本文整理汇总了Python中sympy.core.numbers.ilcm函数的典型用法代码示例。如果您正苦于以下问题:Python ilcm函数的具体用法?Python ilcm怎么用?Python ilcm使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了ilcm函数的7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,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: test_ilcm
def test_ilcm():
assert ilcm(0, 0) == 0
assert ilcm(1, 0) == 0
assert ilcm(0, 1) == 0
assert ilcm(1, 1) == 1
assert ilcm(2, 1) == 2
assert ilcm(8, 2) == 8
assert ilcm(8, 6) == 24
assert ilcm(8, 7) == 56
raises(ValueError, lambda: ilcm(8.1, 7))
raises(ValueError, lambda: ilcm(8, 7.1))
开发者ID:Maihj,项目名称:sympy,代码行数:11,代码来源:test_numbers.py
示例2: test_ilcm
def test_ilcm():
assert ilcm(0, 0) == 0
assert ilcm(1, 0) == 0
assert ilcm(0, 1) == 0
assert ilcm(1, 1) == 1
assert ilcm(2, 1) == 2
assert ilcm(8, 2) == 8
assert ilcm(8, 6) == 24
assert ilcm(8, 7) == 56
开发者ID:101man,项目名称:sympy,代码行数:9,代码来源:test_numbers.py
示例3: solve
#.........这里部分代码省略.........
symbol = symbols[0]
strategy = guess_solve_strategy(f, symbol)
if strategy == GS_POLY:
poly = f.as_poly( symbol )
assert poly is not None
result = roots(poly, cubics=True, quartics=True).keys()
elif strategy == GS_RATIONAL:
P, Q = f.as_numer_denom()
#TODO: check for Q != 0
return solve(P, symbol, **flags)
elif strategy == GS_POLY_CV_1:
args = list(f.args)
if isinstance(f, Add):
# we must search for a suitable change of variable
# collect exponents
exponents_denom = list()
for arg in args:
if isinstance(arg, Pow):
exponents_denom.append(arg.exp.q)
elif isinstance(arg, Mul):
for mul_arg in arg.args:
if isinstance(mul_arg, Pow):
exponents_denom.append(mul_arg.exp.q)
assert len(exponents_denom) > 0
if len(exponents_denom) == 1:
m = exponents_denom[0]
else:
# get the GCD of the denominators
m = ilcm(*exponents_denom)
# x -> y**m.
# we assume positive for simplification purposes
t = Symbol('t', positive=True, dummy=True)
f_ = f.subs(symbol, t**m)
if guess_solve_strategy(f_, t) != GS_POLY:
raise TypeError("Could not convert to a polynomial equation: %s" % f_)
cv_sols = solve(f_, t)
for sol in cv_sols:
result.append(sol**m)
elif isinstance(f, Mul):
for mul_arg in args:
result.extend(solve(mul_arg, symbol))
elif strategy == GS_POLY_CV_2:
m = 0
args = list(f.args)
if isinstance(f, Add):
for arg in args:
if isinstance(arg, Pow):
m = min(m, arg.exp)
elif isinstance(arg, Mul):
for mul_arg in arg.args:
if isinstance(mul_arg, Pow):
m = min(m, mul_arg.exp)
elif isinstance(f, Mul):
for mul_arg in args:
if isinstance(mul_arg, Pow):
m = min(m, mul_arg.exp)
f1 = simplify(f*symbol**(-m))
result = solve(f1, symbol)
# TODO: we might have introduced unwanted solutions
开发者ID:cran,项目名称:rSymPy,代码行数:67,代码来源:solvers.py
示例4: _intersect
def _intersect(self, other):
from sympy.functions.elementary.integers import ceiling, floor
from sympy.functions.elementary.complexes import sign
if other is S.Naturals:
return self._intersect(Interval(1, S.Infinity))
if other is S.Integers:
return self
if other.is_Interval:
if not all(i.is_number for i in other.args[:2]):
return
# In case of null Range, return an EmptySet.
if self.size == 0:
return S.EmptySet
# trim down to self's size, and represent
# as a Range with step 1.
start = ceiling(max(other.inf, self.inf))
if start not in other:
start += 1
end = floor(min(other.sup, self.sup))
if end not in other:
end -= 1
return self.intersect(Range(start, end + 1))
if isinstance(other, Range):
from sympy.solvers.diophantine import diop_linear
from sympy.core.numbers import ilcm
# non-overlap quick exits
if not other:
return S.EmptySet
if not self:
return S.EmptySet
if other.sup < self.inf:
return S.EmptySet
if other.inf > self.sup:
return S.EmptySet
# work with finite end at the start
r1 = self
if r1.start.is_infinite:
r1 = r1.reversed
r2 = other
if r2.start.is_infinite:
r2 = r2.reversed
# this equation represents the values of the Range;
# it's a linear equation
eq = lambda r, i: r.start + i*r.step
# we want to know when the two equations might
# have integer solutions so we use the diophantine
# solver
a, b = diop_linear(eq(r1, Dummy()) - eq(r2, Dummy()))
# check for no solution
no_solution = a is None and b is None
if no_solution:
return S.EmptySet
# there is a solution
# -------------------
# find the coincident point, c
a0 = a.as_coeff_Add()[0]
c = eq(r1, a0)
# find the first point, if possible, in each range
# since c may not be that point
def _first_finite_point(r1, c):
if c == r1.start:
return c
# st is the signed step we need to take to
# get from c to r1.start
st = sign(r1.start - c)*step
# use Range to calculate the first point:
# we want to get as close as possible to
# r1.start; the Range will not be null since
# it will at least contain c
s1 = Range(c, r1.start + st, st)[-1]
if s1 == r1.start:
pass
else:
# if we didn't hit r1.start then, if the
# sign of st didn't match the sign of r1.step
# we are off by one and s1 is not in r1
if sign(r1.step) != sign(st):
s1 -= st
if s1 not in r1:
return
return s1
# calculate the step size of the new Range
step = abs(ilcm(r1.step, r2.step))
s1 = _first_finite_point(r1, c)
if s1 is None:
#.........这里部分代码省略.........
开发者ID:pkgodara,项目名称:sympy,代码行数:101,代码来源:fancysets.py
示例5: solve
def solve(f, *symbols, **flags):
"""Solves equations and systems of equations.
Currently supported are univariate polynomial and transcendental
equations and systems of linear and polynomial equations. Input
is formed as a single expression or an equation, or an iterable
container in case of an equation system. The type of output may
vary and depends heavily on the input. For more details refer to
more problem specific functions.
By default all solutions are simplified to make the output more
readable. If this is not the expected behavior, eg. because of
speed issues, set simplified=False in function arguments.
To solve equations and systems of equations of other kind, eg.
recurrence relations of differential equations use rsolve() or
dsolve() functions respectively.
>>> from sympy import *
>>> x,y = symbols('xy')
Solve a polynomial equation:
>>> solve(x**4-1, x)
[1, -1, -I, I]
Solve a linear system:
>>> solve((x+5*y-2, -3*x+6*y-15), x, y)
{x: -3, y: 1}
"""
if not symbols:
raise ValueError('no symbols were given')
if len(symbols) == 1:
if isinstance(symbols[0], (list, tuple, set)):
symbols = symbols[0]
symbols = map(sympify, symbols)
if any(not s.is_Symbol for s in symbols):
raise TypeError('not a Symbol')
if not isinstance(f, (tuple, list, set)):
f = sympify(f)
if isinstance(f, Equality):
f = f.lhs - f.rhs
if len(symbols) != 1:
raise NotImplementedError('multivariate equation')
symbol = symbols[0]
strategy = guess_solve_strategy(f, symbol)
if strategy == GS_POLY:
poly = f.as_poly( symbol )
assert poly is not None
result = roots(poly, cubics=True, quartics=True).keys()
elif strategy == GS_RATIONAL:
P, Q = f.as_numer_denom()
#TODO: check for Q != 0
return solve(P, symbol, **flags)
elif strategy == GS_POLY_CV_1:
args = list(f.args)
if isinstance(f, Add):
# we must search for a suitable change of variable
# collect exponents
exponents_denom = list()
for arg in args:
if isinstance(arg, Pow):
exponents_denom.append(arg.exp.q)
elif isinstance(arg, Mul):
for mul_arg in arg.args:
if isinstance(mul_arg, Pow):
exponents_denom.append(mul_arg.exp.q)
assert len(exponents_denom) > 0
if len(exponents_denom) == 1:
m = exponents_denom[0]
else:
# get the GCD of the denominators
m = ilcm(*exponents_denom)
# x -> y**m.
# we assume positive for simplification purposes
t = Symbol('t', positive=True, dummy=True)
f_ = f.subs(symbol, t**m)
if guess_solve_strategy(f_, t) != GS_POLY:
raise TypeError("Could not convert to a polynomial equation: %s" % f_)
cv_sols = solve(f_, t)
result = list()
for sol in cv_sols:
result.append(sol**m)
elif isinstance(f, Mul):
result = []
for mul_arg in args:
#.........这里部分代码省略.........
开发者ID:gnulinooks,项目名称:sympy,代码行数:101,代码来源:solvers.py
示例6: poly_lcm
def poly_lcm(f, g, *symbols):
"""Computes least common multiple of two polynomials.
Given two univariate polynomials, the LCM is computed via
f*g = gcd(f, g)*lcm(f, g) formula. In multivariate case, we
compute the unique generator of the intersection of the two
ideals, generated by f and g. This is done by computing a
Groebner basis, with respect to any lexicographic ordering,
of t*f and (1 - t)*g, where t is an unrelated symbol and
filtering out solution that does not contain t.
