12_Matdis_Eng.pdf

Matematika Diskrit

POLITEKNIK TELKOM

BANDUNG

2009

Penyusun dan Editor Adi Wijaya M.Si

Dilarang menerbitkan kembali, menyebarluaskan atau menyimpan baik

sebagian maupun seluruh isi buku dalam bentuk dan dengan cara apapun tanpa izin tertulis dari Politeknik Telkom.

Hak cipta dilindungi undang-undang @ Politeknik Telkom 2009

No part of this document may be copied, reproduced, printed, distributed, modified,

removed and amended in any form by any means without prior written authorization of Telkom Polytechnic.

Politeknik Telkom Matematika Diskrit

Matematika Diskrit iii PAGE 10

Kata Pengantar

Assalamu’alaikum Wr. Wb

Segala puji bagi Allah SWT karena dengan karunia-Nya courseware ini

dapat diselesaikan.

Atas nama Politeknik Telkom, kami sangat menghargai dan ingin

menyampaikan terima kasih kepada penulis, penerjemah dan

penyunting yang telah memberikan tenaga, pikiran, dan waktu sehingga

courseware ini dapat tersusun.

Tak ada gading yang tak retak, di dunia ini tidak ada yang sempurna,

oleh karena itu kami harapkan para pengguna buku ini dapat

memberikan masukan perbaikan demi pengembangan selanjutnya.

Semoga courseware ini dapat memberikan manfaat dan membantu

seluruh Sivitas Akademika Politeknik Telkom dalam memahami dan

mengikuti materi perkuliahan di Politeknik Telkom.

Amin.

Wassalamu’alaikum Wr. Wb.

Bandung, Maret 2009

Christanto Triwibisono

Wakil Direktur I

Bidang Akademik & Pengembangan

Telkom Polytechnic Discrete Mathematics

iv PBO PAGE 10

Daftar Isi

Kata Pengantar.................................................................................... iii Daftar Isi ............................................................................................... iv Daftar Gambar ..........................................Error! Bookmark not defined. Daftar Tabel ...............................................Error! Bookmark not defined. 1 HIMPUNAN ....................................Error! Bookmark not defined. 1.1 Definisi dan Keanggotaan Suatu HimpunanError! Bookmark not

defined. 1.2 Operasi Himpunan .............................. Error! Bookmark not defined. 1.3 Prinsip Dualitas ..................................... Error! Bookmark not defined. 1.4 Multi Set ................................................ Error! Bookmark not defined. 2 RELASI DAN FUNGSI ...................Error! Bookmark not defined. 2.1 Definisi Relasi dan Cara Penyajian ... Error! Bookmark not defined. 2.2 Beberapa Sifat Relasi........................... Error! Bookmark not defined. 2.3 Operasi pada Relasi ............................. Error! Bookmark not defined. 2.4 Relasi Ekivalen dan Relasi Terurut..... Error! Bookmark not defined. 2.5 Fungsi ..................................................... Error! Bookmark not defined. 3 KOMBINATORIK ...........................Error! Bookmark not defined. Prinsip Dasar Menghitung ............................... Error! Bookmark not defined. Permutasi dan Kombinasi ................................ Error! Bookmark not defined. 4 TEORI GRAF ...................................Error! Bookmark not defined. 4.1 Definisi Graf .......................................... Error! Bookmark not defined. 4.2 Terminologi Graf .................................. Error! Bookmark not defined. 4.3 Keterhubungan dan Sub Graf ............ Error! Bookmark not defined. 4.4 Matriks Ketetanggaan (adjacency matrix) dan Matriks Bersisian

(incidency matrix) dari Suatu Graf .. Error! Bookmark not defined. 4.5 Eulerian dan Hamiltonian ................... Error! Bookmark not defined. 4.5.2 Sirkuit Hamilton .................................... Error! Bookmark not defined. 4.6 Graf Isomorfik....................................... Error! Bookmark not defined. 4.7 Beberapa Aplikasi Graf ....................... Error! Bookmark not defined. 5 POHON DAN PEWARNAAN GRAFError! Bookmark not

defined. 5.1 Pohon Merentang Minimum (Minimun Spanning Tree) ................ Error!

Bookmark not defined. 5.2 Pohon Berakar ...................................... Error! Bookmark not defined. 5.3 Penelusuran Pohon Biner..................... Error! Bookmark not defined. 5.4 Pewarnaan Graf ................................... Error! Bookmark not defined. Pewarnaan Peta (Map Coloring) .................. Error! Bookmark not defined. Daftar Pustaka ..................................................................................... vi


Matematika Diskrit v PAGE 10


Himpunan 1 PAGE 10

1 SETS

Overview

In reality, there are many problems related to objects that are grouped together based on certain criteria. This group of objects is a representation

of a condition, either statistically or economically. Such group of objects is defined as set. This chapert will discuss the definition of sets, the elements, and set operation of several types of sets.

Objective

1. Students understand the basic concept of set. 2. Students understand varios type of set operations and property.

3. Students are able to solve problems and phenomena related to set theory.


2 Himpunan PAGE 10

1.1 Definition and Mempership of Sets

A set is a group of different objects that can be clearly defined. The

objects in a set are called elements, or members, of the set. Membership in a

set is described using notation ’’. Example 1 :

A = {x, y, z}

x A : x is the element of set A.

w A : w is not the element of set A.

There are several ways to describe the set, i.e.:

a. By enumerating the elements

In this method, the set is described by listing all its members between braces.

Example 2 :

- The set A of the first four odd numbers can be written as A = {1, 3, 5, 7}.

- The set B of the first five prime numbers can be expressed by

B = {2, 3, 5, 7, 11}. - The set C of natural numbers less than 50 can be written as

C = {1, 2, ..., 50}

- The set of integers can be denoted by {…, -2, -1, 0, 1, 2, …}.

b. By using standard symbol

A set can be expressed using standard symbol that is usually used by academicians.

Example 3 :

N = the set of natural numbers = { 1, 2, ... } Z = the set of integers = { ..., -2, -1, 0, 1, 2, ... } Q = the set of rational numbers

R = the set of real numbers C = the set of complex numbers

Universal set is denoted by U.

Example 4 :

Suppose that U = {1, 2, 3, 4, 5} and A = {1, 3, 5} is the subset of

U.


Himpunan 3 PAGE 10

3. By characterizing all elements in the set

A set can be described by characterizing all elements the elemetnts in the set. The set can be denotated as:

{ x property or properties that x must have }

Example 5 :

(i) Set A of natural numbers less than10 can be expressed by

A = { x | x 10 and x N } or

A = { x N | x 10 } That is equivalent with :

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (ii) M = { x | x is student taking discrete mathematics

subject }

or

M = { x is student | he takes discrete mathematics

subject } 4. By using Venn Diagram

A set can a;so represented by listing all its members in a diagram that is

called Venn diagram.

Example 6 :

Suppose that U = {1, 2, …, 7, 8}, A = {1, 2, 3, 5} and B = {2, 5, 6, 8}.

The Venn Diagram is:

U

1 2

53 6

8

4

7A B


4 Himpunan PAGE 10

In terms of membership of sets, a set can be described as member of other

set. Example 7 :

a. Suppose that M = { students of Politechnic Telkom }

M1 = { students of Computer for Accountancy study progam } M2 = { students of Information System study progam } Thus, M = { M1, M2 }

b. Let P1 = {x, y}, P2 = { {x, y} } or P2={P1},

Meanwhile, P3 = {{{x, y}}}, then x P1 and y P2,

that P1 P2 , while P1 P3, but P2 P3

The number of elements in a set is called the cardinality of the set. For example, cardinality of set A can be denoted by:

n(A) or A

Example 8 :

(i) B = { x | x is prime numbers less than 10 },

or B = {2, 3, 5, 7 } then B = 4

(ii) A = {a, {a}, {{a}} }, then A = 3

If a set does not have any elements, in other words the cardinality of the ser

is zero, the set is called null set. The notation of null set is or {}

Example 9 :

(i) P = {STT Telkom’s Informatic Engineering students who have been to Mars},

then n(P) = 0

Thus, P =

(ii) A = {x | the suare root of x2 + 1 = 0 and x R}, then n(A) = 0

Thus A = {}

(iii) B = {{ }} can also be written as B = {}.

Thus, B is nut null set because it has one element, that is the nul set.


Himpunan 5 PAGE 10

Set A is considered the subset of set B if and only if all elements of set A are

the elements of set B. In this case, B is considered the superset of A.

Subset can be denoted as: A B or A B

Using Venn diagram, the subset can be described as:

U

AB

Example 10 :

(i) N Z R C

(ii) {2, 3, 5} {2, 3, 5}

For every set of set A these conditions apply:

(a) A is the subset of A (yaitu, A A).

(b) The null set is the subset of A ( A).

(c) If A B and B C, then A C

A and A A, then and A is called the improper subsetof set A.

Statement AB is different with statement AB :

A B means that A is the subset of B, but A B. Thus, A is the proper subset ofi B.

Example 11 :

Suppose that A = {1, 2, 3}.

{1} and {2, 3} is the proper subset of A. The power set of set A is a set which elements are the subset of A, including

the null set and set A itself. Poser set is denoted by P(A). The cadinal of a power set depens on the cardinal of the original set (himpunan asal). For

example, the cardinality of set A is m, then P(A) = 2m.

Example 12 :


6 Himpunan PAGE 10

Suppose that A = { x, y }, then P(A) = { , { x }, { y }, { x, y }}

Example 13 :

The power set of null iset is P() = {}, while the poer of set of set

{} is P({}) = {, {}}.

Statement A B is used to described that A is the subset of B the makes A = B.

Two sets are equal if they satisfy these conditions:

A = B if and only if every element of A is the element of B and the other

way around, every element of B is the element of A.

To decide that A = B, it is necessary to prove that A is the subset of B and B

is the subset of A. If the condition is not like that, then A B. or

A = B A B and B A

Example 14 :

(i) Suppose that A = { 0, 1 } and B = { x | x (x – 1) = 0 }, then A = B

(ii) Suppose that A A = { 3, 5, 8, 5 } and B = {5, 3, 8 }, then A = B (iii) Suppose that A A = { 3, 5, 8, 5 } and B = {3, 8},

then A B For three subsets of A, B, and C, this axiom applies:

(a) A = A, B = B, and C = C (b) If A = B, then B = A (c) If A = B and B = C, then A = C

Two subsets are equivalent if ech set has the same cardinality. For example, set A is equivalent with set B. This means that the cardinality of

set A and set B are the same. This can be denoted by A~B

Example 15 :


Himpunan 7 PAGE 10

Suppose that A = { 2, 3, 5, 7 } and B = { a, b, c, d },

then A ~ B because A = B = 4

Two sets of A and B are called disjoint if both sets do not have similar elements. The notation used is A // B . This can be described using Venn

diagram as follows:

U

A B

Example 16 :

Suppose that A = { x | x N, x < 10 } and B = { 11, 12, 13, 14, 15 }, then A // B.

1.2 Set Operations

Some set operations discussed are intersection, union, complement,

difference, and symmetric difference. a. Intersection

The intersection of two sets is denotated by ‘ ‘. Suppose that A and B are not disjoint sets, then

A B = { x x A and x B }

Using Venn diagram, it can be described as:


8 Himpunan PAGE 10

Example 17 :

1. Suppose that A = {2, 3, 5, 7, 11} and B = {3, 6, 9, 12},

then A B = {3}

2. Suppose that A is a set of Industrial Engineering students of STT Telkom and B is a set of old women (50 years above)

then A B = .

It means that A and B are disjoint or A // B.

b. Union

The union of two sets is denoted by ‘‘. Let A and B are sets, then

A B = { x x A or x B }

Using Venn diagram, it can represented as:

Example 18 :

(i) If A = { 2, 3, 5, 7} and B = { 1, 2, 3, 4, 5 }, then A B = {1,2,3, 4,

5, 7}

(ii) A = A

c. Complement

The complement of a set is the elements of all elements of universal set, minus the elements of the set. Suppose that A is a set within universal set U, the complement of set A is denoted by:

A= Ac = { x x U and x A }


Himpunan 9 PAGE 10

It can be represented using Venn diagram as:

Example 19 :

Suppose that U = { 1, 2, 3, ..., 9 },

if A = {1, 3, 7, 9}, then A = {2, 4, 5, 6, 8}

if A = { x U | x is divisible by two}, then A= { 1, 3, 5, 7, 9 }

Example 20 : A = set of students of PolitechnicTelkom

B = set of students living in dormitory C = set of students majoring in Information System D = set of students taking Discrete Mathematics subject

E = set of students riding motorbikes to campus a. The statement

“All students of Politechnic Telkom, majoring in

Information System, and riding motorbikes to campus” can be expressed using the notation of set operation as:

(A C) E

b. The statement

“All students of Politechnic Telkom who live in dormitory and who do not take Discrete Mathematics subject”

can be expressed using the notation of set operation as:

DBA c. The statement

“All students who major in Information System, who do not live in dormitory or who do not ride motorbikes to

campus” can be expressed using the notation of set operation as:

EBC


10 Himpunan PAGE 10

d. Difference

The difference of two sets is denoted by ‘– ‘. Let A and B be sets, the difference between A and B is denoted by

A – B = { x x A and x B } = BA

Example 21 :

Suppose A = { 1, 2, 3, ..., 10 } and B = { 2, 3, 5, 7}, then A – B = { 1,

4, 6, 8, 9 } and B – A = e. Symmetric Difference

The symmetric difference of two sets is denoted by ‘ ‘. Let A and B be sets, the symmetric difference between A and B

dinotasikan oleh :

A B = (A B) – (A B)

= (A – B) (B – A) This can be represented using Venn diagram as:


Himpunan 11 PAGE 10

Example 22 :

Suppose A = { 2, 3, 5, 7} and B = { 1, 2, 3, 4, 5 }, then

A B = { 1, 4, 7 }

Symmetric difference should meet the following requirements:

(a) A B = B A (Commutative laws)

(b) (A B ) C = A (B C ) (Asscociative laws)

f. Cartesian product

The Cartesian product of two sets is denoted by ‘ ‘.

