Question

hi, i understand how fibonacci number works but i don't understand the coding part of it. can someone please explain it to me?

Answer 1

Do you mean you need help with how to build the function, or you're looking at the function on a handout and you need help figuring out what it does?

Answer 2

i'm looking at the function but can't figure out what it does

Answer 3

This one?

def fib(x):
"""Return fibonacci of x, where x is a non-negative int"""
if x == 0 or x == 1: return 1
else: return fib(x-1) + fib(x-2)

Answer 4

yes

Answer 5

well, the first part is easy. the line:

if x == 0 or x == 1: return 1

checks to see if x is 0 or 1

if x isn't 0 or 1, the next line returns

fib(x-1) + fib(x-2)


fib(x-1) is going to go through the same process. If (x-1) is greater than 1, THAT will also get sent to fib. But eventually fib(x-1) will return an integer. When it does, the program will start to evaluate fib(x-2) the same way. when fib(x-2) returns an integer, the problem becomes simple addition.

Answer 6

when you say "If (x-1) is greater than 1, THAT will also get sent to fib" does that mean the function will go through fib(x-1) then fib((x-1)-1 then fib(((x-1)-1)-1) and so on

also when will fib(x-1) return an integer? i thought fib(x-1) will always be an integer

Answer 7

when you say "If (x-1) is greater than 1, THAT will also get sent to fib" does that mean the function will go through fib(x-1) then fib((x-1)-1 then fib(((x-1)-1)-1) and so on


Exactly, UNTIL one of those (x-an accumulating number of ones) terms is in the base case, that is, when the line that says


if x == 0 or x == 1: return 1

is true, the function will stop calling fib(x-whatever) and will just return 1


also when will fib(x-1) return an integer? i thought fib(x-1) will always be an integer

Instead of fib(x-1) think of it as 'the integer that gets returned when fib(x-1) is finished being evaluated' So it is an integer, but it takes some time (the time to evauate fib(x-1)) to figutre out what the integer is.

Answer 8

One way to see this is to consider x = 2. Since x is not 1, the function computes fib(1) + fib(0). But those two values are known to be 1, so the answer is 1+1 = 2. Now consider x = 3. Since x is not 1, the function computes fib(2) + fib(1). fib(2) expands to fib(1) + fib(0), so the answer is 1+1+1 = 3. And so on.

Answer 9

thanks! i think i got
so lets say x = 4
fib(4) = fib(3) + fib(2) = [fib(2) + fib(1)] + [fib(1) + fib(0)]
=[fib(1) + fib(0) + fib(1)] + [fib(1) + fib(0)]

Answer 10

Exactly. Well done.