本文整理汇总了Python中sympy.polys.gcd函数的典型用法代码示例。如果您正苦于以下问题：Python gcd函数的具体用法？Python gcd怎么用？Python gcd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gcd函数的15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: rrs_simple
def rrs_simple(n):
"""
@type n: integer > 0
@rtype: list
@return: a reduced residue system modulo n
"""
return [x for x in range(n) if gcd(x, n) == 1]示例2: weak_normalizer
def weak_normalizer(a, d, DE, z=None):
"""
Weak normalization.
Given a derivation D on k[t] and f == a/d in k(t), return q in k[t]
such that f - Dq/q is weakly normalized with respect to t.
f in k(t) is said to be "weakly normalized" with respect to t if
residue_p(f) is not a positive integer for any normal irreducible p
in k[t] such that f is in R_p (Definition 6.1.1). If f has an
elementary integral, this is equivalent to no logarithm of
integral(f) whose argument depends on t has a positive integer
coefficient, where the arguments of the logarithms not in k(t) are
in k[t].
Returns (q, f - Dq/q)
"""
z = z or Dummy('z')
dn, ds = splitfactor(d, DE)
# Compute d1, where dn == d1*d2**2*...*dn**n is a square-free
# factorization of d.
g = gcd(dn, dn.diff(DE.t))
d_sqf_part = dn.quo(g)
d1 = d_sqf_part.quo(gcd(d_sqf_part, g))
a1, b = gcdex_diophantine(d.quo(d1).as_poly(DE.t), d1.as_poly(DE.t),
a.as_poly(DE.t))
r = (a - Poly(z, DE.t)*derivation(d1, DE)).as_poly(DE.t).resultant(
d1.as_poly(DE.t))
r = Poly(r, z)
if not r.has(z):
return (Poly(1, DE.t), (a, d))
N = [i for i in r.real_roots() if i in ZZ and i > 0]
q = reduce(mul, [gcd(a - Poly(n, DE.t)*derivation(d1, DE), d1) for n in N],
Poly(1, DE.t))
dq = derivation(q, DE)
sn = q*a - d*dq
sd = q*d
sn, sd = sn.cancel(sd, include=True)
return (q, (sn, sd))示例3: _deflation
def _deflation(p):
for y in V:
if not p.has(y):
continue
if _derivation(p) is not S.Zero:
c, q = p.as_poly(y).primitive()
return _deflation(c)*gcd(q, q.diff(y)).as_expr()
else:
return p示例4: deflation
def deflation(p):
for y in V:
if not p.has_any_symbols(y):
continue
if derivation(p) is not S.Zero:
c, q = p.as_poly(y).as_primitive()
return deflation(c)*gcd(q, q.diff(y))
else:
return p示例5: extract
def extract(f):
p = f.args[0]
for q in f.args[1:]:
p = gcd(p, q, *symbols)
if p.is_number:
return S.One, f
return p, Add(*[quo(g, p, *symbols) for g in f.args])示例6: rsa
def rsa(bits):
p = nextprime(randrange(2**(bits // 2 + 1)))
q = nextprime(randrange(2**(bits // 2 + 1)))
n = p * q
phi_n = (p - 1) * (q - 1)
while True:
e = randrange(1, phi_n)
if gcd(e, phi_n) == 1:
break
d = inv_mod(e, phi_n)
return e, d, n示例7: _splitter
def _splitter(p):
for y in V:
if not p.has(y):
continue
if _derivation(y) is not S.Zero:
c, q = p.as_poly(y).primitive()
q = q.as_expr()
h = gcd(q, _derivation(q), y)
s = quo(h, gcd(q, q.diff(y), y), y)
c_split = _splitter(c)
if s.as_poly(y).degree() == 0:
return (c_split[0], q * c_split[1])
q_split = _splitter(cancel(q / s))
return (c_split[0]*q_split[0]*s, c_split[1]*q_split[1])
else:
return (S.One, p)示例8: splitter
def splitter(p):
for y in V:
if not p.has_any_symbols(y):
continue
if derivation(y) is not S.Zero:
c, q = p.as_poly(y).as_primitive()
q = q.as_basic()
h = gcd(q, derivation(q), y)
s = quo(h, gcd(q, q.diff(y), y), y)
c_split = splitter(c)
if s.as_poly(y).degree == 0:
return (c_split[0], q * c_split[1])
q_split = splitter(Poly.cancel((q, s), *V))
return (c_split[0]*q_split[0]*s, c_split[1]*q_split[1])
else:
return (S.One, p)示例9: crack_given_decrypt
def crack_given_decrypt(n, m):
n = int(n)
m = int(m)
while True:
if m % 2 != 0:
break
divide_out = True
for _ in range(5):
a = randrange(1, n)
if gcd(a, n) == 1:
if pow(a, m // 2, n) != 1:
divide_out = False
break
if divide_out:
m //= 2
else:
break
while True:
a = randrange(1, n)
g = gcd(pow(a, m // 2, n) - 1, n)
if g != 1 and g != n:
return g示例10: _split_gcd
def _split_gcd(*a):
"""
split the list of integers ``a`` into a list of integers, ``a1`` having
``g = gcd(a1)``, and a list ``a2`` whose elements are not divisible by
``g``. Returns ``g, a1, a2``
Examples
========
>>> from sympy.simplify.radsimp import _split_gcd
>>> _split_gcd(55, 35, 22, 14, 77, 10)
(5, [55, 35, 10], [22, 14, 77])
"""
g = a[0]
b1 = [g]
b2 = []
for x in a[1:]:
g1 = gcd(g, x)
if g1 == 1:
b2.append(x)
else:
g = g1
b1.append(x)
return g, b1, b2示例11: factor_nc
def factor_nc(expr):
"""Return the factored form of ``expr`` while handling non-commutative
expressions.
