The below function calculates the Fibonacci, why the last calculation needs to include the function name and not just (n-1)+(n-2) ?
function fibonacci(n){
if (n===1){ return 1}
else if (n===0){ return 0}
else return fibonacci(n-1)+fibonacci(n-2) // (n-1)+(n-2) does not work. Why?
}
I know this is a beginner question but couldn't find the answer. I'd appreciate any comments. I understand that I need to use recursion, but that's not the point of my question.
this is recursive solution:
function fibonacci(n){
if (n===1){ return 1}
else if (n===0){ return 0}
else return fibonacci(n-1)+fibonacci(n-2)
}
You are calling this function for nth fibonacci. But you don't know nth fibonacci yet. so you must find (n-1) and (n-2) fibonacci. That is why you must call fibonacci(n-1)+fibonacci(n-2).
And you don't even know n-1th and n-2th fibonacci that is why you must call it until known fibonacci. You know first and second fibonaccis. That is wht when n == 1 or n==0 you return just answer.
for example:
n = 7
fibonacci(7) = fibonacci(6) + fibonacci(5)
fibonacci(6) = fibonacci(5) + fibonacci(4)
fibonacci(5) = fibonacci(4) + fibonacci(3)
fibonacci(4) = fibonacci(3) + fibonacci(2)
fibonacci(3) = fibonacci(2) + fibonacci(1)
fibonacci(2) = fibonacci(1) + fibonacci(0)
fibonacci(1) = 1
fibonacci(0) = 0
(n-1)+(n-2)is the same asn-1+n-2is the same asn+n-1-2is the same as2*n-3. Simply an arithmetic expression whose value is calculated and returned.fibonacci(n-1)andfibonacci(n-2), on the other hand, are function calls.return (n-1)+(n-2)means, and wonder why it doesn't work? Can you explain why you thought it would work? As the question is written, it's a bit like asking "Why islog(n)not the same asn?" or "Why isx+5not doing the same asx?", and I don't know how to answer that.