Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
Setting (p. 2). The -spread sequences are those of Lemma 2.
Definition 3 (p. 2, quoted). "Given , for and , let be equal to , where is chosen to minimize among all choices with . If there are multiple choices for that minimize , let be the smallest one."
Thus , and exactly when some lies in ; the paper also notes that when (p. 2).
Lemma 4 (p. 2, quoted). "There is a universal constant such that the following holds. Let . If , and is an -spread sequence of at least sets of size , is uniformly random, and is a uniformly random set of size at least independent of , then . In particular, ."
The logarithm is to base 2 (p. 3). The paper calls the lemma of independent interest, relevant to applications in theoretical computer science (p. 2).
Source. Anup Rao, Coding for sunflowers, Discrete Analysis 2020:2, 8 pp., doi:10.19086/da.11887 (arXiv:1909.04774v2). Definition 3 and Lemma 4 are on p. 2; the proof is Section 4, pp. 3--7, using Lemma 5 of Section 3 (p. 3). Card: Rao 2020.
Read depth. Claims checked: Definition 3 and the statement were read clause by clause on the printed page. The proof was read for structure only.
Proof pointer
Pages 3--7. Removing sets only increases , so one may take . The proof shows, for a constant and each integer , that a uniformly random of size at least has ; taking and the constant large gives the bound (the print writes for the constant at this step, p. 3). The induction on writes with disjoint and random, fixes , and shows by giving a prefix-free encoding of the pair that is short when is large compared with ; Lemma 5 (p. 3), the converse of Shannon's noiseless coding theorem for the uniform distribution, bounds the average length below by the logarithm of the number of pairs. Two cases are encoded (pp. 4--7), according to whether few or many indices have containing a given part of ; the spread condition controls the second case.
Dependencies
Definition 3 (p. 2); Lemma 5 (p. 3), whose proof uses Kraft's inequality.
Bears on
- Problem 20 indirectly: the paper deduces from Lemma 4 Lemma 2 and from it Theorem 1, whose bound does not answer the problem.