Recursion on Fibonacci Sequence

Question

I need some help in understanding the processing that happens here, so let´s say I call fib(5) I want the fibonacci 5, which is 8. But my brain in trying to understand the algorithm says it´s not. This is how i (wrongly) think:

return fib(4) + fib(3) // Stack frame 1
return fib(3) + fib(1) // Stack frame 2

now cause x is 1 fib(1), the conditional statement if x == 0 or x == 1: causes the recursion to end. Which according to my logic would become 3+1+4+3. Please correct my faulty logic.

def fib(x):
    """assumes x an int >= 0
       returns Fibonacci of x"""
    assert type(x) == int and x >= 0
    if x == 0 or x == 1:
        return 1
    else:
        return fib(x-1) + fib(x-2)

Just write down on a paper the first call and replace each recursive calls of the function with the valid return statement. What you probably don't understand is that `return fib(x - 1) + fib( x - 2)` calls twice your fib function, one with **parameter** x = **current** x decremented once, and the other decremented twice. You should also probably look into function arguments again, since this is a typical misunderstanding when using functions (at first). — Jean-Marie Comets, Dec 01 '12 at 10:36
Just put some print statement in your function and run it to see what it's doing. — martineau, Dec 01 '12 at 10:38

score 6 · Accepted Answer · answered Dec 01 '12 at 10:36

Here is the full expansion of what happens:

fib(5) expands to fib(4)+fib(3)
  fib(4) expands to fib(3)+fib(2)
    fib(3) expands to fib(2)+fib(1)
      fib(2) expands to fib(1)+fib(0)
        fib(1) evaluates to 1
        fib(0) evaluates to 1
      fib(1) evaluates to 1
    fib(2) expands to fib(1)+fib(0)
      fib(1) evaluates to 1
      fib(0) evaluates to 1
  fib(3) expands to fib(2)+fib(1)
    fib(2) expands to fib(1)+fib(0)
      fib(1) evaluates to 1
      fib(0) evaluates to 1
    fib(1) evaluates to 1

If you count the ones, you get 8 as the answer.

score 5 · Answer 2 · answered Dec 01 '12 at 10:37

For all x greater than 1, the fib function calls itself twice:

fib(5) becomes fib(4) + fib(3)
and expands to (fib(3) + fib(2)) + (fib(2) + fib(1))
and expands to ((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + 1)
expands to (((fib(1) + fib(0)) + 1) + (1 + 1)) + ((1 + 1) + 1)
expands to (((1 + 1) + 1) + (1 + 1)) + ((1 + 1) + 1)

which sums up to 8.

Recursion on Fibonacci Sequence

2 Answers2