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REVIEW 3 major objections 5 minor 7 references

An Amalgamation Nim with Restriction

T0 review · 3 major / 5 minor · reviewed 2026-08-12 · deepseek-v4-flash

Pith's one-line read The paper claims a complete, explicit classification of P-positions for a three-pile Amalgamation Nim variant whose merge move requires both piles to have at least two stones.

desk verdict A plausible new P-position formula for a restricted Amalgamation Nim, but the proof as written has load-bearing gaps that need a careful rewrite before the classification is established. read the letter →

arxiv 2411.14488 v1 pith:DCLQC3DS submitted 2024-11-20 math.CO

classification math.CO MSC 91A46
keywords AmalgamationNimP-positionsnim-sumcombinatorialgametheorypilemergingGrundynumbersBouton's
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

Amalgamation Nim allows two kinds of moves: the usual Nim move of reducing one pile, and replacing two non-empty piles by their sum. For three or more piles, no formula for the P-positions (previous-player winning positions) was known. This paper restricts the merge move to pairs of piles each containing at least two stones and claims a complete classification: the P-positions are exactly the set P built from bitwise xor conditions in Definition 6(viii). The interest is that this is the first explicit P-position formula for a version of Amalgamation Nim with more than two piles, and it gives a concrete winning strategy—move into P, and every reply leaves P.

What carries the argument

The argument is carried by the bitwise xor (nim-sum) equation x⊕y⊕z=0 and its per-bit analysis in Lemma 1: for a xor-zero triple, each binary digit satisfies either x_i+y_i=z_i or x_i=y_i=1 and z_i=0, so the ordinary sum x+y is forced to be exactly z, exactly z+2, or at least z+4. This reduces the infinite position space to a few families—N0,1, N0,2, P0,1, P0,2, P1,1, P1,2—defined in Definition 6. The P1 families are one-step shifts of the N0 families (replace z by z±1 according to the parity of x+y), giving nim-sum 1 rather than 0; Lemma 4's bound |Δx|+|Δy|+|Δz|≤1 for moves between P0 and P1 lets the proof rule out cross-family moves.

What would settle it

Compute the set of triples with x,y,z ≤ 16 and x⊕y⊕z = 0, and check whether every such triple lies in P0 ∪ N0 (up to permutation). A single counterexample would break the second half of Theorem 2 as written; alternatively, computing Grundy numbers for all triples up to size 8 and comparing the zero-Grundy set with the paper's P would directly test the classification.

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Extended reading notes

Core claim

The central claim is Theorem 2: the set P defined in Definition 6(viii), i.e. P = P0 ∪ P1, is exactly the set of P-positions of the restricted amalgamation game of Definition 4. Up to permutation of the three piles, a position lies in P0 when its nim-sum x⊕y⊕z is 0 and either x+y=z with one of x,y smaller than 2, or x+y>z+2 with both x,y≤z. It lies in P1 when it is one unit above or below a member of the xor-zero families N0,1 or N0,2, with the direction (z+1 or z−1) decided by the parity of x+y. The proof has two halves: Lemmas 5–9 show that from any P-position every move goes outside P, and the second half of Theorem 2, with Lemmas 10 and 11, claims that from any position outside P there is a move into P.

Load-bearing premise

The proof assumes, without proving it, that every position whose three pile sizes have bitwise xor-sum 0 is covered by one of the named P0 or N0 families, and Lemma 10 (the route into P1) contains an unproved binary-arithmetic identity in one subcase; if either gap cannot be filled, the claim that every non-P position can move into P is incomplete.

Editorial extensions

If this is right

  • From any P-position, every legal move—reducing one pile or merging two piles of size at least 2—lands in an N-position, and from any position outside P there is a move into P.
  • The classification is explicit and checkable by bit operations, so a player can decide in O(log max pile) time whether a three-pile position is winning for the next player.
  • For the xor-zero families N0, the proof gives concrete moves into P: from N0,2 with odd piles, reduce the second pile by one and increase the third by one; from N0,1, merge the two piles of size at least 2 into P0,1.
  • The paper's closing conjecture ties Grundy numbers to nim-sum: a position's Grundy number and its nim-sum both lie in the pair {2n, 2n+1}, so the game values would be almost determined by the nim-sum.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The restriction that merged piles each have at least two stones appears to be what makes the carry analysis finite: with unrestricted merging, P-positions are known only for small heaps, while here Lemma 1 bounds the excess x+y−z to exactly the cases 0, 2, and ≥4.
  • One natural extension is to change the minimum merge size from 2 to k; the same per-bit machinery should yield P-families indexed by the possible excess values of x+y over z, and computing small cases for k=3 or k=4 could reveal the pattern.
  • If Conjecture 1 is true, it yields a fast Grundy-value algorithm using only the nim-sum and a few low-bit checks; a brute-force Grundy computation for piles up to, say, 16 would provide a direct check.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

