update typo and structures

pages/CSE442T/CSE442T_L5.md

@@ -1,11 +1,13 @@
 # Lecture 5
 
+## Chapter 2: Computational Hardness
+
 Proving that there are one-way functions relies on assumptions.
 
-Factoring Assumption: $\forall a, \exist \epsilon (n)$, let $p,q\in prime,p,q<2^n$
+Factoring Assumption: $\forall \mathcal{A}, \exist \epsilon (n)$, let $p,q\in \Pi_n,p,q<2^n$
 
 $$
-P[p\gets \Pi_n;q\gets \Pi_n;N=p\cdot q:a(N)\in \{p,q\}]<\epsilon(n)
+P[p\gets \Pi_n;q\gets \Pi_n;N=p\cdot q:\mathcal{A}(N)\in \{p,q\}]<\epsilon(n)
 $$
 
 Evidence: To this point, best known procedure to always factor has run time $O(2^{\sqrt{n}\sqrt{log(n)}})$
@@ -19,7 +21,7 @@ We want to (guaranteed to) find prime:
 
 $\pi(n)>\frac{2^n}{2n}$
 
-e.g. 
+e.g.
 $$
 P[x\gets \{0,1\}^n:x\in prime]\geq {\frac{2^n}{2n}\over 2^n}=\frac{1}{2n}
 $$
@@ -33,40 +35,40 @@ $$
 Idea: There are enough pairs of primes to make this difficult.
 
 > Reminder: Weak on-way if easy to compute and $\exist p(n)$,
-> $$P[a\ inverts=success]<1-\frac{1}{p(n)}$$
-> $$P[failure]>\frac{1}{p(n)}$$ high enough
+> $P[\mathcal{A}\ \text{inverts=success}]<1-\frac{1}{p(n)}$
+> $P[\mathcal{A}\ \text{inverts=failure}]>\frac{1}{p(n)}$ high enough
 
-## Prove one-way function (under assumptions)
+### Prove one-way function (under assumptions)
 
 To prove $f$ is on-way (under assumption)
 
 1. Show $\exists p.p.t$ solves $f(x),\forall x$.
 2. Proof by contradiction.
    - For weak: Provide $p(n)$ that we know works.
-     - Assume $\exists a$ such that $P[a\ inverts]>1-\frac{1}{p(n)}$
+     - Assume $\exists \mathcal{A}$ such that $P[\mathcal{A}\ \text{inverts}]>1-\frac{1}{p(n)}$
    - For strong: Provide $p(n)$ that we know works.
-     - Assume $\exists a$ such that $P[a\ inverts]>\frac{1}{p(n)}$
+     - Assume $\exists \mathcal{A}$ such that $P[\mathcal{A}\ \text{inverts}]>\frac{1}{p(n)}$
 
-Construct p.p.t B
-which uses $a$ to solve a problem, which contradicts assumption or known fact.
+Construct p.p.t $\mathcal{B}$
+which uses $\mathcal{A}$ to solve a problem, which contradicts assumption or known fact.
 
 Back to Theorem:
 
 We will show that $p(n)=8n^2$ works.
 
-We claim $\forall a$,
+We claim $\forall \mathcal{A}$,
 
 $$
-P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(a(y))=y]<1-\frac{1}{8n^2}
+P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(\mathcal{A}(y))=y]<1-\frac{1}{8n^2}
 $$
 
 For the sake of contradiction, suppose
 
 $$
-\exists a \textup{ such that} P[success]>1-\frac{1}{8n^2}
+\exists \mathcal{A} \textup{ such that} P[\mathcal{A}\ \text{inverts}]>1-\frac{1}{8n^2}
 $$
 
-We will use this $a$ to design p.p.t $B$ which can factor 2 random primes with non-negligible prob.
+We will use this $\mathcal{A}$ to design p.p.t $B$ which can factor 2 random primes with non-negligible prob.
 
 ```python
 def A(y):
@@ -88,27 +90,27 @@ def B(y):
     return A(y)
 ```
 
-How often does B succeed/fail?
+How often does $\mathcal{B}$ succeed/fail?
 
-B fails to factor $N=p\dot q$, if:
+$\mathcal{B}$ fails to factor $N=p\dot q$, if:
 
 - $x$ and $y$ are not both prime
-  - $P_e=1-P(x\in prime)P(y\in prime)\leq 1-(\frac{1}{2n})^2=1-\frac{1}{4n^2}$
-- if $a$ fails to factor
+  - $P_e=1-P(x\in \Pi_n)P(y\in \Pi_n)\leq 1-(\frac{1}{2n})^2=1-\frac{1}{4n^2}$
+- if $\mathcal{A}$ fails to factor
   - $P_f<\frac{1}{8n^2}$
 
 So
 
 $$
-P[B\ fails]\leq P[E\cup F]\leq P[E]+P[F]\leq (1-\frac{1}{4n^2}+\frac{1}{8n^2})=1-\frac{1}{8n^2}
+P[\mathcal{B} \text{ fails}]\leq P[E\cup F]\leq P[E]+P[F]\leq (1-\frac{1}{4n^2}+\frac{1}{8n^2})=1-\frac{1}{8n^2}
 $$
 
 So
 
 $$
-P[B\ succeed]\geq \frac{1}{8n^2}\ (non\ negligible)
+P[\mathcal{B} \text{ succeed}]\geq \frac{1}{8n^2} (\text{non-negligible})
 $$
 
-This contradicting factoring assumption. Therefore, our assumption that $a$ exists was wrong.
+This contradicting factoring assumption. Therefore, our assumption that $\mathcal{A}$ exists was wrong.
 
-Therefore $\forall a$, $P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(a(y))=y]<1-\frac{1}{8n^2}$ is wrong.
+Therefore $\forall \mathcal{A}$, $P[(x_1,x_2)\gets \{0,1\}^{2n};y=f_{mult}(x_1,x_2):f(\mathcal{A}(y))=y]<1-\frac{1}{8n^2}$ is wrong.