Coding for Sunﬂowers

A sunﬂower is a family of sets that have the same pairwise intersections. We simplify a recent result of Alweiss, Lovett, Wu and Zhang that gives an upper bound on the size of every family of sets of size k that does not contain a sunﬂower. We show how to use the converse of Shannon’s noiseless coding theorem to give a cleaner proof of a similar bound.


Introduction
A p-sunflower is a family of p sets whose pairwise intersections are identical.How large can a family of sets of size k be if the family does not contain a p-sunflower?Erdős and Rado [ER60] were the first to pose and answer this question.They showed that any family with more than (p−1) k •k! sets of size k must contain a p-sunflower.This fundamental fact has many applications in mathematics and computer science [ES92, Raz85, FMS97, GM07, GMR13, Ros14, RR18, LZ19, LSZ19].
After nearly 60 years, the correct answer to this question is still not known.There is a family of (p−1) k sets of size k that does not contain a p-sunflower, and Erdős and Rado conjectured that their lemma could be improved to show that this is essentially the extremal example.Recently, Alweiss, Lovett, Wu and Zhang [ALWZ19] made substantial progress towards resolving the conjecture.They showed that (log k) k • (p log log k) O(k) sets ensure the presence of a p-sunflower.Subsequently, Frankton, Kahn, Narayanan and Park [FKNP19] improved the counting methods developed by [ALWZ19] to prove a conjecture of Talagrand [Tal10] regarding monotone set systems.
In this work, we give simpler proofs for these results.Our proofs rely on an encoding argument inspired by a similar encoding argument used in [ALWZ19,FKNP19].The main novelty is our use of Shannon's noiseless coding theorem [Sha48,Kra49] to reason about the efficiency of the encoding, which turns out to avoid complications that show up when using vanilla counting.We show1 : Theorem 1.There is a universal constant α > 1 such that every family of more than (αp log(pk)) k sets of size k must contain a p-sunflower.
Let r(p, k) denote the quantity αp log(pk).We say 2 that a sequence 3 of sets S 1 , . . ., S ⊂ [ ] of size k is r-spread if for every non-empty set Z ⊂ [n], the number of elements of the sequence that contain Z is at most r k−|Z| .We prove that for an appropriate choice of α, the following lemma holds: Lemma 2. If a sequence of more than r(p, k) k sets of size k is r(p, k)-spread, then the sequence must contain p disjoint sets.
As far as we know, it is possible that Lemma 2 holds even when r(p, k) = O(p).Such a strengthening of Lemma 2 would imply the sunflower conjecture of Erdős and Rado.Lemma 2 easily implies Theorem 1: we proceed by induction on k.When k = 1, the theorem holds, since the family contains p distinct sets of size 1.For k > 1, if the sets are not r-spread, then there is a non-empty set Z such that more than r k−|Z| of the sets contain Z.By induction, and since r(p, k) can only increase with k, the family of sets contains a p-sunflower.Otherwise, if the sets are r(p, k)-spread, Lemma 2 guarantees the presence of a p-sunflower.
It only remains to prove Lemma 2. In fact, we prove something much stronger: a small random set is very likely to contain some set of an r-spread family of sets.

Random sets and r-spread families
To prove Lemma 2, we need to understand the extent to which a small random set W ⊆ [n] contains some set of a large family of sets of size k.To that end, it is convenient to use the following definition: Observe that the definition makes sense even if S 1 , . . ., S are not all distinct.When U ⊆ W , we have |χ(x, U )| ≥ |χ(x, W )|.Moreover, χ(x, W ) = ∅ if and only if there is an index y for which S y ⊆ W .Our main technical lemma shows that if a long sequence of sets is r-spread, then |χ(X, W )| is likely to be small for a random X and a random small set W : Lemma 4.There is a universal constant β > 1 such that the following holds.Let This lemma is of independent interest -it is relevant to several applications in theoretical computer science [Ros14,LSZ19].Before we prove Lemma 4, let us see how to use it to prove Lemma 2.
Proof of Lemma 2. Set γ = 1/(2p), = 1/p.Then r = r(k, γ, ) = r(p, k).Let W 1 , . . ., W p be a uniformly random partition of [n] into sets of size at least n/p .So, each set W i is of size at least n/p ≥ γn.By symmetry and linearity of expectation, we can apply Lemma 4 to conclude that is a non-negative integer, there must be some fixed partition W 1 , . . ., W p for which This can happen only if the family contains p disjoint sets.
Next, we briefly describe a technical tool from information theory, before turning to prove Lemma 4.

