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1d
comment Given $g^a, g^b, g^c, g^{1/b}$, is it hard to distinguish $e(g, g)^{abc}$ from a random value?
@cygnusv report link does not work
May
12
comment Possible to check if $a \in \mathrm{QR}_n$?
@DrLecter But since the adversary knows that $Jacobi(a)=1$ it means that there exists for sure $y$ such that $y^2=x \mod n$. So adversary can decide whethera has QR since it known that $a \in \{\mathbb{J}=1\}$
May
5
comment How to calculate if probability is negligible or not
It's weird that the sum of one nn and on n probability is nn while for this for the product does not hold.
May
5
comment security proof in pairing based cryptography
@FlorianBourse even if there is a correlation in $C_x-C_i$, still you have to recover D
May
5
comment security proof in pairing based cryptography
It is more than straightforward i think following the reductionist approach
May
5
comment security proof in pairing based cryptography
what is $C_x$? Of course if you know $C_x$.What is known to the attacker?What is public and secret?
Apr
27
comment Paillier cryptosystem preserve ordering of sums for two integer sequences
@Alexandros The great about Paillier cryptosystem is the fact that users can encrypt with different randomness under the same public key that will decrypt to the correct plaintext under the single decryption key because $r^{N\lambda}=1 , \forall r \in \mathbb{Z}_{N^*}$
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
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
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
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 $\phi(N)$
Dec
28
comment Why is proof-by-reduction needed (for Elgamal proof of security, for example)?
@Maeher i can't get how you start building your distinguisher since you state:compute x. In order to compute x you break DL, which is impossible for computationally bounded adversary.
Dec
10
comment Proving the semantic security of the One Time pad
There is.I edit my answer
Nov
10
comment Best group if one wants the discrete log problem to be hard?
@RickyDemer why #2 is easy?