Unveiling the Fibonacci Mystery: A Matrix Exponentiation Adventure 🌌

Hey Everyone! 👋 Let's embark on a journey through the code that takes the Fibonacci sequence to new heights using matrix exponentiation. I'm Jordan, your code tour guide. We'll break down each step, do a dry run where needed, and unveil the enchanting magic of matrix operations. Ready for the adventure? 💻💫

Understanding the Mission:

The goal of this Java code is to efficiently calculate the A-th Fibonacci number modulo 109+7109+7.

Let's delve into the steps:

public class Solution {
    private static final int Max = 1000000007;
    private static final int[] base = {1, 1, 0};

    public int solve(int A) {
        if (A <= 1) {
            return A;
        }
        int[] fib = base.clone();
        power(fib, A - 1);
        return fib[0];
    }

    // Let's dive deeper...
}

Step 1: Handle the Basics

First, we check if A is 0 or 1. If so, the Fibonacci number is A itself. Simple, right?

if (A <= 1) {
    return A;
}

Step 2: Clone and Empower

Now, we create a clone of the base matrix and raise it to the power of A−1 using the power method.

int[] fib = base.clone();
power(fib, A - 1);
return fib[0];

Matrix Exponential Unvieled

Step 3: The Power Method

Let's dry run the power method for A\=5 to understand how matrix exponentiation unfolds:

public void power(int[] fib, int n) {
    if (n <= 1) {
        return;
    }
    power(fib, n / 2);
    multiply(fib, fib);
    if (n % 2 != 0) {
        multiply(fib, base);
    }
}

Dry Run:

• n\=5: Recursive call with n\=2, then n\=1.

• n\=2: Recursive call with n\=1.

• n\=1: Base case reached.

• Now, backtracking:

  • Multiply fib by itself (matrix multiplication).

  • n\=2 is halved, so no further multiplication in this step.

  • Since 5%2≠0, multiply fib by the base matrix.

Step 4: Multiply Method

Now, let's understand the multiply method:

public void multiply(int[] fib1, int[] fib2) {
    long n2 = 1L * fib1[1] * fib2[1] + 1L * fib1[2] * fib2[2];
    long n1 = 1L * fib1[0] * fib2[1] + 1L * fib1[1] * fib2[2];
    long n = n2 + n1;
    fib1[2] = (int) (n2 % Max);
    fib1[1] = (int) (n1 % Max);
    fib1[0] = (int) (n % Max);
}

Dry Run:

• Compute n2​=1×1+0×0=1.

• Compute n1​=1×1+1×0=1.

• Compute n\=n2​+n1​=2.

• Update fib1 with the results modulo 10^9+7.

Bringing It All Together:

With these steps, the code elegantly calculates the A-th Fibonacci number modulo 10^9+7, showcasing the efficiency of matrix exponentiation.

Conclusion:

Matrix exponentiation isn't just for the experts; it's a powerful technique accessible to all coders. This Java code not only efficiently computes Fibonacci numbers but also highlights the beauty of algorithmic approaches. I hope this exploration into matrices and exponents has been enlightening. #FibonacciMagic #MatrixExponentiationExplained

Happy coding, and may your coding adventures always be filled with wonder! 🌟👩‍💻

Jordan ^_~