组合数学 (Spring 2015)/Extremal set theory

Sunflowers

An set system is a sunflower if all its member sets intersect at the same set of elements.

Definition (sunflower)

A set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] is a sunflower of size [math]\displaystyle{ r }[/math] with a core [math]\displaystyle{ C\subseteq X }[/math] if [math]\displaystyle{ \forall S,T\in\mathcal{F} }[/math] that [math]\displaystyle{ S\neq T }[/math], [math]\displaystyle{ S\cap T=C }[/math].

Note that we do not require the core to be nonempty, thus a family of disjoint sets is also a sunflower (with the core [math]\displaystyle{ \emptyset }[/math]).

The next result due to Erdős and Rado, called the sunflower lemma, is a famous result in extremal set theory, and has some important applications in Boolean circuit complexity.

Sunflower Lemma (Erdős-Rado)

Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math]. If [math]\displaystyle{ |\mathcal{F}|\gt k!(r-1)^k }[/math], then [math]\displaystyle{ \mathcal{F} }[/math] contains a sunflower of size [math]\displaystyle{ r }[/math].

Proof. We proceed by induction on [math]\displaystyle{ k }[/math]. For [math]\displaystyle{ k=1 }[/math], [math]\displaystyle{ \mathcal{F}\subseteq{X\choose 1} }[/math], thus all sets in [math]\displaystyle{ \mathcal{F} }[/math] are disjoint. And since [math]\displaystyle{ |\mathcal{F}|\gt r-1 }[/math], we can choose [math]\displaystyle{ r }[/math] of these sets and form a sunflower. Now let [math]\displaystyle{ k\ge 2 }[/math] and assume the lemma holds for all smaller [math]\displaystyle{ k }[/math]. Take a maximal family [math]\displaystyle{ \mathcal{G}\subseteq \mathcal{F} }[/math] whose members are disjoint, i.e. for any [math]\displaystyle{ S,T\in \mathcal{G} }[/math] that [math]\displaystyle{ S\neq T }[/math], [math]\displaystyle{ S\cap T=\emptyset }[/math]. If [math]\displaystyle{ |\mathcal{G}|\ge r }[/math], then [math]\displaystyle{ \mathcal{G} }[/math] is a sunflower of size at least [math]\displaystyle{ r }[/math] and we are done. Assume that [math]\displaystyle{ |\mathcal{G}|\le r-1 }[/math], and let [math]\displaystyle{ Y=\bigcup_{S\in\mathcal{G}}S }[/math]. Then [math]\displaystyle{ |Y|=k|\mathcal{G}|\le k(r-1) }[/math] (since all members of [math]\displaystyle{ \mathcal{G} }[/math]) are disjoint). We claim that [math]\displaystyle{ Y }[/math] intersets all members of [math]\displaystyle{ \mathcal{F} }[/math], since if otherwise, there exists an [math]\displaystyle{ S\in\mathcal{F} }[/math] such that [math]\displaystyle{ S\cap Y=\emptyset }[/math], then we can enlarge [math]\displaystyle{ \mathcal{G} }[/math] by adding [math]\displaystyle{ S }[/math] into [math]\displaystyle{ \mathcal{G} }[/math] and still have all members of [math]\displaystyle{ \mathcal{G} }[/math] disjoint, which contradicts the assumption that [math]\displaystyle{ \mathcal{G} }[/math] is the maximum of such families. By the pigeonhole principle, some elements [math]\displaystyle{ y\in Y }[/math] must contained in at least [math]\displaystyle{ \frac{|\mathcal{F}|}{|Y|}\gt \frac{k!(r-1)^k}{k(r-1)}=(k-1)!(r-1)^{k-1} }[/math] members of [math]\displaystyle{ \mathcal{F} }[/math]. We delete this [math]\displaystyle{ y }[/math] from these sets and consider the family [math]\displaystyle{ \mathcal{H}=\{S\setminus\{y\}\mid S\in\mathcal{F}\wedge y\in S\} }[/math]. We have [math]\displaystyle{ \mathcal{H}\subseteq {X\choose k-1} }[/math] and [math]\displaystyle{ |\mathcal{H}|\gt (k-1)!(r-1)^{k-1} }[/math], thus by the induction hypothesis, [math]\displaystyle{ \mathcal{H} }[/math]contains a sunflower of size [math]\displaystyle{ r }[/math]. Adding [math]\displaystyle{ y }[/math] to the members of this sunflower, we get the desired sunflower in the original family [math]\displaystyle{ \mathcal{F} }[/math]. [math]\displaystyle{ \square }[/math]

