Download Terminology of Order Statistics in Fundamental of Algorithm | CS 231 and more Study notes Algorithms and Programming in PDF only on Docsity!
Wellesley College ◊ CS231 Algorithms ◊ October 11, 1996 Handout #
ORDER STATISTICS
Reading: CLR Chapter 10
**---------------------------------------------------------------------------------------------
Terminology**
Let S be a set of n elements (necessarily distinct).
The i th order statistic of S is the i th smallest element (i.e., the element that is larger than exactly i - 1 other elements in the set). Such an element is said to have rank i.
The minimum of S is the first order statistic (element with rank 1).
The maximum of S is the nth order statistic (element with rank n).
The median (s) of S is (are) the element(s) with rank (^) (n+1)/2 and (n+1)/2.
Example: B = {43 5 17 91 2 42 19 72 37 3}
- Minimum of B:
- Maximum of B:
- Medians of B:
- Rank of 17:
The Selection Problem
Specification:
Select(A, i) Return the ith order statistic from an array of n (distinct) elements.
Trivial algorithm:
Select(A, i) Sort(A) return A[i]
Can obviously be done in Θ(n lg(n)) time.
The main question of today's lecture: Can we do better?
Important special cases:
- Can find mininum or maximum with n - 1 comparisons.
- Can find minimum and maximum with 3 n/2 comparisons.
- For any fixed k ≥ 1, can select kth or (n - k)th element in Θ(n) time by k applications of minimum/maximum + deletion. --------------------------------------------------------------------------------------------- ------------------------
Selection in Worst Case Linear Time
Presentation in CLR is confusing. I prefer the following explanation.
We will consider the median-of-median-of-c algorithm, where c is an integer constant ≥ 1.
Median-of-Median-of-c(c, A, i)
- View input array A[1..n] as a two-dimensional array B[1..c, 1..n/c]. (If c does not evenly divide n, can always pad the array with extra large elements that won't affect results). Sort each column using any method (even quicksort). Since each column contains c elements, this step is linear, not quadratic. After this step, the row B[ (c+1)/2, 1..n/c] are medians of their respective columns.
- Use Median-of-Median-of-c(c, B[ (c+1)/2, 1..n/c], (n/c+1)/2) to find median of medians mm. At least 1/4 elements are ≥ mm; at least 3/4 elements are < mm. (Can see this by imagining that the columns are sorted from left to right by their medians. Important: we only imagine this sorting, we don't actually do it!)
- Partition A around mm, guaranteeing that mm ends up in the high partition. Let k be the number of elements in the low partition and (n - k) the number of elements in the high partition.
- If i ≤ k, use Median-of-Median-of-c with index i on the low partition; If i > k, use Median-of-Median-of-c with index i - k on the high partition;
Analysis:
