Tournament
$T(V,E)$: $n$ players, each pair has a match, $u$ points to $v$ if and only if $u$ beats $v$.
$k$-paradoxical: For every $k$-subset $S$ of $V$, there is a player in $V \setminus S$ who beats all players in $S$.
Theorem (Erdős 1963): If ${n \choose k} \left( 1 - 2^{-k} \right)^{n-k} < 1$ then there is a $k$-paradoxical tournament of $n$ players.
Proof: Pick a random tournament $T$ on $n$ players $[n]$. (Random $\rightarrow$ each player beat the other by probability $1 / 2$). For any $S \in {[n] \choose k}$, define event $A_S$ as no player in $V \setminus S$ beat all players in $S$.
$$Pr[u \in S, \text{ beaten}] = 1 - (1 / 2)^k$$
$$Pr[A_S] = \left( 1 - 2^{-k} \right)^{n-k}$$
$$Pr[\text{No $k$-paradoxical}] \leqslant \sum_{S \in {[n] \choose k}} Pr[A_S] < 1.$$
The Probabilistic Method
- Pick random bal from a box, $Pr[\text{is blue}] > 0 \Rightarrow$ there is a blue ball.
- Define a probability space $\Omega$, and a property $P$: $$\underset{x}{Pr}[P(x)] > 0$$ $\Rightarrow \exists x \in \Omega$ with the property $P$.
Averaging Principle
- Average height of the students in class is $l$. $\Rightarrow$ there is a student of height $\geqslant l (\leqslant l)$.
- For a random variable $X$,
- $\exists x \leqslant E $, such that $X = x$ is possible,
- $\exists x \geqslant E $, such that $X = x$ is possible.
Hamiltonian Path in Tournament
Hamiltonian path: a path visiting every vertex exactly once.
Theorem (Szele 1943): There is a tournament on $n$ players with at least $n! 2^{-(n-1)}$ Hamiltonian paths.
Proof Pick a random tournament $T$ on $n$ players $[n]$. For every permutation $\pi$ of $[n]$, define $$X_\pi = \begin{cases} 1, & \pi \text{ is a Hamiltonian path}, \\ 0, & \text{otherwise}.\end{cases}$$
$$E[X_\pi] = Pr[X_\pi = 1] = (1 / 2)^{n-1}$$
$$E[X] = \sum_\pi E[X_\pi] = n! 2^{-(n-1)}$$
Large Independent Set
An independent set $S$: no adjacent vertices in $S$.
Max independent set is NP-hard.
A uniform $S$ is very unlikely to be an independent set!
Theorem Graph $G$ has $n$ vertices and $m$ edges, then there exists an independent set $S$ of size $\dfrac{n^2}{4m}$.
Proof
- Sample a random $S$: each vertex is chosen independently with probability $p$
- modify $S$ to $S^*$ (an independent set): $\forall uv \in E$, if $u, v \in S$, then delete one of $u$, $v$ from $S$.
Define $Y$: # of edges in $S$, we have
$$Y = \sum_{uv \in E} Y_{uv}, \ \ \ Y_{uv} = \begin{cases}1, & u, v \in S, \\ 0, &\text{otherwise}.\end{cases}$$
$$E[\vert S \vert] = np, \ \ \ E[Y] = \sum_{uv \in E} E[Y_{uv}] = mp^2$$
$$E[\vert S^* \vert] \geqslant E[\vert S \vert] - E[Y] = np - mp^2$$
Let $p = \dfrac{n}{2m}$, we have $E[\vert S^* \vert] \geqslant \dfrac{n^2}{4m}$.
$\dfrac{n^2}{4m} = \dfrac{n}{2d}$, where $d = \dfrac{2m}{n}$ is the average degree.
Markov’s Inequality
For non-negative $X$, for any $t > 0$, $$Pr[X \geqslant t] \leqslant \dfrac{E[X]}{t}.$$
Proof Define $Y$ as
$$Y = \begin{cases}1, & X \geqslant t, \\ 0, & \text{otherwise}. \end{cases}$$
$$Y \leqslant \lfloor \dfrac{X}{t} \rfloor \leqslant \dfrac{X}{t},$$
$$Pr[X \geqslant t] = E[Y] \leqslant E\left[ \dfrac{X}{t} \right] = \dfrac{E[X]}{t}.$$
Chromatic Number
Graph $G(V,E)$, define
- girth $g(G)$: length of the shortest circle,
- chromatic number $\chi(G)$: mnimum number of color to properly color the vertices of $G$.
Intuition: large cycles are easy to color.
Theorem (Erdős 1959) For all $k$, $l$, there exists a finite graph $G$ with $\chi(G) \geqslant k$ and $g(G) \geqslant l$.
Define independence number $\alpha(G)$: size of the largest independent set in $G$.
$$\text{if } \alpha(G) \leqslant \dfrac{n}{k},$$
$$\text{then } \chi(G) \geqslant \dfrac{n}{\alpha(G)} \geqslant k.$$
Theorem' For all $k$, $l$, there exists a finite graph $G$ with $\alpha(G) \leqslant k$ and $g(G) \geqslant l$.
Define a random graph of $\vert V \vert = n$ vertices $G(n, p)$: forall $u, v \in V$, $Pr[{u, v} \in E] = p$ (independently).
A uniform random graph: $G(n, \dfrac{1}{2})$.
Plan: fix any large $k$, $l$, exists $n$, generate a random graph $G \sim G(n, p)$ and
$$\begin{cases}Pr[\alpha(G) > n / k] < 1 / 2, \\ Pr[g(G) < l] < 1 / 2. \end{cases}$$
Loading Comments By Disqus