Permutation of an array, with repetition, in Java

Question

There are some similar questions on the site that have been of some help, but I can't quite nail down this problem, so I hope this is not repetitive.

This is a homework assignment where you have a set array of characters [A, B, C], and must use recursion to get all permutations (with repetition). The code I have sort of does this:

char[] c = {'A', 'B' , 'C'};

public void printAll(char[] c, int n, int k) {
    if (k == n) {
      System.out.print(c);
      return;
    }
    else {   
      for (int j = 0; j<n; j++) {
        for (int m = 0; m<n; m++) {
           System.out.print(c[k]); 
           System.out.print(c[j]); 
           System.out.print(c[m] + "\r\n");
        }
      }
    }        
    printAll(c, n, k+1);    
}

However, the parameter n should define the length of the output, so while this function prints out all permutations of length 3, it cannot do them of length 2. I have tried everything I can think of, and have pored over Google search results, and I am aggravated with myself for not being able to solve what seems to be a rather simple problem.

Solution 1

If I understand correctly, you are given a set of characters c and the desired length n.

Technically, there's no such thing as a permutation with repetition. I assume you want all strings of length n with letters from c.

You can do it this way:

to generate all strings of length N with letters from C
 -generate all strings of length N with letters from C
     that start with the empty string.

to generate all strings of length N with letters from C
   that start with a string S
 -if the length of S is N
  -print S
 -else for each c in C
  -generate all strings of length N with letters from C that start with S+c

In code:

printAll(char[] c, int n, String start){
  if(start.length >= n){
    System.out.println(start)
  }else{
    for(char x in c){ // not a valid syntax in Java
      printAll(c, n, start+x);
    }
  }
}

Solution 2

I use this java realization of permutations with repetitions. A~(n,m): n = length of array, m = k. m can be greater or lesser then n.

public class Permutations {


    static void permute(Object[] a, int k, PermuteCallback callback) {
        int n = a.length;

        int[] indexes = new int[k];
        int total = (int) Math.pow(n, k);

        Object[] snapshot = new Object[k];
        while (total-- > 0) {
            for (int i = 0; i < k; i++){
                snapshot[i] = a[indexes[i]];
            }
            callback.handle(snapshot);

            for (int i = 0; i < k; i++) {
                if (indexes[i] >= n - 1) {
                    indexes[i] = 0;
                } else {
                    indexes[i]++;
                    break;
                }
            }
        }
    }

    public static interface PermuteCallback{
        public void handle(Object[] snapshot);
    };

    public static void main(String[] args) {
        Object[] chars = { 'a', 'b', 'c', 'd' };
        PermuteCallback callback = new PermuteCallback() {

            @Override
            public void handle(Object[] snapshot) {
                for(int i = 0; i < snapshot.length; i ++){
                    System.out.print(snapshot[i]);
                }
                System.out.println();
            }
        };
        permute(chars, 8, callback);
    }

}

Example output is

aaaaaaaa
baaaaaaa
caaaaaaa
daaaaaaa
abaaaaaa
bbaaaaaa
...
bcdddddd
ccdddddd
dcdddddd
addddddd
bddddddd
cddddddd
dddddddd

Solution 3

I just had an idea. What if you added a hidden character (H for hidden) [A, B, C, H], then did all the fixed length permutations of it (you said you know how to do that). Then when you read it off, you stop at the hidden character, e.g. [B,A,H,C] would become (B,A).

Hmm, the downside is that you would have to track which ones you created though [B,H,A,C] is the same as [B,H,C,A]

Solution 4

Here is c# version to generate the permutations of given string with repetitions:

(essential idea is - number of permutations of string of length 'n' with repetitions is n^n).

string[] GetPermutationsWithRepetition(string s)
        {
            s.ThrowIfNullOrWhiteSpace("s");
            List<string> permutations = new List<string>();
            this.GetPermutationsWithRepetitionRecursive(s, "",
                permutations);
            return permutations.ToArray();
        }
        void GetPermutationsWithRepetitionRecursive(string s, string permutation, List<string> permutations)
        {
            if(permutation.Length == s.Length)
            {
                permutations.Add(permutation);
                return;
            }
            for(int i =0;i<s.Length;i++)
            {
                this.GetPermutationsWithRepetitionRecursive(s, permutation + s[i], permutations);
            }
        }

Below are the corresponding unit tests:

[TestMethod]
        public void PermutationsWithRepetitionTests()
        {
            string s = "";
            int[] output = { 1, 4, 27, 256, 3125 };
            for(int i = 1; i<=5;i++)
            {
                s += i;
                var p = this.GetPermutationsWithRepetition(s);
                Assert.AreEqual(output[i - 1], p.Length);
            }
        }

View more solutions

Comments

[John Dvorak]

If I understand the problem correctly, the required length of permutation is given

[user1788424]

You, sir, are not just a gentleman and a scholar. You are a prince, a gentleman, and a scholar. Some other people online suggested something similar, except using an array and not a string. However, your explanation was much clearer.

[user1788424]

And for reference, if anyone is interested, here is the final function, where the n parameter controls the length of each line that is printed: public void printAll3(char[] c, int n, int k, String s) { if (s.length() >=n) {System.out.print(s + "\r\n"); return;} else if (k<n) {for (int i = 0; i< c.length; i++) { printAll3(c, n, k+1, s + c[i]); //System.out.print(s); } } }

[MichelleJS]

@JanDvorak I know this is old but I had a similar problem I was trying to solve and I modified this and it totally worked.However I don't understand how your final call printAll(c,n,start+x) works. I printed it out and start goes like this for the first few calls (a, aa, aaa, aab, aac, ab, aba). I don't really understand why it doesn't end up being (abc, abcabc, abcabcabc). I was hoping you could explain it. I have been trying to trace it and I understand that each time print all calls itself inside the loop it calls itself n more times. Anyway if you could I would like some more explaination.

[John Dvorak]

@MichelleJS why would it be abcabcabc? You are adding a single character to start, then later backtracking and adding different one on the same place. In the first call (start == ""): start.length == 0, so it goes to the else branch. The first character of C is a, so x == 'a' in the first iteration. "" + 'a' == "a", so the first recursive call is to printAll("abc", 3, "a"). The behavior you describe would occur if you were doing printAll(c, n, start+c) inside the loop - in which case x would end up unused.

[MichelleJS]

Thankyou very much. I drew it all out on a piece of paper in order to understand. I now have to do the other the same thing without repetition. I'm not sure how to modify this code to do that but this has been a very good start.

[John Dvorak]

@MichelleJS one option would be to remove x from the list of candidates (if your language is good enough, then just printAll(c-x, n, start+x)). The other would be to check if x is already is start before recursing. The first option might perform better, but the second option might be easier to implement in Java.

[a.dibacco]

I suppose that they are called not "permutations with repetitions" but "dispositions with repetitions"