Let P be the existing piece alphabet, let P be a single additional piece whose move total T_P(n) is a polynomial in n of degree at most 3, and let P'=PcupP. Strength is measured by the paper's uniformly random-arrow probability on the n× n board. Triple-coincidence conjecture. For every integer n≥4, no three distinct pieces in P' share the same strength on the n× n board. This asks whether the absence of triple strength coincidences persists after adjoining one piece of the stated polynomial complexity; the claim is presented as a further question and no resolution is given.

Triple-coincidence non-occurrence under single-piece extension

Less than 1 year old · traced to 2026

Let $\mathcal{P}$ be the existing piece alphabet, let $P$ be a single additional piece whose move total $T_P(n)$ is a polynomial in $n$ of degree at most $3$ , and let $\mathcal{P}'=\mathcal{P}\cup\{P\}$ . Strength is measured by the paper's uniformly random-arrow probability on the $n\times n$ board.

Triple-coincidence conjecture. For every integer $n\geq4$ , no three distinct pieces in $\mathcal{P}'$ share the same strength on the $n\times n$ board.

This asks whether the absence of triple strength coincidences persists after adjoining one piece of the stated polynomial complexity; the claim is presented as a further question and no resolution is given.

References

Primary source

Frank M. V. Feys, “The Arithmetic of Chess Piece Strength on the n x n Board”, arXiv:2605.20229 (2026).

chess-mathematics combinatorial-game-theory polynomial-identities

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The conjecture is challenged by an unverified reader-proposed construction that claims a single added piece can create three equal strengths on boards of sizes 6, 8, and 12.

Feys (2026) asks whether adjoining one piece whose move total has degree at most $3$ can ever produce three equal strengths for $n \geq 4$ . The paper proves the corresponding non-occurrence result for the original thirteen-piece alphabet, but presents the extended statement only as a conjecture.

Known results

Feys (2026): in the original alphabet, pairwise strength coincidences occur only at $n \in \{6,8,12\}$ .
Feys (2026): no three distinct original pieces have equal strength for any $n \geq 4$ .

Posted attempt

A reader claims three separate counterexamples, using exactly one compound added piece in each case: triple coincidences at $n=6$ , $n=8$ , and $n=12$ . Each proposed move total has degree $2$ , so the constructions would refute the conjecture, but the calculations and interpretation have not been independently verified.

Current status (as of August 2026): The original-alphabet theorem is settled, while the single-piece extension conjecture has an unverified counterexample claim and is not mathematically resolved.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Counterexamples at every original magic board. In each construction below, exactly one new piece is adjoined. Its fixed move set is the disjoint union of several fairy-leaper move sets; the constituent leapers are not separately added to the alphabet. The source explicitly permits such compound pieces.

Let $L_{a,b}$ denote the leaper with all sign changes and coordinate exchanges of displacement $(a,b)$ . Counting starting squares for each displacement gives, whenever $n\ge b$ ,

T_{L_{a,b}}(n)= \begin{cases} 4n(n-b),&a=0,\\ 4(n-a)^2,&a=b>0,\\ 8(n-a)(n-b),&0<a<b. \end{cases}

Different leaper families have disjoint displacement sets, so their totals add.

Board $n=6$ . Adjoin the single compound piece

P_6=L_{1,1}\cup L_{1,2}\cup L_{1,3}.

Its exact polynomial, valid for every $n\ge4$ , is

T_{P_6}(n)=4(n-1)^2+8(n-1)(n-2)+8(n-1)(n-3) =4(n-1)(5n-11).

The source's existing Centaur and Archbishop satisfy

T_{P_6}(6)=T_{\mathrm{Centaur}}(6) =T_{\mathrm{Archbishop}}(6)=380.

Board $n=8$ . Independently adjoin the single compound piece

P_8=L_{0,1}\cup L_{0,2}\cup L_{1,2}\cup L_{2,2}.

Its exact polynomial is

T_{P_8}(n)=20n^2-52n+32,

and

T_{P_8}(8)=T_{\mathrm{Rook}}(8) =T_{\mathrm{Archbishop}}(8)=896.

Board $n=12$ . Independently adjoin

P_{12}=L_{0,3}\cup L_{3,3}\cup L_{4,4}.

Its exact polynomial, again valid on every prescribed board $n\ge4$ , is

T_{P_{12}}(n)=4n(n-3)+4(n-3)^2+4(n-4)^2 =12n^2-68n+100.

Using the source's Bishop convention,

T_{P_{12}}(12)=T_{\mathrm{King}}(12) =T_{\mathrm{Bishop}}(12)=1012.

Each additional piece has one fixed finite move set, is distinct from every original piece, and has a degree-two move-count polynomial. Since all strengths use the same denominator $n^2(n^2-1)$ , equal move totals give equal strengths. Thus Conjecture 9.2 fails separately at all three original magic board sizes $6,8,12$ , always with exactly one added piece.

For completeness, the adjacent Conjecture 9.1 also fails under its stated extension rules: adjoin just the explicitly permitted Wazir $W=L_{0,1}$ . Then

T_W(n)=4n(n-1),\qquad T_{\mathrm{Rook}}(n)=2n^2(n-1) =\frac n2 T_W(n),

giving a second nonconstant linear proportionality, distinct from the Bishop--King proportionality.

Source: Feys, The Arithmetic of Chess Piece Strength on the $n\times n$ Board, Conjectures 9.1 and 9.2.