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pages/CSE442T/CSE442T_L2.md

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+# Lecture 2
+
+## Probability review
+
+Sample space $S=$ set of outcomes (possible results of experiments)
+
+Event $A\subseteq S$
+
+$P[A]=P[$ outcome $x\in A]$
+
+$P[\{x\}]=P(x)$
+
+Conditional probability:
+
+$P[A|B]={P[A\cap B]\over P[B]}$
+
+Assuming $B$ is the known information. Moreover, $P[B]>0$
+
+Probability that $A$ and $B$ occurring: $P[A\cap B]=P[A|B]\cdot P[B]$
+
+$P[B\cap A]=P[B|A]\cdot P[A]$
+
+So  $P[A|B]={P[B|A]\cdot P[A]\over P[B]}$ (Bayes Theorem)
+
+**There is always a chance that random guess would be the password... Although really, really, low...**
+
+### Law of total probability
+
+Let $S=\bigcup_{i=1}^n B_i$. and $B_i$ are disjoint events.
+
+$A=\bigcup_{i=1}^n A\cap B_i$ ($A\cap B_i$ are all disjoint)
+
+$P[A]=\sum^n_{i=1} P[A|B_i]\cdot P[B_i]$
+
+## Back to cryptography
+
+Defining security.
+
+### Perfect Secrecy (Shannon Secrecy)
+
+$K\gets Gen()$ $K\in\mathcal{K}$
+
+$c\gets Enc_K(m)$ or we can also write as $c\gets Enc(K,m)$ for $m\in \mathcal{M}$
+
+And the decryption procedure:
+
+$m'\gets Dec_K(c')$, $m'$ might be null.
+
+$P[K\gets Gen(): Dec_K(Enc_K(m))=m]=1$
+
+#### Shannon Secrecy
+
+Distribution $D$ over the message space $\mathcal{M}$
+
+$P[K\gets Gen;m\gets D: m=m'|c\gets Enc_K(m)]=P[m\gets D: m=m']$
+
+Basically, we cannot gain any information from the encoded message.
+
+Code shall not contain any information changing the distribution of expectation of message after viewing the code.
+
+**NO INFO GAINED**
+
+#### Perfect Secrecy
+
+For any 2 messages, say $m_1,m_2\in \mathcal{M}$ and for any possible cipher $c$,
+
+$P[K\gets Gen:c\gets Enc_K(m_1)]=P[K\gets Gen():c\gets Enc_K(m_2)]$
+
+For a fixed $c$, any message could be encrypted to that...
+
+#### Theorem 
+
+Shannon secrecy is equivalent to perfect secrecy.
+
+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 $D$ be any distribution over messages.
+
+Let $m'\in \mathcal{M}$.
+
+$$
+={P_K[c\gets Enc_K(m')]\cdot P[m=m']\over P_{K,m}[c\gets Enc_K(m)]}\\
+$$
+
+$$
+P[K\gets Gen();m\gets D:m=m'|c\gets Enc_K(m)]={P_{K,m}[c\gets Enc_K(m)\vert m=m']\cdot P[m=m']\over P_{K,m}[c\gets Enc_K(m)]}\\
+P_{K,m}[c\gets Enc_K(m)]=\sum^n_{i=1}P_{K,m}[c\gets Enc_k(m)|m=m_i]\cdot P[m=m_i]\\
+=\sum^n_{i=1}P_{K,m_i}[c\gets Enc_k(m_i)]\cdot P[m=m_i]
+$$
+
+and $P_{K,m_i}[c\gets Enc_K(m_i)]$ is constant due to perfect secrecy
+
+$\sum^n_{i=1}P_{K,m_i}[c\gets Enc_K(m_i)]\cdot P[m=m_i]=\sum^n_{i=1} P[m=m_i]=1$