CS 388C: COMBINATORICS AND GRAPH THEORY
Lecture 18
Scribes: Qiming Yuan, Lu Guo
March 27
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Topics Covered
• Sunflowers
• Modifications
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Sunflowers
Def: A family of sets S1 , S2 , ..., Sk , each of size |Si | = s and for all i, Si ⊆ X is called a sunflower
with k ”petals” and ”core” Y if for all i 6= j Si ∩ Sj = Y and
1. for all i, Y ⊆ Si and Si \Y 6= ∅.
2. The sets Si \Y are nonempty and pairwise disjoint.
Theorem 1 (Sunflower Lemma (Erdős, Rado 1960)). Let F be an s-uniform family of sets (each
of size s). If |F| > s!(k − 1)s then F contains a sunflower with k petals.
Proof. Do induction on s. For base case s = 1, assume |F| > k − 1 which means F contains more
than k − 1 sets. Since we have s = 1 which means F contains at least k singleton sets with no
pairwise intersection. Hence we have Y = ∅ and Si \Y = Si has no pairwise intersection. Thus we
have got a sunflower with k petals and Y = ∅.
Inductive step: Let s ≥ 2 and assume lemma holds for s − 1. Let Ω = {A1 , A2 , ..., At } be a
MAXIMAL family of pairwise disjoint sets from F. Thus we have for all i, j s.t. 1 ≤ i, j ≤ t and
i 6= j, Ai ∩ Aj = ∅. Consider the two cases below:
• If t ≥ k we got a sunflower with Si = Ai and Y = ∅.
• t ≤ k − 1.
The first case is obviously true. For the second case, since t is maximal, we have B = A1 ∪ A2 ∪
. . . ∪ At is a blocking set of F. As we have |Ai | = s, we get |B| ≤ (k − 1)Ai = s(k − 1) By P.P, we
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have there exists x that is included in at least |F|/s(k − 1) sets. Since we have |F| > s!(k − 1)s ,
we get
|F|
s(k − 1)
s!(k − 1)s
>
s(k − 1)
= (s − 1)!(k − 1)s−1
∃x in ≥
Consider Fˆx = {S\x|S ∈ F and x ∈ S}. Each set has size s − 1 and |Fˆx | > (s − 1)!(k − 1)s−1 . By
I.H, we have Fˆx contains a sunflower with k petals and core Y . Putting back x we get sunflower
with k petals and core Y ∪ {x}. QED.
Note: Need more than (k − 1)s sets to ”guarantee” the existence of a sunflower with k petals.
There exists family of (k − 1)s sets that contains no sunflower with k petals. An example is |X|
has s(k − 1) elements and partitioned into s disjoint groups with each group of size k − 1. Let
X = A1 ∪ A2 ∪ . . . ∪ As and F be the family of all s-subsets of X obtained by selecting one element
from each Ai . Thus we have |F| = (k − 1)s . For each k-subset of F, since |Ai | = k − 1 by P.P,
for all i there must exist two subsets in F that intersect at one element of Ai . Thus we have Y
should contain at least one element of each Ai which implies |Y | ≥ s. So Y cannot be core of the
sunflower of s-subsets because it violates the first constraint.
Conjecture: For any fixed k, there is a Ck such that |F| > (Ck )s impliesthe existence of a
sunflower with k petals. The conjecture is open even for k = 3. Note that for k = 3 the sunflower
lemma requires
|F| > s!2s ∼ ss
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Modifications
Weak ∆-system: A weak ∆-system is a family F with sets of size s s.t. for all i 6= j Si ∩ Sj = λ
for some λ.
Theorem 2 (Deza). Let F be an s-uniform weak ∆-system. If |F| ≥ s2 − s + 2, then F is a
sunflower.
Note: For projective plane of order s − 1, the number of lines is s2 − s + 1. This shows that Deza’s
theorem is tight.
Relaxed core: In a sunflower with relaxed core, the relaxed core called common part is defined
as Y = ∪i,j (Si ∩ Sj ) and it is required that the sets Si \Y are nonempty and pairwise disjoint.
Note: If |Y | < mini |Si | then the sets Si \Y will be pairwise disjoint.
Theorem 3 (Füredi). Let s := maxi |Si |. If |F| > (k − 1)s then the common part of some k
members of F is smaller than s.
Flowers: A s-uniform familty of F is ”flower” with k petals and core Y if Y ⊆ Si for Si ∈ F and
τ (FY ) ≥ k where
FY := {S\Y|S ∈ F}
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Theorem 4. If F is an s-uniform family and |F| > (k − 1)s then F contains a flower with k
petals.
Proof. Induction of s. Base case s = 1. Same as the proof above. True.
Induction step: Assume it holds for s − 1 and |F| > (k − 1)s . Consider the two cases below:
• τ (F) ≥ k. Then F itself is a flower with empty core Y and least (k − 1)s + 1 ≥ k petals.
• τ (F) ≤ k − 1. Then there exists a blocking set of size k − 1. By P.P, there exists x which is
included in at least F/(k − 1) members of F. By the same argument as above, let
Fx := {S\{x} : S ∈ F, x ∈ S}
We have
|F|
> (k − 1)s−1
k−1
By I.H, Fx has a flower with k petals and core Y . By putting x back, we get a flower with
k petals and core Y ∪ {x}. QED.
|Fx | ≥
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