update typo and structures

pages/CSE442T/CSE442T_L11.md

@@ -2,13 +2,15 @@
 
 Exam info posted tonight.
 
-## Pseudo-randomness
+## Chapter 3: Indistinguishability and pseudo-randomness
+
+### Pseudo-randomness
 
 Idea: **Efficiently** produce many bits
 
 which "appear" truly random.
 
-### One-time pad
+#### One-time pad
 
 $m\in\{0,1\}^n$
 
@@ -42,29 +44,29 @@ For $1\leq i<j\leq n,P[x_i=1 \textup{ and } x_j=1]=\frac{1}{4}$ (by independence
 
 Let $\{X_n\}_n$ and $\{Y_n\}_n$ be probability ensembles (separate of dist over $\{0,1\}^{l(n)}$)
 
-$\{X_n\}_n$ and $\{Y_n\}_n$ are computationally **in-distinguishable** if for all non-uniform p.p.t adversary $D$ ("distinguishers")
+$\{X_n\}_n$ and $\{Y_n\}_n$ are computationally **in-distinguishable** if for all non-uniform p.p.t adversary $\mathcal{D}$ ("distinguishers")
 
 $$
-|P[x\gets X_n:D(x)=1]-P[y\gets Y_n:D(y)=1]|<\epsilon(n)
+|P[x\gets X_n:\mathcal{D}(x)=1]-P[y\gets Y_n:\mathcal{D}(y)=1]|<\epsilon(n)
 $$
 
 this basically means that the probability of finding any pattern in the two array is negligible.
 
-If there is a $D$ such that
+If there is a $\mathcal{D}$ such that
 
 $$
-|P[x\gets X_n:D(x)=1]-P[y\gets Y_n:D(y)=1]|\geq \mu(n)
+|P[x\gets X_n:\mathcal{D}(x)=1]-P[y\gets Y_n:\mathcal{D}(y)=1]|\geq \mu(n)
 $$
 
-then $D$ is distinguishing with probability $\mu(n)$
+then $\mathcal{D}$ is distinguishing with probability $\mu(n)$
 
-If $\mu(n)\geq\frac{1}{p(n)}$, then $D$ is distinguishing the two $\implies X_n\cancel{\approx} Y_n$
+If $\mu(n)\geq\frac{1}{p(n)}$, then $\mathcal{D}$ is distinguishing the two $\implies X_n\cancel{\approx} Y_n$
 
 ### Prediction lemma
 
 $X_n^0$ and $X_n^1$ ensembles over $\{0,1\}^{l(n)}$
 
-Suppose $\exists$ distinguisher $D$ which distinguish by $\geq \mu(n)$. Then $\exists$ adversary $\mathcal{A}$ such that 
+Suppose $\exists$ distinguisher $\mathcal{D}$ which distinguish by $\geq \mu(n)$. Then $\exists$ adversary $\mathcal{A}$ such that 
 
 $$
 P[b\gets\{0,1\};t\gets X_n^b]:\mathcal{A}(t)=b]\geq \frac{1}{2}+\frac{\mu(n)}{2}
@@ -75,7 +77,7 @@ Proof:
 Without loss of generality, suppose
 
 $$
-P[t\gets X^1_n:D(t)=1]-P[t\gets X_n^0:D(t)=1]\geq \mu(n)
+P[t\gets X^1_n:\mathcal{D}(t)=1]-P[t\gets X_n^0:\mathcal{D}(t)=1]\geq \mu(n)
 $$
 
 $\mathcal{A}=\mathcal{D}$ (Outputs 1 if and only if $D$ outputs 1, otherwise 0.)
@@ -98,15 +100,15 @@ Example:
 
 Building distinguishers
 
-1. $X_n$: always outputs $0^n$, $D$: [outputs $1$ if $t=0^n$]  
+1. $X_n$: always outputs $0^n$, $\mathcal{D}$: [outputs $1$ if $t=0^n$]  
     $$
-    \vert P[t\gets X_n:D(t)=1]-P[t\gets U_n:D(t)=1]\vert=1-\frac{1}{2^n}\approx 1
+    \vert P[t\gets X_n:\mathcal{D}(t)=1]-P[t\gets U_n:\mathcal{D}(t)=1]\vert=1-\frac{1}{2^n}\approx 1
     $$
 2. $X_n$: 1st $n-1$ bits are truly random $\gets U_{n-1}$ nth bit is $1$ with probability 0.50001 and $0$ with 0.49999, $D$: [outputs $1$ if $X_n=1$]  
     $$
-    \vert P[t\gets X_n:D(t)=1]-P[t\gets U_n:D(t)=1]\vert=0.5001-0.5=0.001\neq 0
+    \vert P[t\gets X_n:\mathcal{D}(t)=1]-P[t\gets U_n:\mathcal{D}(t)=1]\vert=0.5001-0.5=0.001\neq 0
     $$
 3. $X_n$: For each bit $x_i\gets\{0,1\}$ **unless** there have been 1 million $0$'s. in a row. Then outputs $1$, $D$: [outputs $1$ if $x_1=x_2=...=x_{1000001}=0$]
    $$
-    \vert P[t\gets X_n:D(t)=1]-P[t\gets U_n:D(t)=1]\vert=|0-\frac{1}{2^{1000001}}|\neq 0
+    \vert P[t\gets X_n:\mathcal{D}(t)=1]-P[t\gets U_n:\mathcal{D}(t)=1]\vert=|0-\frac{1}{2^{1000001}}|\neq 0
    $$