TheAlgorithms-C/misc/fibonacci_formula.c
2022-09-30 14:21:43 +05:30

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/**
* @file
* @brief Finding Fibonacci number of any `n` number using [Binet's closed form formula](https://en.wikipedia.org/wiki/Fibonacci_number#Binet's_formula)
* compute \f$f_{nth}\f$ Fibonacci number using the binet's formula:
* Fn = 1√5 * (1+√5 / 2)^n+1 1√5 * (1√5 / 2)^n+1
* @author [GrandSir](https://github.com/GrandSir/)
*/
#include <math.h> /// for pow and sqrt
#include <stdio.h> /// for printf
#include <assert.h> /// for assert
/**
* @param n index of number in Fibonacci sequence
* @returns nth value of fibonacci sequence for all n >= 0
*/
int fib(unsigned int n) {
float seq = (1 / sqrt(5) * pow(((1 + sqrt(5)) / 2), n + 1)) - (1 / sqrt(5) * pow(((1 - sqrt(5)) / 2), n + 1));
// removing unnecessary fractional part by implicitly converting float to int
return seq;
}
/**
* @brief Self-test implementations
* @returns void
*/
static void test () {
/* this ensures that the first 10 number of fibonacci sequence
* (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89)
* matches with algorithm
*/
assert(fib(0) == 1);
assert(fib(1) == 1);
assert(fib(2) == 2);
assert(fib(3) == 3);
assert(fib(4) == 5);
assert(fib(5) == 8);
assert(fib(6) == 13);
assert(fib(7) == 21);
assert(fib(8) == 34);
assert(fib(9) == 55);
assert(fib(10) == 89);
printf("All tests have successfully passed!\n");
}
/**
* @brief Main function
* @returns 0 on exit
*/
int main() {
test(); // run self-test implementations
for(int i = 0; i <= 10; i++){
printf("%d. fibonacci number is: %d\n", i, fib(i));
}
return 0;
}