Erdős Problem a Daythe open tracker

Erdős Problem #1179

PROVEDdatabase updated 2026-01-26

The statement as posed is true, and a proof exists.

What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).

\(F_A(g) = \#\{ S\subseteq A : g = \sum_{x\in S}x\}\).

`A` is `ε`-balanced: \(\lvert F_A(g)-2^{|A|}/N\rvert \leq \epsilon 2^{|A|}/N\) for all `g`.

The probability that a uniformly random `k`-element subset of `G` is `ε`-balanced.

Let \(0<\epsilon<1\) and let \(g_\epsilon(N)\) be the minimal \(k\) such that if \(G\) is an abelian group of size \(N\) and \(A\subseteq G\) is a uniformly random subset of size \(k\), and \[F_A(g) = \#\left\{ S\subseteq A : g = \sum_{x\in S}x\right\},\] then, with probability \(\to 1\) as \(N\to \infty\), \[\left\lvert F_A(g)-\frac{2^k}{N}\right\rvert \leq \epsilon \frac{2^k}{N}\] for all \(g\in G\). Estimate \(g_\epsilon(N)\) - in particular, is it true that for all \(\epsilon>0\) \[g_\epsilon(N)=(1+o_\epsilon(1))\log_2N?\] It is trivial that \(g_\epsilon(N)\geq \log_2N\) for all \(0<\epsilon<1\). Erdős and Rényi [ErRe65] proved that for all \(0<\epsilon<1\), \(g_\epsilon(N) \leq (2+o(1))\log_2N+O_\epsilon(1)\). Erdős and Hall [ErHa76] proved that, for all \(0<\epsilon<1\), \[g_\epsilon(N)\leq \left(1+O_\epsilon\left(\frac{\log\log\log N}{\log\log N}\right)\right) \log_2N.\] The upper bound \(g_\epsilon(N)\leq (1+o(1))\log_2 N\) is formalised as: there is a choice of sizes \(k(N) = (1+o(1))\log_2 N\) such that, along any sequence of abelian groups with \(|G|\to\infty\), the probability that a random \(k(|G|)\)-subset is \(\epsilon\)-balanced tends to \(1\). The matching lower bound is `erdos_1179.variants.lower_bound`. See
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 1 further variant / known partial result in the Lean file. The authoritative wording is on erdosproblems.com/1179.
Prizenone
Topicsadditive combinatorics, probability
Statement in LeanFormal Conjectures
Solution in Leanno
Original pageerdosproblems.com/1179
NOBODY HAS SENT US AN ATTEMPT YET

If you take a run at this one — with your own AI subscription or without — the guide says how, and the submission form gets you an honest read of what you found.