BC0052-Theory of Computer Science

Fundamentals of Theory of Computer Science Unit 1

Sikkim Manipal University Page No.: 1

Unit 1 Mathematical Preliminaries

for Computer Science

Structure

1.1 Introduction

Objectives

1.2 Sets and Cartesian product of sets

1.3 Functions and Relations

1.4 Equivalence Relations

1.5 Matrix Representations

1.6 Closure Relations

Self Assessment Questions

1.7 Summary

1.8 Terminal Questions

1.9 Answers

1.1 Introduction

The idea in set is fundamental in the study of mathematical structures.

Almost all mathematical objects can be defined in terms of sets and the

language of set theory is used in every mathematical topic. Also, the

concept of relation and function arises when we consider a pair of objects

and compare them. In this unit we study the different relations on a set and

the algebraic structures with single binary operation. The content of this unit

provides a mathematical tool for the later units.

Objectives:

At the end of the unit the student must be able to:

Understand the fundamental idea of certain mathematical concepts.

Learn the various operations on sets.

Understand relations and types of relations

Recognize the Matrix representations

Know the closure relations and functions.



1.2 Sets and Cartesian product of sets

1.2.1 Definition

A set is a collection of objects in which we can say whether a given object is

in the collection. Sets are denoted by capital letters.

The fact that a is a member of a set A is denoted by a A and we call it as

„a belongs to A‟. The members of a set are called elements.

If x is not an element of A we write x A.

1.2.2 Note

There are five ways used to describe a set:

i) Describe a set by describing the properties of the members of the set.

ii) Describe a set by listing its elements.

iii) Describe a set A by its characteristic function, defined as

for all x in U, where U is the universal set, some times called the

“universe of discourse ,” or just universe.

iv) Describe a set by recursive formula. This is to give one or more

elements of the set and a rule by which the rest of the elements of the

set may be generated.

v) Describe a set by an operation (say union, intersection, complement

etc) on some other sets.

1.2.3 Example

Describe the set containing all the nonnegative integers less than or equal

to 4.

Let X denotes the set. Then X can be described in the following ways:

i) X = {x / x is a nonnegative integer less than or equal to 4}

and N = {x / x is a natural number} = {1, 2, 3, 4, …} is an infinite set.



ii) X = {0, 1, 2, 3, 4}.

iii)

iv) X = {xi + 1 = xi + 1, i = 0, 1, 2, 3 where x0 = 0}.

1.2.4 Definition

Suppose A and B are two sets. Then we say that A is a subset of B (written

as if every element of A is also an element of B. A set A is said to be

a proper subset of B if there exists an element of B which is not an element

of A. That A is a proper subset of B if .

1.2.5 Note

The containment of sets has the following properties: Let X, Y and Z be

sets.

1.2.6 Example

i) Let N, Z, Q, R denote the set of natural numbers; the set of integers;

the set of rational numbers; the set of real numbers respectively. Then

RQZN .

ii) If A = {1, 3, 5}, B = {1, 3, 5, 7} then A is a proper subset of B.

1.2.7 Definition

Two sets A and B are said to be equal (denoted by A = B) if A is a subset of

B, and B is a subset of A.

Observation: To show two sets A and B are equal, we must show that each

element of A is also an element of B, and conversely.



1.2.8 Definitions

i) The set that contains no members is called the empty set and it is

denoted by .

ii) A set which contains a single element is called singleton set. X = {2} is

a singleton set.

iii) If A and B are two sets, then the set is denoted

by and we call it the union of A and B.

iv) The set is denoted by and we call it the

intersection of A and B.

1.2.9 Properties

1.2.10 Definition

Let X and Y be two sets. The symmetrical difference of X and Y is defined

as It is denoted as . It is also called

Boolean sum of two sets.

1.2.11 Definition

If A and B are two sets, then the set is denoted by B – A (or

B \ A) and it is called the complement A in B.



1.2.12 Examples

Let A = {1, 2, 3, 4, 5}, B = {a, b, c, d}, C = {2, b, d}, D = {3, a, c} and E = {x /

x is an integer and 1 < x < 2}.

Here we may note that, in roster form, there is no necessity of writing

the same element two times. So we avoid writing the same element

second time in roster form.



1.2.13 Definition

i) If S and T are two sets, then the set {(s, t / s S and t T} is called

the Cartesian product of S and T (here (a, b) = (s, t) a = s and

b = t). The Cartesian product of S and T is denoted by S T. Thus

S T ={(s, t) / s S and t T}.

Note that if S and T are two sets, then S T and T S may not be

equal.

ii) If S1, S2, ..., Sn are n sets, then the Cartesian product is defined as

S1 S2 … Sn = {(s1, s2, …, sn) / si Si for 1 i n}.

Here the elements of S1 S2 … Sn are called ordered n-tuples.

For any two n-tuples, we have (s1, s2, …, sn) = (t1, t2, …, tn) si = ti,

1 i n.

iii) Let Ai be a collection of sets – one for each element i I, where I is

some set (I may be the set of all positive integers). We define

A collection {Ai}i I of sets is said to be mutually disjoint if Ai Aj = for

all i I, j I such that i j.

1.2.14 Example

i) If X = {a, b} and Y = {x, y}, then

X Y = {(a, x), (a, y), (b, x), (b, y)} and Y X = {(x, a) (x, b), (y, a),

(y, b)}.

Note that .

ii) If A = {a, b}, B = {2}, C = {x}, then A B C = {(a, 2, x), (b, 2, x))}.

iii) Let T = {a, b, c, d} and S = {1, 2, 4}. Then the is an

empty set.



iv) Write Ai = {i, i+1, i+2, …} for each I N, the set of natural numbers.

Then it is easy to observe that Ni

iA

= N and Ii

iA

= .

v) If Bi = {2i, 2i +1} for all I N, then {Bi}iN is a collection of mutually

disjoint sets.

1.2.15 Definition

Let A be a set. The set P(A) = the set of all subsets of A, is called the power

set of A.

Observation: If the set A has n elements, then the number of elements in

P(A) is 2n.

Verification: Suppose A has n elements. Let m be an integer such that 0 m

n. We can select m elements from the given set A in nCm ways. So A

contains nCm distinct subsets containing m elements. Therefore the number

of elements in P(A) = number of subsets containing 0 number of elements

+ number of subsets containing only 1 element

+ … + number of subsets containing n elements

= nC0 + nC1 + nC2 + … + nCn = 2n.

1.3 Functions and Relations

Relation describes connections between different elements of the same set,

whereas functions describe connections between two different sets.

Functions give a mathematical precise framework for the intuitive idea of

transformation.

1.3.1 Definitions

Let S and T be sets. A function f from S to T is a subset f of S T such that

Here t is called the image of s; and s is called the preimage of t.



The set S is called the domain of f and T is called the codomain.

The set {f(s) / s S } is a subset of T and it is called the image of S under f

(or image of f). We denote the fact „f is a function from S to T‟ by “ f : S T”.

f: S T is said to be

one-one function (or injective function):

onto function (or surjective function): there exists an element s

S such that f(s) = t.

bijection: if it is both one-one and onto.

Let g: S T and f: T U. The composition of f and g is a function

fog: S U defined by (fog)(s) = f(g(s)) for all s in S.

1.3.2 Definitions

A function f : S T is said to have an inverse if there exists a function g

from T to S such that (gof) (s) = s for all s in S and (fog)(t) = t for all t in

T. We call the function „g‟ the inverse of f. A function f : S S is said to be

an identity function if f(s) = s for all s in S. The identity function on S is

denoted by either I or IS. Inverse of a function f, if it exists, is denoted by f -1.

Two functions f : A B and g : C D are said to be equal if A = C, B = D

and f(a) = g(a) for all elements a in A and C. If two functions f and g are

equal, then we write f = g. The identity function is one-one and onto. A

function g is inverse of f fog and gof are identity functions. A function f

has an inverse f is one-one and onto.

1.3.3 Example

Let the functions f and g defined by f(x) = 2x and g(x) = x + 5 for all x in R

(the set of real numbers). Then (fog)(1) = f(g(1)) = f(1 + 5) = f(6) = 12 and

(gof)(1) = g(f(1)) = g(2) = 2 + 5 = 7. This shows that the two functions are

not equal at 1 and so fog gof.

1.3.4 Definition



Let A be a subset of the Universal set U = {x1, x2, …, xn}. The characteristic

function of A is defined as a function from U to {0, 1} by the following:

fA(xi) = i

i

1 if x A

0 if x A

. For example, if A = {2, 3, 7} and U = {1, 2, …, 10},

then fA(1) = 0, fA(2) = 1, fA(3) = 1, fA(7) = 1. And fA(11) is undefined. It can be

verified that fA is everywhere defined and onto, but not one one.

1.3.5 Definition

A relation R between the sets A1, A2, …, An is a subset of A1 A2 … An.

This relation R is called an n-ary relation.

(two–ary is called binary, three-ary is called ternary). In general, a relation

means binary relation on a set S (means a subset of S S). A relation R on

S is said to be

i) transitive if (a, b) R, (b, c) R implies (a, c) R;

ii) reflexive if (a, a) R for all a S;

iii) anti-symmetric if (a, b) R and (b, a) R a = b;

iv) symmetric if (a, b) R implies (b, a) R;

1.3.6 Note

i) Let the number of elements of A and B be m and n respectively. Then

the number of elements of A B is mn.

ii) The number of elements of the power set of A B is 2mn. A B has

2mn distinct subsets.

iii) Every subset of A B is a relation from A to B. Therefore the number

of different relations A to B is 2mn.

1.4 Equivalence Relations

1.4.1 Definition



A relation is said to be an equivalence relation if it is reflexive, symmetric

and transitive.

If (a, b) is an element of the equivalence relation, then we write a ~ b and

we say that a and b are equivalent.

Let R be an equivalence relation on S and a is an element of S. Then the

set [a] = {s S / s ~ a} is called the equivalence class of a (or equivalence

class containing a).

1.4.2 Example

If A = {a, b, c} and R = {(a, a), (b, b), (c, c), (a, b), (b, a)}, then R is an

equivalence relation on A and [a] = {a, b}, [b] = {a, b}, and [c] = {c}.

1.4.3 Example

i) Let S be the set of all integers. Define for any a, b S, a ~ b a - b

is an even number. Then relation ~ is an equivalence relation.

ii) Let R, the set of all real numbers, x ~ y x – y is an integer. Then ~

is an equivalence relation. The set of all equivalence classes is given

by

{[x] / x (0, 1]}.

1.4.4 Example

For (x1,y1) and (x2,y2) in R2 (Euclidean Plane), define (x1,y1) ~ (x2,y2) if and

only if x12 + y1

2 = x22 + y2

2. Then ~ is an equivalence relation on R2.

1.4.5 Definition

Let n > 0 be a fixed integer. We define a relation namely “Congruence

modulo n” on Z, the set of integers as:

a b mod n n divides (a - b).

Some times we write a b as a b. and we read as “a b mod n” as „a is

congruent to b modulo n‟.



1.4.6 Problem

The relation “a b mod n” defined above is an equivalence relation on Z.

Solution: Let a Z.

Reflexive: Since n divides a - a = 0, we have a a mod n.

Symmetric: Let a b mod n.

n divides a - b

n divides - (a - b)

n divides b - a

b a mod n.

Transitivity: Let a, b, c Z such that a b mod n, b c mod n.

n divides a - b, and n divides b - c

n divides ( a - b ) + ( b - c)

n divides a - c

a c mod n.

Hence the relation is an equivalence relation.

1.4.7 Example

Suppose n = 5. Then

[0] = {x / x 0 mod 5} = {x / 5 divides x – 0 = x} = {…, -10, -5, 0, 5, 10, …}

[1] = {x / x 1 mod 5} = {x / 5 divides x - 1}= {…, -9, -4, 1, 6, …}

[2] = {x / x 2 mod 5} = {x / 5 divides x -2} = {…, -8, -3, 2, 7, 12, …}

[3] = {x / x 3 mod 5} = { x / 5 divides x -3} = {…, -7, -2, 3, 8, 13, …}

[4] = {x / x 4 mod 5} = { x / 5 divides x - 4} = {…, -6, -1, 4, 9, 14, …}

Also it is clear that [0] = [5] = [10] = …

[1] = [6] = [11] = …

[2] = [7] = [12] = …

[3] = [8] = [13] = …

[4] = [9] = [14] = …

Therefore the set of equivalence classes is given by {[0], [1], [2], [3], [4]}.



1.4.8 Definition

A partition of a set S is a set of subsets {Si / Si S and i I where I is

some index set} satisfying Ii

iS

= S and Si Sj = if i j.

1.4.9 Example

i) Write A = {1, 2, 3, 4, 5, a, b, c}, S1 = {1, 2}, S2 = {3}, S3 = {4, 5, a}

and S4 = {b, c}. Then S1, S2, S3, S4 form a partition for A.

ii) Consider R, the set of all real numbers. The collection {(a, b] / a, b Z

and b = a + 1} of subsets of R, forms a partition for R.

1.4.10 Lemma

If R is an equivalence relation on a set S and a, b S, then either [a] = [b]

or [a] [b] = .

Proof. If [a] [b] = , then it is clear.

Now suppose the intersection is non-empty.

Let x [a] [b]

x [a] and x [b]

x ~ a and x ~ b

a ~ x and x ~ b (since ~ is symmetric)

a ~ b (since ~ is transitive).

Now we show that [a] = [b]. For this, let y [a] y ~ a.

Since a ~ b, we get y ~ b (by transitive property) b ~ y y [b].

Hence [a] [b]. Similarly, we get that [b] [a]. Therefore [a] = [b].

From this lemma, we can conclude that any two equivalence classes are

either equal or disjoint.

1.4.11 Problem

Let A be a set and ~ an equivalence relation on A. Then the set of all

equivalence classes forms a partition for A.

Solution: The collection of all equivalence classes is {[a] / a A}.



Observation: Let A be a set and {Ai / i I} be a collection of nonempty

subsets of A, which forms a partition for A. Then there exists an equivalence

relation ~ on A such that the equivalence classes are nothing but the sets of

the partition.

1.5 Matrix Representations

1.5.1 Definition

Let A = {a1, a2, …, an} and B = {b1, b2, … , bn}.

If R is relation from A to B, then R can be represented by matrix MR =

(Mij)mn, defined i j

ij m ni j

1 if (a ,a ) RM

0 if (a ,a ) R

Where Mij is the element in the ith row and jth column. MR can be first obtained

by first constituting a table, whose columns are preceded by a column

consisting of successive elements of A and where rows are headed by row

consisting of successive elements of B. If (ai, bj) R, then we enter 1 in the

ith row and jth column.

1.5.2 Example

Let A = {1, 2, 3} and = {(x, y) | x < y}. Write MR.



Solution: R = {(1, 2), (1, 3), (2, 3)}. Since (1, 2) R, we have m12 = 1;

(1, 3) R, we have m12 = 1; also m23 = 1.

Therefore R

0 1 1

M 0 0 1

0 0 0

1.5.3 Example

Let A = {1, 2, 3, 4}. Define a R b a < b.

Then R

0 1 1 1

0 0 1 1M

0 0 0 1

0 0 0 0

1.5.4 Example

Write the relation for the relation matrix

R

1 0 0

M 0 0 1

1 1 0

Solution: Since M is a 3 3 matrix, take A = {a1, a2, a3} and B = {b1, b2, b3}

Then R = {(a1, b1), (a2, b2), (a2, b3), (a3, b1)}

1.5.5 Definition

A relation R is transitive if and only if MR = [mij] has the property:

mij = 1 and mik = 1 mik = 1

1.5.6 Example

Define a relation R represented by a matrix

110

110

001

MR

Here, m22 = 1, m23 = 1 m23 = 1



m23 = 1, m32 = 1 m22 = 1

m33 = 1, m32 = 1 m32 =1

Therefore the relation R is transitive.

1.6 Closure Relations

1.6.1 Definition

Let R be a relation on a set A. R may or may not have some property P,

such as reflexivity, symmetry or transitivity. If there is a relation S with

property P containing R such that S is a subset of every relation with P

containing R, then S is called the closure of R with respect to P.

1.6.2 Definition

Let R be a relation on a set S. The reflexive closure of R is the smallest

reflexive relation R1which contains R. It is also denoted by R(r).

1.6.3 Note

1.6.4 Definition

The symmetric closure of R is the smallest symmetric relation containing R.

That is R R–1 is symmetric closure R, where R–1 is the inverse of the

relation R. It is denoted by R(s).

1.6.5 Definition

Transitive closure of a relation R is the smallest transitive relation containing

R.

1.6.6 Example

Consider the set S = {1, 2, 3, 4}

i) The relation R = { (1,2), (2, 1), (1, 1), (2, 2)} is not reflexive, since

(3, ,3) S.



Consider = {(1, 1,), (2, 2), (3, 3), (4, 4)}. Now the reflexive closure

R1 = = {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3), (4, 4)}.

Observe that R1 (the reflexive closure of R, sometimes we denote as

R(r)) is obtained by supplementing with exactly essential (no more, no

less) in order to get a reflexive relation containing R.

ii) Consider the relation K = { (1,2), (4, 3), (2, 2), (2, 1), (3, 1)}, which is

not symmetric on S. Now

K –1 = {(2, 1), (3, 4), (2, 2) (1, 2), (1, 3)}.

The symmetric closure K(s) of K is given by

K(s) = K K–1 = { (1, 2), (2, 1), (4, 3), (3, 4), (3, 1), (1, 3)}.

1.6.7 Note

Given a relation R on a set A. To make a relation R transitive, add all pairs

of R2, all pairs of R3, …,all pairs of Rm ( assume that | A| = m ), unless these

pairs are already in R.

Then the transitive closure of R, denoted by R or R(T)

R(T) = R R2 … Rm.

1.6.8 Properties of Transitive closure

i) R(T) is transitive

ii) R R(T)

iii) If S is any other transitive relation that contains R, then R(T) S.

1.6.9 Representation of Closure Relations

Let M be the relation matrix of the relation R. Then

i) the symmetric closure of R, denoted by MS, defined as MS = M M

where M is the transpose of M.

ii) The reflexive closure of R, denoted by MR, defined as MR = M In

where n is the cardinality of the set of for which the relation defined

and In is the identity matrix of size n.



iii) The transitive closure of R, denoted by MT or R

M defined as MT = M

M2 M3

…. Mn.

1.6.10 Example

Take A = {1, 2, 3} and R = {(1, 2), (2, 3), (3, 1)}. Find the reflexive,

symmetric and transitive closure of R, using composition of matrix relation

of R.

Solution: Let M be the matrix relation R then

001

100

010

M

i) The Reflexive closure of R, MR = M I3

101

110

011

100

010

001

001

100

010

One can write the reflexive closure R(r), using the above matrix as

R(r) = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 1), (3, 3)}.

ii) The symmetric closure of R, is MS = M M =

010

001

100

001

100

010

.

One can write the symmetric closure R(S), using the above matrix as

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

iii) To find the transitive closure of R, we first find M2 and M3 (since the

cardinality of the set A = 3).

010

001

100

001

100

010

001

100

010

M2 and

M3 = M2. M =

100

010

001

001

100

010

010

001

100



Therefore the transitive closure of R, MT = M M2 M3

=

111

111

111

100

010

001

010

001

100

001

100

010

.

One can write the transitive closure of R,

RT = { (1,10), (1,2), (1,3), (2,1), (2,1), (2,2), (2,3), (3,1), (3,2) (3,3)}.

1.6.11 Definition

Let R be a relation from A to B, S be a relation from B to C. Then the

relation SoR from A to C is defined by a (SoR) c a R b and b S c for

some b B for all a A and c C.

1.6.12 Example

Take A = {1, 2, 3, 4}

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

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

Then, since (1,1) R, (1,1) S, we have (1,1) SoR.

Since (1,2) R, (2,3) S, we have (1, 3) SoR.

Since (2,3) R, (3,1) S, we have (2, 1) SoR.

Continuing this way we get

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

Similarly,

RoR = {(1,1), (1,2), (1,3), (1,4), (2,4), (2,1), (2,2), (3,1), (3,2), (4,1), (4,2),

(4,3), (4,4)}

1.6.13 Note

a) If R1 and R2 are relations from A to B, R and R4 are relations from B to C,

then



b) If R is a relation from A to B, S is a relation form B to C, and T is a

relation from C to D, then (RoS)o T = Ro(SoT).

1.6.14 Problem

If R is a relation from A to B and S is a relation from B to C, then

(RoS)–1 = S–1o R–1

Solution: Since R is a relation from A to B we have R–1 is a relation from B

to A. Similar way, S–1 is a relation from C to B.

Therefore S–1oR–1 is a relation from C to B.

This is true for any x A and z C. Hence (RoS)–1 = S–1 o R–1.

1.6.15 Problem

If R is a relation on a set A, then R is transitive .

Converse: Suppose R2 R. Take (x, y) R, and (y, z) R.

Then (x, z) RoR = R2 R. Thus R is transitive.


1. Let A = {a, b} and B = {x, y, z}. Then find .

2. Let S = {2, 5, 2 , 25, , 2

5} and T = {4, 25, 2 , 6,

2

3}



3. Find Ac (with respect to the universal set of reals) in the following

cases

4. Let x and y be real numbers with x < y. Find (x, y)c, [x, y)c, (x, )c,

and (-, y]c.

5. Let S = {x, y, z, {x, y}}. Find (i) S \ {x, y}; (ii) S \ ; (iii) ({x, y, z} {S}) \

S; and (iv) S \ {{x, y}}.

6. Find the power set of S = {(x, y), Z }.

7. If a set A has n elements, then the number of relations from A to A is

______

8. If A has m elements and B has n elements, then the number of

relations from A to B is ________

9. Let A = Z+, the set of positive integers. R be the relation defined by

aRb if and only if there exists a k Z+ so that a = bk. Then which of

the following belong to R?

(i) (4, 16), (ii) (1, 3), (iii) (4, 2), (iv) (2, 8), (v) (4, 4) (vi) (2, 16).

10. Given the list of relations among people. For each of the following

relations, state whether the relation is reflexive, symmetric,

antisymmetric, or transitive.

i) xRy stands for x is child of y

ii) xRy stands for x is a spouse of y

iii) xRy stands for x is a wife of y

iv) xRy stands for x is a superior of y

v) xRy stands for x and y have the same parents.

11. Give an example of a relation that is both symmetric and anti

symmetric.



12. Which of the following relations defined in the set of real numbers are

equivalence relation ?

i) aRb if and only if a = b

ii) aRb if and only if a b

13. Let Z be the set of integers. Define a relation R on Z as

R = {(x, y) x, y Z, (x-y) is a multiple of 3}.

Then

i) R is an ____________

ii) The Equivalence class containing 0 is _______

iii) The number of equivalence classes is _________

14. Let S be the set of non-zero integers and let R be the relation on A A

defined by (a, b) R (c, d) if and only if ad = bc.

