Transform & Conquer Lecture 08 ITS033 – Programming & Algorithms Asst. Prof. Dr. Bunyarit Uyyanonvara IT Program, Image and Vision Computing Lab. School of Information and Computer Technology Sirindhorn International Institute of Technology Thammasat University http://www.siit.tu.ac.th/bunyarit bunyarit@siit.tu.ac.th 02 5013505 X 2005

ITS033 Midterm Topic 01 - Problems & Algorithmic Problem Solving Topic 02 – Algorithm Representation & Efficiency Analysis Topic 03 - State Space of a problem Topic 04 - Brute Force Algorithm Topic 05 - Divide and Conquer Topic 06 - Decrease and Conquer Topic 07 - Dynamics Programming Topic 08 - Transform and Conquer Topic 09 - Graph Algorithms Topic 10 - Minimum Spanning Tree Topic 11 - Shortest Path Problem Topic 12 - Coping with the Limitations of Algorithms Power http://www.siit.tu.ac.th/bunyarit/its033.php and http://www.vcharkarn.com/vlesson/showlesson.php?lessonid=7

This Week Overview Introduction Presort Binary Search Tree AVL Tree Problem reduction

Transform & Conquer: Introduction Lecture 08.1 ITS033 – Programming & Algorithms Transform & Conquer: Introduction Asst. Prof. Dr. Bunyarit Uyyanonvara IT Program, Image and Vision Computing Lab. School of Information and Computer Technology Sirindhorn International Institute of Technology Thammasat University http://www.siit.tu.ac.th/bunyarit bunyarit@siit.tu.ac.th 02 5013505 X 2005

Transform and Conquer This group of techniques solves a problem by a transformation to a simpler/more convenient instance of the same problem (instance simplification) to a different representation of the same instance (representation change) to a different problem for which an algorithm is already available (problem reduction)

Introduction Transform-and-conquer Methods work as two-stage procedures. (1) Transformation stage: the problem’s instance is modified to be, for one reason or another, more amenable to solution. (2) Conquering stage, solve it.

Introduction

Transform & Conquer: Instance Simplification Lecture 08.2 ITS033 – Programming & Algorithms Transform & Conquer: Instance Simplification Asst. Prof. Dr. Bunyarit Uyyanonvara IT Program, Image and Vision Computing Lab. School of Information and Computer Technology Sirindhorn International Institute of Technology Thammasat University http://www.siit.tu.ac.th/bunyarit bunyarit@siit.tu.ac.th 02 5013505 X 2005

Instance simplification - Presorting Solve a problem’s instance by transforming it into another simpler/easier instance of the same problem Presorting Many problems involving lists are easier when list is sorted. searching computing the median (selection problem) checking if all elements are distinct (element uniqueness) Also: Topological sorting helps solving some problems for dags. Presorting is used in many geometric algorithms.

Searching with presorting Problem: Search for a given K in A[0..n-1] Presorting-based algorithm: Stage 1 Sort the array by an efficient sorting algorithm Stage 2 Apply binary search Efficiency: O(nlog n) + O(log n) = O(nlog n) Good or bad? Why do we have our dictionaries, telephone directories, etc. sorted?

Element Uniqueness with presorting Presorting-based algorithm Stage 1: sort by efficient sorting algorithm (e.g. mergesort) Stage 2: scan array to check pairs of adjacent elements Efficiency: O(nlog n) + O(n) = O(nlog n) Brute force algorithm Compare all pairs of elements Efficiency: O(n2)

Example 1 Checking element uniqueness in an array ALGORITHM PresortElementUniqueness(A[0..n - 1]) //Solves the element uniqueness problem by sorting the array first //Input: An array A[0..n - 1] of orderable elements //Output: Returns “true” if A has no equal elements, “false” // otherwise Sort the array A for i  0 to n - 2 do if A[i]= A[i + 1] return false return true

