update typo and structures

pages/CSE442T/CSE442T_L6.md

@@ -8,9 +8,11 @@ $$
 
 is a weak one-way.
 
-$P[a\ invert]\leq 1-\frac{1}{8n^2}$ over $x,y\in$ random integers $\{0,1\}^n$
+$P[\mathcal{A}\ \text{invert}]\leq 1-\frac{1}{8n^2}$ over $x,y\in$ random integers $\{0,1\}^n$
 
-## Converting to strong one-way function
+## Chapter 2: Computational Hardness
+
+### Converting weak one-way function to strong one-way function
 
 By factoring assumptions, $\exists$ strong one-way function
 
@@ -22,7 +24,7 @@ $f:\{0,1\}^{8n^4}\to \{0,1\}^{8n^4}$
 
 Idea: With high probability, at least one pair $(x_i,y_i)$ are both prime.
 
-Factoring assumption: $a$ has low chance of factoring $f_{mult}(x_i,y_i)$
+Factoring assumption: $\mathcal{A}$ has low chance of factoring $f_{mult}(x_i,y_i)$
 
 Use $P[x \textup{ is prime}]\geq\frac{1}{2n}$
 
@@ -34,13 +36,13 @@ $$
 P[\forall p,q \in x_i,y_i, p\textup{ and } q \textup{ is not prime }]\leq(1-\frac{1}{4n^2})^{4n^3}\leq (e^{-\frac{1}{4n^2}})^{4n^3}=e^{-n}
 $$
 
-### Proof of strong one-way
+### Proof of strong one-way function
 
 1. $f_{mult}$ is efficiently computable, and we compute it poly-many times.
 2. Suppose it's not hard to invert. Then
-    $\exists n.u.p.p.t.\ a$such that $P[w\gets \{0,1\}^{8n^4};z=f(w):f(a(z))=0]=\mu (n)>\frac{1}{p(n)}$
+    $\exists \text{n.u.p.p.t.}\ \mathcal{A}$such that $P[w\gets \{0,1\}^{8n^4};z=f(w):f(\mathcal{A}(z))=0]=\mu (n)>\frac{1}{p(n)}$
 
-We will use this to construct $B$ that breaks factoring assumption.
+We will use this to construct $\mathcal{B}$ that breaks factoring assumption.
 
 $p\gets \Pi_n,q\gets \Pi_n,N=p\cdot q$
 
@@ -64,11 +66,11 @@ function B:
 
 Let $E$ be the event that all pairs of sampled integers were not both prime.
 
-Let $F$ be the event that $a$ failed to invert
+Let $F$ be the event that $\mathcal{A}$ failed to invert
 
-$P(B\ fails)\leq P[E\cup F]\leq P[E]+P[F]\leq e^{-n}+(1-\frac{1}{p(n)})=1-(\frac{1}{p(n)}-e^{-n})\leq 1-\frac{1}{2p(n)}$
+$P[\mathcal{B} \text{ fails}]\leq P[E\cup F]\leq P[E]+P[F]\leq e^{-n}+(1-\frac{1}{p(n)})=1-(\frac{1}{p(n)}-e^{-n})\leq 1-\frac{1}{2p(n)}$
 
-$P[B\ succeeds]=P[p\gets \Pi_n,q\gets \Pi_n,N=p\cdot q:B(N)\in \{p,q\}]\geq \frac{1}{2p(n)}$
+$P[\mathcal{B} \text{ succeeds}]=P[p\gets \Pi_n,q\gets \Pi_n,N=p\cdot q:\mathcal{B}(N)\in \{p,q\}]\geq \frac{1}{2p(n)}$
 
 Contradicting factoring assumption
 
@@ -87,10 +89,10 @@ $F=\{f_i:D_i\to R_i\},i\in I$, $I$ is the index set.
 1. We can effectively choose $i\gets I$ using $Gen$.
 2. $\forall i$ we ca efficiently sample $x\gets D_i$.
 3. $\forall i\forall x\in D_i,f_i(x)$ is efficiently computable
-4. For any n.u.p.p.t $a$, $\exists$ negligible function $\epsilon (n)$.
-    $P[i\gets Gen(1^n);x\gets D_i;y=f_i(x):f(a(y,i,1^n))=y]\leq \epsilon(n)$
+4. For any n.u.p.p.t $\mathcal{A}$, $\exists$ negligible function $\epsilon (n)$.
+    $P[i\gets Gen(1^n);x\gets D_i;y=f_i(x):f(\mathcal{A}(y,i,1^n))=y]\leq \epsilon(n)$
 
-#### Theorem
+#### An instance of strong one-way function under factoring assumption
 
 $f_{mult,n}:(\Pi_n\times \Pi_n)\to \{0,1\}^{2n}$ is a collection of strong one way function.
 
@@ -107,8 +109,6 @@ Algorithm for sampling a random prime $p\gets \Pi_n$
    - Deterministic poly-time procedure
    - In practice, a much faster randomized procedure (Miller-Rabin) used
 
-        $P[x\cancel{\in} prime|test\ said\ x\ prime]<\epsilon(n)$
+    $P[x\cancel{\in} \text{prime}|\text{test said x prime}]<\epsilon(n)$
 
 3. If not, repeat. Do this for polynomial number of times
-
-> $;$ means and, $:$ means given that. $1$ usually interchangable with $\{0,1\}^n$