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Aug
28
comment In DGHV FHE, why noise $r$ can be in $(-2^{\rho'}, 0)$?
The mistake is that you're just thinking about (elementary ) number theory, $\hspace{1.8 in}$ rather than using the symmetric representation. $\;$
Aug
28
comment In DGHV FHE, why noise $r$ can be in $(-2^{\rho'}, 0)$?
You are using a wrong representation. ​ ​
Aug
28
comment Are there ANY text strings that will generate the same SHA-512 Hash output?
Yes, but it's supposed to be hard to find an example. $\;$
Aug
27
comment Can iterated hashing be used to mitigate collision and preimage weaknesses?
Well, it adds a non-positive amount regarding collisions. $\;$
Aug
27
comment Proving set membership in less than log(N) bandwidth
No; it could be that for honest servers, the client will accept with certainty, but malicious servers can make the clients have an essentially-arbitrary probability of accepting. $\:$ (I don't know of any candidate schemes for which that would be helpful.) $\;\;\;\;$
Aug
27
comment Proving set membership in less than log(N) bandwidth
There could conceivably be probabilistic schemes which also have that property. $\;$
Aug
27
comment Randomness in Public Key Encryption
@mephisto : $\:$ Do you have a proposed definition for "security influencing randomness"? $\;\;\;\;$
Aug
27
comment Proving set membership in less than log(N) bandwidth
Note that what you're doing doesn't operate how you said it needs to - a malicious server can simply send an invalid alleged-branch to one client to get that client to come to a different conclusion as a client it acted honestly with. $\;$
Aug
26
comment Shamir Secret Sharing: Why cannot we recover polynomial's root if we have $t-1$ shares?
@user13676 : $\:$ How is that related to cryptography? $\;\;\;\;$
Aug
26
comment Shamir Secret Sharing: Why cannot we recover polynomial's root if we have $t-1$ shares?
@user13676 : $\:$ Does that mean "What is the probability that given t-1 shares, one's best guess at a root of the polynomial is correct?" or "What is the probability that given t-1 shares, there is a field element x such that those shares force x to be a root of the polynomial?"? $\;\;\;\;$
Aug
26
comment Proving set membership in less than log(N) bandwidth
Also, if you elide the security parameter, then polylog(N) can generally (including in this case) be elided by padding for small N and just showing a break in security for large N. $\;$
Aug
26
comment Proving set membership in less than log(N) bandwidth
No, since the lengths of the hashes should be 2 times the security parameter. $\;$
Aug
26
comment Proving set membership in less than log(N) bandwidth
It looks to me like that takes $\: 2\hspace{-0.04 in}\cdot \hspace{-0.04 in}k\hspace{-0.04 in}\cdot \hspace{-0.04 in}\log(N\hspace{.02 in}) \:$ space (where k is the security parameter), rather than log(N) space.
Aug
26
comment Additive homomorphic encryption scheme without change in operator
On the other hand, if you want $D\hspace{.03 in}(E(m_1)\hspace{-0.04 in}+\hspace{-0.04 in}E(m_2)\hspace{-0.04 in}+\hspace{-0.04 in}E(m_3)+...\hspace{-0.05 in}+E(m_{\hspace{.02 in}n-1})\hspace{-0.04 in}+\hspace{-0.04 in}E(m_{\hspace{.02 in}n})) \: = \: m_1\hspace{-0.06 in}+\hspace{-0.04 in}m_2\hspace{-0.04 in}+\hspace{-0.04 in}m_3+...\hspace{-0.04 in}+m_{\hspace{.02 in}n-1}\hspace{-0.05 in}+\hspace{-0.04 in}m_{\hspace{.02 in}n} \hspace{1.17 in}$ for all $n$, then things will probably be more difficult. $\;\;\;\;$
Aug
24
answered OTP same message encrypted twice (with different keys)
Aug
24
comment Why an upside down path on the evaluation of branching program on encrypted data?
Yes. ${}{}{}\;$
Aug
24
comment Why an upside down path on the evaluation of branching program on encrypted data?
We "can obtain $P(x)$ from a path from the terminal nodes to the initial node" because the value of every non-terminal node is defined in terms of (stuff that's independent of $P$ and) the values of nodes that are closer to being terminal. $\:$ With their approach, those values will already have been computed by the time they are to be used, rather than having to be computed on-the-fly. $\;\;\;\;$
Aug
24
comment Why an upside down path on the evaluation of branching program on encrypted data?
I now realize that memoization is actually independent of the point I was trying to make. $\;$
Aug
24
revised Why an upside down path on the evaluation of branching program on encrypted data?
corrected explanation from memoization to recursion
Aug
24
revised Why an upside down path on the evaluation of branching program on encrypted data?
addressed the PS