Transform & Conquer: Representation Change Lecture 08.3 ITS033 – Programming & Algorithms Transform & Conquer: Representation Change Asst. Prof. Dr. Bunyarit Uyyanonvara IT Program, Image and Vision Computing Lab. School of Information and Computer Technology Sirindhorn International Institute of Technology Thammasat University http://www.siit.tu.ac.th/bunyarit bunyarit@siit.tu.ac.th 02 5013505 X 2005

Searching Problem file size (internal vs. external) Problem: Given a (multi)set S of keys and a search key K, find an occurrence of K in S, if any Searching must be considered in the context of: file size (internal vs. external) dynamics of data (static vs. dynamic) Dictionary operations (dynamic data): find (search) insert delete

Tree Characteristics §- trees hierarchical structures that place elements in nodes along branches that originate from a root. Nodes in a tree are subdivided into levels in which the topmost level holds the root node. Any node in a tree may have multiple successors at the next level. Hence a tree is a non-linear structure.

Tree Structures

Binary Tree: Formal Definition A binary tree T is a finite set of nodes with one of the following properties: a.) T is a tree if the set of nodes is empty. (An empty tree is a tree.) b.) The set consists of a root, R, and exactly two distinct binary trees, the left subtree, TL and the right subtree, TR. c.) The nodes in T consist of node R and all the nodes in TL and TR.

Binary Tree: example

Binary Tree

Binary Tree: Depth

Selected Samples of Binary Trees

Binary Search Trees Binary Search Tree (BSTs) is a binary tree which has following properties: Element to the left of the parent is always smaller than its parents. Element to the right of the parent is always greater than its parent.

Binary Search Tree Arrange keys in a binary tree with the binary search tree property: K <K >K Example: 5, 3, 1, 10, 12, 7, 9

Operations of BSTs: Insert Adds an element x to the tree so that the binary search tree property continues to hold The basic algorithm set the Index to Root Check if the data where index is pointing is NULL, if so Insert x in place of NULL, and finish the process. If the data is not NULL then compare the data with inserting item If the inserting item is equal or bigger then traverse to the right else traverse to the left. Continue with 2nd step again until

Insert Operations: 1st of 3 steps 1)- The function begins at the root node and compares item 32 with the root value 25. Since 32 > 25, we traverse the right subtree and look at node 35.

Insert Operations: 2nd of 3 steps 2)- Considering 35 to be the root of its own subtree, we compare item 32 with 35 and traverse the left subtree of 35.

Insert Operations: 3rd of 3 steps 3)- Create a leaf node with data value 32. Insert the new node as the left child of node 35. newNode = new Node; parent->left = newNode;

Insert in BST Insert Key 52. Start at root. 52 > 51 Go right. 51 14 72 06 33 53 97 13 25 43 64 84 99

Insert in BST Insert Key 52. 52 > 51 Go right. 51 14 72 06 33 53 97 13 25 43 64 84 99

Insert in BST Insert Key 52. 51 14 72 52 < 72 Go left. 06 33 53 97 13 25 43 64 84 99

Insert in BST Insert Key 52. 51 14 72 52 < 72 Go left. 06 33 53 97 13 25 43 64 84 99

Insert in BST Insert Key 52. 51 14 72 06 33 53 97 52 < 53 Go left. 13 25 43 64 84 99

Insert in BST Insert Key 52. 51 14 72 No more tree here. INSERT HERE 06 33 53 97 52 < 53 Go left. 13 25 43 52 64 84 99

Operations of BSTs: Search looks for an element x within the tree The basic algorithm Set the index to root Compare the searching item with the data at index If (the searching item is smaller) then traverse to the left else traverse to the right. Repeat the process untill the index is point to NULL or found the item.

