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Mar 23, 2018 at 5:32 history tweeted twitter.com/StackCrypto/status/977055409486925824
Mar 23, 2018 at 3:02 history edited Blanco CC BY-SA 3.0
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Mar 22, 2018 at 5:47 history edited Blanco CC BY-SA 3.0
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Mar 22, 2018 at 5:16 history edited Blanco CC BY-SA 3.0
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Mar 21, 2018 at 14:54 comment added Geoffroy Couteau I also see no problem with defining $\mathsf{Enc}'(x)$ as $0^n||\mathsf{Enc}(x)$ for every $x$. We do not care whether the fixed prefix $0^n$ is not used in the decryption, nor do we care about the fact that the ciphertext is not random-looking - none of that prevents this "new" encryption scheme from being IND-CCA2...
Mar 21, 2018 at 14:48 answer added Yehuda Lindell timeline score: 1
Mar 21, 2018 at 13:19 comment added Blanco @MaartenBodewes I assume that $\mathcal{C}$ embed $\mathcal{C}'$, then I choose a simple function $f \colon y \to 0^{n} \Vert y$. And the prefix may be useless for a part of ciphertexts instead of all the ciphertexts.
Mar 21, 2018 at 13:10 comment added Maarten Bodewes There is a problem here: the prefix isn't really part of the ciphertext, unless it plays some part during decryption, and you haven't specified that. If it does then the ciphertext may not be indistinguishable from random in the given domain.
Mar 21, 2018 at 12:49 comment added Blanco @user94293 If the form of the ciphertexts is $r \Vert \mathrm{Enc}_{pk}(x)$, then it can not be non-malleability. When you get a ciphertext $r_{1} \Vert y$, you can find another ciphertext $r_{2} \Vert y$ and their plaintexts are the same. And IND-CCA2 $\Leftrightarrow$ NM-CCA2. Actually, I just want to know how to extend the ciphertexts to make it still non-malleability.
Mar 21, 2018 at 12:37 history asked Blanco CC BY-SA 3.0