**examples**
>>> from sympy.core.exprtools import factor_nc
>>> from sympy import Symbol
>>> from sympy.abc import x
>>> A = Symbol('A', commutative=False)
>>> B = Symbol('B', commutative=False)
>>> factor_nc((x**2 + 2*A*x + A**2).expand())
(x + A)**2
>>> factor_nc(((x + A)*(x + B)).expand())
(x + A)*(x + B)
"""
from sympy.simplify.simplify import _mexpand
from sympy.polys import gcd, factor
expr = sympify(expr)
if not isinstance(expr, Expr) or not expr.args:
return expr
if not expr.is_Add:
return expr.func(*[factor_nc(a) for a in expr.args])
expr, rep, nc_symbols = _mask_nc(expr)
if rep:
return factor(expr).subs(rep)
else:
args = [a.args_cnc() for a in Add.make_args(expr)]
c = g = l = r = S.One
hit = False
# find any commutative gcd term
for i, a in enumerate(args):
if i == 0:
c = Mul._from_args(a[0])
elif a[0]:
c = gcd(c, Mul._from_args(a[0]))
else:
c = S.One
if c is not S.One:
hit = True
c, g = c.as_coeff_Mul()
if g is not S.One:
for i, (cc, _) in enumerate(args):
cc = list(Mul.make_args(Mul._from_args(list(cc))/g))
args[i][0] = cc
else:
for i, (cc, _) in enumerate(args):
cc[0] = cc[0]/c
args[i][0] = cc
# find any noncommutative common prefix
for i, a in enumerate(args):
if i == 0:
n = a[1][:]
else:
n = common_prefix(n, a[1])
if not n:
# is there a power that can be extracted?
if not args[0][1]:
break
b, e = args[0][1][0].as_base_exp()
ok = False
if e.is_Integer:
for t in args:
if not t[1]:
break
bt, et = t[1][0].as_base_exp()
if et.is_Integer and bt == b:
e = min(e, et)
else:
break
else:
ok = hit = True
l = b**e
il = b**-e
for i, a in enumerate(args):
args[i][1][0] = il*args[i][1][0]
break
if not ok:
break
else:
hit = True
lenn = len(n)
l = Mul(*n)
for i, a in enumerate(args):
args[i][1] = args[i][1][lenn:]
# find any noncommutative common suffix
for i, a in enumerate(args):
if i == 0:
n = a[1][:]
else:
n = common_suffix(n, a[1])
if not n:
# is there a power that can be extracted?
if not args[0][1]:
break
b, e = args[0][1][-1].as_base_exp()
ok = False
if e.is_Integer:
for t in args:
#.........这里部分代码省略.........
示例12: rsolve_ratio
def rsolve_ratio(coeffs, f, n, **hints):
"""Given linear recurrence operator L of order 'k' with polynomial
coefficients and inhomogeneous equation Ly = f, where 'f' is a
polynomial, we seek for all rational solutions over field K of
characteristic zero.
This procedure accepts only polynomials, however if you are
interested in solving recurrence with rational coefficients
then use rsolve() which will pre-process the given equation
and run this procedure with polynomial arguments.
The algorithm performs two basic steps:
(1) Compute polynomial v(n) which can be used as universal
denominator of any rational solution of equation Ly = f.
(2) Construct new linear difference equation by substitution
y(n) = u(n)/v(n) and solve it for u(n) finding all its
polynomial solutions. Return None if none were found.
Algorithm implemented here is a revised version of the original
Abramov's algorithm, developed in 1989. The new approach is much
simpler to implement and has better overall efficiency. This
method can be easily adapted to q-difference equations case.
Besides finding rational solutions alone, this functions is
an important part of Hyper algorithm were it is used to find
particular solution of inhomogeneous part of a recurrence.
