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Assuming you have the Decryption oracle is the same as assuming you have the key in your example, so you are just saying a OTP is vulnerable if you have the key
by kickling 4y ago
Assuming you have the Decryption oracle is the same as assuming you have the key in your example, so you are just saying a OTP is vulnerable if you have the key. This is true for any encryption scheme that I can think of.
- l33t2328 4y agoThis is not generally true. You can easily imagine a cryptosystem where having a decryption Oracle does not give the key. The fact that it is so easy to get the key given a decryption oracle for OTP just tells us that it’s really easy to show that it’s not CCA secure. The definition of CCA secure allows a decryption oracle with the same key as the challenge ciphertext.
- Robotbeat 4y agoBeing a one-time pad makes it irrelevant if you get the key. So it’s CCA secure because you never reuse the pad. You’ve gathered no useful information, ie you’ve only gathered noise that has no use.
- l33t2328 4y agoIt is not CCA secure because an adversary with access to a decryption oracle may get the key that was used to encrypt a challenge ciphertext via the method I’ve described. In the CCA experiment, the oracle uses the same key as the challenge ciphertext.
- Robotbeat 4y agoBut the OTP is still secure against that attack because the “key” is never reused. It is literally just noise, no useful information whatsoever.
- l33t2328 4y agoIt isn’t secure against the attack because the oracle uses the same key. The oracle is a tool used to formalize our definition. You’re right that the fact that OTP isn’t CCA secure doesn’t matter in practice because the key is only used for one message so such an oracle doesn’t generally exist.