1. Chapter 6 Relations
Relationships are dealed with every day: businesses and their telephone numbers, employees and their salaries, ….
Relationships can be used to solve problems such as: determining which pairs of cities are linked by airline flights in a network, finding a viable order for the different phases of a complicated project.
6.1 Relations and Their Properties
From Definition 1, the binary relation R is a subset of

2. R={(Boulder,Colorado),(Bangor,Maine),(Ann Arbor,Michigan),
Example 1 Let A be the set of students, and let B the set of courses. Define R to be the relation that consists of those pairs (a,b), where a is a student enrolled in course b.
Example 2 Let A be a set of all cities, and let B be the set of the 50 states in the United States. Define the relation R by specifying that (a,b) belongs to R if city a is in state b.
Solution:
R={(Boulder,Colorado),(Bangor,Maine),(Ann Arbor,Michigan),
(Cupertino,California), (Red Bank,New Jersey),…}
Example 3 Let A={0,1,2} and B={a,b}. R={(0,a),(0,b),(1,a),(2,b)} is a relation from A to B. 1 2 a b

3. Functions as Relations
A function f from a set A to a set B assigns a unique element of B to each element of A.
But a relation does not have to be a function.
Example Let A={0,1,2} and B={a,b}. Let R={(0,a),(0,b),(1,a),(2,b)} and R’={(0,a),(1,a),(2,b)} be two relations from A to B. 1 2 a b R 1 2 a b R’
R’ is a function, but R is not a function.

4. Relations on A Set
Definition 2. A relation R on the set A is a relation from A to A.
Example 4. Let A be the set {1,2,3,4}. Which ordered pairs are in the relation R={(a,b) | a divides b}?
Solution R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,4),(3,3),(4,4)}.
Example 5. Consider the following relations on the set of integers.

5. Properties of Relations
Example 6. Consider the following relations on {1,2,3,4}

6. Example 7. Consider the following relations on the set of integers.
Example 8. Is the “divides” relation on the set of a positive integers reflexive?
Solution: It is reflexive, since a|a whenever a is a positive integer.

7. Example 9. Consider the following relations on {1,2,3,4}

8. Example 10. Consider the following relations on the set of integers.
Example 11. Is the “divides” relation on the set of a positive integers symmetric? Is it antisymmetric?
Solution: It is not symmetric, since 1|2, but 2|1. It is antisymmetric, since for any positive integers a and b with a|b and b|a, then a=b.

9. Example 12. Consider the following relations on {1,2,3,4}

10. Example 13. Consider the following relations on the set of integers.
Example 14. Is the “divides” relation on the set of a positive integers transitive?
Solution: It is transitive, since a|b and b|c, then a|c, for any positive integers a,b,c.

11. 6.5 Equivalence Relations
Definition 1. A relation on a set A is called an equivalence relation if it is reflexive, symmetric and transitive.
Two elements are related by an equivalence relation are called equivalent.
Solution: R is reflexive, symmetric, and transitive, it is an equivalence relation.
Example 2 Let R be the relation on the set of real numbers such that aRb if and only if a=b or a=-b. In Section 6.1, we showed that R is reflexive, symmetric, and transitive. It follows that R is an equivalence relation.

12. Note: if two numbers b and c have the property that their difference b-c is integrally divisible by a number m (i.e., (b-c)/m  is an integer), then b and c are said to be "congruent modulo m ." The number m is called the modulus, and the statement “b is congruent to c (modulo m)" is written mathematically as
bc(mod m)
If b-c is not integrally divisible by m, then it is said that “b is not congruent to c (modulo m)," which is written
b c (mod m)

13. Example 3. Let R be the relation on the set of real numbers such that aRb if and only if a-b is an integer. Is R an equivalence relation?
Solution: R is reflexive, symmetric and transitive. It is an equivalence relation.