Let A and B be sets, the Cartesian product of A and B is denoted by:

A B = {(a, b) a A and b B }

Example 23 :

(i) Suppose that C = {1, 2, 3}, and D = { a, b }, then

C D = { (1, a), (1, b), (2, a), (2, b), (3, a), (3, b) } (ii) Suppose that A = B = set of all real numbers, then

A B = set of all dots in a planar shape

Supposet that two sets have finite cardinality, the cardinality of the set as the Cartesian product of the two sets is the cardinality of each set. Thus, if A and B are finite sets, then:

A B = A . B.

An ordered pair (a, b) is different with (b, a), in other words (a, b) (b, a). Referring to this argument, the Cartesian product is not commutative, that is

A B B A In which A or B is not null set.

If A = or B = , then

A B = B A =


12 Himpunan PAGE 10

Laws apply to set operations are:

1. Identity laws:

A = A

A U = A

2. Null/domination laws:

A =

A U = U 3. Complementation laws:

A A = U

A A =

4. Idempotent laws:

A A = A

A A = A

5. Involution laws:

)(A = A

6. Absorption laws:

A (A B) = A

A (A B) = A

7. Commutative laws:

A B = B A

A B = B A

8. Associatitive laws:

A (B C) = (A B) C

A (B C) = (A B) C

9. Distributive laws:

A (B C) = (A B) (A C)

A (B C) = (A B) (A C)

10. De Morgan’s laws:

BA = BA

BA = BA 11. Universal Complementation laws

U

U =


Himpunan 13 PAGE 10

Let A and B be finite sets, then

n(AB) = n(A) + n(B) – n(AB) This is called the principle of inclusion-exclusion. This principle is useful in solving sets and combinatorial.

The principles also apply to three or more finite sets. Let A, B, and C be finite sets. Based on the principle of inclusion-exclusion, the relationship between the cardinalities of the sets’ partitions can be expressed by:

|ABC| = |A| + |B| + |C| – |AB| – |BC| – |AC| + |ABC| The principle of inclusion-exclusion will be explained in chapter on Combinatorics.

1.3 Duality Principle

Duality principle explains that two different concepts can be exchanged. The

exchange provides true answers. Example 24 :

USA steering wheel is placed in front-left Indonesia steering wheel is placed in front-right

Rules: (a) in United State of America,

cars must take the right side of the road,

in the highways, overtaking is done from the left side, when the red light is on, cars are not allowed to directly turn right

(b) in Indonesia,

cars must take the left side of the road, in the highways, overtaking is done from the right side, when the red light is on, cars are allowen to directly turn left

Duality priciples applied to the above case is: The left-right principles can be exchanged. Rules applied in America is used in Indonesia and vice versa.

Duality Principle of Set. Suppose that S is an identity involving set and

operations like , , and complement. If S* is identity that is the dual of S, by

changing , , U, U , while the complement is not changed, when the operations are applied to the identity S*, they provide

correct results. Table 1.1 Duality of Algebraic Set Laws


14 Himpunan PAGE 10

1. Identity law:

A = A

The dual is:

A U = A

2. Null/domination law:

A =

The dual is:

A U = U

3. Complementation law:

A A = U

The dual is:

A A =

4. Idempotent law:

A A = A

The dual is:

A A = A

5. Absorption law:

A (A B) = A

The dual is:

A (A B) = A

6. Commutatitive law:

A B = B A

The dual is:

A B = B A

7. Associative law:

A (B C) = (A B) C

The dual is:

A (B C) = (A B) C

8. Distributive law:

A (B C)=(A B) (A C)

The dual is:

A (B C) = (A B) (A C)

9. De Morgan’s law:

BABA

The dual is:

BABA

10. 0/1 law:

U

The dual is:

U =

Example 25 :

Suppose that A U in which A = BABA

in its dual, suppose that it is U*, applies:

A = BABA


Himpunan 15 PAGE 10

There are several ways in proving the truthfullness of a statement or in

representing a statement using other statement by using sets. They are: a. By using Venn diagram Example 26 :

Let A, B, and C be sets.

Show that A (B C) = (A B) (A C) by using Venn diagram.

Answer :

This way is not for formal proof. By drawing the sets and gradually

shading every operation, the end sets will be obtained.

A (B C) (A B) (A C)

These two Venn diagram provide the same shaded areas.

Statement stating that A (B C) = (A B) (A C) is proven.

b. Some examples in proving statements using algebraic set operation.

Example 27 : Let A and B be sets. Show that:

A (B – A) = A B Anser :

A (B – A) = A (B A) (Definition of difference)

= (A B) (A A ) (Distributive law)

= (A B) U (Complementation law)

= A B (Identity law)


16 Himpunan PAGE 10

Example 28 :

Show that for every sets A and B, these operation can be applied:

(i) BAA = A B and

(ii) A ( A B) = A B

Answer :

(i) BAA = BAAA (Distributive law)

= U (A B) (Complementation law)


(ii) is the dual of (i)

BAA = BAAA (Distributive law)

= (A B) (Complementation law)


1.4 Multiset

A set which elements are allowed to be repeated (do not have to be different) is called multi set.

Example 29 :

A = {1, 1, 1, 2, 2, 3},

B = {2, 2, 2}, C = {2, 3, 4}, D = { }.

The multiplicity of an element in a multiset is the number of the appearance of the element in the multiset. Example 30 :

M = { 1, 1, 1, 2, 2, 2, 3, 3, 1 }, the multiplicity of 1 is 4 and the multiplicity of 2 is 3, while the multiplicity of 3 is 2.

Set is a special example of multiset. In this case, the multiplicity of its element is 0 is 1. A set which multiplicity of ite elemement is 0 is null set. Let P and Q be multisets, the operations applied to those multisets are:

a. P Q is a multiset which multiplicity of elements is the same with maximum multiplicityof the elements is sets P and Q.


Himpunan 17 PAGE 10

Example 31 :

P = { a, a, a, c, d, d } dan Q ={ a, a, b, c, c }, then

P Q = { a, a, a, b, c, c, d, d }

b . P Q is a multiset which multiplicity of the elements is the same as the minimum multiplicity of the elements in sets P and Q.

Example 32 :

P = { a, a, a, c, d, d } and Q = { a, a, b, c, c } then

P Q = { a, a, c }

c. P – Q is a multiset which multiplicity of the elements is the same as the multiplicity of the elements in P substracted by the multiplicity in Q. This is applicable if the difference of the multiplicities is positive. If the

difference is zero or negative, the multiplicity of the elements is zero. Example 33 :

P = { a, a, a, b, b, c, d, d, e } and

Q = { a, a, b, b, b, c, c, d, d, f } then P – Q = { a, e }

d. P + Q, which is defined as the sum of two multisets, is a multiset which

multiplicity of the elements is the same with the sum of the multiplicity of the elements in P and Q.

Example 34 :

P = { a, a, b, c, c } dan Q = { a, b, b, d }, then P + Q = { a, a, a, b, b, b, c, c, d }


18 Himpunan PAGE 10

Summary

1. Set is a group of different objects that can be clearly defined. 2. Set can be described by enumerating all its members, using symbol,

characterizing all elements in the set, or using Venn diagram. 3. The number of elements in set A is called the cardinality of set A and is

denoted by n(A) or A

4. If a set does not have any element, in other words the cardinality of the

set is zero, the set is called null set and is denoted by or { }.

5. Some necessary set operations are: intersection, union, complement, difference, and symetric difference.

6. Let A and B be finite sets, the principle of inklusion-exclusion for thos

two sets id described as:

n(AB) = n(A) + n(B) – n(AB) 7. A set which elements are allowed to be repeated boleh berulang (do not

have to be different) is called multiset. 8. The multiplicity of an element in a multiset is the number of its

appearance in the multiset.


Himpunan 19 PAGE 10

True or False Quiz

(For questions No. 1 – 5) Suppose that A = {2, 3, 5, 7}, B = {2, 4, 6, 8,

10}, and C = {1, 3, 5, 7}

1. CABA

2. BA

3. BB is set of natural numbers

4. CBAA

5. CBACBA 6. Nul set is a set which only has one element. The element is zero.

7. If (A – B) = {1, 2, 3} and (B – A) = {4, 5} then AB = {1, 2, 3, 4, 5}

8. If P Q = R then (R – Q) = P

9. The complement of a set of natural numbers is negative numbers.

10. If a A and b B then A B =


20 Himpunan PAGE 10

Multiple Choices

1. A set which elements may be repeated is called

A. Double set D. Fuzzy set

B. Multiset E. Data set

C. Frequency set

2. If C = { x x A or x B } then C = .....

A. A B D. A B

B. A – B E. A (B A)

C. A B

3. Set operation of A B, is the same as ....

A. A – B D. (A B) – (A B)

B. (A – B) B E. (A – B) (B – A)

C. (A – B) (B – A)

4. D’Morgan’s law is a set is described as:

A. A B = B A D. A U = U

B. A (B C) = (A B) C E. (A B)

C. BABA

5. The frequency of the appearance of an element in a multiset is called ….

A. cardinality D. Multiplicity

B. Multiplication E. Additivity

C. Frequency

6 P = { x x A dan x B } then P = .....

A. A B D. A B

B. A – B E. A (B A)

C. A B


Himpunan 21 PAGE 10

7 Set operation of (A – B) (A B) (B –A) results in

A Set A D Set B – A

B Set B E Set A B

C Set A – B

8 Suppose that U is universal set of A and B, then (A U) = ....

A (B – A)c D (A B)c

B (A – B)c E Bc

C (A B)c

9 There are 10 students who take discrete mathematics subjects, 15

students who take management subject, and 6 students who take the two subjects. If the total number of the studenst is 30, the number of students who do not take the two subjects is...

A 9 D 12

B 10 E 13

C 11

10 Suppose that P = {1, 2, 3, 4, 5} and Q = {a, b, c} then n (A X B) = ....

A 5 D 15

B 3 E 8

C 2


22 Himpunan PAGE 10

Exercises

(For questions no. 1 – 5) Suppose that A = {1,2, 3, 4, 5, 6, 7},

B = {1, 2, 3, 5, 6, 12}, and C = {2, 4, 8, 12, 20}

Find the result of set operations for:

1. (A B) – C

2. (A B) (B C)

3. (A – B) (B – C)

4. (A C) (B C)

5. (A B) – C (A C) – B 6. Find the number of numbers in set A, which is indivisible by by 3 or 5 !

7. Find the number of numbers in set A, which is divisible by 3, but not by 5 !

8. Find the number of numbers in set A, which is divisible by 3, but not by

5 or 7 ! 9. Suppose that the number of students in a class is 60. Twenty of them

like calculus, 30 of them like discrete mathematics, and 10 of them like

linear algebra. Seven of them like 7 calculus and discrete mathematics, 5 of them like discrete mathematics and linear algebra, and 10 of them do not like all the three subjects. a. Find the number of students who like all the three subjects!

b. Find the number of students who like only one subject! (For questions no. 10 – 15)

A survey done to 60 students results in the following data: 25 like to read Kompas 26 like to read Republika

27 like to read Pikiran Rakyat 9 like to read Kompas and Republika 11 like to read Kompas and Pikiran Rakyat

8 like to read Republika and Pikiran Rakyat 3 like to read all the three newspapers

10. Draw the Venn diagram for the data!


Himpunan 23 PAGE 10

11. Find the number of students who never read any of the newspaper!

12. Find the number of students who only read Pikiran Rayat! 13. Find the number of students who exactly read only one kind of

newspaper!


24 Relasi dan Fungsi PAGE 10

2 RELATIONS AND FUNCTIONS

Overview

The relationships between elements/members in a set occur in many

context. The relationships are represented using a structure called a relation.

Relations can be used to solve various problems such as optimization on

communication network, schedulling, and solving database problems.

Objective

1. Students understand the concept of relations and functions.

2. Students understand various operations and properties of relations.

3. Students are able to solve problems and phenomena related to relations and functions.


Relasi dan Fungsi 25 PAGE 10

2.1 The Definition and Representation of Relations

The previous chapter has discussed the Cartesian product, which is

ordered pairs describing the relationship of two sets. All ordered pairs are the elements of a subset as a result of Cartesian product of two sets. Some of the elements of the subset have specific relationship between two

elements of the ordered pair. The relationship occurs according to certain rules. Rules relating two sets are called binary relations. Relation between set A and set B is a set that contains ordered pairs following certain rules. Thus, binary relation R between set A and B is a subset of Cartesian product

A B or R (A B).

The notation used for a ninary relation is a R b or (a, b) R. This

means that a is related to b by R. An element in a Cartesian product is not

the relation of the element is a R b atau (a, b) R, which means that a is not

related to b by relation R. Set A is called the domain of R, and set B is called the range of R.

Example 2.1 : Let A = {2, 3, 4} and B = {2, 4, 8, 9, 15}. If a relation R from A to B is defined by:

(a, b) R if a is the prime factor of b Find the elements of R!

Answer :

As it is previusly explained, the Cartesian product of A B is:

A B = {(2, 2), (2, 4), (2, 8), (2, 9), (2, 15), (3, 2), (3, 4), (3, 8),

(3, 9), (3, 15), (4, 2), (4, 4), (4, 8), (4, 9), (4, 15)} Using the above definition, the relation R from A to B following the rule

is: R = {(2, 2), (2, 4), (2, 8), (3, 9), (3, 15) }

Relation may also happen to only one set, that is the realtion on A. Relation

on set A is the subset of Cartesian product A A.



