20150115, 09:56  #1 
Sep 2011
3·19 Posts 
GCD of Polynomials over a finite field for NFS
I'm stuck at the square root step of the number field sieve. Specifically the part that verifies that primes are inert. I'm supposed to calculate a GCD of two polynomials modulo a prime to verify that such prime is usable for Chinese remaindering. I can't seem to get the code right.
These numbers were taken directly from Briggs paper at page 58 and the code was based on http://stackoverflow.com/a/26178457/3933560 . I've actually implemented my own, but I chose to post this one because it's cleaner. Either way, I get the same wrong results. What am I missing? EDIT: The code is python. Code:
from itertools import izip def normalize(poly): while poly and poly[1] == 0: poly.pop() if poly == []: poly.append(0) def poly_remainder_GFp(num, den, p): #Create normalized copies of the args num = num[:] normalize(num) den = den[:] normalize(den) if len(num) >= len(den): #Shift den towards right so it's the same degree as num shiftlen = len(num)  len(den) den = [0] * shiftlen + den else: return num quot = [] divisor = den[1] for i in xrange(shiftlen + 1): #Get the next coefficient of the quotient. mult = (num[1] * modinv(divisor,p)) % p #quot = ([mult] + quot) % p #Subtract mult * den from num, but don't bother if mult == 0 #Note that when i==0, mult!=0; so quot is automatically normalized. if mult != 0: d = [mult * u for u in den] num = [(u  v)%p for u, v in zip(num, d)] num.pop() den.pop(0) normalize(num) return num def egcd(a, b): if a == 0: return (b, 0, 1) else: g, y, x = egcd(b % a, a) return (g, x  (b // a) * y, y) def modinv(a, m): g, x, y = egcd(a, m) if g != 1: raise Exception('modular inverse does not exist') else: return x % m if __name__ == '__main__': N = [7301, 1477, 7726] D = [8, 29, 15, 1] while (D != [0]): (N,D) = (D,poly_remainder_GFp(N, D,9923)) print "%s, %s" % (N,D) print "" N = [5984,4697,7449] D = [8,29,15,1] while (D != [0]): (N,D) = (D,poly_remainder_GFp(N, D,9929)) print "%s, %s" % (N,D) Last fiddled with by paul0 on 20150115 at 10:07 
20150115, 11:27  #2 
Sep 2011
39_{16} Posts 
PS: I can't seem to edit the post anymore. Anyway, the difference of the code from the stackexchange link is that I reduce the coefficients modulo p, and I used an inverse modulo operation instead of division.
The code that runs first is the one at the bottom. I used a while loop to calculate the GCD. It terminates when the remainder becomes zero. Running this gives me [9858, 7744] and [1548] (which corresponds to 9858 + 7744x and 1548) which are both wrong. And one more thing, I've implemented the actual Chinese remainder part, it's just the prime checking that's missing now. Last fiddled with by paul0 on 20150115 at 11:44 
20150115, 13:03  #3 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{2}×3^{2}×179 Posts 
Running in Pari, I get
Code:
? f=8 + 29*x + 15*x^2 + x^3 %1 = x^3 + 15*x^2 + 29*x + 8 ? f*(Mod(1,9923)) %2 = Mod(1, 9923)*x^3 + Mod(15, 9923)*x^2 + Mod(29, 9923)*x + Mod(8, 9923) ? g=7726*x^2+1477*x+7301 %3 = 7726*x^2 + 1477*x + 7301 ? g*Mod(1,9923) %4 = Mod(7726, 9923)*x^2 + Mod(1477, 9923)*x + Mod(7301, 9923) ? gcd(f*Mod(1,9923),g*Mod(1,9923)) %5 = Mod(7744, 9923)*x + Mod(9858, 9923) Code:
? f2 = f(4214*x)*g %10 = Mod(7581, 9923)*x^2 + Mod(4838, 9923)*x + Mod(8, 9923) ? Mod(7726/7581,9923) %18 = Mod(6081, 9923) ? g2 = gf2*6081 %23 = Mod(3294, 9923)*x + Mod(8268, 9923) ? f3 = f2(6464*x)*g2 %46 = Mod(5764, 9923)*x + Mod(8, 9923) ? 5995*f3g2 %55 = Mod(0, 9923) 
20150115, 23:24  #4 
Sep 2011
71_{8} Posts 
I'm confused now. The two polynomials (7726*x^2+1477*x+7301 and x^3 + 15*x^2 + 29*x + 8) both have the same root 847 mod 9923 since:
(7726*847*847 + 1477*847 + 7301) mod 9923 = 0 (847*847*847 + 15*847*847 + 29*847 + 8) mod 9923 = 0 So, x847 should be the GCD, again from Brigg's paper (page 58, http://scholar.lib.vt.edu/theses/ava...ricted/etd.pdf ). Last fiddled with by paul0 on 20150115 at 23:29 
20150116, 01:07  #5  
Sep 2011
3·19 Posts 
Quote:
EDIT2: It looks like my GCD code return a polynomial degree 0 (a coefficient) if the GCD is 1. The proper GCD is returned otherwise. Last fiddled with by paul0 on 20150116 at 02:00 

20150116, 09:52  #6 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{2}·3^{2}·179 Posts 
The GCD is only meaningfully defined up to multiplication by a unit (IE in this case a nonzero field element); so if it's a degree0 polynomial it might as well be 1. For polynomial GCD, the "greatest" in the names means 'largest degree'.
The degree of the GCD is welldefined, the set of roots of the GCD is welldefined, but you can pick any leading coefficient you want. Last fiddled with by fivemack on 20150116 at 09:53 
20150116, 15:12  #7  
Sep 2011
111001_{2} Posts 
Quote:


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