14. 6.6 Partial Orderings
Definition 1 A relation R on a set S is called a partial ordering or partial order if it is reflexive, antisymmetric, and transitive. A set S together with a partial ordering R is called a partially ordered set, and denoted by (S,R). A partial order R is also denoted as .
Example 1
Show that the “greater than or equal” relation is a partial ordering one the set of integers.
Example 2
Show that the divisibility relation (|) is a partial ordering on the set of positive integers.
Example 3
Show that the inclusion relation ( ) is a partial ordering on the power set of a set S.

15. Definition 2 The elements a and b of a partial ordering set (S, ) is called comparable if either a b or b a. When a and b are elements of S such that neither a b nor b a, a and b are called incomparable.
Example 4
Definition 3 If (S, ) is a partial ordering set and every two elements of S are comparable, S is called a totally ordered or linearly ordered set, and is called a total order or a linear order.
Example 5
Example 6

16. Chapter 7 Graphs
Definition 1. (1) A simple undirected graph G=(V,E) consists of V, a nonempty set of vertices, and E, a set of unordered pairs of distinct elements of V called edges. (2) If more than one edges are allowed between two vertices, G is called multigraph. (3) If there is any edge connection the same vertex, G is called pseudograph.
Detroit
San Francisco
Los Angeles
Denver
Chicago
New York
Washington
(1) Simple undirected graph
V={San Francisco, Los Angles, Denver, Chicago, Detroit, Washington, New York}
E={(San Francisco, Los Angeles), (San Francisco,Denver), (Los Angeles,Denver), (Denver,Chicago), (Chicago,Detroit), (Chicago,New York), (Chicago,Washington), (Detroit,New York), (Washington,New York)}
Detroit
(2) multigraph
San Francisco
Los Angeles
Denver
Chicago
New York
Washington
Detroit
San Francisco
Los Angeles
Denver
Chicago
New York
Washington
(3) pseudograph

17. Example 1 Precedence Graphs and Concurrent Processing
Definition 2. A directed graph G=(V,E) consists of V, a nonempty set of vertices, and E, a set of ordered pairs of distinct elements of V called edges.
New York
San Francisco
Los Angeles
Denver
Chicago
Detroit
Washington
Example 1 Precedence Graphs and Concurrent Processing
A a:=0
B b:=1
C c:=a+1
D d:=b+a
E e:=d+1
F e:=c+d A F C B D E

18. Definition 3.
Let e=(u,v)be an edge of Graph G (undirected or directed). (i)Vertices u and v are called adjacent in G.
(ii)Edge e is called incident with (or connect) the vertices u and v.
The degree of a vertex v in an undirected graph is the number of edges incident with it, denoted by deg(v).
Let e=(u,v) be an edge of a directed graph G. The vertex u is called the initial vertex of (u,v), and v is called the terminal vertex of (u,v).
4. Let v be a vertex of directed graph G. The in-degree of v is the number of edges with v as their terminal vertex. The out-degree of v is the number of edges with b as their initial vertex.

19. Connectivity
Definition 4. Given a graph G=(V,E), a path of G is a sequence of vertices A F C B D E H I A F C B D E H I

20. Definition 5. An undirected graph is called connected if there is a path between every pair of distinct vertices of the graph. A F C B D E H I
Connected graph A F C B D E H I
Unconnected graph
Definition 6. A directed graph is strongly connected if there is a path from a to b whenever a and b are the vertices in the graph. A F C D E
Strongly connected A F C D E
Not Strongly connected

21. Chapter 8 Trees 8.1 Introduction to tree
Definition 1. A tree is a connected undirected graph with no simple circuits.
Example 1
Which of the graphs are trees?
Theorem 1 Au undirected graph is a tree if and only if there is a unique simple path between any two of its vertices.