The Erdős–Ko–Rado Theorem

A set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] is called intersecting, if for any [math]\displaystyle{ S,T\in\mathcal{F} }[/math], [math]\displaystyle{ S\cap T\neq\emptyset }[/math]. A natural question of extremal favor is: "how large can an intersecting family be?"

Assume [math]\displaystyle{ |X|=n }[/math]. When [math]\displaystyle{ n\lt 2k }[/math], every pair of [math]\displaystyle{ k }[/math]-subsets of [math]\displaystyle{ X }[/math] intersects. So the non-trivial case is when [math]\displaystyle{ n\le 2k }[/math]. The famous Erdős–Ko–Rado theorem gives the largest possible cardinality of a nontrivially intersecting family.

According to Erdős, the theorem itself was proved in 1938, but was not published until 23 years later.

Erdős–Ko–Rado theorem (proved in 1938, published in 1961)

Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math] where [math]\displaystyle{ |X|=n }[/math] and [math]\displaystyle{ n\ge 2k }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then [math]\displaystyle{ |\mathcal{F}|\le{n-1\choose k-1} }[/math].

Katona's proof

We first introduce a proof discovered by Katona in 1972. The proof uses double counting.

Let [math]\displaystyle{ \pi }[/math] be a cyclic permutation of [math]\displaystyle{ X }[/math], that is, we think of assigning [math]\displaystyle{ X }[/math] in a circle and ignore the rotations of the circle. It is easy to see that there are [math]\displaystyle{ (n-1)! }[/math] cyclic permutations of an [math]\displaystyle{ n }[/math]-set (each cyclic permutation corresponds to [math]\displaystyle{ n }[/math] permutations). Let

[math]\displaystyle{ \mathcal{G}_\pi=\{\{\pi_{(i+j)\bmod n}\mid j\in[k]\}\mid i\in [n]\} }[/math].

The next lemma states the following observation: in a circle of [math]\displaystyle{ n }[/math] points, supposed [math]\displaystyle{ n\ge 2k }[/math], there can be at most [math]\displaystyle{ k }[/math] arcs, each consisting of [math]\displaystyle{ k }[/math] points, such that every pair of arcs share at least one point.

Lemma

Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math] where [math]\displaystyle{ |X|=n }[/math] and [math]\displaystyle{ n\ge 2k }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then for any cyclic permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ X }[/math], it holds that [math]\displaystyle{ |\mathcal{G}_\pi\cap\mathcal{F}|\le k }[/math].

Proof. Fix a cyclic permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ X }[/math]. Let [math]\displaystyle{ A_i=\{\pi_{(i+j+n)\bmod n}\mid j\in[k]\} }[/math]. Then [math]\displaystyle{ \mathcal{G}_\pi }[/math] can be written as [math]\displaystyle{ \mathcal{G}_\pi=\{A_i\mid i\in [n]\} }[/math]. Suppose that [math]\displaystyle{ A_t\in\mathcal{F} }[/math]. Since [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, the only sets [math]\displaystyle{ A_i }[/math] that can be in [math]\displaystyle{ \mathcal{F} }[/math] other than [math]\displaystyle{ A_t }[/math] itself are the [math]\displaystyle{ 2k-2 }[/math] sets [math]\displaystyle{ A_i }[/math] with [math]\displaystyle{ t-(k-1)\le i\le t+k-1, i\neq t }[/math]. We partition these sets into [math]\displaystyle{ k-1 }[/math] pairs [math]\displaystyle{ \{A_i,A_{i+k}\} }[/math], where [math]\displaystyle{ t-(k-1)\le i\le t-1 }[/math]. Note that for [math]\displaystyle{ n\ge 2k }[/math], it holds that [math]\displaystyle{ A_i\cap C_{i+k}=\emptyset }[/math]. Since [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, [math]\displaystyle{ \mathcal{F} }[/math] can contain at most one set of each such pair. The lemma follows. [math]\displaystyle{ \square }[/math]

