1
$\begingroup$

I am reading the paper A Concrete Security Treatment of Symmetric Encryption and am confused by the reduction from ROR to LOR on page 11. Specifically, when it says:

When $\mathcal{O}_2(\cdot)=\mathcal{E}_K(\mathcal{RR}(\cdot,0))$, we have that $\mathcal{O}_1(\mathcal{LR}(\cdot,\cdot,0))$ and $\mathcal{O}_1(\mathcal{LR}(\cdot,\cdot,1))$ return identically distributed answers.

So, $\Pr[\mathbf{Exp}_\mathcal{SE,A_2}^{ror-atk-0}(k)=1]=1/2$

I'm not sure how the first statement implies the second statement.

$\endgroup$

1 Answer 1

1
$\begingroup$

A key aspect is that whenever we are in experiment $\text{ror-atk-}0$, $\mathcal{A}_1$ receives a ciphertext $c$ corresponding to a random plaintext. In particular, this random plaintext is (by definition) independent of the bit $b$ selected by the $\text{lor}$ adversary $\mathcal{A}_2$. In turn, $\mathcal{A}_1$'s guess $d$ is also independent of $b$ since it only depends on the ciphertext for random plaintexts and the random choices made by $\mathcal{A}_1$. Hence, the probability expression is equivalent to $$\Pr[\mathbf{Exp}_\mathcal{SE,A_2}^{ror-atk-0}(k)=1]= \Pr[d = b] = 1/2.$$

$\endgroup$
1
  • 1
    $\begingroup$ Thanks so much for this answer. It made me realize I had misread the oracle definition and had thought that ror-atk-0 referred to the case where the ciphertext corresponded to the plaintext chosen by A1 and dependent on b. $\endgroup$
    – ax12345
    Commented Nov 14, 2022 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.