Search in BST Search for Key 43. 51 14 72 06 33 53 97 64 25 43 13 99 84

Search in BST Search for Key 43. Start at root. 43 < 51 Go left. 51 14 72 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 43. 43 < 51 Go left. 51 14 72 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 43. 51 14 72 43 > 14 Go right. 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 43. 51 14 72 43 > 14 Go right. 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 43. 51 14 72 06 33 53 97 43 > 33 Go right. 13 25 43 64 84 99

Search in BST Search for Key 43. 51 14 72 06 33 53 97 43 > 33 Go right. 13 25 43 64 84 99

Search in BST Search for Key 43. 51 14 72 06 33 53 97 13 25 43 64 84 99 43 = 43 FOUND

Search in BST Search for Key 52. Start at root. 52 > 51 Go right. 14 72 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 52. 52 > 51 Go right. 51 14 72 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 52. 51 14 72 52 < 72 Go left. 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 52. 51 14 72 52 < 72 Go left. 06 33 53 97 13 25 43 64 84 99

Search in BST Search for Key 52. 51 14 72 06 33 53 97 52 < 53 Go left. 13 25 43 64 84 99

Search in BST Search for Key 52. 51 14 72 06 33 53 97 52 < 53 Go left. 13 25 43 64 84 99 No more tree here. NOT FOUND

Tree Walk Tree walk is a series of steps used to traverse a given tree. inorder tree walk – is a tree walk with following steps + left sub-tree + middle node + right sub-tree

Binary Tree: example

Tree Structures

In-order Tree Walk What does the following code do? TreeWalk(x) TreeWalk(left[x]); print(x); TreeWalk(right[x]); A: prints elements in sorted (increasing) order This is called an inorder tree walk Preorder tree walk: print root, then left, then right Postorder tree walk: print left, then right, then root

Transform & Conquer: problem reduction Lecture 08.4 ITS033 – Programming & Algorithms Transform & Conquer: problem reduction Asst. Prof. Dr. Bunyarit Uyyanonvara IT Program, Image and Vision Computing Lab. School of Information and Computer Technology Sirindhorn International Institute of Technology Thammasat University http://www.siit.tu.ac.th/bunyarit bunyarit@siit.tu.ac.th 02 5013505 X 2005

Problem Reduction This variation of transform-and-conquer solves a problem by a transforming it into different problem for which an algorithm is already available. To be of practical value, the combined time of the transformation and solving the other problem should be smaller than solving the problem as given by another method.

Problem Reduction Problem reduction: If you need to solve a problem, reduce it to another problem that you know how to solve

Reduction to Graph Problems Vertices of a graph typically represent possible states of the problem in question while edges indicate permitted transitions among such states. One of the graph’s vertices represents an initial state, while another represents a goal state of the problem. Such a graph is called a state-space graph. Thus, the transformation just described reduces the problem to the question about a path from the initial-state vertex to a goal-state vertex.

Example A peasant finds himself on a river bank with a wolf, a goat, and a head of cabbage. He needs to transport all three to the other side of the river in his boat. However, the boat has room only for the peasant himself and one other item (either the wolf, the goat, or the cabbage). In his absence, the wolf would eat the goat, and the goat would eat the cabbage. Find a way for the peasant to solve his problem or prove that it has no solution.

Example P, w, g, c stand for the peasant, the wolf, the goat, and the cabbage, respectively; the two bars | | denote the river; there exist two distinct simple paths from the initial state vertex to the final state vertex. Hence, this puzzle has two solutions, each of which requires seven river crossings.

Homework S = {1, 5, 3, 4, 8, 6, 7, 8, 4, 2, 1, 9, 11, 12,15, 14} 1. Construct a Binary Search Tree from the given set S 2. Construct a AVL Tree from the given set S

ITS033 Midterm Topic 01 - Problems & Algorithmic Problem Solving Topic 02 – Algorithm Representation & Efficiency Analysis Topic 03 - State Space of a problem Topic 04 - Brute Force Algorithm Topic 05 - Divide and Conquer Topic 06 - Decrease and Conquer Topic 07 - Dynamics Programming Topic 08 - Transform and Conquer Topic 09 - Graph Algorithms Topic 10 - Minimum Spanning Tree Topic 11 - Shortest Path Problem Topic 12 - Coping with the Limitations of Algorithms Power http://www.siit.tu.ac.th/bunyarit/its033.php and http://www.vcharkarn.com/vlesson/showlesson.php?lessonid=7

End of Chapter 8 Thank you!