Thread: Recursion

I am trying to figure out this problem

consider this recurrence relation

f(1)=2
f(2)=2

f(n)=2*f(n-1)+f(-2) for n>2

write a recursive function to compute f.

I have some code that I have been working on. My question is.

How do u get the function to return a value that it computed back to itself to be used again .

When i run this for f(2) and f(1)it comes out right, but for anything else its wrong ( or at least i think.)

can anyone tell mat f(3) and f(4) are so i know what to look for? Also, give me a hint as to what I am doing wrong in the code?

Code:

#include<iostream>
using namespace std;



int computef(int n)

{
    int answer;
       
    if ((n=2)||(n=1))
    
    return 2;
   
     else
        
    answer= 2*computef(n-1)+computef(n-2);
    
    cout << answer;
    
     
     

}

    int main()
    {
     int x;
     
     cout << "Please input the number for calculation  " ;
     cin >> x;
        
       cout << computef(x);
        
               
        
        system("pause");
        return 0;
        
        }

Holy smokes Batman!!!

That worked. So when do u use something like n=2 vs n==2
also, when u use "return" it return the value to the calling function to use it again if the conditions are not met?

revised code.

Code:

#include<iostream>
using namespace std;



int computef(int n)

{
    int answer;
       
    if ((n==2)||(n==1))
    
    return 2;
   
     else
        
    answer= 2*computef(n-1)+computef(n-2);
    
    return answer;
    
    cout << answer;
     
     

}

    int main()
    {
     int x;
     
     cout << "Please input the number for calculation  " ;
     cin >> x;
        
       cout << computef(x);
        
               
        
        system("pause");
        return 0;
        
        }

Thank you so much foe help. U made my night, i was working on this one problem for like 3 and a half hours