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content/CSE442T/CSE442T_L9.md

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+# Lecture 9
+
+## Chapter 2: Computational Hardness
+
+### Continue on Cyclic groups
+
+$$
+\begin{aligned}
+107^{662}\mod 51&=(107\mod 51)^{662}\mod 51\\
+&=5^{662}\mod 51
+\end{aligned}
+$$
+
+Remind that $\phi(p),p\in\Pi,\phi(p)=p-1$.
+
+$51=3\times 17,\phi(51)=\phi(3)\times \phi(17)=2\times 16=32$, So $5^{32}\mod 1$
+
+$5^2\equiv 25\mod 51=25$  
+$5^4\equiv (5^2)^2\equiv(25)^2 \mod 51\equiv 625\mod 51=13$  
+$5^8\equiv (5^4)^2\equiv(13)^2 \mod 51\equiv 169\mod 51=16$  
+$5^16\equiv (5^8)^2\equiv(16)^2 \mod 51\equiv 256\mod 51=1$  
+
+$$
+\begin{aligned}
+5^{662}\mod 51&=107^{662\mod 32}\mod 51\\
+&=5^{22}\mod 51\\
+&=5^{16}\cdot 5^4\cdot 5^2\mod 51\\
+&=19
+\end{aligned}
+$$
+
+For $a\in \mathbb{Z}_N^*$, the order of $a$, $o(a)$ is the smallest positive $k$ such that $a^k\equiv 1\mod N$. $o(a)\leq \phi(N),o(a)|\phi (N)$
+
+In a general finite group
+
+$g^{|G|}=e$ (identity)
+
+$o(g)\vert |G|$
+
+If a group $G=\{a,a^2,a^3,...,e\}$ $G$ is cyclic
+
+In a cyclic group, if $o(a)=|G|$, then a is a generator of $G$.
+
+Fact: $\mathbb{Z}^*_p$ is cyclic
+
+$|\mathbb{Z}^*_p|=p-1$, so $\exists$ generator $g$, and $\mathbb{Z}$, $\phi(\mathbb{Z}_{13}^*)=12$
+
+For example, $2$ is a generator for $\mathbb{Z}_{13}^*$ with $2,4,8,3,6,12,11,9,5,10,7,1$.
+
+If $g$ is a generator, $f:\mathbb{Z}_p^*\to \mathbb{Z}_p^*$, $f(x)=g^x \mod p$ is onto.
+
+What type of prime $p$?
+
+- Large prime.
+- If $p-1$ is very factorable, that is very bad.
+  - Pohlig-Hellman algorithm
+  - $p=2^n+1$ only need polynomial time to invert
+- We want $p=2q+1$, where $q$ is prime. (Sophie Germain primes, or safe primes)
+
+There are _probably_ infinitely many safe prime and efficient to sample as well.
+
+If $p$ is safe, $g$ generator.
+
+$$
+\mathbb{Z}_p^*=\{g,g^2,..,e\}
+$$
+
+Then $\{g^2,...g^{2q}\}S_{g,p}\subseteq \mathbb{Z}_p^*$ is a subgroup; $g^{2k}\cdot g^{2l}=g^{2(k+l)}\in S_{g,p}$
+
+It is cyclic with generator $g^2$.
+
+It is easy to find a generator.
+
+- Pick $a\in \mathbb{Z}_p^*$
+- Let $x=a^2$. If $x\neq 1$, it is a generator of subgroup $S_p$
+  - $S_p=\{x,x^2,...,x^q\}\mod p$
+
+Example: $p=2\cdot 11+1=23$
+
+we have a subgroup with generator $4$ and $S_4=\{4,16,18,3,12,2,8,9,13,6,1\}$
+
+```python
+def get_generator(p):
+    """
+    p should be a prime, or you need to do factorization
+    """
+    g=[]
+    for i in range(2,p-1):
+        k=i
+        sg=[]
+        step=p
+        while k!=1 and step>0:
+            if k==0:
+                raise ValueError(f"Damn, {i} generates 0 for group {p}")
+            sg.append(k)
+            k=(k*i)%p
+            step-=1
+        sg.append(1)
+        # if len(sg)!=(p-1): continue
+        g.append((i,[j for j in sg]))
+    return g
+```
+
+### (Computational) Diffie-Hellman assumption
+
+If $p$ is a randomly sampled safe prime.
+
+Denote safe prime as $\tilde{\Pi}_n=\{p\in \Pi_n:q=\frac{p-1}{2}\in \Pi_{n-1}\}$
+
+Then
+
+$$
+P\left[p\gets \tilde{\Pi_n};a\gets\mathbb{Z}_p^*;g=a^2\neq 1;x\gets \mathbb{Z}_q;y=g^x\mod p:\mathcal{A}(y)=x\right]\leq \epsilon(n)
+$$
+
+$p\gets \tilde{\Pi_n};a\gets\mathbb{Z}_p^*;g=a^2\neq 1$ is the function condition when we do the encryption on cyclic groups.
+
+Notes: $f:\Z_q\to \mathbb{Z}_p^*$ is one-to-one, so $f(\mathcal{A}(y))\iff \mathcal{A}(y)=x$