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Python EllipticCurve.b_invariants方法代码示例

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, 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)
开发者ID:amitjamadagni,项目名称:sage,代码行数:70,代码来源:ell_local_data.py


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