`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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Answer from Sia

Posted 2 months ago

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Sure, Younglife! Let's break down your questions one by one.
Question 1: Brute Force Algorithm to Find Position p
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Step-by-step Instruction

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For each position, check if the elements before it are in increasing order and the elements after it are in decreasing order

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Return the position p if found

python

python

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.

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Question 2: Divide and Conquer Algorithm to Find Position p
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Step-by-step Instruction

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Check the middle element and determine if it is the position p

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Recursively search in the left or right half based on the order of elements

python

python

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.

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Question 3: Recurrence Relation for Divide and Conquer Algorithm
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Step-by-step Instruction

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Explain the base case and the recursive case

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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.

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- 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
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Step-by-step Instruction

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Apply the iterative method to solve the recurrence relation

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Show that the time complexity is $O(\log n)$

Explanation

Using the iterative method, we can expand the recurrence relation as follows:

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$T(n) = T(n/2) + 1 = T(n/4) + 2 = \ldots = T(1) + \log n = \log n$
```

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.

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I hope this helps! Let me know if you have any further questions.
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