Sieving A Polynomial
May 30, 2017
I watch Stack Overflow and Reddit for questions about prime numbers. Most of the questions are basic, many are confused, a few are interesting, and some are just bizarre. A recent question falls in the latter category:
I need to make a list of the prime factors with multiplicity of all the numbers up to 300,001 that are equivalent to 1 modulo 4. My strategy is to make a list of the numbers 1 modulo 4, then calculate the factors of each. I don’t know how to calculate factors, but I know it’s complicated, so I figure somebody has already built a factoring program and put it on the web. Does anybody know a web API that I can call to determine the factors of a number?
For instance, if we limit the numbers to 50, the desired output is:
5 (5) 9 (3 3) 13 (13) 17 (17) 21 (3 7) 25 (5 5) 29 (29) 33 (3 11) 37 (37) 41 (41) 45 (3 3 5) 49 (7 7)
Calling a web API 75,000 times is a spectacularly wrong way to build the list, but I pointed the questioner to Wolfram|Alpha. Long-time readers of this blog know how to solve the problem directly, which we will do in today’s exercise.
Your task is to write a program to factor all the numbers up to 300,001 that are equivalent to 1 modulo 4. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
Simple Python3 with lots of space for optimization, possibly
from __future__ import division N = 50 def Sieve(m): tmp = [False] * (m + 1) for k in range(3, m + 1, 2): tmp[k] = True i = 3 while i * i <= m: if tmp[i]: j = i * i while j <= m: tmp[j] = False j += i + i i += 2 primes = [k for k in range(3, m + 1, 2) if tmp[k] == True] return primes # if __name__ == "__main__": primes = Sieve(N) for k in range(1, N + 1, 4): f = [] x = k for p in primes: while (x % p == 0): f.append(p) x //= p if x == 1: break print ("%6d %s" % (k, f))Output
1 [] 5 [5] 9 [3, 3] 13 [13] 17 [17] 21 [3, 7] 25 [5, 5] 29 [29] 33 [3, 11] 37 [37] 41 [41] 45 [3, 3, 5] 49 [7, 7]In python:
def factorize(number): # gives a list of all prime factors of number factors = [] while not (number % 2): number = number // 2 factors.append(2) i = 3 while i <= number**0.5 and number-1: if not (number % i): factors.append(i) number = number // i else: i += 2 if number != 1 and i >= number**0.5: factors.append(number) return factors for i in range(1,50,4): print(i, factorize(i)) # outputs # 1 [] # 5 [5] # 9 [3, 3] # 13 [13] # 17 [17] # 21 [3, 7] # 25 [5, 5] # 29 [29] # 33 [3, 11] # 37 [37] # 41 [41] # 45 [3, 3, 5] # 49 [7, 7]Above solution takes 0.42 seconds for all numbers up to 75000 on my machine. Below solution uses a bigger wheel and goes for Brent algorithm trial division for larger numbers. Takes 0.08 seconds for first 75000.
import time from fractions import gcd from random import randint from functools import reduce def brent(N): # brent returns a divisor not guaranteed to be prime, returns n if n prime if N % 2 == 0: return 2 y, c, m = randint(1, N-1), randint(1, N-1), randint(1, N-1) g, r, q = 1, 1, 1 while g == 1: x = y for i in range(r): y = ((y*y) % N + c) % N k = 0 while (k < r and g == 1): ys = y for i in range(min(m,r-k)): y = ((y*y) % N + c) % N q = q*(abs(x-y)) % N g = gcd(q,N) k = k + m r = r*2 if g == N: while True: ys = ((ys*ys) % N + c) % N g = gcd(abs(x-ys), N) if g > 1: break return g def factorize(n, len_wheel, trial_division_bound = 1000000): result = [] if n <= 0: raise if n == 1: return [1] for start_prime in first_primes: while n % start_prime == 0: result.append(start_prime) n //= start_prime if n == 1: return result idx = 0 i = first_primes[-1] wheel[idx] #use trial division for factors below bound while i <= trial_division_bound: inc = wheel[idx] i += inc idx = (idx + 1) % len_wheel if i > n**0.5: result.append(n) return result while n % i == 0: result.append(i) n //= i if n == 1: return result #use brent for factors > bound p = 0 while n > trial_division_bound: p1 = n #iterate until n=brent(n) => n is prime while p1 != p: p = p1 p1 = brent(p) result.append(p1) n //= p1 if n != 1: result.append(n) return result first_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199] wheel = [2, 10, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6, 2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10] start_time = time.time() for i in range(1,75000,4): factorize(i, len(wheel)) print(time.time() - start_time)Good ‘ol shell:
There’s factor in coreutils? That’s cool.
