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Chapter 6: Problem 24

Suppose \(A[1], A[2], \ldots, A[n]\) is a one-dimensional array and \(n \geq 2\). Consider the subarray $$ A[1], A[2], \ldots, A[\lfloor n / 2\rfloor] . $$ a. How many elements are in the subarray (i) if \(n\) is even? and (ii) if \(n\) is odd? b. What is the probability that a randomly chosen array element is in the subarray (i) if \(n\) is even? and (ii) if \(n\) is odd?

Short Answer

Expert verified
a. (i) If $n$ is even, the subarray has $k$ elements. (ii) If $n$ is odd, the subarray has $k$ elements. b. (i) If $n$ is even, the probability of choosing an element from the subarray is $\frac{1}{2}$. (ii) If $n$ is odd, the probability of choosing an element from the subarray is $\frac{k}{2k+1}$.

Step by step solution

01

Part a(i): Number of elements in subarray if n is even

Since n is even, we can write n = 2k for some integer k. Now, let's find the number of elements in the subarray: $$ A[1], A[2], \ldots, A[\lfloor n / 2\rfloor] $$ We can use the floor function to simplify this expression: $$ \lfloor n / 2\rfloor = \lfloor 2k / 2\rfloor = k $$ So, if n is even, there are k elements in the subarray.
02

Part a(ii): Number of elements in subarray if n is odd

Now, if n is odd, we can write n = 2k + 1 for some integer k. Let's find the number of elements in the subarray: $$ A[1], A[2], \ldots, A[\lfloor n / 2\rfloor] $$ Using the floor function again, we have: $$ \lfloor n / 2\rfloor = \lfloor (2k + 1) / 2\rfloor = \lfloor k + 0.5\rfloor = k $$ So, if n is odd, there are also k elements in the subarray.
03

Part b(i): Probability of chosen array element in subarray if n is even

If n is even, we know that there are k elements in the subarray out of a total of 2k elements in the entire array. The probability of choosing an element from the subarray is therefore: $$ P(\text{element in subarray}) = \frac{\text{Number of elements in subarray}}{\text{Total number of elements in array}} = \frac{k}{2k} = \frac{1}{2} $$
04

Part b(ii): Probability of chosen array element in subarray if n is odd

If n is odd, we know that there are k elements in the subarray out of a total of (2k + 1) elements in the entire array. The probability of choosing an element from the subarray is therefore: $$ P(\text{element in subarray}) = \frac{\text{Number of elements in subarray}}{\text{Total number of elements in array}} = \frac{k}{2k + 1} $$ In summary, for part a, there are k elements in the subarray whether n is even or odd. For part b, the probability of choosing an element from the subarray is 1/2 if n is even, and k/(2k + 1) if n is odd.

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