Cho biết độ phức tạp của đoạn mã giả sau dùng ký hiệu big-O.int F(int n)if (n <= 2)return 1elsereturn F(n − 2) * F(n − 2)a.O(2^(n/2))b.O(2^n)c.O(n^2)d.O( n )

Breakdown the Problem

1. We need to analyze the given pseudocode for the function F(n).
2. Determine the time complexity using big-O notation.

Relevant Concepts

1. The function F(n) is defined as:

   • If $n \leq 2$, it returns $1$.
   • Otherwise, it returns $F(n - 2) * F(n - 2)$.

2. This recursive function is essentially splitting into two calls of F(n-2).

Analysis and Detail

1. The recursive case can be rewritten as $F(n) = [F(n-2)]^2$.

2. Each recursive call reduces $n$ by 2, leading to this recurrence relation: $T(n) = 2T(n-2) \quad \text{for} \quad n > 2$ and $T(n) = 1$ for $n \leq 2$.

3. To solve this recurrence, we can expand it: $T(n) = 2T(n-2) = 2(2T(n-4)) = 2^2 T(n-4) = 2^3 T(n-6) = \ldots$ Continuing this until we reach the base case, we have $k$ such that $n - 2k \leq 2$. Thus, $k \approx \frac{n}{2}$.

4. Replacing $T(2)$ (the base case) into the equation, we find: $T(n) = 2^{k} T(2) = 2^{n/2} \cdot 1 = 2^{n/2}$

Verify and Summarize

1. The time complexity is $O(2^{n/2})$, which can also be expressed as $O(\sqrt{2^n})$.

Final Answer

The time complexity of the given pseudocode is a. O(2^(n/2)).