NoteNextra-origin/content/CSE442T/CSE442T_L7.md

Lecture 7

Chapter 2: Computational Hardness

Letter choosing experiment

For 100 letter tiles,

p_1,...,p_{27} (with one blank)

(p_1)^2+\dots +(p_{27})^2\geq\frac{1}{27}

For any p_1,...,p_n, 0\leq p_i\leq 1.

\sum p_i=1

P[\text{the same event twice in a row}]=p_1^2+p_2^2....+p_n^2

By Cauchy-Schwarz: |u\cdot v|^2 \leq ||u||\cdot ||v||^2.

let \vec{u}=(p_1,...,p_n), \vec{v}=(1,..,1), so (p_1^2+p_2^2....+p_n)^2\leq (p_1^2+p_2^2....+p_n^2)\cdot n. So p_1^2+p_2^2....+p_n^2\geq \frac{1}{n}

So for an adversary \mathcal{A}, who random choose x' and output f(x')=f(x) if matched. P[f(x)=f(x')]\geq\frac{1}{|Y|}

So P[x\gets f(x);y=f(x):\mathcal{A}(y,1^n)=y]\geq \frac{1}{|Y|}

Modular arithmetic

For a,b\in \mathbb{Z}, N\in \mathbb{Z}^2

a\equiv b \mod N\iff N|(a-b)\iff \exists k\in \mathbb{Z}, a-b=kN,a=kN+b

Ex: N=23, -20\equiv 3\equiv 26\equiv 49\equiv 72\mod 23.

Equivalent relations for any N on \mathbb{Z}

a\equiv a\mod N

a\equiv b\mod N\iff b\equiv a\mod N

a\equiv b\mod N and b\equiv c\mod N\implies a\equiv c\mod N

Division Theorem

For any a\in \mathbb{Z}, and N\in\mathbb{Z}^+, \exists unique\ r,0\leq r<N.

\mathbb{Z}_N=\{0,1,2,...,N-1\} with modular arithmetic.

a+b\mod N,a\cdot b\mod N

Theorem: If a\equiv b\mod N and$c\equiv d\mod N$, then a\cdot c\equiv b\cdot d\mod N.

Definition: gcd(a,b)=d,a,b\in \mathbb{Z}^+, is the maximum number such that d|a and d|b.

Using normal factoring is slow... (Example: large p,q,r, N=p\cdot q,,M=p\cdot r)

Euclidean algorithm

Recursively relying on fact that (a>b>0)

gcd(a,b)=gcd(b,a\mod b)

def euclidean_algorithm(a,b):
    if a<b: return euclidean_algorithm(b,a)
    if b==0: return a
    return euclidean_algorithm(b,a%b)

Proof:

We'll show d|a and d|b\iff d|b and d|(a\mod b)

\impliedby a=q\cdot b+r, r=a\mod b

\implies d|r, r=a\mod b

Runtime analysis:

Fact: b_{i+2}<\frac{1}{2}b_i

Proof:

Since a_i=q_i\cdot b_i+b_{i+1}, and b_1=q_2\cdot b_2+b_3, b_2>b_3, and q_2 in worst case is 1, so b_3<\frac{b_1}{2}

T(n)=2\Theta(\log b)=O(\log n) (linear in size of bits input)

Extended Euclidean algorithm

Our goal is to find x,y such that ax+by=gcd(a,b)

Given a\cdot x\equiv b\mod N, we do euclidean algorithm to find gcd(a,b)=d, then reverse the steps to find x,y such that ax+by=d

def extended_euclidean_algorithm(a,b):
    if a%b==0: return (0,1)
    x,y=extended_euclidean_algorithm(b,a%b)
    return (y,x-y*(a//b))

Example: a=12,b=43, gcd(12,43)=1

\begin{aligned}
43&=3\cdot 12+7\\
12&=1\cdot 7+5\\
7&=1\cdot 5+2\\
5&=2\cdot 2+1\\
2&=2\cdot 1+0\\
1&=1\cdot 5-2\cdot 2\\
1&=1\cdot 5-2\cdot (7-1\cdot 5)\\
1&=3\cdot 5-2\cdot 7\\
1&=3\cdot (12-1\cdot 7)-2\cdot 7\\
1&=3\cdot 12-5\cdot 7\\
1&=3\cdot 12-5\cdot (43-3\cdot 12)\\
1&=-5\cdot 43+18\cdot 12\\
\end{aligned}

So x=-5,y=18