22. Ordered rooted tree
root
parent
subtree
third child
Binary tree (every node has at most two children)
Left child
Right child
Right subtree
Left subtree

23. 8.3 Tree Traversal
Ordered rooted trees are used to store information. We need procedures for visiting each vertex of and ordered rooted tree to access data. 6 18 30 4 7 13 21 20 17 9 15
Root of T 19 1 5 11
The preorder traversal of T 15,6,9,30,21,4,7,13,11,18,17,1,19,20,19,19,4,5

24. The inorder traversal of T 6 18 30 4 7 13 21 20 17 9 15
Root of T 19 1 5 11
The inorder traversal of T
9,6,21,30,4,11,13,7,15,17,18,1, 20,19,19,19,4,5

25. The postorder traversal of T 6 18 30 4 7 13 21 20 17 9 15
Root of T 19 1 5 11
The postorder traversal of T
9,21,4,30,11,13,7,6,17,1,18,20,19,4,5,19,19,15

26. 8.2 Application of Trees Binary Search Tree
Definition 1 Binary Search Tree is a Binary Tree satisfying the following condition:
Each vertex contains an item with an key which belongs to a total ordering set and two links to its left child and right child, respectively.
In each node, its key is larger than the keys of all vertices in its left subtree and smaller than the keys of all the vertices in its right subtree. 6 18 3 4 7 13 1 20 17 9 15
root
key
Operations in a BST
Search (given a key return the corresponding data if it is in the BST)
Insert (given a data insert it into the BST)
Delete (given a key delete the corresponding data from BST) 26

27. Define a node of BST class Node { item Item; Node Left; Node Right; }
(key/other data)
Right left
A node of BST: item, right child, left child
class Node {
item Item;
Node Left;
Node Right; }
struct item {
int id;
string name;
float grade;
key
65/data
30/data
75/data
25/data /
45/data
40/data
35/data
55/data
42/data
/ /
68/data
80/data
78/data
/ /
88/data
60/data
38/data
20/data
39/data
/ /
36/data
root
A BST 27

28. Visit the nodes of BST in-order
65/data
30/data
75/data
25/data /
45/data
40/data
35/data
55/data
42/data
/ /
68/data
80/data
78/data
/ /
88/data
60/data
38/data
20/data
39/data
/ /
36/data
root
A BST
Visit the nodes of BST in-order
Procedure InOrder(Node root) {
if (root ≠ null))
InOrder(root.Left);
visit root;
InOrder(root.Right); }
In-order visit in the BST (only show keys): 20,25,30,35,36,38,39,40,42,45,55,60,65,68,75,78,80,88
A BST can be visited in three ways: in-order, pre-order and post-order 28

29. Operation 1: Search Operation 1: Search root Example: Search (42)
65/data
30/data
75/data
25/data /
45/data
40/data
35/data
55/data
42/data
/ /
68/data
80/data
78/data
/ /
88/data
60/data
38/data
20/data
39/data
/ /
36/data
root
A BST
Procedure Search(int id) {
Node x = root;
while (x ≠ null and x.Item.id ≠ key )
if (id < x.Item.id) x = x.Left;
else x = x.Right; }
return x; /* x.id = key or x = null */
height
Time Complexity
O(height of BST)
Average case: O(lg n)
Worst case O(n)
where n is the size of the BST
Example:
Search (42)
Search (64) 29

30. Create a new node for new item (e.g., 62/data)
Operation 2: Insert
Procedure Insert( item newItem) {
Node newNode: = new Node(newItem);
Node parent = null;
Node current = root;
while (current ≠ null)
parent: = current;
if (newNode.Item.id < current.Item.id)
current: = current.Left;
else current: = current.Right; }
if (parent = null) root: = newNode;
else
if (newNode.Item.id < parent.Item.id)
parent.Left: = newNode;
else parent.Right: = newNode;
root
65/data
30/data
75/data
25/data /
45/data
40/data
35/data
55/data
42/data
/ /
68/data
80/data
78/data
/ /
88/data
60/data
38/data
20/data
39/data
/ /
36/data
A BST
current
parent
curr
height
paren
curr
paren
curr
paren
curr
paren
curr = null
62/data
/ /
Time Complexity
O(height of BST)
Average case: O(lg n)
Worst case: O(n)
where n is the size of the BST
Insert
Create a new node for new item (e.g., 62/data)
Find the place for insertion
Link the new node 30

