How does matrix exponentiation relate to Fibonacci numbers?

How does matrix exponentiation relate to Fibonacci numbers?

Matrix exponentiation is a powerful method to compute Fibonacci numbers efficiently. By representing Fibonacci numbers in matrix form, we can use matrix exponentiation to find the nth Fibonacci number in logarithmic time.

Can you show me the matrix representation of Fibonacci numbers?

Sure! The Fibonacci sequence can be represented using the following matrix:

[ \begin{pmatrix} F(n) \ F(n-1) \ F(n-1) \ F(n-2) \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix}^{n-1} ]

Here, ( F(n) ) is the nth Fibonacci number.

How do you perform matrix exponentiation?

Matrix exponentiation can be performed using the method of exponentiation by squaring. This involves recursively squaring the matrix and multiplying it when necessary, which reduces the time complexity to O(log n). Would you like a code example?

Yes, please provide a code example for matrix exponentiation.

Here's a Python example for matrix exponentiation:

def matrix_mult(A, B):
    return [[A[0][0] * B[0][0] + A[0][1] * B[1][0], A[0][0] * B[0][1] + A[0][1] * B[1][1]],
            [A[1][0] * B[0][0] + A[1][1] * B[1][0], A[1][0] * B[0][1] + A[1][1] * B[1][1]]]

def matrix_pow(M, p):
    if p == 1:
        return M
    if p % 2 == 0:
        half_pow = matrix_pow(M, p // 2)
        return matrix_mult(half_pow, half_pow)
    else:
        return matrix_mult(M, matrix_pow(M, p - 1))

F = [[1, 1], [1, 0]]
result = matrix_pow(F, n - 1)
print(result[0][0])  # This gives F(n)

This code will compute the nth Fibonacci number.