(And surely the spec includes 1.)
@Rutger: Please time the following function using the same parameters as your other timings:
def prod(xs): p = 1 for n in xs: p *= n return pdef primes(n): # sieve of eratosthenes i, p, ps, m = 0, 3, [2], n // 2 sieve = [True] * m while p <= n: if sieve[i]: ps.append(p) for j in range((p*p-3)/2, m, p): sieve[j] = False i, p = i+1, p+2 return psdef factors(n): # naive trial division f, fs = 2, [] while f * f <= n: if n % f == 0: n /= f fs.append(f) else: f += 1 fs.append(n) return fsdef factors_1mod4(n): from math import sqrt len = (n - 1) // 4 factors = [[] for i in range(len + 1)] def sieve(p, pp): k = pp if pp % 4 == 1 else 3 * pp k = (k - 1) // 4 for i in range(k, len+1, pp): factors[i].append(p) for p in primes(int(sqrt(n)))[1:]: pp = p while pp <= n: sieve(p, pp) pp *= p for k in range(1, len+1): m = 4 * k + 1 f = m / prod(factors[k]) if factors[k] == []: factors[k] = [m] elif f != 1: factors[k].append(f) return factors[1:]@Jussi: Unix has had
factorforever. It uses naive trial division.@Praxis, I never knew.
The info page for the GNU coreutils factor says it uses “the Pollard Rho algorithm” which is “particularly effective for numbers with relatively small factors” (when it’s compiled with the GNU MP library). It recommends other methods for factoring products of large primes.
@Jussi: The V7 source code is in assembler, HERE; the man page is HERE.
I’m not an assembler expert, but I see things like
that looks an awful lot like trial division (there is no division in the Pollard Rho algorithm), and something like
that looks an awful lot like trial division by odd numbers.
I also note that the man page mentions the time complexity as O(sqrt n), which is characteristic of trial division, not Pollard Rho.
I’ve always been under the impression that Unix
factorused trial division. I’m not surprised that the GNU folks changed the algorithm.MUMPS V1 SIEVING ;New routine new ans,flg,i,i2,j,k,max set max=50 for i=5:4:max d . new ans . set i2=i,j=i**.5 . set:j#2=0 j=j-1 . for k=j:-2:3 d . . if i2#k=0,$$isprime(k) do . . . do add(.ans,k) . . . set i2=i2/k . . . for quit:i2#k>0 do add(.ans,k) set i2=i2/k . if i2>1,$$isprime(k) do add(.ans,i2) . write !,i,?9,"(" . set flg=0,j="" . for set j=$o(ans(j)) quit:j="" do . . for k=1:1:ans(j) write $select('flg:j,1:" "_j) set:'flg flg=1 . write ")" quit ; add (ans,n) ; Add n to ans() set ans(n)=$get(ans(n))+1 quit ; isprime (k) ; Is it a prime? new flg,i,sqr if $data(primes(k)) set flg=1 else do . set flg=1,sqr=k**.5 . for i=sqr:-1:3 if k#i=0 set flg=0 q quit flg MCL> d ^SIEVING 5 (5) 9 (3 3) 13 (13) 17 (17) 21 (3 7) 25 (5 5) 29 (29) 33 (3 11) 37 (37) 41 (41) 45 (3 3 5) 49 (7 7) MCL>@Praxis, thanks. I see also an interfance change between V7 and GNU factor: V7 takes one argument (or none), GNU takes any number of arguments and reports on all of them.
To understate the effect, there’s a noticable difference between the running times of these (on a Red Hat server with recent coreutils, but the point should be robust):
@programmingpraxis: tried timing your reply, but it uses an undefined primes() function.
@Rutger: Sorry. I added
primesandprod, both of which are called by the function.Had a little more modifications to the code, but here is the result: 0.0156 seconds.
from math import sqrt from functools import reduce from operator import mul def primes(n): # sieve of eratosthenes i, p, ps, m = 0, 3, [2], n // 2 sieve = [True] * m while p <= n: if sieve[i]: ps.append(p) for j in range((p*p-3)//2, m, p): sieve[j] = False i, p = i+1, p+2 return ps def factors(n): # naive trial division f, fs = 2, [] while f * f <= n: if n % f == 0: n /= f fs.append(f) else: f += 1 fs.append(n) return fs def factors_1mod4(n): len = (n - 1) // 4 factors = [[] for i in range(len + 1)] def sieve(p, pp): k = pp if pp % 4 == 1 else 3 * pp k = (k - 1) // 4 for i in range(k, len+1, pp): factors[i].append(p) for p in primes(int(sqrt(n)))[1:]: pp = p while pp <= n: sieve(p, pp) pp *= p for k in range(1, len+1): m = 4 * k + 1 f = m / reduce(mul, factors[k], 1) if factors[k] == []: factors[k] = [m] elif f != 1: factors[k].append(f) return factors[1:] import time start_time = time.time() n = 75000 x = factors_1mod4(n) print(time.time() - start_time)@Rutger: So to process n = 75,000, it’s 0.42 seconds for trial division by 2 and odds, a five-times improvement to 0.08 seconds for a 2,3,5,7-wheel (since n = 75,000 you never call Brent-rho), and another five-times improvement to 0.0156 for sieving. And sieving has a slower order of growth than the other two, O(log n log log n) compared to O(sqrt n). Sieving always wins!
