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The number of sum-free subsets of lattice cubes

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Pith's one-line read For every fixed dimension $d\ge 3$, the number of sum-free subsets of the lattice cube $[n]^d$ is $2^{M([n]^d)+O_d(n^{d-1})}$, where $M$ is the maximum size of such a subset.

desk verdict New profile-defect machinery gets the sharp boundary-order count for all d≥3; the Lemma 14 stress-test objection is a misreading, and the main caveat is the imported Keevash–Lim preprint. read the letter →

arxiv 2608.23544 v1 pith:GTTWFGKY submitted 2026-08-24 math.CO math.NT

Haoran Luo This is my paper · ORCID
classification math.COmath.NT MSC 05D4011B7505A16
keywords sum-freesetslatticecubeasymptoticenumerationKeevash-LimdualweightsprofilemethodfractionalentropyinequalityCameron-Erdosconjectureextremalcombinatorics
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The pith

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

The reading

This paper proves that for every fixed dimension $d\ge 3$, the number of sum-free subsets of the lattice cube $[n]^d$ is $2^{M([n]^d)+O_d(n^{d-1})}$, where $M$ is the size of the largest such subset. Since every subset of a sum-free set is sum-free, that count is trivially at least $2^M$; the theorem shows the true count exceeds this lower bound only by a boundary-order factor in the exponent. The result settles, for all remaining dimensions, the counting conjecture of Elsholtz and Rackham, after the one-dimensional Cameron–Erdős problem and the two-dimensional case had already been solved. A reader should care because it turns a coarse extremal bound into an exact asymptotic enumeration by a method that avoids container lemmas and stability theorems.

What carries the argument

The central object is the height profile $P(S)$ of a sum-free set, together with its weighted defect $\delta(P)$. The defect is built from three dual weights $\omega_1,\omega_2,\omega_3$ due to Keevash and Lim—supported on triples of fibers $(B,B,D)$, $(A,C_-,C_+)$, and $(C_-,C_-,E)$—and measures, in a weighted sense, how far the profile values are from satisfying the additive relations $x+y=z$ on each relevant triple of fibers. The argument's load-bearing identities are the cancellation identities: four bounded translations of the first or second weight combine so that the signed defect terms cancel the target and one source, isolating the second difference $\Delta_{e_1}^2 P$. That single quantity does double work: it yields the exponential saving in the fixed-profile count through local entropy estimates, and it lets the profile-counting lemma reconstruct any lower profile from two initial values and its second differences. The local entropy estimates are assembled over all fibers by an ordered conditional form of the Madiman–Tetali fractional entropy inequality, with Zhao's bipartite swapping lemma reducing the repeated-source case to the distinct-source case.

What would settle it

Exhibit, for some $d\ge 3$ and infinitely many $n$, a profile with zero weighted defect that is realized by more than $2^{M([n]^d)+C n^{d-1}}$ sum-free sets for every constant $C$; Proposition 6 says this cannot happen, so such an example would refute the theorem. Alternatively, a direct failure of Lemma 12's $O(n^{d-2})$ translation bound would break the profile-counting step.

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Extended reading notes

Core claim

On the paper's own terms, the discovery is that the number of sum-free subsets of $[n]^d$ is controlled, up to a factor $2^{O_d(n^{d-1})}$, by the maximum sum-free set itself: $SF([n]^d)=2^{M([n]^d)+O_d(n^{d-1})}$ for every fixed $d\ge 3$. The proof achieves this by assigning to each sum-free set a height profile—roughly, the lowest occupied point on each fiber in the lower regions and the highest occupied point in the upper regions—and then showing two things: with a fixed profile, the number of sum-free sets realizing it is exponentially smaller than $2^M$ unless the profile's weighted defect is small, and the number of profiles with defect at most $R$ is only $2^{\varepsilon R+O(n^{d-1})}$. These two estimates combine to match the trivial lower bound.

