Algorithm to factor into a few factors as possible

Question

Is there a know algorithm to factor an integer into as few factors as possible (not necessarily prime) where every factor is less than some given constant N?

I don't care about numbers with a prime factor greater than N. Also, I'm not dealing with numbers greater than a few million and the factoring is part of the processing initialization, so I'm not especially worried about computational complexity.

EDIT: Just to be clear. I already have code find the prime factors. I'm looking for a way to combine those factors into as few composite factors as possible while keeping each factor less than N.

Answer 1

You can solve your problem by dividing it into two parts:

1. Factorize your number into primes using any of the standard techniques. For a number of only a few million, trial division would be perfectly fine.

2. Take the logarithm of each factor, and pack them into bins of size log N.

Now, bin packing is NP-hard but in practice it is possible to find good approximate solutions using simple techniques: the first-fit algorithm packs no more than 11/9 times the optimal number of bins (plus one bin).

Here's an implementation in Python:

from math import exp, log, sqrt
import operator

def factorize(n):
    """
    Factorize n by trial division and yield the prime factors.

    >>> list(factorize(24))
    [2, 2, 2, 3]
    >>> list(factorize(91))
    [7, 13]
    >>> list(factorize(999983))
    [999983]
    """
    for p in xrange(2, int(sqrt(n)) + 1):
        while n % p == 0:
            yield p
            n //= p
        if n == 1:
            return
    yield n

def product(s):
    """
    Return the product of the items in the sequence `s`.

    >>> from math import factorial
    >>> product(xrange(1,10)) == factorial(9)
    True
    """
    return reduce(operator.mul, s, 1)

def pack(objects, bin_size, cost=sum):
    """
    Pack the numbers in `objects` into a small number of bins of size
    `bin_size` using the first-fit decreasing algorithm. The optional
    argument `cost` is a function that computes the cost of a bin.

    >>> pack([2, 5, 4, 7, 1, 3, 8], 10)
    [[8, 2], [7, 3], [5, 4, 1]]
    >>> len(pack([6,6,5,5,5,4,4,4,4,2,2,2,2,3,3,7,7,5,5,8,8,4,4,5], 10))
    11
    """
    bins = []
    for o in sorted(objects, reverse=True):
        if o > bin_size:
            raise ValueError("Object {0} is bigger than bin {1}"
                             .format(o, bin_size))
        for b in bins:
            new_cost = cost([b[0], o])
            if new_cost <= bin_size:
                b[0] = new_cost
                b[1].append(o)
                break
        else:
            b = [o]
            bins.append([cost(b), b])
    return [b[1] for b in bins]

def small_factorization(n, m):
    """
    Factorize `n` into a small number of factors, subject to the
    constraint that each factor is less than or equal to `m`.

    >>> small_factorization(2400, 40)
    [25, 24, 4]
    >>> small_factorization(2400, 50)
    [50, 48]
    """
    return [product(b) for b in pack(factorize(n), m, cost=product)]

Answer 2

I don't know if there is an established algorithm, but I would try the following

public static List<Integer> getFactors(int myNumber, int N) {
    int temp=N;
    int origNumber=myNumber;        
    List<Integer> results=new ArrayList<Integer>();
    System.out.println("Factors of "+myNumber+" not greater than "+N);
    while (temp>1) {            
        if (myNumber % temp == 0) {
            results.add(temp);
            myNumber/=temp;                                
        } else {
            if (myNumber<temp) {
                temp= myNumber;                    
            } else {
                temp--;
            }
        }
    }
    for (int div : results) {
        origNumber/=div;
    }
    if (origNumber>1) {
        results.clear();
    }        
    return(results);
}

I hope it helps.