Sum of Fibonacci Numbers in a range

Given a range [l, r], the task is to find the sum fib(l) + fib(l + 1) + fib(l + 2) + ….. + fib(r) where fib(n) is the n^th Fibonacci number.

Examples:

Input: l = 2, r = 5 
Output: 11 
fib(2) + fib(3) + fib(4) + fib(5) = 1 + 2 + 3 + 5 = 11

Input: l = 4, r = 8 
Output: 50  

Naive approach: Simply calculate fib(l) + fib(l + 1) + fib(l + 2) + ….. + fib(r) in O(r – l) time complexity. 
In order to find fib(n) in O(1) we will take the help of the Golden Ratio. 

Fibonacci’s calculation using Binet’s Formula

fib(n) = phiⁿ – psiⁿ) / ?5 
Where, 
phi = (1 + sqrt(5)) / 2 which is roughly equal to 1.61803398875 
psi = 1 – phi = (1 – sqrt(5)) / 2 which is roughly equal to 0.61803398875 

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to return the nth Fibonacci number
int fib(int n)
{
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}

// Function to return the required sum
ll calculateSum(int l, int r)
{

    // To store the sum
    ll sum = 0;

    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);

    return sum;
}

// Driver code
int main()
{
    int l = 4, r = 8;
    cout << calculateSum(l, r);

    return 0;
}

Java

// Java implementation of the approach
import java.lang.Math;
class GFG
{

// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.sqrt(5)) / 2;
    return (int)Math.round(Math.pow(phi, n) / Math.sqrt(5));
}

// Function to return the required sum
static int calculateSum(int l, int r)
{

    // To store the sum
    int sum = 0;

    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);

    return sum;
}

// Driver code
public static void main(String[] args)
{
    int l = 4, r = 8;
    System.out.println(calculateSum(l, r));

}
}

//This code is contributed by Code_Mech.

Python3

# Python3 implementation of the approach

# Function to return the nth
# Fibonacci number
def fib(n):
    phi = ((1 + (5 ** (1 / 2))) / 2);
    return round((phi ** n) / (5 ** (1 / 2)));

# Function to return the required sum
def calculateSum(l, r):

    # To store the sum
    sum = 0;

    # Calculate the sum
    for i in range(l, r + 1):
        sum += fib(i);

    return sum;

# Driver Code
if __name__ == '__main__':
    l, r = 4, 8;
    print(calculateSum(l, r));

# This code contributed by Rajput-Ji

C#

// C# implementation of above approach
using System;

class GFG
{

// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int)Math.Round(Math.Pow(phi, n) / Math.Sqrt(5));
}

// Function to return the required sum
static int calculateSum(int l, int r)
{

    // To store the sum
    int sum = 0;

    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);

    return sum;
}

// Driver code
public static void Main()
{
    int l = 4, r = 8;
    Console.WriteLine(calculateSum(l, r));
}
}

/* This code contributed by PrinciRaj1992 */

PHP

<?php
// PHP implementation of the approach

// Function to return the nth
// Fibonacci number
function fib($n)
{
    $phi = (1 + sqrt(5)) / 2;
    return (int)round(pow($phi, $n) / sqrt(5));
}

// Function to return the required sum
function calculateSum($l, $r)
{

    // To store the sum
    $sum = 0;

    // Calculate the sum
    for ($i = $l; $i <= $r; $i++)
        $sum += fib($i);

    return $sum;
}

// Driver code
$l = 4;
$r = 8;
echo calculateSum($l, $r);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript implementation of the approach

    // Function to return the nth Fibonacci number
    function fib(n) {
        var phi = (1 + Math.sqrt(5)) / 2;
        return parseInt( Math.round
        (Math.pow(phi, n)/ Math.sqrt(5))
        );
    }

    // Function to return the required sum
    function calculateSum(l , r) {

        // To store the sum
        var sum = 0;

        // Calculate the sum
        for (i = l; i <= r; i++)
            sum += fib(i);

        return sum;
    }

    // Driver code

        var l = 4, r = 8;
        document.write(calculateSum(l, r));


// This code contributed by gauravrajput1

</script>

Output:

50

Efficient approach: The idea is to find the relationship between the sum of Fibonacci numbers and n^th Fibonacci number and use Binet’s Formula to calculate its value.

