I think this is a fast algorithm. Im not sure what fast can be just testing
the multiplicative order
El domingo, 20 de abril de 2014 15:00:37 UTC-5, John Cremona escribió:
>
> Any reason not to just test r.multiplicative_order() ?
>
> John Cremona
>
> On 20 April 2014 20:11, Jan Medina >
> wro
Any reason not to just test r.multiplicative_order() ?
John Cremona
On 20 April 2014 20:11, Jan Medina wrote:
> I did this algorithm to find a primitive element of a multiplicative group
> on a finite field. This is a basic algorithm
> def random_primitive(p,h):
> F.=GF(p^h)
> s=p^h-1
>
I did this algorithm to find a primitive element of a multiplicative group
on a finite field. This is a basic algorithm
def random_primitive(p,h):
F.=GF(p^h)
s=p^h-1
r=F.random_element()
j=0
if r!=0:
for t in prime_factors(s):
if r^(s/t)==1:
i know that but this function only give you one generator, always the same
but i want a random generator of a multiplicate group of a finite field.
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Jan Medina wrote:
Hi everybody.
There is a procedure or library in SAGE that allows finding a random
primitive element on a finite field?
http://sagemath.org/doc/reference/finite_rings/sage/rings/finite_rings/finite_field_base.html?highlight=primitive_element#sage.rings.finite_rings.finite_fie
