For loop in recursive function?

Question

When the for loop runs, why does it print all permutations of ABC instead of all 'A's?

def perm(l, n, str_a):
    if len(str_a) == n:
        print str_a
    else:
        for c in l:
            perm(l, n, str_a+c)


perm("ABC", 3, "")

Prints:

AAA AAB AAC ABA ABB ABC ACA ACB ACC BAA BAB BAC BBA BBB BBC BCA BCB...

Answer 1

1. When you call perm("ABC", 3, "") , the else clause is executed (because len("") != 3 ).
2. This result in the calls perm("ABC", 3, "A") , perm("ABC", 3, "B") and perm("ABC", 3, "C") . Let's see what happens with the first one:
3. Again, the else is executed, resulting in the function calls perm("ABC", 3, "AA") , perm("ABC", 3, "AB") and perm("ABC", 3, "AC") .
4. The same thing happens with the other function calls from step 2 (you get the idea).
5. Let's look at perm("ABC", 3, "AA") : When called, the else is executed yet again --> perm("ABC", 3, "AAA") , perm("ABC", 3, "AAB") and perm("ABC", 3, "AAC") .
6. In these calls, the expression len(str_a) finally is == 3 , which means that str_a will be printed.
7. And so on, until CCC .

Answer 2

It does not keep printing 'A's, because, after 3 recursions, it will have formed the string "AAA". Then, the line print str_a will be executed, as the condition len(str_a) == n will be verified.

After that, the execution will go back to the callee function, which was inside the c loop. c had value "A". At the following iteration, c will get value "B", and perm("ABC", 3, "AAB") will be invoked, printing "AAB", and so on.

Maybe the recursion graph could clearen things up (it's incomplete, because it's big)

Answer 3

I have no idea what you are trying to do, but maybe a little bit of debug output would help you figure it out. Try this:

def perm(iter, l, n, str_a):
    print "starting iteration {0}: l='{1}', n='{2}', str_a='{3}'".format(
        iter, l, n, str_a)
    if len(str_a) == n:
        print str_a
    else:
       for c in l:
           perm(iter+1, l, n, str_a+c)

perm(1, "ABC", 3, "")