order less than $14$, up to transposition. Therefore $P_2\square P_7$ is the smallest possible counterexample to the stated conjecture.