# Utility Functions for Cryptography¶

Miscellaneous utility functions for cryptographic purposes.

AUTHORS:

• Minh Van Nguyen (2009-12): initial version with the following functions: ascii_integer, ascii_to_bin, bin_to_ascii, has_blum_prime, is_blum_prime, least_significant_bits, random_blum_prime.

sage.crypto.util.ascii_integer(B)

Return the ASCII integer corresponding to the binary string B.

INPUT:

• B – a non-empty binary string or a non-empty list of bits. The number of bits in B must be 8.

OUTPUT:

• The ASCII integer corresponding to the 8-bit block B.

EXAMPLES:

The ASCII integers of some binary strings:

sage: from sage.crypto.util import ascii_integer
sage: bin = BinaryStrings()
sage: B = bin.encoding("A"); B
01000001
sage: ascii_integer(B)
65
sage: B = bin.encoding("C"); list(B)
[0, 1, 0, 0, 0, 0, 1, 1]
sage: ascii_integer(list(B))
67
sage: ascii_integer("01000100")
68
sage: ascii_integer([0, 1, 0, 0, 0, 1, 0, 1])
69

sage.crypto.util.ascii_to_bin(A)

Return the binary representation of the ASCII string A.

INPUT:

• A – a string or list of ASCII characters.

OUTPUT:

• The binary representation of A.

ALGORITHM:

Let $$A = a_0 a_1 \cdots a_{n-1}$$ be an ASCII string, where each $$a_i$$ is an ASCII character. Let $$c_i$$ be the ASCII integer corresponding to $$a_i$$ and let $$b_i$$ be the binary representation of $$c_i$$. The binary representation $$B$$ of $$A$$ is $$B = b_0 b_1 \cdots b_{n-1}$$.

EXAMPLES:

The binary representation of some ASCII strings:

sage: from sage.crypto.util import ascii_to_bin
sage: ascii_to_bin("A")
01000001
sage: ascii_to_bin("Abc123")
010000010110001001100011001100010011001000110011


The empty string is different from the string with one space character. For the empty string and the empty list, this function returns the same result:

sage: from sage.crypto.util import ascii_to_bin
sage: ascii_to_bin("")

sage: ascii_to_bin(" ")
00100000
sage: ascii_to_bin([])


This function also accepts a list of ASCII characters. You can also pass in a list of strings:

sage: from sage.crypto.util import ascii_to_bin
sage: ascii_to_bin(["A", "b", "c", "1", "2", "3"])
010000010110001001100011001100010011001000110011
sage: ascii_to_bin(["A", "bc", "1", "23"])
010000010110001001100011001100010011001000110011

sage.crypto.util.bin_to_ascii(B)

Return the ASCII representation of the binary string B.

INPUT:

• B – a non-empty binary string or a non-empty list of bits. The number of bits in B must be a multiple of 8.

OUTPUT:

• The ASCII string corresponding to B.

ALGORITHM:

Consider a block of bits $$B = b_0 b_1 \cdots b_{n-1}$$ where each sub-block $$b_i$$ is a binary string of length 8. Then the total number of bits is a multiple of 8 and is given by $$8n$$. Let $$c_i$$ be the integer representation of $$b_i$$. We can consider $$c_i$$ as the integer representation of an ASCII character. Then the ASCII representation $$A$$ of $$B$$ is $$A = a_0 a_1 \cdots a_{n-1}$$.

EXAMPLES:

Convert some ASCII strings to their binary representations and recover the ASCII strings from the binary representations:

sage: from sage.crypto.util import ascii_to_bin
sage: from sage.crypto.util import bin_to_ascii
sage: A = "Abc"
sage: B = ascii_to_bin(A); B
010000010110001001100011
sage: bin_to_ascii(B)
'Abc'
sage: bin_to_ascii(B) == A
True

sage: A = "123 \" #"
sage: B = ascii_to_bin(A); B
00110001001100100011001100100000001000100010000000100011
sage: bin_to_ascii(B)
'123 " #'
sage: bin_to_ascii(B) == A
True


This function also accepts strings and lists of bits:

sage: from sage.crypto.util import bin_to_ascii
sage: bin_to_ascii("010000010110001001100011")
'Abc'
sage: bin_to_ascii([0, 1, 0, 0, 0, 0, 0, 1])
'A'

sage.crypto.util.carmichael_lambda(n)

Return the Carmichael function of a positive integer n.