The Katona's proof of Erdős–Ko–Rado theorem is done by counting in two ways the pairs of member [math]\displaystyle{ S }[/math] of [math]\displaystyle{ \mathcal{F} }[/math] and cyclic permutation [math]\displaystyle{ \pi }[/math] which contain [math]\displaystyle{ S }[/math] as a continuous path on the circle (i.e., an arc).

Katona's proof of Erdős–Ko–Rado theorem (double counting) Let [math]\displaystyle{ \mathcal{R}=\{(S,\pi)\mid \pi \text{ is a cyclic permutation of }X, \text{and }S\in\mathcal{F}\cap\mathcal{G}_\pi\} }[/math]. We count [math]\displaystyle{ \mathcal{R} }[/math] in two ways. First, due to the lemma, [math]\displaystyle{ |\mathcal{F}\cap\mathcal{G}_\pi|\le k }[/math] for any cyclic permutation [math]\displaystyle{ \pi }[/math]. There are [math]\displaystyle{ (n-1)! }[/math] cyclic permutations in total. Thus, [math]\displaystyle{ |\mathcal{R}|=\sum_{\text{cyclic }\pi}|\mathcal{F}\cap\mathcal{G}_\pi|\le k(n-1)! }[/math]. Next, for each [math]\displaystyle{ S\in\mathcal{F} }[/math], the number of cyclic permutations [math]\displaystyle{ \pi }[/math] in which [math]\displaystyle{ S }[/math] is continuous is [math]\displaystyle{ |S|!(n-|S|)!=k!(n-k)! }[/math]. Thus, [math]\displaystyle{ |\mathcal{R}|=\sum_{S\in\mathcal{F}}k!(n-k)!=|\mathcal{F}|k!(n-k)! }[/math]. Altogether, we have [math]\displaystyle{ |\mathcal{F}|\le\frac{k(n-1)!}{k!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k)!}={n-1\choose k-1} }[/math]. [math]\displaystyle{ \square }[/math]

Erdős' shifting technique

We now introduce the original proof of the Erdős–Ko–Rado theorem, which uses a technique called shifting (originally called compression).

Without loss of generality, we assume [math]\displaystyle{ X=[n] }[/math], and restate the Erdős–Ko–Rado theorem as follows.

Erdős–Ko–Rado theorem

Let [math]\displaystyle{ \mathcal{F}\subseteq {[n]\choose k} }[/math] and [math]\displaystyle{ n\ge 2k }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then [math]\displaystyle{ |\mathcal{F}|\le{n-1\choose k-1} }[/math].

We define a shift operator for the set family.

Definition (shift operator)

Assume [math]\displaystyle{ \mathcal{F}\subseteq 2^{[n]} }[/math], and [math]\displaystyle{ 0\le i\lt j\le n-1 }[/math]. Define the [math]\displaystyle{ (i,j) }[/math]-shift [math]\displaystyle{ S_{ij} }[/math] as an operator on [math]\displaystyle{ \mathcal{F} }[/math] as follows: for each [math]\displaystyle{ T\in\mathcal{F} }[/math], write [math]\displaystyle{ T_{ij}=(T\setminus\{j\})\cup\{i\} }[/math], and let [math]\displaystyle{ S_{ij}(T)= \begin{cases} T_{ij} & \mbox{if }j\in T, i\not\in T, \mbox{ and }T_{ij} \not\in\mathcal{F},\\ T & \mbox{otherwise;} \end{cases} }[/math] let [math]\displaystyle{ S_{ij}(\mathcal{F})=\{S_{ij}(T)\mid T\in \mathcal{F}\} }[/math].

