User:Cspan64/Sandbox

This user comes from Germany.

This user comes from Baden-Württemberg.

Key generation

To allow for decryption, the modulus used in Blum-Goldwasser encryption should be a Blum integer. This value is generated in the same manner as an RSA modulus, except that the prime factors $(p,q)$ must be congruent to 3 mod 4. (See RSA, key generation for details.)

Alice generates two large prime numbers $p\,$ and $q\,$ such that $p\neq q$ , randomly and independently of each other, where $(p,q)\equiv 3$ mod $4$ .
Alice computes $N=pq$ .
Alice can also precompute now $q^{-1}\,$ and $p^{-1}\,$ , so that

$ap+bq=1$ (Bezout's identity) using the extended euclidean algorithm.

The public key is $N$ . The secret key is the factorization $(p,q)$ .

Message decryption

Alice receives $(c_{0},\dots ,c_{L-1}),y$ . She can recover $m$ using the following procedure:

Using the prime factorization $(p,q)$ , Alice computes $r_{p}=y^{((p+1)/4)^{L}}~mod~(p-1)$ and $r_{q}=y^{((q+1)/4)^{L}}~mod~(q-1)$ .
Compute the initial seed $x_{0}=q(q^{-1}~{mod}~p)r_{p}+p(p^{-1}~{mod}~q)r_{q}~{mod}~N$
From $x_{0}$ , recompute the bit-vector ${\vec {b}}$ using the BBS generator, as in the encryption algorithm.
Compute the plaintext by XORing the keystream with the ciphertext: ${\vec {m}}={\vec {c}}\oplus {\vec {b}}$ .

Alice recovers the plaintext $m=(m_{0},\dots ,m_{L-1})$ .

Inversion of 3 x 3 matrices

The cofactor equation listed above yields the following result for 3 x 3 matrices. Inversion of these matrices can be done quite easily as follows:

    $A^{-1}={\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\\\end{bmatrix}}^{-1}={\frac {1}{|A|}}{\begin{bmatrix}ei-fh&ch-bi&bf-ce\\fg-di&ai-cg&cd-af\\dh-eg&bg-ah&ae-bd\end{bmatrix}}$