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

n = abs(number);
result = 1;
if (n mod 2 == 0) {
  result = 2;
  while (n mod 2 = 0) n /= 2;
}
for(i=3; i<sqrt(n); i+=2) {
  if (n mod i == 0) {
    result = i;
    while (n mod i = 0)  n /= i;
  }
}
return max(n,result)

There are some modulo tests that are superflous, as n can never be divided by 6 if all factors 2 and 3 have been removed. You could only allow primes for i, which is shown in several other answers here.

You could actually intertwine the sieve of Eratosthenes here:

• First create the list of integers up to sqrt(n).
• In the for loop mark all multiples of i up to the new sqrt(n) as not prime, and use a while loop instead.
• set i to the next prime number in the list.

Also see this question.

Answer 2

Python Iterative approach by removing all prime factors from the number

def primef(n):
    if n <= 3:
        return n
    if n % 2 == 0:
        return primef(n/2)
    elif n % 3 ==0:
        return primef(n/3)
    else:
        for i in range(5, int((n)**0.5) + 1, 6):
            #print i
            if n % i == 0:
                return primef(n/i)
            if n % (i + 2) == 0:
                return primef(n/(i+2))
    return n

Answer 3

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include <time.h>

factor(long int n)
{
long int i,j;
while(n>=4)
 {
if(n%2==0) {  n=n/2;   i=2;   }

 else
 { i=3;
j=0;
  while(j==0)
  {
   if(n%i==0)
   {j=1;
   n=n/i;
   }
   i=i+2;
  }
 i-=2;
 }
 }
return i;
 }

 void main()
 { 
  clock_t start = clock();
  long int n,sp;
  clrscr();
  printf("enter value of n");
  scanf("%ld",&n);
  sp=factor(n);
  printf("largest prime factor is %ld",sp);

  printf("Time elapsed: %f\n", ((double)clock() - start) / CLOCKS_PER_SEC);
  getch();
 }

Answer 4

JavaScript code:

'option strict';

function largestPrimeFactor(val, divisor = 2) { 
    let square = (val) => Math.pow(val, 2);

    while ((val % divisor) != 0 && square(divisor) <= val) {
        divisor++;
    }

    return square(divisor) <= val
        ? largestPrimeFactor(val / divisor, divisor)
        : val;
}

Usage Example:

let result = largestPrimeFactor(600851475143);

Here is an example of the code:

Answer 5

I'm aware this is not a fast solution. Posting as hopefully easier to understand slow solution.

 public static long largestPrimeFactor(long n) {

        // largest composite factor must be smaller than sqrt
        long sqrt = (long)Math.ceil(Math.sqrt((double)n));

        long largest = -1;

        for(long i = 2; i <= sqrt; i++) {
            if(n % i == 0) {
                long test = largestPrimeFactor(n/i);
                if(test > largest) {
                    largest = test;
                }
            }
        }

        if(largest != -1) {
            return largest;
        }

        // number is prime
        return n;
    } 

Answer 6

Found this solution on the web by "James Wang"

public static int getLargestPrime( int number) {

    if (number <= 1) return -1;

    for (int i = number - 1; i > 1; i--) {
        if (number % i == 0) {
            number = i;
        }
    }
    return number;
}