The Carmichael function of $$n$$, denoted $$\lambda(n)$$, is the smallest positive integer $$k$$ such that $$a^k \equiv 1 \pmod{n}$$ for all $$a \in \ZZ/n\ZZ$$ satisfying $$\gcd(a, n) = 1$$. Thus, $$\lambda(n) = k$$ is the exponent of the multiplicative group $$(\ZZ/n\ZZ)^{\ast}$$.

INPUT:

• n – a positive integer.

OUTPUT:

• The Carmichael function of n.

ALGORITHM:

If $$n = 2, 4$$ then $$\lambda(n) = \varphi(n)$$. Let $$p \geq 3$$ be an odd prime and let $$k$$ be a positive integer. Then $$\lambda(p^k) = p^{k - 1}(p - 1) = \varphi(p^k)$$. If $$k \geq 3$$, then $$\lambda(2^k) = 2^{k - 2}$$. Now consider the case where $$n > 3$$ is composite and let $$n = p_1^{k_1} p_2^{k_2} \cdots p_t^{k_t}$$ be the prime factorization of $$n$$. Then

$\lambda(n) = \lambda(p_1^{k_1} p_2^{k_2} \cdots p_t^{k_t}) = \text{lcm}(\lambda(p_1^{k_1}), \lambda(p_2^{k_2}), \dots, \lambda(p_t^{k_t}))$

EXAMPLES:

The Carmichael function of all positive integers up to and including 10:

sage: from sage.crypto.util import carmichael_lambda
sage: list(map(carmichael_lambda, [1..10]))
[1, 1, 2, 2, 4, 2, 6, 2, 6, 4]


The Carmichael function of the first ten primes:

sage: list(map(carmichael_lambda, primes_first_n(10)))
[1, 2, 4, 6, 10, 12, 16, 18, 22, 28]


Cases where the Carmichael function is equivalent to the Euler phi function:

sage: carmichael_lambda(2) == euler_phi(2)
True
sage: carmichael_lambda(4) == euler_phi(4)
True
sage: p = random_prime(1000, lbound=3, proof=True)
sage: k = randint(1, 1000)
sage: carmichael_lambda(p^k) == euler_phi(p^k)
True


A case where $$\lambda(n) \neq \varphi(n)$$:

sage: k = randint(3, 1000)
sage: carmichael_lambda(2^k) == 2^(k - 2)
True
sage: carmichael_lambda(2^k) == 2^(k - 2) == euler_phi(2^k)
False


Verifying the current implementation of the Carmichael function using another implementation. The other implementation that we use for verification is an exhaustive search for the exponent of the multiplicative group $$(\ZZ/n\ZZ)^{\ast}$$.

sage: from sage.crypto.util import carmichael_lambda
sage: n = randint(1, 500)
sage: c = carmichael_lambda(n)
sage: def coprime(n):
....:     return [i for i in range(n) if gcd(i, n) == 1]
sage: def znpower(n, k):
....:     L = coprime(n)
....:     return list(map(power_mod, L, [k]*len(L), [n]*len(L)))
sage: def my_carmichael(n):
....:     if n == 1:
....:         return 1
....:     for k in range(1, n):
....:         L = znpower(n, k)
....:         ones =  * len(L)
....:         T = [L[i] == ones[i] for i in range(len(L))]
....:         if all(T):
....:             return k
sage: c == my_carmichael(n)
True


Carmichael’s theorem states that $$a^{\lambda(n)} \equiv 1 \pmod{n}$$ for all elements $$a$$ of the multiplicative group $$(\ZZ/n\ZZ)^{\ast}$$. Here, we verify Carmichael’s theorem.

sage: from sage.crypto.util import carmichael_lambda
sage: n = randint(2, 1000)
sage: c = carmichael_lambda(n)
sage: ZnZ = IntegerModRing(n)
sage: M = ZnZ.list_of_elements_of_multiplicative_group()
sage: ones =  * len(M)
sage: P = [power_mod(a, c, n) for a in M]
sage: P == ones
True


REFERENCES:

sage.crypto.util.has_blum_prime(lbound, ubound)

Determine whether or not there is a Blum prime within the specified closed interval.

INPUT:

• lbound – positive integer; the lower bound on how small a Blum prime can be. The lower bound must be distinct from the upper bound.

