Blum-Goldwasser Probabilistic Encryption¶

The Blum-Goldwasser probabilistic public-key encryption scheme. This scheme was originally described in [BlumGoldwasser1985]. See also section 8.7.2 of [MenezesEtAl1996] and the Wikipedia article on this scheme.

REFERENCES:

[BlumGoldwasser1985]

M. Blum and S. Goldwasser. An Efficient Probabilistic Public-Key Encryption Scheme Which Hides All Partial Information. In Proceedings of CRYPTO 84 on Advances in Cryptology, pp. 289–299, Springer, 1985.

[MenezesEtAl1996]

(1, 2, 3, 4, 5, 6, 7, 8) A. J. Menezes, P. C. van Oorschot, and S. A. Vanstone. Handbook of Applied Cryptography. CRC Press, 1996.

AUTHORS:

Mike Hogan and David Joyner (2009-9-19): initial procedural version released as public domain software.
Minh Van Nguyen (2009-12): integrate into Sage as a class and relicense under the GPLv2+. Complete rewrite of the original version to follow the description contained in [MenezesEtAl1996].

class sage.crypto.public_key.blum_goldwasser.BlumGoldwasser¶

Bases: sage.crypto.cryptosystem.PublicKeyCryptosystem

The Blum-Goldwasser probabilistic public-key encryption scheme.

The Blum-Goldwasser encryption and decryption algorithms as described in encrypt() and decrypt(), respectively, make use of the least significant bit of a binary string. A related concept is the $k$ least significant bits of a binary string. For example, given a positive integer $n$ , let $b = b_0 b_1 \cdots b_{m-1}$ be the binary representation of $n$ so that $b$ is a binary string of length $m$ . Then the least significant bit of $n$ is $b_{m-1}$ . If $0 < k \leq m$ , then the $k$ least significant bits of $n$ are $b_{m-1-k} b_{m-k} \cdots b_{m-1}$ . The least significant bit of an integer is also referred to as its parity bit, because this bit determines whether the integer is even or odd. In the following example, we obtain the least significant bit of an integer:

sage: n = 123
sage: b = n.binary(); b
'1111011'
sage: n % 2
1
sage: b[-1]
'1'

Now find the 4 least significant bits of the integer $n = 123$ :

sage: b
'1111011'
sage: b[-4:]
'1011'

The last two examples could be worked through as follows:

sage: from sage.crypto.util import least_significant_bits
sage: least_significant_bits(123, 1)
[1]
sage: least_significant_bits(123, 4)
[1, 0, 1, 1]

EXAMPLES:

The following encryption/decryption example is taken from Example 8.57, pages 309–310 of [MenezesEtAl1996]:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser(); bg
The Blum-Goldwasser public-key encryption scheme.
sage: p = 499; q = 547
sage: pubkey = bg.public_key(p, q); pubkey
272953
sage: prikey = bg.private_key(p, q); prikey
(499, 547, -57, 52)
sage: P = "10011100000100001100"
sage: C = bg.encrypt(P, pubkey, seed=159201); C
([[0, 0, 1, 0], [0, 0, 0, 0], [1, 1, 0, 0], [1, 1, 1, 0], [0, 1, 0, 0]], 139680)
sage: M = bg.decrypt(C, prikey); M
[[1, 0, 0, 1], [1, 1, 0, 0], [0, 0, 0, 1], [0, 0, 0, 0], [1, 1, 0, 0]]
sage: M = "".join(map(lambda x: str(x), flatten(M))); M
'10011100000100001100'
sage: M == P
True

Generate a pair of random public/private keys. Use the public key to encrypt a plaintext. Then decrypt the resulting ciphertext using the private key. Finally, compare the decrypted message with the original plaintext.

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: from sage.crypto.util import bin_to_ascii
sage: bg = BlumGoldwasser()
sage: pubkey, prikey = bg.random_key(10**4, 10**6)
sage: P = "A fixed plaintext."
sage: C = bg.encrypt(P, pubkey)
sage: M = bg.decrypt(C, prikey)
sage: bin_to_ascii(flatten(M)) == P
True

If $(p, q, a, b)$ is a private key, then $n = pq$ is the corresponding public key. Furthermore, we have $\gcd(p, q) = ap + bq = 1$ .

sage: p, q, a, b = prikey
sage: pubkey == p * q
True
sage: gcd(p, q) == a*p + b*q == 1
True

decrypt(C, K)¶

Apply the Blum-Goldwasser scheme to decrypt the ciphertext C using the private key K.