Relationship Deduction

1. F(i) refers to the i^th Fibonacci number.
2. S(i) refers to the sum of Fibonacci numbers till F(i). 

We can rewrite the relation F(n + 1) = F(n) + F(n – 1) as below: 
F(n – 1) = F(n + 1) – F(n) 
Similarly, 
F(n – 2) = F(n) – F(n – 1) 
… 
… 
… 
F(0) = F(2) – F(1)
Adding all the equations, on left side, we have 
F(0) + F(1) + … + F(n – 1) which is S(n – 1) 

Therefore, 
S(n – 1) = F(n + 1) – F(1) 
S(n – 1) = F(n + 1) – 1 
S(n) = F(n + 2) – 1

In order to find S(n), simply calculate the (n + 2)^th Fibonacci number and subtract 1 from the result. 
Therefore, 
S(l, r) = S(r) – S(l – 1) 
S(l, r) = F(r + 2) – 1 – (F(l + 1) – 1) 
S(l, r) = F(r + 2) – F(l + 1)

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the nth Fibonacci number
int fib(int n)
{
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}

// Function to return the required sum
int calculateSum(int l, int r)
{

    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}

// Driver code
int main()
{
    int l = 4, r = 8;
    cout << calculateSum(l, r);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.sqrt(5)) / 2;
    return (int) Math.round(Math.pow(phi, n) / Math.sqrt(5));
}

// Function to return the required sum
static int calculateSum(int l, int r)
{

    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}

// Driver code
public static void main(String[] args)
{
    int l = 4, r = 8;
    System.out.println(calculateSum(l, r));

}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
import math

# Function to return the nth
# Fibonacci number
def fib(n):

    phi = (1 + math.sqrt(5)) / 2;
    return int(round(pow(phi, n) /
                         math.sqrt(5)));

# Function to return the required sum
def calculateSum(l, r):

    # Using our deduced result
    sum = fib(r + 2) - fib(l + 1);
    return sum;

# Driver code
l = 4;
r = 8;
print(calculateSum(l, r));

# This code is contributed by mits

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int) Math.Round(Math.Pow(phi, n) /
                            Math.Sqrt(5));
}

// Function to return the required sum
static int calculateSum(int l, int r)
{

    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}

// Driver code
public static void Main()
{
    int l = 4, r = 8;
    Console.WriteLine(calculateSum(l, r));
}
}

// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP implementation of the approach

// Function to return the nth Fibonacci number
function fib($n)
{
    $phi = (1 + sqrt(5)) / 2;
    return (int) round(pow($phi, $n) / sqrt(5));
}

// Function to return the required sum
function calculateSum($l, $r)
{

    // Using our deduced result
    $sum = fib($r + 2) - fib($l + 1);
    return $sum;
}

// Driver code
$l = 4; $r = 8;
echo(calculateSum($l, $r));

// This code is contributed by Code_Mech
?>

Javascript

<script>
// javascript implementation of the approach
// Function to return the nth Fibonacci number
    function fib(n) {
        var phi = (1 + Math.sqrt(5)) / 2;
        return parseInt( Math.round(Math.pow(phi, n) / Math.sqrt(5)));
    }

    // Function to return the required sum
    function calculateSum(l , r) {

        // Using our deduced result
        var sum = fib(r + 2) - fib(l + 1);
        return sum;
    }

    // Driver code

        var l = 4, r = 8;
        document.write(calculateSum(l, r));


// This code contributed by aashish1995
</script>

Output:

50

Time Complexity: O(log r)
Auxiliary Space: O(1)