Example 2.2 :

Let R be the relation on A = {2, 3, 4, 8, 9} that is defined by (x, y)

R if and only if x is divisible by y. Answer :

A relation R on A that follow the rule is:

R = {(2, 2), (4, 4), (4, 2), (8, 8), (8, 2), (8, 4), (3, 3), (9, 9), (9, 3)} Relations can be represented in various ways such as using arrows, table,

matrix, or even using directed graph. In this section, each way of representing relations is discussed. Representing a relation:

a. Representing Relation using Arrows Let A = {2, 3, 4} and B = {2, 4, 8, 9, 15}. If it is defined that reltion R from A to B follows this rule:

(a, b) R if a is the prime factor of b the relation can be represented as the following:

b. Representing Relation usind Ordered Pairs

The example of relation on (a) can be expressed in ordered pairs as: R = {(2, 2), (2, 4), (2, 8), (3, 9), (3, 15)}

c. Representing Relation using Table

The first column of the table represents the domain, while the second column represents the relations. The relation explained in part (a) can be represented using table as the following:

2

3

4

2

4

8

9

15



Table 2.1 Relation of ‘the prime factof of’

A B

2 2

2 4

2 8

3 9

3 15

d. Representing Relation using Matrix

Let R be the relation that relate set A = {a1, a2, …, am} and set B = {b1, b2, …, bn}. The relation can be represented using matrix as follows:

b1 b2 bn

M =

mnmm

n

n

m mmm

mmm

mmm

a

a

a

21

22221

11211

2

1

The elements mij on the matrix have the value of one or zero, depending on whether element ai on set A has relation with element bj on set B. The statement can be represented in the follwing form:

Rba

Rbam

ji

ji

ij),(,0

),(,1

Example 2.3 :

Let A = {2, 3, 4} and B = {2, 4, 8, 9, 15}. If a relation R from A to B is defined as:

(a, b) R if a is the prime factor of b

then the relation can be represented in the for of matrix as follows:

0

1

0

0000

1000

0111

M

e. Representing Relation using Directed Graph The relation on a set can graphically represented using directed

graph or digraph. Directed Graph is only defined to represent a relation



on a set (not between two sets). Every element of the set is represented

by a dot (it is also called vertex), and every ordered pairs is represented

by an edge (arc). If (a, b) R, an edge is drawn from vertex a to vertex b. Vertex a is called initial vertex and vertex b is called terminal vertex.

An ordered pair (a, a) is represented by an edge from vertex a to vertex a itself. Such edge is called loop.

Example 2.4 : Let R = {(a, b), (b, c), (b, d), (c, c) (c, a), (c, d), (d, b)} be the relation on set {a, b, c, d}.

Relation R can be represented using directed graph as follows:

2.2 Some Properties of Relation

A relation on a set has several properties. They are:

1. Reflexive

A relation R on a set A is called reflexive if (a, a) R for every element

a A. In other words, a relation R on a set A is not reflexisive if there

is a A such that (a, a) R.

Example 2.5 :

Suppose that A = {1, 2, 3, 4}, and relation R is a relation ‘’ which is defined on set A, then

R = {(1, 1), (1, 2), (1, 3),(1, 4), (2, 2), (2, 3), (2,4), (3, 3), (3, 4), (4, 4)} It is seen that (1, 1), (2, 2), (3, 3), (4, 4) are the elements of R. Therefore, R is reflexive.

a b

c d



Example 2.6 :

Suppose that A = {2, 3, 4, 8, 9, 15}. If relation R on set A is defined according to:

(a, b) R if a is the prime factor of b

Notice that (4, 4) R . Thus, it is clear that R is not reflexive.

Reflexive property causes several characteristics in the representation of a relation. They are:

Reflexive relation has matrix which main diagonal elements is 1, or

mii = 1, for i = 1, 2, …, n,

1

1

1

1

When a reflexive relation is represented using directed graph, there

will always be loop in every vertex. 2. Transitive

A relation R on set A is called transitive if (a, b)R and (b, c) R,

then (a, c) R, for a, b, c A.

Example 2.7 :

Suppose that A = { 2, 3, 4, 5, 6, 7, 8, 9}, and relation R is defined by:

a R b if and only if b is divisible by a, in which a, b A, Answer :

By referring to the definition for relation R on set A, then: R = {(2, 2), (2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (3, 9), (4, 4), (4, 8)}

When (2, 4) R and (4, 8) R, it is seen that (2, 8) R.

Therefore, R is transitive.

Example 2.8 :

R is a relation on a set of natural number N. R is defined by:



R : a + b = 5, a, b A,

Is R transitive? Answer :

By referring to the definition of relation R on set A, then:

R = {(1, 4), (4, 1), (2, 3), (3, 2) }

Notice that (1, 4) R and (4, 1) R , but (1, 1) R. Therefore, R is not transitive.

Transitive property gives some characteristics in the representation of a relation. They are:

Transitive property on directed graph is described by a condition: if there is an edge from b to c, there will be directed edge from a to c.

When represented in the form of matrix, a transitive relation does not have particular characteristics on its matrix representation.

3. Symmetric and Antisymmetric

A relation R on a set A is called symmetric if (a, b)R, for every

a, b A, then (b, a) R. A relation R on a set A is not simmetric if (a,

b) R, while (b, a) R. A relation R on a set A is called

antisymmetric if for every a, b A, (a, b) R and (b, a) R and it

applies only if a = b. Notice that the term symmetric is not the antonym for the term antisymmetric. This is due to the fact that a relation may have both characteristics at the same time. However, it is impossible

for a relation to have the two characteristics at the same time if it has

ordered pair of (a, b) in which a b.

Example 2.9 :

Let R be the relation on a set of real number. R is defined by :

a R b if and only if a – b Z. Show that the relation R is symmetric!

Answer :

Suppose that a R b, then (a – b) Z, while it is clear that (b – a)

Z. Therefore, R is symmetric.

Example 2.10 :

Show that relation ‘’ on set Z is antisymmetric.



Answer :

It is clear that if a b and b a, then a = b.

Thus, relation ‘’ is antisymmetric.

Example 2.11 :

The relation “dividiblr” on a set of positive interger N is an example of not symmetric relation because yang if b is divisible by a,

a is not divisible byb, except if a = b. Meanwhile, relation “divisible” is an antisymmetric relation because if b is divisible by a and a is divisibyle by b, then a = b.

Example 2.12 :

Suppose that relation R = {(1, 1), (2, 2), (3, 3) }, then relation R is an

symmetric and antisymmetric relation .

The symmetri and antisymmetric property provides several

charcteristics in the representation of the relations either in the form of matrix or graph, i.e.:

Symmetric relation has matrix which elements under the main

diagonal are the reflection of the elements above the main diagonal, or mij = mji = 1, for i = 1, 2, …, n and j = 1, 2, …, n is:

0

1

0

1

A symmetric relation, when represented in the form of directed graph, has a characteristics of: if there is an adge from a to be, there will be an edge from b to a.

Antysymmetric relation has matrix which elements’ characteristic is:

if mij = 1 with i j, then mji = 0. In other words, the matrix of the

antisymmetric relation meets the condition of: is one of mij = 0 or

mji = 0 if i j :



0

1

00

0

1

Meanwhile, the characteristic of a directed graph of an antysemmetric relation is that there will never be two edges with different directions between two different vertices.

Let R be the relation from the set A to the set B. The inverse of relation R, denoted by R–1, is the relation of set B to set A, which is defined by:

R–1 = {(b, a) | (a, b) R }

Example 2.13 : Let P = {2, 3, 4} and Q = {2, 4, 8, 9, 15}. If the relation R of P to Q is defined as:

(p, q) R if and only if q is divisible by p

it will be obtained that: R = {(2, 2), (2, 4), (4, 4), (2, 8), (4, 8), (3, 9), (3, 15)

R–1 is the inverse of relation R, that is a relation from Q to P in the form of:

(q, p) R–1 if q is multiple of p

Thus, it is obtained that: R–1 = {(2, 2), (4, 2), (4, 4), (8, 2), (8, 4), (9, 3), (15, 3) }

If M is a matriix representing a relation R,

M =

00110

11000

00111

then the matrix representing relation R–1, let’s say N, is obtained by finding the transpose of matrix M,



N = MT =

010

010

101

101

001

2.3 Operations of Relations

Relations are sets of oredered pairs. Thus, some algebraic operations that apply to sets also apply to relations. Set operations like intersection,

union, difference, and symmetric difference are also applicable for two

relations. Let R1 and R2 be relations form the set A to set B, R1 R2, R1

R2, R1 – R2, and R1 R2 are also the therelations from A to B. Example 2.14 :

Let A = {a, b, c} and B = {a, b, c, d}. Relation R1 = {(a, a), (b, b), (c, c)} Relation R2 = {(a, a), (a, b), (a, c), (a, d)}

Then :

R1 R2 = {(a, a)}

R1 R2 = {(a, a), (b, b), (c, c), (a, b), (a, c), (a, d)}

R1 R2 = {(b, b), (c, c)}

R2 R1 = {(a, b), (a, c), (a, d)}

R1 R2 = {(b, b), (c, c), (a, b), (a, c), (a, d)}

Suppose that relation R1 and R2 are represented by matrix MR1 and MR2. The Matrices representing the union and intersection of the two relations are:

MR1 R2 = MR1 MR2 and MR1 R2 = MR1 MR2

Example 2.15 :

Suppose that the relations R1 and R2 on set A are represented by the following matrices:



R1 =

011

101

000

and R2 =

001

110

010

then

MR1 R2 = MR1 MR2 =

011

111

010

MR1 R2 = MR1 MR2 =

001

100

000

Let R be the relation from set A to set B, and let T be the relation

from set B to set C. The composition of R and S is denoted by T R. It is the relation from A to C, which is defined by:

T R = {(a, c) a A, c C, for a b B

that (a, b) R and (b, c) S }

Example 2.16:

Let A = {a, b, c}, B = {2, 4, 6, 8} and C = {s, t, u}

Meanwhile, the relation from A ke B is defined as: R = {(a, 2), (a, 6), (b, 4), (c, 4), (c, 6), (c, 8)} while the relation from set B to set C is defined as:

T = {(2, u), (4, s), (4, t), (6, t), (8, u)} Thus, the composition of relation R and T are:

T R = {(a, u), (a, t), (b, s), (b, t), (c, s), (c, t), (c, u) }

The composition of R and T can be graphically represented using arrows as:



1

2

3

2

4

6

8

s

t

u

If relation R1 and R2 are represented by matrices MR1 and MR2, matrix

representing the composition of the two relations are:

MR2 R1 = MR1 MR2

in which MR1 MR2 is the product of the two matrices. It is done by changing

the dot symbol with “” (and), and by changing the addition symbol with “” (or).

Example 2.17 :

Suppose that relations R1 and R2 on set A are represented by the

following matrices:

MR1 =

100

011

101

and MR2 =

101

100

010

the matrix representing R2 R1 is

MR2 R1 = MR1 . MR2

=

)11()10()00()01()00()10()11()00()00(

)10()11()01()00()01()11()10()01()01(

)11()10()01()01()00()11()11()00()01(

=

101

110

111

a

b

c



2.4 Equivalence Relations and Partial Orderings

A relation on a set A is called an equivalence relation if it is

reflexive, symmetric, and transitive. Two elements that are relted by an equivalence relation are called equivalent. Example 2.18 :

Suppose that R is the relation on Z, which is defined by: a R b if and only if a = b atau a = – b .

Is R an equivalence relation? Answer :

It is clear that a = a, in other words, if a R a for every a Z . Thus, R is reflexive.

If a = b and b = c, then a = c. Inother words, if a R b then b R c then a R c. Therefore, R is transitive.

If a = b or a = – b then b = a or b = – a. In other words, if a R b then b R a. Therefore, R is symmetric.

Thus, R is equivalence relation. Example 2.19 :

Let R be the relation on a set of Real numbers, which is definend by:

a R b if and only if a – b Z. Is R an equivalence relation?

Answer :

For every a Real, then a – a = 0 integers. Therefore, R is reflexive.

Suppose that a R b then (a – b) Z, it is obvious that (b – a) Z. Therefore R is symmetric.

If a R b and b R c, it means that (a – b), (b – c) Z then (a – c) = (a – b) + (b – c) are also integers. Therefore a R c. Thus, R is transitive.

In conclusion, R is an equivalence relation.

Example 2.20 :



(Congruence Modulo)

Suppose that m is integers > 1.

Show that R = {(a,b) | a b (mod m)} is an equivalence relation on the set of integers.

Answer :

Notice that a b (mod m) if and only if a – b is divisible by m.

a – a = 0 is divisible by m, that is 0 = 0 m,

Therefore, a a (mod m) that R is reflexive.

a – b is divisible by m that a – b = km, for a k Z. This results in b – a = –km. Thus, the relation is symmetric

If a b (mod m) and b c (mod m),

that a – b and b – c is divisible by m, or

a – b = km and b – c = lm for a k, l Z

By summing them up:

a – c = (a – b) + (b – c) = (k + l) m, then a c (mod m).

This shows that the relation is transitive. Therefore, R is an equivalence relation.

Suppose that R is an equivalence relation on set A. The set of all elements that are related to an element of a of A is called the equivalence class of a. The equivalence class of a with respect to R is denoted by [a]R. When only one relation is under consideration, the notation is [a].

Example 2.21 :

What are the equivalence classes of 0, 1, –2, and –3 for congruence

modulo 4! Answer :

[0] = { . . . , – 12, – 8, – 4, 0, 4, 8, 12, . . . }

[1] = { . . . , – 11, – 7, – 3, 1, 5, 9, . . . } [–2] = { . . . , – 10, – 6, – 2, 2, 6, 10, . . . } [–3] = { . . . , – 11, – 7, – 3, 1, 5, 9, . . . }

A relation R on a set S is called partial ordering or partial order if

it is reflexive, antisymmetric and transitive. A set S together with a partial ordering R is called partially ordered set or poset and is denoted by (S, R).

Example 2.22 :



Show that the relation ‘’ is ordering relation on Z.

Answer :

Since a a for every a Z, the relation ‘’ is reflexive.

If a b and b a, then a = a. Thus, the relation ‘’ is antisymmetric.

If a b and b c, it means that a c. Thus, the relation ‘’ is

transitive.

Therefore, relation ‘’ is an ordering relation on Z.

Eah element in poset (S, ) is called comparable if either a b or b a for

every a, b S. Further, if (S, ) is a poset and every two elements on S are

comparable, S is called totally ordered set (toset) or chain, while is called a total ordering.

Example 2.23 :

1. ( N, ) is a toset. Sebelumnya blm dijelaskan apa itu toset

2. ( N, | ) is not a toset because it is not comparable.

If (S, ) is a toset and each subset, which is not empty subset, of S et

least has one element, (S, ) is called a well-ordered set.

Every partial ordering can be represented using Hasse diagram. The steps in constructing the Hasse diagram of a poset ar:

Present the ordering using directed graph.

Delete all loops (because the ordering is reflexive)

Delete all transitive edges

Example 2.24 :

Draw the Hasse diagram representing a poset ({1,2,3,4}, = {(a, b) |

a < b}}



Answer :

2.5 Function

Let A and B be sets. A function f from A to B is a rule that assign each element of set B to particular element of set A. It is denoted by f(a) =

b, if b is an element of B that is assigned by f to a particular a of A. This means that if f(a) = b and f(a) = c, then b = c. Suppose that f is thefunction from set A to set be B, the statement can be

denoted by:

f : A B It means that f assigns set A to set B.