Then

i) R is an _______

ii) The equivalence class containing (1, 2) is ________

15. Let A = {0, 1, 2, 3, 4}. The relation R = {(0, 0), (0, 4), (1, 1), (1, 3), (2,

2), (3, 1), (3, 3), (4, 0), (4, 4)} is an equivalence relation on A. Find the

distinct equivalence classes of R.

16. The set theoretic union of distinct equivalence classes of an

equivalence relation forms a __________ of the given set.

17. Determine whether or not each of the binary relations defined on the

given sets Si (i = 1, 2, 3) are reflexive, symmetric, anti symmetric or

transitive. If a relation has a certain property, prove this is so;

otherwise, provide a counter example to show that it does not.

i) S1 = {1, 2, 3, 4}; R1 = {(1, 1), (1, 2), (2, 1), (3, 4)}

ii) S2 = Z , (a, b) R2 if and only if ab 0.

iii) S3 = R R, R3 = {((x, y), (u, v)) / x + y u + v}

18. Is R = {(1, 1), (1, 2), (3, 2), (3, 3), (2, 3), (2, 1)} an equivalence relation

on S = {1, 2, 3}?



19. For x, y R \{0}, define x ~ y y

x Q. Whether or not, „‟ is an

equivalence relation.

20. Let A = {x Z + / 1 x 10}. State whether or not each of the

following families of sets is a partition of A.

i) {{1, 3, 5}, {4, 7, 9}, {2, 6,10}};

ii) {{1, 3, 5, 7}, {2, 4, 6}, {3, 8, 9, 10}};

iii) {1, 2, 3}, {5, 8, 9}, {4, 6, 7, 10}}.

21. If |S| = n, then how many relations are there from S to S.

22. Let R be the relation from the set A = {1, 3, 4} on itself. Define R by

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

23. The relations corresponding to the following relation matrices

(i)

011

010

001

, (ii)

0101

1100

0110

1001

on a set A = {1, 2, 3} are __________

24. Let A = {a, b, c, d} and R be a relation defined on A whose matrix

representation is MR =

1000

0110

0110

0001

. Verify whether or not the

relation on A is reflexive and symmetric.

25. Consider the relation R ={(0, 1), (1, 2), (2,3)}defined on A = {0, 1, 2, 3}.

Find the reflexive, symmetric, and transitive closure of R using

composition of relation.

1.7 Summary



In this unit we introduced the basic concept sets and the different properties

of sets. Some properties common to operations on sets and logical

statements were discussed. Cartesian product of sets was studied as

relations between two sets. We also discussed the equivalence relations

and closure relations with few illustrations. The reader can easily apply the

mathematical concepts introduced, in various situations.


1. If R is a relation on the set of integers Z defined by R = {(x, y) x, y Z,

(x-y) is divisible by 6}, then prove that R is an equivalence relation.

2. Represent each of the following relation on {1, 2, 3} with a matrix

(i) R = {(1, 1), (1, 2), (1, 3)}, S = {(1, 2), (2, 1), (2, 3), (3, 3)}, T = {(1, 3),

(3, 2), (3, 3)}.

3. Let A = {1, 2, 3, 4, 5, 6}. Define a relation R = {(x, y) x + y is a divisor

of 24}.

Find the relation matrix M of R. Compute M2 and M and M2 whether or not R

is transitive.

1.9 Answers


1. A B = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}, and

B A = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}. This problems

illustrates that, in general, the sets A B and B A are different.

2. i) S T = { 2 , 25}, S T = {2, 5, 2 , 25, , 2

5, 4, 6,

2

3},

T (S T) = {(4, 2 ), (4, 25), (25, 2 ), (25, 25), ( 2 , 2 ),

( 2 , 25), (6, 2 ), (6, 25), (2

3, 2 ), (

2

3, 25)}.



ii) Z S = { 2 , , 2

5, 0, 1, -1, 2, -2, …. }, Z S = {2, 5, 25}

Z T = { 2 , 2

3, 0, 1, -1, 2, -2, ….}, Z T = {4, 25, 6}.

3. i). c1A = [-2, 1];

5. i) {z, {x, y}}, (ii). S (iii). {x, y, z}, and (iv).{x, y, z}.

6. P (S) = the power set of S = {, S, {(a, b)}, {c}}.

7. 2n

8. 2m n

9. i) No, ii) No, iii) Yes, iv) No, v) Yes, vi) No.

10. i) anti symmetric, ii) symmetric, iii) anti symmetric, iv) anti symmetric

and transitive, v) reflexive, symmetric and transitive.

11. The identity relation is both symmetric and anti-symmetric. Also, define

the relation in which xRy is true if and only if x = y is both symmetric

and anti-symmetric. Take A = {a, b, c}, define R = {(a, a), (b, b)}.

12. i) Equivalence relation, ii) Not an equivalence relation..

13. i) An equivalence relation

ii) Equivalence class containing 0,

[0] = {0, 3, 6, 9, 12, …}

iii) The set of equivalence classes are {[0], [1], [2]}.

14. i) Equivalence relation

ii) [(1, 2)] = {(1, 2), (-1, -2), (2, 4), (-2, -4), (3, 6), (-3, -6), …}

15. [0] = [4] = {0, 4}, [1] = [3] ={1, 3}, [2] = {2}.

16. Partition.

17. i) Not reflexive as (2, 2) R1.

Not symmetric as (3, 4) R1 but (4, 3) R1

Not anti symmetric as (1, 2), (2,1) R1, but 1 2



Not transitive as (2, 1), (1, 2) R1 but (2, 2) R1.

ii) Reflexive: For any a S2, a2 0 and so (a, a) R2

Symmetric: If (a, b) R2, then ab 0, so ba 0 and hence (b, a)

R2.

Not Anti symmetric: (5, 2) R2, since 5(2) = 10 0 and similarly

(2, 5) R2, but

5 2.

Not transitive: (5, 0) R2, (0, -6) R2 but (5, -6) R2, since 5(-6)

≱ 0.

iii) Reflexive:

Not symmetric: ((1, 2), (3, 4)) R3, but ((3, 4), (1, 2)) R, since 3

+ 4 ≰ 1 + 2.

Not anti symmetric: ((1, 2), (0, 3)) R3, ((0, 3), (1, 2)) R3 but (1,

2) (0, 3).

Transitive:

18. No: Since R is not reflexive as (2, 2) R.

19. Yes

20. (i). No; (ii). No; (iii).Yes.

21. 2n.

22. MR =

100

010

011

23. (i) {(1, 1), (2, 2), (3, 1), (3, 2)}

(ii) {(1, 1), (1, 4), (2, 2), (2, 3), (3, 3), (3, 4), (4, 1), (4, 3)}.

24. Reflexive and Symmetric.

25. (i) R(r) = {(0, 0), (1, 1), (2, 2), (3, 3), (0, 1), (1, 2), (2, 3)},

R(s) = {(0, 1), (1, 0), (1, 2), (2, 1), (2, 3), (3, 2)},

R(T) = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)}.



Unit 2 Recursive Functions, Strings

and Growth Functions

Structure

2.1 Introduction

Objectives

2.2 Recursive Functions

2.3 Integer Functions

2.4 Sequences and Strings

2.5 Growth Functions


2.6 Summary


2.8 Answers

2.1 Introduction

Sometimes it is difficult to define an object explicitly. However, it may be

easy to define this object in terms of itself. This process is called recursion.

Recursion relates to several related concepts in computer science.

In computer programming, recursion plays an important role. It is an

important facility in many programming languages.

Recursion is the technique of defining a function, a set or an algorithm in

terms of itself. That is, the definition will be in terms of previous values.

Objectives:


Understand the concept of Recursive Functions

Recognize sequences and Strings

Know the idea of Growth Functions



2.2 Recursive Functions

2.2.1 Definition

A function , where N is the set of non-negative integers is defined

recursively if the value of f at 0 is given and for each positive integer n, the

value of f at n is defined in terms of the values of f at k, where 0 ≤ k < n.

Observation: f defined (above) may not be a function. Hence, when a

function is defined recursively it is necessary to verify that the function is

well defined.

2.2.2 Example

The sequence 1, 4, 16, 64, ... , can be defined explicitly by the formula

f(n) = 4n for all integers n 0.

The same function can also be defined recursively as follows:

f(0) = 1, f(n + 1) = 4f(n), for n > 0

To prove that the function is well defined we have to prove existence and

uniqueness of such function. In this case, existence is clear as f(n) = 4n.

2.2.3 Theorem (Recursion Theorem)

Let F be a given function from a set S into S. Let s0 be fixed element of S.

Then there exists a unique function where N is the set of non-

negative integers satisfying

(i) f(0) = s0

(ii) f(n + 1) = F(f(n)) for all integers .

(Here the condition (i) is called initial condition and (ii) is called the

recurrence relation).

2.2.4 Example

Define n! recursively and compute 5! recursively.



Solution: We have . Then

(i) f(0) = 1

(ii) f(n + 1) = (n + 1)f(n) for all n 0.

Clearly f(n) = n!.

Now we compute 5! recursively as follows:

5! = 5. 4!

= 5. 4. 3!

= 5. 4. 3. 2!

= 5. 4. 3. 2. 1!

= 5. 4. 3. 2. 1. 0!

= 5. 4. 3. 2. 1. 1

=120.

2.2.5 Note

Any sequence in arithmetic progression or geometric progression can be defined

recursively. Consider the sequence a, a + d, a + 2d, …. Then

A(0) = a, A(n + 1) = A(n) + d.

Consider another sequence a, ar, ar2, … . Then

G(0) = a, G(n +1) = r G(n).

2.2.6 Definition

The Fibonacci sequence can be defined recursively as

(i) F0 = 1 = F1

(ii) Fn+1 = Fn + Fn-1 for n > 1.

Then

F2 = F1 + F0 = 2

F3 = F2 + F1 = 3

F4 = F3 + F2 = 5 …..

Here, there are two initial conditions.



2.2.7 Example

Define

f(x) =

oddisxw hen

2

1x

evenisxw hen2

x

.

Solution: Define such that f(0) = 0 and f(x + 1) = x – f(x).

Then f(6) = 5 – f(5) = 5 – [4 – f(4)]

= 5 – 4 + [3 – (3)]

= 5 – 4 + 3 – 2 + [1 – f(1)]

= 5 – 4 + 3 – 2 +1 – [0 – f(0)]

= 3.

and f(5) = 4 – f(4)

= 4 – [3 – f(3)]

= 4 – 3 + 2 – [1– f(1)]

= 4 – 3 + 2 – 1 + [0 – f(0)]

= 2.

2.2.8 Example

Using recursion theorem, verify that the object defined by the recursive

definition is a function. That is.,

(i) g(0) = 1

(ii) g(n + 1) = 3[g(n)]2 + 7 for all n > 0

Solution: We obtain

(i) s0 = 1

(ii) f(k) = 3k2 + 7, where

Then g(0) = s0. And g(n +1) f(g(n)). Thus g is a well-defined function.



The following is an example of a recursive function that does not define a

function.

2.2.9 Example

Consider a recursion function g: Z+ (the set of positive integers) (the set

of integers), for all integers n 1.

Solution: Suppose g is a function. Then by definition of g,

g(1) = 1

g(2) = 1 + g(1) = 1 + 1 = 2.

g(3) = g(8) = 1 + g(4) = 1 + (1 + g(2)) = 1 + (1 + 2) = 4.

g(4) = 1 + g(2) = 1 + 2 = 3.

Now, g(5) = g(14) = 1 = g(7) = 1 + g(20)

= 1 + (1 + g(10))

= 1 + 1(1 + (1 + g(5)))

= 3 + g(5).

Subtracting g(5) we get 0 = 3. Therefore g is not well defined.

2.2.10 Definition

If m and n are two non-negative integers then the (greatest common divisor)

g.c.d. (m, n) is defined as the largest positive integer d such that d divides

both m and n. Euclidean algorithm computes the greatest common divisor

(g.c.d.) of two non-negative integers.

We can find g.c.d. (m, n) recursively as follows:

otherw ise))n,mmod(,n(.d.c.g

0nifm

mnif)m,n(.d.c.g

)n,m(.d.c.g

where mod (m, n) is the remainder obtained when m is divided by n.

Observations:

a) The first part interchanges the order of m and n if n > m.



b) Second part is the initial condition.

c) Third part is the recursive part mod (m, n) will become 0 in a finite

number of steps.

2.2.11 Example

Calculate the g.c.d. (20, 6).

Solution: g.c.d. (20, 6) = g.c.d. (6, mod (20, 6)) (since 20 = 6• 3 + 2)

= g.c.d. (6, 2)

= (2, mod (6, 2))

= g.c.d. (2, 0)

= 2.

2.2.12 Example


Solution: g.c.d. (81, 36) = g.c.d. (36, 9)

= g.c.d. (9, 0)

= 9.

2.2.13 Example


Solution: g.c.d. (22, 8) = g.c.d. (8, mod (22, 8))

= g.c.d. (8, 6)

= g.c.d. (6, mod (8, 6))

= g.c.d. (6, 2)

= g.c.d. (2, 0)

= 2.

Note: The recursive definition can be extended to functions of more than

one variable.

Consider the following example.



2.2.14 Example

Define f(x, y) = x + y recursively.

Solution: Here, we keep x fixed and use recursion on y. We define

(i) f(x, 0) = x

(ii) f(x, y + 1) = f(x, y) + 1.

Take x = 2, y = 3. Now f(2, 3) = f(2, 2) + 1

= f(2, 1) + 1 + 1

= f(2, 0) + 1 + 1 +1

= 2 + 1 + 1 +1

= 5.

2.2.15 Example

Define g(x, 0) = 0, g(x, y + 1) = g(x, y) + x. Take x = 3, y = 4. Then

g(3, 4) = g(3, 3) + 3

= g(3, 2) + 3 + 3

= g(3, 1) + 3 + 3 + 3

= g(3, 0) + 3 + 3 + 3 + 3 = 12 (since g(3, 0) = 0).

2.3 Integer Functions

2.3.1 Definition

For any real number x, we define the floor of x as

2.3.2 Example

Take x = 2.52, then



2.3.3 Definition

2.3.4 Example

Take x = 3.732, then



2.3.5 Geometric Interpretation

Floor and Ceiling functions may be understood from their graphical

(or geometrical) representation. Consider the line f(x) = x, the diagonal on I, III

coordinates, take x=e=2.71828…. we describe floor and ceiling of e as follows:

2.3.6 Properties



2.3.7 Example

The above rules can be illustrated, by taking x = 4.5.

2.3.8 Problem

Prove that for any integer n and real x.



2.3.9 Remark

In general, taking out of a constant factor is not true.

2.3.10 Problem

Find a necessary and sufficient condition that when n is a

positive integer.

2.4 Sequences and Strings

Let us see the simple example, to understand this.



Komal is running a successful business and is planning a short vacation trip.

She is planning to take a cell phone with her so that in the case of an

emergency the manager at work can reach her. To budget her calls, she

looks at various plans and chooses the plan that charges US $ 1.00 for the

connection charge and US $ 10 for each minute. For example, the charges

for a one minute call are US $ 1.10, the charges for a two minute call are US

$ 1.20 and so on. So Komal makes the following table for the first ten

minutes of telephone charges:

Min 1 2 3 4 5 6 7 8 9 10

Charges 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00

From the table we see that for a 30 min call, the charges are 1.00 + 30(0.10)

= 4.00.

In general, for an n-minute call, the charges are

1.00 + n(0.10) = 1.00 + 0.10 n.

Let us list the telephone charges as follows.

1.10, 1.20, 1.30, …, (1.00 + 0.1 n), …

This is an ordered list of real numbers in which the first element is 1.10, the

second element is 1.20, and so on. Such an ordered list of elements is

called a sequence. If the sequence stops after n elements for some positive

integer n then it is called finite otherwise it is called infinite sequence.

Define a function (real numbers) as f(n) = 1.00

+ (0.1)n.

Then f(1) = 1.00 + (0.1)1 = 1.10 = c1

f(2) = 1.00 + (0.1)2 = 1.20 = c2

… … …

f(n) = 1.00 + (0.1)n = cn.



2.4.1 Definition

An infinite sequence or a sequence, on a non empty set X is a function from

the set of positive integers N to X. A sequence whose terms are integers is

called an integer sequence.

2.4.2 Example

Let f: N Z be a function defined by f(n) = n2. Then f(1) = 1, f(2) = 22 = 4,

and so we get a sequence 1, 4, 9, …, n2, … is a sequence on A. Let an

denote the nth term of this sequence. Then a1 = 1, a2 = 4, …, and so on. We

denote this sequence by 1n

2}n{ or simply by {n2}.

2.4.3 Definition

Let 1nn }a{ be a sequence. Consider the following terms of this sequence:

am, am+1, …, an.

Some of the common things we do with these terms are adding them and

multiplying them. Let us first consider the addition.

The sum of the terms am, am+1, …, an is written as

n

mi

ia . There is nothing about the choice of the variable i. We could choose

j or k as the index of the summation and write the sum as

n

mj

ja or

n

mk

ka .

2.4.4 Example

(i)

4

1i

i = 1 + 2 + 3 + 4 = 10.

(ii)

3

1i1i

i = ½ + 2/3 + ¾ = 23/12.



2.4.5 Example

Consider the sums

3

1i

)1i( and

4

2i

j

3

1i

)1i( = (1+1) + (2 + 1) + (3 +1) = 9.

4

2i

j = 2 + 3 + 4 = 9.

Therefore

3

1i

)1i( =

4

2i

j

2.4.6 Note

To change the index variable in a sum we do the following.

1. Calculate the lower limit of the new index variable.

2. Calculate the upper limit of the new index variable.

3. Find the general term of the summation in terms of the new index

variable.

2.4.7 Example

Consider the sum

3

1i

)1i( and change the index variable to j = i + 1.

Solution: Step (i): Lower limit for j: The lower limit for i is 1, so the lower

limit for j is

j = i + 1 = 1 + 1 = 2.

Step (ii): Upper limit for j: The upper limit for i is 3, so the upper limit for j is

j = i + 1 = 3 + 1 = 4.

Step (iii): The general term is i + 1 = j.

Hence, the equivalent sum is

4

2j

j .



2.4.8 Example

Consider the sum

1n

0i

2)i1n( , change the index variable to j = i + 1.

Solution: Step (i): Lower limit for j = i + 1 = 0 + 1 = 1.

Step (ii): Upper limit for j = i + 1 = n – 1 + 1 = n.

Step (iii): The general term for the new summation is given by

n2 + 1 + i = n2 + j.

Hence, the new sum is

n

1j

2 )jn( .

2.4.9 Properties of Summation

Let 1nn }a{ and

1nn }b{ be sequences of real numbers and let c be a real

number. Suppose m and n are integers such that 1 ≤ m ≤ n. Then

(i)

n

mi

ia +

n

mi

ib = )ba(n

mi

ii

,

(ii) c.

n

mi

ia =

n

mi

ica .

2.4.10 Definition

Let A be a nonempty finite set. A string or word, over A is a finite sequence

of elements from A. The set A is called an alphabet.

A string with no element in it is called the empty string or empty word.

If s1 and s2 are two strings over a set A, then the concatenation of s1 and s2

is the string s1s2. That is, to obtain the concatenation of s1 and s2 we list the

elements of s1 followed by the elements of s2.

2.4.11 Example

Suppose s1 = abbabcdb and s2 = caabcdbbd are two strings over the set A =

{a, b, c, d}. Then the concatenation of s1 and s2 is s1s2 = abbabcdbcaabcdbbd.

It follows that if s1 and s2 are strings over a set A, then s1s2 = s1 + s2 .



Example: Let A = {0, 1}.

2.5 Growth Functions

The growth of a function is often described using a special notation, O-

notation (read as “big-oh notation”). It provides a special way to compare

relative sizes of functions that is very useful in the analysis of computer

algorithms. It often happens that the time or memory space requirements for

the algorithms available to do a certain job differ from each other on such a

grand scale that differences of just a constant factor are completely

overshadowed. The O-notation makes use of approximations that highlight

these large-scale differences while ignoring differences of a constant factor

and differences that only occur for small sets of input data.

2.5.1 Definition

Let f and g be functions from the set of integers or the set of real numbers to

the set of real numbers. Then f of order g written as f(x) is O(g(x)), if there

are constants C and k such that f(x) ≤ C g(x) whenever x > k (this is read

as „f(x) is big-oh of g(x)‟).

2.5.2 Remark

To show f(x) is O(g(x)), we need only find one pair of constants C and k

such that f(x) < C(g(x)) if x > k. However, a pair C, k that satisfies the

definition is not unique. Moreover, if one such pair exists, there are infinitely

many such pairs. A simple way to see this is to note that if C, k is one such

pair, any pair C1, k1 with C < C1 and k < k1 also satisfies the definition, since

f (x) < C1g(x) whenever x > k1 > k.

2.5.3 Example



Use O-notation to express , for all real numbers

x > 1.

Solution: Take C = 12 and k = 1, the given statement translates to 2x3 + 2x

+ 7 is O(x3).

2.5.4 Note Order of Polynomial functions

(i) If 1 < x then x < x2 and so x2 < x3. Thus, if 1 < x, then 1 < x < x2 < x3.

(ii) For any rational numbers r and s, if x > 1 and r < s, then xr < xs.

Therefore xs is O(xs).

2.5.5 Example

Show that for any real number x > 1, 2x4 + 4x3 + 5 ≤ 11x4.

Solution: Since x is a real number and x > 1, we have

x3 < x4 and 1 < x4. So

4x3 < 4x4 and 5 < 5x4.

2.5.6 Example

Use the definition of order to show that x2 + 2x + 1 is O(x2).

Solution: The functions f and g referred to in the definition of O-notation are

defined as follows. For all real numbers x, f(x) = x2 + 2x + 1 and g(x) = x2.

2.5.7 Example



Use the definition of order to show that 5x3 – 3x + 4 is O(x3).

Solution: The functions f and g referred to in the definition of O-notation are

defined as follows.

2.5.8 Example

Show that 9x2 is O(x3). Is it true that x3 is O(9x2)?

Solution: We note that 9x2 < x3 is true whenever x > 9 (by dividing both

sides by x2). Hence, 9x2 is O(x3), taking C = 1 and k = 9 in the definition of

big-oh notation.

To determine whether x3 is O(9x2) or not , it is necessary to determine

whether there are constants C and k such that x3 ≤ C (9x2) whenever x > k.

This is equivalent to the inequality x < 9C (we get this by dividing both sides

by x2).

No such x can exist since x can be marked arbitrarily large. Hence, x3 is not

O(9x2).

Now we generalize the above example, to show that any polynomial

function is big –oh of the power function of its highest order term or of any

larger power function.



2.5.9 Theorem

Let f(x) = anxn + an-1x

n-1 + … + a1x + a0 where a0, a1, …, an-1, an are real

numbers then f(x) is O(xn).

