Chapter 9 Medians and order statistics Lee HsiuHui

Chapter 9 Medians and order statistics Lee, Hsiu-Hui Ack: This presentation is based on the lecture slides from Hsu, Lih-Hsing, as well as various materials from the web. 20071102 chap 09 Hsiu-Hui Lee

Order Statistics • The ith order statistic of a set of n element is the ith smallest. • Selection problem Input: a set A of n (distinct) numbers and a number i, with 1 ≤ i ≤ n. Output: the element x A that is larger than exactly i - 1 other elements of A. 20071102 chap 09 Hsiu-Hui Lee 2

Order Statistics Select the ith smallest of n element with rank i i = 1: minimum; i = n: maximum; i= lower or upper median. Naive algorithm: Sort and index ith element. • Worst-case running time = O(nlg n) using merge sort or heapsort (not quicksort). • 20071102 chap 09 Hsiu-Hui Lee 3

9. 1 Minimum and Maximum 20071102 chap 09 Hsiu-Hui Lee 4

Simultaneous min and max • • • Maintain the minimum and maximum of elements seen so far. Don’t compare each element to the minimum and maximum separately. Process elements in pairs. Compare the elements of a pair to each other. Then compare the larger element to the maximum so far, and compare the smaller element to the minimum so far. This leads to only 3 comparisons for every 2 elements. 20071102 chap 09 Hsiu-Hui Lee 5

Simultaneous min and max Setting up the initial values for the min and max depends on whether n is odd or even. (a) If n is odd, set both min and max to the first element. Then process the rest of the elements in pairs. (b) If n is even, compare the first two elements and assign the larger to max and the smaller to min. Then process the rest of the elements in pairs. in either case: 20071102 chap 09 Hsiu-Hui Lee 6

9. 2 Selection in expected linear time 20071102 chap 09 Hsiu-Hui Lee 7

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Example 20071102 chap 09 Hsiu-Hui Lee 9

Analysis of expected time The analysis follows that of randomized quicksort, but it’s a little different. Let T(n) =the random variable for the running time of RAND-SELECTon an input of size n, assuming random numbers are independent. For k= 0, 1, …, n– 1, define the indicator random variable if PARTITION generates a otherwise 20071102 chap 09 Hsiu-Hui Lee split, 10

Analysis (continued) To obtain an upper bound, assume that the ith element always falls in the larger side of the partition: if if if 20071102 chap 09 Hsiu-Hui Lee 11

Analysis (continued) 20071102 chap 09 Hsiu-Hui Lee 12

Taking expected values, we have 20071102 chap 09 Hsiu-Hui Lee 13

Solve this recurrence by substitution: Guess T(n) ≦cn 20071102 chap 09 Hsiu-Hui Lee 14

We choose the constant c so that c/4 - a > 0, i. d. , c>4 a, we can Divide both sides by c/4 – a, giving Thus, if we assume that T(n)=O(1) for n < 2 c/(c-4 a), we have E[T(n)] = O(n). 20071102 chap 09 Hsiu-Hui Lee 15

9. 3 Worst-case linear-time order statistics (According MIT) SELECT(i, n) 1. Divide the n elements into groups of 5. Find the median of each 5 -element group by rote. 2. Recursively SELECT the median x of the n/5 group medians to be the pivot. 3. Partition around the pivot x. Let k= rank(x). 4. If i = k then return x Same as else if i < k RANDthen recursively SELECT the i th SELECT smallest element in the lower part else recursively SELECT the (i-k) th smallest element in the upper part 20071102 chap 09 Hsiu-Hui Lee 16

Choosing the pivot 20071102 chap 09 Hsiu-Hui Lee 17

Choosing the pivot 1. Divide the n elements into groups of 5. 20071102 chap 09 Hsiu-Hui Lee 18

Choosing the pivot 1. Divide the n elements into groups of 5. Find the median of each 5 -element group by rote. lesser greater 20071102 chap 09 Hsiu-Hui Lee 19

Choosing the pivot 1. Divide the n elements into groups of 5. Find the median of each 5 -element group by rote. 2. Recursively SELECT the median x of the n/5 group medians to be the pivot. 20071102 chap 09 Hsiu-Hui Lee lesser greater 20

Analysis At least half the group medians are ≤ x , which is at least group medians. lesser greater 20071102 chap 09 Hsiu-Hui Lee 21

Analysis (Assume all elements are distinct. ) At least half the group medians are ≤ x , which is lesser at least n/5/2= n/10 group medians. • Therefore, at least 3 n/10 elements are ≤ x. greater 20071102 chap 09 Hsiu-Hui Lee 22

Analysis (Assume all elements are distinct. ) At least half the group medians are ≤ x , which is lesser at least n/5/2= n/10 group medians. • Therefore, at least 3 n/10 elements are ≤ x. • Similarly, at least 3 n/10 elements are ≥ x. greater 20071102 chap 09 Hsiu-Hui Lee 23

Minor simplification • For n ≥ 50, we have 3 n/10 ≥ n/4. • Therefore, for n ≥ 50 the recursive call to SELECT in Step 4 is executed recursively on ≤ 3 n/4 elements. • Thus, the recurrence for running time can assume that Step 4 takes time T(3 n/4) in the worst case. • For n< 50, we know that the worst-case time is T(n) = Θ(1). 20071102 chap 09 Hsiu-Hui Lee 24

Developing the recurrence T(n) Θ(n) T(n/5) Θ(n) T(3 n/4) SELECT(i, n) 1. Divide the n elements into groups of 5. Find the median of each 5 -element group by rote. 2. Recursively SELECT the median x of the n/5 group medians to be the pivot. 3. Partition around the pivot x. Let k= rank(x). 4. If i = k then return x else if i < k then recursively SELECT the i th smallest element in the lower part else recursively SELECT the (i-k) th smallest element in the upper part 20071102 chap 09 Hsiu-Hui Lee 25

Solving the recurrence Substitution: if c is chosen large enough to handle both the Θ(n) and the initial conditions. 20071102 chap 09 Hsiu-Hui Lee 26

9. 3 Selection in worst-case linear time (According Text) 20071102 chap 09 Hsiu-Hui Lee 27

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