No need for code. Algorithms are theoretical and please number the steps of the algorithm. Thanks!
No need for code. Algorithms are theoretical and please number the steps of the algorithm. Thanks!
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
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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No need for code. Algorithms are theoretical and please number the steps of the
![Problem 1. Suppose we have an array A[1: n] of n distinct numbers. For any element Ali], we define the
rank of A[i], denoted by rank(A[i]), as the number of elements in A that are strictly smaller than Ali] plus
one; so rank(A[i]) is also the correct position of Ali] 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).
%3D
(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).
(b) Use magic-pivot as a black-box to design an algorithm that given the array A and any integer
1<r<n, finds the element in A that has rank r in O(n) time?.
Hint: Suppose we run partition subroutine in quick sort with pivot p and it places it in position q.
Then, if r < q, we only need to look for the answer in the subarray A[1 : q] and if r > q, we need to
look for it in the subarray A[g +1: n] (although, what is the new rank we should look for now?).
ISuch an algorithm indeed exists, but its descriplion is rather complicated and not relevant to us in this problem.
2Note that an algorithm with runtime O(n log n) follows imnediately from part (a) - sort. the array and return the clement
at position r. The goal however is to obtain an algorithm with runtime O(n).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff26fda3f-4323-495f-b28b-f131dcfa5992%2F3174d9e2-ab42-4656-b2e0-b668448de548%2Ftqarzxu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 1. Suppose we have an array A[1: n] of n distinct numbers. For any element Ali], we define the
rank of A[i], denoted by rank(A[i]), as the number of elements in A that are strictly smaller than Ali] plus
one; so rank(A[i]) is also the correct position of Ali] 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).
%3D
(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).
(b) Use magic-pivot as a black-box to design an algorithm that given the array A and any integer
1<r<n, finds the element in A that has rank r in O(n) time?.
Hint: Suppose we run partition subroutine in quick sort with pivot p and it places it in position q.
Then, if r < q, we only need to look for the answer in the subarray A[1 : q] and if r > q, we need to
look for it in the subarray A[g +1: n] (although, what is the new rank we should look for now?).
ISuch an algorithm indeed exists, but its descriplion is rather complicated and not relevant to us in this problem.
2Note that an algorithm with runtime O(n log n) follows imnediately from part (a) - sort. the array and return the clement
at position r. The goal however is to obtain an algorithm with runtime O(n).
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