Algorithm to find Largest prime factor of a number

Question

What is the best approach to calculating the largest prime factor of a number?

I'm thinking the most efficient would be the following:

1. Find lowest prime number that divides cleanly
2. Check if result of division is prime
3. If not, find next lowest
4. Go to 2.

I'm basing this assumption on it being easier to calculate the small prime factors. Is this about right? What other approaches should I look into?

Edit: I've now realised that my approach is futile if there are more than 2 prime factors in play, since step 2 fails when the result is a product of two other primes, therefore a recursive algorithm is needed.

Edit again: And now I've realised that this does still work, because the last found prime number has to be the highest one, therefore any further testing of the non-prime result from step 2 would result in a smaller prime.

Answer 1

#python implementation
import math
n = 600851475143
i = 2
factors=set([])
while i<math.sqrt(n):
   while n%i==0:
       n=n/i
       factors.add(i)
   i+=1
factors.add(n)
largest=max(factors)
print factors
print largest

Answer 2

Here is the same function@Triptych provided as a generator, which has also been simplified slightly.

def primes(n):
    d = 2
    while (n > 1):
        while (n%d==0):
            yield d
            n /= d
        d += 1

the max prime can then be found using:

n= 373764623
max(primes(n))

and a list of factors found using:

list(primes(n))

Answer 3

Similar to @Triptych answer but also different. In this example list or dictionary is not used. Code is written in Ruby

def largest_prime_factor(number)
  i = 2
  while number > 1
    if number % i == 0
      number /= i;
      i -= 1
    end
    i += 1
  end
  return i
end

largest_prime_factor(600851475143)
# => 6857

Answer 4

Here's the best algorithm I know of (in Python)

def prime_factors(n):
    """Returns all the prime factors of a positive integer"""
    factors = []
    d = 2
    while n > 1:
        while n % d == 0:
            factors.append(d)
            n /= d
        d = d + 1

    return factors


pfs = prime_factors(1000)
largest_prime_factor = max(pfs) # The largest element in the prime factor list

The above method runs in O(n) in the worst case (when the input is a prime number).

EDIT:
Below is the O(sqrt(n)) version, as suggested in the comment. Here is the code, once more.

def prime_factors(n):
    """Returns all the prime factors of a positive integer"""
    factors = []
    d = 2
    while n > 1:
        while n % d == 0:
            factors.append(d)
            n /= d
        d = d + 1
        if d*d > n:
            if n > 1: factors.append(n)
            break
    return factors


pfs = prime_factors(1000)
largest_prime_factor = max(pfs) # The largest element in the prime factor list

Answer 5

All numbers can be expressed as the product of primes, eg:

102 = 2 x 3 x 17
712 = 2 x 2 x 2 x 89

You can find these by simply starting at 2 and simply continuing to divide until the result isn't a multiple of your number:

712 / 2 = 356 .. 356 / 2 = 178 .. 178 / 2 = 89 .. 89 / 89 = 1

using this method you don't have to actually calculate any primes: they'll all be primes, based on the fact that you've already factorised the number as much as possible with all preceding numbers.

number = 712;
currNum = number;    // the value we'll actually be working with
for (currFactor in 2 .. number) {
    while (currNum % currFactor == 0) {
        // keep on dividing by this number until we can divide no more!
        currNum = currNum / currFactor     // reduce the currNum
    }
    if (currNum == 1) return currFactor;    // once it hits 1, we're done.
}

Answer 6

I think it would be good to store somewhere all possible primes smaller then n and just iterate through them to find the biggest divisior. You can get primes from prime-numbers.org.

Of course I assume that your number isn't too big :)