Explain Fibonacci code - c

Question

#include<stdio.h>
#define max 2000
int arr1[max], arr2[max], arr3[max];
void fib();
int main()
{
int num, i, j, flag = 0;
for(i = 0; i<max; i++)
arr1[i] = arr2[i] = arr3[i] = 0;
arr2[max - 1] = 1;
printf("Enter the term : ");
scanf("%d", &num);
for(i = 0; i<num; i++)
{
fib();
if(i == num - 3)
break;
for(j = 0; j<max; j++)
arr1[j] = arr2[j];
for(j = 0; j<max; j++)
arr2[j] = arr3[j];
}
for(i = 0; i<max; i++)
{
if(flag || arr3[i])
{
flag = 1;
printf("%d", arr3[i]);
}
}
getch();
return 1;
}
void fib()
{
int i, temp;
for(i = 0; i<max; i++)
arr3[i] = arr1[i] + arr2[i];
for(i = max - 1; i>0; i--)
{
if(arr3[i]>9)
{
temp = arr3[i];
arr3[i] %= 10;
arr3[i - 1] += (temp / 10);
}
}
}
The above code generates the nth Fibonacci number. I am not able to understand how this works. Basically the Fibonacci number get stored in a very large array arr3[].
Please explain the logic involved in this code.
How does the fib() function work as well?

Answer 1

Here is a simple Fibonacci loop.
#include <stdio.h>
int main()
{
int term = 20, last2=0, last1=1, fib, i;
for (i=0; i<term; i++) {
fib = last2 + last1;
last2 = last1;
last1 = fib;
}
printf ("Term %d = %d\n", i, fib);
return 0;
}
Program output:
Term 20 = 10946
Although there is more than one idea as to where the sequence starts.

Answer 2

The example code in the original post is dealing with large numbers by storing 1 decimal digit per element in each of the arrays. It initializes arr[3] = arr2[] = arr1[] = 0, then arr2[] = 1. In the loop, fib() performs one instance of arr3[] = arr1[] + arr2[], handling the carries, then the loop does arr[1] = arr2[], arr2[] = arr3[]. If num < 3, the for loop exits on the loop condition i < num, if n >= 3, the loop exit when i == (num-3). (This could be avoided). The print loop skips leading zeroes in arr3[], setting flag once a non-zero value is found. The code needs some minor fixes. Here is a fixed example. Note that getch() may be _getch() in some environments (from conio.h). The second example below only uses two arrays. Fibonacci numbers starting from 0 are 0 1 1 2 3 5 8 ...
#include <conio.h>
#include <stdio.h>
#define max 2000
int arr1[max], arr2[max], arr3[max];
void fib();
int main()
{
int num, i, j;
for(i = 0; i<max; i++)
arr1[i] = arr2[i] = arr3[i] = 0;
arr1[max - 1] = 1;
printf("Enter the term : ");
scanf("%d", &num);
for(i = 0; i<num; i++)
{
fib();
for(j = 0; j<max; j++)
arr1[j] = arr2[j];
for(j = 0; j<max; j++)
arr2[j] = arr3[j];
}
for(i = 0; i < max-1; i++)
if(arr3[i])
break;
for( ; i < max; i++)
printf("%d", arr3[i]);
getch();
return 0;
}
void fib()
{
int i, temp;
for(i = 0; i<max; i++)
arr3[i] = arr1[i] + arr2[i];
for(i = max - 1; i>0; i--)
{
if(arr3[i]>9)
{
temp = arr3[i];
arr3[i] %= 10;
arr3[i - 1] += (temp / 10);
}
}
}
This example only uses two arrays, by alternating which array contains the sum (a1 += a0, a0 += a1). It uses Duff's device to enter the loop. Since the largest sum from digit + digit + carry is < 20, the carry loop in fib() was simplified.
#include <conio.h>
#include <stdio.h>
#define max 2000
void fib(unsigned char *a0, unsigned char *a1);
int main()
{
unsigned char a0[max], a1[max];
size_t i;
int n;
printf("Enter the term : ");
scanf("%d", &n);
for(i = 0; i < max; i++)
a0[i] = a1[i] = 0;
a0[max-1] = n & 1; /* if n even, a0=0=fib(0), a1=1=fib(-1) */
a1[max-1] = 1 - a0[max-1]; /* if n odd, a1=0=fib(0), a0=1=fib(-1) */
switch(n&1){
do{
fib(a0, a1);
case 1:
fib(a1, a0);
case 0:
continue;
}while(0 <= (n -= 2));
}
for(i = 0; i < max-1; i++)
if(a0[i])break;
for( ; i < max; i++)
printf("%d", a0[i]);
getch();
return 0;
}
void fib(unsigned char *a0, unsigned char *a1)
{
size_t i;
for(i = 0; i < max; i++)
a1[i] += a0[i];
for(i = max - 1; i > 0; i--){
if(a1[i] >= 10){
a1[i] -= 10;
a1[i-1] += 1;
}
}
}

Answer 3

Here's a much better implementation of the Fibonacci series
#include<iostream>
using namespace std;
main()
{
int n, c, first = 0, second = 1, next;
cout << "Enter the number of terms of Fibonacci series you want" << endl;
cin >> n;
cout << "First " << n << " terms of Fibonacci series are :- " << endl;
for ( c = 0 ; c < n ; c++ )
{
if ( c <= 1 )
next = c;
else
{
next = first + second;
first = second;
second = next;
}
cout << next << endl;
}
return 0;
}