Base case of recursion function for Fibonacci sequence - c

Question

I'm trying to write the function void fib(int arr[], int n), which would fill the array with Fibonacci numbers until index n.
I've tried to find base cases, and chose these:
void fib(int arr[], int num){
int arrLength = num + 1;
if(num<0){
return;
}else if(num == 0){
arr[num] = 1;
}else if(num == 1){
arr[num-1] = 1;
arr[num] = 1;
}
}
But, as you can see, I did not find recursive method itself.
Here's sample output, for example, for call fib(arr, 5):
0 1 2 3 4 5
1 1 2 3 5 8
My main function for testing case:
int main(){
int n = 10, i;
int arr[n+1];
fib(arr, n);
for(i=0;i<=10;i++){
printf("%i ", arr[i]);
}
return 0;
}
Is there any other way to make base cases more "elegant"? Also, I would truly appreciate hints using which I could fill the array with numbers starting from 2 with recursive option.

Answer 1

You question is asking for recursion but the program you write is just using function, because of this reason I am writing very basic code for your better understanding, you can improve this after understanding the flow and functionality or ask new question with some work.
Below one is a working code tested on TurboC, I am sharing complete test code.
#include <stdio.h>
#include<conio.h>
#define MAX 100
void fib(int *arr, int num, int a, int b, int term){
if(term == 0 && term <= num){
arr[term] = 1;
term++;
fib(arr,num,a,b,term);
}else if(term ==1 && term <= num){
arr[term] = 1;
term++;
fib(arr,num,a,b,term);
}else if(term <= num){
arr[term] = a+b;
term++;
fib(arr,num,b,a+b,term);
}
}
void main()
{
int firstTerm = 1;//First term of fibbo series
int secondTerm = 1;//Second term of fibbo series
int tracker = 0; // Tracker to track how much term we printed
int i;//To run loop here to check array after recursive function
int ar[MAX],n=5;// n is number of term we want to print
clrscr();
fib(ar,n,firstTerm,secondTerm,tracker);//recursive function call
// below is printing array to check
for(i=0;i<=n;i++){
printf("%d\t",ar[i]);
}
getch();
}
One thing I have to suggest is, if n is 5 then you just get 1 1 2 3 5, In code I did according to your requirement, so here it will print 1 1 2 3 5 8

Answer 2

I'd state that the "elegant" solution should be a simple loop, without any recursion, but let's see how it could be done in the less efficient and more error prone way.
// I'll assume that the function signature can't be changed
void fib(int arr[], int num)
{
// In the general case, use the well known recurrence relation.
if ( num > 1 )
{
// Use recursion here to calculate the previous elements of the array.
fib(arr, /* ... */);
// ^^^^^^^^^ Can you figure out what index should be passed here?
// Then, calculate the element at index num using the recurrence relation.
arr[num] = arr[num - 1] + arr[num - 2];
// ^^^^^^^ ^^^^^^^ Note the indices.
// Are those values alredy known?
}
// When num is 0 or 1, we can't use the general formula. Can you tell why?
else if ( num >= 0 )
{
fib(arr, /* ... */);
// ^^^^^^^^^ Same as before.
arr[num] = 1;
}
// If num is less than 0, it just do nothing.
}