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pages/CSE442T/CSE442T_L7.md

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+# Lecture 7
+
+## Letter choosing experiment
+
+For 100 letter tiles,
+
+$p_1,...,p_{27}$ (with oe blank)
+
+$(p_1)^2+\dots +(p_{27})^2\geq\frac{1}{27}$
+
+For any $p_1,...,p_n$, $0\leq p_i\leq 1$.
+
+$\sum p_i=1$
+
+$P[$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 $A$, who random choose $x'$ and output $f(x')=f(x)$ if matched. $P[f(x)=f(x')]\geq\frac{1}{|Y|}$
+
+So $P[x\gets f(x);y=f(x):f(a(y,1^n))=y]\geq \frac{1}{|Y|}$
+
+## 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 $(a>b>0)$
+
+$gcd(a,b)=gcd(b,a\mod b)$
+
+```python
+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)