A is called the domain of f and B is the codomain of f. Another term for function is mapping or transformation.

Suppose that f(a) = b, b is called the image of a and a is the pre-image

of b. A set that contains all values of the mapping f is called the range of f. Notice that the range of f is the subset (perhaps the proper subset) of B.

4

3

2

1

4

3

2

1



Example 2.25 :

Suppose that f : R R is defined by f(x) = x2.

The domain and codomain of f is a set of Real numbers, while the range of f is a set of non-negative Real numbers.

Example 2.26 :

Below is an example of a relation, but it is not a function:

Below are some examples of representing function: 1. Set of Ordered pair

Suppose that f is a square function of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, the funtion can be denoted by f = {(2, 4), (3, 9)}

2. Assignment

Example 2.27 : f(x) = x2 + 10, f(x) = 5x,

a b

A B

f

b = f(a) a

a 1

A B

2

3

b

c

cd 4



3. Words

Example 2.28 : “f is a function that assigns the sum product of integers to its square”. 4. Source Code Example 2.29 :

Counting function |x| (the absolute value of). function abs(x:integer):integer; begin

if x > 0 then abs := x else

abs := –x; end;

Suppose that g is the function from set A to set B, and f is the function from set B to set C. The composition of the function f and g,

denoted by f g, is the function from A to C and is defined by:

(f g)(a) = f(g(a)), for an a of A. Pay attention to the following illustration of the compoditiond of functions

below:

1

2

3

2

4

6

8

s

t

u

Example 2.30 :

Let f : Z Z and g : Z Z. given the function f(x) = x + 1 and

g(x) = x2, find f g and g f!

Answer :

(i) (f g)(x) = f(g(x)) = f(x2 ) = x2 + 1 .

a

b

c

g f

A B

C



(ii) (g f)(x) = g(f(x)) = g(x + 1) = (x + 1)2 = x2 + 2x + 1.

A function f form set A to set B is said to be one-to-one or injective, if

set A do not have two elements with similar reflection in set B.

Example 2.31 :

Suppose that f : Z Z and g : R R.

Determine whether the functions f(x)=x2 and g(x)=x+1 are one-to-one!

Answer :

a. Function f(x) = x2 is not one-to-one since f(2) = f(–2) = 4, while –2

2.

b. Function g(x)=x+1 is one-to-one sincesince for a b, a +1 b+1. Suppose that for x = 1, g(1)=2. Meanwhile, for x=2, g(2) = 3.

A function f form set A to set B is said to be onto or surjective, if every element of set B is the the reflection of one or more of the elements of set A. In other words, all elements of set B are the range of f. The function f is

called the onto function of set B. Example 2.32:

Let f : Z Z and g : R R. Determine whether f(x) = x2 and g(x) = x + 1 are onto functions!

Answer :

a. Function f(x) = x2 is not onto since not all integers are the range of f, like the negative integers.

b. Function g(x) = x +1 is onto since for for every real number y

there is a real number x such that y = x + 1.

A function f form set A to set B is said to beone-to-one correspondence, or

bijection, if it is both one-to-one and onto.

Pay attention to the following illustration for better understanding of the above explanation.



A one-to-one function, An onto function, but not onto pada but not one-to-one

Suppose that f is a function from set A to set B that is one-to-one

correspondence, we will always be able to find the invers function of f. Inverse function is denotted by f –1. Suppose that a is an element of set A and b is an element of set B, then f -1(b) = a jika f(a) = b. A one-t-one

correspondence is called invertible since we can can define an invers of this function. A function is not invertible if it is not a one-to-one correspondence, since the inverse of such a function does not exist.

Example 2.33 :

What is the invers of functionf(x) = x + 1?

Answer : Function f(x) = x + 1 is a one-to-one correspondence function that we can find its inverse.

Suppose that f(x) = y, that y = x + 1, then x = y – 1. Thus, the inverse function is f-1(y) = y – 1.

Example 2.34 : What is the invers of function f(x) = x2.

Answer :

Like the previous example, f(x) = x2 is not a one-to one corespondence that the inverse does not exist. Thus, f(x) = x2 is not invertible.

Recursively Defined Functions

a1

AB

2

3b

c4

a1

AB

2

3

b

c

cd



A function is a specific form of a relation. A function is called

recursive if it is defined in terms of itself. The component that construct the recursively defined function are:

1. Basis Value This component is the innitial value of the function.

2. Recurence This component definies the argument of the function in terms of itself.

An example of simple recursively defined function is factorial formula. Refer to the formula for factorial function:

a. Basis Value

n! = 1, untuk n = 0 b. Recurence

n! = n x (n-1) !, for n 1.

Therefore, when we try to define the value of function 4!, then: 4 ! = 4 . 3 ! = 4 . 3 . 2! = 4 . 3 . 2 . 1! = 4 . 3 . 2. 1 . 0! = 24

In an algorithm, usually the appearance of a recursively defined function occurs in a looping. Suppose that the function when t = 0 is a, then function f(k) = 2 . f(k – 1) + 3. Thus, in simple way, the variable of recursively defined function is a function of the previous iteration. This is the reason why

we need an innitial value of the function.



Summary

1. Rule that relate two sets is called binary relation.

2. Binary relation R between set A and set B is the subset of a cartesian

product A B or R (A B).

3. The notasi of a binary relation biner is a R b or (a, b) R.

4. Relation can be presented using arrows, ordered pairs, table, matrix, and directed graph.

5. If R1 and R2 are the relations from set A to set B, then R1 R2, R1 R2,

R1 – R2, and R1 R2 are alos the relation from A to B.

6. A relation R on a set A is called reflexive, if (a, a) R for every a A.

7. A relation R on a set A is called transitive, if (a, b)R and (b, c) R,

then (a, c) R, for a, b, c A.

8. A relation R on a set A is called symmetric, if (a, b)R, for every a, b

A, then (b, a) R.

9. A relation on a set A is called equivalence, if it is reflexive, symmetric and transitive.

10. A function f from A to B is a rule that relate the each element of set A with each element of B.

11. A function f from set A to set B is calledone-to-one, or injective, if set A

doesnot have two elements that have similar reflections in set B. 12. A function f from set A to set B is said onto, or surjective, if each

element of set B is the reflection of one or more elements of set A.

13. A function f from set A to set B is said tobe one-to-one correspondence, or bijection, if it is both one-to-one and onto.



True or False Quiz

(For questions no. 1 – 5) Given that A = {2, 3, 5, 7}, the relation on A is defined in sets:

R = {(2, 2), (3, 3), (3, 5), (5, 3), (7, 3), (7, 7)}

11. R is reflexive 12. R is not symmetric 13. R is transitive 14. R is not antisymmetric

15. R R2 16. The relation of ‘the factor of’ of set A = {1, 2, 4, 8} is transitive.

17. A relation that is not symmetric is called antisymmetric. 18. Equivalence relation is a relation that is symmetric and transitive. 19. If each element of set B is the reflection of one or more elements of set

A, the function from A to B is injective. 20. Let R and S be the relations of A = {a, b, c}. If R and S are symmetric,

then R S is symmetric.



Multiple Choices

1. An equivalence relation must meet the following properties:

A. Reflexive and symmetric D.

Reflexive, antisymmetric,

transitive

B. Reflexive, symmetric, transitive E. Reflexive, symmetric , antisymmetric

C. Symmetric and transitive

2. Function f(x) = x3 is

A. Injectie Function D. Recursive Function

B. Surjective Function E. Antisymmetric Function

C. Bijective Function

3. A relation defined by “a R b jif and only if + b Z”, meet the following

properties, EXCEPT:

A. Reflexive D. Anti Symmetric

B. Recursive E. Transitive

C. Symmetric

4. If relation R = {(1,2)} then R R-1 is:

A. not transitive D. not symmetric

B. reflexive E. Recursive

C. not antisymmetric

5. Given that U1 = 5 and Un = 2Un-1+ 3, then U4 = ....

A. 29 D. 73

B. 58 E. 0



C. 61

6 A relation is called partial ordering if it satisfies the following properties:

A. Reflexive and symmetric D. Reflexive, antisymmetric, and transitive

B.

Reflexive, symmetric, and

transitive E.

Reflexive, symmetric , and

antisymmetric

C. Symmetric and transitive

7 Suppose that R is a relation of a finite set. If the matrix of the relation is

an identity matrix, the relation is:

A Reflexive, but not transitive D Symmetric and transitive

B Recursive and transitive E Reflexive and total ordering

C Reflexsive, but not symmetric

(For questions no. 8 – 10) Let R and S be the relations of A = {1, 2, 3},

R = {(1,1), (1,2), (2,3), (3,1), (3,3)} and S = {(1,2), (1,3), (2,1), (3,3)}

8 The sets of relation Rc = ....

A {(1,3), (2,1)} D {(1,3), (2,2), (3,2)}

B {(1,3), (2,1), (2,2), (3,2)} E {(1,1), (2,2), (3,3)}

C {(1,2), (3,3)}

9 The sets of composition relation R S = ....

A {(1,3), (2,1), (2,2), (3,2)} D {(1,1), (1,3), (2,2), (3,2)}

B {(1,3), (2,2), (3,2)} E {(1,1),(1,2),(1,3) (2,3), (3,2), (3,3)}

C {(1,2), (1,3), (2,3), (3,2)}

10 The sets of composition relation S2 = ....

A {(1,3), (2,1), (2,2), (3,2)} D {(1,1), (1,3), (2,2), (3,2)}

B {(1,3), (2,2), (3,2)} E {(1,1),(1,2),(1,3) (2,3), (3,2), (3,3)}

C {(1,2), (1,3), (2,3), (3,2)}



Exercise

(For questions no. 1 – 3) Let A = {a, b, c, d, e} and B = {x, y, z}. Let R be the relation of A to B in the:

R = {(a, y), (a, z), (c, y), (d, x), (d, z)} 1. Find the matrix for relation R 2. Draw the directed graph for relation R

3. Find the inverse relation of relation R 4. Find whether the following relations (in the forms of ordered pairs) are

equivalence:

a. {(0,0), (1,1), (2,2), (3,3) } b. {(0,0), (1,1), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3) } 5. Suppose that the matrix of a relation is represented as:

MR =

111

110

111

Find whether the relation is reflexsive, symmetric, antisymmetric, and transitive.

6. Proove that the relation represented in the following matrix is equiivalence relation.

MR =

1000

0111

0111

0111

7. Suppose that a relation R is represented in the following matrix:



1000

0100

1010

0111

RM

Find whether the relation is an ordered relation!

8. Let A be the set of integers taknol and R is the relation on the set A X A, which is defined by:

(a, b) R (c, d) if ad = bc Show that R is equivalence relation!

9. Suppose that a relation R is represented in the following matrix:

1000

0100

1010

0111

RM

Find two matrixes that represent the inverse relations R–1

and che omposition of R R–1 !

10. Draw the Hasse diagram of poset {B , }

in which B = {1, 2, 3, 4, 6, 8, 12} and = {(a,b) | b is divisible by a}}


Kombinatorik 51 PAGE 10

3 COMBINATORICS

Overview

Combinatorics is animportant part of discrete mathemathics. This chapter

will discuss the techniques of counting, permutation, and comniation. One of the advantages of counting technique is to determine the complexity in an algorithm. It is expected that basic knowledge about combinatorics will

provide basis in understanding optimization or in developing or using the applications of combinatorics related to computerization.

Objective

1. Students understand the basic concept of combinatorics.

2. Students are able to differentiate between permutation and combination.

3. Students are able to solve various problems related to combinatorics.


52 Kombinatorik PAGE 10

3.1 Basic Principle of Counting

Two basic principles used in counting are the sum and product rules.

The Sum Rule

If a set A is devided into subsets A1, A2, …, An, the number of the

elements of set A will be the same with the number of all elements

of every subset A1, A2, …, An.

Indirectly, according to the sum rule, the subsets A1, A2, …, An do not

overlap (disjoint). The sum rule does not apply to sets that overlie one

another t. Such case must be solved using the principle of inclusion-exclusion

that will be explained in the next section.

Example 1 :

An elementary teacher in a village teaches 4th, 5th, and 6th graders. If

the number of students in grade 4 is 25, the number of students in

grade 5 is 27, and the number of students in grade 6 is 20, the total

number of students taught by the teacher is 25 + 27 + 20 = 72

students.

Example 2 :

A university students wants to by a motorcycle. He has to choose one

type over three kinds of motorcycle. Honda offers 3 types of

motorcycles, Suzuki offers 2 types of motorcycles, and Yamaha

offers 2 pili types of motorcycles. Thus, the number of choices for the

student is 3 + 2 + 2 = 7 choices.

The Product Rule

Suppose that a procedure can be broken down into two tasks.

There are n1 ways to do the first task and n2 ways to do the

second task after the first task has been done. Thus, there are (n1 x

n2) ways to do the procedures.



In product rule, idirectly, the sets operated do not have to be disjoint.

Example 1:

How many strings are there of length 7 out of two bits (0 and 1)

Answer :

Each of the seven bits has two possibilities, i.e. 0 or 1.

Terefore, the product rule shows there are a total of

2 x 2 x 2 x 2 x 2 x 2 x 2 = 27 = 128 different bit strings of

length seven

Example 2 :

An elementary teacher in a village teaches students of grade 4, 5, and 6. Suppose that the number of students in grade 4 is 25, the number of students in grade 5 is 27, the number of students in

grade 6 is 20. If the teacher wants to choose three students, one from each class, out of all students he taught, the teacher has 25 x 27 x 20 = 13.500 ways in choosing the three students.

Example 3 :

How many odd numbers are there between 1000 and 9999

(including 1000 and 9999) in which (a) all numbers are different (b) there are repeated numbers.

Answer :

(a) the first digit from the right: 5 possible numbers (they are 1, 3, 5,

7 and 9); the fourth digit from the right: 8 possible numbers (1 to 9 except

numbers that have been chosen)

the third digit from the right: 8 possible numbers the second digit from the right: 7 possible numbers Thus, the total number of odd numbers is (5)(8)(8)(7) = 2240.



