本文整理匯總了Python中sage.modular.cusps.Cusp.parent方法的典型用法代碼示例。如果您正苦於以下問題：Python Cusp.parent方法的具體用法？Python Cusp.parent怎麽用？Python Cusp.parent使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類sage.modular.cusps.Cusp的用法示例。
在下文中一共展示了Cusp.parent方法的1個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: _reduce_cusp
# 需要導入模塊: from sage.modular.cusps import Cusp [as 別名]
# 或者: from sage.modular.cusps.Cusp import parent [as 別名]
def _reduce_cusp(self, c):
r"""
Compute a minimal representative for the given cusp c.
Returns a pair (c', t), where c' is the minimal representative
for the given cusp, and t is either 1 or -1, as explained
below. Largely for internal use.
The minimal representative for a cusp is the element in `P^1(Q)`
in lowest terms with minimal positive denominator, and minimal
positive numerator for that denominator.
Two cusps `u1/v1` and `u2/v2` are equivalent modulo `\Gamma_H(N)`
if and only if
- `v1 = h*v2 (mod N)` and `u1 = h^(-1)*u2 (mod gcd(v1,N))`
or
- `v1 = -h*v2 (mod N)` and `u1 = -h^(-1)*u2 (mod gcd(v1,N))`
for some `h \in H`. Then t is 1 or -1 as c and c' fall into
the first or second case, respectively.
EXAMPLES::
sage: GammaH(6,[5])._reduce_cusp(Cusp(5,3))
(1/3, -1)
sage: GammaH(12,[5])._reduce_cusp(Cusp(8,9))
(1/3, -1)
sage: GammaH(12,[5])._reduce_cusp(Cusp(5,12))
(Infinity, 1)
sage: GammaH(12,[])._reduce_cusp(Cusp(5,12))
(5/12, 1)
sage: GammaH(21,[5])._reduce_cusp(Cusp(-9/14))
(1/7, 1)
"""
c = Cusp(c)
N = int(self.level())
Cusps = c.parent()
v = int(c.denominator() % N)
H = self._list_of_elements_in_H()
# First, if N | v, take care of this case. If u is in \pm H,
# then we return Infinity. If not, let u_0 be the minimum
# of \{ h*u | h \in \pm H \}. Then return u_0/N.
if not v:
u = c.numerator() % N
if u in H:
return Cusps((1,0)), 1
if (N-u) in H:
return Cusps((1,0)), -1
ls = [ (u*h)%N for h in H ]
m1 = min(ls)
m2 = N-max(ls)
if m1 < m2:
return Cusps((m1,N)), 1
else:
return Cusps((m2,N)), -1
u = int(c.numerator() % v)
gcd = get_gcd(N)
d = gcd(v,N)
# If (N,v) == 1, let v_0 be the minimal element
# in \{ v * h | h \in \pm H \}. Then we either return
# Infinity or 1/v_0, as v is or is not in \pm H,
# respectively.
if d == 1:
if v in H:
return Cusps((0,1)), 1
if (N-v) in H:
return Cusps((0,1)), -1
ls = [ (v*h)%N for h in H ]
m1 = min(ls)
m2 = N-max(ls)
if m1 < m2:
return Cusps((1,m1)), 1
else:
return Cusps((1,m2)), -1
val_min = v
inv_mod = get_inverse_mod(N)
# Now we're in the case (N,v) > 1. So we have to do several
# steps: first, compute v_0 as above. While computing this
# minimum, keep track of *all* pairs of (h,s) which give this
# value of v_0.
hs_ls = [(1,1)]
for h in H:
tmp = (v*h)%N
if tmp < val_min:
val_min = tmp
hs_ls = [(inv_mod(h,N), 1)]
elif tmp == val_min:
hs_ls.append((inv_mod(h,N), 1))
if (N-tmp) < val_min:
val_min = N - tmp
#.........這裏部分代碼省略.........