Problem Suppose we have an array A[1 : n] of n distinct numbers. For any element A[i], we define the rank of A[i], denoted by rank(A[i]), as the number of elements in A that are strictly smaller than A[i] plus one; so rank(A[i]) is also the correct position of A[i] in the sorted order of A. Suppose we have an algorithm magic-pivot that given any array B[1 : m] (for any m > 0), returns an index i such that rank(B[i]) = m/4 and has worst-case runtime of O(n)'.
Problem Suppose we have an array A[1 : n] of n distinct numbers. For any element A[i], we define the rank of A[i], denoted by rank(A[i]), as the number of elements in A that are strictly smaller than A[i] plus one; so rank(A[i]) is also the correct position of A[i] in the sorted order of A. Suppose we have an algorithm magic-pivot that given any array B[1 : m] (for any m > 0), returns an index i such that rank(B[i]) = m/4 and has worst-case runtime of O(n)'.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
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it's not a
![Problem
Suppose we have an array A[1: n] of n distinct numbers. For any element A[i], we define the
rank of A[i], denoted by rank(A[i]), as the number of elements in A that are strictly smaller than A[i] plus
one; so rank(A[i]) is also the correct position of A[i] in the sorted order of A.
Suppose we have an algorithm magic-pivot that given any array B[1 : m] (for any m > 0), returns an index
i such that rank(B[i]) = m/4 and has worst-case runtime of O(n)'.
Example: if B = [1,7,6, 3, 13, 4, 5, 11], then magic-pivot(B) will return index 4 as rank of B[4] = 3 is 2
which is m/4 in this case (rank of B[4]
3 is 2 since there is only one element smaller than 3 in B).
(a) Use magic-pivot as a black-box to obtain a deterministic quick-sort algorithm with worst-case running
time of O(n log n).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ae8a81d-a7d2-4710-8da9-0de0f94786a9%2Fc82e1ca7-166b-4d68-93c5-898168eae109%2Fj8y6509j_processed.png&w=3840&q=75)
Transcribed Image Text:Problem
Suppose we have an array A[1: n] of n distinct numbers. For any element A[i], we define the
rank of A[i], denoted by rank(A[i]), as the number of elements in A that are strictly smaller than A[i] plus
one; so rank(A[i]) is also the correct position of A[i] in the sorted order of A.
Suppose we have an algorithm magic-pivot that given any array B[1 : m] (for any m > 0), returns an index
i such that rank(B[i]) = m/4 and has worst-case runtime of O(n)'.
Example: if B = [1,7,6, 3, 13, 4, 5, 11], then magic-pivot(B) will return index 4 as rank of B[4] = 3 is 2
which is m/4 in this case (rank of B[4]
3 is 2 since there is only one element smaller than 3 in B).
(a) Use magic-pivot as a black-box to obtain a deterministic quick-sort algorithm with worst-case running
time of O(n log n).
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