(b) the first digit from the right: 5 possible numbers (they are 1, 3, 5, 7 and 9);

the fourth digit from the right: 9 possible numbers (1 to 9)

the third digit from the right: 10 possible numbers (0 to 9) the second digit from the right: 10 possible numbers (0 to 9)

Thus, the total number of odd numbers is (5)(9)(10)(10) = 4500. Example 5 :

A login password of a computer system is five to seven characters long. Every character may be a letter (can be upper or lower cases)

or number. How many passwords are there for a login ?

Answer :

The number of alphabets is 26 (A – Z) and the number of numbers

is 10 (0 – 9). Thus, the total number of the characters is 36.

The number of possible 5-character passwords is

(36)(36)(36)(36)(36) = 365 = 60.466.176,

the number of possible 6-character passwords is

(36)(36)(36)(36)(36)(36)(36) = 366 = 2.176.782.336,

and The number of possible 8-character passwords is

(36)(36)(36)(36)(36)(36)(36)(36) = 367 = 78.364.164.096.

Thus, the total number of all possible passwords is:

60.466.176 + 2.176.782.336 + 78.364.164.096 =

80.601.412.608.

In other words, one login will have 80.601.412.608 possible passwords.



When two processes are done at the same time, the sum rule

cannot be used to count the number of ways of choosing one out of the two

processes. Such case must be solved by using the principle of inclusion-

exclussion.

Example :

How many bytes of 8 bits are there if the byte starts with ‘11’ or

ends with ‘00’?

Answer :

Suppose that A is a set of byte starting with ‘11’, and B is a set of byte ending with ‘00’,

A B is a set of byte starting with ‘11’ and ending with ‘00’,

and

A B is a set of byte starting with ‘11’ or ending with ‘00’.

The number of possible byte that can be arranged on set A is (1)(1)(2)(2)(2)(2)(2)(2) = 26.

It can be denoted by A = 26 = 64

Meanwhile, the number of possible byte that can be arranged on the set B is (2)(2)(2)(2)(2)(2)(1)(1) = 26.

Thus, B = 26 = 64,

Using similar ways, the number of possible byte that can be arranged

on the set A B is (1)(1)(2)(2)(2)(2)(1)(1) = 24

that A B = 24 = 16. Then,

A B = A + B – A B = 64 + 64 – 16 = 112.

Therefore, the number of 8-bit bytes that starts with ‘11’ or ends with ‘00’ is 112.



3.2 Permutations and Combinations

Permutations

A Suatu permutation is possible arrangement done according to

certain order. In other words, permutation is a specific application of

product rule. Let A be a set with n, number of elements. The ordered arrangement that consits of r number of elements is called r-permutation of A and is denoted by P(n, r). To make this clearer, pay attention to this

explanation: If r > n, it is obvious that P(n, r) = 0 becuase it is impossible to

arrange r elements of A that only contains n number of elements in which n < r.

If r n,

The first element of the permutation can be chosen in n ways, since there are n elements in the set. There are n – 1 ways to choose the second element of the permutation, since there are n – 1 elements left in the set

after using the element picked for the first posisiton. Similarly, there are n – 2 ways to choose the third element, and so on until there are exactly n – r + 1 ways to choose the r element. Consequently, by the product rule, there

are n(n – 1) (n – 2) … (n – r + 1) r-permutation of the set.

Therefore, r-permutation of n objects is the number of possible order of r

objects chosen from n objects, with r n, and for each possible arranggement of r objects there will be no similar arragement of objects.

This can be expressed by:

P(n, r) = n(n – 1) (n – 2) … (n – r + 1)= )!(

!

rn

n

Example 1 :

Suppose that S = {p, q, r}. How many wasy are there to arrange two

letters on S that there will not be any similar arrangement? Answer :

The possible arragements of two letters are:

pq, pr, qr, qp, rp, rq Thus, the arrangement can be done in six ways. In this case, the definition of permutation can be used. It states that:



6

1

1.2.3

!23

!3)2,3(

P

By using the definition of permutation, the arrangement can be done in

six ways.

Example 2 :

Suppose that we have five balls with different colors and three boxes. We will put the balls into the boxes. Each box is only for one ball. How many order of different-clolor balls that can be made by placing

the balls into the boxes? Answer :

Box 1 can be filled with one of the 5 balls (there are 5 choices);

kotak 2 dapat diisi oleh salah satu dari 4 bola (there are 4 choices); kotak 3 dapat diisi oleh salah satu dari 3 bola (there are 3 choices). The number of different orders is = (5)(4)(3) = 60

By using the definition of permutation, then:

60

1.2

1.2.3.4.5

!35

!5)3,5(

P

Combination Suppose that r is an element of infinite negative integers. An r-

combination of n different elements of a set B is the number of subsets of set

B with r elements. Another interpretation of combination is ordering or choosing r out of n elements.

Example 1 : Suppose that A = {p, q, r }, find all subsets of set A with cardinality of 2.

Answer :



The subsets are: {p, q}, {p, r}, and {q, r}.

Thus, the combinations are: pq, pr, and qr

In the case of sets, the order of element is not considered. Therefore, 2 combinations of th set A (combining two letters without considering the

order) is 3, i.e., pq, pr, and qr. This is different with permutation that considers order. Using permutation, there are six ways in combining process the letters, i.e., yaitu pq, pr, qr, qp, rp, and rq.

Example 2 :

Suppose that there are 2 balls with the same color and and 3 boxes.

The balls will be put in the boxes in such a way that each box will contain only one ball. How many ways are there to put the balls into the boxes ?

Answer :

Suppose that the boxes are placed horizontally, one box is placed next

to another. There will be 3 ways to put the 2 balls into the boxes. They are: First method : the first and second balls are put into the first and

second boxes. Second method: the first and second balls are put intothe first and

third boxes.

Third method : the first and second balls are put into the second and third boxes.

Generally, the number of ways to put r balls of the same color into n boxes is:

)!(!

!

!

))1()...(2)(1(

rnr

n

r

rnnnn

This is the formula of combination, that is denoted by C(n,r) or

r

n

Suppose that there are n balls (some balls are differnt in color and

some are similar in color). They will be put into n boxes. If the compostions

of the balls are: n1 balls are colored 1,



n2 balls are colored 2,

nk balls are colored 1k bola diantaranya berwarna k,

then n1 + n2 + … + nk = n.

How many ways are there to put the n balls into the boxes (each box contains only one ball?

If all the n balls are considered different in color, there number of ways

to put the n balls into n buah boxes is: P(n, n) = n!. Therefore,

there are n1! ways of putting balls having color 1 there are n2! ways of putting balls having color 2

there are nk! ways of putting balls having color k

The permutation of n balls in which n1 consists of some balls having

color 1, n2 consists of some balls having color 2, …, nk consists of some balls havin color k is:

!!...!

!

!!...!

),(),...,,;(

212121

kkk

nnn

n

nnn

nnPnnnnP

Other method:

There are C(n, n1) ways to put n1 balls which color is 1. There are C(n – n1, n2) ways to put n2 balls which color is 2. There are C(n – n1 – n3, n3) ways to put n3 balls which color is 3.

There are C(n – n1 – n2 – … – nk-1, nk ) ways to put nk balls which color

is k. The total number of ways to put the balls into the boxes is:

C(n; n1, n2, …, nk) = C(n, n1) C(n – n1, n2) C(n – n1 – n2 , n3)

… C(n – n1 – n2 – … – nk-1, nk)

= )!(!

!

11 nnn

n

)!(!

)!(

212

1

nnnn

nn

)!(!

)!(

213

21

knnnnn

nnn

… )!...(!

)!...(

121

121

kkk

k

nnnnnn

nnnn

= !!...!!

!

321 knnnn

n

Conclusion:



!!...!

!),...,,;(),...,,;(

212121

kkk

nnn

nnnnnCnnnnP

Combinations with Repetition Suppose that there are r balls with the same color and n boxes.

Each box must contain only one ball. The number of ways to put the balls into the boxes is C(n, r).

If each box may contains more than one ball (no limitation on the

number of balls), the number of ways to put the balls into the boxes is:

C(n + r – 1, r) = C(n + r –1, n – 1).

For example :

Twenty apples and 15 oranges are going to be shared out to 5

children. Each child may have more than one aple or orangge, or none of them. How many ways are there to share out the aples and oranges?

Answer :

n = 5, r1 = 20 (apel) and r2 = 15 (jeruk)

Sharing out 20 apples to 5 children: C(5 + 20 – 1, 20) ways, Sharing out 15 oranges to 5 children: C(5 + 15 – 1, 15) ways. The total number of ways to share out the apples and oranges is:

C(5 + 20 – 1, 20) C(5 + 15 – 1, 15) = C(24, 20) C(19, 15)

Binomial Coefficients

Suppose that n is postive integers, using biinomial theorem, the power function of (x + y)n can be represented in Pascal’s triangle as follows: (x + y)0 = 1

(x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 Generaly, it will result in the following formula:

(x + y)n = C(n, 0) xn + C(n, 1) xn-1 y +…+ C(n, k) xn-k yk +…+ C(n, n)yn

= kknn

k

yxknC

0

),(

C(n, k) is the coefficient for x(n–k)yk and it is called binomial coefficient.



Example :

What is the expansion of (2x + y)3. Answer : Let a = 2x and b = y, (a + b)3 = C(3, 0) a3 + C(3, 1) a2b1 + C(3, 2) a1b2 + C(3, 3) b3

= 1 (2x)3 + 3 (2x)2 (y) + 3 (2x) (y)2 + 1 (y)3 = 8 x3 + 12x2 y + 6x y2 – y3

Example : Find the expansion of (2x – 3)3.

Answer :

Let a = 2x and b = –3, (a + b)3 = C(3, 0) a3 + C(3, 1) a2b1 + C(3, 2) a1b2 + C(3, 3) b3

= 1 (2x)3 + 3 (2x)2 (–3) + 3 (2x) (–3)2 + 1 (–3)3

= 8x3 – 36x2 + 54x – 27 Example :

Find the fifth order from the expansion of (x – y)5. Answer :

(x – y)5 = (x + (–y))5.

The fifth order obtained from the expansion is: C(5, 4) x5– 4 (–y)4 = –10 x y4.



Summary

1. Two basic principles of counting are the sum and product rules.

2. A permutation is a possible arrangement according to certaine order. 3. Suppose that B has n different elements. The r-combination of the set B

is the sum of the subset of B that contains elements of r objects.

4. The formula of r-permutation of n objets is:

P(n, r) = n(n – 1) (n – 2) … (n – r + 1)= )!(

!

rn

n

5. The formula of r-combination of n elements is denoted by C(n,r) =

)!(!

!

rnr

n

r

n

6. On polynom (x – y)n, the number C(n, k) is coeficient for x(n–k)yk and is called binomial coefficient.



True or False Quiz

21. The sum principle is the same with the product principle. 22. The value of 5! is 120.

23. A permutation is an arragement that is possible to be made without regarding order.

24. Choosing the formation of 3 futsal teams out of 15 players can be done

using combination principle. 25. The value of P(5, 3) is 15. 26. The value of C(5, 2) is 10 27. (2x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

28. P (10, 5) = C (10, 2) 29. Permutation can counted by using product rule. 30. C(n + r – 1, r) = C(n + r –1, n – 1)



Multiple Choices

1. Suppose that there are 3 male lecturers and 2 female lecturers. A student can choose one out of those five lecturers in ….

A. 1 way D. 5 ways B. 2 ways E. 6 ways C. 3 ways

2. How many ways are there for a set of 50 students to choose one chief,

one secretary, and one treasurer, in condition that no student has more than one position?

A. 25 ! ways D. 3 ways B. 25 ways E. 25 x 24 x23 ways

C. 25x3 ways

3. ....

2

8

A. 2 D. 56

B. 8 E. 128

C. 28 4. A farmer bought 3 cows, 4 sheep, and 5 chickens from a person who

have 4 cows, 5 sheep, and 7 chickens. How many are there for the farmer to choose the three kinds of animals?

A. 420 ways D. 7 ways

B. 140 ways E. 1 way

C. 35 ways

5. There are 9 kinds of toys to be distributed to 4 children. Each child will



be given 2 kinds of toys, except the youngest. The youngest will be given

3 kinds of toys. How many ways are there to distribute the toys?

A. !9 ways D. )!2!.2.!3(/!9 ways

B. !2/!9 ways E. )!2.!2!.2.!3(/!9 ways

C. )!3.!2(/!9 ways

Excercises

1. Find the value of :

a.

13

15 b.

9

11

2. Find the value of: a. P(6, 3) b. C(5, 1)

3. Show that

6

16

5

16

6

17

4. Find n if : a. P(n,2) = 72 b. P(n,4) = 42 P(n,2)

5. Twenty students will be divided into three teams. How many formation

of teams are there?

6. There are five people who are attending a seminar. How many wasy are there for the pople to take a seat if:

a. 5 seats are placed in one row. b. 5 seats are placed in circle behind a round table.

7. ‘Duny donut’ store offers four different kinds of donuts (the stock of

each kind is 10 donuts). How many ways are there to choose six donuts?

8. How many strings of lenght ten are there that can be formed out of two

bits (0 and 1), that contain exactly seven 1s.



9. There 12 racers (it is assumed that all racers are able to reach the finish

line) in a horse race Dalam suatu pacuan kuda dengan 12 peserta. How many possible arrangement of winners (first, second, and third) are there at the end of the race?

10. Using binomial theorm, find: a. coeficient x5y8 in (x + y)13 b. coeficient x7 in (1 + x)11

c. coeficient x9 in (1 – x)19


Teori Graf 67 PAGE 10

4 GRAPHS THEORY

Overview

Graphs is applied to solve different kinds of problems, for example in determining the shortest path related to communication as well as transportation, frequency assignment in telecommunication, schedule

optimation, and so on. The explanation of graphs in this chapter covers definition and terminology related to graphs, essential problems of the shortest path, which are used in application.

