Tag Info

Hot answers tagged

5

Well, the problem is with logical OR and subtraction (which Pallier can also do), you've got FHE; that is, you can compute any combinatorial function of encrypted (binary) inputs. Here's how it works, you can construct the NAND function: $NAND(x, y) = (Enc(1) - x)\ OR\ (Enc(1) - y)$ If we limit $x$ and $y$ to being either encrypted 0, or encrypted 1, ...


4

No, exactly equal length of primes $p$ and $q$ is not mandatory for security (or proper functioning) in the Pailler cryptosystem. Sufficient requirements are that $p$ and $q$ are prime, $N=p q$ is hard to factor, and $\gcd(p q,(p-1)(q-1))=1$. The requirement that $p$ and $q$ are of exactly equal size is usually made in the Pailler cryptosystem because this ...


3

The requirement is that your element $g$ is in $\mathbb{Z}_{n^2}^*$ and not in $(\mathbb{Z}_{n}^*)^2$. The set $\mathbb{Z}_{n^2}^*$ is the set of integers smaller than $n^2$ that are relatively prime to $n^2$, i.e., you require an element $g$ from $\mathbb{Z}_{n^2}$ such that $\gcd(g,n^2)=1$. $(\mathbb{Z}_{n}^*)^2$ on the other hand is the set of pairs ...


1

In Paillier, as you note, multiplication in the ciphertext domain translates to addition in the plaintext domain. Thanks to the algebraic structure behind Paillier what you can do to get subtraction is use the multiplicative. This works fine when the result is positive. When the result is negative, however, you would like to return that value, but what ...


1

In your question, you already pointed out, that the necessary condition is More generally largest prime minus one does not consists of smallest prime as a prime factor therefore, it is sufficient to just check if $p$ divides $q-1$ by computing division. You can just verify this condition during the key generation. I don't know of a more efficient ...


1

Since $p$ and $q$ are primes, the only factors you needs to rule out are those two numbers. Suppose $p$ divides $(p-1)(q-1)$. Then it divides either $p-1$ (clearly not true) or $q-1$. The latter means $q-1 = p \cdot x$, for some $x \ge 2$ (if $x = 1$ either $p$ or $q$ is even, which is only possible if the numbers are 2 and 3). However, then $q \ge 2p+1$, ...



Only top voted, non community-wiki answers of a minimum length are eligible