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# Lecture 22 # Lecture 22
## Chapter 7: Types of Attacks
So far we've sought security against
$$
c\gets Enc_k(m)
$$
Adversary knows $c$, but nothing else.
### Known plaintext attack (KPA)
Adversary has seen $(m_1,Enc_k(m_1)),(m_2,Enc_k(m_2)),\cdots,(m_q,Enc_k(m_q))$.
$m_1,\cdots,m_q$ are known to the adversary.
Given new $c=Enc_k(m)$, is previous knowledge helpful?
### Chosen plaintext attack (CPA)
Adversary can choose $m_1,\cdots,m_q$ and obtain $Enc_k(m_1),\cdots,Enc_k(m_q)$.
Then adversary see new encryption $c=Enc_k(m)$. with the same key.
Example:
In WWII, Japan planned to attack "AF", but US suspected it means Midway.
So US use Axis: $Enc_k(AF)$ and ran out of supplies.
Then US know Japan will attack Midway.
### Chosen ciphertext attack (CCA)
Adversary can choose $c_1,\cdots,c_q$ and obtain $Dec_k(c_1),\cdots,Dec_k(c_q)$.
Capture these ideas with the adversary having oracle access.
$$
\Pi=(Gen,Enc,Dec)
$$
private key encryption scheme.
$$
IND_b^{O_1,O_2}(\Pi,\mathcal{A},n)
$$
where $O_1$ and $O_2$ are the round 1 and round 2 oracle access.
$b$ is zero or one denoting the real scheme or the adversary's challenge.
$n$ is the security parameter.
is the following experiment:
- Key $k\gets Gen(1^n)$
- Adversary $\mathcal{A}^{O_1(k)}(1^n)$ queries oracles
- $m_0,m_1\gets \mathcal{A}^{O_2(k)}(1^n)$
- $c\gets Enc_k(m_b)$
- $\mathcal{A}^{O_2(c)}(1^n,c)$ queries oracles
- $\mathcal{A}$ outputs bit $b'$ which is either zero or one
$\Pi$ is CPA/CCA1/CCA2 secure if for all PPT adversaries $\mathcal{A}$,
$$
\{IND_0^{O_1,O_2}(\Pi,\mathcal{A},n)\}_n\approx\{IND_1^{O_1,O_2}(\Pi,\mathcal{A},n)\}_n
$$
where $\approx$ is statistical indistinguishability.
|Security|$O_1$|$O_2$|
|:---:|:---:|:---:|
|CPA|$Enc_k$|$Enc_k$|
|CCA1|$Enc_k,Dec_k$|$Enc_k$|
|CCA2 (or full CCA)|$Enc_k,Dec_k$|$Enc_k,Dec_k^*$|
Note that $Dec_k^*$ will not allowed to query decryption of a functioning ciphertext.
#### Theorem: Our mms private key encryption scheme is CPA, CCA1 secure.
Have a PRF family $\{f_k\}:\{0,1\}^|k|\to\{0,1\}^{|k|}$
$Gen(1^n)$ outputs $k\in\{0,1\}^n$ and samples $f_k$ from the PRF family.
$Enc_k(m)$ samples $r\in\{0,1\}^n$ and outputs $(r,f_k(r)\oplus m)$. For multi-message security, we need to encrypt $m_1,\cdots,m_q$ at once.
$Dec_k(r,c)$ outputs $f_k(r)\oplus c$.
Familiar Theme:
- Show the R.F. version is secure.
- $F\gets RF_n$
- If the PRF version were insecure, then the PRF can be distinguished from a random function...
$IND_b^{O_1,O_2}(\Pi,\mathcal{A},n), F\gets RF_n$
- $Enc$ queries $(m_1,(r_1,m_1\oplus F_k(r_1))),\cdots,(m_{q_1},(r_{q_1},m_{q_1}\oplus F_k(r_{q_1})))$
- $Dec$ queries $(s_1,c_1),\cdots,(s_{q_2},c_{q_2})$, where $m_i=c_i-F_k(s_i)$
- $m_0,m_1\gets \mathcal{A}^{O_2(k)}(1^n)$, $Enc_F(m_b)=(R,M_b+F(R))$
- Query round similar to above.
As long as $R$ was never seen in querying rounds, $P[\mathcal{A} \text{ guesses correctly}]=1/2$.
$P[R\text{ was seen before}]\leq \frac{p(n)}{2^n}$ (by the total number of queries in all rounds.)