Python converting recursive permutation function to iterative

Question

I have an unknown number of Integer variables say that can range from [0,9] I want to iterate over all permutations of these values.

If the number of variables was constant, it would be easy to write nested for loops. I came up with a recursive function that does what I want, but was curious if there was a way to do it iteratively.

def nested(array,index):
    n = len(array)
    for i in range(10):
        array[n-index]=i
        #len(array-1) end of array
        if(index == 1):
            print(array)
            #do something later
        else:
            nested(array,index-1)

#generate all permutations, change n to change size of list.            
n = 4
array = [0]*n
nested(array,len(array))

I tried using the so called "Simple Method" found here -> http://blog.moertel.com/posts/2013-05-11-recursive-to-iterative.html But I couldn't get it to work.

Answer 1

As was mentioned by another commenter, the key is to simulate your tail recursion using a stack.

Take note I append() a tuple of (array, index) into the stack , which mirrors the call to recursive function in your original recursive solution. At the start of the iteration, it does stack.pop() , which mimics the body of your recursive function. Recursive call becomes stack.append() .

def nested(array):
    stack = []
    n = len(array)
    stack.append((array.copy(), n))
    while(stack):
        array, index = stack.pop()        
        for i in range(10):
            array[n-index]=i
            #len(array-1) end of array
            if(index == 1):
                print(array)
                #do something later
            else:
                stack.append((array.copy(), index-1)) 

#generate all permutations, change n to change size of list.            
n = 4
array = [0]*n
nested(array)

Answer 2

Please refer to itertools . There is a Class "permutations" that could solve your problem perfectly.