Fibonacci sequence using only 1 recursive call C

Question

Trying to calculate the Fibonacci numbers using a recursive function, but my code is using 2 recursive calls. Is it possible to do it using just one? Would saving the fib number of n-1 into an array or similar and then adding the numbers at end of function, would that work?

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

int fib(int n) {
    assert(n >= 0);
    if (n > 1) 
        return fib(n - 1) + fib(n - 2);
    else if (n == 1)
        return 1;
    else 
        return 0;
}

int main(void) {
    int n, f;
    printf("the nth number: ");
    scanf("%d", &n);
    f = fib(n);
    printf("%d \n", f);
    return 0;
}

KamilCuk · Accepted Answer · 2019-11-04 17:14:54Z

2

The following seems to work:

int fib_in(int n, int cur, int prev) {
    if (n == 1 || n == 2) {
        return cur;
    }
    return fib_in(n - 1, cur + prev, cur);
}
int fib(int n) {
    fib_in(n, 1, 1);
}

answered Nov 4, 2019 at 17:14

KamilCuk

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hleme · Accepted Answer · 2019-11-04 17:13:08Z

0

The function "fib" must return 2 values.

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

void fib(int n, int * ra, int * rb) {
    assert(n>=0);
    if (n>1) {
        int ya,yb;
        fib (n-1, &ya, &yb);
        *rb = ya;
        *ra = ya + yb;
    } else if (n==1) {
        *ra = 1;
        *rb = 0;
    } else {
        *ra = 0;
        *rb = 0;
    }
}
int main(int argc, const char * argv[]) {
    int n;
    printf("the nth number: ");
    scanf("%d",&n);
    int ya,yb;
    fib(n, &ya, &yb);
    printf("%d \n", ya);
    return 0;
}

answered Nov 4, 2019 at 17:13

hleme

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