Write a
recursive function to find the Fibonacci series
The
Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..
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.
Given
a number n, print n-th Fibonacci Number.
Examples:
Input
: n = 2
Output : 1
Input
: n = 9
Output : 34
Recommended
Problem
Nth
Fibonacci Number
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
For n = 9
Output:34
The following are different methods to get
the nth Fibonacci number.
Method
1 (Use recursion)
A simple method that is a direct recursive implementation mathematical
recurrence relation is given above.
|
//
Fibonacci Series using Recursion #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; } |
Output
34
Time
Complexity: Exponential, as every function
calls two other functions.
If the original recursion tree were to be
implemented then this would have been the tree but now for n times the
recursion function is called
Original tree for recursion
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)
Optimized tree for recursion for code above
fib(5)
fib(4)
fib(3)
fib(2)
fib(1)
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 in method 1 by storing the Fibonacci
numbers calculated so far.
|
//Fibonacci
Series using Dynamic Programming #include<stdio.h> int fib(int n) { /*
Declare an array to store Fibonacci numbers. */ int f[n+2]; // 1 extra to
handle case, n = 0 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; } |
Output
34
Time
complexity: O(n) for given n
Auxiliary
space: O(n)
Method
3: (Space Optimized 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 Fibonacci number in
series.
|
//
Fibonacci Series using Space Optimized Method #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; } |
Output
34
Time
Complexity: O(n)
Extra Space: O(1)
Method
4: Using power of the matrix {{1, 1}, {1, 0}}
This is another O(n) that 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 matrices 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 designed 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; } |
Output
34
Time
Complexity: O(n)
Extra Space: O(1)
Method
5: (Optimized Method 4)
Method 4 can be optimized to work in O(Logn) time complexity. We can do
recursive multiplication to get power(M, n) in the previous 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; } |
Output
34
Time
Complexity: O(Logn)
Extra Space: O(Logn) if we consider the function
call stack size, otherwise O(1).
Method
6: (O(Log n) Time)
Below is one more interesting recurrence formula that can be used to find n’th
Fibonacci Number in O(Log n) time.
If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)
If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)
How
does this formula work?
The formula can be derived from the above matrix equation.
Taking determinant on both sides, we get
(-1)n = Fn+1Fn-1 - Fn2
Moreover, since AnAm = An+m for any
square matrix A,
the following identities can be derived
(they are obtained
from two different coefficients of the
matrix product)
FmFn + Fm-1Fn-1 = Fm+n-1 ---------------------------(1)
By putting n = n+1 in equation(1),
FmFn+1 + Fm-1Fn = Fm+n --------------------------(2)
Putting m = n in equation(1).
F2n-1 = Fn2 + Fn-12
Putting m = n in equation(2)
F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source:
Wiki) --------
( By putting Fn+1 = Fn + Fn-1 )
To get the formula to be proved, we simply
need to do the following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2
Below is the implementation of the above
idea.
|
// C++
Program to find n'th fibonacci Number in // with
O(Log n) arithmetic operations #include
<bits/stdc++.h> using namespace std; const int
MAX = 1000; //
Create an array for memoization int f[MAX] = {0}; // Returns
n'th fibonacci number using table f[] int fib(int n) { //
Base cases if (n == 0) return 0; if (n == 1 || n == 2) return (f[n] = 1); //
If fib(n) is already computed if (f[n]) return f[n]; int k = (n & 1)? (n+1)/2 : n/2; //
Applying above formula [Note value n&1 is 1 //
if n is odd, else 0. f[n]
= (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1)) :
(2*fib(k-1) + fib(k))*fib(k); return f[n]; } /*
Driver program to test above function */ int main() { int n = 9; printf("%d
", fib(n)); return 0; } |
Output
34
Time
complexity: O(Log n), as we divide the problem in
half in every recursive call.
Method
7: (Another approach(Using Binet’s formula))
In this method, we directly implement the formula for the nth term in the
Fibonacci series.
Fn =
{[(√5 + 1)/2] ^ n} / √5
Note: Above Formula gives correct result only upto for
n<71. Because as we move forward from n>=71 , rounding error becomes
significantly large . Although , using floor function instead of round function
will give correct result for n=71 . But after from n=72 , it also fails.
Example: For N=72 , Correct result is 498454011879264 but above
formula gives 498454011879265.
|
// C
Program to find n'th fibonacci Number #include<stdio.h> #include<math.h> int fib(int n) { double phi = (1 + sqrt(5)) / 2; return round(pow(phi, n) / sqrt(5)); } int main () { int n = 9; printf("%d",
fib(n)); return 0; } |
Output
34
Time
Complexity: O(logn), this is because
calculating phi^n takes logn time
Auxiliary Space: O(1)
Method
8: DP using memoization(Top down approach)
We can avoid the repeated work done in
method 1 by storing the Fibonacci numbers calculated so far. We just need to
store all the values in an array.
|
#include
<bits/stdc++.h> using namespace std; int dp[10]; int fib(int n) { if (n <= 1) return n; //
temporary variables to store //
values of fib(n-1) & fib(n-2) int first, second; if (dp[n - 1] != -1) first
= dp[n - 1]; else first
= fib(n - 1); if (dp[n - 2] != -1) second
= dp[n - 2]; else second
= fib(n - 2); //
memoization return dp[n] = first + second; } //
Driver Code int main() { int n = 9; memset(dp,
-1, sizeof(dp)); cout
<< fib(n); getchar(); return 0; //
This code is contributed by Bhavneet Singh } |
Output
34
Tirthankar Pal
MBA from IIT Kharagpur with GATE, GMAT, IIT Kharagpur Written Test, and Interview
2 year PGDM (E-Business) from Welingkar, Mumbai
4 years of Bachelor of Science (Hons) in Computer Science from the National Institute of Electronics and Information Technology
Google and Hubspot Certification
Brain Bench Certification in C++, VC++, Data Structure and Project Management
10 years of Experience in Software Development out of that 6 years 8 months in Wipro
Selected in Six World Class UK Universities:-
King's College London, Durham University, University of Exeter, University of Sheffield, University of Newcastle, University of Leeds
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