1. 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
X 2005

2. 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
and

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

4. 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
X 2005

5. 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)

6. 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.

7. Introduction

8. 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
X 2005

9. 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.

10. 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?

11. 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)

12. 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

13. 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
X 2005

14. 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

15. 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.

16. Tree Structures

17. 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.

18. Binary Tree: example

19. Binary Tree

20. Binary Tree: Depth

21. Selected Samples of Binary Trees

22. 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.

23. 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

24. 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

25. 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.

26. 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.

27. 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;

28. Insert in BST Insert Key 52. Start at root. 52 > 51 Go right. 51 1472 06 33 53 97 13 25 43 64 84 99

29. Insert in BST Insert Key 52. 52 > 51 Go right. 51 14 72 06 33 53 9713 25 43 64 84 99

30. Insert in BST Insert Key 52. 51 14 72 52 < 72 Go left. 06 33 53 9713 25 43 64 84 99

31. Insert in BST Insert Key 52. 51 14 72 52 < 72 Go left. 06 33 53 9713 25 43 64 84 99

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

33. Insert in BST Insert Key 52. 51 14 72 No more tree here. INSERT HERE06 33 53 97
52 < 53
Go left. 13 25 43 52 64 84 99

34. 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.

35. Search in BST Search for Key 43. 51 14 72 06 33 53 97 64 25 43 13 9984

36. Search in BST Search for Key 43. Start at root. 43 < 51 Go left. 5114 72 06 33 53 97 13 25 43 64 84 99

37. Search in BST Search for Key 43. 43 < 51 Go left. 51 14 72 06 33 5397 13 25 43 64 84 99

38. Search in BST Search for Key 43. 51 14 72 43 > 14 Go right. 06 3353 97 13 25 43 64 84 99

39. Search in BST Search for Key 43. 51 14 72 43 > 14 Go right. 06 3353 97 13 25 43 64 84 99

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

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

42. Search in BST Search for Key 43. 51 14 72 06 33 53 97 13 25 43 64 8499
43 = 43
FOUND

43. 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

44. Search in BST Search for Key 52. 52 > 51 Go right. 51 14 72 06 3353 97 13 25 43 64 84 99

45. Search in BST Search for Key 52. 51 14 72 52 < 72 Go left. 06 33 5397 13 25 43 64 84 99

46. Search in BST Search for Key 52. 51 14 72 52 < 72 Go left. 06 33 5397 13 25 43 64 84 99

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

48. 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

49. 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

50. Binary Tree: example

51. Tree Structures

52. 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

53. 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
X 2005

54. 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.

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

56. 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.

57. 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.

58. 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.

59. 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

60. 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
and

61. End of Chapter 8
Thank you!