Python 获得基本判别式
Sage中是否有函数返回与给定判别式相关的基本判别式Python 获得基本判别式,python,sage,Python,Sage,Sage中是否有函数返回与给定判别式相关的基本判别式 sage: fundamental_discriminant(1) 1 sage: fundamental_discriminant(3) 12 sage: fundamental_discriminant? Signature: fundamental_discriminant(D) Docstring: Return the discriminant of the quadratic extension K=Q(sqrt{
sage: fundamental_discriminant(1)
1
sage: fundamental_discriminant(3)
12
sage: fundamental_discriminant?
Signature: fundamental_discriminant(D)
Docstring:
Return the discriminant of the quadratic extension K=Q(sqrt{D}),
i.e. an integer d congruent to either 0 or 1, mod 4, and such that,
at most, the only square dividing it is 4.
INPUT:
* "D" - an integer
OUTPUT:
* an integer, the fundamental discriminant
看
这是我编写的函数,没有找到现有函数:
def getFund(D):
if D % 4 == 2 or D % 4 == 3:
raise ValueError("Not a discriminant.")
if D == 0:
raise ValueError("There is no fundamental associated to 0.")
P = sign(D)
for p in factor(D):
if p[1] % 2 == 1:
P *= p[0]
if P % 4 == 2 or P % 4 == 3:
P *= 4
return P
我不知道这个函数是否用SageMath实现 但如果我正确理解了定义,您可以如下定义函数:
def fundamental_discriminant(d):
if d % 4 == 1:
return d.squarefree_part()
if d % 4 == 0:
k = d.valuation(4)
dd = d // 4^k
if dd % 4 == 1:
return dd.squarefree_part()
if dd % 4 == 2:
return 4 * dd.squarefree_part()
if dd % 4 == 3:
return 4 * dd.squarefree_part()
raise ValueError("Not a discriminant.")
基本判别法有什么问题
sage: fundamental_discriminant(1)
1
sage: fundamental_discriminant(3)
12
sage: fundamental_discriminant?
Signature: fundamental_discriminant(D)
Docstring:
Return the discriminant of the quadratic extension K=Q(sqrt{D}),
i.e. an integer d congruent to either 0 or 1, mod 4, and such that,
at most, the only square dividing it is 4.
INPUT:
* "D" - an integer
OUTPUT:
* an integer, the fundamental discriminant
谢谢我还编写了一个自己的实现,因为我在最初的几个小时里没有得到答案。我将把它添加到我的问题中。a//b是否表示楼层(a/b)?@主要理想域——确切地说:在Python中,
a//b
是“商和余数”除法中的商,a%b
是余数。在Sage中,一个人可以同时用quo_-rem
,例如23。quo_-rem(7)
给出(3,2)
,实际上23=3*7+2
。我认为没有问题。谢谢你提到这个方法。