Range is too large Python

This is what I would do:

def prime_factors(x):
    factors = []
    while x % 2 == 0:
        factors.append(2)
        x /= 2
    i = 3
    while i * i <= x:
        while x % i == 0:
            x /= i
            factors.append(i)
        i += 2
    if x > 1:
        factors.append(x)
    return factors

>>> prime_factors(600851475143)
[71, 839, 1471, 6857]

It's pretty fast and I think it's right. It's pretty simple to take the max of the factors found.

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2017-11-08

Returning to this 5 years later, I would use yield and yield from plus faster counting over the prime range:

def prime_factors(x):
    def diver(x, i):
        j = 0
        while x % i == 0:
            x //= i
            j += 1
        return x, [i] * j
    for i in [2, 3]:
        x, vals = diver(x, i)
        yield from vals
    i = 5
    d = {5: 2, 1: 4}
    while i * i <= x:
        x, vals = diver(x, i)
        yield from vals
        i += d[i % 6]
    if x > 1:
        yield x

list(prime_factors(600851475143))

The dict {5: 2, 1: 4} uses the fact that you don't have to look at all odd numbers. Above 3, all numbers x % 6 == 3 are multiples of 3, so you need to look at only x % 6 == 1 and x % 6 == 5, and you can hop between these by alternately adding 2 and 4, starting from 5.

I would definitely stick with xrange since creating a list between 0 and what looks like a number rivaled by infinity would be taxing for memory. xrange will generate only the numbers when asked. For the number too large problem, you might want to try a "long". This can be achieved by writing a L on the end of the number. I made my own version to test it out. I put in a small sleep as to not destroy my computer into virtually a while(1) loop. I was also impatient to see the program come to a complete end, so I put in print statements

from time import sleep

x = 600851475143L
maxPrime = 0

for i in xrange(1,x):
    isItPrime = True
    if (x%i) == 0:
        for prime in xrange(2,i-1):
            if (i%prime) == 0:
                isItPrime = False
                break
        if isItPrime:
            maxPrime = i
            print "Found a prime: "+str(i)
    sleep(0.0000001)


print maxPrime

Hope this helps!

EDIT: I also did a few more edits to yield this version. It is fairly efficient and I checked quite a few numbers this program provides (it seems to check out so far):

from time import sleep

x = 600851475143L

primes = []

for i in xrange(2,x):
    isItPrime = True
    for prime in primes:
        if (i%prime) == 0:
            isItPrime = False
            break
    if isItPrime:
        primes.append(i)
        print "Found a prime: "+str(i)
    sleep(0.0000001)


print primes[-1]

The accepted answer suggests a drop-in replacement for xrange, but only covers one case. Here is a more general drop-in replacement.

def custom_range(start=0,stop=None,step=1):
    '''xrange in python 2.7 fails on numbers larger than C longs.
    we write a custom version'''
    if stop is None:
        #handle single argument case. ugly...
        stop = start
        start = 0
    i = start
    while i < stop:
        yield i
        i += step

xrange=custom_range

In old (2.x) versions of Python, xrange can only handle Python 2.x ints, which are bound by the native long integer size of your platform. Additionally, range allocates a list with all numbers beforehand on Python 2.x, and is therefore unsuitable for large arguments.

You can either switch to 3.x (recommended), or a platform where long int (in C) is 64 bit long, or use the following drop-in:

import itertools
range = lambda stop: iter(itertools.count().next, stop)

Equivalently, in a plain form:

def range(stop):
   i = 0
   while i < stop:
       yield i
       i += 1