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comment Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
Is the "secret" suppose to be known only to the 5,000 nodes? Or can it be generated by anyone, just not generated until the right time? In other words, is your P2P network closed?
2m
comment Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
OP, it looks like you have multiple accounts. You need to merge them so you can edit your own question and comment on answers.
3m
reviewed Approve Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
4m
comment Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
@MikeOunsworth I see your point. There are some ambiguities in the question, in my opinion. For example, is the network open to all? If so, why not use public data? If not, you are correct. I'll ask for clarification.
42m
revised Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
added 213 characters in body
51m
answered Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
1h
reviewed Approve Possible to generate a one time secret which all nodes on the distributed network can know but cannot pre compute
4h
comment RSA private key finding method
Maybe she beats Bob with $5 wrench until he tells her the private key.
7h
comment Fast hash for 64-bit inputs
What is a hash collision DoS? Are you worried that an attacker will send you a bunch of values that all result in the same hash, making your hash table inefficient?
17h
comment Fast hash for 64-bit inputs
For hash tables, you don't really need cryptographic hashes. So this should probably be migrated to Computer Science.
23h
revised How do substution-permutation networks work?
edited title
1d
comment Need CryptDeriveKey AES implementation in C#
I don't really understand what your question is.
Jan
30
comment When all shares of a secret are given to adversary as a permuted matrix
@user153465 simply multiplying the shares with the same index would give you a sharing of the product. There are 2 problems with this, however. 1) The threshold has doubled. So if you were doing $t=n/2$ out of $n$ secret sharing, you would now need all $n$ shares to reconstruct. 2) The coefficients of the polynomial that results in the sharing of the product are not independent. The only way we know around this is to use interaction.
Jan
30
comment When all shares of a secret are given to adversary as a permuted matrix
@user153465 you mean so that you end up with shares of the product of the two secrets?
Jan
28
comment When all shares of a secret are given to adversary as a permuted matrix
Is the permutation fixed or how is the permutation generated?
Jan
28
comment When all shares of a secret are given to adversary as a permuted matrix
Are you doing $n$-out-of-$n$ secret sharing or $t$-out-of-$n$ with $t<n$?
Jan
26
comment How can comparision like (>=,==) be done using homomorphic encryption?
So both operands are encrypted and the result is an encrypted bit? Is that the scenario?
Jan
25
comment Hamiltonian Path Zero Knowledge Proof using Commitments to a Series of Edges
Is your method really zero-knowledge? It leaks $|E|$.
Jan
23
comment Why reveal MAC-keys after using them?
no, I'm not the author.
Jan
22
comment Plaintext and ciphertext block sizes
Both plaintexts and ciphertexts have an upper bound of $n-1$.