chapter 6 relations relationships are dealed with every day: businesses and their telephone numbers,...
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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
. ofsubset a is
to from relation binary A sets. be and Let 1. Definition
BA
BARBA
• From Definition 1, the binary relation R is a subset of},|),{( BbAaba
. denote to and , denote to use weSometimes,
.by torelated be tosaid is ,When
R(a,b)aRbR(a,b)aRb
RbaR(a,b)
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.
} coursein enrolled student ,,A |),{( is,That baBbabaR
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
0
2
a
b
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.
. as denoted isIt . to unique aassign ,each for is,That bf(a)aBbfAa
}.)(,,|),{( :relation a as considered becan bafBbAabaff
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
0
2
a
b
R
1
0
2
a
b
R’
R’ is a function, but R is not a function.
Relations on A Set
Definition 2. A relation R on the set A is a relation from A to A. . ofsubset a is is,That A}A, b{(a,b) |aR
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.
(2,2)? and (1,-1)(2,1),(1,2),(1,1), ofeach contain relations theseofWhich
}3|),{(
}1|),{(
}|),{(
}or |),{(
}|),{(
}|),{(
6
5
4
3
2
1
babaR
babaR
babaR
bababaR
babaR
babaR
.)2,2(;)1,1(
;(2,1) ;(1,2);(1,1) :Solution
431632
652616431
,R,RR,R,RR
,R,RR,RR ,R,R,RR
Properties of Relations
.element every for
if reflexive called is set aon relation A 3. Definition
Aa
R(a,a)AR
Example 6. Consider the following relations on {1,2,3,4}
?reflextive are relations theseofWhich
)}4,3{(
)}4,4(),4,3(),3,3(),4,2(),3,2)(2,2(),4,1(),3,1(),2,1(),1,1{(
)}3,4(),2,4(),1,4(),2,3(),1,3(),1,2{(
)}4,4)(1,4(),3,3(),2,2(),1,2(),4,1(),2,1(),1,1{(
)}1,2(),2,1(),1,1{(
)}4,4(),1,4(),4,3(),2,2(),1,2(),2,1(),1,1{(
6
5
4
3
2
1
R
R
R
R
R
R
reflexive.not are Others e.reflexitiv are and :Solution 53 RR
Example 7. Consider the following relations on the set of integers.
reflexive? are relations theseofWhich
}3|),{(
}1|),{(
}|),{(
}or |),{(
}|),{(
}|),{(
6
5
4
3
2
1
babaR
babaR
babaR
bababaR
babaR
babaR
. and ,integer andfor since
,reflextive are and , :Solution
431
431
RRRa, (a,a)
RRR
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.
. called is ,any for
, ifonly )( and )(such that set aon relation A
.any for
,)( whenever)( if called is set aon relation A
4. Definition
ricantisymmetAa,b
baRb,aRa,bAR
Aa,b
Ra,bRb,asymmetricAR
Example 9. Consider the following relations on {1,2,3,4}
ric?antisymmet are which and symmetric are relations theseofWhich
)}4,3{(
)}4,4(),4,3(),3,3(),4,2(),3,2)(2,2(),4,1(),3,1(),2,1(),1,1{(
)}3,4(),2,4(),1,4(),2,3(),1,3(),1,2{(
)}4,4)(1,4(),3,3(),2,2(),1,2(),4,1(),2,1(),1,1{(
)}1,2(),2,1(),1,1{(
)}4,4(),1,4(),4,3(),2,2(),1,2(),2,1(),1,1{(
6
5
4
3
2
1
R
R
R
R
R
R
ric.antisymmet are and R, symmetric, are and :Solution 65432 RRRR
Example 10. Consider the following relations on the set of integers.
ric?antisymmet are which and symmetric are relations theseofWhich
}3|),{(
}1|),{(
}|),{(
}or |),{(
}|),{(
}|),{(
6
5
4
3
2
1
babaR
babaR
babaR
bababaR
babaR
babaR
ric.antisymmet are
and,, and symmetric, are and , :Solution 5421643 R RRRRRR
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.
