In Lecture 13 while explaining memoisation, the prof. actually takes fib(0) to be 1.
Though Fibonacci series is 0 1 2 3 5 8 13 etc., fib(0) shoule be 0, fib(1) should be 1 and so on. I know I'm missing something, please correct me.

Question

In Lecture 13 while explaining memoisation, the prof. actually takes fib(0) to be 1.
Though Fibonacci series is 0 1 2 3 5 8 13 etc., fib(0) shoule be 0, fib(1) should be 1 and so on. I know I'm missing something, please correct me.

lopus · Answer

http://commons.wikimedia.org/wiki/File:FibbonacciRecurisive.png?uselang=es

lopus · Answer

Fibonacci both 0 and 1 are two cases separately and apart from 2 starts with the algorithm: $$F _{n}=F _{n-1}+F _{n-2}$$

base case:
f(0)=0
f(1)=1

lopus · Answer

fib(0)=0   fib(1)=1   fib(2)=1   fib(3)=2

if ( n == 0 || n == 1)
		return n;
	else
		return (fibonacci(n-1) + fibonacci(n-2));

anonymous · Answer

@lopus This exactly what I mean but acc to the lecture, 
if n<=1: return 1
which returns 1 in both 0 and 1. Also, he clearly specifies fib(0) = 1 and fib(1) = 1 in the memory as {0:1, 1:1}

lopus · Answer

is an error, the mathematical definition is as you wrote it

anonymous · Answer

Okay then! thanks!

lopus · Answer

http://www.youtube.com/watch?v=WbWb0u8bJrU&feature=player_embedded minute 42 he explained why take fib(0)==1