20200916, 02:01  #1 
Sep 2020
7 Posts 
Difficulty of SNFS polynomials
I have recently come across SNFS factorization, and I have yet to find an answer to the following questions: How is the SNFS difficulty calculated? Is there an equation for calculating it?

20200916, 03:16  #2 
"Curtis"
Feb 2005
Riverside, CA
3·1,579 Posts 
It's generally the size of the input, or the size of the original composite that generates the polynomial (for example, you might have a 25 digit factor already, but doing SNFS on the original number is faster than doing GNFS on the cofactor, so the SNFS difficulty doesn't match the composite size).
Say I'm trying to factor 2^901  1. That can be written 2x^6  1, with x equal to 2^150. That nice poly is what SNFS would use so even if a couple of factors are known and the cofactor is 200 digits, this SNFS job at ~270 digits would be easier than a GNFS of 200. 
20200916, 06:54  #3 
Jul 2003
So Cal
2×3×347 Posts 
SNFS requires two polynomials, a degree polynomial , and a linear polynomial, , which share a common root modulo the number you are factoring. The difficulty is given by the size of .
In the example above, and . , , and . So the difficulty is given by the size of , the number being factored. Difficulty is usually expressed as the common log of the number, . 
20200916, 07:41  #4  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
24626_{8} Posts 
Quote:
A linear polynomial is almost always used (but not exclusively) because of the difficulty of finding good polynomials when neither are linear. 

20200916, 14:34  #5  
Aug 2006
1761_{16} Posts 
Quote:


20200916, 20:02  #6 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·29·101 Posts 
Are there any good examples of snfs with nonlinear polynomials? As far as I know only gnfs has been done with nonlinear.

20200916, 22:40  #7 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×5,323 Posts 

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