.any for ,)( then ,)( and )( whenever if
e transitivcalled is set aon relation A 5. Definition
Aa,b,cRa,cRb,cRa,b
AR
Example 12. Consider the following relations on {1,2,3,4}
e? transitivare relations theseofWhich
)}4,3{(
)}4,4(),4,3(),3,3(),4,2(),3,2)(2,2(),4,1(),3,1(),2,1(),1,1{(
)}3,4(),2,4(),1,4(),2,3(),1,3(),1,2{(
)}4,4)(1,4(),3,3(),2,2(),1,2(),4,1(),2,1(),1,1{(
)}1,2(),2,1(),1,1{(
)}4,4(),1,4(),4,3(),2,2(),1,2(),2,1(),1,1{(
6
5
4
3
2
1
R
R
R
R
R
R
not. are others The e. transitivare and , :Solution 654 RRR
Example 13. Consider the following relations on the set of integers.
e? transitivare relations theseofWhich
}3|),{(
}1|),{(
}|),{(
}or |),{(
}|),{(
}|),{(
6
5
4
3
2
1
babaR
babaR
babaR
bababaR
babaR
babaR
.not are others and e, transitivare and ,, :Solution 4321 RRRR
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.
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.
relation? eequivalencan Is
}. and | ){(relation aLet . string theof
length thebe and lettersEnglish of strings ofset a be Let 1 Example
R
l(b) l(a)Aa,ba,bRx
l(x)A
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.
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.
integers. ofset on therelation eequivalencan is
)}mod({
relation that theShow 1.an greater thinteger positive a beLet
Modulo Congruence 4. Example
mb(a,b) | aR
m
m
).(mod then ,)(mod and )(mod if since e, transitivisIt
).(mod then )(mod if since symmetric, isIt
).mod(since reflexive, isIt
:Solution
mcamcbmba
mabmba
ma a
6.6 Partial Orderings
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.
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 .
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
?comparable 7 and 5 Are
?comparable 9 and 3 integers theare |),,(Zset ordering partial In the
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
ordered. totally is ) (Z,set ordering partial The
Example 6
ordered.y not totall is |),(Zset ordering partial The
6.7 n-ary Relations and Database
Definition 1
Let be the sets. An n-ary relation on these sets is a subset of
are called as the domains of the relation and n is called as its degree.
nAAA ,...,, 21
.21 nAAA nAAA ,...,, 21
A database consists of records, which are n-tuples, made up of fields.
Example 1
A student record are represented by 4-tuples of the form (student name, ID, Major, GPA). A sample database of six such record is:
(Ackermann, 231455, Computer Sci, 3.88)
(Adams, 888323, physics, 3.45)
(Chou, 102147, Computer Sci, 3.49)
(Goodfriend, 453876, Mathematics, 3.45)
(Rao, 678543, Mathematics, 3.90)
(Stevens, 786576, Psychology, 2.99)
Student_name ID Major GPA
Ackermann
Adams
Chou
Goodfriend
Rao
Stevens
231455
888323
102147
453876
678543
786576
Computer Sci
Physics
Computer Sci
Mathematics
Mathematics
Psychology
3.88
3.45
3.49
3.45
3.90
2.99
Table 1 Students
Primary key (no same value in the domain)
Student_name ID Major GPA
Ackermann
Adams
Chou
Goodfriend
Rao
Stevens
231455
888323
102147
453876
678543
786576
Computer Sci
Physics
Computer Sci
Mathematics
Mathematics
Psychology
3.88
3.45
3.49
3.45
3.90
2.99
Student_name ID Major GPA
Ackermann
Rao
231455
678543
Computer Sci
Mathematics
3.88
3.90
Operations on n-ary relations
Definition 2
Let R be an n-ary relation and C a condition that elements in R may satisfy. Then the selection operator maps the n-ary relation R to the n-ary relation of all n-tuples from R that satisfy the condition C.
