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update typo and structures

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Trance-0
2024-12-16 13:41:24 -06:00
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pages/CSE442T/CSE442T_L2.md
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@@ -2,13 +2,13 @@
## Probability review
Sample space $S=$ set of outcomes (possible results of experiments)
Sample space $S=\text{set of outcomes (possible results of experiments)}$
Event $A\subseteq S$
$P[A]=P[$ outcome $x\in A]$
$P[\{x\}]=P(x)$
$P[\{x\}]=P[x]$
Conditional probability:
@@ -32,27 +32,27 @@ $A=\bigcup_{i=1}^n A\cap B_i$ ($A\cap B_i$ are all disjoint)
$P[A]=\sum^n_{i=1} P[A|B_i]\cdot P[B_i]$
## Back to cryptography
## Chapter 1: Introduction
Defining security.
### Defining security
### Perfect Secrecy (Shannon Secrecy)
#### Perfect Secrecy (Shannon Secrecy)
$K\gets Gen()$ $K\in\mathcal{K}$
$k\gets Gen()$ $k\in K$
$c\gets Enc_K(m)$ or we can also write as $c\gets Enc(K,m)$ for $m\in \mathcal{M}$
$c\gets Enc_k(m)$ or we can also write as $c\gets Enc(k,m)$ for $m\in M$
And the decryption procedure:
$m'\gets Dec_K(c')$, $m'$ might be null.
$m'\gets Dec_k(c')$, $m'$ might be null.
$P[K\gets Gen(): Dec_K(Enc_K(m))=m]=1$
$P[k\gets Gen(): Dec_k(Enc_k(m))=m]=1$
#### Shannon Secrecy
#### Definition 11.1 (Shannon Secrecy)
Distribution $D$ over the message space $\mathcal{M}$
Distribution $D$ over the message space $M$
$P[K\gets Gen;m\gets D: m=m'|c\gets Enc_K(m)]=P[m\gets D: m=m']$
$P[k\gets Gen;m\gets D: m=m'|c\gets Enc_k(m)]=P[m\gets D: m=m']$
Basically, we cannot gain any information from the encoded message.
@@ -60,15 +60,15 @@ Code shall not contain any information changing the distribution of expectation
**NO INFO GAINED**
#### Perfect Secrecy
#### Definition 11.2 (Perfect Secrecy)
For any 2 messages, say $m_1,m_2\in \mathcal{M}$ and for any possible cipher $c$,
For any 2 messages, say $m_1,m_2\in M$ and for any possible cipher $c$,
$P[K\gets Gen:c\gets Enc_K(m_1)]=P[K\gets Gen():c\gets Enc_K(m_2)]$
$P[k\gets Gen:c\gets Enc_k(m_1)]=P[k\gets Gen():c\gets Enc_k(m_2)]$
For a fixed $c$, any message could be encrypted to that...
For a fixed $c$, any message (have a equal probability) could be encrypted to that...
#### Theorem
#### Theorem 12.3
Shannon secrecy is equivalent to perfect secrecy.
@@ -76,22 +76,22 @@ Proof:
If a crypto-system satisfy perfect secrecy, then it also satisfy Shannon secrecy.
Let $(Gen, Enc,Dec)$ be a perfectly secret crypto-system with $\mathcal{K}$ and $\mathcal{M}$.
Let $(Gen,Enc,Dec)$ be a perfectly secret crypto-system with $K$ and $M$.
Let $D$ be any distribution over messages.
Let $m'\in \mathcal{M}$.
Let $m'\in M$.
$$
={P_K[c\gets Enc_K(m')]\cdot P[m=m']\over P_{K,m}[c\gets Enc_K(m)]}\\
={P_k[c\gets Enc_k(m')]\cdot P[m=m']\over P_{k,m}[c\gets Enc_k(m)]}\\
$$
$$
P[K\gets Gen();m\gets D:m=m'|c\gets Enc_K(m)]={P_{K,m}[c\gets Enc_K(m)\vert m=m']\cdot P[m=m']\over P_{K,m}[c\gets Enc_K(m)]}\\
P_{K,m}[c\gets Enc_K(m)]=\sum^n_{i=1}P_{K,m}[c\gets Enc_k(m)|m=m_i]\cdot P[m=m_i]\\
P[k\gets Gen();m\gets D:m=m'|c\gets Enc_k(m)]={P_{k,m}[c\gets Enc_k(m)\vert m=m']\cdot P[m=m']\over P_{k,m}[c\gets Enc_k(m)]}\\
P_{k,m}[c\gets Enc_k(m)]=\sum^n_{i=1}P_{k,m}[c\gets Enc_k(m)|m=m_i]\cdot P[m=m_i]\\
=\sum^n_{i=1}P_{K,m_i}[c\gets Enc_k(m_i)]\cdot P[m=m_i]
$$
and $P_{K,m_i}[c\gets Enc_K(m_i)]$ is constant due to perfect secrecy
and $P_{k,m_i}[c\gets Enc_k(m_i)]$ is constant due to perfect secrecy
$\sum^n_{i=1}P_{K,m_i}[c\gets Enc_K(m_i)]\cdot P[m=m_i]=\sum^n_{i=1} P[m=m_i]=1$
$\sum^n_{i=1}P_{k,m_i}[c\gets Enc_k(m_i)]\cdot P[m=m_i]=\sum^n_{i=1} P[m=m_i]=1$