English

The sum-capture problem for abelian groups

Discrete Mathematics 2014-01-07 v2 Combinatorics

Abstract

Let GG be a finite abelian group, let 0<α<10 < \alpha < 1, and let AGA \subseteq G be a random set of size Gα|G|^\alpha. We let μ(A)=maxB,C:B=C=A{(a,b,c)A×B×C:a=b+c}. \mu(A) = \max_{B,C:|B|=|C|=|A|}|\{(a,b,c) \in A \times B \times C : a = b + c \}|. The issue is to determine upper bounds on μ(A)\mu(A) that hold with high probability over the random choice of AA. Mennink and Preneel \cite{BM} conjecture that μ(A)\mu(A) should be close to A|A| (up to possible logarithmic factors in G|G|) for α1/2\alpha \leq 1/2 and that μ(A)\mu(A) should not much exceed A3/2|A|^{3/2} for α2/3\alpha \leq 2/3. We prove the second half of this conjecture by showing that μ(A)A3/G+4A3/2ln(G)1/2 \mu(A) \leq |A|^3/|G| + 4|A|^{3/2}\ln(|G|)^{1/2} with high probability, for all 0<α<10 < \alpha < 1. We note that 3α1(3/2)α3\alpha - 1 \leq (3/2)\alpha for α2/3\alpha \leq 2/3. In previous work, Alon et al.. have shown that μ(A)O(1)A3/G\mu(A) \leq O(1)|A|^3/|G| with high probability for α2/3\alpha \geq 2/3 while Kiltz, Pietrzak and Szegedy show that μ(A)A1+2α\mu(A) \leq |A|^{1 + 2\alpha} with high probability for α1/4\alpha \leq 1/4. Current bounds on μ(A)\mu(A) are essentially sharp for the range 2/3α12/3 \leq \alpha \leq 1. Finding better bounds remains an open problem for the range 0<α<2/30 < \alpha < 2/3 and especially for the range 1/4<α<2/31/4 < \alpha < 2/3 in which the bound of Kiltz et al.. doesn't improve on the bound given in this paper (even if that bound applied). Moreover the conjecture of Mennink and Preneel for α1/2\alpha \leq 1/2 remains open.

Cite

@article{arxiv.1309.5582,
  title  = {The sum-capture problem for abelian groups},
  author = {John P Steinberger},
  journal= {arXiv preprint arXiv:1309.5582},
  year   = {2014}
}
R2 v1 2026-06-22T01:31:43.299Z