Objective

1. Students understand the concept and terminology related to graphs. 2. Students model problems in the form of graphs.

3. Students are able to solve problems related to graphs theory.


68 Teori Graf PAGE 10

4.1 Definiton of Graphs

Graphs is a discrete structure consisting of a collection of vertices or ‘nodes’ and a collection of edges that connec pairs of vertices. Graphs is used

to represent discrete objects and the relationship among these objects. A notation of a graphs is G = (V, E), in which:

V is a set, whose elements are called vertices or nodes, i.e. V

= { v1 , v2 , ... , vn } E is a set of pairs (unordered) of distinct vertives called edges

or line, i.e. E = {e1 , e2 , ... , en }

For example:

Graphs of Königsberg bridge problem is presented below:

Suppose that the graph is called G(V, E) with: V = { A, B, C, D }

E = { (A, C), (A, C), (A, B), (A, B), (B, D), (A, D), (C, D)} = { e1, e2, e3, e4, e5, e6, e7}

On the graph above, edge e1 = (A, C) and edge e2 = (A, C) are called multiple edges or paralel edges as both of the edges connect two same vertices: vertex A and vertex C. Edge e3 and edge e4 are likewise. However, there is no loop on it. Loop is an edge (directed or undirected)

which starts and ends on the same vertex.

e2

e3 e4 e5

e6

e7 e1

B

A

C

D



Considering graph definition, edge (E) collection enables null

collection. If a graph contains null collection, it is called null graph or empty graph. For example : Null graph with 3 vertices (graph 3)

By considering graph’s edges, a graph is classified into directed and undirected graph. Undirected graph has been explained on the Bridge of

Königsberg’s example. Directed graph or digraph is a graph having directed

edges. It means a vertex connected by the edges is an initial vertex and other vertex is called terminal vertex.

Several worth knowing graphs are shown below: I. Simple graph is an undirected graph with no loops and multiple

edges

For example : Simple graph

Furthemore, the word “graph”in this book represents simple graph unless there is additional word such as pseudograph, directed graph,

etc.

2. Multigraph

v1

v2 v3

P

S Q

R



Multigraph is an undirected graph without loop.

For example :

Multigraph

Thus, a simple graph is a multigraph as well.

3. Pseudo graph

Pseudo graph is a graph which is allowed to contain loop.

For example :

Pseudo graph:

P

S Q

R

P

S Q

R



4. Directed graph or digraph

Directed graph or digraph is a graph in which its edges have direction but do not have two contradicted edges between two vertices. In other words, they do not have multiple edges.

For example :

a. Directed graph or digraph:

b. Directed multigraph Directed multigraph is a graph which allows multiple edges on it. (It is allowed to have contradicted edges between two vertices).

Based on the explanation above, we can make conclusions as follow :

P

S Q

R

R

P

S Q



Table 4.1 Types of Graphs

Types Edges Multiple edges

Loop

Simple graph Multigraph

Pseudograph

Directed graph Directed multigraph

Undirected Undirected Undirected

Directed Directed

Disallowed Allowed Allowed

Disallowed Allowed

Disallowed Disallowed

Allowed

Allowed Allowed

For example : The following graph is a directed graph:

It can be seen that e1 = (P, S), e3 = (R, Q), and e5 = (Q, Q) Vertex P is an initial vertex for edge e1 and vertex S is a terminal

vertex for edge e1.

4.2 Graph Terminology

There are several graph terminologies which are worthy to know, such as: two vertices adjacency, incidency, vertex degree, etc.

1. Adjacent

Two vertices are said to be adjacent if they are directly connected by an

edge. For example:

e6

P

S Q

R

e1 e4

e3 e2



Observe the graph below:

On the graph above: vertex P is adjacent to vertex Q and vertex S. However, vertex P is not adjacent to vertex R.

2. Incidency Edge e is said to be incident to vertex v1 and v2 if e connects both of the

vertices. In other words, e = (v1, v2).

For example :

Observe the graph of Bridges of Königsberg as follow:

edge e1 is incident to vertex A and C, but this edge is not incident to vertex B.

3. Isolated Vertex

If a vertex does not have edge which is adjacent to it then the vertex is called isolated vertex.

For example :

P

S Q

R

e2

e3 e4 e5

e6

e7 e1

B

A

C

D



Observe the following graph:

Vertex T and U are isolated vertices

5. Degree A vertex’s degree is the number of edges which are adjacent to the vertex. For example, vertex v has 3 edges which are adjacent to it. Then, it can be

said that the vertex has the degree of 3 or it is denoted by d(v) = 3.

For example 1: Observe the graph presented below:

On the graph above : d(P) = d(Q) = d (S)= 5, while d(R) = 3.

The degree of a vertex on a directed graph can be explained as follow::

din(v) is the number of arc coming into the vertex v dout(v) is the number of arc coming out of the vertex v

Therefore, the degree of the vertex is:

d(v) = din(v) + dout(v)

P

S Q

R

T

U

P

S Q

R



For example 2 :

Observe the following directed graph:

On the graph above : din (P) = 1 and dout (P) = 3 then d (P) = 4

din (Q) = 4 and dout (Q) = 1 then d (Q) = 5 din (R) = 1 and dout (R) = 1 then d (R) = 2 din (S) = 1 and dout (S) = 2 then d (S) = 3

The number of all vertices’ degree on a graph is always even which is twice of the edges number. If G = (V, E) is a graph, then:

Evd

Vv

2)(

For example 3 :

Observe the graph on Example 1. It has 9 edges. Therefore, the degree is :

18

9.2

.2)(

Evd

Vv

or

18

3555

)()()()()(

SdRdQdPdvd

Vv

Observe the graph on Example 2. It has 7 edges, thus the number of degree is:

147.2.2)(

EvdVv

P

S Q

R



or

143254

)()()()()(

SdRdQdPdvdVv

Hence, we cannot draw a graph which the degre of its edges is known but its total degree is odd.

6. Path 31. Lane of initial vertex v0 to destination vertex vT on a graph is a series of

an edge or more (x0, x1), (x1, x2), (x2, x3), …, (xn-1, xn) on G, which x0 = v0 and xn = vT. A lane does not experienced edge repetition. Lane is denoted by vertices passed by it:

x0, x1, x2, x3, …, xn If a lane does not experience vertex repetition, it is called path. A path is said to have length n, if it contains n edges passed by from the initial

vertex v0 to destination vertex vT on a graph G. A lane which starts and ends at the same vertex is called Circuit. While, a path which starts and ends at the same vertex is called cycle.

For example : Observe the following graph :

On the graph above, the length of path P, Q, R is 2. The length of

path P, Q, S, R is 3. The path P, Q, R, S, P is called cycles with the length of 4. There is no path between vertex P, U and T.

The shortest cycle length on a simple graph is 3. It means that a cycle has to pass 3 edges. Meanwhile, the shortest cycle length on a pseudo graph is 1. It means that this cycle can be a loop. Graph’s diameter

is the longest path.

P

S Q

R

T

U



The graphs presented below are the most common ones: a. Complete Graph

A complete graph is a simple graph in which its vertices are connected to each other. In other words, its vertices are adjacent to each other.

A complete graph with n vertices is denoted by Kn. The number of edges on a complete graph with n vertices is n(n – 1)/2. Example :

K1 K2 K3 K4 K5 K6

Figure 4.3 Complete Graphs Kn, 1 n 6

b. Cycle Graph A cycle graph is a simple graph which each vertices degree is 2. Cycle graph with n vertex is denoted by Cn.

C3 C4 C5 C6

Figure 4.4 Cycle Graphs Cn, 3 n 6

c. Wheels Graph A wheels graph is made by adding a vertex to a cycle graph Cn and connecting the new vertex with all vertices on it.



W3 W4 W5

Figure 4.5 Wheels Graph Wn, 3 n 5

d. Regular Graphs

A regular graphs is a graph with each vertex has similar degree. If the degree of each vertex is r, then the graph is called regular graph with r

degree. The number of edges on a regular graph with n vertex is 2

nr .

Figure 4.5 Regular Graph with 3 Degree e. Planar Graphand Plane Graph

A graph which can be drawn on a flat plane without its edges

crossing each other is called planar graph. If it is not so, the graph is called non-planar graph. Several examples of planar graphs:

- All cycle graphs are planar graph. - Complete graphs K1, K2, K3, K4 are planar graphs.

However, a complete graph Kn with n 5 is non-planar graph.

Planar graph K4 illustration is presented below.

Figure 4.6 K4 is planar graph



Planar graph which is drawn without its edges crossing each other

is called plane graph.. Observe the following illustration on graph K4 to differentiate planar graph and a plane graph:

(a) (b) (c)

Figure 4.7 Three Planar Graphs. Graph(b) and (c) are plane

graph

Several features of planar graph G(V, E) are: (Formula Euler) For example G is planar graph connected to

edges e and vertices v, and r is the number of area on the graph, then r = e – v + 2.

If G is a planar graph connected to edges e and vertices v (v

3) then e 3v – 6 (unsimilarity Euler).

If G is a planar graph connected to edges e and vertices v (v

3) but it does not contain circuit with the length of 3 then e

2v – 4.

f. Bipartite Graph

A simple grap G is said to be bipartite graph if its vertices collection can be divided into two non-null disjoint collection V1 dan V2... Therefore, each edge on G connects a vertex on V1 and a vertex on V2. On

bipartite graph, there is no edge connecting two vertices on V1 or V2. Bipartite graph is denoted by G(V1, V2).



For example :

The following graph G is a bipartite graph:

The graph above can be represented by bipartite graph G(V1, V2), with V1= {a, b} and V2 = {c, d, e}

Bipartite graph representation based on the graph above is presented below:

Figure 4.7 Bipartite Graph

V1 V2

a

c

d

e

b

a

c

d

e

b



g. Labeled Graph

Labeled graph is a graph with each edge is given a label (weight).

Figure 4.8 Graph K5 with its labeled edges A graph can be labeled on its vertex. It depends on given label

representation.

4.3 Connection and Sub Graph

Vertex v1 and v2 on a graph are said to be connected if there is a path between them. If vertex vi and vj in V collection on graph G have a path from vi to vj then the graph is said to be connected graph. If it is not so, the graph

G is called disconnected graph.

For example 1 :

Wheel graph is one of connected graph:

For example 2 :

p

t

s r

q

8 9

10

13 12

11

7



Observe the following graph:

(i) (ii) (iii)

Obviously, (i) C3 and (ii) C4 are connected graph. Meanwhile, graph

(iii) is disconnected graph because there is no path connecting vertices on {p, q, r} and vertices on {a, b, c, d}.

Furthermore, we will study a connection on a directed graph. A directed graph is said to be connected if we eliminate its direction (undirected graph) so it is called connected graph. Two vertices, u and v,

on graph G are strongly connected if there is a directed path from u to v and vice versa. If u and v are not strongly connected or in other words the graph is only connected on its undirected one, then u and v are said to be

weakly connected. If each pair of vertices on a directed graph G is strongly connected, then graph G is called strongly connected graph. If it is not so, the graph is called weakly connected graph.

For example :

Directed graph which is strongly connected Directed graph which is weakly connected

For example G = (V, E) is a graph, then G1 = (V1, E1) is called

subgraph of G if V1 V dan E1 E. Complement of subgraph G1 to

graph G is graph G2 = (V2, E2) in such a way that E2 = E – E1 and V2 is

p

q r

a

b

c

d

a

b

c

d

p

q r

p

q r

p

q

r

c



vertices collection which E2 members are incident to it. For example, G1 =

(V1, E1) is a subgraph of graph G = (V, E). If V1 =V (G1 contains all G’s vertices) then G1 is called Spanning Subgraph Example :

(a) Graph G1 (b) Subgraph (c) Spanning subgraph

Figure 4.9 Subgraph and Spanning Subgraph of a Graph

4.4 Adjacency Matrix and Incidency Matrix of a Graph

On previous discussion, we have introduced that two vertices are said to be adjacent if both of them are directly connected by an edge.

Adjacency matrix of a simple graph is a square matrix in which its elements consist of two numbers: 0 (zero) and 1 (one). The line and column of the matrix represent each vertex of the graph. For

example, aij is element of the matrix, then: If aij = 1 then vertex i and vertex j are adjacent. If aij = 0 then vertex i and vertex j are not adjacent.

p

t

s r

q

p

t

r

q

p

t

s r

q



For example :

Observe a simple graph presented below:

Adjacency matrix of the graph above is presented below :

SRQP

S

R

Q

P

0111

1010

1101

1010

It is obvious that the matrix is symmetrical and its diagonal element

is zero (0).

Meanwhile, an edge e is said to be incident to vertex v1 and v2 if e connects both of them. In other words, e = (v1, v2). Like adjacency matrix,

the elements of incidency matrix consist of two numbers, 0 (zero) and 1 (one). However, it must not be a square because the line and column on incidency matrix represent vertex and edge of the mentioned graph. For

example, aij is the element of the matrix, then: If aij = 1, it means vertex i and edge j are incident. If aij = 0, it means vertex i and edge j are not incident..

For example :

Observe the graph presented below :

P

S Q

R



The form of incidency matrix of the graph is :

7654321 eeeeeee

D

C

B

A

1110000

1000011

0011100

0101111

4.5 Eulerian and Hamiltonian

4.5.1 Eulerian Circuit Eulerian circuit is a circuit passing each edge just once. A graph

containing Eulerian circuit is called Eulerian graph. While a graph which

contains an Eulerian lane is called semi-Eulerian graph. For example :

Observe the graph presented below :

p q

r s

t

G1

e2

e3 e4 e5

e6

e7 e1

B

A D

C



Graph G1 is an Eulerian graph because it has a lane forming a circuit

which is : pr–rt– ts – sq – qt – tp.

It can be seen that graph G2 is a semi-Eulerian graph because it has a lane passing by each edge just once. The lane is : pq – qs – st – tp – pr – rt – tq.

Several characteristics of Eulerian Graph and Semi-Eulerian Graph :

A graph is called Eulerian graph (having Eulerian circuit) only if each

vertex has even degree. A directed graph is called semi-Eulerian graph (having Eulerian lane)

only if the graph contains two vertices with odd degree.

A connected directed graph G is called Eulerian graph only if each vertex has the same input and output degree.

A connected directed graph G is called Semi-Eulerian graph only if G is

connected to each vertex and it has the same input and output degree except the first vertex (initial vertex) has output degree more than input degree and the second vertex (terminal vertex) has input degree

more than output degree. The degree differences is only one point.