For more information on the implemented algorithm refer to:
[1] S. A. Abramov, Rational solutions of linear difference
and q-difference equations with polynomial coefficients,
in: T. Levelt, ed., Proc. ISSAC '95, ACM Press, New York,
1995, 285-289
"""
f = sympify(f)
if not f.is_polynomial(n):
return None
coeffs = map(sympify, coeffs)
r = len(coeffs)-1
A, B = coeffs[r], coeffs[0]
A = A.subs(n, n-r).expand()
h = Symbol('h', dummy=True)
res = resultant(A, B.subs(n, n+h), n)
if not res.is_polynomial(h):
p, q = res.as_numer_denom()
res = exquo(p, q, h)
nni_roots = roots(res, h, filter='Z',
predicate=lambda r: r >= 0).keys()
if not nni_roots:
return rsolve_poly(coeffs, f, n, **hints)
else:
C, numers = S.One, [S.Zero]*(r+1)
for i in xrange(int(max(nni_roots)), -1, -1):
d = gcd(A, B.subs(n, n+i), n)
A = exquo(A, d, n)
B = exquo(B, d.subs(n, n-i), n)
C *= Mul(*[ d.subs(n, n-j) for j in xrange(0, i+1) ])
denoms = [ C.subs(n, n+i) for i in range(0, r+1) ]
for i in range(0, r+1):
g = gcd(coeffs[i], denoms[i], n)
numers[i] = exquo(coeffs[i], g, n)
denoms[i] = exquo(denoms[i], g, n)
for i in xrange(0, r+1):
numers[i] *= Mul(*(denoms[:i] + denoms[i+1:]))
result = rsolve_poly(numers, f * Mul(*denoms), n, **hints)
if result is not None:
if hints.get('symbols', False):
return (simplify(result[0] / C), result[1])
else:
return simplify(result / C)
else:
return None示例13: rsolve_ratio
def rsolve_ratio(coeffs, f, n, **hints):
"""
Given linear recurrence operator `\operatorname{L}` of order `k`
with polynomial coefficients and inhomogeneous equation
`\operatorname{L} y = f`, where `f` is a polynomial, we seek
for all rational solutions over field `K` of characteristic zero.
This procedure accepts only polynomials, however if you are
interested in solving recurrence with rational coefficients
then use ``rsolve`` which will pre-process the given equation
and run this procedure with polynomial arguments.
The algorithm performs two basic steps:
(1) Compute polynomial `v(n)` which can be used as universal
denominator of any rational solution of equation
`\operatorname{L} y = f`.
(2) Construct new linear difference equation by substitution
`y(n) = u(n)/v(n)` and solve it for `u(n)` finding all its
polynomial solutions. Return ``None`` if none were found.
Algorithm implemented here is a revised version of the original
Abramov's algorithm, developed in 1989. The new approach is much
simpler to implement and has better overall efficiency. This
method can be easily adapted to q-difference equations case.
Besides finding rational solutions alone, this functions is
an important part of Hyper algorithm were it is used to find
particular solution of inhomogeneous part of a recurrence.
Examples
========
>>> from sympy.abc import x
>>> from sympy.solvers.recurr import rsolve_ratio
>>> rsolve_ratio([-2*x**3 + x**2 + 2*x - 1, 2*x**3 + x**2 - 6*x,
... - 2*x**3 - 11*x**2 - 18*x - 9, 2*x**3 + 13*x**2 + 22*x + 8], 0, x)
C2*(2*x - 3)/(2*(x**2 - 1))
References
==========
.. [1] S. A. Abramov, Rational solutions of linear difference
and q-difference equations with polynomial coefficients,
in: T. Levelt, ed., Proc. ISSAC '95, ACM Press, New York,
1995, 285-289
See Also
========
rsolve_hyper
"""
f = sympify(f)
if not f.is_polynomial(n):
return None
coeffs = map(sympify, coeffs)
r = len(coeffs) - 1
A, B = coeffs[r], coeffs[0]
A = A.subs(n, n - r).expand()
h = Dummy('h')
res = resultant(A, B.subs(n, n + h), n)
if not res.is_polynomial(h):
p, q = res.as_numer_denom()
res = quo(p, q, h)
nni_roots = roots(res, h, filter='Z',
predicate=lambda r: r >= 0).keys()
if not nni_roots:
return rsolve_poly(coeffs, f, n, **hints)
else:
C, numers = S.One, [S.Zero]*(r + 1)
for i in xrange(int(max(nni_roots)), -1, -1):
d = gcd(A, B.subs(n, n + i), n)
A = quo(A, d, n)
B = quo(B, d.subs(n, n - i), n)
C *= Mul(*[ d.subs(n, n - j) for j in xrange(0, i + 1) ])
denoms = [ C.subs(n, n + i) for i in range(0, r + 1) ]
for i in range(0, r + 1):
g = gcd(coeffs[i], denoms[i], n)
numers[i] = quo(coeffs[i], g, n)
denoms[i] = quo(denoms[i], g, n)
for i in xrange(0, r + 1):
numers[i] *= Mul(*(denoms[:i] + denoms[i + 1:]))
#.........这里部分代码省略.........
示例14: nc_gcd
def nc_gcd(aa, bb):
a, b = [i.as_coeff_Mul() for i in [aa, bb]]
c = gcd(a[0], b[0]).as_numer_denom()[0]
g = Mul(*(a[1].args_cnc(cset=True)[0] & b[1].args_cnc(cset=True)[0]))
return _keep_coeff(c, g)示例15: gcd
def gcd(f, g):
from sympy.polys import gcd
return f.__class__(gcd(f.ex, f.__class__(g).ex))