本文整理汇总了Python中sage.rings.all.Integer.factor方法的典型用法代码示例。如果您正苦于以下问题：Python Integer.factor方法的具体用法？Python Integer.factor怎么用？Python Integer.factor使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sage.rings.all.Integer的用法示例。
在下文中一共展示了Integer.factor方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: solve_mod
# 需要导入模块: from sage.rings.all import Integer [as 别名]
# 或者: from sage.rings.all.Integer import factor [as 别名]
def solve_mod(eqns, modulus, solution_dict = False):
r"""
Return all solutions to an equation or list of equations modulo the
given integer modulus. Each equation must involve only polynomials
in 1 or many variables.
By default the solutions are returned as `n`-tuples, where `n`
is the number of variables appearing anywhere in the given
equations. The variables are in alphabetical order.
INPUT:
- ``eqns`` - equation or list of equations
- ``modulus`` - an integer
- ``solution_dict`` - bool (default: False); if True or non-zero,
return a list of dictionaries containing the solutions. If there
are no solutions, return an empty list (rather than a list containing
an empty dictionary). Likewise, if there's only a single solution,
return a list containing one dictionary with that solution.
EXAMPLES::
sage: var('x,y')
(x, y)
sage: solve_mod([x^2 + 2 == x, x^2 + y == y^2], 14)
[(4, 2), (4, 6), (4, 9), (4, 13)]
sage: solve_mod([x^2 == 1, 4*x == 11], 15)
[(14,)]
Fermat's equation modulo 3 with exponent 5::
sage: var('x,y,z')
(x, y, z)
sage: solve_mod([x^5 + y^5 == z^5], 3)
[(0, 0, 0), (0, 1, 1), (0, 2, 2), (1, 0, 1), (1, 1, 2), (1, 2, 0), (2, 0, 2), (2, 1, 0), (2, 2, 1)]
We can solve with respect to a bigger modulus if it consists only of small prime factors::
sage: [d] = solve_mod([5*x + y == 3, 2*x - 3*y == 9], 3*5*7*11*19*23*29, solution_dict = True)
sage: d[x]
12915279
sage: d[y]
8610183
For cases where there are relatively few solutions and the prime
factors are small, this can be efficient even if the modulus itself
is large::
sage: sorted(solve_mod([x^2 == 41], 10^20))
[(4538602480526452429,), (11445932736758703821,), (38554067263241296179,),
(45461397519473547571,), (54538602480526452429,), (61445932736758703821,),
(88554067263241296179,), (95461397519473547571,)]
We solve a simple equation modulo 2::
sage: x,y = var('x,y')
sage: solve_mod([x == y], 2)
[(0, 0), (1, 1)]
.. warning::
The current implementation splits the modulus into prime
powers, then naively enumerates all possible solutions
(starting modulo primes and then working up through prime
powers), and finally combines the solution using the Chinese
Remainder Theorem. The interface is good, but the algorithm is
very inefficient if the modulus has some larger prime factors! Sage
*does* have the ability to do something much faster in certain
cases at least by using Groebner basis, linear algebra
techniques, etc. But for a lot of toy problems this function as
is might be useful. At least it establishes an interface.
TESTS:
Make sure that we short-circuit in at least some cases::
sage: solve_mod([2*x==1], 2*next_prime(10^50))
[]
Try multi-equation cases::
sage: x, y, z = var("x y z")
sage: solve_mod([2*x^2 + x*y, -x*y+2*y^2+x-2*y, -2*x^2+2*x*y-y^2-x-y], 12)
[(0, 0), (4, 4), (0, 3), (4, 7)]
sage: eqs = [-y^2+z^2, -x^2+y^2-3*z^2-z-1, -y*z-z^2-x-y+2, -x^2-12*z^2-y+z]
sage: solve_mod(eqs, 11)
[(8, 5, 6)]
Confirm that modulus 1 now behaves as it should::
sage: x, y = var("x y")
sage: solve_mod([x==1], 1)
[(0,)]
sage: solve_mod([2*x^2+x*y, -x*y+2*y^2+x-2*y, -2*x^2+2*x*y-y^2-x-y], 1)
[(0, 0)]
#.........这里部分代码省略.........