Nested For Loops Dynamic Depth Java

Question

Hello I'm new to programming and registered to this forum :)

So I created a little program with nested for loops that prints out all combinations of five numbers which can have a value from 0 to 5. With nested for-loops this works fine. But isn't there a cleaner solution? I tried it with calling the for loop itself, but my brain doesn't get the solution.. :(

//my ugly solution
int store1, store2, store3, store4, store5;
        for (int count = 0; count <= 5; count++) {
            store1 = count;
            for (int count2 = 0; count2 <= 5; count2++) {
                store2 = count2;
                for (int count3 = 0; count3 <= 5; count3++) {
                    store3 = count3;
                    for (int count4 = 0; count4 <= 5; count4++) {
                        store4 = count4;
                        System.out
                                .println(store1 + " " + store2 + " " + store4);
                }
 //I'm trying around with something like this 
    void method1() {
        for (int count = 0; count <= 5; count++) {
                    list.get(0).value = count;
            count++;
            method2();
        }
    }
    void method2() {
        for (int count = 0; count <= 5; count++) {
                    list.get(1).value = count;
            count++;
            method1();
        }
    }

Yeah, by the search today I stumbled it sometimes, but I didn't really understand how to implement it. — harfe
– harfe, Commented Dec 13, 2012 at 17:35
Why do you increment count INSIDE the loop when it is being incremented in BY the for loop? — Scott Hunter
– Scott Hunter, Commented Dec 13, 2012 at 17:39

Jarno · Accepted Answer · 2019-06-19 17:02:04Z

3

Usually when people try to use recursion or functional, using a loop is simpler or faster. However, in this case recursion is the simpler option in combination with a loop.

public static void method(List<Integer> list, int n, int m) {
   if (n < 0) {
       process(list);
   } else {
      for(int i = 0; i < m; i++) {
         list.set(n, i);
         method(list, n-1, m);
      }
   }
}

edited Jun 19, 2019 at 17:02

Jarno

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answered Dec 13, 2012 at 17:42

Peter Lawrey

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1 Comment

harfe Over a year ago

Thank you, I'm trying it out.

John Eipe · Accepted Answer · 2012-12-13 18:04:54Z

2

I know that you are trying combinations but this might help.

Permutation with repetitions

When you have n things to choose from ... you have n choices each time!

When choosing r of them, the permutations are:

n × n × ... (r times) = n^r

//when n and r are known statically

class Permutation
{
    public static void main(String[] args)
    {
        char[] values = {'a', 'b', 'c', 'd'};
        int n = values.length;
        int r = 2; 

        int i = 0, j = 0;
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                System.out.println(values[j] + " " + values[i]);
            }
        }
    }
}


//when n and r are known only dynamically

class Permutation
{
    public static void main(String[] args)
    {
        char[] values = {'a', 'b', 'c', 'd'};
        int n = values.length;
        int r = 2; 
        int i[] = new int[r];
        int rc = 0;
        for(int j=0; j<Math.pow(n,r); j++)
        {

            rc=0;
            while(rc<r)
            {
                System.out.print(values[i[rc]] + " ");
                rc++;
            }
            System.out.println();
            rc = 0;
            while(rc<r)
            {
                if(i[rc]<n-1)
                {
                    i[rc]++;
                    break;
                }
                else
                {
                    i[rc]=0;
                }
                rc++;
            }
        }
    }
}

answered Dec 13, 2012 at 18:04

John Eipe

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2 Comments

harfe Over a year ago

Thank you :) For a question with poor grammar (I'm learning English), I get six instructive answers in 30 minutes. Just amazing. :)

jbiWeisendorf Over a year ago

for me - it is the solution!

Scott Hunter · Accepted Answer · 2012-12-13 17:41:37Z

0

Something like this?

// Print all sequences of len(list)+n numbers that start w/ the sequence in list
void method( list, n ) {
    if ( list.length == n )
        // print list
    else for ( int c=0; c<=5; c++ ) {
        // add c to end of list
        method( list, n );
        // remove c from end of list
    }
}

Initial call would be method( list, 5 ) where list is initially empty.

answered Dec 13, 2012 at 17:41

Scott Hunter

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1 Comment

harfe Over a year ago

Thank you, I'm trying it out.

GameDroids · Accepted Answer · 2012-12-13 17:46:32Z

0

here another interative but less elegant version

while (store1 < 6) {
        store5++;
        if (store5 == 6) {
            store5 = 0;
            store4++;
        }
        if (store4 == 6) {
            store4 = 0;
            store3++;
        }
        if (store3 == 6) {
            store3 = 0;
            store2++;
        }
        if (store2 == 6) {
            store2 = 0;
            store1++;
        }
        System.out.println(store1 + " " + store2 + " " + store3 + " " + store4 + " " + store5 + " ");
    }

answered Dec 13, 2012 at 17:46

GameDroids

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Comments

Durandal · Accepted Answer · 2012-12-13 18:00:24Z

The simplest code I can think of would tackle the problem with an entirely different approach:

public class TestA {
    public static void main(String[] argv) {
        for (int i=0; i<(6 * 6 * 6 * 6 * 6); ++i) {
            String permutation = Integer.toString(i, 6);
            System.out.println("00000".substring(permutation.length()) + permutation);
        }
    }
}

From your text (not your code) I gather you have 5 places and 6 symbols, which suggests there are 6 to the 5th power combinations. So the code just counts through those numbers and translates the number to the output combination.

Since this can also be viewed as a number system with base 6, it makes use of Integer.toString which already has formatting code (except the leading zeros) for this. Leading zeros are added where missing.

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