It is easy to verify the following propositions of shifts.

Proposition

[math]\displaystyle{ |S_{ij}(T)|=|T|\, }[/math] and [math]\displaystyle{ |S_{ij}(\mathcal{F})|=\mathcal{F} }[/math]; if [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then so is [math]\displaystyle{ S_{ij}(\mathcal{F}) }[/math].

Proof. (1) is quite obvious. Now we prove (2). Consider any [math]\displaystyle{ A,B\in\mathcal{F} }[/math]. If [math]\displaystyle{ A\cap B }[/math] includes any element other than [math]\displaystyle{ j }[/math] then [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] are still intersecting after [math]\displaystyle{ (i,j) }[/math]-shift. Thus without loss of generality, we may consider the only unsafe case where [math]\displaystyle{ A\cap B=\{j\} }[/math], and [math]\displaystyle{ B }[/math] is successfully shifted to [math]\displaystyle{ B_{ij}=(B\setminus\{j\})\cup\{i\} }[/math] but [math]\displaystyle{ A }[/math] fails to shift to [math]\displaystyle{ A_{ij}=(A\setminus\{j\})\cup\{i\} }[/math]. Since [math]\displaystyle{ B }[/math] is successfully shifted to [math]\displaystyle{ B_{ij} }[/math], we know that it must hold that [math]\displaystyle{ i\not\in B }[/math] and [math]\displaystyle{ B_{ij}\not\in\mathcal{F} }[/math]. And the only two reasons for which [math]\displaystyle{ A }[/math] may fail to shift to [math]\displaystyle{ A_{ij} }[/math] are: (1) [math]\displaystyle{ i\in A }[/math] and (2) [math]\displaystyle{ i\not\in A }[/math] but [math]\displaystyle{ A_{ij}\in\mathcal{F} }[/math]. Case.1: [math]\displaystyle{ i\in A }[/math]. In this case, since [math]\displaystyle{ i\in B_{ij} }[/math], we have [math]\displaystyle{ S_{ij}(A)\cap S_{ij}(B)=A\cap B_{ij}=\{i\} }[/math], i.e. [math]\displaystyle{ B }[/math] is still intersecting with [math]\displaystyle{ A }[/math] after shifting. Case.2: [math]\displaystyle{ i\not\in A }[/math] but [math]\displaystyle{ A_{ij}\in\mathcal{F} }[/math]. In this case, it is easy to verify that [math]\displaystyle{ A_{ij}\cap B=(A\cap B)\setminus\{j\}=\emptyset }[/math]. Recall that we assume [math]\displaystyle{ A_{ij}\in\mathcal{F} }[/math]. This contradicts to that [math]\displaystyle{ \mathcal{F} }[/math] is intersecting. In conclusion, in all cases, [math]\displaystyle{ S_{ij}(\mathcal{F}) }[/math] remains intersecting. [math]\displaystyle{ \square }[/math]

Repeatedly applying [math]\displaystyle{ S_{ij}(\mathcal{F}) }[/math] for any [math]\displaystyle{ 0\le i\lt j\le n-1 }[/math], since we only replace elements by smaller elements, eventually [math]\displaystyle{ \mathcal{F} }[/math] will stop changing, that is, [math]\displaystyle{ S_{ij}(\mathcal{F})=\mathcal{F} }[/math] for all [math]\displaystyle{ 0\le i\lt j\le n-1 }[/math]. We call such an [math]\displaystyle{ \mathcal{F} }[/math] shifted.

The idea behind the shifting technique is very natural: by applying shifting, all intersecting families are transformed to some special forms, and we only need to prove the theorem for these special form of intersecting families.