Proof: Using the triangle inequality, if x > 1, we have that

2.5.10 Example

Use big-oh notation to estimate the sum of the first n positive integers?

Solution: Each positive integer n is greater than every positive integer that

precedes it. Therefore, for each positive integer n,

1 + 2 + 3 + … + n ≤ timesn

nnn

... = n n = n2.

Therefore 1 + 2 + 3 + … + n is O(n2).

2.5.11 Note

“big O-estimates will be developed for the factorial function and its

logarithm. These estimates will be important in the analysis of the number of

steps used in sorting procedures.

2.5.12 Example

Give big-O estimates for the factorial function and the logarithm of the

factorial function, where the factorial function f(n) = n = 1 2 3 … n

where n is a positive integer, and 0 = 1.

For example, 1 = 1, 2 = 2, 3 = 6, 4 = 24.

Note that the function n grows rapidly.

20 = 2, 432, 902, 008, 176, 640, 000.



Solution: A big-O estimate for n can be obtained by noting that each term

in the product does not exceed n. Hence,

n = 1 2 … n ≤ n n … n = nn.

This inequality shows that n is O(nn), taking C = 1 and k = 1.

Taking logarithm both sides, we get

log n ≤ log nn = n log n.

This shows that log n is O(n log n), again taking C = 1 and k =1.

2.5.13 Theorem

Suppose that f1(x) is O(g1(x)) and f2(x) is O(g2(x)).

2.5.14 Theorem

Suppose that f1(x) and f2(x) are both O(g(x)). Then (f1 + f2)(x) is O(g(x)).

2.5.15 Theorem

Suppose that f1(x) is O(g1(x)) and f2(x) is O(g2(x)). Then (f1f2) (x) is O(max

(g1(x)g2(x)).

2.5.16 Example

Give a big O-estimate for f(n) = 3n log(n!) + (n2 + 3) log n, where n is a

positive integer.

Solution: First we estimate the product 3n log (n). From the example, we

have log n is O(n log n). Also 3n is O(n).

Using these two estimates, and the theorem, we have estimation that 3n

log (n) is

O(n2 log n).

Next the product (n2 + 3) log n will be estimated.

Therefore (n2+3)log n is O(n2 log n). Using theorem, to combine the two big O-



estimates for the products shows that f(n) = 3nlog (n) + (n2 + 3)log n is O(n2 log n).

2.5.17 Example

Give a big – O estimate for f(x) = (x + 1)log(x2 + 1) + 3x2.

Solution: A big-O estimate for (x + 1) is O(x).

When x > 1, x2 + 1 ≤ 2x2. Therefore,

log (x2 + 1) ≤ log(2x2) = log 2 + log x2 = log 2 + 2 log x ≤ 3 log x, if x > 2.

Therefore log(x2 + 1) is O(log x).

By theorem, it follows that (x + 1)log(x2 + 1) is O(x logx).

Further, 3x2 is O(x2), by theorem, f(x) is O(max (x logx, x2 )).

Since x log x ≤ x2, for x > 1, it follows that f(x) is O(x2).

2.5.18 Big-Omega and Big-Theta Notations

If f(x) is O(g(x)), all that one can conclude is that except for constants and a

finite number of exceptions, f is bounded above by g, so g grows at least as

fast as f. For example, if f(x) = x and g(x) = 2x, then f(x) is O(g(x)), but g

grows considerable faster than f. The statement f(x) is g(x) says nothing

about lower bund for f. For this, we use big-Omega notation. When we want

to give both upper and a lower bound on the size of a function f(x) relative to

a reference function g(x), we use big-Theta notation.

2.5.19 Definition


the set of real numbers. We say that f(x) is , if there are positive

constants C and k such that whenever x > k which is

read as “f(x) is big-Omega g(x)”.


the set of real numbers. We say that f(x) is g(x) if f(x) is O(g(x)) and f(x) is

. When f(x) is g(x), we say that “f is big-Theta of g(x)” and we say

f(x) is of order g(x).



Therefore, f(x) is O(g(x)), then f(x) is an upper bound for f(x) and whenever

f(x) is g(x), g(x) is a lower bound for f(x). The big-O notation compares the

rate of growth of functions rather than their values, so when f(x) is g(x),

f(x) and g(x) have the same rates of growth, but can be very different in their

values.

2.5.20 Example

Let f(x) = x and g(x) = 1, 000, 000x, then f(x) ≤ C g(x) for C = 1 and k = 1, so

that f(x) is O(g(x)). Also Cf(x) g(x) for C = 1, 000, 000 and k = 1, so f(x) is

O(g(x)). Therefore f(x) is g(x).


2.6 Summary

In this unit we discussed the recursive functions which are useful to write

efficient algorithms. There are special integer functions defined with enough

illustrations. The concepts, sequences, strings and growth are introduced.

These are useful in approximating the time complexity and construction of

DFA, NDFA.


1. Explain the geometrical meaning of floor and ceiling functions.

2. Practice to find order of different algebraic polynomials.



3. Practice problems on Big-O, Big- notations.

2.8 Answers


1. 2

2. 2

3. 0

4. 2

5. 1

6. The concatenation of 1110 and 0111 is: 11100111.

7. The concatenation of aaabbb and bbabab is: aaabbbbbabab.



Unit 3 Methods of Proof

Structure

3.1 Introduction

Objectives

3.2 Proof Techniques


3.3 Summary


3.5 Answers

3.1 Introduction

In this unit we discuss various methods of proofs and few examples. The

techniques will give an idea to analyze and solve the problems.

Objectives:


Understand the proof techniques

Apply in various problems

Analyse various proof in automata

3.2 Proof Techniques

A significant requirement for reading this subject is the ability to follow

proofs. In mathematical arguments, we employ the accepted rules of

deductive reasoning, and many proofs are simply a sequence of such steps.

Direct Proof: Consider a set of hypothesis H1, H2, …, Hn from which we

want to infer a conclusion C.

Consider the example: Prove that if x and y are rational numbers then x +

y is rational.



Solution: Since x and y are rational numbers, we can find integers p, q, m,

n such that x = p/q and y = m/n. Then x + y = p/q + m/n = (pn + mq)/qn.

Since pn + mq and qn are both integers, we conclude that x + y is a rational

number.

Indirect Proof: Proofs which are not direct are called indirect. Two main

types of indirect proof, uses the negation and conclusion, so they are often

suitable when that negation is easy to state. The first type of proof is contra-

positive proof.

Consider the example: Prove that if m + n 73, then m 37 or n 37, m

and n being positive integers.

Solution: We prove this by taking contra-positive: not “m 37 or n 37”

implies not “m + n 73”. By De morgan law, the negation of “m 37 or n

37” is “not m 37 and n 37”. That is,

“m ≤ 36 and n ≤ 36” so that the contrapositive proposition is if m ≤ 36 and n

≤ 36 then m + n ≤ 72. This follows from the inequalities: a ≤ c and b ≤ d

imply that

a + b ≤ c + d for all real numbers a, b, c, d.

Few special proof techniques are used so frequently that it is appropriate to

review them briefly.

1. Proof by induction

2. Proof by contradiction

3. The pigeonhole principle, and

4. The Digitalization Principle

5. Proof by Contradiction

6. Exhaustive Proof and Proof by cases



3.2.1 Proof by Induction

Let A be the set of all natural numbers such that

Then A = N.

In other words: The principle of mathematical induction states that any set

of natural numbers containing zero, and with the property that it contains n +

1 whenever it contains all the numbers up to and including n, must in fact be

the set of all natural numbers.

In practice, induction is used to prove assertions of the following form:

“For all natural numbers n, property P is true.”

The above principle is applied to the set A = {n: P is true of n} in the

following way.

(1) In the basis step we show that 0 A, that is, that P is true of 0.

(2) The induction hypothesis is the assumption that for some fixed but

arbitrary n 0, P holds for each natural number 0,1,... , n.

(3) In the induction step we show, using the induction hypothesis, that P is

true for n + 1. By the induction principle, A is then equal to N, that is, P

holds for every natural number.

3.2.2 Example

Prove by mathematical induction that the sum of the first n natural numbers

is 2

1nn .

Solution:

That is to prove that 1 + 2 + 3 + …. + n

2

1nn

(i) Base Step: Let n = 0. Then the sum on the left is zero, since there is



nothing to add. The expression on the right is also zero.

For 1n , left side = 1, right side

12

111

. Hence the result is

true for 1n

(ii) Induction Hypothesis: Assume that the result to be true for m ≤ n

and n 0. Then 1 + 2 + 3 + … + 2

1mmm

(iii) Induction Step: We now show that the above result is true for 1mn .

Adding the th1m term viz., 1m to both sides we obtain.

1 + 2 + 3 + ... +

1m2

1mm1mm

2

2m1m1

2

m1m

2

11m1m ,

which is the same as the given result for 1mn

Hence by mathematical induction, the result is true for all positive integral

values of n.

3.2.3 Example

Prove by mathematical induction that

6

1n21nnn....321

2222

Solution:

(i) Base Step: Let n = 0. Then the sum on the left is zero, since there is

nothing to add. The expression on the right is also zero.

If 1n , left side 112

.



Right side

16

3.2.1

6

11.2111

.

Hence the result is true for 1n .

(ii) Induction Hypothesis: Assume that the result to be true for mn

Then

6

1m21mmm...321

2222 .

Adding the th1m term i.e. 21m to both sides of the above

equation, we get,

222221m

6

1m21mm1mm...21

1m61m2m6

1m

6m7m2

6

1m 2

6

3m22m1m

6

11m211m1m

Therefore the result is true for 1mn . Hence by mathematical induction the

given result is true for all positive integers n.

3.2.4 Example

For any finite set A, the cardinality of the power set of A is 2 raised to a

power equal to the cardinality of A.

Solution:



Now the power set of A can be divided into two parts, those sets containing

the element a and those sets not containing a. The latter part is just 2B, and

the former part is obtained by introducing a into each member of 2B. Thus

This division, in fact partitions 2A into two

disjoint equinumerous parts, so the cardinality of the whole is twice 2B,

which, by the induction hypothesis, is 2 2n = 2 n+1. This completes the

proof.

3.2.5 Example

(Refer unit 4 for definition of binary tree) A binary tree is a tree in which no

parent can have more than two children. Prove that a binary tree of height n

has at most 2n leaves.

Solution: If we denote the maximum number of leaves of a binary tree of

height n by l(n), then we want to show that l(n) ≤ 2n.

Basic Step: Clearly l(0) = 1 = 20 since a tree of height 0 can have no nodes

other than the root, that is , it has at most one leaf.

Inductive Hypothesis: l(i) ≤ 2i for i = 0, 1, …, n.

Inductive step: To get a binary tree of height n +1 from one of height n, we

can create, at most, two leaves in place of each previous one. Therefore

l(n + 1) = 2l(n).

Now, using the inductive assumption, we get



Therefore our claim is true for n + 1. Since n is

arbitrary, we can conclude that the statement is true for all n.

3.2.6 Example

(Refer unit 4 for the definition of tree) A tree G with „n‟ vertices has (n -1)

edges.

Proof : We prove this theorem by induction on the number vertices n.

Basic step: If n = 1, then G contains only one vertex and no edge. So the

number of edges in G is n –1 = 1 – 1 = 0.

Induction hypothesis: The statement is true for all trees with less than „n‟

vertices. Induction step: Now let us consider a tree with „n‟ vertices. Let „ek‟

be any edge in T whose end vertices are vi and vj. Since T is a tree, by

Theorem 6.5, there is no other path between vI and vj. So by removing ek

from T, we get a disconnected graph. Furthermore, T- ek consists of exactly

two components (say T1 and T2). Since T is a tree, there were no circuits in

T and so there were no circuits in T1 and T2. Therefore T1 and T2 are also

trees.

It is clear that |V(T1)| + |V(T2)| = |V(T)| where V(T) denotes the set of vertices

in T.

Also |V(T1)| and |V(T2)| are less than n.

Therefore by the induction hypothesis, we have

|E(T1)| = |V(T1)| - 1 and |E(T2)| = |V(T2)| - 1.

|E(T1)| = |V(T1)| - 1 and |E(T2)| = |V(T2)| - 1.

3.2.12 Problem

Prove by mathematical induction that n2 n for all positive integer n.



Solution: Let P n be the given proposition. Now P 1 implies 2 > 1 which

is true. Hence P 1 is true

Induction hypothesis: Let us assume that P m is true. That is m2 m

Now m 1 m2 2 . 2 2m . We know that 2m m m m 1 for all

m N . m 1Therefore 2 m 1 . Hence P m 1 is true.

Therefore by induction P n is true for all n.

3.2.13 Example

Show by induction that n n 1 2n 1 is divisible by 6.

Solution: Let P n n n 1 2n 1

Now P 1 1 . 1 1 2 1 6 , this is divisible by 6.

Assume that P m is divisible by 6.

That is, m m 1 2m 1 is divisible by 6.

Therefore m m 1 2m 1 6k for some integer k.

Now

P(m+1) = (m+1) [(m+1) + 1] [2 (m+1)+1]

m 1 m 2 2m 3

m 1 m 2 2m 1 2

m 1 m 2 2m 1 2 m 1 m 2

m m 1 2m 1 2 m 1 2m 1 2 m 1 m 2

6k 2 m 1 3m 3 by induction hypothesis

2

6k 6 m 1



Since each term on the R.H.S is divisible by 6 their sum is also divisible by 6.

Hence P m 1 is divisible by 6. Therefore by induction P n is divisible by

6 for all n N

3.2.7 The Pigeonhole Principle

If A and B are finite sets and , then there is no one-to-one function

from A to B.

(In other words, if we attempt to pair off the elements of A (the “pigeons”)

with elements of B (the “pigeonholes”), sooner or later we will have to put

more than one pigeon in a pigeonhole).

Proof:

Basis Step: Suppose B = 0, that is, B = . Then there is no function f: A

B and so no one to one function.

Induction Hypothesis: Suppose that f is not one-to-one, provided that f: A

B, A> B, and B ≤ n, where n 0.

Induction Step: Suppose that f: A B and A> B = n + 1. Choose some

a A (since A > B = n + 1 1, A is nonempty, and therefore such a

choice is possible). If there is another element a a1 A, such that f(a) =

f(a1), then obviously f is not a one-to-one function, and we are done.

So, suppose that a is the only element mapped by f to f(a).

Consider then the sets A – {a}, B-{f(a)}.

The function g: A-{a} B-{f(a)} that agrees with f on all elements of A-{a}.

Now the induction hypothesis applies, because B-{f(a)} has n elements, and

A -{a} = A -1 > B -1 = B-{f(a)}.

Therefore, there are two distinct elements of A-{a} that are mapped by g

(and therefore by f) to the same element of B-{b}. Hence f is not one-to-one.



3.2.8 The Diagonalization Principle

Let R be a binary relation on a set A, and let D, the diagonal set for R, be {a

aA and (a, a) R}. For each a A, let

Ra = {b: b A and (a, b) R}. Then D is distinct from each Ra.

If A is a finite set, then R can be pictured as a square array; the rows and columns

are labeled with the elements of A and there is a cross in the box with row labeled

a and column labeled b, just in case (a, b) B. The diagonal set D corresponds to

the complement of the sequence of boxes along the main diagonal, boxes with

crosses being replaced by boxes without crosses, and vice versa. The sets Ra

correspond to the rows of the array. The diagonalization principle can then be

rephrased: the complement of the diagonal is different from each row.

3.2.9 Example

Let us consider the relation R = {(a, b), (a, d), (b, b), (b, c), (c, c), (d, b), (d,

c), (d, e), (d, f), (e, e), (e, f), (f, a), (f, c), (f, d), (f, e)}; notice that Ra = {b, d},

Rb = {b, c}, Rc = {c}, Rd = {b, c, e, f}, Re = {a, e}, and Rf = {c, d, e}. All in all, R

may be pictured like this:

The sequence of boxes along the diagonal is

x x x



Its complement is

x x x

which corresponds to the diagonal set D = {a, d, f}. Indeed, D is different

from each row of the array; for D, because of the way it is constructed,

differs from the first row in the first position, from the second row in the

second position, and so on.

Mathematical induction is the process of proving a general theorem or

formula involving the positive integer n from particular cases.

A proof by mathematical induction consists of the following two steps.

(i) Show by actual substitution that the theorem is true for 1n

(ii) Assuming the theorem to be true for mn , prove that it is also true for

1mn

Note that here m is a particular value of n . From (i) the theorem is true for

1n and from (ii) it is true for 211n ; since it is true for 2n it follows from

(iii) that it is also true for 312n and so on. Hence theorem is true for

all positive integral values of n .

3.2.10 Proof by Contradiction

Proof by contradiction is sometimes a very useful technique to prove that

some statements are true. In this technique, let us assume that property P

is not true. Using logical reasoning we have to get a conclusion that

contradicts the given conditions.

3.2.11 Example

Prove by contradiction, that is not a rational number.

Solution: A rational number is of the form p/q where , and p, q are not

having any common factors.

Assume that is a rational number. So it can be written as



If p is even, then it can be written as p = 2k. Therefore 4k2 = 2q2. Therefore

q is even.

This is a contradiction to our assumption that p and q have no common

factors. Therefore is not a rational number.

3.2.12 Example

Give a proof by contradiction of “if 3n + 2 is odd, then n is odd”.

Solution: Let p: 3n+2 is odd

q: n is odd.

To construct a proof by contradiction, assume that both p and q are true.

That is, assume that 3n + 2 is odd and that n is not odd.

Since n is not odd, it is even.

Now we can show that if n is even, then 3n + 2 is even.

(Verification: n is even n = 2k for some integer k. Substituting 2k for n,

we get 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) 3n + 2 is even).

Now the statement “3n + 2 is even” is p. Now since p and p are true, we

have a contradiction. This completes the proof by contradiction, proving that

if 3n + 2 is odd, then n is odd.

3.2.13 Exhaustive Proof and Proof by Cases

Some theorems can be proved by examining a relatively small number of

examples. Such proofs are called exhaustive proofs, since these proofs

proceed by exhausting all possibilities. An exhaustive proof is a special type

of proof by cases where each case involves checking a single example.



3.2.14 Example

Prove that (n + 1)3 3n if n is a positive integer with n ≤ 4.

Solution: We use a proof by exhaustion. We only need verify the inequality

(n + 1)3 3n when n = 1, 2, 3, 4.

For n = 1, we have (n + 1)3 = 23 = 8 and 3n = 31 = 3;

for n = 2, we have (n + 1)3 = 33 = 27 and 3n = 32 = 9;

for n = 3, we have (n + 1)3 = 43 = 64 and 3n = 33 = 27; and

for n = 4, we have (n + 1)3 = 53 = 125 and 3n = 34 = 81;

Therefore, (n + 1)3 3n for all positive integers n ≤ 4.

3.2.15 Example

(Proof by Cases) Prove that if n is an integer, then .


1. Prove by mathematical induction that

4

1nnn...321

223333 .

2. Prove that is not a rational number (by the method of contradiction).

3. Prove that the product of two odd integers is an odd integer.



3.3 Summary

We introduced a variety of different methods of proof and illustrated how

each method is used. This unit is useful for several other important proof

methods, where we consider different cases separately and proof where we

prove the existence of objects with desired properties.


1. What is wrong with the following purported proof that all horses are the

same color?

The proof is by induction on the number of horses.

Basic step: There is only one horse. Then clearly all horses have the same

color.

Induction Hypothesis: In any group of upto n horses, all horses have the

same color.

Induction Step: Consider a group of n+1 horses. Discard one horse; by

induction hypothesis, all the remaining horses have the same color. Now

put that horse back and discard another; again all the remaining horses

have the same color. So all the horses have the same color as the ones that

were not discarded either time and so they all have the same color.

(Hint: The induction proof fails for n=2).

3.5 Answers


1. (i) For 1n , left side 113

right side

14

4.1

4

11122

Hence it is true for 1n .

(ii) Assume the result to be true for mn



Then

4

1mmm...321

223333 (induction

hypothesis)

Adding the th1m term viz., 31m to both sides,

322

33331m

4

1mm1mm...21

4m4m

4

1m 22

4

2m1m22

4

11m1m22

Therefore the result is true for 1mn . Hence by mathematical induction

the given result is established for all positive integers.

2. Proof by Contradiction method .

3. Take two odd integers m and n. Then there exist two integers r and t so

that m = 2r + 1 and n = 2t + 1.



Unit 4 Graphs and Binary Trees

Structure

4.1 Introduction

Objectives

4.2 Definitions and Examples

4.3 Isomorphism

4.4 Trees

4.5 Rooted Trees


4.6 Summary


4.8 Answers

4.1 Introduction

Graph theory was invented in 1736 with Euler‘s paper in which he solved

the Kongsberg Bridges problem. In 1847, Kirchhoff (1824 - 87) developed

the theory of trees to applications in electrical networks. The last three

decades have witnessed more interest in Graph Theory, particularly among

applied mathematicians and engineers. Graph Theory has a surprising

number of applications in many developing areas. The Graph Theory is also

intimately related to many branches of mathematics including Group Theory,

Matrix Theory, Automata and Combinatorics. Graph Theory serves as a

mathematical model for any system involving a binary relation. The

development of high-speed computers is also one of the reasons for the

recent growth of interest in Graph Theory.

Objectives:

At the end of the unit the student must be able to

appreciate the relevance of Graph Theory in real life situation



observe the difference between different concepts defined through the

examples presented

understand some techniques used in proving simple theorems

know the isomorphism between two graphs.

Understand the trees and rooted trees.

4.2 Definitions and Examples

4.2.1 Definition

(i) A linear graph (or simply a graph). G = (V, E) consists of a nonempty

set of objects, V = {v1, v2, …} called vertices and another set, E = {e1,

e2, …} of elements called edges such that each edge ‗ek‘ is identified

with an unordered pair {vi, vj} of vertices. The vertices vi, vj associated

with edge ek are called the end vertices of ek.

(ii) An edge associated with a vertex pair {vi, vi} is called a loop (or)

selfloop.

(iii) If there are more than one edge associated with a given pair of

vertices, then these edges are called parallel edges (or) multiple

edges.

4.2.2 Example

Consider the graph given here.

e6

e7

e1

e2 e4

v3 v4

v5 e5

v2

v1



This is a graph with five vertices and six edges. Here G = (V, E) where

V = {v1, v2, v3, v4, v5} and E = {e1, e2, e4, e5, e6, e7}.

The identification of edges with the unordered pairs of vertices is given by

e1 {v2, v2}, e2 {v2, v4}, e4 {v1, v3}, e5 {v1, v3}, e6

{v3, v4}.

Here ‗e1‘ is a loop and e4, e5 are parallel edges.

4.2.3 Definition

A graph that has neither self-loops nor parallel edges is called a simple

graph. Graph containing either parallel edges or loops is also referred as

general graph. A graph ‗G‘ with a finite number of vertices and a finite

number of edges is called a finite graph. A graph ‗G‘ that is not a finite

graph is said to be an infinite graph.

