import math
def summation_2(numbers, start, stop):
"""Computes the sum of numbers[start:stop]"""
if stop <= start:
return 0
if stop - start == 1:
return numbers[start]
middle = math.floor((start + stop) / 2)
left_sum = summation_2(numbers, start, middle)
right_sum = summation_2(numbers, middle, stop)
return left_sum + right_sum

### Analyzing the Time Complexity

The `summation_2` function is a recursive algorithm that computes the sum of a subarray `numbers[start:stop]`. Let's determine the time complexity by analyzing the recurrence relation.

#### 1. **Base Case:**
The base cases are:
- If `stop <= start`, the function returns `0`. This operation takes \( O(1) \) time.
- If `stop - start == 1`, the function returns `numbers[start]`. This also takes \( O(1) \) time.

#### 2. **Recursive Case:**
For subarrays with more than one element, the function divides the subarray into two halves:
- **Splitting the array:** The array is split into two subarrays at the midpoint `middle = math.floor((start + stop) / 2)`. Calculating the midpoint and splitting the array into two halves is an \( O(1) \) operation.
- **Recursive calls:** The function makes two recursive calls on the subarrays `[start:middle]` and `[middle:stop]`.

The time complexity of the function can be described by the following recurrence relation:

\[
T(n) = 2T\left(\frac{n}{2}\right) + O(1)
\]

Here:
- \( 2T\left(\frac{n}{2}\right) \) accounts for the two recursive calls on the two halves of the array.
- \( O(1) \) accounts for the time taken to compute the midpoint and sum the results of the recursive calls.

#### 3. **Solving the Recurrence Relation:**
This recurrence relation can be solved using the **Master Theorem**. The recurrence is in the standard form:

\[
T(n) = aT\left(\frac{n}{b}\right) + O(n^d)
\]

Where:
- \( a = 2 \)
- \( b = 2 \)
- \( d = 0 \) (since the non-recursive work is \( O(1) \)).

Now, compute the critical exponent \( \log_b{a} \):

\[
\log_2{2} = 1
\]

Since \( d < \log_b{a} \) (here \( d = 0 \) and \( \log_b{a} = 1 \)), this falls into the first case of the Master Theorem, which states:

\[
T(n) = O\left(n^{\log_b{a}}\right) = O(n)
\]

### Conclusion
The worst-case time complexity of the `summation_2` function is \( O(n) \).

### Recurrence Relation:
\[
T(n) = 2T\left(\frac{n}{2}\right) + O(1)
\]

This result is intuitive because the algorithm processes every element of the array exactly once, leading to a linear time complexity.

Would you like to delve deeper into any part of this analysis, or explore any related concepts? Here are five related questions:

1. How does this time complexity compare to that of the first `summation` function?
2. What are the space complexity implications of this recursive approach?
3. How would the time complexity change if we used a different splitting strategy?
4. Could this algorithm be optimized for parallel processing?
5. How would the time complexity change if the base case was modified to handle larger subarrays?

**Tip:** When analyzing recursive algorithms, always pay attention to both the number of recursive calls and the amount of work done at each level of recursion.

Explore the time complexity analysis of the recursive summation algorithm in Python. Understand how the algorithm computes the sum of a subarray efficiently using recursion. This analysis includes the application of the Master Theorem and implications for algorithmic complexity. Suitable for advanced students and professionals interested in algorithm analysis.

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