The given diagram shows the flowchart for a recursive function $\text{A}(n)$. Assume that all statements, except for the recursive calls, have $O(1)$ time complexity. If the worst case time complexity of this function is $O(n^\alpha)$, then the least possible value (accurate up to two decimal positions) of $\alpha$ is _______.

To determine the worst-case time complexity, we trace all execution paths through the flowchart to find the path that executes the maximum number of recursive calls of size $n/2$:

1. From Start, there is $1$ recursive call $\text{A}(n/2)$.

2. At the first decision diamond:

   • Left branch: $1$ recursive call $\text{A}(n/2)$ and returns $\implies 1 + 1 = 2$ calls total.

   • Right branch: $2$ recursive calls $\text{A}(n/2)$, bringing the total so far to $1 + 2 = 3$ calls, followed by the second decision diamond.

3. At the second decision diamond:

   • Right branch: Returns immediately $\implies 3$ calls total.

   • Down branch: $1$ recursive call $\text{A}(n/2)$, followed by the third decision diamond $\implies 3 + 1 = 4$ calls so far.

4. At the third decision diamond:

   • Left branch: $1$ recursive call $\text{A}(n/2)$ and returns $\implies 4 + 1 = 5$ calls total.

   • Right branch: Returns immediately $\implies 4$ calls total.

The worst-case path executes a total of $5$ recursive calls of size $n/2$. Thus, the recurrence relation for the worst-case time complexity is:

Using the Master Theorem for $T(n) = aT(n/b) + f(n)$ where $a = 5$, $b = 2$, and $f(n) = O(1)$:

Since $f(n) = O(n^0)$ and $0 < \log_2 5$, the worst-case time complexity is $\Theta(n^{\log_2 5}) = \Theta(n^{2.32})$.

Therefore, the least possible value of $\alpha$ is 2.32 (accepted range: 2.2 to 2.4).