algorithmmathprimesfactorshamming-numbers

nᵗʰ ugly number


Numbers whose only prime factors are 2, 3, or 5 are called ugly numbers.

Example:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... 

1 can be considered as 2^0.

I am working on finding nth ugly number. Note that these numbers are extremely sparsely distributed as n gets large.

I wrote a trivial program that computes if a given number is ugly or not. For n > 500 - it became super slow. I tried using memoization - observation: ugly_number * 2, ugly_number * 3, ugly_number * 5 are all ugly. Even with that it is slow. I tried using some properties of log - since that will reduce this problem from multiplication to addition - but, not much luck yet. Thought of sharing this with you all. Any interesting ideas?

Using a concept similar to Sieve of Eratosthenes (thanks Anon)

    for (int i(2), uglyCount(0); ; i++) {
        if (i % 2 == 0)
            continue;
        if (i % 3 == 0)
            continue;
        if (i % 5 == 0)
            continue;
        uglyCount++;
        if (uglyCount == n - 1)
            break;
    }

i is the nth ugly number.

Even this is pretty slow. I am trying to find the 1500th ugly number.


Solution

  • A simple fast solution in Java. Uses approach described by Anon..
    Here TreeSet is just a container capable of returning smallest element in it. (No duplicates stored.)

        int n = 20;
        SortedSet<Long> next = new TreeSet<Long>();
        next.add((long) 1);
    
        long cur = 0;
        for (int i = 0; i < n; ++i) {
            cur = next.first();
            System.out.println("number " + (i + 1) + ":   " + cur);
    
            next.add(cur * 2);
            next.add(cur * 3);
            next.add(cur * 5);
            next.remove(cur);
        }
    

    Since 1000th ugly number is 51200000, storing them in bool[] isn't really an option.

    edit
    As a recreation from work (debugging stupid Hibernate), here's completely linear solution. Thanks to marcog for idea!

        int n = 1000;
    
        int last2 = 0;
        int last3 = 0;
        int last5 = 0;
    
        long[] result = new long[n];
        result[0] = 1;
        for (int i = 1; i < n; ++i) {
            long prev = result[i - 1];
    
            while (result[last2] * 2 <= prev) {
                ++last2;
            }
            while (result[last3] * 3 <= prev) {
                ++last3;
            }
            while (result[last5] * 5 <= prev) {
                ++last5;
            }
    
            long candidate1 = result[last2] * 2;
            long candidate2 = result[last3] * 3;
            long candidate3 = result[last5] * 5;
    
            result[i] = Math.min(candidate1, Math.min(candidate2, candidate3));
        }
    
        System.out.println(result[n - 1]);
    

    The idea is that to calculate a[i], we can use a[j]*2 for some j < i. But we also need to make sure that 1) a[j]*2 > a[i - 1] and 2) j is smallest possible.
    Then, a[i] = min(a[j]*2, a[k]*3, a[t]*5).