def test_chinese_preconditioned(self):
moduli = [3, 5, 7]
preconditioning_data = crt_preconditioning_data(moduli)
system = [(2, 3), (3, 5), (2, 7)]
solution = chinese_preconditioned(system, preconditioning_data)
self.assertEqual(solution, 23)
system = [(1, 3), (2, 5), (4, 7)]
solution = chinese_preconditioned(system, preconditioning_data)
self.assertEqual(solution, 67)
def test_chinese_preconditioned(self):
moduli = [3, 5, 7]
preconditioning_data = crt_preconditioning_data(moduli)
system = [(2, 3), (3, 5), (2, 7)]
solution = chinese_preconditioned(system, preconditioning_data)
self.assertEqual(solution, 23)
system = [(1, 3), (2, 5), (4, 7)]
solution = chinese_preconditioned(system, preconditioning_data)
self.assertEqual(solution, 67)
def sqrts_mod_n(a, n, n_factorization=None):
"""Returns the solutions x to the congruence x**2 = a (mod n).
Given integers a and n, this function returns all solutions to the
congruence x**2 = a (mod n).
Input:
* a: int
A square modulo n.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of n.
Returns:
* roots: list
A sorted list of all square roots of a modulo n.
Examples:
>>> sqrts_mod_n(3, 11**3*23**3)
[16152263, 28026114, 150110933, 161984784]
>>> sqrts_mod_n(49, 53**3*61**4)
[7, 721465236980, 1339862033577, 2061327270550]
>>> sqrts_mod_n(-1, 5**3*7**2)
[]
"""
if n_factorization is None:
n_factorization = factor.factor(n)
congruences = []
moduli = []
for (p, k) in n_factorization:
roots = _sqrts_mod_pk(a, p, k)
pk = p**k
pairs = [(r, pk) for r in roots]
congruences.append(pairs)
moduli.append(pk)
if len(congruences) == 1:
values = [r for (r, _) in congruences[0]]
else:
preconditioning_data = crt_preconditioning_data(moduli)
values = []
for system in itertools.product(*congruences):
values.append(chinese_preconditioned(system, preconditioning_data))
values.sort()
return values
def sqrts_mod_n(a, n, n_factorization=None):
"""Returns the solutions x to the congruence x**2 = a (mod n).
Given integers a and n, this function returns all solutions to the
congruence x**2 = a (mod n).
Input:
* a: int
A square modulo n.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of n.
Returns:
* roots: list
A sorted list of all square roots of a modulo n.
Examples:
>>> sqrts_mod_n(3, 11**3*23**3)
[16152263, 28026114, 150110933, 161984784]
>>> sqrts_mod_n(49, 53**3*61**4)
[7, 721465236980, 1339862033577, 2061327270550]
>>> sqrts_mod_n(-1, 5**3*7**2)
[]
"""
if n_factorization is None:
n_factorization = factor.factor(n)
congruences = []
moduli = []
for (p, k) in n_factorization:
roots = _sqrts_mod_pk(a, p, k)
pk = p**k
pairs = [(r, pk) for r in roots]
congruences.append(pairs)
moduli.append(pk)
if len(congruences) == 1:
values = [r for (r, _) in congruences[0]]
else:
preconditioning_data = crt_preconditioning_data(moduli)
values = []
for system in itertools.product(*congruences):
values.append(chinese_preconditioned(system, preconditioning_data))
values.sort()
return values
def sqrts_mod_n(a, n, n_factorization=None):
"""
Returns all solutions x, 1 <= x <= n, to the congruence x^2 = a (mod n).
Input:
* a: int
A square modulo n.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of n.
Output:
* roots: list
A list of all square roots of a modulo n.
"""
if n_factorization is None:
n_factorization = rosemary.number_theory.factorization.factor(n)
congruences = []
moduli = []
for (p, k) in n_factorization:
roots = sqrts_mod_pk(a, p, k)
pk = p**k
pairs = [(r, pk) for r in roots]
congruences.append(pairs)
moduli.append(pk)
if len(congruences) == 1:
values = [r for (r, _) in congruences[0]]
else:
preconditioning_data = crt_preconditioning_data(moduli)
values = []
for system in itertools.product(*congruences):
values.append(chinese_preconditioned(system, preconditioning_data))
values.sort()
return values