Reputation
1,490
Top tag
Next privilege 1,500 Rep.
Approve tag wiki edits
Badges
1 9 22
Newest
 Caucus
Impact
~71k people reached

Apr
16
revised What is a Key Derivation Function?
added 82 characters in body
Apr
16
revised What is a Key Derivation Function?
added 82 characters in body
Apr
16
answered What is a Key Derivation Function?
Apr
8
comment Altering the message space of Paillier
@FlorianBourse this can be done by any homomorphic scheme since they are not supposed to be CCA secure. Meaning, the adversary can alter on its will the ciphertext
Apr
8
accepted Altering the message space of Paillier
Apr
7
asked Altering the message space of Paillier
Apr
7
comment Fault-based transition for crypto proof (a la Shoup) with big probability of fault - does it work?
Well, the assumption is that as long the fault event does not happen then adversary cannot identify that the simulator cheats by not giving her the true transcripts of the original game. This event F is often of this form: [F: A breaks a problem as DDH, CDH, DL, etc] I cannot get why you want to make an assumption that this even happens not negligibly since if this is the case then everything is broken....
Apr
1
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
@Thomas you mean that you can find the inverse moduli $N$ but you do not which one is the correct one.
Apr
1
accepted Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
Apr
1
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
But $v \equiv u^{-1} \mod N$ also which is easy computable
Apr
1
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
@Aleph Then why finding $x \mod N$ from $x^u \mod N$ is difficult? You compute $v=u^{-1}$ and then $x= (x^{u})^v=x^{uu^{-1}} \mod N$
Apr
1
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
Does this imply that finding inverses mod $N=pq$ is also hard as long you don't know the factorization of $N$
Mar
31
revised Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
deleted 9 characters in body
Mar
31
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
The adversary knows $N^2$, so i guess it is not hard to learn the inverse of $u$ with the extended Euclidean.
Mar
31
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
Yes but my concern is that i want the inverse not moduli $\phi(N^2)$ but $(N^2)$
Mar
31
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
The solution to the RSA problem as you said is the inverse of $e$ moduli $\phi(N)=(p-1)(q-1)$. And it is believed that if the factorization $(p,q)$ is not known you cannot find the inverse.
Mar
31
comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
RSA problem states that it is difficult to compute multiplicative inverses moduli $\phi(N)$
Mar
31
revised Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
deleted 47 characters in body
Mar
31
revised Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
deleted 15 characters in body
Mar
31
revised Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$?
edited body