Bound for quantum adversary in CRYSTALS-Kyber

P_Gate

6/15/24, 5:10 PM

In Kyber the following estimation is given for a quantum adversary:

Theorem 4. For any quantum adversary $A$ that makes at most $q_{RO}$ many queries to quantum random oracles $H$ and $G$, and at most $q_D$ many (classical) queries to the decryption oracle, there exists a quantum adversary $B$ such that $$\text{Adv}_{\text{Kyber}}^{\text{cca}}(A) \leq 8 q_{RO}^2 \cdot \delta + 4q_{RO} \cdot \sqrt{\text{Adv}_{\text{Kyber.CPA}}^{\text{pr}}(B)}$$

In accordance to @lamontap's comment, I have also found this estimation in 2:

$$\text{Adv}_{\text{KEM}}^{\text{IND-CCA}}(B) \leq 8q_{RO} \left( \sqrt{q_{RO}^2 \cdot \delta + q_{RO} \cdot \sqrt{\text{Adv}_{\text{PKE}}^{\text{OW-CPA}}}} + q_{RO}\cdot \delta \right)$$

However, it remains open, how the exact way is to rewrite the double root, so that the result from Theorem 4 follows. I have not yet been able to find anything suitable in the reference 69 to rewrite the expression.

My question is therefore, how is the estimation in Theorem 4 actually created?

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post-quantum-cryptography

kyber

lamontap

6/15/24, 9:10 PM

Note that the bound from the HHZ paper is as a function of the OW-PCA advantage, which is not the same as OW-CPA. The HHZ bound would actually lead to a double square root over the advantage of adversary B (see the very last page of their paper). The Kyber paper says they use ref [25] (the HHZ paper) for the correctness error. The main reduction seems to be done using the techniques of ref [69].

Elon Musk

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EN: Bound for quantum adversary in CRYSTALS-Kyber

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