Comments on: 2009 is not a prime number. A python program to compute factors.

By: David Sterry

David Sterry — Mon, 29 Jun 2009 05:35:06 +0000

@Jaime Nice job. I’m surprised it took someone so long to mention this algorithm. When I wrote some code in C to do this, I was trying to generate all primes up to a certain number and I stored the primes I found just as you said. IIRC it took about 8 minutes to generate all primes under 10^7 on a P3 866Mhz.

By: Gurjeet

Gurjeet — Sun, 31 May 2009 08:44:52 +0000

How aboout going with 6K + 1 where K is a Whole Number..
This gives all prime numbers with some numbers whose factors are prime numbers, like 25, 35, etc… So this is also not 100% correct, but can b helpful…

By: Hmm

Hmm — Fri, 22 May 2009 19:55:00 +0000

Why does it output 1? The canonical way to write the prime factors of 4 is [2,2], not [1,2,2].

From Wikipedia, “Ω(n) for n = 1, 2, 3, … is 0, 1, 1, 2, …”

Ω(1) = 0 –> []
Ω(2) = 1 –> [2]
Ω(3) = 1 –> [3]
Ω(4) = 2 –> [2,2]
…

By: Jaime

Jaime — Thu, 26 Feb 2009 12:16:44 +0000

I hope you people realize how inefficient the code you've written is… After you've divided out all the 2's from the number, you still try to find out if it is divisible by 4,6,8,10,12… If nothing else, you should only do trial division by odd numbers. Or even better, if you can afford the memory cost, generate first a list of all prime numbers below sqrt(n), and then do trial division by those only…

By: deeptext

deeptext — Sun, 04 Jan 2009 21:08:19 +0000

Yes, and I needed to know. What is the smallest prime factor?
I guessed around: 7

(Also, your “copy to clipboard” links don't work for me).

By: Gerry B

Gerry B — Fri, 02 Jan 2009 22:45:00 +0000

There are supposed to be tab indentations. Not sure how to keep the tabs in a post

By: Gerry B

Gerry B — Fri, 02 Jan 2009 22:42:37 +0000

I just wrote this code (works in python 3.0 with the special "//" for division of integers). This code is much faster and will find your prime factors. I take advantage of the fact that 30=2*3*5, so any primes above 30 must be 1,7,11,13,17,23,or 29 modulo 30, otherwise the number would be divisible by 2,3 or 5. Here is the function:

def pfast(x):
remder = x
outcat="1"
while 1:
if remder%2==0:
outcat=outcat+"*2"
remder=remder//2
else:
break
while 1:
if remder%3==0:
outcat=outcat+"*3"
remder=remder//3
else:
break
while 1:
if remder%5==0:
outcat=outcat+"*5"
remder=remder//5
else:
break
while 1:
if remder%7==0:
outcat=outcat+"*7"
remder=remder//7
else:
break
while 1:
if remder%11==0:
outcat=outcat+"*11"
remder=remder//11
else:
break
while 1:
if remder%13==0:
outcat=outcat+"*13"
remder=remder//13
else:
break
while 1:
if remder%17==0:
outcat=outcat+"*17"
remder=remder//17
else:
break
while 1:
if remder%19==0:
outcat=outcat+"*19"
remder=remder//19
else:
break
while 1:
if remder%23==0:
outcat=outcat+"*23"
remder=remder//23
else:
break
while 1:
if remder%29==0:
outcat=outcat+"*29"
remder=remder//29
else:
break
c=0
while 1:
top = remder**0.5+1
c=c+30
if c>top:
return outcat+"*"+str(remder)
while 1:
if remder%(c+1)==0:
outcat=outcat+"*"+str(c+1)
remder=remder//(c+1)
else:
break
while 1:
if remder%(c+7)==0:
outcat=outcat+"*"+str(c+7)
remder=remder//(c+7)
else:
break
while 1:
if remder%(c+11)==0:
outcat=outcat+"*"+str(c+11)
remder=remder//(c+11)
else:
break
while 1:
if remder%(c+13)==0:
outcat=outcat+"*"+str(c+13)
remder=remder//(c+13)
else:
break
while 1:
if remder%(c+17)==0:
outcat=outcat+"*"+str(c+17)
remder=remder//(c+17)
else:
break
while 1:
if remder%(c+23)==0:
outcat=outcat+"*"+str(c+23)
remder=remder//(c+23)
else:
break
while 1:
if remder%(c+29)==0:
outcat=outcat+"*"+str(c+29)
remder=remder//(c+29)
else:
break

By: Dhananjay Nene

Dhananjay Nene — Fri, 02 Jan 2009 22:40:21 +0000

Blame it on google .. it got me the other link first and of course the fact that I wanted to get it done quickly.

By: wj32

wj32 — Fri, 02 Jan 2009 22:28:39 +0000

You didn't see my prime factorization code, then? http://wj32.wordpress.com/2007/12/08/prime-fact...

By: audax

audax — Fri, 02 Jan 2009 22:10:36 +0000

[i for i in factor(int(sys.argv[index]))]
should be just list([factor(int(sys.argv[index])))

By: Gerry B

Gerry B — Fri, 02 Jan 2009 22:02:48 +0000

sorry I just realized I am using Python 3.0, which changed the rule for division compared to old Python. Now, to get what you did, the division sign is // and not simply /

By: Gerry B

Gerry B — Fri, 02 Jan 2009 21:30:31 +0000

How do you get Python to do more than 18 decimal places of accuracy in a division? When I try your algo I get the following factors:

[1037468051.0, 863693, 7, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 29]

By: Dhananjay Nene

Dhananjay Nene — Fri, 02 Jan 2009 20:07:43 +0000

I could've .. just chose n**0.5 instead of sqrt(n). Thats already in the code.

By: varun thacker

varun thacker — Fri, 02 Jan 2009 19:32:49 +0000

why dont you keep the limit variable to like sqrt(n).That can be the highest factor other than itelsf

regards,
Varun