CODEWITHSUNDEEP
pythonalgorithmpermutation
Biu
Biu

Reputation: 143

Non-recursive algorithm for full permutation with repetitive elements?

I usually use the recursive algorithm as follows in Python:

def permutate(array, t, n):             

    if t == n:
       for i in range(n):
            print array[i]

       return

   for j in range(t,n):
      flag = 1
      for r in range(t,j):
          if array[r] == array[j]:
              flag = 0
              break

      if flag == 0:
          continue
      else:
          array[j],array[t] = array[t],array[j]
          permutate(array,t+1,n)
          array[j],array[t] = array[t],array[j]

This one is neat. But I hope to find a convenient, non-recursive algorithm to do full permutation with repetitive elements?

Upvotes: 2

Views: 1516

Answers (1)

Brainless
Brainless

Reputation: 1758

Here is a generic way to "un-recursify" a recursive function : Let's say we have the following recursive function :

RecFunc (parameters)
    ...
    ...
    var x = RecFunc (other parameters)
    ...
    ...
EndFunc

To "un-recursify" it, you can use a stack like this :

NonRecFunc (parameters)
    stack MyStack;
    MyStack.push (InitialValue);
    While (MyStack is non empty)
       var S = MyStack.pop;
       # begin operations with S
       ....
       # results are x_1, ..., x_n
       for x_i = x_1 to x_n
           MyStack.push (x_i);
       endfor
   endWhile
EndFunc

In your case, the stack contains a pair consisting of an array and an int. The initial value is the array in input and the int m=0. The operations could look like this 

for i = m to n
    for j = i+1 to n
       if array[i] == array[j]
           continue
       endif
       array_c = copy of array
       permute entries i and j in array_c
       push (array_c, m+1) in the stack
    endfor
endfor

Good luck !

Upvotes: 4

Related Questions