Recursive Fibonacci

Question

I\'m having a hard time understanding why

#include 

using namespace std;

int fib(int x) {
    if (x == 1) {
        return 1;
    } else {
         


        

Answer 1

int fib(int x) 
{
    if (x < 2)
      return x;
    else 
      return (fib(x - 1) + fib(x - 2));
}

Answer 2

By definition, the first two numbers in the Fibonacci sequence are 1 and 1, or 0 and 1. Therefore, you should handle it.

#include <iostream>
using namespace std;

int Fibonacci(int);

int main(void) {
    int number;

    cout << "Please enter a positive integer: ";
    cin >> number;
    if (number < 0)
        cout << "That is not a positive integer.\n";
    else
        cout << number << " Fibonacci is: " << Fibonacci(number) << endl;
}

int Fibonacci(int x) 
{
    if (x < 2){
     return x;
    }     
    return (Fibonacci (x - 1) + Fibonacci (x - 2));
}

Answer 3

Why not use iterative algorithm?

int fib(int n)
{
    int a = 1, b = 1;
    for (int i = 3; i <= n; i++) {
        int c = a + b;
        a = b;
        b = c;
    }           
    return b;
}

Answer 4

int fib(int n) {
    if (n == 1 || n == 2) {
        return 1;
    } else {
        return fib(n - 1) + fib(n - 2);
    }
}

in fibonacci sequence first 2 numbers always sequels to 1 then every time the value became 1 or 2 it must return 1

Answer 5

The reason is because Fibonacci sequence starts with two known entities, 0 and 1. Your code only checks for one of them (being one).

Change your code to

int fib(int x) {
    if (x == 0)
        return 0;

    if (x == 1)
        return 1;

    return fib(x-1)+fib(x-2);
}

To include both 0 and 1.

Answer 6

This is my solution to fibonacci problem with recursion.

#include <iostream>
using namespace std;

int fibonacci(int n){
    if(n<=0)
        return 0;
    else if(n==1 || n==2)
        return 1;
    else
        return (fibonacci(n-1)+fibonacci(n-2));
}

int main() {
    cout << fibonacci(8);
    return 0;
}