Thank you for doing the timing comparison.
May 30th, 2017.c:
#include "seal_bool.h" /* <http://GitHub.com/sealfin/C-and-C-Plus-Plus/blob/master/seal_bool.h> */ #include "printDateAndTime.h" /* <http://Gist.GitHub.com/sealfin/6d35f3a3958bd6797a0f> */ #include <stdio.h> #include <stdlib.h> #define N 300002L #if N <= 0 #error "N ≤ 0." #endif bool g_isPrime[ N ]; long g_numberOfPrimes = 0, *g_primes; bool f_IsPrime( const long p ) { if( p % 2 == 0 ) return p == 2; else return g_isPrime[ p ]; } void p_RecursivelyPrintPrimeFactorsOfNumber( const long p_number, const bool p_firstPrimeFactor, FILE *p_fileToPrintTo ) { if( !p_firstPrimeFactor ) fprintf( p_fileToPrintTo, " " ); if( f_IsPrime( p_number )) fprintf( p_fileToPrintTo, "%ld", p_number ); else { long i = 0; for( ; i < g_numberOfPrimes; i ++ ) if( p_number % g_primes[ i ] == 0 ) { fprintf( p_fileToPrintTo, "%ld", g_primes[ i ] ); p_RecursivelyPrintPrimeFactorsOfNumber( p_number / g_primes[ i ], false, p_fileToPrintTo ); break; } } } #define p_PrintPrimeFactorsOfNumber( p_number, p_fileToPrintTo ) p_RecursivelyPrintPrimeFactorsOfNumber( p_number, true, p_fileToPrintTo ) void main( void ) { p_PrintDateAndTime(); { long i = 0, k; FILE *f; /* Firstly, let's determine the prime numbers in the range [ 0, N ) so that later we can avoid trying to determine the prime factors of a prime number. */ for( ; i < N; i ++ ) g_isPrime[ i ] = true; if( N >= 1 ) g_isPrime[ 0 ] = false; if( N >= 2 ) g_isPrime[ 1 ] = false; if( N >= 3 ) g_numberOfPrimes ++; for( i = 3; i < N; i += 2 ) if( g_isPrime[ i ] ) { g_numberOfPrimes ++; for( k = i + i; k < N; k += i ) g_isPrime[ k ] = false; } /* Secondly, let's create a list of the prime numbers in the range [ 0, N ) so that later we can test if a prime number is a factor of a number. */ g_primes = ( long* )malloc( sizeof( long ) * g_numberOfPrimes ); k = 0; if( N >= 3 ) g_primes[ k ++ ] = 2; for( i = 3; k < g_numberOfPrimes; i += 2 ) if( f_IsPrime( i )) g_primes[ k ++ ] = i; /* Now, for every number i in the range [ 2, N ), let's determine if i % 4 == 1, and print the number and its prime factors if it is. */ f = fopen( "May 30th, 2017.out", "w" ); for( i = 2; i < N; i ++ ) if( i % 4 == 1 ) { int n; fprintf( f, "%ld%n", i, &n ); for( ; n < 6; n ++ ) fprintf( f, " " ); fprintf( f, "\t(" ); p_PrintPrimeFactorsOfNumber( i, f ); fprintf( f, ")\n" ); } fclose( f ); free( g_primes ); } p_PrintDateAndTime(); }May 30th, 2017.out:
On an Apple Power Mac G4 (AGP Graphics) (450MHz processor, 1GB memory) to run the solution took:
approximately forty-six seconds on Mac OS 9.2.2 (International English) (the solution interpreted using Leonardo IDE 3.4.1);
approximately one second on Mac OS X 10.4.11 (the solution compiled using Xcode 2.2.1).
(I’m just trying to solve the problems posed by this ‘site whilst I try to get a job; I’m well aware that my solutions are far from the best – but, in my defence, I don’t have any traditional qualifications in computer science :/ )