# CS2955-Prob&Comp-Notes

## Intro

### Why Randomness?

• Simplicity
• Efficiency
• Better in adversarial situation: 由于随机性很难造数据卡掉

### Toy Example

array $$A: A[i]=0$$ for $$\frac{n}{3}$$ indices $$i$$. Output some $$i$$ s.t. $$A[i]=0$$.

• Deterministic: check $$\frac{2n}{3}-1$$ positions. Runtime $$\Omega(n)$$.
• Randomized: choose a uniformly random $$i$$. $$Pr\{A[i]=0\} = \frac{1}{3}$$. Repeat 10 times.
• Failure Prob $$(\frac{1}{3})^{10}$$ Runtime $$O(1)$$.

### Monte-Carlo & Las Vegas

• Monte-Carlo: finite time, but the output could be WRONG.
• Las Vegas: output is always CORRECT. (Runtime could be infinite, but the $$\mathbb{E}(\text{runtime})$$ must be finite).

$$ZPP \subseteq BPP$$ (Las Vegas 无限长运行时间，可以设一个 Threshold 截断. )

e.g. 一个 Las Vegas 算法有期望运行时间 $$\mathbb{E}[T]$$, 设置一个 Threshold $$T_0 = 3\mathbb{E}[T]$$ ，那么有
$$Pr(T > T_0) \le \frac{1}{3}$$

## Polynomial Identity Testing (P.I.T)

### Problem Description

Alice has
$$f(x) = \sum_{i=0}^{n-1} a_i x^i$$
Bob has
$$g(x) = \sum_{i=0}^{n-1} b_i x^i$$
Alice send sth to Bob, and Bob answer the question "whether f(x) and g(x) are identical".

#### A Simple Deterministic Algorithm

Alice send all the coefficients $$a_i$$ to Bob.

Assumption: $$a_i$$, $$b_i$$ are of size poly($$n$$), so we can use $$\log(n)$$ bits to represent them. Totally it takes

$$O(n \log n)$$.

#### Randomized Algorithm

• Idea: Select a random $$r$$. check if $$f(r) = g(r)$$.
• Analysis:
• Depends on $$r$$ (trade off): large range of $$r$$ $$\Longrightarrow$$ smaller failure prob; large communication complexity.
• If $$f(x)=g(x)$$, this algorithm is always correct.
• If $$f(x) \neq g(x)$$, the algorithm fails only when $$r$$ is a root of $$f(x)-g(x)$$.$$f(x)-g(x)$$ has at most $$n$$ roots hence if we choose $$r$$ from $$\{1,2, \dots, 100n\}$$, the failure prob is $$\frac{1}{100}$$.
• Communication Complexity:
• $$r$$ : $$O(\log n)$$
• $$f(r)$$: $$O(n \log n)$$

#### Fingerprinting

Choose a random prime $$p$$, if $$a \equiv b \mod p$$ then there is a good chance $$a = b$$.

Fingerprints: $$a\ \text{mod}\ p$$, $$b\ \text{mod}\ p$$

#### New Algorithm (Random with Fingerprinting)

Check if
$$f(r) \equiv g(r) \mod p$$
Only need to pass $$f(r)\ \text{mod}\ p$$.

• Analysis: Failure probability comes from
• Case1: $$r$$ is a root of $$f(x)-g(x)$$
• Case 2: $$p$$ is a factor of $$f(r)-g(r)$$
• $$f(r)-g(r)$$ is a number with size $$n^n$$, has $$\le n \log n$$ prime factors.
• There are $$O(\frac{n^2}{\ln n})$$ primes $$\le n^2$$.Then choose a prime $$p \le n^2$$.
$$Pr{p \mid f(r)-g(r)} = \frac{n \ln n}{\frac{n^2}{\ln n}} = O(\frac{\ln^2 n}{n})$$
• Facts used in Analysis:
• (Prime Number Thm) 设 $$\pi(x)$$ 为不超过 $$x$$ 的素数个数,
$$\pi(x) \sim \frac{x}{\ln x}$$
• $$N$$ has $$\le \ln N$$ prime factors. 考虑前 $$m$$ 个质数连乘, 数量级为
$$e^{(1+o(1))m \ln m} = \#m$$
$$m \sim \frac{\ln N}{\ln m} \le \ln N$$.
• 上面的部分简单推导 设 $$\theta(x) := \sum_{k=1}^{\pi(x)} \ln p_k$$ (Chebyshev Function), 素数定理等价于
$$\theta(n) \sim n$$
又有
$$p_m \sim m \ln m$$
那么 $$\theta(p_m) \sim p_m$$, $$\ln (\#m) \sim p_m \sim m \ln m$$
• Communication Complexity
• $$r$$ : $$O(\log n)$$
• $$p$$: $$O(\log n)$$
• $$f(r)\ \text{mod}\ p$$: $$O(\log n)$$

Repeat $$O(\log (\frac{1}{\delta}))$$ times to achieve suc prob $$\ge 1-\delta$$.

### Multiple Variables

Input: $$f, g \in \mathbb{F}[x_1, x_2, \dots, x_n]$$ of total degree $$d$$.
$$f(x_1, \dots, x_n) = \sum_{i_1, \dots, i_n \ge 0,\ i_1+ \dots+i_n \le d} a_{i_1, \dots, i_n} x_1^{i_1} x_2^{i_2} \dots x_n^{i_n}$$
Output: $$f$$ and $$g$$ are identical?

### Schwartz-Zippel Lemma

Let $$P \in \mathbb{F}[x_1, x_2, \dots, x_n$$, $$P \neq 0$$, total degree $$d$$ over field $$\mathbb{F}$$. Let $$S$$ be a finite subset of $$\mathbb{F}$$ and let $$r_1, r_2, \dots, r_n$$ be selected at random independently and uniformly from $$S$$. Then
$$Pr[P(r_1, r_2, \dots, r_n) = 0] \le \frac{d}{|S|}$$

### Application: Bipartite Matching

check whether G has a perfect matching (algebra method)

$$A: n \times n$$ Tutte matrix:
$$A_{ij} = \begin{cases} x_{ij} & (i,j) \in E \\ 0 & (i, j) \not\in E \end{cases}$$
Claim: $$\text{det} A = 0$$ iff G has no perfect matching. 因为把行列式写开（对排列 $$\pi$$ sigma），每个排列描述了一个完美匹配。

### Finding a Perfect Matching

#### Isolation Lemma

Let $$S$$ be a family of subsets on a ground set $$U$$ with $$|U|=n$$. For each $$x \in U$$, assign a weight

$$w(x)$$ from $$\{1, 2, \dots, 100n\}$$ uniformly.

Then there exists a unique minimum weight subset in $$S$$ w.p. $$\ge 0.99$$.

#### Parallel Alg

$$\text{det}(A') \frac{2^{w_{ij}}}{2^W}$$

TODO

TODO

TODO

TODO