Observation: The two graphs given below are one and the same.

4.2.4 Example

Consider the following three graphs

4 3 2

1

3

2

4

1

u

1

u1

u2

u3

e1

e2

e4

e3

e5

u4

u2

u3

e3

e1

e5

e6

e7 e2

e7

u4

e4



It can be observed that the number of vertices, and the number of edges

are finite. Hence these three graphs are finite graphs.

4.2.5 Example

Consider the two graphs given here. It can be understood that the number

of vertices of these two graphs is not finite. So we conclude that these two

figures represent infinite graphs.

4.2.6 Example

The diagrams of fig. (a) and (b) illustrate two non directed graphs. The

graph G, shown in Fig. (a) is not simple since there is a loop incident on

vertex c.

The graph G1 shown in Fig. (b) is simple since there are no self loops and

parallel edges.

The graph G11 in fig. (c) represents a multi-graph since there are three

edges between the vertices b and c.

u1

e4

e6

e6

e5

e2

e3

u2

u3

u5 e1

u4

Infinite Graphs



V(G) = {a, b, c, d) and E(G) = {{a, b},{a, c},{b, c},{c, c},{a, d},{c, d}}.

Therefore G is of order 4 and size 6.

Similarly, the graph G1 has order 4 and size 5, and the multi-graph G11 has

order 4 and size 7.

Observation: The non-directed graphs may be viewed as symmetric

directed graphs, in which for every edge (u, v) between two vertices in

direction there is also an edge (v, u) between the same vertices in the other

direction.

4.2.7 Definition

If a vertex v is an end vertex of some edge e, then v and e are said to be

incident with (or on, or to) each other.

4.2.8 Example

Consider the graph given here. Here the edges e2, e6, e7 are incident with

the vertex u4.

e6

e7

e1

e2

e3

e4

u3 u4

u5 e5

u2

u1



4.2.9 Definition

(i) Two non-parallel edges are said to be adjacent if they are incident on

a common vertex.

(ii) Two vertices are said to be adjacent if they are the end vertices of the

same edge.

4.2.10 Example

Consider the graph given in example 4.2.8. Here the vertices u4, u5 are

adjacent. The vertices u1 and u4 are not adjacent. The edges e2 and e3 are

adjacent.

4.2.11 Definition

The number of edges incident on a vertex v is called the degree (or

valency) of v. The degree of a vertex v is denoted by d(v). It is to be noted

that a self-loop contributes two to the degree of the vertex.

4.2.12 Example

Consider the graph given here.

Here d(u1) = 2; d(u2) = 1; d(u3) = 3; d(u4) = 2; d(u5) = 2; d(u6) = 2; d(u7) = 1;

d(u8) = 3; d(u9) = 2; d(u10) = 2.

u

10

u

1

u

2

u

3

u

4

u

5

u

6

u

7

u

8

u

9



4.2.13 Example

Consider the graph.

Here d(u1) = 2, d(u3) = d(u4) = 3; d(u2) = 3; d(u5) = 1

So, d(u1) + d(u2) + d(u3) + d(u4) + d(u5) = 2 + 3 + 3 + 3 + 1 = 12 = 2(6) =

2e, where e denotes the number of edges. Hence we can observe that

d(u1) + d(u2) + d(u3) + d(u4) + d(u5) = 2e

(that is, the sum of the degrees of all vertices is equal to twice the number of

edges).

4.2.14 Theorem

The sum of the degrees of the vertices of a graph G is twice the number of

edges. That is, Viv

i )v(d = 2e. (Here e is the number of edges).

Proof: (The proof is by induction on the number of edges ‗e‘).

Case-(i): Suppose e = 1. Suppose f is the edge in G with f = uv.

Then d(v) = 1, d(u) = 1. Therefore

Vx

)x(d = }v,u{\Vx

)x(d + d(u) + d(v) = 0 + 1 + 1 = 2 = 2 1

= 2 (number of edges).

Hence the given statement is true for n = 1.

Now we can assume that the result is true for e = k - 1.

Take a graph G with k edges.

u1

e5

e4

u3 e6

u5

u2

e2

u4

e7



Now consider an edge ‗f‘ in G whose end points are u and v.

Remove f from G.

Then we get a new graph G* = G - {f}.

Suppose d*(v) denotes the degree of vertices v in G*.

Now for any x {u, v}, we have d(x) = d*(x), and

d*(v) = d(v) -1, d*(u) = d(u) - 1.

Now G* has k - 1 edges. So by induction hypothesis

Viv

i*

)v(d = 2(k - 1).

Now 2(k - 1) = Viv

i*

)v(d = }v,u{iv

i*

)v(d + d*(u) + d*(v)

= }v,u{iv

i )v(d + (d(u) - 1) + (d(v) - 1)

= }v,u{iv

i )v(d + d(u) + d(v) - 2 = Viv

i*

)v(d - 2

2(k - 1) + 2 = Viv

i*

)v(d 2k = Viv

i )v(d

Hence by induction we get that ―the sum of the degrees of the vertices of the

graph G is twice the numbers of edges‖.

4.2.15 Theorem

The number of vertices of odd degrees is always even.

Proof: We know that the sum of degrees of all the ‗n‘ vertices (say, vi, 1 i

n) of a graph G is twice the number of edges (e) of G. So we have

n

1i

i )v(d = 2e --------- (i)

If we consider the vertices of odd degree and even degree separately, then



n

1i

i )v(d = isevenjv

j )v(d + isoddkv

k )v(d -------- (ii)

Since the L.H.S of (ii) is even (from (i)) and the first expression on the RHS

side is even, we have that the second expression on RHS is always even.

Therefore

isoddkv

k )v(d --------- (iii)

is an even number.

In (iii), each d(vk) is odd. The number of terms in the sum must be even to

make the sum an even number. Hence the number of vertices of odd

degree is even.

4.2.16 Example

Show that the number of people who dance (at a dance where the dancing

is done in couples) an odd number of times is even.

Solution: Suppose the people are vertices. If two people dance together,

then we can consider it as an edge. Then the number of times a person v

danced is (v). By Theorem 9.5.9, the number of vertices of odd degree is

even. Therefore the number of people who dance odd number of times is

even.

4.2.17 Definition

A vertex having no incident edge is called an isolated vertex. In other

words, a vertex v is said to be an isolated vertex if the degree of v is equal

to zero.

4.2.18 Example

Consider the graph given here. The vertices v4 and v7 are isolated vertices.



4.2.19 Definition

A vertex of degree one is called a pendent vertex or an end vertex. Two

adjacent edges are said to be in series if their common vertex is of degree

two.

4.2.20 Definition

In a non-directed graph G a sequence P of zero more edges of the form {v0,

v1}, {v1, v2},…, {vn-1, vn}, (in this repetition of vertex is allowed) is called a

path from v0 to vn. The vertex v0 is called the initial vertex and vn is the

terminal vertex, and they both are called endpoints of path P.

We denote this path P as a v0 — vn path. If v0 = vn then it is called a closed

path, and if v0 vn then it is called an open path.

4.2.21 Definition

A path P may have no edges at all, in which case, the length of P is zero, P

is called a trivial path, and V(P) = {v0}.

A path P is simple if all edges and vertices on the path are distinct except

possibly the endpoints.

Two paths in a graph are said to be edge-disjoint if they share no common

edges; they are vertex-disjoint if they share no common vertices.

v6

v2

v7 v1 v5

v3

v4



4.2.22 Note

(i) An open simple path of length n has n + 1 distinct vertices and n distinct

edges, while a closed simple path of length n has n distinct vertices and n

distinct edges. The trivial path is taken to be a simple closed path of length

zero.

4.2.23 Definition

A path of length 1 with no repeated edges and whose endpoints are equal

is called a circuit. A circuit may have repeated vertices other than the

endpoints; a cycle is a circuit with no other repeated vertices except its

endpoints.

Observation:

(i) A cycle is a simple circuit, and, in particular, a loop is a cycle of length

1.

(ii) In a graph, a cycle that is not a loop must have length at least 3, but

there may be cycles of length 2 in a multi-graph.

4.2.24 Example

Consider the graphs in example 4.2.6.

(i) The path {c, c} is a cycle of length 1; the sequence of edges {a, b}, {b,

c}, {c, a} and {a, d}, {d, c}, {c, a} form cycles of length 3.

(ii) The path {a, b}, {b, c}, {c, d}, {d, a} is a cycle of length 4.

(iii) The sequence {a, b}, {b, c}, {c, c}, {c, a} is a circuit of length 4; it is not

a cycle because the sequence of vertices a-b-c-c-a includes more than

one repeated vertex. Similarly the sequence of edges {a, b},{b, c},{c,

a},{a, d},{d, c},{c, a} forms a closed path of length 6, but this path is not

a circuit because the edge {c, a} is repeated twice.

Observation: A simple path is certainly a path and the converse statement

need not be true.



4.3 Isomorphism

4.3.1 Definition

Two graphs G and G1 are said to be isomorphic to each other if there is a

one-to-one correspondence between their vertices and a one-to-one

correspondence between their edges such that the incident relation ship

must be preserved.

That is, two graphs G = (V, E) & G1 = (V1, E1) are said to be isomorphic if

there exist one- one and onto functions f : V V1 and g :E E1 such that

g(vivj) = f(vi)f(vj) for any edge vivj in G.

4.3.2 Example

(i) Consider the two graphs given in Figures A and B. Observe that these

are isomorphic. The correspondence between these two graphs is as

follows. f(ai) = vi for 1 i 5 and g(i) = ei for 1 i 6. Except the labeling of

their vertices and edges of the isomorphic graphs, they are same, perhaps

may be drawn differently.

Observations: If there is an isomorphism between two graphs G and H,

then G and H must have:

i) The same number of vertices,

ii) The same number of edges,

iii) An equal number of vertices of a given degree, and

iv) The incident relationship must be preserved.

a5

a4

a3

a2

a1

6

5

3

2 4

1

Fig A

v3

v1 v2

v5

v4

e5

e1

e3

e6 e2

e4

Fig B



4.3.3 Example

For each pairs of graphs shown below, either label the graphs so as to

exhibit an isomorphism or explain why the graphs are not isomorphic.

(i).

(ii).

(iii).

(iv).

Solution:

(i) The graphs are isomorphic, by the labeling shown below.

• •

• •

• •

• •

• •

• •

•

• •

•

• •

• •

•

•

• •

• •

• •

• •

• •

v1 v2

v3 v4

• •

• •

• •

• • v2

v1 v4

v3



(ii) The graphs are not isomorphic, since second graph has a vertex of

degree 1 and the first graph does not have.

(iii) The graphs are not isomorphic

(iv) the graphs are not isomorphic, since the two graphs differ by number

of edges.

4.3.4 Definition

A graph H is said to be a sub-graph of a graph G if all the vertices and all

the edges of H are in G, and each edge of H has the same end vertices in H

as in G.

4.3.5 Example

The graphs H and K are subgraphs of graph G.

Observations: Here we can observe the following facts:

i) Every graph is a sub graph of itself;

ii) A subgraph of a subgraph of G is a subgraph of G;

iii) A single vertex in a graph G is a subgraph of G; and

iv) A single edge in G together with its end vertices is a subgroup of G.

1

2

3

6

i b

d

j

Graph K

3

Graph H

4

1

b

d

c

a

e

2 g

4

1

2

3

6

a

5

b c

d e

f

h

j

i

Graph G

v1

v1 v2

v3 v4

• •

• •

•

• •

•

v2

v4

v3



4.4 Trees

Trees are extensively used as models in areas like computer science,

chemistry, geology, electrical networks and botany etc. Trees are also

useful in design of wide range of algorithms.

The concept of a ‗tree‘ plays a vital role in the theory of graphs. First we

introduce the definition of ‗tree‘, study some of its properties and its

applications. We shall also provide equivalent conditions for a tree.

4.4.1 Definition

A connected graph without circuits is called a tree.

4.4.2 Example

Trees with one, two, three and four vertices are given in the fig.

4.4.3 Example

Consider the two trees G1 = (V, E1) and G2 = (V, E2) where

V = {a, b, c, d, e, f, g, h, i, j}

E1 = {{a, c}, {b, c}, {c, d}, {c, e}, {e, g}, {f, g}, {g, i}, {h, i}, {i, j}}

E2 = {(c, a), (c, b), (c, d), (c, f), (f, e), (f, i), (g, d), (h, e), (j, g)}

Neither of these two trees is a directed tree.



If vertex c is designated as the root of each tree, vertex j is a level 4 in G1

and at level 3 in G2.

4.4.4 Example

A directed tree T is shown in the following fig. Here T = (V, E) where V = {a,

b, c, d, e, f, g, h} and E = {(a, b), (a, c), (a, d), (b, e), (d, f), (e, g), (e, h)}.

The root of T is the vertex a and the vertices at level 2 are e and f.

4.4.5 Note

Directed trees are conventionally drawn with the root at the top and all

edges going from the top of the page toward the bottom so that the direction

of edges is sometimes not explicitly shown.

4.4.6 Note

i) Since a tree is a graph, we have that a tree contains at least one vertex.

ii) A tree without any edge is referred to as a null tree.

iii) Since we are considering only finite graphs, we have that the trees

considered are also finite.

iv) A tree is always a simple graph.

v) A vertex of degree of 1 is called a pendent vertex.



4.4.7 Note

Let G = (V, E) be a disconnected graph. We define a relation ~ on the set

of vertices as follows: v ~ u there is a walk from v to u.

Then this relation ~ is an equivalence relation.

Let {Vi}i be the collection of all equivalence classes. Now V = i

iV .

Write Ei = {e E / an end point of e is in Vi} for each i.

Then (Vi, Ei) is a connected subgraph of G for every i.

This connected subgraph (Vi, Ei) of G is called a connected component (or

component) of G for every i .

The collection {(Vi, Ei)}i of subgraphs of G is the collection of all

connected components of G.

Observations:

i) If G is a connected graph, then G is the only connected component

of G.

ii) A disconnected graph G consists of two or more connected components.

iii) Connected component of a graph G is a maximal connected sub graph

of G.

iv) A graph is connected if it has exactly one component.

v) Consider the graph given in Fig.

This graph is a disconnected graph with two components.

4.4.8 Problem

T is a tree there is one and only one path between every pair of vertices.

Solution:

Part 1: Suppose T is a tree. Then T is a connected graph and contains no

circuits.



Since T is connected, there exists at least one path between every pair of

vertices in T.

Suppose that between two vertices a and b of T, there are two distinct

paths.

Now, the union of these two paths will contain a circuit in T, a contradiction

(since T contains no circuits).

This shows that there exists one and only one path between a given pair of

vertices in T.

Part 2: Let G be a graph.

Assume that there is one and only one path between every pair of vertices

in G.

This shows that G is connected.

If possible suppose that G contains a circuit.

Then there is at least one pair of vertices a, b such that there are two

distinct paths between a and b. But this is a contradiction to our assumption.

So G contains no circuits. Thus G is a tree.

4.4.9 Problem

A tree G with ‗n‘ vertices has (n-1) edges.

(Refer unit 3 for the solution)

Problem: If T is a tree (with two or more vertices), then there exists at least

two pendant (a vertex of degree 1) vertices.

Solution: Let n = the number of vertices in G. Then G has n-1 edges. Now

1

deg( ) 2 2( 1) (2 2)n

i

i

v E n n

.

Now if there is only one vertex, say v1 of degree 1, then

deg(vi) 2 for i = 2, 3, …, n and

1 2

deg( ) 1 deg( ) 1 2 2 2 1.n n

i i

i i

v v n n



But 2n-2 2n-1 or -2 -1, a contradiction.

Therefore there are at least two vertices of degree 1.

Note: If two nonadjacent vertices of a tree T are connected by adding an

edge, then the resulting graph will contain a cycle.

4.4.11 Problem

Any connected graph with ‗n‘ vertices and n –1 edges is a tree.

Solution: Let ‗G‘ be a connected graph with n vertices and n – 1 edges.

It is enough to show that G contains no circuits.

If possible suppose that G contains a circuit.

Let ‗e‘ be an edge in that circuit.

Since ‗e‘ in a circuit, we have that G – e is still connected.

Now G - e is connected with ‗n‘ vertices, and so it should contain at least n –

1 edges, a contradiction (to the fact that G - e contain only (n-2) edges).

So G contains no circuits. Therefore G is a tree.

4.4.12 Problem

If a graph G contains n vertices, n -1 edges and no circuits, then G is a

connected graph.

Proof: Let G be a graph with ‗n‘ vertices, n – 1 edges and contains no circuits.

In a contrary way, suppose that G is disconnected.

G consists of two or more circuitless components (say, g1, g2, …, gk).

Now k 2. Select a vertex vi in gi, for 1 i k.

Add new edges e1, e2, …, ek-1 where ei = 1ii vv to get a new graph G*.

It is clear that G* contains no circuits and connected, and so G* is a tree.

Now G* contains n vertices and (n - 1) + (k - 1) = (n + k - 2) n edges, a

contradiction (since a tree contains (n - 1) edges).

This shows that G is connected.

This completes the proof.

We summarize the above problems as a following theorem.



4.4.13 Theorem

For a given graph G, with n vertices the following conditions are equivalent:

(i) G is connected and is circuitless;

(ii) G is connected and has n -1 edges;

(iii) G is circuitless and has n -1 edges;

(iv) There is exactly one path between every pair of vertices in G;

(v) G is a tree.

4.5 Rooted Trees

We study binary trees and applications of binary trees. Since the natural

way their vertices correspond to an initial segment of the positive integers,

complete binary trees can be represented very efficiently on computers.

They are applied in a number of excellent algorithms, including ―Heap Sort,‖

priority queue implementation, and algorithms for the efficient ordering of

data in hash tables. We also considered Height balanced trees which are

important in computer science.

4.5.1 Definition

A tree in which one vertex (called the root) is distinguished from all the

other vertices, is called a rooted tree. In a rooted tree, the root is generally

marked in a small triangle.

4.5.2 Example

Distinct rooted trees with four vertices were given in Fig.

Generally, the term ‗tree‘ means trees without any root. However they are

sometimes called free trees (or) non-rooted trees. A verity of rooted trees

(called the Binary rooted trees) is of particular interest (since they are



extensively used in the computer search methods, binary identification

problems, and variable length binary codes).

4.5.3 Definition

A tree in which there is exactly one vertex of degree 2, and all other

remaining vertices are of degree one or three, is called a binary tree.

(i) Clearly the following graph represents a binary tree (because the only

vertex ‗v1‘ is of degree 2, and all other vertices are of degree either 1 or 3).

(ii) Since the vertex of degree 2 (that is, v1) is distinct from all other

vertices, this vertex v1 is the root.

(iii) In a binary tree, the vertex with degree 2 serves as a root. So every

binary tree is a rooted tree.

4.5.4 Properties of Binary trees

Property (i): The number of vertices n, in a binary tree is always odd.

Property (ii): The number of pendent vertices is 2

1n .

Property (iii): Number of vertices of degree 3 is = n - p - 1 = n - (2

1n )

- 1 = 2

3n .

v3

v2

v4

v6

v8

v5

v1

1

v7

v1

v1

0 v13

v9

v12



4.5.5 Definition

A non-pendent vertex in a tree is called an internal vertex.

4.5.6 Note

(i) The number of internal vertices in a Binary tree is 2

1n = (p -1)

where p = the number of pendent vertices.

(ii) In the binary tree 4.5.3, the internal vertices are v1, v3, v4, v5, v6, v9.

These are 6 (= 7 - 1 = p - 1) in number.

4.5.7 Definition

Let v be a vertex in a binary tree. Then v is said to be at level i if v is at a

distance of i from the root.

4.5.8 Definition

The sum of path lengths from the root to all pendent vertices is called the

path length (or) external path length of a tree.

4.5.9 Example

(i) A 13-vertex, 4-level binary tree was given in Fig.

level 1

level 4

level 0

level 2

level 3



Here the number of vertices at levels 0, 1, 2, 3, 4 is 1, 2, 2, 4 and 4

respectively.

4.5.10 Definition

A binary tree is a directed tree T = (V, E), together with an edge-labeling f: E

{0, 1} such that every vertex has at most one edge incident from it

labeled with 0 and at most one edge incident from it labeled with 1.

Each edge (u, v) labeled with 0 is called a left edge; in this case u is called

the parent of v and v is called the left child of u.

Each edge (u, v) labeled with 1 is called a right edge; in this case u is also

called the parent of u, but v is called the right child of u. The subtrees of

which the left and right children of a vertex u are the roots are called the left

and right subtrees of u, respectively. We represent a binary tree by a triple

(V, E, f).

Observation:

(i) Every vertex in a binary tree has a unique parent, a unique left child,

and a unique right child, if it has any at all. That each vertex has a

unique parent (if any) follows from the definition of tree, where it is

required that there be a unique path from the root to each vertex.

(ii) Each vertex has a unique left child and a unique right child (if any)

follows from the labeling of the edges of the tree with 0‘s and l‘s. (At

most one edge from the parent can have a 0 label and at most one

edge can have a 1.)

(iii) Every vertex other than the root has a parent. Since every vertex u in a

tree must have a path to it from the root and the last vertex before v on

such a path must be the parent of v.

(iv) See the following figure for an illustration.



4.5.11 Example

Details such as edge labels and the direction of edges are usually

represented only implicitly in drawings of binary trees. The convention is that

for each vertex v the root of u‘s left subtree lies below v and to its left on the

page, whereas the root of u‘s right subtree lies below v and to its right on the

page. Figure 5-46 shows an example of a binary tree drawn with and

without edge labels and directed edges.

There are a few special kinds of binary trees that are important in computer

applications; one of these is the complete binary tree.

4.5.12 Definition

A complete binary tree is a binary tree for which the level-order indices of

the vertices form a complete interval 1, … , n of the integers. That is, if such

a tree has n vertices there is a vertex in the tree with index i for every i from

1 to n.



4.5.13 Example

Consider the two binary trees shown in the figure. Only the first one is

complete. In particular, the second tree has ten vertices but has no vertex

with index 10.

4.5.14 Lemma

In a complete binary tree with n vertices the indices of the vertices in the kth

level comprise the complete interval 2k through 2k+1 - 1,or from 2k through n

if n is less than 2k+1 -1.

Proof: The proof is by induction on k.

For k = 0 and n = 0 the lemma holds vacuously.

For k = 0 and n > 0 there is exactly one vertex with index 20 = 20+1 - 1 = 1,

and that is the root, which is also the only vertex at level 0.

Induction Hypothesis: For larger values of k, we assume the lemma holds

for k-1,

If n < 2k, the lemma holds vacuously. Otherwise, we invoke the definition of

level.

The vertices in level k are exactly those at distance k from the root.

The vertices in level k - 1 are exactly those at distance k-1 from the root.