INPUT:

C – a ciphertext resulting from encrypting a plaintext using the Blum-Goldwasser encryption algorithm. The ciphertext $C$ must be of the form $C = (c_1, c_2, \dots, c_t, x_{t+1})$ . Each $c_i$ is a sub-block of binary string and $x_{t+1}$ is the result of the $t+1$ -th iteration of the Blum-Blum-Shub algorithm.
K – a private key $(p, q, a, b)$ where $p$ and $q$ are distinct Blum primes and $\gcd(p, q) = ap + bq = 1$ .

OUTPUT:

The plaintext resulting from decrypting the ciphertext C using the Blum-Goldwasser decryption algorithm.

ALGORITHM:

The Blum-Goldwasser decryption algorithm is described in Algorithm 8.56, page 309 of [MenezesEtAl1996]. The algorithm works as follows:

Let $C$ be the ciphertext $C = (c_1, c_2, \dots, c_t, x_{t+1})$ . Then $t$ is the number of ciphertext sub-blocks and $h$ is the length of each binary string sub-block $c_i$ .
Let $(p, q, a, b)$ be the private key whose corresponding public key is $n = pq$ . Note that $\gcd(p, q) = ap + bq = 1$ .
Compute $d_1 = ((p + 1) / 4)^{t+1} \bmod{(p - 1)}$ .
Compute $d_2 = ((q + 1) / 4)^{t+1} \bmod{(q - 1)}$ .
Let $u = x_{t+1}^{d_1} \bmod p$ .
Let $v = x_{t+1}^{d_2} \bmod q$ .
Compute $x_0 = vap + ubq \bmod n$ .
For from 1 to , do:
1. Compute $x_i = x_{t-1}^2 \bmod n$ .
2. Let $p_i$ be the $h$ least significant bits of $x_i$ .
3. Compute $m_i = p_i \oplus c_i$ .
The plaintext is $m = m_1 m_2 \cdots m_t$ .

EXAMPLES:

The following decryption example is taken from Example 8.57, pages 309–310 of [MenezesEtAl1996]. Here we decrypt a binary string:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: p = 499; q = 547
sage: C = ([[0, 0, 1, 0], [0, 0, 0, 0], [1, 1, 0, 0], [1, 1, 1, 0], [0, 1, 0, 0]], 139680)
sage: K = bg.private_key(p, q); K
(499, 547, -57, 52)
sage: P = bg.decrypt(C, K); P
[[1, 0, 0, 1], [1, 1, 0, 0], [0, 0, 0, 1], [0, 0, 0, 0], [1, 1, 0, 0]]

Convert the plaintext sub-blocks into a binary string:

sage: bin = BinaryStrings()
sage: bin(flatten(P))
10011100000100001100

Decrypt a longer ciphertext and convert the resulting plaintext into an ASCII string:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: from sage.crypto.util import bin_to_ascii
sage: bg = BlumGoldwasser()
sage: p = 78307; q = 412487
sage: K = bg.private_key(p, q)
sage: C = ([[1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0], \
...   [1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1], \
...   [0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0], \
...   [0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1], \
...   [1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0], \
...   [0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1], \
...   [1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0], \
...   [1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1], \
...   [0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0], \
...   [1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1], \
...   [1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1], \
...   [1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0], \
...   [0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1]], 3479653279)
sage: P = bg.decrypt(C, K)
sage: bin_to_ascii(flatten(P))
'Blum-Goldwasser encryption'

TESTS:

The private key $K = (p, q, a, b)$ must be such that $p$ and $q$ are distinct Blum primes. Even if $p$ and $q$ pass this criterion, they must also satisfy the requirement $\gcd(p, q) = ap + bq = 1$ .

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: C = ([[0, 0, 1, 0], [0, 0, 0, 0], [1, 1, 0, 0], [1, 1, 1, 0], [0, 1, 0, 0]], 139680)
sage: K = (7, 7, 1, 2)
sage: bg.decrypt(C, K)
...
ValueError: p and q must be distinct Blum primes.
sage: K = (7, 23, 1, 2)
sage: bg.decrypt(C, K)
...
ValueError: a and b must satisfy gcd(p, q) = ap + bq = 1.
sage: K = (11, 29, 8, -3)
sage: bg.decrypt(C, K)
...
ValueError: p and q must be distinct Blum primes.

encrypt(P, K, seed=None)¶

Apply the Blum-Goldwasser scheme to encrypt the plaintext P using the public key K.

INPUT:

P – a non-empty string of plaintext. The string "" is an empty string, whereas " " is a string consisting of one white space character. The plaintext can be a binary string or a string of ASCII characters. Where P is an ASCII string, then P is first encoded as a binary string prior to encryption.
K – a public key, which is the product of two Blum primes.
seed – (default: None) if $p$ and $q$ are Blum primes and $n = pq$ is a public key, then seed is a quadratic residue in the multiplicative group $(\ZZ/n\ZZ)^{\ast}$ . If seed=None, then the function would generate its own random quadratic residue in $(\ZZ/n\ZZ)^{\ast}$ . Where a value for seed is provided, it is your responsibility to ensure that the seed is a quadratic residue in the multiplicative group $(\ZZ/n\ZZ)^{\ast}$ .

