20080408, 03:08  #1 
Nov 2006
Singapore
3·5^{2} Posts 
Prime Factoring Algorithm
Looking for comments/criticisms before I proceed further. The idea seems (at least to me) too promising to abandon but as of now is still "half baked".I am hoping other eyes and brains can help me see the things that I have missed.
Visu Last fiddled with by Visu on 20080408 at 03:10 
20080408, 12:11  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
of Fermat's method. 

20080410, 15:01  #3 
Nov 2006
Singapore
3·5^{2} Posts 
Trial division? There is no trial division anywhere.

20080410, 15:08  #4 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·3·17·31 Posts 

20080410, 17:22  #5 
∂^{2}ω=0
Sep 2002
República de California
13·29·31 Posts 
Bit late to the party, and have a possiblyverystupid question, which is orthogonal to the estimatedruntime discussion:
Why do we need a "prime factoring algorithm"? 
20080410, 21:19  #6 
"William"
May 2003
New Haven
23·103 Posts 

20080410, 21:38  #7 
"William"
May 2003
New Haven
23×103 Posts 
Have you tried your method on the Zimmermann challenges?
Paul Zimmermann's page: "If you want me to look at your algorithm, please first factor one of the numbers below. These small challenges are hard enough so that naive algorithms will not be able to solve them, and easy enough so that an implementation on a personal computer should be able to solve them (if the corresponding algorithm is really efficient)." http://www.loria.fr/~zimmerma/records/rsa.html 
20080410, 22:29  #8  
Nov 2006
Singapore
3·5^{2} Posts 
Quote:


20080410, 22:35  #9  
Nov 2006
Singapore
3·5^{2} Posts 
Quote:
I could have just called it a factoring algorithm though And seems I'm in good company if Bill Gates is making the same mistake I did in a more public forum. We non mathematicians have a lot to learn. 

20080410, 22:41  #10 
Nov 2006
Singapore
75_{10} Posts 
I thought for a minute that some requirement for trial division managed to sneak in. I was fairly certain that I left out trial divisions, iterations that loop indefinitely and sievings the usual road bumps in the way of an amateur mathematicians' road to glory. ( I might have divided something by 0 though.)

20080410, 22:44  #11 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
13·787 Posts 
My understanding is that Mathematica or other program can be used to perform the need calculations.

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