For more information on the implemented algorithm refer to:
[1] D. Cox, J. Little, D. O'Shea, Ideals, Varieties and
Algorithms, Springer, Second Edition, 1997, pp. 187
"""
if not isinstance(f, Poly):
f = Poly(f, *symbols)
elif symbols:
raise SymbolsError("Redundant symbols were given")
f, g = f.unify_with(g)
symbols, flags = f.symbols, f.flags
if f.is_monomial and g.is_monomial:
monom = monomial_lcm(f.LM, g.LM)
fc, gc = f.LC, g.LC
if fc.is_Rational and gc.is_Rational:
coeff = Integer(ilcm(fc.p, gc.p))
else:
coeff = S.One
return Poly((coeff, monom), *symbols, **flags)
fc, f = f.as_primitive()
gc, g = g.as_primitive()
lcm = ilcm(int(fc), int(gc))
if f.is_multivariate:
t = Symbol('t', dummy=True)
lex = { 'order' : 'lex' }
f_monoms = [ (1,) + monom for monom in f.monoms ]
F = Poly((f.coeffs, f_monoms), t, *symbols, **lex)
g_monoms = [ (0,) + monom for monom in g.monoms ] + \
[ (1,) + monom for monom in g.monoms ]
g_coeffs = list(g.coeffs) + [ -coeff for coeff in g.coeffs ]
G = Poly(dict(zip(g_monoms, g_coeffs)), t, *symbols, **lex)
def independent(h):
return all(not monom[0] for monom in h.monoms)
H = [ h for h in poly_groebner((F, G)) if independent(h) ]
if lcm != 1:
h_coeffs = [ coeff*lcm for coeff in H[0].coeffs ]
else:
h_coeffs = H[0].coeffs
h_monoms = [ monom[1:] for monom in H[0].monoms ]
return Poly(dict(zip(h_monoms, h_coeffs)), *symbols, **flags)
else:
h = poly_div(f * g, poly_gcd(f, g))[0]
if lcm != 1:
return h.mul_term(lcm / h.LC)
else:
return h.as_monic()
开发者ID:gnulinooks,项目名称:sympy,代码行数:77,代码来源:algorithms.py
示例7: intersection_sets
def intersection_sets(a, b):
from sympy.solvers.diophantine import diop_linear
from sympy.core.numbers import ilcm
from sympy import sign
# non-overlap quick exits
if not b:
return S.EmptySet
if not a:
return S.EmptySet
if b.sup < a.inf:
return S.EmptySet
if b.inf > a.sup:
return S.EmptySet
# work with finite end at the start
r1 = a
if r1.start.is_infinite:
r1 = r1.reversed
r2 = b
if r2.start.is_infinite:
r2 = r2.reversed
# this equation represents the values of the Range;
# it's a linear equation
eq = lambda r, i: r.start + i*r.step
# we want to know when the two equations might
# have integer solutions so we use the diophantine
# solver
va, vb = diop_linear(eq(r1, Dummy()) - eq(r2, Dummy()))
# check for no solution
no_solution = va is None and vb is None
if no_solution:
return S.EmptySet
# there is a solution
# -------------------
# find the coincident point, c
a0 = va.as_coeff_Add()[0]
c = eq(r1, a0)
# find the first point, if possible, in each range
# since c may not be that point
def _first_finite_point(r1, c):
if c == r1.start:
return c
# st is the signed step we need to take to
# get from c to r1.start
st = sign(r1.start - c)*step
# use Range to calculate the first point:
# we want to get as close as possible to
# r1.start; the Range will not be null since
# it will at least contain c
s1 = Range(c, r1.start + st, st)[-1]
if s1 == r1.start:
pass
else:
# if we didn't hit r1.start then, if the
# sign of st didn't match the sign of r1.step
# we are off by one and s1 is not in r1
if sign(r1.step) != sign(st):
s1 -= st
if s1 not in r1:
return
return s1
# calculate the step size of the new Range
step = abs(ilcm(r1.step, r2.step))
s1 = _first_finite_point(r1, c)
if s1 is None:
return S.EmptySet
s2 = _first_finite_point(r2, c)
if s2 is None:
return S.EmptySet
# replace the corresponding start or stop in
# the original Ranges with these points; the
# result must have at least one point since
# we know that s1 and s2 are in the Ranges
def _updated_range(r, first):
st = sign(r.step)*step
if r.start.is_finite:
rv = Range(first, r.stop, st)
else:
rv = Range(r.start, first + st, st)
return rv
r1 = _updated_range(a, s1)
r2 = _updated_range(b, s2)
# work with them both in the increasing direction
if sign(r1.step) < 0:
r1 = r1.reversed
if sign(r2.step) < 0:
r2 = r2.reversed
# return clipped Range with positive step; it
# can't be empty at this point
#.........这里部分代码省略.........
开发者ID:bjodah,项目名称:sympy,代码行数:101,代码来源:intersection.py
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