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Old 2020-02-21, 20:08   #1
bhelmes
 
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Default pollard rho with bilinear form and eigenvektor ?

A peaceful evening,


if i make a pollard rho algorithm with a suitable bilinear form instead

of the function f(x)=x²+1, is it then better to use the eigenvektor of the matrix or not.


I only know the concept of eigenvektors with a matrix and one singular vektor, but i thought that the "radius" of the bilinearform with the eigenvektor is smaller.


Any ideas, suggestion or mathematical clarification ?


Greetings from the eliptic curves

Bernhard
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Old 2020-02-26, 19:46   #2
bhelmes
 
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Quote:
Originally Posted by bhelmes View Post
A peaceful evening,

if i make a pollard rho algorithm with a suitable bilinear form instead
of the function f(x)=x²+1, is it then better to use the eigenvektor of the matrix or not.

this was mathematical nonsens:
if A is the matrix of the bilinearform and v the eigenvektor,
than Av=rv, multiplication with the vektor v on the left side gives the result
vAv=v*r*v = r *v *v where the bilinearform is disappearing



Nevertheless you can reduce C on the unit circle,
is it possible to find a subgroup of a bilinearform such as



A= (1, 1)
(1, -1) ?



Thanks for your patience

Bernhard

Last fiddled with by bhelmes on 2020-02-26 at 20:42
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Old 2020-03-04, 20:52   #3
bhelmes
 
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Quote:
Originally Posted by bhelmes View Post

Nevertheless you can reduce C on the unit circle,
is it possible to find a subgroup of a bilinearform such as

A= (1, 1)
(1, -1) ?

It is always possible to reduce the 2x2 bilinearform to a quadratic polynom
by setting y=x, this is trivial.


If I choose a linear substituion for y depending on x (for example y=2x)
can i make any mathematical consequnences by it ?


Greetings

Bernhard
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