4.5.2 Hamilton Circuit

In 1859, Sir Wiliam Hamilton created a game namely dodecahedron. He offered it to a toys factory in Dublin. The game consisted of 12 pentagon

and it had 20 angles (each angle is named after the capital of a country). It provided a travel around the world visiting each capital city just once and coming back to the starting point city. It is a Hamilton circuit searching. The

problem can be illustrated by the following graph.

p q

r s

t

G2



Figure 4.10 Hamilton Circuit of a Graph

On the illustration above, Hamilton circuit is denoted by bold type path. If the path is back to the initial vertex to make closed path (circuit), then it is

called Hamilton circuit. Therefore, Hamilton circuit is a circuit passing by each edge just

once. A graph containing Hamilton circuit is called Hamiltonian graph. Meanwhile, a graph containing Hamiltonpath is called semi- Hamiltonian graph.

Example :

Observe 3 graphs below:

p q

r s

G1 p q

r s

t

G3

t

p q

r s

G2



Graph G1 is a semi-Hamiltonian graph. Its Hamilton path is: s – r – p – q – r.

While graph G2 is a Hamiltonian graph. Its Hamilton circuit is t – p –

r – q – p – s – q – t . Meanwhile, there is no Hamilton path or circuit on graph G3 .

For example, G is a simple graph containing n edges (at least n=3). If each

vertex’s degree is at least n/2 vertex then graph G is a Hamiltonian graph. Several things about Hamiltonian graph:

Every complete graph is a Hamiltonian graph.

On a complete graph G with n vertices (n 3), there are 2

!1n

Hamilton circuits.

On a complete graph G with n vertices (n 3 and n is odd), there

are 2

1n Hamilton circuits which are independent. If n is even and

n 4, then on a graph G there are 2

1n Hamilton circuits which

are independent.

4.6 Isomorphic Graph

Observe the graphs presented below:

Figure 4.10 Hamilton Circuit of a Graph The two graphs above consist of 4 vertices which its degree is 3. Viewed from geometry, the two graphs are different. However, both of them are not

different based on the graph principles. It can be proved by pulling the middle vertex on the second graph. It will be the same as the first graph. They are called isomorphic. In addition to the graphs above, there are many

isomorphic graphs.



Two graphs are said to be isomorphic if there is one on one

correspondence between each vertex and edge. If edge e is incident to vertex u and v on G then edge e’ on G2 is also incident to vertex u’ and v’. Two graphs are said to be isomorphic if they meet the three following requirements:

a. They have similar vertices number. b. They have similar edges number. c. They have similar number of vertices which have certain degree.

To understand easily the concept of two isomorphic graphs, observe the following graph.

Example :

Check whether the graphs above are isomorphic. If yes, decide the

vertices which correspond between G1 and G2

Answer:

Yes, the two graphs are isomorphic. The graphs containing vertices in which each vertex has the degree of 3. The vertices which correspond to each other are:

Vertex u1 with vertex v1






u1 u2 u3

u4 u5 u6 G1

v1 v2

v6

v5 v4

v3

G2



Two isomorphic graphs have the same adjacency matrix. Surely, it is decided

after corresponding matrix is sorted out by the same sequence. Observe adjacency matrix of the two graphs. Here is the adjacency matrix of G1:

654321 uuuuuu

MG1 =

6

5

4

3

2

1

u

u

u

u

u

u

000111

000111

000111

111000

111000

111000

While, the adjacency matrix of G2 is presented below: 246531 vvvvvv

MG2 =

2

4

6

5

3

1

v

v

v

v

v

v

000111

000111

000111

111000

111000

111000

It can be seen that the two graphs have the same adjacency matrix. They are

MG1 = MG2.

4.7 Several Graph Applications

a. The shortest path and lane For example: G is a weighted graph. Each edge has certain weight as the following illustration depicts:

a b c

d e f

12

28

3 10 172

132

9 11

25

8

9

9

102



Figure 4.11 The Shortest Path and Lane Ilustration on a Graph

The shortest path from a to d is 22, with the path a – b – d. If we use the path a – d, a – e – d, dan a – b – c – d then the path weight will be 28, 26, dan 29.

A common thing is to decide the shortest path of the graph or to decide the path which has the minimum weight. For example:

Determine the shortest distance, the fastest time, the cheapest fare

between two cities. Determine the fastest time of sending a message between two

terminal on a computer network.

Several problems related to the shortest path are presented below:

The shortest path between two certain vertices.

The shortest path between all couples of vertices. The shortest path from certain vertex to all vertices. The shortest path between two vertices passing by certain vertices.

Dijkstra Shortest Path Algorithm Dijkstra Algorithm is an algorithm used to determine the shortest path from

a vertex to all vertices. To make it easier in understanding Dijkstra Algorithm, observe the graph below which its vertices represent the cities in the U.S. The graph edges represent the distance between two cities (in

kilometers) For example:

800

1200

1500

1000

1700

1000300

1400

250

900

1000

Boston(5)

New

York(6)

Miami(7)New

Orleans(8)

Chicago(4)

Denver(3)

Los

Angeles

(1)

San

Fransisco

(2)



Using Dijkstra Algorithm, determine the shortest distance from Boston to another cities.

Then, the shortest path from: 5 to 6 is 5, 6 with the distance = 250 km

5 to 7 is 5, 6, 7 with the distance = 1150 km 5 to 4 is 5, 6, 4 with the distance = 1250 km 5 to 8 is 5, 6, 8 with the distance = 1650 km

5 to 3 is 5, 6, 4, 3 with the distance = 2450 km 5 to 2 is 5, 6, 4, 3, 2 with the distance = 3250 km 5 to 1 is 5, 6, 8, 1 with the distance = 3350 km

b. Travelling Salesperson Problem (TSP)

It is the same as the example on (a), for example you are given a number of cities and their distances. Determine the shortest circuit which has to

be passed by a salesperson if s/he goes from a city and s/he has to stop by in each city just once and goes back to the city where s/he starts. It is a problem of determining Hamilton circuit with the minimum weight.

For example :

Determine the circuit having the shortest path from a complete graph K4:

a b

c d

20

8

25

10 15 18



The number of Hamilton circuit on a complete graph with n vertex: (n -

1)!/2. The graph above has (4 – 1)!/2 = 3 Hamilton circuits, they are:

Circuit 1 = (a, b, c, d, a) has the length of = 20 + 8 + 25 + 10 = 53

Circuit 2 = (a, c, d, b, a) has the length of = 18 + 8 + 15 + 10 = 51 Circuit 3 = (a, c, b, d, a) has the length of = 20 + 15 + 25 + 18 = 73 Then, the shortest Hamilton circuit is circuit 2 = (a, c, d, b, a) or (a, d, b,

c, a) with the circuit length of 51. c. Chinese Postman Problem

This problem was firstly proposed by Mei Gan (a Chinese) in 1962. It is about a postman who will deliver some letters to the addresses along the street of an area. How does he plan his traveling route in order that he

pass each street just once and go back to his starting point. It is a problem about determining Eulerian circuit on a graph.

For example : Determine the lane for the postman so that he can pass all the streets.

The lane passed by the postman is A, B, C, D, E, F, C, E, B, F, A.

a b

c d

20

8

25

10

a b

c d

8 10 15 18

a b

c d

20

25

15 18

B C

E F

7

5

4 A D

9

2

3

5

4 4

3



Summary

1. A graph is a discrete structure consisting of a collection of vertices or ‘nodes’ and a collection of edges that connect pairs of vertices.

2. Two vertices are said to be adjacent if they are directly connected by an edge.

3. An edge e is said to be incident to vertex v1 and vertex v2 if e connects

both of the vertices, in other words e = (v1, v2). 4. The vertex’s degree is the number of edges which are incident to the

vertex.

5. Lane of initial vertex v0 to destination vertex vT on a graph is a series of an edge or more (x0, x1), (x1, x2), (x2, x3), …, (xn-1, xn) on G, which x0 = v0 and xn = vT. A lane does not experienced edge repetition.

6. If a lane does not experience vertex repetition, it is called path. A path is said to have length n, if it contains n edges passed by from the initial vertex v0 to destination vertex vT on a graph G.

7. A lane which starts and ends at the same vertex is called Circuit. While, a path which starts and ends at the same vertex is called cycle.

8. Eulerian circuit is a circuit passing each edge just once. A graph

containing Eulerian circuit is called Eulerian graph. While a graph which contains an Eulerian lane is called semi-Eulerian graph.

9. Hamiltonian path of a graph is a path passing by every vertex just once. If

the path comes back to the initial vertex forming a closed path (circuit) then it is called Hamiltonian circuit.

10. Two graph are said to be isomorphic if they meet the three requirements below:

a. They have similar vertices number. b. They have similar edges number. c. They have similar number of vertices which have certain degree.



True or False Quiz

1. We cannot draw a simple graph with 5 vertices which each vertex’s

degree are 4, 3, 3, 2, 1 2. We can draw a simple graph with 5 vertices which each vertex’s

degree are 6, 4, 4, 2, 2

3. Every eulerian graph is a semi euler graph. 4. Graph K5 is Hamilton graph. 5. Graph C4 is bipartit graph.

6. A reguler graph with 3 degree and 4 vertices is a complete graph. 7. On a semi euler graph, all vertices have even degree. 8. W3 is a regular graph with vertex’s degree of 3.

9. The shortest path on a graph is the graph’s diameter. 10. Two isomorphic graph always have the same adjacency matrix.



Multiple Choices

1. Graph C4 is isomorphic with the graph….

A. K4 D. K3,3

B. W3 E. P4

C. K2,2

2. How many Hamiltonian circuits are there on a complete graph having 5 vertices?

A. 1 D. 4

B. 2 E. 5

C. 3

3. Which one is the non planar graph?

A. Graph C4 D. K2,2

B. Graph K4 E. K5

C. Graph W4

4. The number of edges on a complete graph is…

A. 21 D. 15

B. 20 E. 12

C. 18

5. The diameter of a wheel graph W10 is

A. 1 D. 4

B. 2 E. 5

C. 3



Exercises

1. Draw a graph containing five vertices. Each vertex’s degree is 2, 3, 4, 1,

and 3 !

2. Check the following graph whether it is an Eulerian graph, semi-Eulerian graph or its is none of them! (Explain it.)

(For question number 3 – 7) Observe the graph presented below :

3. Determine the graphs’ adjacency matrix.

4. Give an example of spanning subgraph of the graph above. 5. Mention a complete graph which is its subgraph of the graph above. 6. Determine the shortest lane of the label graph presented below :

u1 u2 u3

u4 u5 u6

b

c

d e

f g

6 2

3 4

1

5

6 2

4

a



7. Check whether the graph above is a Hamiltonian graph.

8. See the following graph :

Check if the graph above is a planar graph. If yes, write its plane graph. If no, explain its reason.

(For Question no. 9 – 13) Observe the following graphs :

G1 G2

a

b

c d

e g

f

h

j

i

o p q r

s t u v

a

c d

b a

e f t u

s r

q p



9. Are the two graphs above an Eulerian Graph, Semi Eulerian Graph or

not both of them? Explain it! 10. Check if graph G1 or G2 is a Hamiltonian Graph, Semi-Hamiltonian

Graph or not both of them. Explain it! Write one of Hamiltonian circuit or Hamiltonian path, if any.

11. Check if G1 or G2 is a complete graph or regular graph or not both of them. Explain it!

12. Check if G1 or G2 is a Bipartite graph. Explain it!

13. Is graph G1 isomorphic to graph G2? If it is not, explain its reason. If it is yes, explain and prove it by providing adjacency matrix.


100 Pohon dan Pewarnaan

Graf PAGE 10

TREE AND GRAPH COLORING

Overview

Tree is an important part on a graph theory which does not have a cycle. It is usually used in binary theory starting from binary tree expression up to

binary tree tracking. Tree application discussed in this chapter are expression tree and its use on decision tree and Hufman coding. Meanwhile, coloring is one of aplication on optimization field. Graph coloring discussion covers

vertex coloring and area coloring. Graph coloring is widely used in scheduling problem optimization to determine minimum color on a graph which represents scheduling problems.

Objective

1. Students understand the tree concept and graph coloring. 2. Students understand minimum application of spanning tree and

graph coloring.

3. Students are able to understand and solve problems and phenomena related to tree and graph coloring.


Pohon dan Pewarnaan Graf 101 PAGE 10

Tree is a special form of a graph structure. For example A is a

collection of vertices on a connected graph G. For each pair of vertices on A, we can determine a path connecting pairs of vertices. A connected graph with its pair of vertices that can be connected only by certain path is called tree. In other words, a tree is undirected graph which is connected but it

does not have cycle or circuit.

Figure 5.1 G1 and G2 are trees, G3 is not a tree

A forest is a collection of independent trees. In other words, a forest is a disconnected graph which does not have a circuit. Every

component in connected graph is the trees. Figure 6. 1 G4 is an example of a forest consisting of two trees.

Several trees’ characteristics are presented below :

For example G is a graph with n vertices and n – 1 edges. If G does not

have a circuit, then it is a tree.

A tree with n vertices has n – 1 edges.

Every pairs of vertices in a tree connects to a single path.

Suppose that G is a simple graph with n vertices. If G does not contain

circuit then one edge addition on the graph will make only one circuit.

G1 G2 G3

a b

c d

e f

a b

c d

e f

a b

c d

e f



Graf PAGE 10

4.8 Minimum Spanning Tree

Spanning Tree of a connected graph is a spanning subgraph in the

form of tree. Spanning tree is obtained by eliminating the circuit on the graph.

G T1 T2 T3 T4

Figure 5.2 Graph and Spanning Tree

It is obvious that T1, T2, T3, T4 are spanning trees of graph G. Bear in mind that every connected graph with the minimum weight has a spanning tree. A

tree having minimum weight is called minimum spanning tree. One of spanning tree applications is to determine series of streets with a very minimum total distance to connect all cities so that every city is connected

to each other. To determine a spanning tree of a connected graph, we can use two ways e.g. Prim algorithm and Kruskal algorithm. Prim algorithm has several steps below:

Choose a minimum weighted edge of the graph G. Insert it into T. Choose edge (u, v) on G which has a minimum weight and is

incident to vertex on T. This edge does not form a circuit on T.

Insert (u, v) into T. Repeat step 2 for n – 2.

The number of steps in Prim algorithm is equivalent to the number of edges in the spanning tree with n vertices which is (n – 1).