CS (C: GPA>3.5)
CS
Definition 3
The projection maps the n-tuple to the m-tuple
, where miiiP ,...,, 21
),...,,( 21 naaa
),...,,(21 miii aaa .nm
Student_name ID Major GPA
Ackermann
Adams
Chou
Goodfriend
Rao
Stevens
231455
888323
102147
453876
678543
786576
Computer Sci
Physics
Computer Sci
Mathematics
Mathematics
Psychology
3.88
3.45
3.49
3.45
3.90
2.99
Student_name ID GPA
Ackermann
Adams
Chou
Goodfriend
Rao
Stevens
231455
888323
102147
453876
678543
786576
3.88
3.45
3.49
3.45
3.90
2.99
4,2,1P
Definition 4
Let R be a relation of degree m and S a relation of degree n. The join where is a relation of degree m+n-p that consist of all (m+n-p)-tuples , where belongs to R and belongs to S.
),,( SRJ p
, and npmp ),...,,,,...,,,,...,,( 212121 pnppm bbbcccaaa ),...,,,,...,,( 2121 ppm cccaaa
),...,,,,...,,( 2121 pnp bbbccc
Student_name ID Major GPA
Ackermann
Adams
Chou
Goodfriend
Rao
Stevens
231455
888323
102147
453876
678543
786576
Computer Sci
Physics
Computer Sci
Mathematics
Mathematics
Psychology
3.88
3.45
3.49
3.45
3.90
2.99
3.88
3.45
3.49
3.45
3.90
2.99
Computer Sci
Physics
Computer Sci
Mathematics
Mathematics
Psychology
231455
888323
102147
453876
678543
786576
GPAMajorID
3.88
3.45
3.49
3.45
3.90
2.99
231455
888323
102147
453876
678543
786576
Ackermann
Adams
Chou
Goodfriend
Rao
Stevens
GPAID Student_name
R S
),(2 SRJ
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.
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)}
(1) Simple undirected graph DetroitSan Francisco
Los AngelesDenver
ChicagoNew York
Washington
Detroit
(2) multigraphSan Francisco
Los AngelesDenver
ChicagoNew York
WashingtonDetroitSan Francisco
Los AngelesDenver
ChicagoNew York
Washington(3) pseudograph
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 YorkSan Francisco
Los Angeles Denver
ChicagoDetroit
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
Definition 3.
1. 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.
2. The degree of a vertex v in an undirected graph is the number of edges incident with it, denoted by deg(v).
3. 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.
Connectivity
Definition 4. Given a graph G=(V,E), a path of G is a sequence of vertices .),(),....,,(),,(),,(x where,,...,, 132211010 Exxxxxxxxxx nnn
circle. a is
simple.not but path a is
path. simple a is
FEHBF
ACFEHBF
ACFEH
A
F
C
B
D
E
H I
A
F
C
B
D
E
H I
circle. a is
simple.not but path a is
path. simple a is
FADBHEF
FADBHEB
FADBH
Definition 5. An undirected graph is called connected if there is a path between every pair of distinct vertices of the 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
B
D
E
H I
Connected graph
A
F
C
B
D
E
H I
Unconnected graph
A
F
C D
EStrongly connected
A
F
C D
ENot Strongly connected
Chapter 8 Trees
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.
8.1 Introduction to tree
Ordered rooted tree root
subtree
parent
third child
Binary tree (every node has at most two children)
Left subtree
Right subtree
Left child Right child
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.
preorder.in traversedis until ...
then preorder,in g traversin(3)
then ,preorderin g traversin(2)
then , visiting(1)
by begins of raversalpreorder t the
right, left to from at subtrees the
are ,...,, that supposing Otherwise,
. of traversal
preorder theis then, ofonly contains If
root with treerooted orderedan be Let
1 Definition
2
1
21
n
n
T
T
T
r
T
r
TTT
T
rrT
r.T 6 18
30
4
7
1321
20179
15Root of T
19
1
4
19
19 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
inorder.in traversedis until ...
