Problem 1.
Determine the worst case time complexity of each of the recursive algorithms below. In each case, state the
recurrence relation describing the runtime. Solve the recurrence relation, either by unrolling it or showing
that it is the same as a recurrence we have encountered in lecture.
a) import math
def summation(numbers):
"""Given a list, returns the sum of the numbers in the list."""
n = len(numbers)
if n == 0:
return 0
if n == 1:
return numbers[0]
middle = math.floor(n / 2)
return summation(numbers[:middle]) + summation(numbers[middle:])
Hint: remember that slicing causes a copy to happen!

### Analyzing the Time Complexity

The given algorithm is a recursive function that computes the sum of a list of numbers. Let’s break down the time complexity by analyzing the recurrence relation step by step.

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

#### 2. **Recursive Case:**
When the list has more than one element, the function divides the list into two halves and recursively sums each half. The key steps are:
- **Slicing the list:** Slicing `numbers[:middle]` and `numbers[middle:]` creates two new lists, each of size approximately \( n/2 \). Slicing an array takes \( O(k) \) time, where \( k \) is the size of the new array. Here, the slicing operations together take \( O(n) \) time.
- **Recursive calls:** The function makes two recursive calls on these subarrays of size \( n/2 \).

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

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

Here:
- \( 2T\left(\frac{n}{2}\right) \) represents the time taken by the two recursive calls.
- \( O(n) \) accounts for the time taken to slice the array.

#### 3. **Solving the Recurrence Relation:**
This recurrence is similar to the standard "divide and conquer" recurrence that appears in many algorithms, such as merge sort.

We can solve it using the **Master Theorem**. The recurrence is in the form:

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

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

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

\[
\log_2{2} = 1
\]

Since \( d = \log_b{a} \), this falls into the second case of the Master Theorem, which states:

\[
T(n) = O(n^d \log n) = O(n \log n)
\]

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

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

Would you like to explore more details about how the Master Theorem was applied or any other aspect of this problem? Here are five related questions:

1. How does the time complexity change if we avoid slicing the list?
2. Can you provide an example of another algorithm with a similar \( O(n \log n) \) time complexity?
3. How would the time complexity differ if the list is split into three parts instead of two?
4. What are the space complexity implications of this recursive approach?
5. Could the algorithm be optimized further, and if so, how?

**Tip:** Understanding the Master Theorem is crucial when dealing with divide-and-conquer algorithms, as it simplifies solving many common recurrences.

Learn how to analyze the time complexity of recursive algorithms using recurrence relations. Explore a detailed solution to determining the worst-case time complexity of recursive functions, such as the summation function. This problem involves applying the Master Theorem and understanding divide-and-conquer algorithms.

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