图论与组合数学证明技巧-优快云博客

本文链接：https://blog.youkuaiyun.com/jeremy43/article/details/53390896

本文通过概率方法和Cauchy-Schwarz不等式证明了独立集定理，并探讨了太阳花引理的应用，同时利用反证法证明了集合族的相关性质。

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1.
(1)Show that this theorem is a corollary to the Turan theorem for cliques.
According to the third proof (the probabilistic method),we can change the Claim:

ω (G) \geq \sum v \in V 1 n - d v .

$\omega(G)\ge\sum_{v\in V}\frac{1}{n-d_v}.$
to

σ (G) \geq \sum v \in V 1 1 + d v .

$\sigma(G)\ge\sum_{v\in V}\frac{1}{1+d_v}.$

σ(G) $\sigma(G)$ repensents the number of vertices in an independent set.
So, it’s a corollary to the Turan theorem.

(2)Prove the theorem directly for the independent sets by the probabilistic method along with the Cauchy-Schwartz theorem, without using the Turan theorem.

We prove this by the probabilistic method. Fix a random ordering of vertices in $V,$ say $v_1,v_2,\ldots,v_n.$ We construct an independent set as follows:
$\sigma(G)=\emptyset$
while $G!= \emptyset$
{
$\forall v_i \in G$
$\sigma(G)+=v_i$
$G=G-v_i$ -all neighbour of $v_i$
}
Let $X_v$ be the random variable that indicates whether $v\in \sigma(G),$ i.e.,
$\sigma(G)=S$
Note that a vertex $v\in S$ The probability that v is ranked is $\frac{1}{1+d_v}.$ Thus,

E [X v] = Pr [v \in S] \geq 1 1 + d v .

$\mathbf{E}[X_v]=\Pr[v\in S]\ge\frac{1}{1+d_v}.$
Observe that

|S|=∑v∈VXv. $|S|=\sum_{v\in V}X_v.$ Due to linearity of expectation,

E [| S |] = \sum v \in V E [X v] \geq \sum v \in V 1 1 + d v .

$\mathbf{E}[|S|]=\sum_{v\in V}\mathbf{E}[X_v]\ge\sum_{v\in V}\frac{1}{1+d_v}.$
Apply the Cauchy-Schwarz inequality

(\sum v \in V a v b v) 2 \leq (\sum v \in V n a 2 v) (\sum v \in V n b 2 v) .

$\left(\sum_{v\in V}a_vb_v\right)^2\le\left(\sum_{v\in V}^na_v^2\right)\left(\sum_{v\in V}^nb_v^2\right).$
Set

av=1+dv‾‾‾‾‾‾√ $a_v=\sqrt{1+d_v}$ and

bv=11+dv√, $b_v=\frac{1}{\sqrt{1+d_v}},$ then

avbv=1 $a_vb_v = 1$ and so

n 2 \leq \sum v \in V (1 + d v) \sum v \in V 1 1 + d v \leq σ (G) \sum v \in V (1 + d v) .

$n^2\le\sum_{v\in V}(1+d_v)\sum_{v\in V}\frac{1}{1+d_v}\le\sigma(G)\sum_{v\in V}(1+d_v).$
so we have

|S|≥n22m+n $|S|\ge \frac{n^2}{2m+n}$

2 according to Sunflower Lemma (Erdős-Rado)
Let $\mathcal{F}\subseteq {X\choose k}.$ If $|\mathcal{F}|>k!(r-1)^k,$ then contains a sunflower of size r.
Edge implies $k=2$
because $\mathcal{F}>2(r-1)^2$
$\mathcal{F}$ contains a sunflower of size r
if core $\emptyset$ ,then it’s a match, else it’s a star.

3.
(1)proof by contradiction：
fix $B$ ,suppose $S= \{A\cap B\mid A\in\mathcal{F}, A\neq B\}$ is comparable, and for any $x_i,x_j \in S$ , we have $x_i=A_i\cap B, x_j=A_j\cap B$
suppose $x_i \subset x_j$ , we have $x_i \subset A_j,$ which contradicts with $A\cap B \not\subset C$ for any $A,B,C\in\mathcal{F}.$
so,S is an anti chain.

(2)because the k-uniform ${X\choose k}$ is an antichain
$\mathcal{F}$ is antichain
According to theorem (Sperner)
Let $\mathcal{F}\subseteq 2^X$ where $| X | = n.$ If $\mathcal{F}$ is an antichain, then
$|\mathcal{F}|\le{n\choose \lfloor n/2\rfloor}.$
because n=k, we have $|\mathcal{F}|\le 1+{k\choose \lfloor k/2\rfloor}.$

4.
Consider that
for every item $c$ in $\mathcal{F}，\mathcal{F_1}=\mathcal{F_1} \cup \{|X|-c\}$
$|\mathcal{F_1}|=|\mathcal{F}|\subseteq{[n]\choose n-k}$
we have $t=n-k, n>2t$
since $A\cup B\neq [n]$ for all $A,B\in\mathcal{F}$
$A_1 \cap B_1 \neq \emptyset$ for all $A_1,B_1 \in \mathcal{F_1}$ ,
$\mathcal{F_1}$ is intersecting
according to Erdős–Ko–Rado theorem
Let $\mathcal{F_1}\subseteq {[n]\choose t}$ and $n\ge 2t.$ If $\mathcal{F}$ is intersecting, then $|\mathcal{F_1}|\le{n-1\choose t-1}.$
so we have $|\mathcal{F_1}|=|\mathcal{F}|\le{n-1\choose t-1}$
${n-1\choose t-1}=\left(1-\frac{k}{n}\right){n\choose k}.$
so, $|\mathcal{F}|\le\left(1-\frac{k}{n}\right){n\choose k}$