OUTPUT:

The ciphertext resulting from encrypting P using the public key K. The ciphertext $C$ is of the form $C = (c_1, c_2, \dots, c_t, x_{t+1})$ . Each $c_i$ is a sub-block of binary string and $x_{t+1}$ is the result of the $t+1$ -th iteration of the Blum-Blum-Shub algorithm.

ALGORITHM:

The Blum-Goldwasser encryption algorithm is described in Algorithm 8.56, page 309 of [MenezesEtAl1996]. The algorithm works as follows:

Let $n$ be a public key, where $n = pq$ is the product of two distinct Blum primes $p$ and $q$ .
Let $k = \lfloor \log_2(n) \rfloor$ and $h = \lfloor \log_2(k) \rfloor$ .
Let $m = m_1 m_2 \cdots m_t$ be the message (plaintext) where each $m_i$ is a binary string of length $h$ .
Choose a random seed $x_0$ , which is a quadratic residue in the multiplicative group $(\ZZ/n\ZZ)^{\ast}$ . That is, choose a random $r \in (\ZZ/n\ZZ)^{\ast}$ and compute $x_0 = r^2 \bmod n$ .
For from 1 to , do:
1. Let $x_i = x_{i-1}^2 \bmod n$ .
2. Let $p_i$ be the $h$ least significant bits of $x_i$ .
3. Let $c_i = p_i \oplus m_i$ .
Compute $x_{t+1} = x_t^2 \bmod n$ .
The ciphertext is $c = (c_1, c_2, \dots, c_t, x_{t+1})$ .

The value $h$ in the algorithm is the sub-block length. If the binary string representing the message cannot be divided into blocks of length $h$ each, then other sub-block lengths would be used instead. The sub-block lengths to fall back on are in the following order: 16, 8, 4, 2, 1.

EXAMPLES:

The following encryption example is taken from Example 8.57, pages 309–310 of [MenezesEtAl1996]. Here, we encrypt a binary string:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: p = 499; q = 547; n = p * q
sage: P = "10011100000100001100"
sage: C = bg.encrypt(P, n, seed=159201); C
([[0, 0, 1, 0], [0, 0, 0, 0], [1, 1, 0, 0], [1, 1, 1, 0], [0, 1, 0, 0]], 139680)

Convert the ciphertext sub-blocks into a binary string:

sage: bin = BinaryStrings()
sage: bin(flatten(C[0]))
00100000110011100100

Now encrypt an ASCII string. The result is random; no seed is provided to the encryption function so the function generates its own random seed:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: K = 32300619509
sage: P = "Blum-Goldwasser encryption"
sage: bg.encrypt(P, K)  # random
([[1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0], \
[1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1], \
[0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1], \
[1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0], \
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1], \
[1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0], \
[1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1], \
[0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0], \
[1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1], \
[1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1], \
[1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0], \
[0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1]], 3479653279)

TESTS:

The plaintext cannot be an empty string.

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: bg.encrypt("", 3)
...
ValueError: The plaintext cannot be an empty string.

private_key(p, q)¶

Return the Blum-Goldwasser private key corresponding to the distinct Blum primes p and q.

INPUT:

p – a Blum prime.
q – a Blum prime.

OUTPUT:

The Blum-Goldwasser private key $(p, q, a, b)$ where $\gcd(p, q) = ap + bq = 1$ .

Both p and q must be distinct Blum primes. Let $p$ be a positive prime. Then $p$ is a Blum prime if $p$ is congruent to 3 modulo 4, i.e. $p \equiv 3 \pmod{4}$ .

EXAMPLES:

Obtain two distinct Blum primes and compute the Blum-Goldwasser private key corresponding to those two Blum primes:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: from sage.crypto.util import is_blum_prime
sage: bg = BlumGoldwasser()
sage: P = primes_first_n(10); P
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
sage: map(is_blum_prime, P)
[False, True, False, True, True, False, False, True, True, False]
sage: bg.private_key(19, 23)
(19, 23, -6, 5)

Choose two distinct random Blum primes, compute the Blum-Goldwasser private key corresponding to those two primes, and test that the resulting private key $(p, q, a, b)$ satisfies $\gcd(p, q) = ap + bq = 1$ :

sage: from sage.crypto.util import random_blum_prime
sage: p = random_blum_prime(10**4, 10**5)
sage: q = random_blum_prime(10**4, 10**5)
sage: while q == p:
...       q = random_blum_prime(10**4, 10**5)
...
sage: p, q, a, b = bg.private_key(p, q)
sage: gcd(p, q) == a*p + b*q == 1
True

TESTS:

Both of the input p and q must be distinct Blum primes.

