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Median Finding, Order Statistics & Quick Sort. Vanithadevi Devarajan. 3. 4. 13. 14. 23. 27. 41. 54. 65. 75. 7th order statistic. The lower median is the - th order statistic The upper median is the - th order statistic
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Median Finding, Order Statistics & Quick Sort Vanithadevi Devarajan
3 4 13 14 23 27 41 54 65 75 7th order statistic Thelower medianis the- th order statistic Theupper medianisthe - th order statistic If n is odd, lower and upper median are the same 3 4 13 14 23 27 41 54 65 75 lower median upper median Order Statistics Thek-th order statisticis the k-th smallest element of an array.
Given an unsorted array, how quickly one can find the i-th order statistics = A[i] ? By Sorting the array and retrieving the i-th value takes (n lg n), even though max (i=n) and min(i=1) can be done in (n). Can we do better in Linear Time? - Solved affirmatively in 1972 by (Manuel)Blum, Floyd, Pratt, Rivest, and Tarjan. - Described two Linear-Time algorithms - One Randomized and One Deterministic
Selection ( Deterministic and Randomized ) – Finding the Median in Linear Time Randomized Algorithm • After the partitioning step, one can tell which subarray has the • item, one is looking for, by just looking at their sizes. • So, only need to recursively examine one subarray, not two. • For instance, • 1. If one is looking for the 87th-smallest element in • an array of size 200, and after partitioning the “LESS” • subarray (of elements less than the pivot), then just • need to find the 87th smallest element in LESS. • 2. On the other hand, if the “LESS” subarray has size 40, then • just need to find the 87−40−1 = 46th smallest element in • GREATER. • 3. And if the “LESS” subarray has size exactly 86 then, just • return the pivot. • (
Randomized/QuickSelect: • Given an array A of size n and integer k ≤ n, • Pick a pivot element p at random from A. • Split A into subarrays - LESS and GREATER, by comparing • each element to p as in Quicksort. While we are at it, count the number of elements ( L ) going into LESS. • 3. (a) If L = k − 1, then output p. • (b) If L > k − 1, output QuickSelect(LESS, k). • (c) If L < k − 1, output QuickSelect(GREATER, k − L − 1) • The expected number of comparisons for QuickSelect is O(n).
Deterministic Linear Time Algorithm • Deterministic Select • Given an array A of size n and integer k ≤ n, • Group the array into n/5 groups of size 5 and find the median • of each group. • 2. Each group is then sorted and its median is selected. • 3. Recursively, find the median of the medians. Call this as ‘p’. • Use p as a pivot to split the array into subarrays - LESS and • GREATER. • Recurse on the appropriate array to find the K-th smallest • element. • DeterministicSelect makes O(n) comparisons to find the kth smallest in an array of size n.
Quick Sort • Given array of some length n, • Pick an element p of the array as the pivot (or halt if the • array has size 0 or 1). • 2. Split the array into sub-arrays LESS, EQUAL, and GREATER • by comparing each element to the pivot. (LESS has all • elements less than p, EQUAL has all elements equal • to p, and GREATER has all elements greater than p). • Recursively sort LESS and GREATER. • Worst-case running time is O(n^2 ) • Randomized-Quicksort • Run the Quicksort algorithm as given above, each time • picking a random element in the array as the pivot. • Worst-case Expected-Time bound is O(n log n)
Project Implementations Implementation of QuickSort , where the pivot is chosen from the previous Order Statistics Algorithm. Implementation of Randomized QuickSort, that chooses a random element as the Pivot. • Implementation of Real World Quick Sort with the following heuristics. • If L and R partitions are of unequal sizes, sort the smallest • partition first. • 2. If array size <= 7, use insertion sort • 3. Use the following idea for getting the pivot. • middle = (start+end)/2 • if array size > 40 • length = array size / 8 • pivot1 = median(start, start - length, start - (2*length)) • pivot2 = median(end, end+length, end + 2*length) • pivot3 = median(middle,middle-length, middle+length) • pivot = median(pivot1,pivot2,pivot3) • else • pivot = median(start,middle,end)
