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...
perm("ABC", 3, ""), the else clause is executed (because len("") != 3).perm("ABC", 3, "A"), perm("ABC", 3, "B") and perm("ABC", 3, "C"). Let's see what happens with the first one:else is executed, resulting in the function calls perm("ABC", 3, "AA"), perm("ABC", 3, "AB") and perm("ABC", 3, "AC").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").len(str_a) finally is == 3, which means that str_a will be printed.CCC.
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)
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, "")