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: bg.private_key(78307, 78307)
...
ValueError: p and q must be distinct Blum primes.
sage: bg.private_key(7, 4)
...
ValueError: p and q must be distinct Blum primes.

public_key(p, q)¶

Return the Blum-Goldwasser public key corresponding to the distinct Blum primes p and q.

INPUT:

p – a Blum prime.
q – a Blum prime.

OUTPUT:

The Blum-Goldwasser public key $n = pq$ .

Both p and q must be distinct Blum primes. Let $p$ be a positive prime. Then $p$ is a Blum prime if $p$ is congruent to 3 modulo 4, i.e. $p \equiv 3 \pmod{4}$ .

EXAMPLES:

Obtain two distinct Blum primes and compute the Blum-Goldwasser public key corresponding to those two Blum primes:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: from sage.crypto.util import is_blum_prime
sage: bg = BlumGoldwasser()
sage: P = primes_first_n(10); P
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
sage: map(is_blum_prime, P)
[False, True, False, True, True, False, False, True, True, False]
sage: bg.public_key(3, 7)
21

Choose two distinct random Blum primes, compute the Blum-Goldwasser public key corresponding to those two primes, and test that the public key factorizes into Blum primes:

sage: from sage.crypto.util import random_blum_prime
sage: p = random_blum_prime(10**4, 10**5)
sage: q = random_blum_prime(10**4, 10**5)
sage: while q == p:
...       q = random_blum_prime(10**4, 10**5)
...
sage: n = bg.public_key(p, q)
sage: f = factor(n)
sage: is_blum_prime(f[0][0]); is_blum_prime(f[1][0])
True
True
sage: p * q == f[0][0] * f[1][0]
True

TESTS:

The input p and q must be distinct Blum primes.

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: bg.public_key(3, 3)
...
ValueError: p and q must be distinct Blum primes.
sage: bg.public_key(23, 29)
...
ValueError: p and q must be distinct Blum primes.

random_key(lbound, ubound, ntries=100)¶

Return a pair of random public and private keys.

INPUT:

lbound – positive integer; the lower bound on how small each random Blum prime $p$ and $q$ can be. So we have 0 < lower_bound <= p, q <= upper_bound. The lower bound must be distinct from the upper bound.
ubound – positive integer; the upper bound on how large each random Blum prime $p$ and $q$ can be. So we have 0 < lower_bound <= p, q <= upper_bound. The lower bound must be distinct from the upper bound.
ntries – (default: 100) the number of attempts to generate a random public/private key pair. If ntries is a positive integer, then perform that many attempts at generating a random public/private key pair.

OUTPUT:

A random public key and its corresponding private key. Each randomly chosen $p$ and $q$ are guaranteed to be Blum primes. The public key is $n = pq$ , and the private key is $(p, q, a, b)$ where $\gcd(p, q) = ap + bq = 1$ .

ALGORITHM:

The key generation algorithm is described in Algorithm 8.55, page 308 of [MenezesEtAl1996]. The algorithm works as follows:

Let $p$ and $q$ be distinct large random primes, each congruent to 3 modulo 4. That is, $p$ and $q$ are Blum primes.
Let $n = pq$ be the product of $p$ and $q$ .
Use the extended Euclidean algorithm to compute integers $a$ and $b$ such that $\gcd(p, q) = ap + bq = 1$ .
The public key is $n$ and the corresponding private key is $(p, q, a, b)$ .

Note

Beware that there might not be any primes between the lower and upper bounds. So make sure that these two bounds are “sufficiently” far apart from each other for there to be primes congruent to 3 modulo 4. In particular, there should be at least two distinct primes within these bounds, each prime being congruent to 3 modulo 4.

EXAMPLES:

Choosing a random pair of public and private keys. We then test to see if they satisfy the requirements of the Blum-Goldwasser scheme:

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: from sage.crypto.util import is_blum_prime
sage: bg = BlumGoldwasser()
sage: pubkey, prikey = bg.random_key(10**4, 10**5)
sage: p, q, a, b = prikey
sage: is_blum_prime(p); is_blum_prime(q)
True
True
sage: p == q
False
sage: pubkey == p*q
True
sage: gcd(p, q) == a*p + b*q == 1
True

TESTS:

Make sure that there is at least one Blum prime between the lower and upper bounds. In the following example, we have lbound=24 and ubound=30 with 29 being the only prime within those bounds. But 29 is not a Blum prime.

sage: from sage.crypto.public_key.blum_goldwasser import BlumGoldwasser
sage: bg = BlumGoldwasser()
sage: pubkey, privkey = bg.random_key(24, 30)
...
ValueError: No Blum primes within the specified closed interval.

Blum-Goldwasser Probabilistic Encryption¶

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