Load-bearing premise

The load-bearing premise is that the Keevash–Lim dual weights $\omega_1,\omega_2,\omega_3$ exist with exactly the supports, marginal sums, and covering properties stated in Lemmas 8–11; the paper verifies the refinements and boundary estimates but imports the recursive coupling construction itself from [21], and if any of those weight properties failed, both counting propositions would collapse.

Editorial extensions

If this is right

  • For every fixed $d\ge 3$, the asymptotic count is settled: $SF([n]^d)=2^{M([n]^d)+O_d(n^{d-1})}$, so the extremal size determines the total number up to a boundary-order factor in the exponent.
  • The $d=1$ and $d=2$ results together with this theorem give a complete dimensional picture for the number of sum-free subsets of integer lattice cubes.
  • The method shows that container lemmas and stability theorems are not necessary for this counting problem; the direct profile-and-defect route suffices.
  • Parity of $n$ is not essential: the even case transfers to odd $n$ with only $O(n^{d-1})$ loss, so the parity assumption is a technical convenience.
  • Profiles with large defect are exponentially rare, and sum-free sets with a fixed profile are exponentially few unless its defect is small; the two counting propositions quantify this tradeoff explicitly.

Reading between the lines

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

  • The same weighted-defect scheme could likely enumerate subsets avoiding other translation-invariant linear equations on grids, whenever the extremal linear program has a dual certificate with the needed marginal-sum properties; this generalization is not in the paper.
  • A byproduct of the defect bookkeeping is a potential route to stability in dimensions $d\ge 3$: tracing near-equality in the Keevash–Lim weights may show that near-extremal sum-free sets are close to the optimal stripe, a question the paper leaves open.
  • If the anti-collision estimates discussed in the concluding remarks are verified, the repeated-source case could be handled without the entropy inequality and swapping lemma, possibly simplifying the proof; the paper reports only a preliminary, unverified check.
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Formalized claims in Lean

  1. Claim #1: On the paper's own terms, the discovery is that the number of sum-free subsets of $[n]^d$ is controlled, up to a factor $2^{O_d(n^{d-1})}$, by the maximum sum-free set itself: $SF([n]^d)=2^{M([n]^d)+O_d(n^{d-1})}$ for every fixed $d\ge 3$. The proof achieves this by assigning to each sum-free set a height profile—roughly, the lowest occupied point on each fiber in the lower regions and the highest

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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

1 major / 3 minor

Summary. The paper proves that for every fixed dimension d at least 3, the number SF([n]^d) of sum-free subsets of the d-dimensional lattice cube is 2^{M([n]^d)+O_d(n^{d-1})}, where M([n]^d) is the maximum size of a sum-free subset. This confirms the Elsholtz-Rackham conjecture in all remaining dimensions. The proof introduces height profiles for subsets of Q=[0,n]^d, defines weighted defects using the Keevash-Lim dual weights, and proves two counting propositions: one bounding the number of sum-free sets with a fixed profile in terms of its defect, and one bounding the number of profiles with defect at most R. The fixed-profile estimate uses local entropy inequalities and the Madiman-Tetali ordered fractional Shearer inequality, while the profile count uses cancellation identities for second differences along e1-lines and weighted geometric-series counting.

Significance. If correct, this is a substantial result: it settles the Elsholtz-Rackham counting conjecture for all remaining dimensions, matching the trivial lower bound and extending the one-dimensional Cameron-Erdos theorem and Ghosal's two-dimensional result. The argument is methodologically interesting because it avoids both container lemmas and stability theorems. The paper also has real strengths in presentation: the one-dimensional estimate of Ghosal is given with a self-contained proof, the coordinate sums and translation effects of the dual weights are computed explicitly in Appendix A, and the numerical constants are tracked carefully. I specifically checked the potential gap in Lemma 14 raised during review: it does not land. The translated weight omega_i^{r,s} is supported on triples that themselves lie in T(B,B,D) or T(A,C-,C+), so the identity (11) applies directly; the proof of Lemma 14 is sound on this point. The main external dependency is the existence and support properties of the Keevash-Lim dual weights imported from [21], but the manuscript verifies the refinements it needs without reproducing the underlying recursive coupling construction.

major comments (1)
  1. [Section 3, Lemma 8] The existence of the three weight functions omega_1, omega_2, omega_3 with the stated support, covering, and coordinate-sum properties is imported from Keevash-Lim [21, Lemma 4.3 and Corollary 2.4]. Since these weights are the foundation for both Proposition 6 and Proposition 7, this is a load-bearing black box. The appendix verifies the refinements and coordinate sums, but it does not reproduce the construction itself. This is not a mathematical error, but the authors should either ensure that [21] is available in final published form or include enough of the construction to make the paper self-contained on this point.
minor comments (3)
  1. [Section 3.2] To avoid a possible misreading, it would be helpful to state explicitly that the support of omega_1^{r,s} is contained in T(B,B,D) and the support of omega_2^{r,s} is contained in T(A,C-,C+); this makes the application of (11) in Lemma 14 immediate.
  2. [Lemma 17] The sentence beginning 'Indeed, Lemma 2.1 in [30] actually gives a size-preserving bijection...' is informal and not needed for the proof; it could be deleted or expanded into a precise statement.
  3. [References] References [15] and [21] are arXiv preprints; if any of them appear in final form before this paper is published, the references should be updated to the published versions.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: the counting proof is built from external dual-weight, entropy, and profile-count inputs that are independent of the target count.

full rationale

The derivation chain is not circular. The paper takes as inputs the Keevash–Lim dual weights for M([n]^d), Ghosal's one-dimensional counting estimate, Zhao's bipartite swapping lemma, and the Madiman–Tetali fractional entropy inequality. None of these inputs is defined in terms of SF([n]^d), and none is fitted to any counting data from the present paper. The quantity M([n]^d) appears only as an externally established maximum-size parameter, and the trivial lower bound SF([n]^d) ≥ 2^{M([n]^d)} is not used as an assumption anywhere in the upper-bound proof; it merely supplies the matching direction after the upper bound is proved. The profile defect δ(P) is a purely structural function of a profile and is not fitted to the number of sum-free sets; Proposition 6 and Proposition 7 are separate counting statements whose combination gives the theorem. The Appendix verifies the needed refinements of the Keevash–Lim weight properties rather than importing the theorem being proved. The apparent concern about Lemma 14 applying equation (11) to translated supports is resolved by the definition of the translated weights: ω_i^{r,s}(x,y,z) is nonzero only when (x-r,y-s,z-r-s) lies in G_i^{r,s}, which by definition forces (x,y,z) itself to lie in the same region product T(B,B,D) or T(A,C^-,C^+), so the fiber-size identity (11) applies termwise. The acknowledgment of generative-AI assistance and the discussion of alternative methods in Section 6 are not circularity, because the author states that all mathematical statements were independently checked and the final proof does not rely on those alternatives. Overall, the central claim is self-contained given its stated external ingredients, so no step reduces by construction to its own inputs.

Assumptions & free parameters 0 free parameters · 6 assumptions · 0 invented entities

No free parameters are fitted to data; all constants are absolute or dimension-dependent implicit constants. The proof depends on external domain assumptions from Keevash-Lim [21] and standard entropy/swap inequalities; no new physical or combinatorial entities are postulated beyond the proof-defined profiles and defects.

