The median m of a sequence of n elements is the element that would fall in the middle if the sequence was sorted. That is, e ≤ m for half the elements, and m ≤ e for the others. Clearly, one can obtain the median by sorting the sequence, but one can do quite a bit better with the following algorithm that finds the kth element of a sequence between a (inclusive) and b (exclusive). (For the median, use k = n/2, a = 0, and b = n.) select(k, a, b) Pick a pivot p in the subsequence between a and b. Partition the subsequence elements into three subsequences: the elements p Let n1, n2, n3 be the sizes of each of these subsequences. if k < n1 return select(k, 0, n1). else if (k > n1 + n2) return select(k, n1 + n2, n). else return p. c++

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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The median m of a sequence of n elements is the element that would fall in the middle if the sequence was sorted. That is, e ≤ m for half the elements, and m ≤ e for the others. Clearly, one can obtain the median by sorting the sequence, but one can do quite a bit better with the following algorithm that finds the kth element of a sequence between a (inclusive) and b (exclusive). (For the median, use k = n/2, a = 0, and b = n.)

select(k, a, b)
Pick a pivot p in the subsequence between a and b.
Partition the subsequence elements into three subsequences: the elements <p, =p, >p 
Let n1, n2, n3 be the sizes of each of these subsequences.
if k < n1 
   return select(k, 0, n1).
else if (k > n1 + n2) 
   return select(k, n1 + n2, n).
else 
   return p. c++

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