• ubound – positive integer; the upper bound on how large a Blum prime can be. The lower bound must be distinct from the upper bound.

OUTPUT:

• True if there is a Blum prime p such that lbound <= p <= ubound. False otherwise.

ALGORITHM:

Let $$L$$ and $$U$$ be distinct positive integers. Let $$P$$ be the set of all odd primes $$p$$ such that $$L \leq p \leq U$$. Our main focus is on Blum primes, i.e. odd primes that are congruent to 3 modulo 4, so we assume that the lower bound $$L > 2$$. The closed interval $$[L, U]$$ has a Blum prime if and only if the set $$P$$ has a Blum prime.

EXAMPLES:

Testing for the presence of Blum primes within some closed intervals. The interval $$[4, 100]$$ has a Blum prime, the smallest such prime being 7. The interval $$[24, 28]$$ has no primes, hence no Blum primes.

sage: from sage.crypto.util import has_blum_prime
sage: from sage.crypto.util import is_blum_prime
sage: has_blum_prime(4, 100)
True
sage: for n in range(4, 100):
....:     if is_blum_prime(n):
....:         print(n)
....:         break
7
sage: has_blum_prime(24, 28)
False

sage.crypto.util.is_blum_prime(n)

Determine whether or not n is a Blum prime.

INPUT:

• n a positive prime.

OUTPUT:

• True if n is a Blum prime; False otherwise.

Let $$n$$ be a positive prime. Then $$n$$ is a Blum prime if $$n$$ is congruent to 3 modulo 4, i.e. $$n \equiv 3 \pmod{4}$$.

EXAMPLES:

Testing some integers to see if they are Blum primes:

sage: from sage.crypto.util import is_blum_prime
sage: from sage.crypto.util import random_blum_prime
sage: is_blum_prime(101)
False
sage: is_blum_prime(7)
True
sage: p = random_blum_prime(10**3, 10**5)
sage: is_blum_prime(p)
True

sage.crypto.util.least_significant_bits(n, k)

Return the k least significant bits of n.

INPUT:

• n – an integer.

• k – a positive integer.

OUTPUT:

• The k least significant bits of the integer n. If k=1, then return the parity bit of the integer n. Let $$b$$ be the binary representation of n, where $$m$$ is the length of the binary string $$b$$. If $$k \geq m$$, then return the binary representation of n.

EXAMPLES:

Obtain the parity bits of some integers:

sage: from sage.crypto.util import least_significant_bits
sage: least_significant_bits(0, 1)

sage: least_significant_bits(2, 1)

sage: least_significant_bits(3, 1)

sage: least_significant_bits(-2, 1)

sage: least_significant_bits(-3, 1)



Obtain the 4 least significant bits of some integers:

sage: least_significant_bits(101, 4)
[0, 1, 0, 1]
sage: least_significant_bits(-101, 4)
[0, 1, 0, 1]
sage: least_significant_bits(124, 4)
[1, 1, 0, 0]
sage: least_significant_bits(-124, 4)
[1, 1, 0, 0]


The binary representation of 123:

sage: n = 123; b = n.binary(); b
'1111011'
sage: least_significant_bits(n, len(b))
[1, 1, 1, 1, 0, 1, 1]

sage.crypto.util.random_blum_prime(lbound, ubound, ntries=100)

A random Blum prime within the specified bounds.

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

INPUT:

• lbound – positive integer; the lower bound on how small a random Blum prime $$p$$ can be. So we have 0 < lbound <= p <= ubound. The lower bound must be distinct from the upper bound.

• ubound – positive integer; the upper bound on how large a random Blum prime $$p$$ can be. So we have 0 < lbound <= p <= ubound. The lower bound must be distinct from the upper bound.

• ntries – (default: 100) the number of attempts to generate a random Blum prime. If ntries is a positive integer, then perform that many attempts at generating a random Blum prime. This might or might not result in a Blum prime.

OUTPUT:

• A random Blum prime within the specified lower and upper bounds.

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 Blum primes within the specified bounds.

EXAMPLES:

Choose a random prime and check that it is a Blum prime:

sage: from sage.crypto.util import random_blum_prime
sage: p = random_blum_prime(10**4, 10**5)
sage: is_prime(p)
True
sage: mod(p, 4) == 3
True