Prefix-free encodings
A prefix-free encoding is a map E : [t] → {0, 1} * into the set of all binary strings, such that if i = j, E(i) is not a prefix of E(j).Another way to view such an encoding is as a map from the set [t] to the vertices of the infinite binary tree.The encoding is prefix-free if E(i) is never an ancestor of E(j) in the tree.
Shannon [Sha48] proved that one can always find a prefix-free encoding such that the expected length of the encoding of a random variable X ∈ [t] exceeds the entropy of X by at most 1.Conversely, every encoding must have average length that is at least as large as the entropy.For our purposes, we only need the converse under the uniform distribution.The proof is short, so we include it here.All logarithms are taken base 2.
Lemma 5. Let E : [t] → {0, 1} * be any prefix-free encoding, and i be the length of E(i).Then (1/t) • t i=1 i ≥ log t.Proof.We have where the inequality follows from the concavity of the logarithm function.The fact that this last quantity is at most 0 is known as Kraft's inequality [Kra49].Consider picking a uniformly random binary string longer than all the encodings.Because the encodings are prefix-free, the probability that this random string contains the encoding of some element of [t] as a prefix is exactly t i=t 2 − i .So, this number is at most 1, and the above expression is at most 0.

Proof of Lemma 4
We shall prove that there is a constant κ such that the following holds.For each integer m with 0 ≤ m ≤ rγ/κ, if W is a uniformly random set of size at least κmn/r, then E [|χ(X, W )|] ≤ k(12/13) m .By the choice of r(k, γ, ), setting m = rγ/κ , we get that when W is a set of size at least γn, We prove the bound by induction on m.When m = 0, the bound holds trivially.When m > 0, sample W = U ∪ V , where U, V are uniformly random disjoint sets, |U | = u = κ(m − 1)n/r , and It is enough to prove that for all fixed choices of U , The number of possible pairs (V, X) is at least r k • n−u v .Our bound will follow from using Lemma 5. We give a prefix-free encoding of (V, X) as follows: The first case is that for all A, |τ (A, X, V )| ≤ φ(X, V ).In this case, the first bit of the encoding is set to 0, and we proceed to encode (V, X) like this: (a) Encode |χ(X, U )|.It suffices to use a trivial encoding of this integer: we encode it with the string 0 |χ(X,U )| 1, which has length |χ(X, U )| + 1.(b) Encode W ∪ χ(X, U ). Since U has been fixed, there are choices for this set.So, the encoding has length at most (c) Let j be such that χ(j, U ) ⊆ W ∪χ(X, U ), and |χ(j, U )| is minimized.If there are multiple choices for j that achieve the minimum, let j be the smallest one.X is a potential candidate for j, so we must have |χ(j, We have already encoded χ(j, U ) ∩ χ(X, U ) ⊆ S X .We claim that this set must have size at least |χ(X, W )|. Indeed, χ(j, U ) = S h \ U for some set S h of the r-spread sequence.
We have By the definition of χ(X, W ), this implies that as claimed.Now, since |τ (A, X, V )| ≤ φ(X, V ) for all A of size |χ(X, W )|, we can encode X using a binary string of length at most log (e) Because X has been encoded, χ(X, U ) is also determined.Encode W ∩χ(X, U ). Together with W ∪ χ(X, U ), this determines W , and so V .This last step takes |χ(X, U )| bits.
Combining all of the above steps, and using the fact that |χ(X, U )| ≥ 1 and vr/n ≥ κ/2, the total length of the encoding in this case is at most for κ chosen large enough.
2. In the second case there is a set A ⊆ χ(X, U ) of size χ(X, W ) such that |τ (A, X, V )| > φ.
Then the first bit of the encoding is set to 1, and we proceed like this: (a) Encode X.This takes at most log r k + 1 bits.
(b) Now χ(X, U ) is determined.Encode the set A promised above.This takes at most |χ(X, U )| bits.
(c) We claim that at this point, the number of candidates for V is at most n−u v •16 −|χ(X,U )| .Indeed, consider the following random experiment.Choose a set B uniformly at random from the collection of sets satisfying A ⊆ B ⊆ χ(X, U ), and then sample V ⊆ [n] \ U uniformly at random.Consider the collection of y ∈ τ (A, X, V ) for which This is because there are at most r k−|B| sets of size |χ(X, U )| containing B, and for each one, the probability that it is included in Note that this last bound is exactly the same as the bound we obtained in the case that τ (A, X, V ) ≤ φ(X, V ).Now, it only remains to apply Lemma 5.The expected length of the encoding cannot be less than log r k n−u v .This implies that 13 E [|χ(X, W )|] ≤ 12 E [|χ(X, U )|], as required.
there are multiple choices for y that minimize |S y \ W |, let y be the smallest one.