Lecture 15 - 2025 / 6 / 3

$EQ(x, y) = [x = y]$, $x, y \in \{0, 1\}^*$

1. randomized algorithm
2. high probability correctness

Prime Number Theorem (PNT): $\text{\#primes} \le m \sim \dfrac{m}{\ln m}$

Protocol:

Alice has $x \in \{0, 1\}^n$ i.e. $x = x_0x_1\cdots x_{n-1}$, and Bob has $y \in \{0, 1\}^n$ i.e. $y = y_0y_1\cdots y_{n-1}$. They want to comute $EQ(x, y)$.

1. Choose a prime $p \in [n^{100}, n^{101}]$
2. Alice: randomly choose an integer $t \in \{0, 1, \cdots, p - 1\}$
3. Alice: Compute the polynomial $f(t) = (x_{n-1} t^{n-1} + x_{n-2} t^{n-2} + \cdots + x_0) \bmod p$
4. Alice: Send $t$ and $f(t)$ to Bob (using $O(\log p) = O(\log n)$ bits)
5. Bob: Receiving $t$ and $f(t)$, compute $g(t) = (y_{n-1} t^{n-1} + y_{n-2} t^{n-2} + \cdots + y_0) \bmod p$
6. Bob: Compare $f(t)$ and $g(t)$, if they are equal, output $1$, otherwise output $0$.

Error occurs when $f \ne g$ but $f(t) = g(t)$, i.e. $t$ is a root of $f - g$. The probability that $t\in \{0, 1, \cdots, p - 1\}$ is a root of $f - g$ is at most $\dfrac{n-1}{p} = O(n^{-99})$.

Max Entropy

$X$ is a $d$-dim random vector, $E(X) = 0$, $Cov(X) = E(X X^T) = \Sigma$. Max entropy distribution is the multivariate Gaussian distribution $\mathcal N(0, \Sigma)$.

Theorem. Let $X$ be a $d$-dim random vector, $E(X) = 0$, $Cov(X) = \Sigma$. Let $Y\sim \mathcal N(0, \Sigma)$. Then $h(X) \le h(Y)$.

$$\begin{aligned} D(X \| Y) & = \int f_X(t) \ln \frac{f_X(t)}{f_Y(t)} \text d t \\ & = - h(X) - \int f_X(t) \ln f_Y(t) \text d t \\ & = - h(X) - \ln \frac{1}{(2\pi)^{d/2}}+ \int f_X(t) \frac{t^T \Sigma^{-1} t}{2} \text d t \\ & = - h(X) - \ln \frac{1}{(2\pi)^{d/2}} +\int f_Y(t) \frac{t^T \Sigma^{-1} t}{2} \text d t \\ & = - h(X) + h(Y) \ge 0 \end{aligned}$$