Question

Students are discussing variations of algorithms to find Fibonacci numbers, like fib(6). What is the best possible time complexity that they can get for this algorithm?(1 point) Responses O(n2) upper O left parenthesis n squared right parenthesis O(1) upper O left parenthesis 1 right parenthesis O(logn) upper O left parenthesis log n right parenthesis O(n)

Answer 1

The best possible time complexity for an algorithm that computes the Fibonacci numbers efficiently is \( O(\log n) \).

This can be achieved using matrix exponentiation. The Fibonacci numbers can be represented in terms of matrix multiplication, allowing the computation of Fibonacci numbers in logarithmic time.

Here's a brief overview of how this works:

The nth Fibonacci number can be computed using the following matrix relation:

\[ \begin{pmatrix} F(n) \ F(n-1) \end{pmatrix}

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

By exponentiating the matrix using methods like exponentiation by squaring, we can compute the nth Fibonacci number in \( O(\log n) \) time.

Therefore, the best answer from the options provided is \( O(\log n) \).