then inorder,in g traversin(4)
then ,inorderin g traversin(3)
then, visiting(2)
then,inorder in ng traversi(1)
by begins of aversalinorder tr the
right, left to from at subtrees the
are ,...,, that supposing Otherwise,
. of aversalinorder tr the
is then, ofonly contains If
root with treerooted orderedan be Let
2 Definition
3
2
1
21
n
n
T
T
T
r
T
T
r
TTT
T
rrT
r.T
6 18
30
4
7
1321
20179
15Root of T
19
1
4
19
19 5
11
The inorder traversal of T9,6,21,30,4,11,13,7,15,17,18,1, 20,19,19,19,4,5
. visiting(3)
then postorder,in traversedis until ...
then ,postorderin g traversin(2)
thenpostorder,in ng traversi(1)
by begins of traversalpostorder the
right, left to from at subtrees the
are ,...,, that supposing Otherwise,
. of traversalpostorder the
is then, ofonly contains If
root with treerooted orderedan be Let
3 Definition
2
1
21
r
T
T
T
T
r
TTT
T
rrT
r.T
n
n
6 18
30
4
7
1321
20179
15Root of T
19
1
4
19
19 5
11
The postorder traversal of T9,21,4,30,11,13,7,6,17,1,18,20,19,4,5,19,19,15
Operations in a BST1. Search (given a key return the corresponding
data if it is in the BST)2. Insert (given a data insert it into the BST)3. Delete (given a key delete the corresponding data
from BST)
8.2 Application of Trees
Binary Search TreeDefinition 1 Binary Search Tree is a Binary Tree satisfying the following
condition:
(1) 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.
(2) 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
131
2017
9
15root
key
item (key/other data)
Right left
A node of BST: item, right child, left childDefine a node of BST
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/ /
rootA BST
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/ /
rootA BST
Visit the nodes of BST in-orderProcedure InOrder(Node root)
{ if (root ≠ null)) { InOrder(root.Left); visit root; InOrder(root.Right); } }
A BST can be visited in three ways: in-order, pre-order and post-order
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
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/ /
rootA BST
Operation 1: Search
Example:
Search (42)
Search (64)
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 */ }
Operation 1: Search
height
Time Complexity
O(height of BST)Average case: O(lg n)Worst case O(n) where n is the size of the BST
Insert 1.Create a new node for new item (e.g., 62/data)2.Find the place for insertion 3.Link the new node
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; } }
62/data/ /
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 currentparent
currparen
currparen
currparen
currparen
curr = 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
Delete 1. Find the deleting node (current) and its parent2. Delete currentThe first step is similar to Search. We only consider the step of 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/ /
rootA BST
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; }
Case 1: Deleting node (current) has no child
current
parent
height
Time Complexity 1.Find the deleting node (current) and its parent: O(height of BST) 2. 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
Example: Delete the node with item (36/data)
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/ /
rootABST 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; else if (parent.Left = current) parent.Left: = current.Right; else parent.Right: = current.Right;}
Case 2: Deleting node (current) has one child
current
parent
Example: delete the node with item (35/data)
height
Delete 1. Find the deleting node (current) and its parent2. Delete currentThe first step is similar to Search. We only consider the step of delete current.
Time Complexity 1. Find the deleting node (current) and its parent: O(height of BST) 2. 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
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/ /
rootABST
if (!(successor == current.Right)) sucessorParent.Right = successor.Right;current.item = successor.item
Case 3: Deleting node (current) has two children
current
Delete 1.Find the deleting node (current) and its parent2.Find successor which is the node with the smallest key in current’s left, and successor’s parent. 3.Link successorParent to successor.Right4.Replace the current.item by successor.item
successor
Succ Parent
35/data
height
Example: delete the node with item (30/data)
Step 1 and 2 are similar to Search. We only consider step 3 and 4.
Time Complexity 1.Find the deleting node (current) and its parent: O(height of BST)2.Find successor and successor’s parent: O(height of BST)3.Link successor’s parent to successor.Right: O(1) 4.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
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.
Example 1
Form a binary search tree for the words mathematics, physics, geography, zoology, meteorology, geology, psychology, and chemistry (using alphabetical order).