3 major / 5 minor

Summary. The paper introduces a variant of Amalgamation Nim in which two piles may be amalgamated only when both piles contain at least two stones, and it proposes an explicit formula for the set of P-positions of the three-pile game. The claimed set P is defined in Definition 6(viii) as the union of P0 and P1, where P0 and P1 are built from families of triples satisfying xor-zero conditions and simple additive relations. The proof proceeds by showing in Lemmas 5-9 that a position in P cannot move to another position in P, and in Lemmas 10-11 that every position outside P can move into P. Theorem 2 states that P is exactly the set of P-positions.

Significance. If Theorem 2 is correct, it would give the first complete formula for the P-positions of a nontrivial Amalgamation Nim variant with more than two piles, a problem for which the paper notes no general formula was previously known. The formula is explicit, elementary, and falsifiable, and the associated Grundy-number conjecture offers a possible direction for future work. The proof strategy is natural and the individual lemmas are mostly simple binary-arithmetic arguments. However, the proof of the key lemmas 10 and 11 is incomplete and contains several concrete errors, and Theorem 2 relies on an unproved partition of xor-zero positions. These issues are repairable in principle, but they block acceptance in the current form.

major comments (3)
  1. [Theorem 2, proof [II] and [I.1]] The proof assumes, without statement or proof, that every position with xor sum 0 belongs to P0 or N0. In particular, the line "If x⊕y⊕z=0, then (x,y,z)∈N0" is false as written: for example, (1,1,0) has xor sum 0 and belongs to P0 (via a cyclic permutation of (0,1,1)∈P0,1), not to N0. A correct proof needs a lemma establishing the trichotomy of xor-zero positions into P0, N0, or neither, with the "neither" case proved impossible whenever it is needed. This partition is load-bearing for the second half of Theorem 2, because it justifies the invocation of Lemmas 10 and 11 in the two branches of the proof.
  2. [Lemma 10] Lemma 10 is not rigorously proved as written. Several concrete errors occur. In case [II.1], the sentence "Since z > w+1, we reach (u, v, w+1) ∈ P1" uses variables u, v, w that are not defined in that case; the case is about the triple (x, v, z), so the sentence is meaningless. In the same case, the proof asserts "If we have (75), x+(v+1)=z+1" from (75), which is x+v=z+2; the correct identity is x+(v+1)=z+3=(z+1)+2. This matters because the conclusion that (x, v+1, z+1) ∈ N0,2 depends on it. In case [II.3], the proof asserts (x, v+1, z−1) ∈ N0,2 from the additive relation x+(v+1)=(z−1)+2, but it does not prove the required xor identity x⊕(v+1)⊕(z−1)=0. In case [II.4.2], the proof derives (x, v−1, z+1) ∈ N0,1 and then states "by (90) (x, v−1, z+1) ∈ P1,1"; this is inconsistent with Definition 6(iii), which would place (x, v−1, z) in P1,1 when x+(v−1) is odd. Finally, in case [I.2], the proof says "we reach (x, v, w−1)" with an undefined v; the intended target appears to be (x, y, w−1). These errors are not merely typographical, because Lemma 10 is the link that shows a non-P position with a move to N0 can also reach P1.
  3. [Lemma 11] Lemma 11's statement and proof are inconsistent. The lemma claims that from a position in N0 one can reach a position in P1, but its case [II] for N0,1 amalgamates x and y and reaches (x+y, 0, z) ∈ P0,1, which is a position in P0, not P1. Thus the lemma as stated is false for N0,1. The proof of Theorem 2 only needs a move into P, so the lemma could be repaired by changing the conclusion to "can reach a position in P", but as written it is incorrect. In case [I] for N0,2, the proof derives (x, y−1, z+1) ∈ N0,1 and then concludes (x, y−1, z) ∈ P1,2; according to Definition 6(iv), the base family for P1,2 is N0,2, not N0,1, so the conclusion should be P1,1, not P1,2. These mismatches must be fixed for Lemma 11 to support Theorem 2.
minor comments (5)
  1. [Lemma 4] In both (i) and (ii), the last term is written as |z−x|; it should be |z−w|. As stated, the lemma is false, and the applications in Lemmas 5 and 9 implicitly use the corrected version.
  2. [Lemma 7, proof [II]] The sentence "By Lemma 4, for any position (u,v,w) ∈ P1, u,v,w≥1" is incorrect: Lemma 4 is about the L1 distance after a move. The lower bounds on the coordinates should be cited from Lemma 2(iii) and (iv).
  3. [Lemma 6, proof [II]] The text "i.e., (1), (1), (1) of Definition 5" should read "(1), (2), (3)", and the phrase "By We assume" contains a stray "By".
  4. [Lemma 2(ii), proof] In the proof of Lemma 2(ii), the sentence "By (16), x, y are odd and z is odd" should say "z is even" (the same proof later uses z even), and "by (1 and (16)" is an incomplete citation.
  5. [Introduction and general exposition] There are several typographical errors, including "we Maximum Amalgamation Nim" in the introduction, and Lemma 11's statement is grammatically incomplete because it lacks a main clause after "If ... and we can reach ...".

