The reverse-mex characterization for three-element subtraction Nim

Consider subtraction Nim with subtraction set $S=\{s_1,s_2,s_3\}$ , where $s_1<s_2<s_3$ and $s_1,s_2,s_3\in\mathbb{N}$ . Suppose that there exist $p,q\in\mathbb{N}$ such that $\mathcal{G}(x+p)=\mathcal{G}(x)$ for every $x\geq q+1$ . For $w\in\mathbb{Z}_{\geq 0}$ and $x\in\mathbb{N}$ , define $w-x\mathbin{\bmod}p$ by the eventual-periodic reduction determined by $q$ and $p$ , and for a $\mathcal{P}$ -position $w$ define

dist(w)=\min\{k:(w-ks_1)\mathbin{\bmod}p\text{ is a }\mathcal{P}\text{-position}\}.

The reverse-mex characterization. The subtraction Nim has the reverse- $mex$ property if and only if, for every $\mathcal{P}$ -position $w$ with $dist(w)=2m$ for some $m\in\mathbb{N}$ with $m\geq2$ , the positions $w+s_3-s_1\mathbin{\bmod}p$ and $w+s_3-3s_1\mathbin{\bmod}p$ are respectively not a $\mathcal{P}$ -position and a $\mathcal{P}$ -position. This is presented as a computer-based prediction; no proof or resolution is supplied in the source.

Progress summary

Solved

An unverified counterexample claims to disprove the proposed characterization, while a published paper proves the reverse rule only for one special family.

Larsson, Manabe, and Miyadera proposed the characterization in 2026 as a computer-based conjecture for three-move subtraction games. It asserts that a condition on periodic losing positions exactly determines when the usual game values can be recovered by looking at future positions.

Known results

Larsson, Manabe, and Miyadera (2026): the reverse- $\operatorname{mex}$ property is proved for $S=\{2,4n,4n+2\}$ with $n\ge3$ , including the one-time-pass variant; this does not settle the general characterization.

Posted attempt

A reader-written computation claims a complete counterexample: $S=\{3,6,10\}$ satisfies the conjectured condition nonvacuously but fails reverse- $\operatorname{mex}$ , and an infinite scaling family is also claimed. The calculation has not been independently verified.

Current status (as of August 2026): The general characterization is not settled; a complete counterexample is claimed but unverified, while the published special-family theorem remains valid.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Urban Larsson, Hikaru Manabe and Ryohei Miyadera, “A Subtraction Nim with a Pass”, arXiv:2605.14321 (2026).

combinatorial-games subtraction-nim

1 week ago

A nonvacuous counterexample to the proposed equivalence.

Take the subtraction set

S=\{3,6,10\}, \qquad g(x)=\operatorname{mex}\{g(x-s):s\in S,\ s\le x\}.

The exact initial values are

\begin{array}{c|rrrrrrrrrrrr} x&0&1&2&3&4&5&6&7&8&9&10&11\\ \hline g(x)&0&0&0&1&1&1&2&2&2&0&3&3 \end{array}

followed, starting at $x=12$ , by the repeating block

(1,0,0,2,1,1,0,2,2,1,0,3,2).

This is an exact eventual period, not an extrapolation: the defining recurrence has memory $10$ , and direct evaluation gives

(g(12),\ldots,g(21)) =(g(25),\ldots,g(34)).

Determinism of the mex recurrence therefore proves

g(x+13)=g(x)\qquad(x\ge12).

Thus the source definitions apply with $q=11$ , $p=13$ , and $s_1=3$ .

The prefix $P$ -positions are $0,1,2,9$ , and representatives of all eventual periodic $P$ -positions are $13,14,18,22$ . Computing the source's periodic distance gives

\begin{array}{c|rrrr|rrrr} w&0&1&2&9&13&14&18&22\\ \hline \operatorname{dist}(w)&4&3&2&3&4&3&3&3. \end{array}

Therefore the condition $\operatorname{dist}(w)=2m$ , $m\ge2$ , occurs nonvacuously, precisely for $w=0,13,26,39,\ldots$ . For every such $w$ , the source's eventual-periodic representatives give

g\bigl((w+s_3-s_1)\bmod p\bigr) =g(20)=2\ne0,

and

g\bigl((w+s_3-3s_1)\bmod p\bigr) =g(14)=0.

Consequently condition (a) holds at every required $P$ -position.

Nevertheless, at heap size $x=5$ ,

g(5)=1, \qquad \operatorname{mex}\{g(8),g(11),g(15)\} =\operatorname{mex}\{2,3,2\}=0.

Hence reverse mex fails although condition (a) holds nonvacuously, disproving the conjectured “if and only if.”

Source: Larsson, Manabe, and Miyadera, A Subtraction Nim with a Pass, Definitions 12–13 and Conjecture 1, https://arxiv.org/html/2605.14321 .

Infinite strengthening. For every integer $d\ge1$ , take

S_d=\{3d,6d,10d\}.

Direct induction in the mex recurrence gives, for all $j\ge0$ and $0\le r<d$ ,

g_{S_d}(dj+r)=g_S(j).

Consequently the preperiod is $12d$ , the eventual period is $13d$ , and one may take $q=12d-1$ , $p=13d$ . The source's periodic reduction preserves the remainder $r$ , while its distance uses the step $3d$ ; hence every $P$ -position has exactly the distance of the corresponding base position. All positions with even distance at least four belong to the blocks $j\equiv0\pmod{13}$ , have distance four, and satisfy

g_{S_d}(w+7d)=2\ne0, \qquad g_{S_d}(w+d)=0.

Thus condition (a) holds nonvacuously for all $d$ . Nevertheless, for every $0\le r<d$ ,

g_{S_d}(5d+r)=1 \ne\operatorname{mex}\{2,3,2\}=0.

Therefore there are infinitely many distinct three-element subtraction sets satisfying the conjectured condition while failing reverse mex.

0 endorsements

Shivam Patel · 2 days ago · edited