Reduction from Real-Or-Random to Left-Or-Right

ax12345

11/11/23, 12:35 AM

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.

1 + 0

provable-security

Score:1

Crypto

Marc Ilunga

11/12/23, 9:53 PM

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

+ 1

ax12345

11/14/23, 11:10 PM

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.

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Reduction from Real-Or-Random to Left-Or-Right

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