Fibonacci Expression

Given a number n, evaluate the following expression. f(n-1)*f(n+1) - f(n)*f(n) where f(n) is the n-th Fibonacci number with n >= 1. Fibonacci Sequence is 0, 1, 1, 2, 3, 5, 8,13,… (here 0 is the 0th Fibonacci number).

Examples:

Input: n = 1
Output: -1
Explanation: f(n+1)*f(n-1) - f(n)*f(n) = 1*0 - 1*1 = -1.

Input: n = 2
Output: 1
Explanation: f(n+1)*f(n-1) - f(n)*f(n) = 2*1 - 1*1 = 1.

[Naive Approach] Calculate Fibonacci Numbers - O(n) Time and O(1) Space

The idea is to calculate the Fibonacci numbers F(n-1), F(n) and F(n+1) iteratively. Then, use these values to evaluate the given expression.

Working of Approach:

• Initialize F(0) = 0 and F(1) = 1.
• Generate Fibonacci numbers up to F(n).
• Keep F(n-1) and F(n) during the iteration.
• Calculate F(n+1) = F(n-1) + F(n).
• Substitute the values in the given expression and return the result.

#include <cmath>
#include <iostream>
using namespace std;

int fibExpression(int n)
{
    // Handle the base case n = 1.
    if (n == 1)
        return -1;

    // Initialize F(0) and F(1).
    long long a = 0, b = 1;

    // Generate Fibonacci numbers up to F(n).
    for (int i = 2; i <= n; i++)
    {
        long long c = a + b;
        a = b;
        b = c;
    }

    // a = F(n-1) and b = F(n).
    long long fnMinus1 = a;
    long long fn = b;
    long long fnPlus1 = fnMinus1 + fn;

    // Evaluate the given Fibonacci expression.
    return fnMinus1 * fnPlus1 - fn * fn;
}

int main()
{

    int n = 2;

    cout << fibExpression(n) << endl;

    return 0;
}

import java.util.*;

public class GFG {
    // Handle the base case n = 1.
    public static int fibExpression(int n)
    {
        if (n == 1)
            return -1;

        // Initialize F(0) and F(1).
        long a = 0, b = 1;

        // Generate Fibonacci numbers up to F(n).
        for (int i = 2; i <= n; i++) {
            long c = a + b;
            a = b;
            b = c;
        }

        // a = F(n-1) and b = F(n).
        long fnMinus1 = a;
        long fn = b;
        long fnPlus1 = fnMinus1 + fn;

        // Evaluate the given Fibonacci expression.
        return (int)(fnMinus1 * fnPlus1 - fn * fn);
    }

    public static void main(String[] args)
    {
        int n = 2;
        System.out.println(fibExpression(n));
    }
}

def fibExpression(n):
    # Handle the base case n = 1.
    if n == 1:
        return -1

    # Initialize F(0) and F(1).
    a = 0
    b = 1

    # Generate Fibonacci numbers up to F(n).
    for i in range(2, n + 1):
        c = a + b
        a = b
        b = c

    # a = F(n-1) and b = F(n).
    fnMinus1 = a
    fn = b
    fnPlus1 = fnMinus1 + fn

    # Evaluate the given Fibonacci expression.
    return int(fnMinus1 * fnPlus1 - fn * fn)

if __name__ == '__main__':
    n = 2
    print(fibExpression(n))

using System;

public class GFG {
    // Handle the base case n = 1.
    public static int fibExpression(int n)
    {
        if (n == 1)
            return -1;

        // Initialize F(0) and F(1).
        long a = 0, b = 1;

        // Generate Fibonacci numbers up to F(n).
        for (int i = 2; i <= n; i++) {
            long c = a + b;
            a = b;
            b = c;
        }

        // a = F(n-1) and b = F(n).
        long fnMinus1 = a;
        long fn = b;
        long fnPlus1 = fnMinus1 + fn;

        // Evaluate the given Fibonacci expression.
        return (int)(fnMinus1 * fnPlus1 - fn * fn);
    }

    public static void Main()
    {
        int n = 2;
        Console.WriteLine(fibExpression(n));
    }
}

function fibExpression(n)
{
    // Handle the base case n = 1.
    if (n === 1)
        return -1;

    // Initialize F(0) and F(1).
    let a = 0, b = 1;

    // Generate Fibonacci numbers up to F(n).
    for (let i = 2; i <= n; i++) {
        let c = a + b;
        a = b;
        b = c;
    }

    // a = F(n-1) and b = F(n).
    let fnMinus1 = a;
    let fn = b;
    let fnPlus1 = fnMinus1 + fn;

    // Evaluate the given Fibonacci expression.
    return fnMinus1 * fnPlus1 - fn * fn;
}

// Driver Code
let n = 2;

console.log(fibExpression(n));

1

[Expected Approach] Using Cassini's Identity - O(1) Time and O(1) Space

The idea is to use Cassini's Identity, which directly simplifies the given expression to (-1)^n. So we only need to compute parity of n.

By Cassini's Identity: F_{n-1}F_{n+1} - F_n^2 = (-1)^n. Hence:

• n odd -> -1
• n even -> 1

How does above formula work? The formula is based on matrix representation of Fibonacci numbers.

Let us understand with an example:
Input: n = 2

• n % 2 = 0, so n is even.
• Condition (n % 2 == 1) is false.
• Therefore, the function returns 1.

#include <cmath>
#include <iostream>
using namespace std;

int fibExpression(int n)
{

    // If n is odd, return -1, otherwise return 1
    return ((n % 2 == 1) ? -1 : 1);
}

int main()
{

    int n = 2;

    cout << fibExpression(n) << endl;

    return 0;
}

public class GFG {
    // If n is odd, return -1, otherwise return 1
    public static int fibExpression(int n)
    {
        return (n % 2 == 1) ? -1 : 1;
    }

    public static void main(String[] args)
    {
        int n = 2;
        System.out.println(fibExpression(n));
    }
}

def fibExpression(n):
    # If n is odd, return -1, otherwise return 1
    return -1 if n % 2 == 1 else 1

if __name__ == "__main__":
    n = 2
    print(fibExpression(n))

using System;

class GFG {
    // If n is odd, return -1, otherwise return 1
    public static int fibExpression(int n)
    {
        return (n % 2 == 1) ? -1 : 1;
    }

    static void Main()
    {
        int n = 2;
        Console.WriteLine(fibExpression(n));
    }
}

function fibExpression(n)
{
    // If n is odd, return -1, otherwise return 1
    return (n % 2 === 1) ? -1 : 1;
}

// Driver Code
let n = 2;
console.log(fibExpression(n));

1