It follows that the vertices at level k are precisely the children of the vertices

at level k -1, which the inductive hypothesis asserts are those with indices

2k-1 through 2k-1. By the definition of level-order index, the children have



indices in the range 2k through 2k+1-1. This complete interval, or the initial

segment of it up through n, must be in T, by the definition of complete binary

tree.


1. Write the degrees of all vertices in the graph given in example 4.2.8.

2. Can a simple graph exist with 15 vertices each of degree five.

3. How many vertices does a regular graph of degree 4 with 10 edges

have ?

4. Consider the following graph: Write the paths, lengths of path, cycles,

and also specify whether or not simple path/closed path.

5. Verify whether the following pairs of graphs are isomorphic. If not,

explain the relation..

(i)

(ii).

•

•

• •

•

•

•

• •

•

•

•

•

•

•

•

• •

• •

•

•



(iii).

(iv).

(v).

6. Which of the following graphs are trees?

7. Draw all trees with five vertices.

8. Consider the following graph.

The number of components in this graph is _____

9. How many non-isomorphic trees have 6 vertices ?

•

• • •

• • •

•

• •

•

•

•

• •

• • •

•

•

• •

•

•

• •

•

•

•

• •

•

G1 G2



10. In the graph 4.5.3, the number of vertices of degree 3 is _______

11. Draw the 11-vertex binary trees, also find its path lengh.

4.6 Summary

This Unit is meant for beginning your process of learning Graph Theory. It

started with the definition of Graph and moved on to illustrate the concepts

of finite and infinite graph, incidence, degree, isolated vertex, pendent vertex

and null graph. We also discussed the isomorphism between graphs and

subgraphs of a given graph with appropriate illustrations. We dealt with a

special type of graph called trees and studied some properties. We

considered the binary search tree and their properties. We presented some

illustrations on binary search tree and complete binary search trees.


1. Define the terms: Graph, finite graph, infinite graph, incidence, degree,

isolated vertex, pendent vertex, null graph

2. Explain the Koingsberg Bridges problem.

3. Explain the Seating Arrangement Problem.

4. Show that the sum of the degrees of the vertices of a finite graph G is

twice the number of edges.

5. Show that the number of vertices of odd degree is always even.

6. Show that an infinite graph with finite number of edges must have an

infinite number of isolated vertices.

7. Show that the maximum degree of any vertex in a simple graph is (n -

1).

8. Show that the maximum number of edges in a simple graph with n

vertices is 2

)1n(n .



9. Define an isomorphism between two graphs and give an example of it.

1. Draw all regular binary trees

(a) with exactly 7 vertices

(b) with exactly 9 vertices

2. Draw all distinct binary tree with (i) 3 vertices, (ii) 4 vertices.

3. Draw binary search trees for the following lists.

(i) 18, 44, 2, 5, 73, 45, 14, 6, 8, 10, 20, 11

(ii) 2, 1, 5, 6, 8, 9, 7, 3, 4.

4. Define a binary tree, complete binary tree and give examples of

each.

4.8 Answers


1. Here d(u1) = 3; d(u2) = 4; d(u3) = 3; d(u4) = 3; and d(u5) = 1

Now

5

1i

iu = 3 + 4 + 3 + 3 + 1 = 14. E = 7.

So

5

1i

i )u(d = 2 E

Therefore the sum of degrees of all the vertices of a graph G is twice

the number of edges in G

2. No, since the sum of the degrees of the vertices cannot be odd.

3. Let G be a regular graph of degree 4 with 10 edges and let 'n' be the

number of vertices in G. Then Vu

)u(d = 2 10 = 20.

n 4 = 20. n = 5.



4.

Path Length Simple

(yes/no)

Closed (yes/no)

Circuit (yes/no)

Cycle (yes/no)

a-d-c-e-g-j-d-a 7 no yes no no

b-c-e-f-g-j-f-b 7 no yes yes no

a-b-a 2 No Yes No No

a-d-c-b-a 4 Yes Yes Yes Yes

i-I 1 Yes Yes Yes Yes

a 0 Yes Yes No No

e-f-g-j-f-b 5 No No No No

d-b-c-d 3 yes yes yes Yes

5. (i) Isomorphic.

(ii) Isomorphic

(iii). Not isomorphic, since they do not have the same number of

vertices.

(iv). Not isomorphic, since the first graph has a vertex of degree 2

but second does not..

(v). Not isomorphic, since the first graph has two vertices of degree

2 and the second has one vertex of degree 1.

6. G1 is a tree, since it is a connected graph without circuits. G2 is not a

tree (since it is not connected).

7.

G1 G2 G3



First draw five vertices. Then connect them, so that no cycles are

created. In this process, we must be careful not to repeat trees since

two trees which appear different may just be drawn differently. Here

there are three trees with five vertices as shown above.

8. Three components

9. Six non-isomorphic trees.

10. We have that n = 13, p = 2

1n =

2

113 =

2

14= 7.

Therefore number of vertices of degree 3 is 2

3n =

2

313 = 5.

11. (i)

level -1

level - 0

level -2

level -3



The path length: 2 + 2 + 3 + 3 + 3 + 3 = 16.

(ii)

The path length: 1 + 2 + 3 + 4 + 5 + 5 = 20.

level 5

level 0

level 1

level 2

level 4

level 3



Unit 5 Formal Languages and Grammars

Structure

5.1 Introduction

Objectives

5.2 Grammars and Languages

5.3 Classification of Grammars


5.4 Summary


5.6 Answers

5.1 Introduction

The basic machine instructions of a digital computer are very primitive

compared with the complex operations that must be performed in various

disciplines such as engineering, science, management and mathematics.

Even though a complex procedure can be programmed in machine

language, it is desirable to use a high level language that contains

instructions similar to those required in a particular application. The

specification of a programming language involves the set of symbols and set

of correct programs.

Objectives:


Learn to construct the language using grammar.

Construct the grammar, for a given language.

5.2 Grammars and Languages

A language L can be considered as a subset of the free monoid on an

alphabet. It is a set of strings or sentences over some finite alphabet. Finite



languages can be specified by exhaustively enumerating all their sentences.

Any device which specifies a language should be finite. A simple method of

specification which satisfies this requirement using a generative device is

referred as grammar. Precisely, a grammar consists of a finite set of rules or

productions which specify the syntax of the language.

The theory of formal languages exclusively involves the study of the

language syntax and this theory incepts from the works of well-known

linguist Noam Chomsky.

The goal of Chomsky was to define the syntax of natural languages using

simple and clear mathematical rules in order to precisely characterize the

structure of language. The primary idea behind this concept was to define a

formal grammar for describing the natural languages like English so that

language translation using a computer would become easy. Chomsky

developed the mathematical model of grammar in 1956. However, rather

than becoming useful for natural languages, this model turned out to be

suitable for the grammar of computer languages.

In order to simplify the concepts of grammar in computer languages, you

need have some basic idea of two types of sentences used in English and.

how these sentences are constructed. The first type of sentence contains

only noun and verb such as ‘Jack sang’, while the other type contains noun,

verb and adverb such as ‘Sandy ran swiftly’. The sentence ‘Sandy ran

swiftly’ has the words ‘Sandy’, ‘ran’, ‘swiftly’ in the order of noun, verb and

adverb. You can replace ‘Sandy’ with any other noun such as ‘Tom’ and

‘Jim’, ‘ran’ by any verb in the past tense like ‘jumped’ and ‘drank’ and ‘swiftly’

by any adverb like ‘quickly’ and ‘fast’. These replacements can give other

grammatically correct sentences like ‘Sandy ran swiftly’. So, the structure of

this type of sentence can be defined <noun> <verb> <adverb>. Here, for

<noun>, you substitute any name ‘Sandy’, ‘Jack’, ‘Hana’, etc. Similarly,



<verb> can be substituted by ‘ate’, ‘ran’ ‘drank’, etc., while the <adverb> can

be substituted by ‘slowly’, ‘‘quickly’, etc. In the same way, you can define the

structure of the first t sentence <‘Jack sang’> by <noun> <verb>.

It is seen that the sequence <noun> <verb> <adverb> does not exactly

represents a sentence; rather, it provides description of a particular type of

sentence. Replacing <noun>, <verb> and <adverb> with appropriate words,

you can produce grammatically correct sentences. The terms <noun>,

<verb> and <adverb>, are known as variables and the suitable words that

can form sentences are known as terminals. This signifies that a sentence

is nothing but a string of terminals. Now, if the variable S denotes a

sentence then you can form the rules below to generate two different types

of sentences:

In the above representation, each of the arrows specifies a rule, which

depicts that the work on the left side of the arrow can be replaced by the

word on the right side of the arrow. These rules may be called production

rules and let denotes these production rules. Now if you consider that

your vocabulary is restricted to ‘Sandy’, ‘Jack’, ‘Sam’, ‘ran’, ‘jumped’, ‘ate’,



‘swiftly’ and ‘quickly’ to form sentences of the form <noun> <verb>

<adverb>, then you can describe the grammar as 4-tulple.

Before presenting the formal definition of grammar, we review some

preliminary notations and definitions.

5.2.1 Definition

Let S denote a nonempty set of symbols, called an alphabet. We assume

that S to be finite. The elements of the set are called letters. A word or a

string on the set S is a finite sequence of the elements.

5.2.2 Example

Take S = {a, b}. Then x = abab, y = aaab, z = aaabb are strings on S.

5.2.3 Definition

The length of the string is the number of symbols in the string.

5.2.4 Properties of strings

Let S be the set of symbols, and S* denote the set of all strings (including

empty string).



5.2.5 Notation

VT = Finite non empty set of symbols (alphabet), called terminal

symbols.

(The strings of terminal symbols denoted by lower case

letters x, y , z, …)

VN = Set of non – terminal symbols, which are used to define the

syntax (or structure) of the language (A, B, C, …, X, Y, Z, …)

VN VT = Consisting of non terminal and terminal symbols, called

vocabulary of the language. (Strings of symbols over the

vocabulary are given by , , , … ).

,N TV V empty set (assumption).

5.2.6 Definition

A grammar (phrase structure) is defined by a 4 – tuple G = (VN, VT, S, )

where S is a distinguished element of VN (called the starting symbol), is a

finite subset of the relation from

* * *

.T N N T N T NV V V V V to V V

In general, an element (, ) is written as (called a production rule or

a rewriting rule).

5.2.7 Example

Let the symbols

L : letter

D : digit

I : identifier

Write the grammar G = , , ,N TV V S



Where VN = {I, L, D}

VT = {a, b, c, … , x, y, z}

S = I

5.2.8 Definition

5.2.9 Example

Consider the above example, the direct derivatives are as follows:

5.2.10 Definition



5.2.11 Definition

A sentential form is any derivative of the unique non terminal symbol S.

The language L generated by a grammar G is the set of all sentential forms

whose symbols are terminal.

That is, *

*

TL G S and V

This means that, the language is a subset of all terminal strings over VT.

5.2.12 Example

Let G = , , , , , , , , ,E T F a E

Where the variables E (expression), T (term), and F (factor) used in

conjunction with arithmetic expressions.



We wish to derive the expression a * a + a as follows: Starting with the

symbol E.

5.2.13 Problem

Generate the language L (G) = 1n n na b c n by the following grammar.

Solution:



We generate the language for n = 2. That is, we derive the string a2b2c2.

5.2.14 Example

Consider the grammar , , , , ,G S C a b S

Solution:

Derivation for n = 2. i.e., the string a2ba2

5.2.15 Problem



Solution:

Take n = 2, m = 3. We generate the string a2ba3

5.2.16 Problem

5.2.17 Problem



Suppose G = ({S, A, B}, {0, 1}, , S) where consists of productions: S

0AB0, A 10AB1, B A01, 0A 100 and 1B1 0101. Show that w =

100110100011010 is in L(G).

Solution: To prove that the given w L(G), we need to start with an S-

production and subsequently apply the suitable productions in order to

derive w. The following sequences show the derivation of w.

In this sequence, the strings that can be replaced are underlined.

5.2.18 Problem

Solution: Here, we can follow the proof by starting with an S-production and

subsequently applying the suitable productions in order, we can derive w.

The following sequences show the derivation of w:



5.3 Classification of Grammars

Every language is specified by a particular grammar. The classification of

languages is based on the classification of the grammar used to specify

them. Grammars are classified accordingly to the types of productions.

5.3.1 Definition

i) A grammar in which there are no restrictions on its productions is called

type – 0 grammar or unrestricted grammar 0L T .

ii) A grammar that contains only productions of the form where

is called type – 1 grammar or context sensitive grammar.

The language generated by this grammar is called context sensitive

language 1L T .

iii) A grammar that contains only productions of the form where

and NV is called type – 2 grammar or context free

grammar. The language generated by this grammar is called context

free language 2L T .

iv) A grammar that contains only productions of the form , where

, NV and has the form a, B or a, where a VT, B VN is



called type -3 grammar or regular grammar. The language generated

by this grammar is called a regular language 3L T .

v) A grammar G(VN, VT, S, ) is called monotonic when every

production in is of the form having ≤ or S . In the

second situation, S does not appear on the right hand side of any of the

production of G.

In other words, in any production, the left hand string is always a single non

– terminal and right hand string is either a terminal or a terminal followed by

a non – terminal.

5.3.2 Theorem

If G be type 0 grammar, then we can find an equivalent grammar G1 where

each production is either of the form or A a. Here, and are the

strings variables, A is a variable and a is a terminal.

Proof: To construct G1, consider a production in G with or having

the same terminals. Let in both and , a new variable Ca replace each of

the terminals to produce ’s and ’s.

Now, for every , where and have same terminals, we can get a

corresponding with productions of the form Ca a for each terminal

that appears on or . Therefore, the new productions obtained from the

above constriction are the new productions for G1. Also, the variables of G

along with the new variables of the form Ca are the variables of G1.

Similarly, the terminals and the start symbol of G1 are also same as those of

G. Hence, G1 satisfies the required conditions for a grammar and it is

equivalent to G. Therefore L(G) = L(G1).

5.3.3 Note

i) The above theorem also holds for grammars of type 1, 2 and 3.



ii) 3 2 1 0L T L T L T L T .

5.3.4 Theorem

Every monotonic grammar G is equivalent to type 1 grammar.

5.3.5 Problem

Construct a grammar for the language.

L = {aaaa,aabb, bbaa, bbbb}

Solution:

Since L has a finite number of strings, we can list all strings in the language.

Let VT = {a, b} be the set of terminals.

VN = {S}, non terminal (starting symbol)

We simplify the [productions as follows.

Let VN = {S, A}

: , , .S AA A aa A bb

Therefore the Grammar , , , , ,T NG V a b V S A S

5.3.6 Problem


*, , ' 3L x x a b the number of a s in x is a multiple of



Solution:

Therefore the grammar , , , , , ,T NG V a b V S A B S

5.3.7 Problem

Find the highest type number that can be applied to the following productions:

.

Solution:



Notation: Let L0, Lcs, Lcf and Lr are the family of type 0, context sensitive,

context free and regular languages respectively.

5.3.8 Operations on Languages

In formal languages, there are certain common operations. These

operations include standard set operations such as intersection, union and

complementation operations. There are other operations such as string

operations that are applied element-wise on the languages. For example, if

we consider two languages L1 and L2 over some common alphabets, then

we can define the following operations.

Concatenation: It combines the two languages to produce the

concatenated language denoted by L1L2. Here, L1L2 consists of all the

strings of type xy, where x is a string in L1 and y is a string in L2.

Intersection: It produces the language L1 L2 which consists of all the

strings that are contained in both the languages L1 and L2.

Union: It produces the languages L1 L2 which consists of all the strings

that are contained in either of the languages L1 and L2.

Complement: It produces the language L1 from the language L1. Here,

L1 is known as the complement of the language L1 with respect to an

alphabet, where L1 consists of all the strings over the alphabets that are

not in the language L1.



These operations are generally used in determining the closure properties of

the classes of languages. A class of languages is called closed under some

operation, when applying the operation to the class of languages always

produces a language in the same class. Refer the units 10, 11 for detailed

proofs of closed operations.

5.3.9 Theorem

The languages L0, Lcs, Lcf and Lr are closed under the operations

concatenation and union.

5.3.10 Problem


, 1,i jL a b i j i j

Solution:

We decompose 1 2L L L where

1 2

i j i jL a b i j and L a b i j

Grammar for L1: Set of production for L1

where VT = {a, b} , VN = {A, B}

A is a starting symbol.

Grammar for L2:

VT = {a, b}, VN = {C, D}, C is starting symbol.



5.3.11 Problem

Obtain a grammar to generate the language

Solution: It is clear from the statement that if a string has n number of 0’s

as the prefix, this prefixed string should not be followed by n number of 1’s,

that is, we should not have equal number of 0’s and 1’s. At the same time

0’s should precede 1’s. The grammar for this can be written as:

G = (VN, VT, , S) where

VN = {S, A, B, C}

VT = {0, 1}

Productions

: S 0S1 (generates 0i1j recursively)

S A (to generate more 0’s than 1’s)

S B (to generate more 1’s than 0’s)

A 0A 0 (at least one 0 is generated)

B 1B 1 (at least one 1 is generated); and S is the starting symbol.

5.3.12 Problem



Obtain a grammar to generate the language L = {x / x mod 3 = 0} on the

set VT = {a}.

Solution: The language accepted by the grammar can also be written as

L = {, aaa, aaaaaa, aaaaaaaaa, …}.

It is clear from this definition that any string generated should have the

length multiple of 3 which can be easily done by the production rule:

S aaaS .

Therefore the final grammar is

G = (VT, VN, , S), where

VT = {S}

VT = {a}, and the set of productions

: S aaaS ,

S is the starting symbol.


1. Suppose G = ({S, A, B}, {a, b}, , S) where consists of the following

productions: S abAB, A aBb, B abA, bab, aB aaa. Then

verify whether or not w = abaaababaaab L(G).

2. Consider the string x = well, find all prefixes and suffixes of x. Also find

all subwords of x.

3. Let x = 0100, y = 11. Find xy and yx.

4. Given the strings u = a2bab2 and v = bab2, find the strings uv, vu, v2, u.

Also find their lengths.

5. Let A = {ab, bc, ca}. Find whether the following strings

i) abc, ii) ababab, iii) abba, iv) bcabbab belong to A*.

6. Find the language for the grammar.

0,1 , , ,T NG V V S S

where the set of productions



: 11 , 0.S S S

7. Find the language L (G), generated by the grammar.

, , , , , ,T NG V x y z V S A S

where : , , , .S xS S yA A yA A Z

8. Find the language L (G), generated by the grammar

, , , ,T NG V a b V S S where

: , , .S aaS S a S b

5.4 Summary

In this unit, we study the formal languages and develop mathematical

expressions, phrase structure grammar, a simple device for the construction

of useful formal languages. Some types of grammars depending on their

productions, were discussed. These are useful for generating algorithms.


1. Construct the grammar which generates the following language and also

specify their types.

i) 1 , 3n mL a b n m

(Hint: (i)

Where is : , , , .S aS S bbB B bB B b

It is a regular language).

ii) 1n nL a ba n

(Hint: , , , ,T NG V a b V S S

Where is : , .S aSa S b

It is a context – free – language).



iii) 1 , 1n mL a ba n m

(Hint: , , , , , ,T NG V a b V S A S where

: , , , , , .S aAb S bAa A bAa A aAb A ab A ba


5.6 Answers


1. Yes, w L(G).

2. , w, we, wel, well;

, l, ll, ell, well;

, w, e, l, we, el, ll, wel, ell, well.

3. xy = 010011, yx = 110100.

4. uv = a2bab4ab2, uv = 11

vu = bab2a2bab3, vu = 11;

v2 = bab3ab2, v2 = 8;

u = a2bab2, u = 6

5. No, Yes, No, No.

6. ( ) 0, 110 ,11110, 1111110,L G

7. ( ) 0, 1n mL G x y z n m

8. 2 1 2( ) 0 0n nL G a n a b n



Unit 6 Deterministic Finite Automata

Structure

6.1 Introduction

Objectives

6.2 Basic Terms

6.3 Deterministic Finite Automaton (DFA)

6.4 Transition System (Transition graph)

6.5 Language accepted by a DFA

Self Assessment Questions.

6.6 Summary


6.8 Answers

6.1 Introduction

A study of finite automaton is their applicability to the design of several

common types of computer algorithms and programs. For example, the

lexical analysis phase of a compiler (in which program units such as „begin‟

and „+‟ are identified) is often based on the simulation of a finite automaton.

Also, the problem of finding an occurrence of a string within another, for

example, whether any of the strings air, water, earth, and fire occur in the

text of Elements of the Theory of Computation, can also be solved efficiently

by methods originating from the theory of finite automata.



Let us now describe the operation of a finite automaton in more detail.

Strings are fed into the device by means of an input tape, which is divided

into squares, with one symbol inscribed in each tape square (see figure).

The main part of the machine itself is a “black box” with innards that can be,

at any specified moment, in one of a finite number of distinct internal states.

This black box - called the finite control - can sense what symbol is written

at any position on the input tape by means of a movable reading head.

Initially, the reading head is placed at the leftmost square of the tape and

the finite control is set in a designated initial state.

At regular intervals the automaton reads one symbol from the input tape and

then enters a new state that depends only on the current state and the

symbol just read. This is why we shall call this device a deterministic finite

automaton. After reading an input symbol, the reading head moves one

square to the right on the input tape so that on the next move it will read the

symbol in the next tape square. This process is repeated continuously; a

symbol is read, the reading head moves to the right, and the state of the

finite control changes. Eventually the reading head reaches the end of the

input string. The automaton then indicates its approval or disapproval of

what it has read by the state it is in at the end: if it winds up in one of a set of

final states the input string is considered to be accepted. The language

accepted by the machine is the set of strings it accepts.

Objectives:


Understand the idea of DFA.

Draw transition diagram.

Find the language accepted by a DFA.

Apply the techniques to various finite automata problems.

Know applications of DFA.



6.2 Basic Terms

Input: The various inputs i1, i2, …, ip applied at the input side of the model

are the elements of an input set, , also called the input alphabet.

Output: The various outputs o1, o2, …, oq generated at the output side of

the model are the elements of an output set O, also called the output

alphabet.

States: The entire automaton system, at any given instant of time, is in any

one of the states q1, q2, …, qn. (These are labeled with circles)

State relation: State relation helps determine the next state that the

automaton system is going to attain. State relation takes into consideration

the present input and the present state of the system in determining its next

state.

Output relation: It helps to determine the next output of the automaton

system. The output relation may take into consideration only the current

input or both the current input and the current state for determining the next

output.

6.2.1 Definition

An automaton system in which the output depends only on the present input

is called a Moore machine. Alternatively, an automaton system in which the

output depends both on the present input and the present state is called

Mealy machine.

6.3 Deterministic Finite Automaton (DFA)

6.3.1 Definition

A DFA is 5-tuple or quintuple M = (Q, , , q0, F) where

Q is non-empty, finite set of states.

is non-empty, finite set of input alphabets.



is transition function, which is a mapping from Q to Q. For this

transition function the parameters to be passed are state and input symbol.

