Consider the three-element subtraction Nim setup above, including the eventual period p and the associated distance function dist(w). The reverse-mex condition. If s_3=s_1+s_2, then the subtraction Nim satisfies condition (a) of the reverse-mex characterization: for every P-position w with dist(w)=2m for some m∈N with m≥2, w+s_3-s_1 mathbinbmodp is not a P-position and w+s_3-3s_1 mathbinbmodp is a P-position. This is a special-case prediction supporting the preceding characterization, based on the authors’ computer calculations; the source gives no proof or resolution.

The reverse-mex condition for subtraction Nim with s_3=s_1+s_2

Less than 1 year old · traced to 2026

Consider the three-element subtraction Nim setup above, including the eventual period $p$ and the associated distance function $dist(w)$ . The reverse-mex condition. If $s_3=s_1+s_2$ , then the subtraction Nim satisfies condition $(a)$ of the reverse- $mex$ characterization: for every $\mathcal{P}$ -position $w$ with $dist(w)=2m$ for some $m\in\mathbb{N}$ with $m\geq2$ , $w+s_3-s_1\mathbin{\bmod}p$ is not a $\mathcal{P}$ -position and $w+s_3-3s_1\mathbin{\bmod}p$ is a $\mathcal{P}$ -position. This is a special-case prediction supporting the preceding characterization, based on the authors’ computer calculations; the source gives no proof or resolution.

References

Primary source

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

combinatorial-games subtraction-nim

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-posted claim of a complete proof has appeared, but it has not been independently checked, while published work covers only a narrower family.

The problem asks whether the stated reverse- $\operatorname{mex}$ pattern always holds for additive subtraction sets with $s_3=s_1+s_2$ . The primary paper reports computer-supported predictions for this setting but does not prove the general distance condition.

Known results

Larsson, Manabe, and Miyadera (2026) prove reverse- $\operatorname{mex}$ behavior for the narrower family $S=\\{2,4n,4n+2\\}$ with $n\ge3$ , including a one-time-pass variant; this does not establish the stated general condition.

Posted attempt

A reader-posted argument claims a complete proof for every additive set $S=\\{a,b,a+b\\}$ and all relevant positions, including positions before the eventual-period tail. The argument has not been independently verified, so it establishes no confirmed resolution.

Current status (as of August 2026): A complete-proof claim is publicly posted but unverified; the general reverse- $\operatorname{mex}$ condition remains mathematically unsettled, while the narrower family above is proved.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Complete proof for all additive subtraction sets and all positions.

Larsson–Manabe–Miyadera, arXiv:2605.14321, Conjecture 2, considers the ordinary subtraction game with

S=\{a,b,a+b\},\qquad 1\le a<b.

Let $p$ be any eventual period of its Sprague–Grundy sequence. For a losing position $w$ , define

d=\operatorname{dist}(w) =\min\{j\ge1:(w-ja)\bmod p\text{ is losing}\}.

The conjecture asserts that whenever $d\ge4$ is even,

(w+b)\bmod p\text{ is winning}, \qquad (w+b-2a)\bmod p\text{ is losing}.

A prefix issue must first be handled: the conjecture includes losing positions before the eventual Grundy preperiod. We use the prior result of Bhagat–Larsson–Manabe–Yamashita, arXiv:2601.18715, equation (1), Theorem 3, and Corollary 4. Their equation (1) gives

o_{\mathrm{wall}}(x) =o_{\mathrm{sink}}(x+\max S+1) \qquad(x\ge0),

and their Theorem 3 proves that the additive sink sequence is purely periodic. Consequently the ordinary additive-game outcome has a global period $P$ :

o_{\mathrm{wall}}(x+P)=o_{\mathrm{wall}}(x) \qquad(x\ge0).

Although $p$ is initially assumed to be only an eventual Grundy period, it is in fact a global outcome period. For any $x\ge0$ , choose $t$ so large that $x+tP$ and $x+p+tP$ lie in the eventual Grundy tail. Then

o(x+p)=o(x+p+tP)=o(x+tP)=o(x).

Thus all source-prescribed residue classes, including those represented by initial-prefix positions, have well-defined outcomes. Choose sufficiently large representatives so every subtraction below is legal. The ordinary recursion on the cycle is

x\text{ losing}\iff x-a,\ x-b,\ x-a-b\text{ are all winning}.

Now fix a losing $w$ and set

z_j=w+b-ja\pmod p.

The first queried position $z_0=w+b$ is winning: its $b$ -move reaches the losing position $w$ .

By minimality of $d$ ,

w-ja\text{ is winning}\quad(1\le j<d), \qquad w-da\text{ is losing}.

The three followers of $z_j$ are

z_j-a=z_{j+1}, \qquad z_j-b=w-ja, \qquad z_j-(a+b)=w-(j+1)a.

Therefore $z_{d-1}$ is winning, since its $(a+b)$ -move reaches $w-da$ . For $1\le j\le d-2$ , its latter two followers are both winning, so the entire recursion reduces to

z_j\text{ losing}\iff z_{j+1}\text{ winning}.

Backward induction yields the stronger exact alternating-strip formula

\boxed{ z_j\text{ losing} \iff d-1-j\text{ is odd} \qquad(1\le j\le d-1).}

When $d\ge4$ is even, $d-3$ is odd, and therefore

\boxed{z_2=w+b-2a\text{ is losing},\qquad z_0=w+b\text{ is winning}.}

These are exactly both parts of the conjectured condition for every $a<b$ , every permitted eventual Grundy period, and every losing position, including the entire initial preperiod.

The earlier sink-subtraction paper supplies global outcome periodicity; the alternating-strip argument above proves the distinct subsequent conjecture.