Timeline for How can fully homomorphic encryption ever be secure?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2017 at 19:59 | comment | added | 111 | I don't think that is obvious. The notation you used is not very sound since the cloud can not compute $E(F(m)),$ unless you clearly state what you mean. Also, if you feel that you are wasting your time, you have the option not to answer to the comments! | |
| Feb 20, 2017 at 17:06 | comment | added | MickLH | $F'(x) = \text{Eval}(F, x)$ Everything you've said is extremely obvious and you clearly have not taken the time to understand why I've given a more detailed answer than your extremely obvious rambling there. You're wasting my time and I'm feeling quite silly for sketching a rigorous argument when clearly the community isn't even at Wikipedia level on the topic yet. :( | |
| Feb 19, 2017 at 19:14 | comment | added | 111 | Not sure, what do you mean. But a FHE scheme has an evaluator, ${\it Eval}$ which is an efficient algorithm and a function (usually implemented as a circuit) say F, such that if $c=Enc(m)$ then $Dec(Eval(F,c))=F(m).$ So with $E(F(m))$ perhaps you mean ${\it Eval}(F,c)?$ Cloud can not work directly with the plaintext $m.$ | |
| Feb 19, 2017 at 4:10 | comment | added | MickLH | No. That's the mistake of the asker that I've corrected. It is actually $F'(E(m)) = E(F(m))$, the actual function applied to the ciphertext may be strange. It is the effective function that we care about. | |
| Feb 19, 2017 at 0:19 | comment | added | 111 | you mean $F(E(m))$? | |
| Feb 17, 2017 at 21:30 | history | answered | MickLH | CC BY-SA 3.0 |