Dissecting a permutation algorithm in Python

It does skip the if at the top level. It drops into the else and iterates j through the list. The first iteration has i == j == 0, so the swap does nothing, and you recur with ([1, 2, 3], 1).

This process repeats for the that instance, having i == j == 1. That recurs with ([1, 2, 3], 2) That instance is the one that print [1, 2, 3] as the first line of output.

Does that clear it up?

If not, learn how to insert useful print statements to trace execution. Perhaps this makes it more clear.

indent = ""

def perm(n, i):
    global indent
    indent += "  "
    print indent, "ENTER", n, i

    if i == len(n) - 1: 
        print n
    else:
        for j in range(i, len(n)):
            print indent, "RECUR", i, j
            n[i], n[j] = n[j], n[i]
            perm(n, i + 1)
            n[i], n[j] = n[j], n[i] # swap back, for the next loop

    indent = indent[2:]

perm([1, 2, 3], 0)

Output:

   ENTER [1, 2, 3] 0
   RECUR 0 0
     ENTER [1, 2, 3] 1
     RECUR 1 1
       ENTER [1, 2, 3] 2
[1, 2, 3]
     RECUR 1 2
       ENTER [1, 3, 2] 2
[1, 3, 2]
   RECUR 0 1
     ENTER [2, 1, 3] 1
     RECUR 1 1
       ENTER [2, 1, 3] 2
[2, 1, 3]
     RECUR 1 2
       ENTER [2, 3, 1] 2
[2, 3, 1]
   RECUR 0 2
     ENTER [3, 2, 1] 1
     RECUR 1 1
       ENTER [3, 2, 1] 2
[3, 2, 1]
     RECUR 1 2
       ENTER [3, 1, 2] 2
[3, 1, 2]