从Sage详细程度输出中提取值

Question

我是一个数学家，在使用Sage时需要一些帮助。我使用下面的代码，通过2同源下降，找到不同椭圆曲线的秩的上界。

下面是一个例子：

 from sage.schemes.elliptic_curves.descent_two_isogeny import two_descent_by_two_isogeny
 E = EllipticCurve([0, 0, 0, 17, 0])
 S = two_descent_by_two_isogeny(E, verbosity =1)

以下哪项输出：

2-isogeny

Results:
2 <= #E(Q)/phi'(E'(Q)) <= 2
2 <= #E'(Q)/phi(E(Q)) <= 8
#Sel^(phi')(E'/Q) = 2
#Sel^(phi)(E/Q) = 8
1 <= #Sha(E'/Q)[phi'] <= 1
1 <= #Sha(E/Q)[phi] <= 4
1 <= #Sha(E/Q)[2], #Sha(E'/Q)[2] <= 4
0 <= rank of E(Q) = rank of E'(Q) <= 2

现在我感兴趣的是最后的数字，'2‘，它是排名的上界。

我想要做的是找到一种方法来从输出中提取这个数字，这样我就可以在许多不同的椭圆曲线上运行代码，每次只收到一个秩的上界的值，而不是大的文本输出。

这个是可能的吗？

Answer 1

有吗？

E.rank_bounds()

给你想要的答案？它输出(0, 2)，文档说：“使用”simon_two_descent()“返回上下界。”(我是一名数学家，但这不是我的专业领域，所以我不知道simon_two_descent是否给出了与two_descent_by_two_isogeny相同的排名信息。我建议您输入E.<TAB>并浏览这些选项，看看是否有合适的选项。)

Answer 2

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最终简短答案的详细信息：

two_descent_by_two_isogeny内部的代码，在ipython解释器中检查，例如通过...

sage: two_descent_by_two_isogeny??

说明在(给定椭圆曲线的参数a1、a2、a3、a4、a6的)准备之后，“工作”被委托给two_descent_by_two_isogeny_work……

if a1==0 and a3==0:
    s2=a2; s4=a4; s6=a6
else:
    s2=a1*a1+4*a2; s4=8*(a1*a3+2*a4); s6=16*(a3*a3+4*a6)
f = ((x_ZZ + s2)*x_ZZ + s4)*x_ZZ + s6
x_list = f.roots() # over ZZ -- use FLINT directly?
x0 = x_list[0][0]
c = 3*x0+s2;  d = (c+s2)*x0+s4
return two_descent_by_two_isogeny_work(c, d,
    global_limit_small, global_limit_large, verbosity, selmer_only, proof)

后一个函数的返回值也可以检查...

sage: from sage.schemes.elliptic_curves.descent_two_isogeny import  \
....:      two_descent_by_two_isogeny
sage: two_descent_by_two_isogeny_work??

(或者直接打开导入的py-file，它是开源的！)最后几行，包括return行，是：

if verbosity > 0:
    print("\nResults:")
    print(n1, "<= #E(Q)/phi'(E'(Q)) <=", n2)
    print(n1_prime, "<= #E'(Q)/phi(E(Q)) <=", n2_prime)
    print("#Sel^(phi')(E'/Q) =", n2)
    print("#Sel^(phi)(E/Q) =", n2_prime)
    print("1 <= #Sha(E'/Q)[phi'] <=", n2/n1)
    print("1 <= #Sha(E/Q)[phi] <=", n2_prime/n1_prime)
    print("1 <= #Sha(E/Q)[2], #Sha(E'/Q)[2] <=", (n2_prime/n1_prime)*(n2/n1))
    a = Integer(n1*n1_prime).log(Integer(2))
    e = Integer(n2*n2_prime).log(Integer(2))
    print(a - 2, "<= rank of E(Q) = rank of E'(Q) <=", e - 2)

return n1, n2, n1_prime, n2_prime

现在回到OP中的示例代码，让我们也打印4元组S

sage: from sage.schemes.elliptic_curves.descent_two_isogeny import \
....:      two_descent_by_two_isogeny
....: E = EllipticCurve([0, 0, 0, 17, 0])
....: S = two_descent_by_two_isogeny(E, verbosity =1)
....: S

2-isogeny

Results:
2 <= #E(Q)/phi'(E'(Q)) <= 2
2 <= #E'(Q)/phi(E(Q)) <= 8
#Sel^(phi')(E'/Q) = 2
#Sel^(phi)(E/Q) = 8
1 <= #Sha(E'/Q)[phi'] <= 1
1 <= #Sha(E/Q)[phi] <= 4
1 <= #Sha(E/Q)[2], #Sha(E'/Q)[2] <= 4
0 <= rank of E(Q) = rank of E'(Q) <= 2
(2, 2, 2, 8)

返回的元组对应于值return n1, n2, n1_prime, n2_prime，打印中的“最终2”是从print命令获得的：

    e = Integer(n2*n2_prime).log(Integer(2))
    print(a - 2, "<= rank of E(Q) = rank of E'(Q) <=", e - 2)

所以你想要那个e - 2。

长话短说：

使用：

sage: from sage.schemes.elliptic_curves.descent_two_isogeny import \
....:      two_descent_by_two_isogeny
....: E = EllipticCurve([0, 0, 0, 17, 0])
....: S = two_descent_by_two_isogeny(E)
....: n1, n2, n1_prime, n2_prime = S
....: e = Integer(n2*n2_prime).log(Integer(2))
....: print(f"e = {e} so the needed upper bound for rank E'(Q) is e - 2 = {e-2}")
e = 4 so the needed upper bound for rank E'(Q) is e - 2 = 2

注意:也请考虑John Palmieri的回答！