# Integer factorization, the fast way?

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Ok, i've been building a small integer factorizer. It is currenly very slow. It takes 2.8 seconds to factorize 102,812,541,257 into 53 * 6,199 * 312,931. It is currently using a few things, First, it checks the common divisors (2,3,5,7 and 11). if it is divisible by those, then it has the factors almost instantly. Then it sqrt's the number. if it is < 200 then it does a trial division search (all numbers < 200, takes < 1 second) And finally, if it is > 200, then it uses a modified sieve of Eratosthenes. Basically, it checks n. if n isn't a factor, then 2n, 3n, 4n, 5n, on. will not be a factor, so you can rule them off your list. It is quite slow for big numbers, but spends most of its time removing all the common factors (2, 3, 5,7 and 11) and i don't know how to get rid of them beforehand. My question is: What is a method, which is faster then my sieve, and which is not too hard (without explanation, some things i just don't understand yet...), to understand or code. From, Nice coder

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That is actually the sieve of Eratosthenese[edit can't spell]..;)

To make it as fast as possible, implement an Extended Euclidean Algorithm. That is about as fast as you'll find. It's also what is used for a vast majority of encryption/decryption techniques, so you know it is fast.

Here's a link for javascript implementation:
http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

Here's a link with a more algorithmic-orientation:
http://www-math.cudenver.edu/~wcherowi/courses/m5410/exeucalg.html

Hope this helps ya.

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Quote:
 Original post by Alex SwinneyThat is actually the sieve of Eratosthenese[edit can't spell]..;)To make it as fast as possible, implement an Extended Euclidean Algorithm. That is about as fast as you'll find. It's also what is used for a vast majority of encryption/decryption techniques, so you know it is fast.You will have to adapt it to your situation, though.Here's a link for javascript implementation:http://en.wikipedia.org/wiki/Extended_Euclidean_algorithmHere's a link with a more algorithmic-orientation:http://www-math.cudenver.edu/~wcherowi/courses/m5410/exeucalg.htmlHope this helps ya.

Thanks, it does help me a bit :)

I'm limited to 100 trillion (15 digits before vb stuffs up the numbering system.) so, any algorith which isn't too slow should work.

What i was wondering about, tho is how to use that to get the factors...

I also wouldn't mind dumping the seive, for something a bit more elegent (my sieve is a little hacky... well, maybe more then a little).

From,
Nice coder

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? i was logged in before...
Above post is mine

From,
Nice coder

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You can get rid of them by taking their product and checking +/-1 and +/- all the primes greater than the highest prime in the product and less than half the product. To eliminate all multiples of the first five primes takes a list of 184 primes. So for example say your product was 2 then you check 2n+1, for 2*3 you check 6n+/-1, for 2*3*5 you check 30n+/-{1,7,11,13}, for 2*3*5*7 you check 210n+/-{1,11,13,...,97,101,103).

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Quote:
 Original post by LilBudyWizerYou can get rid of them by taking their product and checking +/-1 and +/- all the primes greater than the highest prime in the product and less than half the product. To eliminate all multiples of the first five primes takes a list of 184 primes. So for example say your product was 2 then you check 2n+1, for 2*3 you check 6n+/-1, for 2*3*5 you check 30n+/-{1,7,11,13}, for 2*3*5*7 you check 210n+/-{1,11,13,...,97,101,103).

Thank you!

Now if only i could find a way to implement one of the faster approaches... (so that i wouldn' need to do this at all)

From,
Nice coder

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This should do the trick.
#include <stdlib.h>#include <iostream.h>int main(){__int64 number, j, k;short int again; system("cls"); cout << "\n\n\t\t\t Prime Finder\n\n"; cout << "\n\n\t\t\t By: Jared Stewart.\n\n\n\n\n\t\t "; system("PAUSE"); system("cls"); do { system("cls"); cout << "Please enter a number to get the prime factors: "; cin >> number; j = 2; system("cls"); cout << "The prime factors are: "; do { if (number == 2) break; if (number % j == 0 ) //check if its a factor { cout << j << " "; number = number/j; //make the number to check smaller } else j++; } while (j < 3); for (j =3; j <= sqrt(number); ) { if (number % j == 0 ) //check if its a factor { cout << j << " "; number = number/j; //make the number to check smaller } else j+=2; } cout << number; cout << "\n\n"; cout << "To test a new number, press 1. To quit, press 2.\n\n\n"; cin >> again;} while (again ==1);}

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Quote:
 Original post by Jareds411This should do the trick.

Trial division,
very slow!

ok, i got rid of my sieve (its slower then trial division!).

And i replaced it with very simple trial division

Dim sn As DoubleDim i As DoubleDim j As DoubleDim fin As DoubleDim t As DoubleDim k As DoubleDim out(1) As Doublesn = Int(Sqr(n)) + 1fin = Int(sn / 2) + 1i = 12j = sn + 1Doj = j - 1i = i + 1k = number / it = number / jIf Int(k) = k Then    out(0) = k    out(1) = i    factorize = out    Exit FunctionEnd IfIf Int(t) = t Then    out(0) = t    out(1) = j    factorize = out    Exit FunctionEnd IfDoEventsLoop Until j < finout(0) = 1out(1) = numberfactorize = out

Very simple, not very effective (but it still finds factors that are close to 13 and close to sqr(n), quite fast.

Is there any faster way?
From,
Nice coder

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There are faster ways, but they involve fun and games with number theory. An example is Quadratic Sieve.

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!!!!!

Thats a little... (ok way) over my head.

First, what is p?

Second, How Do you solve the congruencies on the factor base?

What is Dixons factorization?
What is Gausian elemination?

The quadratic prime sieve is nice...
But the factorization sieve is very complicated.

From,
Nice coder

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