How to find 3 numbers in all combinations possible that in a sum the result equals n?

Question

I am working in an Authomata and need three numbers that suumed equal n For example, if n = 2 the numbers I need are:

It doesn´t matter if the combination are repeated.

If n = 3 I need:

So I read thar this was similar to the partition in the theory of numbers but I can get the particular case where only 3 numbers give me the target value (n). (The code is from a example I get here)

package automata2;

import java.util.ArrayList;

/**
 *
 * @author jaime
 */
public class Automata2 {


    /**
     * @param args the command line arguments
     */

    public static void main(String[] args) {
        int N = 14;
        partition(N);
    }

    public static void partition(int n) {
        partition(n, n, "");
    }
    public static void partition(int n, int max, String prefix) {
        if (n == 0) {
            System.out.println(prefix);
            return;
        }

        for (int i = Math.min(max, n); i >= 1; i--) {
            partition(n-i, i, prefix + " " + i);
        }
    }
}

But I only need al the combinations with three digits, not all since 14 until 1 1 1 1 1 1 1 1 1 1 1 1 1 1

I am not pretty sure on how to do it, I tried a switch structure but the conditions where a pretty hard, I am new in programming, sorry if this is a easy question. — jaime_briones, Aug 03 '14 at 18:49
looks like you need 2 nested for loops to generate all options, and for each option, you can test if the node satisfies your condition or not. It's called brute force because you generate all possible solutions just to find those few ones that you actually want. — Leo, Aug 03 '14 at 18:52
no it dosen´t, is there a way where you can use three variables iny type, because if n = 14, 14 is not a result I need, but 14 0 0 it is a result I need — jaime_briones, Aug 03 '14 at 19:03
You *really* could just google "java partition implementation"... — djechlin, Aug 03 '14 at 19:15
@Leo actually, two loops are enough. The third number is better calculated as `N-n1-n2`. Additional bonus: much more efficient. — Henry, Aug 03 '14 at 19:27
Maybe you would like to know what I am working on, here is it http://stackoverflow.com/questions/25109050/how-to-recursive-the-authomata-strange-planet-exercise I need some help also with recursive — jaime_briones, Aug 03 '14 at 21:01

sujithvm · Answer 1 · 2014-08-03T19:17:23.480

One solution is to check number of numbers is three and then print it

    public static void main(String[] args) {
        int N = 14;
        partition(N);
    }

    public static void partition(int n) {
        partition(n, n, "");
    }
    public static void partition(int n, int max, String prefix) {
        // added condition here
        if (n == 0 && prefix.trim().split(" ").length == 3) {
            System.out.println(prefix);
            return;
        }

        for (int i = Math.min(max, n); i >= 1; i--) {
            partition(n-i, i, prefix + " " + i);
        }
    }

Output :

Using nested loops - only for 3 number solution

    int N = 14;
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= N; j++) {
            for (int k = 0; k <= N; k++) {
                if (i + j + k == N) {
                    System.out.println(i + " " + j + " " + k);              
                }           
            }       
        }   
    }

Output :

score 0 · Accepted Answer · edited Aug 03 '14 at 19:13

0

The algorithm is like this to generate the numbers. This code will generate all nos.

public class main1 {
    public static void main(String args[]) {

        int N = 5, n1, n2, n3;

        for (n1 = 0; n1 <= N; n1++) {
            for (n2 = 0; n2 <= N; n2++) {
                for (n3 = 0; n3 <= N; n3++) {
                    if ( (n1+n2+n3)==N ) {
                        System.out.println(n1 + " " + n2 + " " + n3);
                    }
                }
            }
        }
    }
}

what you can do is instead of System.out.println(n1+" "+n2+" "+n3), you can store the numbers in an array list then arrange the list.

edited Aug 03 '14 at 19:13

Michael Yaworski

13,410
19
69
97

answered Aug 03 '14 at 19:05

Mohammed Ali

2,758
5
23
41

this only partitions into 3 numbers. – djechlin Aug 03 '14 at 19:14
1

the partition in three numbers is what I need – jaime_briones Aug 03 '14 at 19:17
you can also use the numbers n1 n2 n3 and convert the numbers into a string, but i think better to use array list. – Mohammed Ali Aug 03 '14 at 19:25
1

You could use only two nested loop and do `System.out.println(i + " " + j + " " + (N-i-j) );` to save a complexity degree. – Lærne Aug 03 '14 at 21:06

score 0 · Answer 3 · answered Aug 03 '14 at 21:01

One by one checking with skipping unnecessary partitions.

public static void main(String[] args) {
    int n =14;
    ArrayList<String> temp = partition(n);
    for (int z =0; z<temp.size(); z++){
        System.out.println(temp.get(z));
    }

}
public static ArrayList<String> partition(int n){
    ArrayList<String> out = new ArrayList<String>();
    int a=0, b=0, c=0;
    for (c=n; c>=0; c--){
        for(int j =a; j<=n;j++){
            if((c+j)>n){
                j=n+1;
                continue;
            }
            for (int i=b;i<=n; i++){
                int p=c+j+i;
                if (p>n) i=n+1;
                if(p==n){
                    String s = Integer.toString(c)+" "+ Integer.toString(j)+" "+ Integer.toString(i); 
                    out.add(s);
                }
            }

        }

    }
    return out;
}

Out put:

score 0 · Answer 4 · edited Dec 13 '14 at 12:23

import java.util.Scanner;

public class ExampleMinNumber {

public static void main(String args[]) 
{

    System.out.println("enter number which comibnation of sum you want");
    Scanner input=new Scanner(System.in);
    int num=input.nextInt();
    for(int i=0;i<=3;i++)
    {
        int sum=0;
        for(int j=0;j<=3;j++)
        {
            for(int k=0;k<=3;k++)
            {
                sum=i+j+k;
                if(sum==num)
                {
                    System.out.println(+i+""+j+""+k);
                }

            }
        }
    }

    input.close();  
}
}

How to find 3 numbers in all combinations possible that in a sum the result equals n?

4 Answers4