# [R] Multiset Permutations

Stropharia stevenworthington at hotmail.com
Sun Apr 6 22:57:07 CEST 2008

```Dear R users,

I want to perform an exact permutation of multisets. I have looked at the
coin package, but it doesn't seem to offer what I am looking for. I want to
perform permutations (exact - without duplications) on multisets of scalars
(e.g., the multiset 0,0,1,2,2). I want to output all the possible
permutations into a data frame so that each multiset permutation occupies a
row (or column). The output would look something like this:

00122
01220
01210
20201
10202
12200
etc...

There seem to be numerous algorithms in the primary statistical literature
for doing this, but they haven't been implemented in R. I've been searching
for weeks online and in books for a way of doing this, but the only thing
i've come across is a Java script posted anonymously on a forum. I don't
know Java at all, so can't make enough sense of it to try a translation into
R, but i thought i'd include it below in case someone can glean from this a
way of doing it in R. Thanks in advance for any help on this, it is greatly
appreciated.

Here are the original (anonymous) poster's comments and code:

"We definitely need to use recursion. There can be up to n! permutations.
For example we'll use recursion to generate a tree of n! leaves. Each node
is a run of the recursive function. The root node has n children, the nodes
at the second level have each n-1 children, the nodes at the third level
have n-2 children, etc. Each leaf of this tree will be a permutation. The
idea is to add an element to the permutation, remove it from the set, and
call the recursive function on that set. Because it's a multiset i'm using a
Hashtable to eliminate duplicate entries. A pointer to the Hashtable gets
passed along with each recursive call.
The recursive function will know that it's on a "leaf" when the size of the
set is 0, at which point it will insert its permutation in the Hashtable.
Since identical permutations map to the same "bucket" in the hashtable, they
overwrite eachother to leave only unique permutations. Here's the algorithm
in Java:"

-----------------------------  JAVA CODE  --------------------------------
import java.util.*;

public class Test {
public static void PermutationsRecursive(char[] s, String p, Hashtable
perms){
for(int x=0; x<s.length; x++){
String np = new String(p);
char[] ns = new char[s.length-1];
int y = 0;
for(int z=0; z<s.length; z++){
if(z != x) ns[y++] = s[z];
}
np = np + s[x];
if(ns.length == 0) perms.put(np, new Boolean(true));
else PermutationsRecursive(ns, np, perms);
}
}
public static String[] GetPermutations(char[] in){
int fact = Factorial(in.length);
Hashtable perms = new Hashtable(fact);
PermutationsRecursive(in, "", perms);
Enumeration e = perms.keys();
String[] out = new String[perms.size()];
int i = 0;
while(e.hasMoreElements()){
out[i++] = (String) e.nextElement();
}
return out;
}
private static int Factorial(int n){
int fact = 1;
for(int i=2; i<=n; i++){
fact *= i;
}
return fact;
}
public static void main(String[] args){
char[] set = new char[]{'A', 'A', 'B', 'C'};
String[] perms = GetPermutations(set);
Arrays.sort(perms, String.CASE_INSENSITIVE_ORDER);
for(int i=0; i<perms.length; i++){
System.out.println(perms[i]);
}
}
}
-------------------------  END JAVA CODE  -----------------------------

This code prints the following:
AABC
AACB
ABAC
ABCA
ACAB
ACBA
BAAC
BACA
BCAA
CAAB
CABA
CBAA

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```