Please explain this code

Question

def fib(n):
if n == 0:
return 0
elif n == 1:
return 1
elif n == 2:
return 1
else:
return fib(n-1) + fib(n-2)
print (fib(10))

Why 55?

Answer 1

It is a recursive code for calculating the numbers of Fibonacci sequence.

fib(1)=1
fib(2)=1
fib(3)=fib(2)+fib(1)=1+1=2
fib(4)=fib(3)+fib(2)=2+1=3
fib(5)=fib(4)+fib(3)=3+2=5
fib(6)=fib(5)+fib(4)=5+3=8
fib(7)=fib(6)+fib(5)=8+5=13
fib(8)=fib(7)+fib(6)=13+8=21
fib(9)=fib(8)+fib(7)=21+13=34
fib(10)=fib(9)+fib(8)=34+21=55

Answer 2

You probably know the fibonacci sequence: 1, 1, 2, 3, 5, 8...
The formula to the nth number is:
an = a(n - 1) + a(n - 2).
It does the same: It returns the formula if it isn't 1 or 2, the other ones are predefined values.
Let's say you want the 3rd element. You get fib(n - 1) and fib(n - 2), which are both one, and 1 + 1 is 2. If you want the 4th number, you get fib(2) and fib(3). fib(2) is 1, and fib(3) is calculated like above, so it's 2 + 1, which is 3.
You can calculate it up to 10

Answer 3

fib is a recursive function which can call itself.

when you call 'fib(10)' the function will be called 2 more times once with the value 9 (10-1) and 8 (10-2), which will call themself until the number 0, 1 or 2 are reached.

step by step:
fib(10)

fib(9) + fib(8)

(fib(8) + fib(7)) + (fib(7) + fib(6))

(fib(7) + fib(6) + (fib(6) + fib(5)) + (fib(6) + fib(5)) + (fib(5) + fib(4))

(fib(6) + fib(5)) + (fib(5) + fib(4)) + (fib(5) + fib(4)) + (fib(4) + fib(3)) + (fib(5) + fib(4)) + (fib(4) + fib(3)) + (fib(4) + fib(3)) + (fib(3) + fib(2))

... (this is getting quite enoying... 😅😅)

Answer 4

thaks