Based on the current state and input symbol, the machine may enter into

another state.

q0 Q is the start state.

F Q is set of accepting or final states.

6.3.2 Note

For each input symbol a, from a given state there is exactly one transition

(there can be no transitions from a state also) and we are sure (or can

determine) to which state the machine enters. So, the machine is called

Deterministic machine. Since it has finite number of states the machine is

called Deterministic finite machine (automaton).

6.3.3 Illustration

Let us take the pictorial representation of DFA shown in figure and

understand the various components of DFA.

It is clear from this diagram that, the DFA is represented using circles,

arrows and arcs labeled with some digits, concentric circles etc. The circles

are nothing but the states of DFA. In the DFA shown in figure, there are

three states viz., q0, q1 and q2. An arrow enters into state q0 and is not

originating from any state and so it is quite different from other states and is

called the start state or initial state. The state q2 has two concentric circles

and is also a special state called the final state or accepting state. In this

DFA, there is, only one final state. Based on the language accepted by DFA,

there can be more than one final state also.

The states other than start state and final states are called intermediate



states. Always the machine initially will be in the start state. It is clear from

the figure that, the machine in state q0, after accepting the symbol 0, stays in

state q0 and after accepting the symbol 1, the machine enters into state q1.

Whenever the machine enters from one state to another state, we say that

there is a transition from one state to another state. Here we can say that

there is a transition from state q0 to q1 on input symbol 1.

In state q1, on input symbol 0, the machine will stay in q1 and on symbol 1,

there is a transition to state q2. In state q2, on input symbol 0 or 1, the

machine stays in state q2 only. This DFA has three states q0, q1 and q2 and

can be represented as Q = {q0, q1, q2}, the possible input symbols set = {0,

1}, which is set of input symbols (alphabets) for the machine.

There will be a transition from one state to another based on the input

alphabets. If there is a transition from vi, to vj on an input symbol a, it can be

represented as (vi, a) = vj.

The transitions from each state of the machine shown in figure based on the

input alphabets {0, 1} are shown in table.

State input Output Transition Representation

q0 0 q0 (q0, 0) = q0

q0 1 q1 (q0, 1) = q1

q1 0 q1 (q1, 0) = q1

q1 1 q2 (q1, 1) = q2

q2 0 q2 (q2, 0) = q2

q2 1 q2 (q2, 1) = q2

6.4 Transition System (Transition graph)

A finite directed labeled graph in which each node or vertex of the graph

represents a state and the directed edges from one node to another

represent transition of a state. All the edges of the transition graph are

labeled as input/output. For example, an edge labeled 1/0 specifies that for

a certain initial state if the input is 1, then the output is 0.



Consider the following diagram:

In the transition graph as shown in the figure,

The initial state, q0, of the system is represented by a circle with an

arrow pointing towards it.

The final state, q1, is represented by two concentric circles.

The directed edges from the initial state to the final state are labeled as

input/output.

6.4.1 Example

The graph represents the DFA,

Representation of DFA using Transition table:

In this method, the DFA is represented in the tabular form. This table is

called transitional table. There is one row for each state, and one column for

each input. Since, in the transition diagram shown in the fig., there are three



states, there are three rows for each state. The input symbols are only 0

and 1 so, there are two columns for the input symbols. The transitional

table for the diagram is given below.

6.5 Language accepted by a DFA

Consider the transition diagram or DFA shown in figure. The start state is q0

and the final state is q2. To start with the machine will be in start state q0.

Verification of acceptance of the string 1011:

Let us assume that the string 1011 is the input. On first input symbol 1, the

machine enters into state q1. In state q1, on input symbol 0, the machine

stays in state q only. In state q1, on input symbol 1, the machine enters into

state q2. In state q2, on the input symbol 1, the machine stays in state q2.

Now we encounter end of the input and note that we are in the accepting

state q2. The moves made by the DFA for the string 1011. Therefore after

scanning the input string 1011, the machine finally stays in state q2.

Verification of non-acceptance of the string 1011:

Take the string 0100: The moves made by the machine for the string 0100

are clear from the following figure. Note that after scanning the string 0100



the machine stays in state q1 which is not a final state. Therefore the string

0100 is rejected by the machine.

6.5.1 Definition

Let M = (Q, , , q0, F) be a DFA where


is a non-empty, finite set of input alphabets.

is a transition function, which is a mapping from Q to Q. For this

transition function the parameters to be passed are state and input symbol.

Based on the current state and input symbol, the machine may enter into

another state.

q0 Q is the start (or initial) state.


The string (also called language) w accepted by a DFA can be defined as

follows.

L(M) = {w w * and

(q0, w) F}.

Non-acceptance means that after the string is processed, the DFA will not

be in the final state and so the string is rejected. The non-acceptance of the

string w by a DFA can be defined in notation as:

)M(L = {w w * and

(q0, w) F}, where

: Q * Q, is an extended

transition function. The second argument of

is a string, rather than a

single symbol, and its value gives the state the automaton will be in after

reading that string.

(For example, in the above figure, (q0, 1) = q1 and (q1, 1) = q2. So,

(q0,

11) = q2).



6.5.2 Properties

6.5.3 Example

Solution:

Property 1: This means that when the current state of the machine is q and

when there is no input ( means no input), the machine will not move to any

new state, instead, it stays in the same state q.

6.5.4 Example

Obtain a DFA to accept strings of a‟s and b‟s starting with the string ab.

Solution: It is clear that the string should start with ab and so, the minimum



string that can be accepted by the machine is ab.

To accept the string ab, we need three states and the machine can be

written as

where q2 is the final or accepting state. In state q0, if the input symbol is b,

the machine should reject b (note that the string should start with a). So, in

state q0, on input b, we enter into the rejecting state q3. The machine for this

can be of the form

The machine will be in state q1, if the first input symbol is a. If this a is

followed by another a, the string aa should be rejected by the machine. So,

in state q1, if the input symbol is a, we reject it and enter into q3 which is the

rejecting state. The machine for this can be of the form



Whenever the string is not starting with ab, the machine will be in state q3

which is the rejecting state. So, in state q3, if the input string consists of a‟s

and b‟s of any length, the entire string can be rejected and can stay in state

q3 only.

The resulting machine can be of the form

The machine will be in state q2, if the input string starts with ab. After the

string ab, the string containing any combination of a‟s and b‟s, can be

accepted and so remain in state q2 only. The complete machine to accept

the strings of a‟s and b‟s starting with the string ab is shown in figure.

The state q3 is called trap state or rejecting state.



In the set notation, the language accepted by DFA can be represented as

L = {ab(a + b)n n 0}

Or

L = {ab(a + b)*}

Therefore the DFA which accepts strings of a‟s and b‟s starting with the

string ab is given by M = (Q, , , q0, F), where

Q = {q0, q1, q2, q3}, = {a, b}, q0: initial state, F = {q2}, and the transition

function is defined as

To accept the string abab: The string is accepted by the machine.



To reject the string aabb:

Therefore the string aabb is not accepted by the machine.

6.5.5 Note

Sometimes we ignore the extended notion

and we use only (assuming

that the reader is well acquainted with it).

6.5.6 Example

Consider a finite automation that will accept the set of natural numbers

which are divisible by 3,



6.5.7 Example

Obtain a DFA to accept even number of a‟s, and odd number of a‟s.



Solution: Observe the following transition diagrams.

Consider the string aa: (q0, aa) = ((q0, a), a) = (q1, a) = q0, which is a

final state (acceptable state).

Consider the string aaa: (q0, aaa) = ( (q0, aa), a) = ( ( (q0, a), a),

a) = ( (q1, a), a) = (q0, a) = q1, which is an acceptable state.

6.5.8 Problem

Obtain DFA to accept strings of a‟s and b‟s having exactly one a, at least

one a, not more than three a‟s.

Solution:

To accept exactly one a:

To accept exactly one a, we need two states q0 and q1 and make q1 as the

final state. The machine to accept one a is shown below.

In q0, on input symbol b, remain q0 only so that any number of b‟s can end



with one a. The machine for this can be of the form

In q1, on input symbol b, remains q1 then machine can take the form

But, in state q1, if the input symbol is a, the string has to be rejected as the

machine can have any number of b‟s but exactly one a. So, the string has to

be rejected and we enter into a trap state q2. Once the machine enters into

trap state, there is no way to come out of the state and the string is rejected

by the machine. The complete machine is shown in the figure.

In the set notation, the language accepted by DFA can be represented as

L = {bmabn m, n 0}.

The machine M = (Q, , , q0, F), where

Q = {q0, q1, q2}, = {a, b}, q0: initial state, F = {q1}, and the transition function

is defined as



The machine to accept at least one a: The minimum string that can be

accepted by the machine is a. For this, we need two states q0 and q1 where

q1 is the final state. The machine for this is shown below.

In state q0, if the input symbol is b, remain in q0. Once the final state q1is

reached, whether the input symbol is a or b, the entire string has to be

accepted. The machine to accept at least one a is shown in fig.

In set notation, the language accepted DFA can be represented as

L = {bma(a + b)n m, n 0}.

The machine M = (Q, , , q0, F), where

Q = {q0, q1}, = {a, b}, q0 : initial state, F = {q1}, and the transition function

is defined as



The machine to accept not more than three a‟s: The machine should accept

not more than three a‟s means,

It can accept zero a‟s

It can accept one a

It can accept two a‟s

It can accept 3 a‟s

But, it cannot accept more than three a‟s.

In this machine maximum of three a‟s can be accepted (that is, the machine

can accept zero a‟s, one a, two a‟s). So, we need maximum four states q0,

q1, q2 and q3 where all these states are final states and q0 is the start state.

In state q3, if the input symbol is a, the string has to be rejected and we

enter into a trap state q4. Once this trap state is reached, whether the input

symbol is a or b, the entire string has to be rejected and remain in state q4.

Now, the machine can take the form as shown below.



In state q0, q1, q2 and q3, if the input symbol is b, stay in their respective

states and the final transition diagram is shown below.

In set notation, the language accepted DFA can be represented as

L = {biabjabkabl i, j, k, l 0}

The DFA is M = (Q, , , q0, F) where Q = {q0, q1, q2, q3, q4}, = {a, b}, q0 is

the start state, F = {q0, q1, q2, q3}, and is the transition function.



6.5.9 Problem

Solution: The number of symbols in a string consisting of a‟s and b‟s should

not have multiples of 5. The machine to accept the corresponding language

is shown below.


1. The symbol is used for _______

2. The symbol O is used for ________

3. In an automaton system, the states are represented by _________

4. State relation helps to determine _________

5. An automaton system in which the output depends only on the present



input is called a ________

An automaton system in which the output depends both on the present

input and the present state is called ________

6. Consider the example 6.3.3.

Write the states, input alphabet, final states, starting state.

7. Construct the state table for the following DFA.

8. Draw a DFA to accept strings of a‟s and b‟s with even number of a‟s and

even number of b‟s. Also find the language accepted by DFA.

9. Consider the example 6.5.8,

(i) The set of states ___________

(ii) The set of final states _________

(iii) Starting state ____________

(iv) S(q2, b) = _________, S(q0, b) = _______

6.6 Summary

The concept of finite automata is used in wide applications. Large natural

vocabularies can be described using finite automaton which includes the

applications such as spelling checkers and advisers, multi-language

dictionaries, to indent and documents, in calculators to evaluate complex

expressions based on the priority of an operator etc. In this unit we gave a

comprehensive idea about the DFA and a graphical representation of DFA.

Further we discuss the language accepted by DFA with certain examples.




1. What is DFA ? Explain with example.

2. When do we say that a language is accepted by the machine? Illustrate

with example.

3. Obtain a DFA to accept strings of 0‟s and 1‟s starting with at least two

0‟s and ending with at least two 1‟s. Also find the language accepted by

this.

(Hint:

The language accepted by DFA can be represented as

L = {w w 00(0+1)*11}

4. Obtain a DFA to accept strings of a‟s and b‟s with at most two

consecutive b‟s.

6.8 Answers




1. Input alphabet.

2. Output alphabet.

3. These are labeled with circles.

4. The next state that the automaton system is going to attain.

5. Moore machine; Mealy machine.

6. States: {q0, q1, q2}, Input alphabet: {0, 1}, final state: {q2}, starting state

{q0}.

7.

8. The language accepted by DFA is

L = {w w (a + b)* and total number of strings in both a and b are

even}.



9. (i) Set of states {q0, q1, q2, q3, q4}

(ii) Set of final states {q0,q1, q2, q3}

(iii) Starting state {q0}

(iv) S(q2, b) = q2, S(q0, b) = q0



Unit 7 Non Deterministic Finite Automata

Structure

7.1 Introduction

Objectives

7.2 Non-Deterministic Finite Automata

7.3 Language accepted by a NDFA

7.4 Conversion from NDFA to DFA

7.5 Moore and Mealy Machines


7.6 Summary


7.8 Answers

7.1 Introduction

Consider the transition diagram shown in the figure. To start with the machine

will be in q0. If the first input symbol is a, the machine can enter into either

state q1 or q2 (since there are two transitions on input symbol a from state q0).

In state q0, if the input symbol is b, the machine enters into state q2. Similarly

from state q1, there are multiple transitions on an input symbol a to the states

q1 and q2. That is, the machine can either enter into state q2 or state q1. At this

point of time, we cannot determine exactly in which state the machine will be.

So, this is called Non-Deterministic Finite Automaton (NDFA).



The transitions from each state of the machine shown below based on the

input alphabets {a, b}.

Let Q is the set of states, then we define 2Q to represents the set of subsets

of Q.

Objectives:


Understand the idea of NDFA.

Draw transition diagram for NDFA.

Find the language accepted by a NDFA.

Know applications of NDFA.

Learn Moore and Mealy Machines

7.2 Non-Deterministic Finite Automata



There is another type of finite state automaton in which there may be

several possible next states for each of input value and state. Such

machines are called nondeterministic.

7.2.1 Definition

A NDFA is 5-tuple or quintuple M = (Q, , , q0, F) where



is transition function, which is a mapping from Q to set of subsets of Q

(that is, 2Q). This function accepts two arguments as the input with first

argument being q Q and the symbol argument as the symbol a and

returns a set of states which are reachable from q on input a. This function

shows change of state from one state to a set of states based on the input

symbol.



7.2.2 Note

Suppose (p, ) = A where q Q, A 2Q and a . Here there will be a

transition from state q to the set of states A on an input symbol a. The

transition function can be extended to

whenever string operations are

involved. The transition from state q to the set of states A, on the input string

w can be written as

(q, w) =A.

7.2.3 Properties



7.3 Language Accepted by NDFA

A string is a sequence of symbols obtained from . The set of all strings

recognized by automaton is called language. The language L accepted by

an NFA M = (Q, , , q0, F) is defined as follows.

7.3.1 Definition

Let M = (Q, , , q0, F) be an NDFA where Qs is the set of finite states, is

set of input alphabets (from which a string can be formed), is transition

from Q to 2Q, q0 is the start state and F is the final states. The string

(also called language) w accepted by an NDFA can be defined in formal

notation as:

L (M) = {w w * and

(q0, w) = P with at least on component of P in F}.

Here, P is set of states.

(In other words, the language consists of all strings w for which there is a

walk labeled w from the initial vertex of the transition graph to some final

vertex).

7.3.2 Note (Need for Non-deterministic finite automaton)

Digital computers are deterministic machines. Given the input, the state of

the machine is predictable. Sometimes, constructing deterministic machine

is difficult compared to non-deterministic machine. In such cases, there is a

need to construct a machine very easily which can be achieved by

constructing an NDFA. After constructing an NDFA, DFA can be easily



constructed. This is an efficient mechanism to describe some complicated

languages concisely. So, practically non-deterministic machines will not

exist. But, one can construct an NDFA easily and later that can be

converted into DFA.

7.3.3 Example

Obtain an NDFA to accept the following language L = {w w ababn or aban

where n 0}.

The machine to accept ababn where n 0 is shown below.

The machine to accept aban where n 0 is shown below.

Since both the machines accept a as the first input symbol, the states q1 and

q5 can be merged into a single state and the machine to accept either ababn

or aban where n 0 is shown below.

7.4 Conversion from NDFA to DFA

Let MN = (QN, N, N, q0, FN) be an NDFA and accept the language L(MN).

There should be an equivalent DFA, MD = (QD, D, D, q0, FD) such that

L(MD) = L(MN). The procedure to convert an NDFA to its equivalent DFA is

shown below.

Step 1: The start state of MN is the start state of MD. So, add q0 (which is

the start state of MN) to Q and find the transitions from this state. The way to

obtain different transitions is shown in the next step.

Step 2: For each state {qi, qj, …, qk} QD, the transitions for each input

symbol in can be obtained as shown below.

1. D({qi, qj, …, qk}, a) = N(qi, a) N(qj, a) … N(qk, a) = {ql, qm, …,

qn}(say)

2. Add the state {ql, qm, …, qn} to Q, if it is not already in QD.



3. Add the transition from {qi, qj, …, qk} to {ql, qm, …, qn} on the input

symbol a if and only if the state {ql, qm, …, qn} is not added to QD in the

previous step.

Step 3: The state {qa, qb, …, qc} Q is the final state, if at least one of the

state in qa, qb, …, qc FN (that is, at least one component in {qa, qb, …, qc}

should be the final state of NDFA.

7.4.1 Example

Convert the following NDFA to its equivalent DFA.

Solution:

Step 1: q0 is the start DFA, so QD = {{q0}} _______________(i).

Step 2: Find the new states from each state in QD and obtain the

corresponding transitions.

Consider the state {q0}:

When a = 0, D ({q0}, 0) = N ({q0}, 0) = {q0, q1}

When a = 1, D ({q0}, 1) = N ({q0}, 1) = {q1}.

Since the states obtained above are not in QD, add these two states to QD

so that

QD = {{q0, q1}, {q0, q1}, {q1}} ____________(ii).

The corresponding transition on a = 0 and a = 1 are shown below.

Consider the state {q0, q1}:



Since the states obtained above are the not defined in QD (refer (ii)), add

these two states to QD so that

QD = {{q0, q1}, {q0, q1}, {q1}, {q0, q1, q2}, {q1, q2}} ____________(iii).

Consider the {q0}:

Since the states obtained above are same and {q2} is not in QD, add the

state {q2} so that

QD = {{q0, q1}, {q0, q1}, {q1}, {q0, q1, q2}, {q1, q2}, {q2}} ____________(iii).



Consider the state {q0, q1, q2}:

Since the states obtained above are not new states (are already in QD), do

not add these two states to QD. But, the transitions on a = 0 and a = 1

should be added to the transitional table.

Consider the state {q1, q2}:



Since the states obtained above are not new states (are already in QD), do

not add these two states to QD. But, the transitions on a = a0 and a = 1

should be added to the transitional table as shown.

Consider the state {q2}:

Since the states obtained above are not new states (already in QD), do not

add these two states to QD. But, the transition on a = 0 and a = 1 should be

added to the transitional table. The final transitional table and the final DFA

are shown below.



7.4.2 Example:

Construct a DFA corresponding to the NDFA as shown below.

Solution: The DFA corresponding to the given NDFA is

7.5 Moore and Mealy Machines

The automaton systems we have discussed so far are limited to binary

output. That is, the systems can either accept or do not accept a string. In

those systems, this acceptability is decided based on the reachability from

the initial state to the final state. This property of the system produces

restrictions in choosing outputs from some other alphabet, then output. You



can remove this constraint using both the Moore and Mealy machine, which

help in generating an output from a certain output alphabet.

Let us denote the output function with the symbol . Thus, when the output

of an automation system is dependent only on the present state, then,

Output = (q(t)), where q(t) is the present state.

The above automaton system is called a Moore machine.

Alternatively, when the output of the automaton system is dependent on

both the present input and the present state, then,

Output = (q(t), x(t)), where q(t) is the present state and x(t) is the present

input.

The above automaton system is called a Mealy machine.

Since the output of a Mealy machine is dependent on both the input and the

present state, no output is generated when the input is a null string. This

implies that when the input is , the output is also . However, in case of the

Moore machine, there is some output of the Moore machine which is only

dependent on the present state and not on the present input. Hence, when

the input to a Moore machine is , the output is (q0).

7.5.1 Definition:

A Moore machine can be with a 6-tuple (Q, , , , , q0) where:



is the output alphabet

is transition function, which is a mapping from Q into Q.

is the output function mapping Q into




From the state table of Moore machine as shown in the table, it is clear that

the output is common for both the inputs, which means the output is only

dependent on the present state and not on the present input.

7.5.2 Definition:

A Moore machine can be with a 6-tuple (Q, , , , , q0) where:



is the output alphabet

is transition function, which is a mapping from Q into Q.

is the output function mapping E Q into


From the state table of Mealy machine as shown in the table, it is clear that

the output is dependent on both the present state and present input in a

Mealy machine.

7.5.3 Conversion of Moore Machine into Mealy machine:

The procedure of converting a Moore machine into a Mealy Machine is very

simple. From the given Moore machine state table, you need to transform



each column of inputs into the pairs of the next state and the output. The

output for a particular state in a pair can be determined by observing the

outputs in first table (Moore machine). If, after converting table to this

format, it is observed that for any two of the present state the other values in

the row are identical, then we can delete one of the states. Let us now

consider a few examples to convert a given Moore machine into a Mealy

machine.

7.5.4 Example:

Convert the Moore machine M1 whose state table is given in table into and

equivalent mealy machine.

Solution: Let us first transform each column of inputs into the pairs of the

next state and the output as shown in the following table.

Present State Input a = 0 Input a = 1

State Output State Output

q0 q1 0 q2 1

q1 q3 1 q2 1

q2 q2 1 q1 0

q3 q0 1 q3 1

Since there are no two states in the above state table which have identical

row elements, the state table as shown represents the equivalent Mealy

machine for the Moore machine depicted above.

7.5.5 Example:

Convert the Moore machine whose state table is given into an equivalent

Mealy machine.



Solution: Let us first transform each column of inputs into the pairs of the

next state and the output as shown in the following table.

From the state table shown above, we observe that for the states for the

states q1 and q2 the rows are identical; hence, we can delete one of these

states to generate the equivalent Mealy machine as shown below.

In this table, we replaced the occurrence of q2 in the remaining rows with q1.

7.5.6 Conversion of Mealy machine into Moore Machine:

Consider the following steps

Step 1: For a state qi determine the number of different outputs that are

available in state table of the Mealy machine.



Step 2: If the outputs corresponding to state qi in the next state columns are

same, then retain state qi as it is. Else, break qi into different states with the

number of new states being equal to the number of different outputs of qi.

Step 3: Rearrange the states and outputs in the format of a Moore machine.

The common output of the new state table can be determined by examining

the outputs under the next state columns of the original Mealy machine.

