Question

I was studying tower of hanoi recursive implementation in python. In my prgram I gave print at different points to know it better like

def hanoi(n, src, inm, dest):
    print "n=",n,"src=",src,"inm=",inm,"dest=",dest
    if n == 0:
        return
    hanoi(n-1, src, dest, inm)
    print src, '->', dest
    print n 
    hanoi(n-1, inm, src, dest)

hanoi(2,'A','B','C')

The answer is printed like:

n= 2 src= A inm= B dest= C
n= 1 src= A inm= C dest= B
n= 0 src= A inm= B dest= C
A -> B
1
n= 0 src= C inm= A dest= B
A -> C
2
n= 1 src= B inm= A dest= C
n= 0 src= B inm= C dest= A
B -> C
1
n= 0 src= A inm= B dest= C

I could understand upto

   1
    n= 0 src= C inm= A dest= B

I couldnt understand how A -> C is printed after this. After the call with n= 0 src= A inm= B dest= C, I know function will be returned. There the active function is n= 1 src= A inm= C dest= B. What happens to that?

Please explain the trace

Answer 1

If you add two more prints:

print "r1"   # before first return
print "r2"   # at the end of the function, before second, implicit return

Then you will see that there were two returns in a row:

n= 2 src= A inm= B dest= C
n= 1 src= A inm= C dest= B
n= 0 src= A inm= B dest= C
r1
A -> B
1
n= 0 src= C inm= A dest= B
r1
r2
A -> C
2
n= 1 src= B inm= A dest= C
n= 0 src= B inm= C dest= A
r1
B -> C
1
n= 0 src= A inm= B dest= C
r1
r2
r2