本文整理汇总了Python中constructor.EllipticCurve.b_invariants方法的典型用法代码示例。如果您正苦于以下问题：Python EllipticCurve.b_invariants方法的具体用法？Python EllipticCurve.b_invariants怎么用？Python EllipticCurve.b_invariants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constructor.EllipticCurve的用法示例。
在下文中一共展示了EllipticCurve.b_invariants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: _tate
# 需要导入模块: from constructor import EllipticCurve [as 别名]
# 或者: from constructor.EllipticCurve import b_invariants [as 别名]
#.........这里部分代码省略.........
def _pcubicroots(b, c, d):
r"""
Local function returning the number of roots of `x^3 +
b*x^2 + c*x + d` modulo `P`, counting multiplicities
"""
return sum([rr[1] for rr in PolynomialRing(F, 'x')([F(d), F(c), F(b), F(1)]).roots()],0)
if p == 2:
halfmodp = OK(Integer(0))
else:
halfmodp = pinv(Integer(2))
A = E.a_invariants()
A = [0, A[0], A[1], A[2], A[3], 0, A[4]]
indices = [1,2,3,4,6]
if min([pval(a) for a in A if a != 0]) < 0:
verbose("Non-integral model at P: valuations are %s; making integral"%([pval(a) for a in A if a != 0]), t, 1)
e = 0
for i in range(7):
if A[i] != 0:
e = max(e, (-pval(A[i])/i).ceil())
pie = pi**e
for i in range(7):
if A[i] != 0:
A[i] *= pie**i
verbose("P-integral model is %s, with valuations %s"%([A[i] for i in indices], [pval(A[i]) for i in indices]), t, 1)
split = None # only relevant for multiplicative reduction
(a1, a2, a3, a4, a6) = (A[1], A[2], A[3], A[4], A[6])
while True:
C = EllipticCurve([a1, a2, a3, a4, a6]);
(b2, b4, b6, b8) = C.b_invariants()
(c4, c6) = C.c_invariants()
delta = C.discriminant()
val_disc = pval(delta)
if val_disc == 0:
## Good reduction already
cp = 1
fp = 0
KS = KodairaSymbol("I0")
break #return
# Otherwise, we change coordinates so that p | a3, a4, a6
if p == 2:
if pdiv(b2):
r = proot(a4, 2)
t = proot(((r + a2)*r + a4)*r + a6, 2)
else:
temp = pinv(a1)
r = temp * a3
t = temp * (a4 + r*r)
elif p == 3:
if pdiv(b2):
r = proot(-b6, 3)
else:
r = -pinv(b2) * b4
t = a1 * r + a3
else:
if pdiv(c4):
r = -pinv(12) * b2
else:
r = -pinv(12*c4) * (c6 + b2 * c4)
t = -halfmodp * (a1 * r + a3)