Question 1(27 marks) Consider the following problem. Given an array A consisting of n distinct integers A[1], ... A[n]. It is known that there is a position p (1≤ p Sn), such that A[1], .., A[p] is in increasing order and A[p], A[p+1],..., A[n] is in decreasing order. Write a brute force algorithm to find the position p. What is the time complexity of your algorithm? 2. Devise a "divide and conquer" algorithm to find the position p. 3. Set up a recurrence relation for the number of comparisons made by your algorithm and explain it. 4 Based on the recurrence relation, show the complexity of your algorithm in big-O notation and prove it using either the iterative method or the substitution method, i.e., Mathematical Induction (for simplicity, you can assume that n = 2*).

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Sure, Younglife! Let's break down your questions one by one. ### Question 1: Brute Force Algorithm to Find Position p ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Define a function to iterate through the array and check for the position p. ‖ ‖ ▪ ⋮ For each position, check if the elements before it are in increasing order and the elements after it are in decreasing order. ‖ ‖ ▪ ⋮ Return the position p if found. ‖ ```python # Define a function to find the position p using brute force def find_position_brute_force(A): n = len(A) # Get the length of the array for p in range(1, n): # Iterate through each possible position p increasing = all(A[i] < A[i + 1] for i in range(p - 1)) # Check if A[1] to A[p] is increasing decreasing = all(A[i] > A[i + 1] for i in range(p, n - 1)) # Check if A[p] to A[n] is decreasing if increasing and decreasing: return p # Return the position p if both conditions are met return -1 # Return -1 if no such position is found ``` ```python # Main function to test the brute force algorithm def test_find_position_brute_force(): # Test case 1 A1 = [1, 3, 5, 4, 2] assert find_position_brute_force(A1) == 3 # Expected position is 3 # Test case 2 A2 = [1, 2, 3, 4, 5, 3, 1] assert find_position_brute_force(A2) == 5 # Expected position is 5 test_find_position_brute_force() ``` ∻Explanation∻ ⚹The brute force algorithm iterates through each possible position p and checks if the elements before p are in increasing order and the elements after p are in decreasing order.⚹ ∻Time Complexity∻ ⚹The time complexity is $O(n^2)$ because for each position p, we check the order of elements before and after p, which takes linear time.⚹ ∻Space Complexity∻ ⚹The space complexity is $O(1)$ as we are not using any extra space apart from a few variables.⚹ ### Question 2: Divide and Conquer Algorithm to Find Position p ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Define a function to use binary search to find the position p. ‖ ‖ ▪ ⋮ Check the middle element and determine if it is the position p. ‖ ‖ ▪ ⋮ Recursively search in the left or right half based on the order of elements. ‖ ```python # Define a function to find the position p using divide and conquer def find_position_divide_and_conquer(A, low, high): if low == high: return low # Base case: only one element left mid = (low + high) // 2 # Find the middle index if A[mid] > A[mid + 1]: return find_position_divide_and_conquer(A, low, mid) # Search in the left half else: return find_position_divide_and_conquer(A, mid + 1, high) # Search in the right half ``` ```python # Main function to test the divide and conquer algorithm def test_find_position_divide_and_conquer(): # Test case 1 A1 = [1, 3, 5, 4, 2] assert find_position_divide_and_conquer(A1, 0, len(A1) - 1) == 2 # Expected position is 2 # Test case 2 A2 = [1, 2, 3, 4, 5, 3, 1] assert find_position_divide_and_conquer(A2, 0, len(A2) - 1) == 4 # Expected position is 4 test_find_position_divide_and_conquer() ``` ∻Explanation∻ ⚹The divide and conquer algorithm uses binary search to find the position p by recursively searching in the left or right half based on the order of elements.⚹ ∻Time Complexity∻ ⚹The time complexity is $O(\log n)$ because we are dividing the search space in half at each step.⚹ ∻Space Complexity∻ ⚹The space complexity is $O(\log n)$ due to the recursive call stack.⚹ ### Question 3: Recurrence Relation for Divide and Conquer Algorithm ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Define the recurrence relation for the number of comparisons. ‖ ‖ ▪ ⋮ Explain the base case and the recursive case. ‖ ‖ ▪ ⋮ Solve the recurrence relation to find the time complexity. ‖ ∻Explanation∻ ⚹Let $T(n)$ be the number of comparisons made by the divide and conquer algorithm.⚹ - Base case: $T(1) = 0$ (no comparisons needed for a single element) - Recursive case: $T(n) = T(n/2) + 1$ (one comparison to decide which half to search in) ### Question 4: Complexity Analysis Using Recurrence Relation ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Use the recurrence relation $T(n) = T(n/2) + 1$. ‖ ‖ ▪ ⋮ Apply the iterative method to solve the recurrence relation. ‖ ‖ ▪ ⋮ Show that the time complexity is $O(\log n)$. ‖ ∻Explanation∻ ⚹Using the iterative method, we can expand the recurrence relation as follows:⚹ $$ T(n) = T(n/2) + 1 = T(n/4) + 2 = \ldots = T(1) + \log n = \log n $$ ∻Time Complexity∻ ⚹The time complexity is $O(\log n)$ because the recurrence relation $T(n) = T(n/2) + 1$ solves to $\log n$.⚹ ∻Space Complexity∻ ⚹The space complexity is $O(\log n)$ due to the recursive call stack.⚹ I hope this helps! Let me know if you have any further questions.

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