Augmenting Data Structures. General approach Dynamic order statistics Interval trees

Transcription

1 Augmenting Data Structures General approach Dynamic order statistics Interval trees 1

2 General Approach In some applications custom data structures will be necessary. In others, a well-known data structure will be modified or augmented. Augmenting a data structure: choose an underlying data structure add additional data or to the data structure modify basic operations to maintain the additional data Modified operations must maintain the key properties of the original data structure. 2

3 Order Statistics Dynamic Order Statistics: The i th order statistic of a set of n elements, where i {1, 2,, n}, is the element in the set with i th smallest key. The rank of a given element from a set of n elements is it s position in the linear order of the set. Note that in the presence of duplicates, the notion of rank is not well-defined. Augmented red-black trees can be used to find the i th order statistics and rank of a given element in O(lg n) time. 3

4 Order Statistics An order statistics tree is a red-black tree that has a size value at each node, where the size of node x is the number of nodes in the subtree rooted at x. If we define T.nil.size = 0 then: x.size = x.left.size + x.right.size

5 Example #1 Retrieving an element with a given rank: i = 13 rank = 13 5

6 Example #2 Retrieving an element with a given rank: i = 17 rank = 13 i = = 4 rank = 6 i = 4 rank = 2 i = 4 2 = 2 rank = 2 6

7 OS-Select Running Time Retrieving an element with a given rank: Running time: recurses from root to leaf does a constant amount of work at each level since it is a red-black tree, it has height O(lgn) total time is therefore O(lgn) 7

8 Calculating Rank of a Node Determining the rank of a given node: note that x could be any node in T OS-Rank returns the position of x in a linear order determined by an inorder traversal of T. Note how this removes the ambiguity of rank introduced by duplicates. 8

9 Calculating Rank of a Node How does it work? testing for a right ancestor go up one level The rank of a given node x consists of: 1. 1 for each node in it s left subtree 2. 1 for the node itself 3. 1 for each ancestor for which x is a right descendant 4. 1 for each node in a left subtree of an ancestor for which x is a right descendant 9

10 Calculating Rank of a Node Running time: iterates from leaf to root does a constant amount of work at each level since it is a red-black tree, it has height O(lgn) total time is therefore O(lgn) 10

11 Maintaining Subtree Sizes Red-black insert and delete operations must be modified to maintain the size values stored at each node. Recall that insert worked in two steps: 1. traverse from root to leaf using the BST algorithm 2. recoloring and rotating from leaf to root Step #1 can be modified to add 1 to the size of each node visited during the BST search, and the size of the new node is 1. Step #2 can be modified by adjusting the size of two nodes after rotation. 11

12 Maintaining Subtree Sizes Step #2 size adjustment: How do the size values of the nodes change during a right-rotation? If the pointers have already been adjusted: x.size = y.size y.size = y.left-size + y.right.size + 1 These can be added to the very end of the rotation procedure. 12

13 Maintaining Subtree Sizes Running time step #1: iterates from root to leaf does a constant amount of work at each level since it is a red-black tree, it has height O(lgn) time for this step is therefore O(lgn) Running time step #2: iterates from leaf to root does a constant amount of work at each level performs at most 2 rotations total, each doing a constant amount of work since it is a red-black tree, it has height O(lgn) time for this step is therefore O(lgn) Total time is therefore O(lgn) 13

14 Maintaining Subtree Sizes Deletion also consists of two steps, and the analysis is similar (but with 3 rotations, rather than 2). Total time is therefore O(lgn) 14

15 Interval Trees A closed interval is an ordered pair of real numbers [t 1, t 2 ], where t 1 t 2. The interval represents the set t R t 1 t t 2 Open and half-open omit both or one of the endpoints. The results and algorithms in this section ban be extended to work with open and half-open intervals. 15

16 Interval Trees Intervals can be used to represent many things. One set of applications uses them to represent events occurring over a continuous span of time. Example #1 - Car Rental: suppose a company rents cars record each interval of time during which a particular car is rented out (id# attached) store all the intervals in a database Many simple questions reduce to interval overlap or counting problems: which cars were rented out from noon to 3pm on Monday? how many cars were rented out from noon to 3pm on Monday? was there a period of time when no cars were rented out? 16

17 Interval Trees Example #2 Students sleeping in class: suppose we record each time interval during which a student nods off store all the intervals in a database (names attached) Questions: was any student sleeping between 2:15 and 2:30? was Mary sleeping between 2:15 and 2:30? how many students were sleeping between 2:15 and 2:30? how many times did Mary nod off in class? what is the largest interval of time during which a student was sleeping? who nodded off the largest number of times on Monday? 17

18 Interval Trees Each interval [t 1, t 2 ] is an object i with attributes i.low = t 1 and i.high = t 2 Intervals i and i are said to overlap if i i, in other words, if i.low i.high and i.low i.high Any two intervals i and i satisfy the interval trichotomy; that is exactly one of the following three properties holds: i and i overlap i is to the left of i (i.e., i.high < i.low) i is to the right of i (i.e., i.high < i.low) 18

19 Interval Trees An interval tree is a red-black tree where each node x in the tree represents an interval x.int Operations on interval trees: Interval-Insert (T, x) adds element x, which has interval x.int, to tree T. Interval-Delete (T, x) deletes element x from tree T. Interval-Search(T, i) returns a pointer to an element x in T such that x.int overlaps interval i, or a pointer to T.nil if no such element exists. 19

20 Interval Trees Each node x stores x.int and x.max, the maximum interval endpoint in the subtree rooted at x. z.int.left is used as the key during insert, where z is the new node. Thus, an in-order traversal outputs intervals in order of x.int.left 20

21 Interval Trees Red-black insert and delete operations must be modified to maintain the max values stored at each node. Recall that insert worked in two steps: 1. traverse from root to leaf using the BST algorithm 2. recoloring and rotating from leaf to root Step #1 is modified in two ways: (z is the new node) use z.int.low as the key during insertion update the max values of each node x as insert descends x.max = max(x.max, z.int.right) or Step #2 can be modified by adjusting z.max of two nodes after rotation. 21

22 Interval Trees Insert: (just step #1) [10, 12] [22, 25] [33, 35] [18, 26] [27,28] 28 [27,28] 28 22

23 Interval Trees Step #2 max adjustment: How do the max values of the nodes change during a right-rotation? If the pointers have already been adjusted: y.max = max(y.int.high, y.left.max, y.right.max) x.max = max(x.int.high, x.left.max, x.right.max) These can be added to the end of the rotation procedure. 23

24 Interval Trees Running time step #1: iterates from root to leaf does a constant amount of work at each level since it is a red-black tree, it has height O(lgn) time for step #1 is therefore O(lgn) Running time step #2: iterates from leaf to root does a constant amount of work at each level performs at most 2 rotations total, each doing a constant amount of work since it is a red-black tree, it has height O(lgn) Time for step #2 is therefore O(lgn) Total time is therefore O(lgn) 24

25 Interval Trees Deletion also consists of two steps, and the analysis is similar (but with 3 rotations, rather than 2). Total time is therefore O(lgn) 25

26 Interval Trees How does search work? Interval-Search(T, i) finds a node x in T such that x.int overlaps with i T is a tree, and i is a given interval note that the answer may not be unique returns T.nil if no overlapping interval exists in T 26

27 Interval Trees Search: [17, 23] [22, 25] [33, 35] [24, 25] [27,28] 28 [27,28] 28 27

28 Interval Trees Running time: iterates from root to leaf does a constant amount of work at each level since it is a red-black tree, it has height O(lgn) total time is therefore O(lgn) 29