assumptions (6)
  • domain assumption Keevash-Lim dual weights ω1,ω2,ω3 exist with supports T(B,B,D), T(A,C-,C+), T(C-,C-,E), W≤1 on C, coordinate-sum identities, and O_d(n^{d-2}) boundary losses (Lemmas 8-11).
    This is the foundation of the defect accounting in Propositions 6 and 7; the paper verifies refinements but not the original recursive construction in [21, Lemma 4.3].
  • domain assumption Keevash-Lim additive couplings (Corollary 2.4 in [21], restated as Lemma 24): for all j,r,s there is a coupling X+Y=Z with uniform marginals on slices.
    Used in Appendix A to derive coordinate sums for the three weights.
  • standard math The extremal stripe S_{d,n} is sum-free and has fiber sums 2(n+1) on the weight supports (Lemma 13).
    Direct calculation from the region definitions; used in Lemma 14 to bound the dual objective by M(Q).
  • standard math Ordered conditional fractional Shearer inequality (Lemma 19, Madiman-Tetali) and Zhao's bipartite swapping lemma (Lemma 17).
    External published inequalities used to assemble local entropy savings and to handle repeated source fibers.
  • standard math Ghosal's one-dimensional counting estimate g(t) ≤ K(2-η)^t (Lemma 16).
    A nontrivial auxiliary estimate; the paper gives a self-contained proof in Appendix B.
  • domain assumption The threshold u*_d is the unique maximizer of the stripe volume with d/3<u*_d<d/2, and m=floor(u*_d n) defines the five regions.
    Taken from Keevash-Lim Section 3; sets up the boundary geometry used everywhere.

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Pith. "Pith review of The number of sum-free subsets of lattice cubes." pith.science (2026). https://pith.science/paper/GTTWFGKY

@misc{pith2026260823544,
  author       = {Pith},
  title        = {Pith review of: The number of sum-free subsets of lattice cubes},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/GTTWFGKY}},
  note         = {Machine review of arXiv:2608.23544}
}
abstract

A subset of the $d$-dimensional lattice cube $[n]^d$ is sum-free if it contains no solution to the equation $x+y=z$. We study the total number of such subsets. For $d=1$, Cameron and Erd\H{o}s conjectured that the number of sum-free subsets of $[n]$ is $O(2^{n/2})$, and this was proved independently by Green and Sapozhenko. A recent work by Ghosal solved the case $d = 2$. In this paper, we consider all remaining dimensions and prove that for every fixed integer $d \geqslant 3$, the number of sum-free subsets of $[n]^d$ is $2^{M([n]^d) + O_d(n^{d-1})}$, where $M([n]^d)$ is the maximum possible size of a sum-free subset of $[n]^d$. This verifies a conjecture of Elsholtz and Rackham. Our proof combines the dual weights constructed by Keevash and Lim in their work for $M([n]^d)$, a one-dimensional counting estimate due to Ghosal, a bipartite swapping lemma of Zhao, and a strong fractional entropy inequality of Madiman and Tetali, and it avoids the use of the container lemma or deriving a stability theorem first.

Figures

Figures reproduced from arXiv: 2608.23544 by the authors.

Figure 1
Figure 1. The regions in the two-dimensional base V when d = 3. (ω, β) is a feasible solution to the dual program in (Dual) and gives the required asymptotic upper bound M(Q) + Od(n d−1 ) for the primal program. For our counting argument, we need slightly more information from their construction. The reason is that the region C plays different roles in ω2 and ω3, and this requires us to use different information about C. Ther… view at source ↗
Figure 2
Figure 2. Schematic picture for d = 3. The profile on each line can be determined from the first two values and the second difference. There are (n + 1)d−2 choices for u ∈ [0, n] d−2 . For each u, there are at most three intervals, corresponding to A1, B, C−, respectively. Each interval requires at most two initial profile values, 19 [PITH_FULL_IMAGE:figures/full_fig_p019_2.png] view at source ↗
Figure 3
Figure 3. A schematic picture for d = 3 showing how the upper profile on D is counted after the lower profile on B has been fixed. For a fixed z ∈ D, every representation z = x + y with x, y ∈ B gives the known P(x) + P(y). Thus, the weighted sum ψz(P(z)) controls the possible value of P(z). Multiplying by the corresponding weight and summing over all x, y with x + y = z, we obtain ψz(t1) +ψz(t2) ⩾ λ(z)|t1 −t2|. For every z ∈… view at source ↗

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Works this paper leans on

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