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When could CPA work against non-deterministic encryption schemes?

ba flag

Is there a non-deterministic encryption scheme that is vulnerable to a CPA? I got asked this in an interview and I answered with no but the interviewer said that my answer is wrong. Would somebody clarify if this is possible?

poncho avatar
my flag
The obvious way to think about this is to try to answer the question "can you devise a non-deterministic encryption scheme that is vulnerable to a CPA"?
Daniel S avatar
ru flag
A little history: https://en.wikipedia.org/wiki/Transport_Layer_Security#BEAST_attack
poncho avatar
my flag
@DanielS: actually, the problem with CBC mode encryption in TLS 1.0 was that it *wasn't* nondeterministic :-) Instead, it was a deterministic function if you included the previous ciphertext.
Daniel S avatar
ru flag
@poncho I'd personally describe the issue as allowing the plaintext to be chosen after the non-deterministic information, but this is semantics.
cn flag
What about $\mathsf{Enc}(k,m)=(m,r)$ for uniformly chosen $r$ (from some space). Definitely not deterministic. How secure do you think it is?
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