CSE442T Introduction to Cryptography (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”)

this basically means that the probability of finding any pattern in the two array is negligible.

If there is a $\mathcal{D}$ such that

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

Proof:

Without loss of generality, suppose

$\mathcal{A}=\mathcal{D}$ (Outputs 1 if and only if $D$ outputs 1, otherwise 0.)

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 $x_1=x_2=...=x_{1000001}=0$] $$\vert P[t\gets X_n:\mathcal{D}(t)=1]-P[t\gets U_n:\mathcal{D}(t)=1]\vert=|0-\frac{1}{2^{1000001}}|\neq 0$$