121 lines
2.9 KiB
Markdown
121 lines
2.9 KiB
Markdown
# CSE442T Introduction to Cryptography (Lecture 7)
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## Chapter 2: Computational Hardness
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### Letter choosing experiment
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For 100 letter tiles,
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$p_1,...,p_{27}$ (with one blank)
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$(p_1)^2+\dots +(p_{27})^2\geq\frac{1}{27}$
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For any $p_1,...,p_n$, $0\leq p_i\leq 1$.
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$\sum p_i=1$
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$P[\text{the same event twice in a row}]=p_1^2+p_2^2....+p_n^2$
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By Cauchy-Schwarz: $|u\cdot v|^2 \leq ||u||\cdot ||v||^2$.
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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 $p_1^2+p_2^2....+p_n^2\geq \frac{1}{n}$
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So for an adversary $\mathcal{A}$, who random choose $x'$ and output $f(x')=f(x)$ if matched. $P[f(x)=f(x')]\geq\frac{1}{|Y|}$
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So $P[x\gets f(x);y=f(x):\mathcal{A}(y,1^n)=y]\geq \frac{1}{|Y|}$
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### Modular arithmetic
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For $a,b\in \mathbb{Z}$, $N\in \mathbb{Z}^2$
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$a\equiv b \mod N\iff N|(a-b)\iff \exists k\in \mathbb{Z}, a-b=kN,a=kN+b$
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Ex: $N=23$, $-20\equiv 3\equiv 26\equiv 49\equiv 72\mod 23$.
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#### Equivalent relations for any $N$ on $\mathbb{Z}$
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$a\equiv a\mod N$
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$a\equiv b\mod N\iff b\equiv a\mod N$
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$a\equiv b\mod N$ and $b\equiv c\mod N\implies a\equiv c\mod N$
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#### Division Theorem
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For any $a\in \mathbb{Z}$, and $N\in\mathbb{Z}^+$, $\exists unique\ r,0\leq r<N$.
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$\mathbb{Z}_N=\{0,1,2,...,N-1\}$ with modular arithmetic.
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$a+b\mod N,a\cdot b\mod N$
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Theorem: If $a\equiv b\mod N$ and$c\equiv d\mod N$, then $a\cdot c\equiv b\cdot d\mod N$.
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Definition: $gcd(a,b)=d,a,b\in \mathbb{Z}^+$, is the maximum number such that $d|a$ and $d|b$.
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Using normal factoring is slow... (Example: large $p,q,r$, $N=p\cdot q,,M=p\cdot r$)
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##### Euclidean algorithm
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Recursively relying on fact that $(a>b>0)$
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$gcd(a,b)=gcd(b,a\mod b)$
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```python
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def euclidean_algorithm(a,b):
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if a<b: return euclidean_algorithm(b,a)
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if b==0: return a
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return euclidean_algorithm(b,a%b)
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```
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Proof:
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We'll show $d|a$ and $d|b\iff d|b$ and $d|(a\mod b)$
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$\impliedby$ $a=q\cdot b+r$, $r=a\mod b$
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$\implies$ $d|r$, $r=a\mod b$
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Runtime analysis:
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Fact: $b_{i+2}<\frac{1}{2}b_i$
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Proof:
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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 $b_3<\frac{b_1}{2}$
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$T(n)=2\Theta(\log b)=O(\log n)$ (linear in size of bits input)
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##### Extended Euclidean algorithm
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Our goal is to find $x,y$ such that $ax+by=gcd(a,b)$
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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 $ax+by=d$
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```python
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def extended_euclidean_algorithm(a,b):
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if a%b==0: return (0,1)
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x,y=extended_euclidean_algorithm(b,a%b)
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return (y,x-y*(a//b))
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```
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Example: $a=12,b=43$, $gcd(12,43)=1$
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$$
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\begin{aligned}
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43&=3\cdot 12+7\\
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12&=1\cdot 7+5\\
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7&=1\cdot 5+2\\
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5&=2\cdot 2+1\\
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2&=2\cdot 1+0\\
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1&=1\cdot 5-2\cdot 2\\
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1&=1\cdot 5-2\cdot (7-1\cdot 5)\\
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1&=3\cdot 5-2\cdot 7\\
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1&=3\cdot (12-1\cdot 7)-2\cdot 7\\
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1&=3\cdot 12-5\cdot 7\\
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1&=3\cdot 12-5\cdot (43-3\cdot 12)\\
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1&=-5\cdot 43+18\cdot 12\\
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\end{aligned}
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$$
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So $x=-5,y=18$
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