def test_ext_gcd(self):
self.assertEqual(ext_gcd(5, 7), (3, -2, 1))
self.assertEqual(ext_gcd(12, 10), (1, -1, 2))
self.assertEqual(ext_gcd(12, -18), (-1, -1, 6))
self.assertEqual(ext_gcd(-3, -1), (0, -1, 1))
for i in xrange(20):
x = random.randint(10, 100)
y = random.randint(10, 100)
(a, b, d) = ext_gcd(x, y)
self.assertEqual(a*x + b*y, d)
def test_ext_gcd(self):
self.assertEqual(ext_gcd(5, 7), (3, -2, 1))
self.assertEqual(ext_gcd(12, 10), (1, -1, 2))
self.assertEqual(ext_gcd(12, -18), (-1, -1, 6))
self.assertEqual(ext_gcd(-3, -1), (0, -1, 1))
for i in xrange(20):
x = random.randint(10, 100)
y = random.randint(10, 100)
(a, b, d) = ext_gcd(x, y)
self.assertEqual(a*x + b*y, d)
def linear_congruence(a, b, n):
"""Returns a list of the solutions x to the congruence a*x = b (mod n).
Input:
* a: int
* b: int
* n: int (n > 1)
Returns:
* solutions: list
Raises:
* ValueError: if n <= 1.
Examples:
>>> linear_congruence(10, 6, 12)
[3, 9]
>>> linear_congruence(12, 9, 15)
[2, 7, 12]
>>> linear_congruence(10, 3, 12)
[]
>>> linear_congruence(10, 3, 0)
Traceback (most recent call last):
...
ValueError: linear_congruence: Must have n >= 2.
Details:
The linear congruence a*x = b (mod n) has a solution if and only if d =
gcd(a, n) divides b. This function uses a straightforward application of
this theorem. See Theorem 3.7 from "Elementary Number Theory" By Jones
and Jones for details.
"""
if n < 2:
raise ValueError("linear_congruence: Must have n >= 2.")
(u, v, d) = ext_gcd(a, n)
# The congruence has a solution if and only if gcd(a, n) | b.
if b % d != 0:
return []
# x0 is our particular solution.
# There will be exactly d incongruent solutions modulo n.
x0 = b * u // d
solutions = [(x0 + k * n // d) % n for k in xrange(d)]
solutions.sort()
return solutions
def linear_congruence(a, b, n):
"""Returns a list of the solutions x to the congruence a*x = b (mod n).
Input:
* a: int
* b: int
* n: int (n > 1)
Returns:
* solutions: list
Raises:
* ValueError: if n <= 1.
Examples:
>>> linear_congruence(10, 6, 12)
[3, 9]
>>> linear_congruence(12, 9, 15)
[2, 7, 12]
>>> linear_congruence(10, 3, 12)
[]
>>> linear_congruence(10, 3, 0)
Traceback (most recent call last):
...
ValueError: linear_congruence: Must have n >= 2.
Details:
The linear congruence a*x = b (mod n) has a solution if and only if d =
gcd(a, n) divides b. This function uses a straightforward application of
this theorem. See Theorem 3.7 from "Elementary Number Theory" By Jones
and Jones for details.
"""
if n < 2:
raise ValueError("linear_congruence: Must have n >= 2.")
(u, v, d) = ext_gcd(a, n)
# The congruence has a solution if and only if gcd(a, n) | b.
if b % d != 0:
return []
# x0 is our particular solution.
# There will be exactly d incongruent solutions modulo n.
x0 = b*u//d
solutions = [(x0 + k*n//d) % n for k in xrange(d)]
solutions.sort()
return solutions