Please explain this algorithm to get all permutations of a String

Question

The following code generates all the permutations for a string:

def permutations(word):
    if len(word)<=1:
        return [word]

    #get all permutati         


        

Answer 1

I've written the steps out for a string of length 2 and a string of length 3 below.

permutations('ab')

len('ab') is not <= 1 
perms = permutations of 'b'
len('b') <= 1 so return 'b' in a list
perms = ['b']
char = 'a'
result = []
for 'b' in ['b']:
    for 0 in [0,1]:
        result.append('' + 'a' + 'b')
    for 1 in [0,1]:
        result.append('b' + 'a' + '')
result = ['ab', 'ba'] 

permutations('abc')

len('abc') is not <= 1
perms = permutations('bc')
perms = ['bc','cb']
char = 'a'
result =[]
for 'bc' in ['bc','cb']:
    for 0 in [0,1,2]:
        result.append('' + 'a' + 'bc')
    for 1 in [0,1,2]:
        result.append('b' + 'a' + 'c')
    for 2 in [0,1,2]:
        result.append('bc' + 'a' + '') 
for 'cb' in ['bc','cb']:
    for 0 in [0,1,2]:
        result.append('' + 'a' + 'cb')   
    for 1 in [0,1,2]:
        result.append('c' + 'a' + 'b')   
    for 2 in [0,1,2]:
        result.append('cb' + 'a' + '')
result = ['abc', 'bac', 'bca', 'acb', 'cab', 'cba']  

Answer 2

The algorithm is:

• Remove the first letter
• Find all the permutations of the remaining letters (recursive step)
• Reinsert the letter that was removed in every possible location.

The base case for the recursion is a single letter. There is only one way to permute a single letter.

Worked example

Imagine the start word is bar.

• First remove the b.
• Find the permuatations of ar. This gives ar and ra.
• For each of those words, put the b in every location:
  • ar -> bar, abr, arb
  • ra -> bra, rba, rab