The Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ……..
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
Fn = Fn-1 + Fn-2
with seed values
F0 = 0 and F1 = 1.
We strongly recommend that you click here and practice it, before moving on to the solution.
Write a function int fib(int n) that returns Fn. For example, if n = 0, then fib() should return 0. If n = 1, then it should return 1. For n > 1, it should return Fn-1 + Fn-2
Following are different methods to get the nth Fibonacci number.
Method 1 ( Use recursion )
A simple method that is a direct recusrive implementation mathematical recurance relation given above.
#include<stdio.h>
int
fib(
int
n)
{
if
(n <= 1)
return
n;
return
fib(n-1) + fib(n-2);
}
int
main ()
{
int
n = 9;
printf
(
"%d"
, fib(n));
getchar
();
return
0;
}
|
Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.
We can observe that this implementation does a lot of repeated work (see the following recursion tree). So this is a bad implementation for nth Fibonacci number.
fib(5)
/ \
fib(4) fib(3)
/ \ / \
fib(3) fib(2) fib(2) fib(1)
/ \ / \ / \
fib(2) fib(1) fib(1) fib(0) fib(1) fib(0)
/ \
fib(1) fib(0)
Extra Space: O(n) if we consider the function call stack size, otherwise O(1).
Method 2 ( Use Dynamic Programming )
We can avoid the repeated work done is the method 1 by storing the Fibonacci numbers calculated so far.
#include<stdio.h>
int
fib(
int
n)
{
/* Declare an array to store Fibonacci numbers. */
int
f[n+1];
int
i;
/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;
for
(i = 2; i <= n; i++)
{
/* Add the previous 2 numbers in the series
and store it */
f[i] = f[i-1] + f[i-2];
}
return
f[n];
}
int
main ()
{
int
n = 9;
printf
(
"%d"
, fib(n));
getchar
();
return
0;
}
|
Time Complexity: O(n)
Extra Space: O(n)
Method 3 ( Space Otimized Method 2 )
We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibannaci number in series.
#include<stdio.h>
int
fib(
int
n)
{
int
a = 0, b = 1, c, i;
if
( n == 0)
return
a;
for
(i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return
b;
}
int
main ()
{
int
n = 9;
printf
(
"%d"
, fib(n));
getchar
();
return
0;
}
|
Time Complexity: O(n)
Extra Space: O(1)
Method 4 ( Using power of the matrix {{1,1},{1,0}} )
This another O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n )), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.
The matrix representation gives the following closed expression for the Fibonacci numbers:
#include <stdio.h>
/* Helper function that multiplies 2 matricies F and M of size 2*2, and
puts the multiplication result back to F[][] */
void
multiply(
int
F[2][2],
int
M[2][2]);
/* Helper function that calculates F[][] raise to the power n and puts the
result in F[][]
Note that this function is desinged only for fib() and won't work as general
power function */
void
power(
int
F[2][2],
int
n);
int
fib(
int
n)
{
int
F[2][2] = {{1,1},{1,0}};
if
(n == 0)
return
0;
power(F, n-1);
return
F[0][0];
}
void
multiply(
int
F[2][2],
int
M[2][2])
{
int
x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int
y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int
z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int
w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
void
power(
int
F[2][2],
int
n)
{
int
i;
int
M[2][2] = {{1,1},{1,0}};
// n - 1 times multiply the matrix to {{1,0},{0,1}}
for
(i = 2; i <= n; i++)
multiply(F, M);
}
/* Driver program to test above function */
int
main()
{
int
n = 9;
printf
(
"%d"
, fib(n));
getchar
();
return
0;
}
|
Time Complexity: O(n)
Extra Space: O(1)
Method 5 ( Optimized Method 4 )
The method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the prevous method (Similar to the optimization done in this post)
#include <stdio.h>
void
multiply(
int
F[2][2],
int
M[2][2]);
void
power(
int
F[2][2],
int
n);
/* function that returns nth Fibonacci number */
int
fib(
int
n)
{
int
F[2][2] = {{1,1},{1,0}};
if
(n == 0)
return
0;
power(F, n-1);
return
F[0][0];
}
/* Optimized version of power() in method 4 */
void
power(
int
F[2][2],
int
n)
{
if
( n == 0 || n == 1)
return
;
int
M[2][2] = {{1,1},{1,0}};
power(F, n/2);
multiply(F, F);
if
(n%2 != 0)
multiply(F, M);
}
void
multiply(
int
F[2][2],
int
M[2][2])
{
int
x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int
y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int
z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int
w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
/* Driver program to test above function */
int
main()
{
int
n = 9;
printf
(
"%d"
, fib(9));
getchar
();
return
0;
}
|
Time Complexity: O(Logn)
Extra Space: O(Logn) if we consider the function call stack size, otherwise O(1).
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
References:
http://en.wikipedia.org/wiki/Fibonacci_number
http://www.ics.uci.edu/~eppstein/161/960109.html
