All info comes from David Wu’s Lecture and Boneh-Shoup Book.

This note will be focusing mainly on **perfect security**, **semantics security** and **PRG (Pseudo Random Generator)**.

The overall goal of cryptography is to secure communication over untrusted network. Two things must be achieved:

**Confidentiality**: No one can eavesdrop the communication
**Integrity**: No one can tamper with communication

# Perfect Security

A cipher $(Enc, Dec)$ satisfies **perfect secure** if $\forall m_0, m_1 \in M$ and $\forall c\in C$, $\Pr[k\overset{R}{\longleftarrow} K: Enc(k, m_0) = c] = \Pr[k\overset{R}{\longleftarrow} K:Enc(k,m_1) = c]$.

$k$ in two $\Pr$ might mean different $k$, the $\Pr$ just indicate the possibility of $\dfrac{\text{number of }k\text{ that }Enc(k, m) = c}{|K|}$.

## OTP is Perfect Secure

For every fixed $m = \lbrace 0, 1\rbrace^n$ there is $k, c = \lbrace 0, 1\rbrace^n$ uniquely paired that $m \oplus k = c$.

Considering perfect security definition, only one $k$ can encrypt $m$ to $c$. Thus $\Pr = \dfrac{1}{|K|} = \dfrac{1}{2^n}$ and equation is satisfied.

## Shannon “Bad News” Theorem

If a cipher is perfect secure, then $|K| \ge |M|$.

Assume $|K| < |M|$, we want to show it is not perfect secure. Let $k_0 \in K$ and $m_0 \in M$, then $c \leftarrow Enc(k_0, m_0)$. Let $S = \lbrace Dec(k, c): k \in K\rbrace$, we can see $|S| \le |K| < |M|$.

We can see that $\Pr\lbrack k \overset{R}{\longleftarrow} K: Enc(k, m_0) = c\rbrack > 0$, if we choose $m_1 \in M \backslash S$, then $\not\exists k \in K: Enc(k, m_1) = c$. Thus it is not perfect secure. $\square$