# Tag Info

## Hot answers tagged paillier

4

I'm not sure if I understand what you are asking, so I'll clarify what I am about to answer. We are given two ciphertexts and we want to know if they encrypt the same plaintext or if they encrypt different plaintexts, and we want to do this without revealing anything but this fact. Then, using the additive homomorphism, it's possible to compute $c=enc(r\... 1 Let$c_a$be the encrypted version of$a$, and$c_b$be the encrypted version of$b$. What you want to compute is$c_c$which is the encrypted version of$a-b$, so that when you decrypt$c_c$you get$c=a-b$. Paillier supports a homomomorphic addtion of ciphertexts to get an encrypted version of the sum. The actual mathematical operation it takes to get ... 1 What you're missing is the fact that your$c$value can get waaay beyond what the library is expecting there and thus issues an error which can be read as "your value is too large". The solution is simple: Reduce the multiplication result$\bmod N^2$, where$N=pq\$ is the actual value of your modulus. The code-line which you would need to add there would ...

1

I do not think what you want is possible with any simple solution - by simple, I mean computationally less expensive than downloading and decrypting all the ciphertexts. Basically, multiplicative homomorphism allows you to check some algebraic "OR": if there is a single 0, then the product of all the plaintexts will be 0. Additive homomorphism, on the other ...

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