update typo and structures

pages/CSE442T/CSE442T_L7.md

@@ -1,10 +1,12 @@
 # Lecture 7
 
-## Letter choosing experiment
+## Chapter 2: Computational Hardness
+
+### Letter choosing experiment
 
 For 100 letter tiles,
 
-$p_1,...,p_{27}$ (with oe blank)
+$p_1,...,p_{27}$ (with one blank)
 
 $(p_1)^2+\dots +(p_{27})^2\geq\frac{1}{27}$
 
@@ -12,17 +14,17 @@ For any $p_1,...,p_n$, $0\leq p_i\leq 1$.
 
 $\sum p_i=1$
 
-$P[$the same event twice in a row$]=p_1^2+p_2^2....+p_n^2$
+$P[\text{the same event twice in a row}]=p_1^2+p_2^2....+p_n^2$
 
 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 $A$, who random choose $x'$ and output $f(x')=f(x)$ if matched. $P[f(x)=f(x')]\geq\frac{1}{|Y|}$
+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 $P[x\gets f(x);y=f(x):f(a(y,1^n))=y]\geq \frac{1}{|Y|}$
+So $P[x\gets f(x);y=f(x):\mathcal{A}(y,1^n)=y]\geq \frac{1}{|Y|}$
 
-## Modular arithmetic
+### Modular arithmetic
 
 For $a,b\in \mathbb{Z}$, $N\in \mathbb{Z}^2$
 
@@ -30,7 +32,7 @@ $a\equiv b \mod N\iff N|(a-b)\iff \exists k\in \mathbb{Z}, a-b=kN,a=kN+b$
 
 Ex: $N=23$, $-20\equiv 3\equiv 26\equiv 49\equiv 72\mod 23$.
 
-### Equivalent relations for any $N$ on $\mathbb{Z}$
+#### Equivalent relations for any $N$ on $\mathbb{Z}$
 
 $a\equiv a\mod N$
 
@@ -38,7 +40,7 @@ $a\equiv b\mod N\iff b\equiv a\mod N$
 
 $a\equiv b\mod N$ and $b\equiv c\mod N\implies a\equiv c\mod N$
 
-### Division Theorem
+#### Division Theorem
 
 For any $a\in \mathbb{Z}$, and $N\in\mathbb{Z}^+$, $\exists unique\ r,0\leq r<N$.
 
@@ -52,7 +54,7 @@ Definition: $gcd(a,b)=d,a,b\in \mathbb{Z}^+$, is the maximum number such that $d
 
 Using normal factoring is slow... (Example: large $p,q,r$, $N=p\cdot q,,M=p\cdot r$)
 
-#### Euclidean algorithm.
+##### Euclidean algorithm
 
 Recursively relying on fact that $(a>b>0)$
 
@@ -81,4 +83,38 @@ 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)
+$T(n)=2\Theta(\log b)=O(\log n)$ (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 $ax+by=d$
+
+```python
+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 $x=-5,y=18$