New answers tagged

0 votes
Accepted

BV FHE Scheme symbolic polynomial

Note that $$\left(b-\sum a[i]x[i]\right)\cdot\left(b'-\sum a'[i]x[i]\right) =bb'-b\sum a'[i]x[i]-b'\sum a[i]x[i] +\sum\sum a[i]a'[j]x[i]x[j]$$ and we can extract the coefficients $$h_0=bb'$$ $$h_i=-ba'...
  • 18.1k
1 vote

Paillier cryptosystem break with random number

Yes, it would be possible to decrypt the ciphertext of the Paillier cryptosystem if the random number $r$ leaked. Encryption if per $c=(g^m)(r^n)\bmod n^2$. We know that $g=n+1$. Thus $c=((n+1)^m)(r^n)...
  • 132k
2 votes

Paillier cryptosystem break with random number

Once you know $r$, you can remove it from the ciphertext. If $c=(1+n)^m\bmod n^2$ then $m=(c-1)/n\mod n.$

Top 50 recent answers are included