defining functions

[Michael Chermside]

Patio87 at aol.com wrote:

>        Thanks for your reply, but I had another question. I dont 
> understand how factorial(n) gets the value of two? And does The 'n' have 
> 1 assigned to by default?

Okay, try this simpler example:

 >>> def threeTimes(x):
	print 'x is now %i' % x
	if x == 0:
		return 0
	else:
		return 3 + threeTimes(x-1)

After writing this, what is the value of 'x'?
	
 >>> print x
Traceback (most recent call last):
   File "<pyshell#9>", line 1, in ?
     print x
NameError: name 'x' is not defined

As you can see, x is not defined... it doesn't have ANY value. But if we 
try CALLING the function, it'll print out x DURING the call:

 >>> threeTimes(0)
x is now 0
0

Great... so x was 0 then. But what about now, AFTER we called it?

 >>> print x
Traceback (most recent call last):
   File "<pyshell#11>", line 1, in ?
     print x
NameError: name 'x' is not defined

Okay... what's going on here is that 'x' took on a value TEMPORARILY, 
while the function was being evaluated, but then it went back to being 
undefined. Actually, a more precise definition would be that the 
variable 'x' was "local" to the function, and so the 'x' that had been 
set to 0 was destroyed when the function finished.

What will happen if we call threeTimes() with a bigger number? Let's 
start with 1:

 >>> threeTimes(1)
x is now 1
x is now 0
3

First, it starts up the function "threeTimes", which means that it has 
to create an 'x' -- this gets set to 1 since that's what you used when 
you called it. Then it prints it out, so you get the line "x is now 1". 
It tries the 'if', and decides to go with the 'else' clause... so it 
needs to return "3 + threeTimes(x-1)", or "3 + threeTimes(0)". But NOW 
it has to work out what "threeTimes(0)" is.

So it starts up ANOTHER COPY of the function "threeTimes". The old copy 
is still there (in which x=1), but the new copy has to set x=0. It 
prints out x (so we get the line "x is now 0"), and then it tries the if 
and finds out that all it has to do is return an answer of 0.

So the SECOND copy of "threeTimes" has finished, and the FIRST copy can 
now continue. It now knows that "3 + threeTimes(0)" is just "3 + 0", and 
so it returns "3", which then gets printed out.

I imagine you can now guess what happens when we call "threeTimes(6)"... 
right?

 >>> threeTimes(6)
x is now 6
x is now 5
x is now 4
x is now 3
x is now 2
x is now 1
x is now 0
18

There are many situations for which this technique (recursion) is very 
powerful and useful, so it's worth learning even if it warps your brain 
a little bit at first.

-- Michael Chermside