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: the P-position formula is derived from the game rules by explicit arithmetic definitions, and the proof gaps identified by the reader are incompleteness, not circular reasoning.

full rationale

The paper's central claim, Theorem 2, asserts that the explicitly defined set P in Definition 6(viii) is exactly the set of P-positions of the restricted amalgamation game in Definition 4. The definition of P is given purely in terms of xor and addition conditions, independent of the outcome classes, and the proof proceeds by showing that positions in P cannot move to P (Lemmas 5-9) and that positions outside P can move into P (Lemmas 10-11). No fitted parameters are renamed as predictions, no target result is assumed in the definition of P, and no load-bearing self-citation chain is present. The skeptical concerns about Lemma 10's undefined variables and unproved binary identities, and about the unstated trichotomy in Theorem 2's second half, are genuine proof-completeness issues: they mean the derivation is not fully rigorous as written. However, these are missing derivations or arithmetic gaps, not cases where an equation reduces to its own input by construction. There is no evidence that the claimed P set was defined in terms of the outcome class or that a cited earlier work is doing the mathematical work. Therefore the correct circularity score is 0.

Assumptions & free parameters 0 free parameters · 2 assumptions · 0 invented entities

The paper relies on standard Nim facts and on an unstated partition property of its own sets; no free parameters or invented entities appear.

assumptions (2)
  • standard math Standard Nim property: a position with xor sum not equal to 0 can be moved to xor sum 0 by reducing one pile, and a position with xor sum 0 cannot be moved to xor sum 0 by reducing one pile.
    Used in Theorem 2 [II] without proof; this is well-known.
  • ad hoc to paper Every position with xor sum 0 belongs to either P0 or N0 (the union of the paper's defined sets).
    Assumed in Theorem 2 [II] when it states 'If x⊕y⊕z=0, then (x,y,z)∈N0'; never explicitly stated or proved, though it follows from Lemma 1 and the definitions.

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Cite this review

Pith. "Pith review of An Amalgamation Nim with Restriction." pith.science (2026). https://pith.science/paper/DCLQC3DS

@misc{pith2026241114488,
  author       = {Pith},
  title        = {Pith review of: An Amalgamation Nim with Restriction},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/DCLQC3DS}},
  note         = {Machine review of arXiv:2411.14488}
}
read the original abstract

In an amalgamation Nim, players are allowed to use a move from the traditional form of Nim or to amalgamate two piles when they are not empty. No formula that describes the set of P-positions of Amalgamation Nim is known. The author gives a condition on the amalgamation of two piles. Players can amalgamate two piles only when the number of stones in these two piles is equal to or more than 2. Then, we get a formula that describes the set of P-positions for this variant of amalgamation Nim.

Discussion (0). Continue with ORCID to comment.

Reference graph

Works this paper leans on

7 extracted references · 7 canonical work pages

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    D. Richards, Personal Communication, (2001)

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Reviewed August 12, 2026 · model on record in the stance chip above.