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Statement

Setting (p. 1). A $p$-sunflower is a family of $p$ sets whose pairwise intersections are identical.

Theorem 1 (p. 1, quoted). "There is a universal constant $\alpha>1$ such that every family of more than $(\alpha p\log(pk))^k$ sets of size $k$ must contain a $p$-sunflower."

The paper declares on p. 3 that all its logarithms are to base 2; in Theorem 1 the base only rescales $\alpha$. The constant $\alpha$ is not made explicit. The introduction (p. 1) presents the result as a simpler proof of a bound similar to that of Alweiss, Lovett, Wu and Zhang, who showed that $(\log k)^k\cdot(p\log\log k)^{O(k)}$ sets of size $k$ force a $p$-sunflower; for comparison it recalls Erdős and Rado's bound $(p-1)^k\cdot k!$ and their family of $(p-1)^k$ sets of size $k$ with no $p$-sunflower.

Source. Anup Rao, Coding for sunflowers, Discrete Analysis 2020:2, 8 pp., doi:10.19086/da.11887 (arXiv:1909.04774v2). Theorem 1 is on p. 1, its deduction from Lemma 2 on p. 2. Card: Rao 2020.

Read depth. Claims checked: the statement and the definition of a $p$-sunflower were read clause by clause on the printed page. The deduction on p. 2 was read for structure only.

Proof pointer

Page 2, by induction on $k$, with $r(p,k)=\alpha p\log(pk)$. For $k=1$ a family of more than $r(p,1)$ distinct singletons contains $p$ of them, which form a $p$-sunflower. For $k>1$, if the family is not $r(p,k)$-spread, some nonempty $Z$ lies in more than $r^{k-|Z|}$ of its sets; removing $Z$ from those sets and applying the induction hypothesis, using that $r(p,k)$ does not decrease in $k$, gives a $p$-sunflower. If the family is $r(p,k)$-spread, Lemma 2 gives $p$ pairwise disjoint sets, which form a $p$-sunflower.

Dependencies

Lemma 2 (p. 2), itself deduced from Lemma 4.

Bears on

• Problem 20: in the problem's notation, with $n$ the size of the sets and $k$ the size of the sunflower, Theorem 1 gives $f(n,k)\le(\alpha k\log(kn))^n+1$. The base $\alpha k\log(kn)$ grows with $n$, so the bound is not of the form $c_k^n$ the problem asks for, and the theorem does not answer it.