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*). 不要python写伪代码

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⚹ Let's address each part of your question step by step. ⚹ ### 1. Brute Force Algorithm to Find Position p ⚹ To find the position $p$ using a brute force approach, you can iterate through the array and check for the condition where the sequence changes from increasing to decreasing. ⚹ ⚹ Here is the pseudocode for the brute force algorithm: ⚹ ``` function findPositionP(A, n): for p from 1 to n-1: if A[p] > A[p+1]: return p return -1 // if no such position is found ``` ⚹ The time complexity of this brute force algorithm is $O(n)$ because it involves a single pass through the array. ⚹ ### 2. Divide and Conquer Algorithm to Find Position p ⚹ A more efficient way to find the position $p$ is to use a divide and conquer approach. ⚹ ⚹ Here is the pseudocode for the divide and conquer algorithm: ⚹ ``` function findPositionP(A, low, high): if low == high: return low mid = (low + high) / 2 if A[mid] > A[mid + 1]: return findPositionP(A, low, mid) else: return findPositionP(A, mid + 1, high) ``` ⚹ The time complexity of this divide and conquer algorithm is $O(\log n)$ because it divides the problem into two halves at each step. ⚹ ### 3. Recurrence Relation for the Number of Comparisons ⚹ The recurrence relation for the number of comparisons made by the divide and conquer algorithm can be set up as follows: ⚹ ⚹ Let $T(n)$ be the number of comparisons made by the algorithm for an array of size $n$. ⚹ ⚹ The recurrence relation is: ⚹ $$ T(n) = T\left(\frac{n}{2} ight) + 1 $$ ⚹ This is because the algorithm makes one comparison at each step and then recursively calls itself on one half of the array. ⚹ ### 4. Complexity of the Algorithm in Big-O Notation ⚹ To solve the recurrence relation, we can use the iterative method. ⚹ ⚹ Starting with the recurrence relation: ⚹ $$ T(n) = T\left(\frac{n}{2} ight) + 1 $$ ⚹ Let's expand this relation: ⚹ $$ T(n) = T\left(\frac{n}{2} ight) + 1 \ = T\left(\frac{n}{4} ight) + 2 \ = T\left(\frac{n}{8} ight) + 3 \ \vdots \ = T\left(\frac{n}{2^k} ight) + k $$ ⚹ When $\frac{n}{2^k} = 1$, we have $k = \log n$. ⚹ ⚹ Therefore, the recurrence relation simplifies to: ⚹ $$ T(n) = T(1) + \log n $$ ⚹ Since $T(1)$ is a constant, we have: ⚹ $$ T(n) = O(\log n) $$ ⚹ Thus, the time complexity of the divide and conquer algorithm is $O(\log n)$. ⚹ ⚹ Conclusion: ⚹ ⚹ 1. The brute force algorithm has a time complexity of $O(n)$. ⚹ ⚹ 2. The divide and conquer algorithm has a time complexity of $O(\log n)$. ⚹ ⚹ 3. The recurrence relation for the divide and conquer algorithm is $T(n) = T\left(\frac{n}{2} ight) + 1$. ⚹ ⚹ 4. The time complexity of the divide and conquer algorithm is $O(\log n)$, as shown using the iterative method. ⚹

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