NoteNextra-origin/content/CSE442T/CSE442T_L11.md

Lecture 11

Exam info posted tonight.

Chapter 3: Indistinguishability and pseudo-randomness

Pseudo-randomness

Idea: Efficiently produce many bits

which "appear" truly random.

One-time pad

m\in\{0,1\}^n

Gen(1^n):k\gets \{0,1\}^N

Enc_k(m)=m\oplus k

Dec_k(c)=c\oplus k

Advantage: Perfectly secret

Disadvantage: Impractical

The goal of pseudo-randomness is to make the algorithm, computationally secure, and practical.

Let \{X_n\} be a sequence of distributions over \{0,1\}^{l(n)}, where l(n) is a polynomial of n.

"Probability ensemble"

Example:

Let U_n be the uniform distribution over \{0,1\}^n

For all x\in \{0,1\}^n

P[x\gets U_n]=\frac{1}{2^n}

For 1\leq i\leq n, P[x_i=1]=\frac{1}{2}

For 1\leq i<j\leq n,P[x_i=1 \textup{ and } x_j=1]=\frac{1}{4} (by independence of different bits.)

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 \mathcal{D} ("distinguishers")

|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 \mathcal{D} such that

|P[x\gets X_n:\mathcal{D}(x)=1]-P[y\gets Y_n:\mathcal{D}(y)=1]|\geq \mu(n)

then \mathcal{D} is distinguishing with probability \mu(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 \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}

Proof:

Without loss of generality, suppose

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.)

\begin{aligned}
    &~~~~~P[b\gets \{0,1\};t\gets X_n^b:\mathcal{A}(t)=b]\\
    &=P[t\gets X_n^1;\mathcal{A}=1]\cdot P[b=1]+P[t\gets X_n^0;\mathcal{A}(t)=0]\cdot P[b=0]\\
    &=\frac{1}{2}P[t\gets X_n^1;\mathcal{A}(t)=1]+\frac{1}{2}(1-P[t\gets X_n^0;\mathcal{A}(t)=1])\\
    &=\frac{1}{2}+\frac{1}{2}(P[t\gets X_n^1;\mathcal{A}(t)=1]-P[t\gets X_n^0;\mathcal{A}(t)=1])\\
    &\geq\frac{1}{2}+\frac{1}{2}\mu(n)\\
\end{aligned}

Pseudo-random

\{X_n\} over \{0,1\}^{l(n)} is pseudorandom if \{X_n\}\approx\{U_{l(n)}\}. i.e. indistinguishable from the true randomness.

Example:

Building distinguishers

1. X_n: always outputs 0^n, \mathcal{D}: [outputs 1 if $t=0^n$]

   
   \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:\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:\mathcal{D}(t)=1]-P[t\gets U_n:\mathcal{D}(t)=1]\vert=|0-\frac{1}{2^{1000001}}|\neq 0