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Given "n"(n>0) number of dices with "m"(m>0) sides. Print all possible combinations

Posted on 2015-02-17
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Last Modified: 2015-02-18
Hi,
  Write a Java program to print all possible combinations in below case
   "n"(n>0) number of dices with "m"(m>0) sides. Print all possible combinations
  for example.
     2 dices with 3 sides should print out this
 (1,1)
 (1,2)
 (1,3)
(2,1)
(2,2)
(2,3)
(3,1)
(3,2)
(3,3)
   I gave it a sincere try before posting this. My code doesn't work if number of dices is more than 2 which is not right. Thanks in advance for looking.
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Question by:koppcha
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Expert Comment

by:mccarl
Comment Utility
Can you post your code?

Also, I assume that this is for a homework/assignment exercise, but can you please confirm if that is the case? This allows everyone to provide the correct level of assistance.
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by:koppcha
Comment Utility
This is definitely not homework/assignment.  It's been long time since I have been to school(which could be the reason I spent half day figuring this out )

import java.util.Arrays;

public class DiceHistogram {
	public void getHistogram(int dices, int sides) {
		int diceArray[] = new int[dices];
		
		//Initialize Array
		for (int i = 0; i< diceArray.length; i++) {
			diceArray[i] = 1;
		}
		
		//System.out.println(Arrays.toString(diceArray));
			for(int k=dices-2; k>=0; k--) {
				for(int j=1; j<= sides; j++) {
				   diceArray[k] = j;
			       for(int i=1; i<=sides; i++ ) {
				       diceArray[dices-1] = i;
				       System.out.println(Arrays.toString(diceArray));
				       
			       }
			       
				}   
			}

		}
		
		//for ( int j)
		
		
	
	public static void main(String[] args) {
		DiceHistogram dh = new DiceHistogram();
		dh.getHistogram(2, 3);
	}

}

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Expert Comment

by:phoffric
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Often, when the term "combinations" is used in an exercise, the order does not matter, so (2,1) is considered the same as (1,2).
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Author Comment

by:koppcha
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In this case they should be considered different. "Combination" is my own way of explaining it not part of original problem. Example should be taken for reference. They are considered different.
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Expert Comment

by:ozo
Comment Utility
Can we assume that pow(m,n) fits in an int?
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LVL 26

Accepted Solution

by:
dpearson earned 300 total points
Comment Utility
This is actually a fairly tricky problem.  You can't just use one or two or even 5 loops to do it, but you really need 1 loop for each dice - which is a parameter to the problem.

That sort of problem is usually easiest to solve with recursion - where a function calls to itself.

Something like this.

Recursive functions tend to be not a lot of code, but can be hard to understand if you've not seen them before.  I added some comments to help explain what's going on.

Doug

	public static List<String> dice(int nDice, int nSides) {
		List<String> combinations = new ArrayList<>();

		if (nDice == 1) {
			// For one dice we get a list with just 1 entry
			// 1
			// 2
			// 3
			for (int side = 1; side <= nSides; side++) {
				combinations.add(Integer.toString(side));
			}
		} else {
			// When we have more than one dice we add the values for this dice
			// to all possible combinations from "nDice-1"
			// So add "1," to all subsequent combinations (e.g. "1,1", "1,2", "2,1", "2,2" etc.)
			List<String> following = dice(nDice-1, nSides) ;
			for (int side = 1; side <= nSides; side++) {
				for (String follow : following) {
					combinations.add(side + ", " + follow);
				}
			}
		}

		return combinations ;
	}

	public static void listCombinations() {
		List<String> combinations = dice(4, 3) ;

		for (String combo : combinations) {
			System.out.println("(" + combo + ")") ;
		}
	}

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Author Comment

by:koppcha
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Hi Dpearson,
  I tried your program with different inputs and it works as expected. I am still trying to understand. will get back to you. Thanks for your help.

Hi Ozo,
  Yes I think it's fair assumption.
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Assisted Solution

by:ozo
ozo earned 100 total points
Comment Utility
if pow(sides,dices) fits in an int, another method might be
static void getHistogram(int dices, int sides) {
        for( int i=(int)Math.pow(sides,dices);--i>=0;){
          System.out.print("(");
          int n=i;
          for( int d=0;d<dices;++d ){
            if( d>0 ){ System.out.print(","); }
            System.out.print(sides-(n%sides));
            n/=sides;
          }
          System.out.println(")");
       }      
}
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Assisted Solution

by:rrz
rrz earned 100 total points
Comment Utility
I think what your asking for here is permutations.  Look at  
http://www.mathsisfun.com/combinatorics/combinations-permutations.html    
Here is my version.  What you call a side, I call a face.
import java.util.*;
 public class Dice{
   public static void main(String[] args) {
     int numberOfDice = 3; 
	 int numberOfFaces = 3;
     int numberOfOutcomes = (int)Math.pow(numberOfFaces,numberOfDice);   
     int[] roll = new int[numberOfOutcomes]; //all initially 0 (roll to the next face)
     StringBuilder[] outcomes = new StringBuilder[numberOfOutcomes];
     for(int j=0; j<numberOfOutcomes; j++){
                                  outcomes[j] = new StringBuilder("("); 
     }
     for(int k = 1; k <= numberOfDice; k++){
                int numberOnFace = 1; 
                for(int y = 0; y < numberOfOutcomes; y++){
                                   if(roll[y] == 0){
                                            outcomes[y].append(numberOnFace);
					    if(k != numberOfDice)outcomes[y].append(",");
                                            numberOnFace++;
                                            roll[y] = 1; // don't roll, just stay on the same face
                                            if(numberOnFace == numberOfFaces + 1){
                                                              numberOnFace = 1;
                                                              roll[y] = 0; // roll to the next face
                                            }
                                   }else{
                                         outcomes[y].append(numberOnFace);
					 if(k != numberOfDice)outcomes[y].append(",");
                                        }
                }            
     }
     for(int z = 0; z < numberOfOutcomes; z++){
                                           System.out.println(outcomes[z].toString() + ")");
    }
   }
}

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Expert Comment

by:dpearson
Comment Utility
I am still trying to understand. will get back to you. Thanks for your help.

The key for a recursive solution is to understand that the larger problem can be seen as a repetition of a smaller problem.

E.g. If I gave you a list of all combinations of 2 dice with 3 sides:

 (1,1)
 (1,2)
 (1,3)
(2,1)
(2,2)
(2,3)
(3,1)
(3,2)
(3,3)

then you can use this list to create the list of all combinations of 3 dice by adding "1, ...", "2, ..." and "3, ..." to the front of each of the earlier combinations:

1 +  (1,1) => (1,1,1)
1+  (1,2) => (1,1,2)
1 + (1,3) => (1,1,3)
1 + (2,1) => (1,2,1)
...
3 + (1,1) => (3,1,1)
3 + (1,2) => (3,1,2)
...
etc..

Once you recognize the problem has that property, then you can write a function that calls to itself to get it to compute this.

Anyway - hope you can figure it out :)

Doug
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Author Comment

by:koppcha
Comment Utility
Hi Doug,
  I got it now. Thanks for the help.

Ozo/rrz,
  Thanks guys. Your code works too.

Doug's solution is easier for me to understand.
I am going to close this question now.
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