Example 5.1 :

Determine minimum spanning tree of the graph presented below :

Answer :

Choose edge fg so that we have T ({f, g}, fg)

Next steps can be choosing edge ef because this edge has a minimum weight and it is adjacent to vertex f.

Next, choose edge ae or as these edges have minimum weight and

they are incident to vertex T, which are e and g. If the process continues, we will obtain the following minimum spanning tree :

It can be noticed that the spanning tree has the total weight 2 + 3 + 4 + 4 + 4 + 4 + 3 = 24.

Kruskal algorithm steps are slightly different from Prim algorithm. In Kruskal

algorithm, all edges having minimum weight are put into T consecutively.

a

b

c

d

e

f

g

h

5

5

4

4

5

4 5

4

3 2

3

4

a

b

c d

e

f

g

h 4

4

4

3 2

3

4



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The steps to determine minimum spanning tree by Kruskal algorithm are as follows : Step I : T form is like the following tree

Step II : Inserting edges weighted 3 so the form of T is

Step III : Inserting edges weighted 4 so finally we get minimum spanning tree as the following picture:

4.9 Rooted Tree

A tree which its edges are given direction is similar to directed graph.

All vertices connected to all vertices on the tree are called root. A tree with

f

g

2

c

e

f

g

3 2

3

a

b

c d

e

f

g

h 4

4

4

3 2

3

4



its vertex treated as a root is called rooted tree. See Figure 5.3 (a). A vertex

role as a root has input degree 0. A vertex which functions a root has input-degree equal to 0. While, other vertices has input-degree equal to one. On a rooted tree, vertices whose output-degree equal to 0 is called leaf. Later on, direction component are abandoned so a rooted tree is depicted as an

undirected graph 5.3 (b)

(a) (b)

Figure 5.3 Rooted Tree

Several terminologies on rooted tree are as follow: a. Child or children and parents

If there is an edge between two vertices, the vertex which is closer to the root is called parents while other edges are called children. It can be noted on Figure 5.3 that b, c, and d are the children of vertex a, and a is the parent of the children. G and h are the children of d, while d is their

parents. Furthermore, a is called the ancestor of e, f, g and h. While e, f, g and h are called descendant of a. In addition, f is e sibling, but g is not e sibling,

because their parents are different.

b. Path

The path from a to h is a, d, h with its length is 2. On a tree, the path between two random vertices are unique as there is only one path.

a

b

c d

e f g

a

b

c d

e f g h h



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c. Subtree

For example d is a vertex on a tree, then subgraph (tree) consisting of d with all descendants are called subtree. On the following example, the inside part is the subtree of the main tree.

c. Degree The degree of the vertex is the number of children on it. On Figure 5.3 :

Vertices having 0 degree are vertices of c, e, f, g, and h

No vertices having 1 degree.

Vertices having 2 degree are vertices b and d.

Vertex having 3 degree is vertex a.

Degree on this context is output-degree. Maximum degree of all vertices is the degree of the tree itself. Thus, the tree on Figure 5.3 having 3 degree.

d. Leaf

Vertex with 0 degree (or having no children) is called leaf. Vertices c, e, f, g and h are leaves.

e. Internal vertex

Vertex (except root) having children is called internal vertex. Vertices b

and d are called internal vertex.

a

b

c d

e f g h



f. Level

Root has the same level with 0, while other vertex depends on its position. For example, on Figure 5.3, it can be seen that b, c and d are on level 2. While e, f, g and h are on level 3.

Rooted tree whose children order are important is called ordered tree. While, rooted tree which its branch vertex having maximum n children is called n-ary tree. If n = 2, its tree is called binary tree.

Example 5.2 :

The followings are several examples of binary tree:

1. Expression Tree Aritmetic expression (a * b) – ((c + d) / e) can be stated in a binary tree, in which the alteration is as the leaf and aritmetic

operator as the internal vertex and root.

2. Decision Tree

a b

c

e

d

+

* /

–



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A tree in which its internal vertex corresponds to a decision is

called decision tree. One function of the decision tree is to sort complexity from any kinds of algorithm.

Figure 5.4 Decision Tree to Sort Three Diferent Elements [4]

3. Prefix code Prefix code is code collections (one of them is binary code) in such a way that there is no collection member which is the prefix of

other codes. Example :

{ 001, 010, 011, 11} is a prefix code. If it is stated in binary

tree, it will be :

a : b

a : c b : c

b : c c > a > b a : c c > b > b

a > b > c a > c > b b > a > c b > c > a

b>c a>c

a>b

b>c

b<c a<c

b<c a>c a<c

A<b

1

11

0

0

0

1

11

1

010000 011



4. Hufman Code

Hufman coding is often used in data compression. Observe the table of ASCII code:

Symbol ASCII Code

A 01000001 B 01000010

C 01000011 D 01000100

Then bit series for string ‘ADABACA’ , can be represented inthe form of :

01000001010001000100000101000010010000010100001101000001

Code length of the string is

7 8 = 56 bit (7 byte).

Table 5.1 Table of Huffman Code for string ’ADABACA’

Symbol Frequency Possibility Huffman Code

A 4 4/7 0

B 1 1/7 10

C 1 1/7 11

D 1 1/7 110

Thus, bit series for string ’ ADABACA’ is :

01100100110 or the 56 bit can be written in 11 bit.

4.10 Binary Tree Tracking

For example, the following picture is a binary tree in which A is the root. S and T are subtree of the binary tree.

S T

A



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There are 3 kinds of binary tree tracking :

1. Preorder : A, S, T - visit A - visit S preorderly - visit T preorderly

2. Inorder : S , A, T - visit S inorderly - visit A

- visit T inorderly 3. Postorder : S, T , A - visit S post-orderly

- visit T post-orderly - visit A

For example : Determine the result of preorder, inorder, and postorder tracking of the tree presented below:

Answer :

preorder : – * a b / + c d e (prefix)

inorder : a * b – c + d / e (infix) postorder : a b * c d + e / – (postfix)

a b

c

e

d

+

* /

–



4.11 Graph Coloring

Graph coloring is an assignment of colors to several vertices on graph

G in such a way that no two adjacent vertices share the same color. In addition to vertex coloring, there is edge coloring. However, this chapter will be focused on vertex coloring.

A graph is said to have color n if there is n color in graph G

coloring. Minimum colors required in graph coloring are called chromatic

numbers denoted as )(G ( : read or pronounced chi).

For example :

Chromatic number of a complete graph -n (Kn) is n. It is caused by

every vertex on a complete graph is adjacent. Then (Kn) = n. Observe a complete graph with 5 vertices presented below :

Its graph coloring needs 5 colors.

Welch-Powell algorithm in graph coloring can be illustrated as follow : 1. Sort all vertices on graph G based on their degree starting from the

biggest to the smallest. This order is not unique because several

vertices perhaps have similar degree. 2. Use first color to color first vertex and other vertex which is on the

order as long as the vertex is not adjacent to the previous vertex.

3. Give second color to color vertex on the highest order (which has not been colored). Do it just like the previous step.

4. It is like the third step. Do it continuously until every vertex on the

graph becomes colorful.

a

b c

d e



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5. Welch-Powell algorithm only gives top limit of chromatic number.

Thus, this algorithm does not always give the number of minimum color needed in graph coloring.

Example :

Use Welch-Powell algorithm for graph coloring :

The order of vertex degree is as follow: a b c d e f 4 3 3 3 2 1

Thus, graph coloring can be done as follow : Color I for vertices : b, f

Color II for vertices : a, d, e Color III for vertex : c

For example G ia a graph, the following statements are euivalent :

G is a bipartite graph

Chromatic number of G is two ( )(G = 2 )

Every circuit of G has an even length

For example :

a

b c

d e

f



Observe the following bipartite graph K3,3 :

Graph coloring on the graph above can be done by two colors, such as :

Color I for vertices a, b, c

Color II for vertices d, e, f If we want to make a circuit on the graph, the circuit will pass 3 or 5 vertices before it comes back to the initial vertex. Thus, the circuit has

an even length.

Map Coloring

Before discussing area coloring on a planar graph, observe the following definitions related to planar graph :

Area r1, r2, r3, r4, and r5 are called region of the planar graph. Two regions

on a planar graph are said to be adjacent if they have at least one similar vertex.

The example of adjacent regions:

r1 and r2

a b c

d e f

p

q r

t

s

u

r1 r2 r5 r3 r4



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r2 and r3

r2 and r5 r4 and r5 r1 and r5 r2 and r4

The example of non-adjacent regions: r1 and r4 r5 and r3

r3 and r4 The number of regions adjacent to a region on a graph can be obtained by calculating the number of regions which have at least an edge with the region

itself. Thus, each region has adjacent region as follow :

r1 has 2 adjacent regions : r2 and r5

r2 has 3 adjacent regions : r1, r3 dan r5 r3 has 1 adjacent region : r2 r4 has 2 adjacent regions : r2 and r5

r5 has 3 adjacent regions r1, r2 and r4 Region coloring (map) on a planar graph G is a mapping of color collection to

several regions on a planar graph G in such a way that the adjacent regions do not share the same colors.

For example : Observe the graph presented below :

Conduct a region coloring using : a. 3 colors b. 2 colors

Answer :

a. Graph coloring using 3 colors :

Color I for regions r1 and r4

p

q r

t

s

u

r1 r21 r51 r31 r41



Color II for region r2

Color III for regions r3 and r5 b. Graph coloring using 2 color is impossible to do. It is because

region r2 , r4 and r5 are adjacent to each other.

The duality of map coloring is vertex coloring of a planar graph. Bear in mind that graph coloring will connect to vertex coloring of dual G*. In other words, a map G is colored n if only planar graph of dual G with n color. To

make it clearer, observe the following graph:

Choose a vertex on the graph below. Connect the two vertices with an edge if the two regions are adjacent.

If we draw a graph, it will be as follow :

r1 r2

r3

r4

r1 r2

r3

r4

r4



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Map coloring can be represented in vertex coloring. The most important thing is a given graph model represents real problems.

r1

r2

r3



Summary

1. A connected graph with its pair of vertices that can be connected only by certain path is called tree. In other words, a tree is undirected graph which is connected but it does not have cycle or circuit.

2. Spanning Tree of a connected graph is a spanning subgraph in the form of tree. Spanning tree is obtained by eliminating the circuit on the graph.

3. A tree having minimum weight is called minimum spanning tree.

4. There are several terminologies worthed to know such as root, leaf, subtree, the shortest path, etc.

5. There are two algorithms to determine minimum spanning tree, Prim and Kruskal.

6. Graph coloring is an assignment of colors to several vertices on graph G in such a way that no two adjacent vertices share the same color. In addition to vertex coloring, there is edge coloring.

7. Minimum colors required in graph coloring are called chromatic

numbers denoted as )(G ( : dibaca chi).

8. Map coloring is the duality of vertex coloring of a graph.



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True or False Quiz

1. Tree is a subgraph of a graph.

2. Two vertices on a tree are only connected by a path. 3. Chromatic number of a tree is 2. 4. If two vertices having the degree of 1 are connected by an edge, then

the graph is still a tree. 5. Minimum spanning tree of a graph K10 is a tree with 10 edges. 6. For example, G is a graph, the complement of minimum spanning tree

graph G is a tree. 7. The number of colors used to color wheel graph W7 is 3. 8. Hufman code is usually used in graph coloring. 9. An aritmetic expression can only be stated on a expression tree.

10. Map coloring can be analogized with common vertex coloring.



Multiple Choices

1. It is impossible to find it on a tree.

A. Root D. Cycle

B. Leaf E. Children

C. Path

2. Chromatic number of graph C5 is ….

A. 2 D. 5

B. 3 E. 6

C. 4

3. The minimum number of color for coloring bipartite graph is ….

A. 1 D. 5

B. 2 E. 4

C. 3

4. A complete bipartite graph K3,3 is not a tree because

A. Disconnected D. Vertex’s degree is the same

B. Impossible to color E. Must have cycle

C. Divided into two

5. The following is the example of the use of binary tree, except….

A. Expression tree D. Decision Tree

B. Minimum spanning tree E. Prefix code

C. Hufman code



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Exercise

1. Determine all spanning trees of the following graph :

2. Observe the following graph data : a. Graph G1

b. Graph G2

Determine minimum spanning tree using Prim Algorithm and Kruskal

Algorithm

B C

EF

8

5

3A D

8

2

1

6

44

2

p q

r s

t

a b

c

d e

f g

3 4

1

5

6 2

4



3. Make a sketch of binary graph (expression tree) representing expression

: a. p / (q – r )*(s + t) b. (p + q) / r – (s + t * u)

4. Determine the result of expression tree tracking on question number 3 in the form of preorder, inorder, and postorder.

5. On the following graph, vertices collection defines a collection of villages in a district. A budget for village street development is created as weight (in million rupiahs) on each edge. Determine minimum cost prepared for

the village street development so that all villages in the district will be connected. (Bear in mind the definition of ’connected’ in a graph).

6. Use Welch-Powell algorithm to color the following graph :

a

f e d c

h

b

j i

g

a

f e d

c

h

b

j i

g

6

5

7

3

5

4

6

3 5 4

7 8

6

6



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7. In a semester, the schedule of final exam will be created. The schedule

covers Calculus, Discrete Mathematics, Physic, English, Bahasa Indonesia, Religion, Pancasila and Chemistry. There is no students who take the following pair of subjects on a smilar time in the same semester :

- Calculus & Chemistry

- Discrete Mathematics & Chemistry

- English & Bahasa Indonesia - English & Religion

- Calculus & Discrete Mathematics - Calculus & Physic

- Physic & English However, there are students who take a combination of subject at similar time in the semester.

How many minimum time slot required to create the final exam schedule so that the students will not encounter two or more exams at the same time?

8. How many colors required for map coloring (map) below?

p q

r s

t


Daftar Pustaka

Kolman, B., Busby, R. C., Discrete mathemetical Structures for Computer

Science, 2nd edition, Prentice Hall, New Delhi, 1992

Munir, R., Matematika Diskrit, Edisi kedua, Informatika, Bandung, 2003

Lipschutz S., Lipson M., Discrete Mathematics, McGraw Hill USA, 1997

Rosen, K. H., Discrete Mathematics and Its Applications, 5th edition,

McGraw-Hill, Singapore, 2003

Slamet, S., Makaliwe, H., Matematika Kombinatorik, Elek Media

Komputindo, Jakarta, 1991

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