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library/additive_combinatorics/dubroff_2021_note_erdos_distinct_subset_sums_problem

Full paper in Markdown.


Dubroff, Q., Fox, J., and Xu, M. W., "A note on the Erdős distinct subset sums problem," SIAM Journal on Discrete Mathematics 35 (2021), 322--324. The retained local source is arXiv:2006.12988v2, 20 July 2020. See the Full paper in Markdown and the arXiv record.

Result

If

0<a1<⋯<an0<a_1<\cdots<a_n

are integers whose subset sums are all distinct, the paper proves the exact finite bound

an≥(n⌊n/2⌋).a_n\geq \binom{n}{\lfloor n/2\rfloor}.

This is stated immediately after Theorem 1 and obtained in its second proof (local PDF pp. 1--2; Markdown paragraphs beginning "The second proof" and "A second proof of Theorem 1"). By the central-binomial asymptotic, it implies the displayed Theorem 1:

an≥(2π−o(1))n−1/22n.a_n\geq \left(\sqrt{\frac{2}{\pi}}-o(1)\right)n^{-1/2}2^n.

The asymptotic constant matches an unpublished bound of Elkies and Gleason and improves Aliev's previously published constant 3/(2π)\sqrt{3/(2\pi)}. The paper also records Bohman's construction an≤0.22002 2na_n\leq 0.22002\,2^n for the opposite direction of the distinct-subset-sums problem.

The two proofs

Berry--Esseen proof. In the proof of Theorem 1 (local PDF p. 2; Markdown paragraph beginning "Proof of Theorem 1"), take independent uniform signs and put

X=∑i=1nϵiai,σ2=∑i=1nai2.X=\sum_{i=1}^n\epsilon_i a_i, \qquad \sigma^2=\sum_{i=1}^n a_i^2.

Distinct subset sums make the 2n2^n values of XX distinct and of one parity, each with mass 2−n2^{-n}. Choose δ→0\delta\to0 sufficiently slowly. If σ≤an/δ\sigma\leq a_n/\delta, Moser's variance bound

σ2≥12+22+⋯+22(n−1)=4n−13\sigma^2\geq 1^2+2^2+\cdots+2^{2(n-1)}=\frac{4^n-1}{3}

already gives more than the required asymptotic lower bound for ana_n. If σ>an/δ\sigma>a_n/\delta, apply the quoted Berry--Esseen theorem (Theorem 2, local PDF pp. 1--2) to Xi=ϵiaiX_i=\epsilon_i a_i. Since

∑iE∣Xi∣3=∑iai3≤anσ2,\sum_i\mathbb E|X_i|^3=\sum_i a_i^3\leq a_n\sigma^2,

the distribution of XX is within O(an/σ)=O(δ)O(a_n/\sigma)=O(\delta) of the centered normal distribution with variance σ2\sigma^2. For ℓ=ασ\ell=\alpha\sigma, where α→0\alpha\to0 and δ=o(α)\delta=o(\alpha), compare

Pr⁡(∣X∣≤ℓ)≤(ℓ+1)2−n\Pr(|X|\leq\ell)\leq(\ell+1)2^{-n}

with the normal estimate

Pr⁡(∣X∣≤ℓ)∼2πℓσ.\Pr(|X|\leq\ell) \sim\sqrt{\frac{2}{\pi}}\frac{\ell}{\sigma}.

This gives σ≥(2/π−o(1))2n\sigma\geq(\sqrt{2/\pi}-o(1))2^n; the inequality σ2≤nan2\sigma^2\leq n a_n^2 finishes the first proof.

Harper-isoperimetric proof. Theorem 3 and the second proof of Theorem 1 (local PDF p. 2; the correspondingly labelled passages in the Markdown copy) work on the half-cube {−1/2,1/2}n\{-1/2,1/2\}^n. The family

F={ϵ:a⋅ϵ<0}\mathcal F=\{\epsilon:a\mathbin{\cdot}\epsilon<0\}

has size 2n−12^{n-1}. Harper's vertex-isoperimetric inequality gives

∣∂F∣≥(n⌊n/2⌋).|\partial\mathcal F|\geq\binom{n}{\lfloor n/2\rfloor}.

Every η∈∂F\eta\in\partial\mathcal F satisfies 0<a⋅η<an0<a\mathbin{\cdot}\eta<a_n. These values are distinct: equality for two boundary points would give equal sums of two distinct subsets. Their pairwise differences are integers, so they occupy distinct points of one unit-spaced lattice inside an interval of length ana_n. Consequently ∣∂F∣≤an|\partial\mathcal F|\leq a_n, proving the exact central-binomial bound.

Consequence for the interval competitor in Problem 963

Write d(A)d(A) for the greatest size of a dissociated subset of AA, in the notation of Problem 963, and put m=d([N])m=d([N]). A largest dissociated B⊆[N]B\subseteq[N] is an mm-element set of positive integers with distinct subset sums and max⁡B≤N\max B\leq N. Applying the exact bound to BB gives

N≥max⁡B≥(m⌊m/2⌋).N\geq\max B\geq\binom{m}{\lfloor m/2\rfloor}.

Thus the finite interval-side upper bound is

d([N])≤max⁡{m:(m⌊m/2⌋)≤N},d([N])\leq \max\left\{m:\binom{m}{\lfloor m/2\rfloor}\leq N\right\},

and Stirling's formula yields

d([N])≤log⁡2N+12log⁡2log⁡2N+O(1).d([N])\leq \log_2 N+\frac12\log_2\log_2 N+O(1).

This constrains one admissible competitor; it does not prove the universal lower bound asked for in Problem 963. Indeed, if f(N)=min⁡∣A∣=Nd(A)f(N)=\min_{|A|=N}d(A), then choosing A=[N]A=[N] gives f(N)≤d([N])f(N)\leq d([N]), the opposite direction from the proposed f(N)≥⌊log⁡2N⌋f(N)\geq\lfloor\log_2N\rfloor. The paper's theorem also uses positivity and integrality after a dissociated subset has already been selected. Problem 963 ranges over every NN-element set of reals, so a lower bound for f(N)f(N) must produce a large dissociated subset uniformly for all such sets.