Proof of Erdős-Ko-Rado theorem (The original proof of Erdős-Ko-Rado by shifting) By the above lemma, it is sufficient to prove the Erdős-Ko-Rado theorem holds for shifted [math]\displaystyle{ \mathcal{F} }[/math]. We assume that [math]\displaystyle{ \mathcal{F} }[/math] is shifted. First, it is trivial to see that the theorem holds for [math]\displaystyle{ k=1 }[/math] (no matter whether shifted). Next, we show that the theorem holds when [math]\displaystyle{ n=2k }[/math] (no matter whether shifted). For any [math]\displaystyle{ S\in{X\choose k} }[/math], both [math]\displaystyle{ S }[/math] and [math]\displaystyle{ X\setminus S }[/math] are in [math]\displaystyle{ {X\choose k} }[/math], but at most one of them can be in [math]\displaystyle{ \mathcal{F} }[/math]. Thus, [math]\displaystyle{ |\mathcal{F}|\le\frac{1}{2}{n\choose k}=\frac{n!}{2k!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k)!}={n-1\choose k-1} }[/math]. We then apply the induction on [math]\displaystyle{ n }[/math]. For [math]\displaystyle{ n\gt 2k }[/math], the induction hypothesis is stated as: the Erdős-Ko-Rado theorem holds for any smaller [math]\displaystyle{ n }[/math]. Define [math]\displaystyle{ \mathcal{F}_0=\{S\in\mathcal{F}\mid n\not\in S\} }[/math] and [math]\displaystyle{ \mathcal{F}_1=\{S\in\mathcal{F}\mid n\in S\} }[/math]. Clearly, [math]\displaystyle{ \mathcal{F}_0\subseteq{[n-1]\choose k} }[/math] and [math]\displaystyle{ \mathcal{F}_0 }[/math] is intersecting. Due to the induction hypothesis, [math]\displaystyle{ |\mathcal{F}_0|\le{n-2\choose k-1} }[/math]. In order to apply the induction, we let [math]\displaystyle{ \mathcal{F}_1'=\{S\setminus\{n\}\mid S\in\mathcal{F}_1\} }[/math]. Clearly, [math]\displaystyle{ \mathcal{F}_1'\subseteq{[n-1]\choose k-1} }[/math]. If only it is also intersecting, we can apply the induction hypothesis, and indeed it is. To see this, by contradiction we assume that [math]\displaystyle{ \mathcal{F}_1' }[/math] is not intersecting. Then there must exist [math]\displaystyle{ A,B\in\mathcal{F} }[/math] such that [math]\displaystyle{ A\cap B=\{n\} }[/math], which means that [math]\displaystyle{ |A\cup B|\le 2k-1\lt n-1 }[/math]. Thus, there is some [math]\displaystyle{ 0\le i\le n-1 }[/math] such that [math]\displaystyle{ i\not\in A\cup B }[/math]. Since [math]\displaystyle{ \mathcal{F} }[/math] is shifted, [math]\displaystyle{ A_{in}=A\setminus\{n\}\cup\{i\}\in\mathcal{F} }[/math]. On the other hand it can be verified that [math]\displaystyle{ A_{in}\cap B=\emptyset }[/math], which contradicts that [math]\displaystyle{ \mathcal{F} }[/math] is intersecting. Thus, [math]\displaystyle{ \mathcal{F}_1'\subseteq{[n-1]\choose k-1} }[/math] and [math]\displaystyle{ \mathcal{F}_1' }[/math] is intersecting. Due to the induction hypothesis, [math]\displaystyle{ |\mathcal{F}_1'|\le{n-2\choose k-2} }[/math]. Combining these together, [math]\displaystyle{ |\mathcal{F}|=|\mathcal{F}_0|+|\mathcal{F}_1|=|\mathcal{F}_0|+|\mathcal{F}_1'|\le {n-2\choose k-1}+{n-2\choose k-2}={n-1\choose k-1} }[/math]. [math]\displaystyle{ \square }[/math]