The parity-pattern conjecture for generalized OOOOOOB positions

Consider a position with $k\geq 3$ piles of tokens, and let

m=\left\lceil\frac{k}{3}\right\rceil.

Assume that every pile has more than $m$ tokens. Parity-pattern conjecture. The position is a $\mathcal{P}$ -position precisely under the following conditions: if $k$ is odd, the number of odd piles is even; if $k\equiv 0\pmod 4$ , the number of odd piles is one of $0,2,4,\ldots,\frac{k}{2}-2,\frac{k}{2}+1,\ldots,k-5,k-3,k-1$ ; and if $k\equiv 2\pmod 4$ , the number of odd piles is one of $0,2,4,\ldots,\frac{k}{2}-1,\frac{k}{2}+2,\ldots,k-5,k-3,k-1$ . The pattern was verified by computer for up to $11$ piles and up to $10$ tokens per pile; its status beyond that verification remains open.

Progress summary

Solved

The conjecture was checked only in small cases, and a reader-posted computation now claims a counterexample at twelve piles, but nobody has independently verified it.

Danai, Ellis, and Thanatipanonda formulated the parity characterization as Conjecture 2.1 in 2026. Their paper reports computer verification through $11$ piles and $10$ tokens per pile, not a proof beyond that range.

Posted attempt

A posted finite backward-induction computation claims that the conjecture fails at $k=12$ : a position with six odd piles is losing although the pattern predicts winning, while a position with seven odd piles is winning although the pattern predicts losing. This is presented as a complete disproof, but the computation has not been independently verified.

Current status (as of August 2026): The published work establishes only bounded computational verification, while a purported $12$ -pile counterexample would settle the conjecture negatively but remains unverified.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Alon Danai, Paul Ellis and Thotsaporn Aek Thanatipanonda, “Generalizing OOOOOOB”, arXiv:2605.23213 (2026).

combinatorial-game-theory impartial-games

1 week ago

Counterexample at the first pile count beyond the source's verification range.

In Version B, a move removes one token from one nonempty pile or removes one token from every nonempty pile. Consider the twelve-pile position

(5,5,5,5,5,5,6,6,6,6,6,6).

Here

k=12,\qquad m=\left\lceil\frac{k}{3}\right\rceil=4,

and every pile contains strictly more than $m$ tokens, exactly as required. There are six odd piles.

For $k=12\equiv0\pmod4$ , the conjecture asserts that a position is losing precisely when the number of odd piles belongs to

\{0,2,4,7,9,11\}.

Since $6$ is absent, it predicts that the displayed position is winning.

In fact the position is losing. Encode a position by the multiplicity tuple $a=(a_1,\ldots,a_s)$ , where $a_i$ counts size- $i$ piles and trailing zeros are removed. Its followers are exactly

a-e_i+e_{i-1}\quad(a_i>0,\ e_0=0),

together with the all-pile follower

(a_2,a_3,\ldots,a_s).

Every follower has strictly smaller token count. Thus the following complete finite backward-induction certificate uses no conjectural shortcut:

from functools import cache

def norm(a):
    while a and not a[-1]:
        a = a[:-1]
    return a

def followers(a):
    for i, count in enumerate(a):
        if count:
            b = list(a)
            b[i] -= 1
            if i:
                b[i - 1] += 1
            yield norm(tuple(b))
    if a:
        yield norm(a[1:])

@cache
def losing(a):
    return not any(losing(b) for b in followers(a))

a = (0, 0, 0, 0, 6, 6)
assert losing(a)
assert all(not losing(b) for b in followers(a))
assert losing.cache_info().currsize == 17101

The three distinct follower types are obtained by decreasing a size-five pile, decreasing a size-six pile, or decreasing every pile; all are winning. Therefore the twelve-pile position is losing despite having six odd piles, disproving the proposed parity classification.

The paper checked at most eleven piles. This counterexample occurs at $k=12$ , the first untested pile count, while satisfying every strict size assumption.

Source: Danai, Ellis, and Thanatipanonda, Generalizing OOOOOOB, Conjecture 2.1, https://arxiv.org/html/2605.23213 .

The reverse implication fails as well. Consider the second twelve-pile position

(5,5,5,5,5,5,5,6,6,6,6,6).

All piles again exceed $m=4$ , but there are now seven odd piles. Since

7\in\{0,2,4,7,9,11\},

the conjecture predicts a losing position. In fact it is winning: remove one token from every pile to obtain

(4,4,4,4,4,4,4,5,5,5,5,5),

which the same exact backward-induction recursion certifies as losing. In multiplicity notation,

\operatorname{losing}(0,0,0,0,7,5)=\mathrm{false}, \qquad \operatorname{losing}(0,0,0,7,5)=\mathrm{true}.

Consequently both directions of the proposed parity characterization fail at $k=12$ , the first pile count beyond the source's verification.

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Shivam Patel · 2 days ago · edited