31. Find the deleting node (current) and its parent Delete current
Operation 3: Delete
65/data
30/data
75/data
25/data /
45/data
40/data
35/data
55/data
42/data
/ /
68/data
80/data
78/data
/ /
88/data
60/data
38/data
20/data
39/data
/ /
36/data
root
Case 1: Deleting node (current) has no child
A BST
Example: Delete the node with item (36/data)
height
if (current.Left = null and current.Right: = null) {
if (current = root) root: = null;
else
if (parent.left = current) parent.Left: = null;
else parent.Right = null; }
Time Complexity
Find the deleting node (current) and its parent: O(height of BST)
Delete current: O(1)
Therefore, the time complexity for case 1 is O(height of BST)
Average case: O(lg n)
Worst case: O(n)
where n is the size of the BST
parent
current
Delete
Find the deleting node (current) and its parent
Delete current
The first step is similar to Search. We only
consider the step of delete current. 31

32. Example: delete the node with item (35/data) root
Case 2: Deleting node (current) has one child
Operation 3: Delete
Example: delete the node with item (35/data)
65/data
30/data
75/data
25/data /
45/data
40/data
35/data
55/data
42/data
/ /
68/data
80/data
78/data
/ /
88/data
60/data
38/data
20/data
39/data
/ /
36/data
root
ABST
if (current.Left = null or current.Right = null) {
if (current.Right = null)
if (current = root) root: = current.Left;
else if (parent.Left = current)
parent.Left: = current.Left;
else
parent.Right: = current.Left;
else if (current.Left = null)
if (current = root) root: = current.Right;
parent.Left: = current.Right;
parent.Right: = current.Right; }
height
parent
current
Time Complexity
Find the deleting node (current) and its parent: O(height of BST)
Delete current: O(1)
Therefore, the time complexity for case 1 is O(height of BST)
Average case: O(lg n)
Worst case: O(n)
where n is the size of the BST
Delete
Find the deleting node (current) and its parent
Delete current
The first step is similar to Search. We only
consider the step of delete current. 32

33. Operation 3: Delete root
Case 3: Deleting node (current) has two children
root
ABST
65/data
height
Example: delete the node with item (30/data)
current
30/data
75/data
Step 1 and 2 are similar to Search. We only consider step 3 and 4.
25/data /
45/data
68/data
/ /
80/data
if (!(successor == current.Right))
sucessorParent.Right = successor.Right;
current.item = successor.item
Succ Parent
20/data
/ /
40/data
55/data /
78/data
/ /
88/data
/ /
successor
Time Complexity
Find the deleting node (current) and its parent: O(height of BST)
Find successor and successor’s parent: O(height of BST)
Link successor’s parent to successor.Right: O(1)
Replace the current.item by successor.item: O(1)
Therefore, the time complexity for case 1 is O(height of BST)
Average case: O(lg n)
Worst case: O(n)
where n is the size of the BST
35/data
35/data /
42/data
/ /
60/data
/ /
38/data
36/data
/ /
39/data
/ /
Delete
Find the deleting node (current) and its parent
Find successor which is the node with the smallest key in current’s left, and successor’s parent.
Link successorParent to successor.Right
Replace the current.item by successor.item 33

34. Example 1
Form a binary search tree for the words mathematics, physics, geography, zoology, meteorology, geology, psychology, and chemistry (using alphabetical order).
Example 2
Find the item with key Zoology in the tree of Example 1.
Example 3
Insert the item with key English in the tree of Example 2.
Example 4
Delete the item with key English in the tree of Example 3.