Ejemplo n.º 1
0
    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)
Ejemplo n.º 2
0
    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)
Ejemplo n.º 3
0
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
Ejemplo n.º 4
0
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