Fibonacci series recursion error

[harrismh777]

Thomas Rachel wrote:
>> ... because each recursion level 'return' calls fib() twice, and each of
>> those calls fib() twice, and you get the point...
>
> yes - but they are called one after the other, so the "twice" call
> counts only for execution speed, not for recursion depth.
 >>> def fib(n):
 >>>     if n == 1:
 >>>         return(n)
 >>>     else:
 >>>         return (fib(n-1)+fib(n-2))

They *are not* called one after the other in the sense that there is 
ever only one level of recursion with each call (not at all). Take a 
closer look. Each call to fib() forces a double head, and each of those 
forces another double head (now four), and so on...  the recursion depth 
exception of the OP testifies against your theory.

The thing about this problem that puzzles me, is why we might consider 
recursion for a possible solution in the first place. In other words, 
normally we consider using recursion when we need information further 
down the stack then we have now... so that recursion is necessary 
because each head call will not have the information it needs for 
completion until the tail clean-up (coming back up the stack) provides 
that information.

In the case of the fib() numbers (or the fib sequence) recursion is not 
necessary for correctly handling of the problem. The simple 
straight-forward obvious way to handle the problem is to calculate the 
sequence outright in a straight-forward way. On the other hand, Otten's 
use of yield (making a generator) makes some sense too, in the case that 
we want to get the fib numbers over time without taking the time to 
calculate the entire sequence up-front.
Just saying...

kind regards,
m harris