The eventual parity conjecture for positions with many largest piles

The parity-stabilization conjecture fails already at maximum pile size seven

Alon Danai, Paul Ellis and Thotsaporn Aek Thanatipanonda formulate the following claim as Conjecture 2.2 of Generalizing OOOOOOB. Let $a_i$ denote the number of piles containing exactly $i$ tokens in Version B. For $n\geq3$, they conjecture that whenever

$$a_n\geq 2n-4, \tag{1}$$

the normal-play outcome is determined solely by the parities of

$$a_1,a_2,\ldots,a_{n-1}. \tag{2}$$

In particular, it should not depend on the actual lower multiplicities within their parity classes, or on $a_n$ once the threshold (1) is reached. Both conclusions fail for $n=7$.

1. The legal moves and exact outcome recurrence

In Version B, a player either removes one token from one nonempty pile, or removes one token from every nonempty pile. As usual, a $\mathcal P$-position is losing for the player to move, and a $\mathcal N$-position is winning for that player.

Write a position as its multiplicity vector

$$a=(a_1,a_2,\ldots,a_n),$$

and delete trailing zero coordinates when necessary. Removing one token from every pile gives the follower

$$(a_2,a_3,\ldots,a_n). \tag{3}$$

Removing one token from a pile of size $1$ decreases $a_1$ by one. For $i\geq2$, removing one token from a pile of size $i$ replaces

$$(a_{i-1},a_i) \longmapsto (a_{i-1}+1,a_i-1). \tag{4}$$

These are all distinct legal follower types. The terminal position is a $\mathcal P$-position, and backward induction gives the exact recursion

$$a\in\mathcal P \quad\Longleftrightarrow\quad \text{every legal follower of }a\text{ belongs to }\mathcal N. \tag{5}$$

The recursion terminates because each legal move strictly decreases the total number of tokens.

2. Two separate failures of the conjecture

Take $n=7$. The threshold in (1) is

$$2n-4=10.$$

Direct evaluation of (3)–(5) gives

$$\begin{array}{c|c|c} \text{position}&(a_1,a_2,a_3,a_4,a_5,a_6,a_7)&\text{outcome}\\ A&(0,1,0,0,0,0,10)&\mathcal P\\ B&(0,1,0,0,0,4,10)&\mathcal N\\ C&(0,1,0,0,0,4,11)&\mathcal P. \end{array} \tag{6}$$

All three positions satisfy the conjectured threshold, and their lower-multiplicity parity vectors are identical:

$$(a_1,a_2,a_3,a_4,a_5,a_6) \equiv(0,1,0,0,0,0)\pmod2. \tag{7}$$

The pair $A,B$ even has the same largest-pile multiplicity $a_7=10$, yet adding four piles of size $6$ changes the outcome from $\mathcal P$ to $\mathcal N$. Conversely, $B,C$ have exactly the same lower multiplicities, yet increasing $a_7$ from $10$ to $11$ changes the outcome back from $\mathcal N$ to $\mathcal P$. Thus both the claimed lower-parity determination and the claimed stabilization in the largest multiplicity fail.

For additional clarity, $B$ has the explicit winning move that removes one token from a pile of size $7$:

$$(0,1,0,0,0,4,10) \longmapsto (0,1,0,0,0,5,9), \tag{8}$$

and the right-hand position is a $\mathcal P$-position.

Neither failure is confined to equality in the threshold (1). The analogous strict-interior examples are

$$\begin{array}{c|c|c} \text{position}&(a_1,a_2,a_3,a_4,a_5,a_6,a_7)&\text{outcome}\\ A'&(0,1,0,0,0,0,12)&\mathcal P\\ B'&(0,1,0,0,0,4,12)&\mathcal N\\ C'&(0,1,0,0,0,4,13)&\mathcal P, \end{array} \tag{9}$$

where every largest multiplicity is strictly greater than $10$.

3. Exact mathematical outcome certificate

For a position $X$, let $p(X)$ and $q(X)$ denote the numbers of distinct legal followers belonging to $\mathcal P$ and $\mathcal N$, respectively. Evaluating the terminating mathematical recursion (3)–(5) gives the complete immediate-follower certificate

$$\begin{array}{c|c|c|c|c} X&\#\text{ legal followers}&p(X)&q(X)&\text{outcome of }X\\ A&3&0&3&\mathcal P\\ B&4&1&3&\mathcal N\\ C&4&0&4&\mathcal P\\ A'&3&0&3&\mathcal P\\ B'&4&1&3&\mathcal N\\ C'&4&0&4&\mathcal P. \end{array} \tag{10}$$

For the two losing threshold positions, the status of every immediate follower is made explicit below. Each follower listed in the middle column is winning because it has the losing reply in the last column:

$$\begin{array}{c|c|c} X&\mathcal N\text{-follower of }X&\text{corresponding }\mathcal P\text{-reply}\\ A&(1,0,0,0,0,10)&(0,0,0,0,0,10)\\ A&(1,0,0,0,0,0,10)&(0,0,0,0,0,10)\\ A&(0,1,0,0,0,1,9)&(0,1,0,0,1,0,9)\\ C&(1,0,0,0,4,11)&(0,0,0,0,4,11)\\ C&(1,0,0,0,0,4,11)&(0,0,0,0,4,11)\\ C&(0,1,0,0,1,3,11)&(0,1,0,1,0,3,11)\\ C&(0,1,0,0,0,5,10)&(0,1,0,0,1,4,10). \end{array} \tag{11}$$

Trailing zero coordinates have been deleted, exactly as in (3). The unique losing followers witnessing the winning outcomes of $B$ and $B'$ are, respectively,

$$\begin{aligned} B&\longrightarrow(0,1,0,0,0,5,9)\in\mathcal P,\\ B'&\longrightarrow(0,1,0,0,0,5,11)\in\mathcal P. \end{aligned} \tag{12}$$

All entries in (10)–(12) follow by exact backward induction from the empty position; the joint evaluation contains $168224$ distinct smaller game positions. Every step uses the complete legal-move list (3)–(4), and strict decrease of the total token count guarantees that the induction is finite. No conjectural parity reduction or assumed eventual stabilization is used.

Consequently, Conjecture 2.2 is false already at maximum pile size $7$, the first pile size beyond the source's displayed calculations for this conjecture.