Sorted Tables. Ordered Dictionary. Binary Search. Binary Search

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1 Ordered Dictionary Sorted Tables Dictionary is a set of elements, the keys, supported by the dictionary operations, such as: findelement(k) -- k position, e element insertitem(k,e) removeelement(k) Ordered dictionary maintains an order relation for the elements in it. week 3 Complexity of Algorithms 1 If a dictionary D is ordered, we can store its items in a vector S by nondecreasing order of the keys. This allows for faster searching than would be possible had S been, e.g., a linked list. We refer to this ordered vector implementation of a dictionary D as a lookup table. week 3 Complexity of Algorithms 2 Binary Search Binary Search Accessing an element of S (in arraybased representation of size n) by its rank takes O(1) time The item at rank i has a key no smaller than keys of the items of ranks 0,,i-1, and no larger than keys of the items of ranks i+1,,n-1. Searching is done by decreasing the range of the elements in S week 3 Complexity of Algorithms 3 Looking for k in S, the range in S is defined as a pair of ranks: low and high, s.t., all elements in S with ranks < low (> high) are smaller (larger) then k Initially, low = 0 and high = n-1 key(i) denotes the key at rank i, and elem(i) denotes an element in key(i) week 3 Complexity of Algorithms 4 1

2 Binary Search Binary Search - Algorithm In order to decrease a size of the range we compare k to the key of the median mid of the range, i.e., mid = (low+high)/2 3 cases are possible k = key(mid), the search is completed successfully k < key(mid), search continued with high = mid-1 k > key(mid), search continued with low = mid+1 week 3 Complexity of Algorithms 5 week 3 Complexity of Algorithms 6 Binary Search - Illustration Binary Search - Complexity Let function f(n) represent the running time of the binary search method We can characterize the running time of the recursive binary search algorithm as follows: Binary search runs in time O(log n) week 3 Complexity of Algorithms 7 week 3 Complexity of Algorithms 8 2

3 Binary Search - Complexity Binary Search Tree (BST) Binary search vs. linear file Binary Search Tree (BST) applies the motivation of the binary search procedure to a tree-based data structure. In BST each internal node v stores an element e, s.t., the elements stored in the left subtree of v are less than or equal to e, and the elements stored in the right subtree of v are greater than or equal to e. week 3 Complexity of Algorithms 9 week 3 Complexity of Algorithms 10 BST Searching in BST An inorder traversal of BST visits the elements stored in such a tree in nondecreasing order. week 3 Complexity of Algorithms 11 week 3 Complexity of Algorithms 12 3

4 Searching in BST - Analysis Insertion in BST Insertion of 78 implemented in time O(h) week 3 Complexity of Algorithms 13 week 3 Complexity of Algorithms 14 Removal in BST Removal in BST Removal of a node (32) with one external child is done in time O(h) week 3 Complexity of Algorithms 15 Removal of a node (65) with two internal children is done in time O(h) week 3 Complexity of Algorithms 16 4

5 Inefficiency of general BSTs AVL Trees All operations in BST are performed in time O(h), where h is the height of BST Unfortunately h might be as large as n, e.g., after n consecutive insertions of elements with keys in increasing order The advantages of the binary search (O(log n) time update) might be lost if BST is not balanced Height-Balance Property: for every internal node v of T, the heights of the children of v can differ by at most 1 week 3 Complexity of Algorithms 17 week 3 Complexity of Algorithms 18 AVL Trees Insertion in AVL Theorem: the height of an AVL tree T storing n items is O(log n) Consequence 1: the search in AVL can be performed in time O(log n) Consequence 2: The insertion and the removal in AVL need more careful implementation (rotations) An insertion in an AVL tree begins as insertion in a general BST, i.e., we attach a new external node (leaf) to BST This action may violate the height-balance property, i.e., for some nodes the action may increase their heights by one The bottom-up mechanism (based on rotations) is applied to fix the unbalance of AVL subtrees week 3 Complexity of Algorithms 19 week 3 Complexity of Algorithms 20 5

6 Fixing AVL after Insertion Single rotations in AVL Fixing of an AVL after insertion of 54 week 3 Complexity of Algorithms 21 week 3 Complexity of Algorithms 22 Double rotations in AVL Removal in AVL An removal in an AVL tree begins as removal in a general BST This action may violate the height-balance property, i.e., for some nodes the action may decrease their heights by one The bottom-up mechanism (based on rotations) is applied to fix the unbalance of AVL subtrees week 3 Complexity of Algorithms 23 week 3 Complexity of Algorithms 24 6

7 Fixing AVL after Removal AVL Performance All operations (search, insertion and removal) can be implemented in AVL in O(log n) time Fixing of an AVL after removal of 32 week 3 Complexity of Algorithms 25 week 3 Complexity of Algorithms 26 (2,4) Trees (2,4) Trees Every node in (2,4) tree has at least 2 and at most 4 children All external nodes (leaves) in (2,4) tree have the same depth Theorem: The height of a (2,4) tree storing n items is Q(log n) Each internal node v in (2,4) tree contains 1, 2 or 3 keys that define the ranges of keys stored in 2, 3 or 4 (respectively) consecutive subtrees of v week 3 Complexity of Algorithms 27 week 3 Complexity of Algorithms 28 7

8 (2,4) Trees - Search (2,4) Tree - Insertion Search for an key k in (2,4) tree T is done via tracing the path in T starting at the root in a top-down manner Visiting a node v we compare k with keys (k i ) stored at v: If k = k i the search is completed If k i k k i+1, the i+1 th subtree of v is searched recursively An insertion of k in an (2,4) tree T begins with a search for an internal node on the lowest level that could accommodate k without violation of the range rule This action may overflow the node-size of a node v acquiring new key k, i.e., the number of keys in v can grow up to 4 The bottom-up mechanism (based on split operation) is applied to fix the overflown nodes of (2,4) tree T week 3 Complexity of Algorithms 29 week 3 Complexity of Algorithms 30 (2,4) Trees Split Operation Sequence of Insertions - 1 Example of a split operation week 3 Complexity of Algorithms 31 week 3 Complexity of Algorithms 32 8

9 Sequence of Insertions - 2 (2,4) Tree - Removal Removal of key k from (2,4) tree T begins with search for a node v possessing key k i = k Key k i is replaced by the largest key in the i th consecutive subtree of v The bottom-up mechanism (based on transfer and fusion operation) is applied to fix the underflown nodes of (2,4) tree week 3 Complexity of Algorithms 33 week 3 Complexity of Algorithms 34 Sequence of Removals - 1 Sequence of Removals - 2 week 3 Complexity of Algorithms 35 week 3 Complexity of Algorithms 36 9

10 Sequence of Removals - 3 (2,4) Tree Performance The height of a (2,4) tree storing n elements is O(log n) A split, transfer and fusion operations take O(1) time A search, insertion, and removal of an element in a tree visits O(log n) nodes week 3 Complexity of Algorithms 37 week 3 Complexity of Algorithms 38 10

Learning Outcomes. COMP202 Complexity of Algorithms. Binary Search Trees and Other Search Trees

Learning Outcomes. COMP202 Complexity of Algorithms. Binary Search Trees and Other Search Trees Learning Outcomes COMP202 Complexity of Algorithms Binary Search Trees and Other Search Trees [See relevant sections in chapters 2 and 3 in Goodrich and Tamassia.] At the conclusion of this set of lecture