Step 4: If the output in the constructed state table corresponding to the

initial state is 1, then this specifies the acceptance of the null string by

Mealy machine. Hence, to make both the Mealy and Moore machines

equivalent, we either need to ignore the output corresponding to the null

string or we need to insert a new initial state at the beginning whose output

is 0; the other row elements in this case would remain the same.

Consider the following example, to convert a given mealy machine into a

Moore machine.

7.5.7 Example:

Consider the Mealy machine:

After assessing the state table given above, we observe that the outputs for

the states q1 and q4 are same while the outputs for the states q2 and q3,

have two different values. Hence, after splitting states q2 and q3, we obtain

the state table shown below.



Now, rearranging the state table as shown in table, into the Moore machine

format, we obtain the state table as shown below.

Now, from the state table as shown above, we observe that the output

corresponding to the initial state q1 is 1. Hence, to ensure that the Mealy

and Moore machines are

equivalent, we need to insert a row at the beginning of the state table with

present initial state q0.




1. Obtain NDFA to recognize the strings abc, abd and aacd assuming

= {a, b, c, d}.

7.6 Summary

In this unit, we learnt the concept of Non-deterministic finite automata.

Construction of equivalent DFA from a given NDFA was discussed. We also

discussed a method to convert a Moore Machine into a mealy machine

(vice-versa). Some illustrations are presented on these aspects.


1. Draw the transition diagram of the nondeterministic finite-state

automaton whose next state function is given in the table.

Hint:



7.8 Answers


1. The machine to accept the string abc is shown below.

The machine to accept the string abd is shown below.

The machine to accept the string aacd



Unit 8 Further Problems on DFA and NDFA

Structure

8.1 Introduction

Objectives

8.2 Problems on DFA

8.3 Problems on NDFA

8.4 Difference between DFA and NDFA


8.5 Summary


8.7 Answers

8.1 Introduction

In this unit we study the different problems on DFA and NDFA. We have

given corresponding transition diagrams to understand the concepts easily.

Objectives:


Understand more precisely the idea of DFA.

Draw transition diagrams.

Obtain the DFA to accept a given string.

Apply the techniques to various DFA and NDFA.

Know applications of DFA and NDFA.

8.2 Problems on DFA

8.2.1 Problem

A finite state automaton is defined by a transition diagram shown below.



Find

(i) Its states

(ii) Its input symbols

(iii) Its initial state

(iv) Its accepting states

(v) (q1, 1), (q0, 0)

(vi) write the next state table.

Solution:

(i) The states are Q = {q0, q1, q2}, these are labeled with circles.

(ii) The input symbols of Q are 0 and 1, these are the labels of the arrows.

(iii) The initial state of Q is q0, since the unlabeled arrows points to q0.

(iv) The final or accepting state is q2, since this state is marked by double

circle.

(v) Since there is arrow from q1 to q2 labeled 1, we have (q1, 1) = q2, and

similarly, (q0, 0) = q1.

(vi) The next state table is



8.2.2 Problem:

Consider the finite state automaton defined by the following next table.

Find

(i) the states,

(ii) the input symbols,

(iii) the initial state,

(iv) the accepting states

(v) (q0, c) = ____

(vi) the transition diagram.

Solution:

i) States of M = {q0, q1, q2, q3}

ii) Set of input symbols I = {a, b, c}

iii) Initial state : q0

iv) 20 qc,q

v) State diagram as follows.



8.2.3 Example

Consider the finite state automaton given by the transition diagram

(i) To what state does the automaton go when the symbols of the following

strings are input to it in sequence starting from the initial state ?

(ii) 00, 0010, 10101, 010100

(iii) Which of the strings in part (i) are accepted by an automaton?

(iv) What is the language accepted by an automaton?

Solution:



(i) The respective states are q2, q1, q0, q2.

(ii) Since q2 is an accepting state, the string 00 and 010100 are sent to an

accepting state and hence they are accepted by an automaton.

(iii) We note that if x is any string that ends in 00, then x is accepted by an

automaton. For if the length of x 2, then the first n-2 symbols of x

have been input, the DFA is in one of the three states q0, q1 and q2.

From any of these three states, the input of 00 in sequence sends DFA

first q1 and then the accepting state q2. Hence, any strings of 0‟s and

1‟s that ends in 00 is accepted by DFA.

8.2.4 Problem:

Draw the state diagram for the finite automation F,,q,,QM 0 where

= {a, b}, 210 q,q,qQ , F = {q0, q1}, QxQ: defined by

Verify whether or not the strings “aaab” and “bbaa” are acceptable to M.

Solution:

i) Consider the string: aaab



Therefore the string „aaab’ is not accepted by M.

ii) Consider the string: bbaa

Therefore „bbaa’ is not accepted by M.

iii) State diagram:



Example: Construct deterministic finite state automata that recognize each

of this language.

a) the set of bit strings that begin with two 0‟s.

b) the set of bit strings that contain two consecutive 0‟s.

c) the set of bit strings that do not contain two consecutive 0‟s.

d) the set of bit strings that end with two 0‟s.

Solution: (a) our aim is to construct a deterministic finite state automaton

that recognizes the set of bit strings with two 0‟s. Besides that start state s0,

we include a nonfinal state s1; we move to s1 from s0 if the first bit is a 0.

Next, we add a final state s2, which we move to from s1 if the second bit is a

0. When we have reached s1 we know that the first two input bits are both

0‟s, so we stay in the state s2 (no matter what the succeeding bits, if any

are). We move to a nonfinal state s3 from s0 if the first bit is a 1 and from s1

if the second bit is a 1. We can easily verify that the finite state automaton

recognizes the set of bit strings that begin with two 0‟s.



To construct a deterministic finite-state automaton that recognizes that

contain two consecutive 0‟s. Besides the start state s0, we include a nonfinal

state s1, which tells us that the last input bit seen is a 0, but either the bit

before it was a 1, or this bit was the initial bit of the string. We include a final

state s2 that we move to from s1 when the next input bit after a 0 is also a 0.

If a 1 follows a 0 in the string (before we encounter two consecutive 0s), we

return to s0 and begin looking for consecutive 0s all over again.

In this, our goal is to construct a deterministic finite-state automaton that

recognizes the set of bit strings that end with two 0s. Besides the start state

s0, we include a nonfinal state s1, which we move to the first bit is 0. We

include a final state s2, which we move to from s1 if the next input bit after a

0 is also 0. If an input of 0 follows a previous 0, we stay in state s2 because

the last two input bits are still 0s. Once we are in state s2, an input bit of 1

sends us back to s0 and we begin looking for consecutive 0s all over again.



8.3 Problems on NDFA:

8.3.1 Example: For NDFA whose transition diagram is shown below

Find

(i) initial state

(ii) set of states

(iii) input set

(iv) state table defining the next state function.

Solution:

8.3.2 Problem:

Construct a DFA which accepts strings of 0‟s and 1‟s where the value of

each string is represented as a binary number. Only the strings

representing zero modulo five should be accepted.

For instance, 0000, 0101, 1010, 1111 etc should be accepted.

Solution: Each string represents a modulo 5, the modulo 5 integer may

represent either 0, 1, 2, 3 or 4.

For each integer we shall represent states S, A, B, C and D respectively.

First we list states corresponding to binary numbers below.

Decimal Binary States



0 0000 S

1 0001 A

2 0010 B

3 0011 C

4 0100 D

5 0101 S

6 0110 A

7 0111 B

8 1000 C

9 1001 D

10 1010 S

11 1011 A

12 1100 B

13 1101 C

14 1110 D

15 1111 S

The transition diagram corresponding to this shown below.

8.3.3 Problem

Construct a DFA which accepts the set of all strings beginning with a 1 that

when interpreted as a binary integer, is a multiple of 5. For example, 101,

1010, 1111 etc are multiples of 5.

(Observe that the string 0101 is not beginning with 1 and it should not be

accepted).

Solution:



8.3.4 Problem

Obtain deterministic finite automata to accept the language

L = {w(ab + ba) w {a, b}*}

Solution: The set of states: {q0, q1, q2, q3, q4};

The set input symbols: {a, b}

The set of final states: {q2, q4}

Starting state: {q0}.

8.4 Difference between DFA and NDFA



1. Find the transition diagram for DFA to accept the language

L = {wbab w {a, b}*}

2. Find the transition diagram for DFA to accept the language “only odd

number of a‟s and any number of b‟s”.

8.5 Summary

This unit deals with some problems on DFA and NDFA, we also study the

difference between DFA and NDFA here. The different state diagrams are

illustrated in the form of pictures. We also come to know how to accept a

given string and get the DFA. The different applications of DFA and NDFA

are illustrated here


1. Obtain deterministic finite automata to accept the language

L = {w(ab+ba) | w {a, b}* }

2. Construct a deterministic finite state automata that recognizes each of

this language.

a) the set of bit strings that begin with two 0‟s .

b) the set of bit strings that contain two consecutive 0‟s.

c) the set of bit strings that do not contain two consecutive 0‟s .

d) the set of bit strings that end with two 0‟s

8.7 Answers


1. The transition diagram to accept the given language is



Here the set of states: {q0, q1, q2, q3};

The set input symbols: {a, b}

The set of final states: {q3}

Starting state :{q0}.

2. The transition diagram to accept the given language is

Here the set of states: {q0, q1};

The set input symbols: {a, b}; The set of final states: {q1}; Starting state

:{q0}.



Unit 9 Regular Expressions and

Regular Languages

Structure

9.1 Introduction

Objectives

9.2 Regular expressions

9.3 Regular Expressions accepted by the Language

9.4 Finite Automaton from Regular Grammar

9.5 Regular Grammar from Finite Automata


9.6 Summary


9.8 Answers

9.1 Introduction

In this unit, you will learn about regular expressions along with finite

automata, which act as a device for computing regular expressions. A

regular expression is a set of strings of symbols that can be generated by a

regular grammar using certain operations such as union, intersection and

concatenation. A regular expression also follows different identities that is

based on common mathematical operations such as addition and

multiplication. These identities help simplify the regular expression. A

regular expression can be accepted both by deterministic as well as non-

deterministic automata.

Objectives:

After going through this unit, you will be able to

explain the concept of regular expressions

understand the regular expression accepted by the language.

Convert finite automata from regular grammar.

Convert regular grammar from finite automata.



9.2 Regular Expressions

In computing, regular expressions are used to represent a set of strings and

include symbols that are arranged using certain syntax rules. We can de

regular expression R1 using terminal symbols such as and that are

elements of . Some of the algebraic operations defined with regular

expression are:

1. Union: The union of two regular expressions is also a regular

expression. For example, if R1 and R2 are the two regular expressions,

then the union R1 + R2 is also a regular expression.

2. Concatenation: The concatenation of two regular expressions is a

regular expression. For example, if R1 and R2 are the two regular

expressions, then the concatenation R1R2 is also a regular expression.

3. Iteration: The iteration of a regular expression is also a regular

expression. For example, if R1 is a regular expression, then the iteration

1R is also a regular expression.

4. Order of evolution: The order of evolution of a regular expression is a

regular expression. For example, if R1 is a regular expression, then

order of evolution (R1) is also a regular expression.

9.2.1 Definition:

A regular expression is recursively defined as follows.

1. is a regular expression denoting an empty language.

2. is a regular expression which indicates the language containing an

empty string.

3. a is a regular expression which indicates the language containing only

{a}

4. If R is a regular expression denoting the language LR and S is a regular

expression denoting the language LS, then



a. R+S is a regular expression corresponding to the language

LR LS.

b. RS is a regular expression corresponding to the language LR.LS.

c. R* is a regular expression corresponding to the language LR.

5. The expressions obtained by applying any of the rules from 1 to 4 are

regular expressions.

Note: If parentheses are not present in the regular expressions, then

precedence of the operands is as follows: iteration, concatenation and

union. First you need to perform the iteration operation, then the

concatenation operation and finally the union operation.

Note: Any set, which is represented by using a regular expression, is known

as regular set. If the regular expression is R, then the regular set of R is

L(R).

9.2.2 Example:

Let x, y , where,

x represents the set {x}

x + y represents the set {x, y}

xy represents the set {xy}

x* represents the set {, x, xx, xxx, …}

(x + y)* represents the set{x + y}*

9.3 Regular Expressions accepted by the Language

9.3.1 Example:

Some examples of regular expressions and the language corresponding to

these regular expressions are given here.



Regular Expressions

Meaning

(a+b)* Set of strings of a‟s and b‟s of any length including the NULL string.

(a+b)*abb Set of strings of a‟s and b‟s ending with the string abb

ab(a+b)* Set of strings of a‟s and b‟s starting with the string ab.

(a+b)*aa(a+b)* Set of strings of a‟s and b‟s having a sub string aa.

a*b*c*

Set of string consisting of any number of a‟s(may be empty string also) followed by any number of b‟s(may include empty string) followed by any number of c‟s(may include empty string).

abc

Set of string consisting of at least one „a‟ followed by string consisting of at least one „b‟ followed by string consisting of at least one „c‟.

aa*bb*cc*

Set of strings consisting of at least one „a‟ followed by string consisting of at least one „b‟ followed by string consisting of at least one „c‟.

(a+b)* (a + bb) Set of strings of a‟s and b‟s ending with either a or bb

(aa)* (bb)* b Set of strings consisting of even number of a‟s followed by

odd number of b‟s.

9.3.2 Example

Obtain a regular expression to accept a language consisting of strings of a‟s

and alternate a‟s and b‟s.

Solution: The alternate a‟s and b‟s can be obtained by “concatenating the

string ab zero or more times” which can be represented by the regular

expression

(ab)*

and adding an optional b to the front and adding an optional a at the end as

shown below:

( + b) (ab)* ( + a).

Thus, the complete expression is given by

( + b) (ab)* ( + a)



9.3.3 Note

The expression can also be obtained as shown below:

The a‟s and b‟s can be generated using one of the following ways:

i) (ab)*

ii) b(ab)*

iii) (ba)*

iv) a(ba)*

Therefore the expression to generate alternate a‟s and b‟s can be obtained

by taking the union of regular expressions as shown below:

(ab)* + b(ab)* + (ba)* + a(ba)*

9.3.4 Example

Obtain a regular expression to accept a language consisting of strings of 0‟s

and 1‟s with at most one pair of consecutive 0‟s.

Solution: It is clear from the statement that the string consisting of at most

one pair of consecutive 0‟s may

o begin with combination of any number of 1‟s and 01‟s represented by (1

+ 01)*

o end with any number of 1‟s represented by 1*.

Therefore the complete regular expression which consists of strings 0‟s and

1‟s with at most one pair of consecutive 0‟s is given by

(1 + 01)*00 1*.

9.3.5 Example

Obtain a regular expression to accept a language containing at least one a

and at least one b where = {a, b, c}.

Solution: Strings of a‟s, b‟s and c‟s can be generated using the regular

expression

(a + b + c)*.



But this string should have at least one „a‟ and at least one „b‟. There are

two cases to be considered:

First „a‟ preceding „b‟ which can be represented using

c*a(a + c)*b

First „b‟ preceding „a‟ which can be represented using

c*b(b + c)*a

The regular expression (a + b + c)* can be preceded by one of the regular

expressions considered in the two cases just discussed.

Therefore the final regular expression is

c*a(a + c)*b(a + b + c)* +c*b(b + c)*a(a + b + c)*

This expression can also be written as shown below:

[c*a(a+c)*b + c*b(b4c)*a] (a+b+c)*

9.3.6 Example


and b‟s of even length.

Solution: String of a‟s and b‟s of even length can be obtained by the

combination of the strings aa, ab, ba and bb.

The language may even consist of an empty string denoted by .

Therefore the regular expression can be of the form

(aa + ab + ba + bb)*

The * closure includes the empty string.

The language corresponding to the regular expression is denoted by

L(R)={(aa + ab + ba + bb)n n 0}.



9.3.7 Example


and b‟s of odd length.

Solution: String of a‟s and b‟s of odd length can be obtained by the

combination of the strings aa, ab, ba and bb followed by either a or b.

Therefore the regular expression can be of the form

(aa + ab + ba + bb)* (a + b)

String of a‟s and b‟s of odd length can also be obtained by the combination

of the strings aa, ab, ba and bb preceded by either a or b.

Therefore the regular expression can also be represented as

(a + b) (aa + ab + ba + bb)*.

Observation: Even though these two expressions seem to be different, the

language corresponding to these two expressions is same.

9.3.8 Example

Obtain a regular expression such that L(R) = {w w {0, 1}* with at least

three consecutive 0‟s.

Solution: A string consisting of 0‟s and l‟s can be represented by the

regular expression

(0 + 1)*

This arbitrary string can precede three consecutive zeros and can follow

three consecutive zeros.

Therefore the regular expression can be written as

(0 +1)* 000(0+1)*.

The language corresponding to the regular expression can be written as

L(R) = { (0 + 1)m000(0+1)n m 0 and n 0}.



9.4 Finite Automaton from Regular Grammar

9.4.1 Definition

A grammar G = (VN, VT, S, ) is said to be regular grammar the

grammar is right regular or left regular.

A grammar G is said to be right regular if all the productions are of the form

A wB and / or A w, where A, B VT and w VT*.

A grammar G is said to be left regular if all the productions are of the form

A Bw and / or A w, where A, B VT and w VT*.

9.4.2 Example

(i) The grammar with the set of productions

S aaB bbA

A aA b

B bB a

is a right linear grammar.

(ii) The grammar with the set of productions

S Baa Abb

A Aa b

B Bb a

is a left linear grammar.

9.4.3 Definition

A grammar which has at most one non terminal on the right side of any

production without restriction on the position of this non terminal (observe

that: non terminal can be leftmost or rightmost) is called linear grammar.

9.4.4 Theorem

Let G = (VN, VT, S, ) be a right linear grammar. Then there exists a

language L(G) which is accepted by a finite automata, that is, the language

generated from the regular grammar is a regular language.



Proof: Let V = {q0, q1, …} be the variables and S = q0 be the start state.

Let the productions in the grammar be

q0 x1q1

q1 x2q2

q2 x3q3

… … …

qn xn+1

Assume that the language L(G) generated from these productions is w.

Corresponding to each production in the grammar we can have equivalent

transitions in the FA to accept the string w.

After accepting the string wm the FA will be in the final state.

The procedure to obtain FA from these productions is given below.

Step 1: The start symbol q0 in the grammar is the start state of FA.

Step 2: For each production of the form qI wqj the corresponding

transition defined will be of the form

*(qi, w) = qj.

Step 3: For each production of the form qi w, the corresponding transition

defined will be of the form

*(qi, w) = qf, where qf is the final state.

Since the string w L(G) is also accepted by FA, by applying the transitions

obtained in step 1 through 3, the language is regular.



9.4.5 Problem: Construct a DFA and the transition diagram, to accept the

language generated by the following grammar.

S 01A

A 10B

B 0A 11

Solution: Observe that each production of the form

A wB

the corresponding transition will be (A, w) = B

Also, for each production of the form A w, we can introduce the transition

(A, w) = qf, where qf is the final state.

The transitions obtained from grammar G is shown in the table.

The transition diagram is shown below.

The DFA is

M = (Q, , , q0, F) where



Q = {S, A, B, qf, q1, q2, q3},

= {0, 1}, q0 = S (start state), F = {qf}, is shown in the table. Here, the

additional vertices (states) introduced are q1, q2, q3.

9.4.6 Problem:

Construct DFA and the corresponding transition diagram to accept the

language generated by the following grammar.

S aA

A aA bB

B bB

Solution: Observe that each production of the form

A wB

the corresponding transition will be

(A, w) = B

Also, for each production of the form

A w,

we can introduce the transition

(A, w) = qf, where qf is the final state.

The transitions obtained from grammar G is shown in the table.



Observe that for each production of the form , make A as the final

state.

The transition diagram corresponding to this is shown below.

9.5 Regular Grammar from Finite Automata

9.5.1 Theorem:

Let M = (Q, , , q0, F) be a finite automata. If L is the regular language

accepted by FA, then there exists a right linear grammar G = (VN, VT, S, )

so that L = L(G).

Proof: Let M = (Q, , , q0, F), where Q = {q0, q1, …, qn}, = {a1, a2, …, am}.

A regular grammar G = (VN, VT, S, ) can be constructed where

VN = {q0, q1, …, qn}, VT = , S = q0.

The set of productions can be obtained as shown below.

Step 1: For each transition of the form (qi, a) = qj the corresponding production is

qi aqj

Step 2: If q F, the final state in FA, then introduce the production q .

Since these productions are obtained from the transitions defined for FA, the

language accepted by FA is also accepted by the grammar.



9.5.2 Example:

Obtain a regular grammar from the following DFA given by the transition diagram.

Solution: For each transition of the form (A, a) = B, introduce the

production A aB. If q F (the final state), introduce the production A .

The productions obtained from the transitions defined for DFA is shown

below.

From the diagram, it is clear that the state B is a final state.

Therefore we introduce the production .

The grammar G corresponding to the productions obtained is shown below.



9.5.3 Example

Construct a regular grammar for the following DFA given by the transition

diagram.

Solution: For each transition of the form (A, a) = B, introduce the

production A aB.

If q F (the final state), introduce the production A . The productions

obtained from the transitions defined for DFA is shown below.



Since the set of final states: {S, A, B}, we introduce the productions S ,

A , and B .

Therefore the grammar G is:

G = (VN, VT, S, ) where

VN = {S, A, B, C}

VT = {a, b}

Observation: The finite automaton in this problem accepts strings of a‟s

and b‟s except those containing the substring abb. Therefore from the

grammar G we can obtain a regular language which consist of strings of a‟s

and b‟s without the substring abb.

9.5.4 Example

Obtain a right linear grammar for the regular expression ((aab)* ab)*, given

by the transition diagram.

The right linear grammar is given by


VN = {S, A, B}

VT = {a, b}



9.5.5 Note

The left linear grammar can be obtained from FA as follows.

Step 1: Obtain the reverse of given DFA.

Step 2: Obtain the right linear grammar from the reversed DFA.

Step 3: Obtain the left linear grammar from right linear grammar.

9.5.6 Example

Obtain a left linear grammar for the DFA shown below.

Step 1: Reverse the DFA. That is, A as the final state and C as the start

state and reverse the direction of the arrow. The reversed DFA is shown

below.

Step 2: obtain the right linear grammar for the above DFA. The

corresponding productions are shown below.



Step 3: Reverse the productions of right linear grammar to get left linear

grammar.

If A abcdB is the production in right linear grammar, after reversing the

production will be of the form

A Bdcba.

The conversion of right linear grammar to the left linear grammar is shown

below.

Therefore the final left linear grammar is


VN = {C, A, B}

VT = {0, 1}

Now we show that the string 10101 is accepted by DFA.



Hence the left linear grammar obtained is equivalent to the given FA.


1. The regular expression (11)* stands for _______

2. The regular expression (01)* + 1 stands for _____

3. The regular expression (0 + 10)*1* stands for ______

4. Obtain a left linear grammar for the regular expression ((aab)* ab)*.

9.6 Summary

In this unit special type of grammar called regular grammars were

considered. Different forms of regular expressions and the regular

expressions accepted by the language are given. We provided a method of



obtaining a regular grammar from the finite and automaton (and vice versa).

Sufficient number of examples were given.


1. Obtain a right linear grammar for the language L = {anbm n 2, m 3}.

2. Obtain the left linear grammar for the right linear grammar shown below.

9.8 Answers


1. Set of strings consisting of even number of 1‟s.

2. The language consists of a string 1 or strings of (01)‟s that repeat zero

or more times.

3. Stings of 0‟s and 1‟s ending with any number of 1‟s (possible none).

4. G = (VN, VT, S, ) where VN = {A, B, S}, VT = {a, b}



Unit 10 Properties of Regular Languages

and Pumping Lemma

Structure

10.1 Introduction

Objectives

10.2 Closure Properties of Regular Sets

10.3 Pumping Lemma

10.4 Applications of Pumping Lemma:


10.5 Summary


10.7 Answers

10.1 Introduction

This unit deals with closure properties of regular languages. The different

closure properties covered in this unit are union, concatenation, closure,

intersection, complementation etc. We also cover the Pumping lemma and

some of its applications to check whether the given language is regular or

not.

Objectives


Apply the closure property on regular sets

Understand the applications of Pumping Lemma

10.2 Closure Properties of Regular Sets

When two real numbers are multiplied, the resultant product also becomes a

real number. Thus, we can say that the real numbers are closed under the

operation of multiplication. A regular set is closed under the following



operations such as union, concatenation, closure, transpose, intersection

and complementation.

10.2.1 Theorem

If L1 and L2 are regular, then L1 L2, L1.L2 and L1* also denote the regular

languages and we say that the regular languages are closed under union,

concatenation, start-closure.

Proof: Let L1 and L2 be regular languages corresponding to the regular

expressions R1 and R2.

By definition, R1 + R2, R1.R2 and R1* are regular expressions and so L1 L2,

L1.L2, L1* denote the regular languages and so regular languages are closed

under union, concatenation and star- closure.

10.2.2 Theorem

If L1 and L2 are regular, then the regular language is closed under

complementation.

Proof: Let M = (Q, , , q0, F) be a DFA which accepts the language L1.

Now, let us define the machine M1 = (Q, , , q0, Q-F).

(Observe that there is no difference between M and M1 except the final

states).

The non-final states of M are the final states of M1 and final states of M are

the non-final states of M1. So, the language which is rejected by M is

accepted by M1.

Also, a language accepted by a DFA is regular. So, the language accepted

by M1 is regular. So, a regular language is closed under complementation.

10.2.3 Theorem


intersection.



Proof: Let us consider M1 = (Q1, , 1, q1, F1) which accepts L1 and

M2 = (Q2, , 2, q2, F2) which accepts L2.

It is clear from these two machines that the alphabets of both machines are

same.

Assume both the machines are DFAs.

To accept the language L1 L2, let us construct the machine M that

simulates both M1 and M2 where the states of the machine M are the pairs

(p, q) where p Q1 and q Q2.

The transition for the machine M from the state (p, q) on input symbol a

is the

((p, a), (q, a)).

That is, if, 1(p, a) = r and 2(q, a) = s, then the machine moves from the

state (p, q) to the state (r, s) on input symbol a.

In this manner, the machine M can simulate the effect of M1 and M2.

Now, the machine M = (Q, , , q, F) recognizes L1 L2 where

Q = Q1 Q2

q = (q1, q2) where q1 and q2 are the start states of machine M1 and M2

respectively.

F = {(p, q) p F1 and q F2}

: Q Q, defined by ((p, q), a) = (1(p, a), 2(q, a))

It is clear from that

((p, q), w) = (

1 (p, w),

2 (q, a)) and the string w is accepted only if

((q1, q2), w) F, which implies that (

1 (q1, w),

2 (q2, w)) F

1 (q1, w) F1 and

2 (q2, w)) F2.

w L1 L2.

Therefore, the regular language is closed under intersection.



10.2.4 Theorem


difference.

Proof is straightforward as above.

10.2.5 Definition

Let and be the set of alphabets. The function f: * is called

homomorphism (that is, substitution where a single letter is replaced by a

string) if w = a1a2…an then f(w) = f(a1)f(a2)…f(an).

If L is made of alphabets from, then f(L) = {f(w) w L} is called

homomorphic image.

10.2.6 Example

Let = {0, 1}, = {0, 1, 2} and f(0) = 01, f(1) =112. Find f(010). Write the

homomorphic image of L = {00, 010}.

Solution: By definition we have

f(w) = f(a1)f(a2)…f(an).

So, f(010) = f(0) f(1)f(0) = 0111201.

L (00, 010) = L( f(00), f(010)) = L( f(0)f(0), f(0)f(1)f(0)) = L(0101, 0111201).

Therefore,

f(010) = 0111201

L (00, 010) = L(0101, 0111201)

10.2.7 Example

If = {0, 1}, = {1, 2, 3}, f(0) = 3122, f(1) = 132, then find (0 +1 )*(00)*.

Solution: By definition we have

f(w) = f(a1)f(a2)…f(an).

So, (0+1)*(00)* = (f(0) + f(1))* (f(0)f(0))* = (3122 + 132)* (31223122)*



10.2.8 Theorem

If L is regular and f is homomorphism, then homomorphic image f(L) is

regular.

Proof: Let R be the regular expression and L(R) be the corresponding

regular language. We can easily find f(R) by substituting f(a) for each a .

By definition of regular expression, f(R) is a regular expression and so f(L) is

regular language. So, the regular language is closed under homomorphism.

Remark:

Any finite language can be expressed using regular expression and for any

language which can be represented using regular expression, we can have

a DFA. Even some of the infinite languages can be represented using

regular expressions and for these languages also we can construct DFA

and so are regular.

But, some of the infinite languages are not regular. Though the regular

languages are important, there are non-regular languages which are very

interesting and important.

Some of the non-regular languages are:

1. {w {0, 1} w contains an equal number of 0’s and 1’s}

2. {0n1n {0, 1}* n 0}

3. {ap {a}* p 2 is a prime number}

4. Check for the matching parentheses (not possible using DFA).

5. Count number of a’s and then the number of b’s (not possible using

DFA) and so it can not be used as a counter.

10.3 Pumping Lemma

It is clear from these examples that it is not possible to obtain corresponding

DFA for these and so they are not regular. So, we face problems whenever

we come across non regular languages. The easiest way to prove that a



language is regular is to construct a DFA. Then, what is the easiest way to

prove that a language is not regular?

In this section, we discuss the way to prove that certain languages are

infinite using Pumping Lemma.

The principle used in Pumping Lemma is similar to the Pigeonhole principle.

Pigeonhole principle (Refer unit 3): The pigeonhole principle is based on

a simple observation. Suppose there are n objects and m boxes where

number of objects n, are greater than number of boxes m. In this case, if all

n objects are placed into m boxes, then at least one box will have more than

one object.

The pumping lemma is based on this principle.

The pumping lemma is stated as follows:

10.3.1 Lemma

Let M = (Q,, , q0, F) be an FA and has n number of states. Let L be the

language accepted by M and assume the language is regular. Let x L and

x n, that is, length of the string x is greater than the number of states of

FA. If the string x can be decomposed into combinations of strings

x = uvw

such that uv ≤ n, v 1, then uvi w L for i = 0, 1, 2, ….

Proof: Let M = (Q, , , q0, F) be an FA and Let x = a1a2a3 … am is the input

which is accepted by the machine. Assume the machine has n states and

assume that m n.

Observe that, if exactly one symbol is accepted by an FA then there are two

distinct states in it. Similarly to accept exactly two symbols. FA should have

three distinct states. In general, to accept a string x where x = n, then the

FA should have n + 1 states.



Let those states be q0, q1, …, qm.

Since we have m input symbols, naturally we should have m + 1 states in

the sequence q0, q1, …, qm, where q0 will be the start state and qm will be the

final state as shown below:

It is clear from the figure that (qi-1, ai) = qi for each 1 ≤ i ≤ m. In this chain of

states from q0 to qm, suppose a state q appears more than once, say q = qi

and q = qj where i < j.

In this case, the chain should have a loop and we can split into three

groups:

1. The first group is the string prefix from a1a2a3 … ai

2. The second group is the loop string from ai+1ai+2 … aj-1aj

3. The third group is the string suffix from aj+1aj+2 … am, shown below.

Note that the machine cannot remember the previous state or it does not

know how it has reached the current state.



So, whenever we say that the machine has reached the state q, the

machine does not know whether the state is reached after the end of the

prefix or after it has processed the loop. Since the machine cannot

distinguish these states, after the prefix (u), the machine may be in a loop

for zero or more time (accepting the string v zero or more times) and then

accept the suffix string (w). In general, if the string x is split into sub strings

uvw, then for all i 0,

uvi w L.

If F = {qm}, a1a2a3 … am L(M), then

a1a2a3 … ai-1ai ai+1ai+2 … aj-2aj-1aj aj+1aj+2 … am L(M).

This can be expressed using the transition as shown below.

(q0, a1a2a3 … ai-1ai ai+1ai+2 … aj-2aj-1aj aj+1aj+2 … am)

= ( (q0, a1a2a3 … ai-1ai ), ai+1ai+2 … aj-2aj-1aj aj+1aj+2 … am)

= (q, aj+1aj+2 … am)

= (qk, ak+1ak+2 … am)

= qm.

Also, after the string

a1a2a3 … ai-1ai

the machine will be in state qi. Since qi and qj are same, we can input the

string

ai+1ai+2 … aj-2aj-1aj

any number of times and the machine will stay in qj only. Finally, if the input

string is

aj+1aj+2 … am

the machine enters into final state qm.



10.4 Applications of Pumping Lemma

All languages are not-regular. A non-regular language is that language for

which a FA cannot be constructed. So, to prove that certain languages are

not regular, we use pumping lemma. The typical applications of Pumping

Lemma are:

1. To prove that certain languages are not regular. The pumping lemma

cannot be used to prove that a given language is regular.

2. To check whether the language is infinite. If there is a string x such that

x the number of states accepted by DFA M, then L(M) is infinite.

Otherwise, L(M) is finite. So, using Pumping lemma we can check

whether the language is finite or infinite.

10.4.1 Steps to prove that certain language is not regular:

Step 1: 1. Assume that the language L is regular and the number of states

in FA be n.

Step 2: Select the string say x and break it into substrings u, v and w such

that x = uvw with the constraints x n, uv ≤ n and v 1.

Step 3: Find any i such that uviw L. According to pumping lemma, uviw

L. So, the result is a contradiction to the assumption that the language is

regular.

Conclusion: The given language L is not regular.

10.4.2 Example

Show that L = {an bn n 0} is not regular.

Solution:

Step 1: Let L be regular and n be the number of states in FA. Consider the

string x = anbn.

Step 2: Note that x = 2n and is greater than n. So, we can split x into uvw

such that uv ≤ n and v 1 as shown below.



x =

n

vu

aaaaaaa

n

w

bbbbbbb

where u = n – 1 and v = 1 so that uv = u + v = n -1 + 1 = n and

w = n.

According to Pumping lemma, uviw L for i = 0, 1, 2, ….

Step 3: If i = 0, that is, v does not appear and so the number of a’s will be

less than the number of b’s and so the string x does not contain same

number of a’s followed by same number of b’s (equal to that of a’s).

Similarly, if i = 2, 3, ..., then number of a’s will be more than the number of

b’s and so number of a’s followed by equal number of b’s does not exist.

But, according to pumping lemma, n number of a’s should be followed by n

number of b’s which is a contradiction to the assumption that the language

is regular.

So, the language L = {an bn n 0} is not regular.

10.4.3 Example

Show that L = {aibj i > j}is not regular.

Solution:

Step 1: Let L be regular and n be the number of states in FA.

Consider the string x = an+1 bn.

Step2: Note that x = 2n+1 n. So, we can split x into uvw such that uv ≤

n and v 1 as shown below.

x = an+1bn = u

ja v

ka w

nab

where u = j and v = k 1 and so that uv = u + v = j + k ≤ n.

Step 3: According to pumping lemma, uviw L for i = 0

That is, niki abaa )( L for i = 0.



Now, if we choose i = 0, number of a’s in string u will not be more than the

number of b’s in w which is contradiction to the assumption that number of

a’s are more than the number of b’s.

So, the language L = {aibj i > j} is not regular.

10.4.4 Example

Show that L = {anblcn+l n, l 0}is not regular.

Solution:

Step 1: Let L be regular and n be the number of states in FA.

Step2: Since L is regular, it is closed under homomorphism, so, we can take

f(a) = a, f(b) = a and f(c) = c.

Now, the language L is reduced to

L = {analcn+l n + l 0}

which can be written as

L = {an+1cn+l n + l 0}

We know that the above language is not regular (proved above), which is a

contradiction to the assumption that the language is regular. So, the given

language is not regular.

10.4.5 Example

Show that L = {0n n is prime} is not regular.

Solution: The language generated from this can take the following form

L = {00, 000, 00000, …}

Step 1: Let L be regular and n be the number of states in FA. Let us choose

the value of x which depends on n.

Let x = 0n L where n is prime.

Step2: x = n and so, we can split x into uvw such that uv ≤ n and v 1

as shown below.



x = 0n = u

j0 v

k0 w

kjn 0


Step 3: According to pumping lemma, uviw L for i = 0, 1, 2, ...

That is, u

j0 v

k0 w

kjn 0 L.

This means that j + ki + n – j – k = n + k(i -1) is prime for all i 0.

Now, if we choose i = n + 1, then

n + k(i-1) = n + kn = n(k + 1)

is also a prime for each k 1,

which is a contradiction (because if k = 1, it will not be a prime) to the

assumption that the language is regular.

Therefore the language L = {0n n is prime} is not regular.

10.4.8 Example

Show that L = {ww w {a, b}* } is not regular.

Solution

Step 1: Let L be regular and n be the number of states in FA. Let us choose

the value of x which depends on n. Let x = anban b L.

Step2: Note that x = 2n + 2 n and so, we can split x into uvw such that

uv ≤ n and v 1, as shown below.

x = anban b = u

ja v

ka w

nbba



That is, bbaaa nikj )( L for i = 0, 1, 2, ...

Now, if we choose i = 0, number of a’s on the left of first b will be less than

the number of a’s after the first b which is contradiction to the assumption

that they are not equal. Therefore, the language L = {ww w {a, b}*} is not

regular.



10.4.9 Example

Show that L = {an n = k2 for k 0} is not regular.

Equivalently: The above language can also be defined as: L = {an is a

perfect square}

Solution:

Step1: Let L is regular and ii be the number of states in FA.

Let us choose the value of x which depends on n. Let x = am L (where m =

n2)

Step 2: Note that x n and so, we can split x into uvw such that uv ≤ n

and v 1 as shown below.

x = am = u

ja v

ka w

kjma



That is, kjmkij aaa L for i = 0, 1, 2, …

Now, if we choose i = 2, we have

uv2w L.

That is, kjmij aaa 2 L for i = 0, 1, 2, …

Therefore am+k L. Since k 1, we have,

am+k = m + k = n2 + k (since m = n2).

Note that n2 < n2 + k < n2 + 1 (when k = 1) < (n2 + 2n + 1) = (n + 1)2.

This means that n2 < (n + 1)2.

Since n2 + k lies between n2 and (n + 1)2, it is not a perfect square which is a

contradiction to the assumption that it should be a perfect square.

Therefore the language L = {an n = k2 for k 0} is not regular.




1. The homomorphic image of a regular language is _______

Answer: Regular.

2. Intersection of two regular languages is ______

Answer: regular

3. Difference of two regular languages is _____

Ans: Regular.

4. State advantages of pumping lemma.

5. Verify whether the language L = {ww w {a, b}*} is regular.

10.5 Summary

In this unit we discussed certain properties (called closed properties) of

regular languages. We gave the definition of homomorphism and obtained

that the regular languages are closed under homomorphic images. Further

we discussed the way to prove that certain languages are infinite using

Pumping Lemma. Some applications of Pumping lemma were discussed.


1. Show that L = {w number of a’s in w is less than the number of b’s in

w} is not regular.

(Hint: Use Pumping Lemma).

2. Prove that the regular languages are closed under intersection and

difference.

3. State and prove Pumping Lemma.



10.7 Answers

Self Assessment Question

1. Regular.

2. Regular

3. Regular.

4. We can easily prove that certain languages are non-regular.

It is possible to check whether a language accepted by FA is finite or

infinite.

5. By Pumping Lemma L is not regular.



Unit 11 Context Free Grammars

Structure

11.1 Introduction

Objectives

11.2 CFG for various types of CFL

11.3 Derivations

11.4 Ambiguous Grammar


11.5 Summary


11.7 Answers

11.1 Introduction

In the regular grammar there is restriction on the number of non-terminals

on the left hand side as well as on the right hand side. On the left hand side

of the production, there will be only one symbol and that symbol must be a

non-terminal (variable). But, the right hand side of the production may

contain zero or more terminals. If a non-terminal (variable) is present, that

non- terminal should be the left most symbol or the right most symbol. But,

in a context free grammar there is only one restriction. The symbol on the

left hand side of the production should be a single non-terminal but, there is

no restriction on the right hand side of the production. Any number of non-

terminals and terminals may be present (including the symbol ).

The context free grammar is defined as follows.

(Refer unit 4, definition of type 2 grammar and few simple illustrations).



Objectives:


Understand the concept of Context free grammars

Know the ideas in derivation trees

Test the ambiguity in context free grammar

11.2 Context -free Grammars

11.2.1 Definition

A grammar G is a quadruple G = (VN, VT, , S) where

VN is set of variables or non-terminals

VT is set of terminal symbols

is set of productions

S is the start symbol,

is said to be type 2 grammar or context free grammar (CFG) if all the

productions are of the form A where

(VN VT)* and A VN. The symbol (indicating NULL string) can

appear on the right hand side of any production.

The language generated from this grammar is called type-2 language or

context free language (CFL).

Observations:

1. There is only one symbol A on the left hand side of the production and

that symbol must be a non-terminal.

2. (VN VT)* implies that right hand side of the string may contain any

number of terminals and non-terminals including (NULL string).

3. Every regular grammar is a CFG and hence a regular language is also

context free language but the reverse is not true always.

4. Notation: nx(w) = number of x’s in the string w.



11.2.3 Example

Let G = (VN, VT, , S) be a CFG where

VN = {S}

VT = {a, b}

: S aSa bSb

S: Starting symbol.

Find the language generated by this grammar.

Solution: The null string can be obtained by applying the production

Therefore, by applying the productions

S aSa and S bSb, and any number of times and in any order, finally

applying the production S , we get a string w followed by reverse of it

(say wR).

Hence the language generated by this grammar is

L = {wwR w {a + b}*}.

Since this language is generated from the context free grammar, it is a

context free language.

11.2 4 Example

Show that the language L = {ambn m n} is context free.

Solution: It is clear from the given language that if number of a’s are

followed by n number of b’s and number of a’s and b’s are not equal.



As a first attempt we can have the production

S aSb

using which n number of a’s are followed by one S followed by n number of

b’s are generated.

Now, if we replace the non-terminal S by the production

S

we get n number of a’s followed by n number of b’s.

But, we should see that number of a’s and b’s are different. So, we should

be in a position to generate either one or more extra a’s or one or more

extra b’s. Hence, instead of the production

S

we can have productions

S A B

From the non-terminal A we can generate one or more a’s and from non-

terminal B we can generate one or more b’s as shown below:

So, the context free grammar G = (VN, VT, , S) where

VN = {S, A, B}, VT = {a, b},

S: starting symbol,

which generates the language L = {ambn m n}. Since a CFG exists for the

language, the language is context free.

11.2.5 Example

Obtain a Context free grammar on {a, b} to generate a language

.



Solution: Here, the string wwR must be enclosed between an and bn where

n 1. The final grammar is

VN = (S}

VT = {a, b}

: S aSb aAb

A aAa bAb , and

S is the start symbol.

We will verify, by taking n = 2, so that we generate a string aabaabbb (here

w = ba, and that wR = ab). Consider the following sequence of productions

to get the string aabaabbb.

11.2.6 Problem

Obtain the context free grammar for the regular expression

(011 + 1)*(01)*.

Solution: The expression (011 + 1)*(01)* is of the form A*B* where A = 001

or 1 and B = 01. The regular expression A*B* means that any number of

A’s (possibly none) are followed by any number of B’s (possibly none). Any

number of A’s (that is, 011’s or 1’s) can be generated using the productions

A 011A 1A

Any number of B’s (that is, 01’s) can be generated using the productions

B 01B



Now, the language generated from the regular expression (011 + 1)*(01)*

can be obtained by concatenating A and B using the production

S AB

Therefore, the final grammar G = (VN, VT, , S) where

VN: {S, A, B}

VT: {0, 1}

: S AB,

A 011A 1A

B 01B

S: start symbol.

11.2.7 Backus – Naur Form (BNF)

This is an alternative method of displaying productions of context – free –

grammars.

In a context – free – grammar, left hand side of all productions is single non-

terminal symbols.

i) Every non terminal symbol is enclosed in angle brackets < >.

ii) The terminal symbols are written without any special making.

iii) The symbol : : = is used instead of and should be read as “ is

defined as “.

iv) All the productions with the same non terminal left hand side are

combined into one statement with all the right hand sides listed on the

right of : : = , separated by vertical bars.

For instance, the production A B is written as

< A > : : = < B >

production of the form

< A > : : = < B1 > , < A > : : = < B2 > , …, < A > = < Bn >

In BNF it may be combined as :



< A > : : = <B1>|<B2>|…|Bn>

For example, productions A Aa, A a and A AB can be combined in

BNF as

::A A a a A B

11.2.8 Example

Consider the grammar which generates the decimal numbers

, , ,N TG V V S where

VN = {Decimal number, Decimal fraction, Unsigned integer, Digit}.

VT = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

The productions are given in BNF notations as follows :

1. <decimal numbers > : : = < unsigned integer > | < decimal fraction > |

< unsigned integer> < decimal fraction>

2. < decimal fraction > : : = < unsigned integer >

3. < unsigned integer > : : = <digits > | <digit > <unsigned integer >

4. < digit > : : = 0 |1 | 2 | 3 | 4 | 5 | 6 | 7 |8 | 9.

11.2.9 Example

Consider the grammar



, , , , , ,G S A B a b S where

: , , , .S aB A aB B bA B b

These productions can be written in BNI notation as:

::

::

:: |

S a B

A a b

B b A b

11.3 Derivation Trees

A derivation tree is an ordered tree in which each vertex is labeled with the

left sides of a production and in which the children of a vertex represents its

corresponding right sides. At the root of the tree, we put the non terminal

symbol with which we begin the derivation. Interior vertex of the tree

corresponds to a non terminal symbol that arises in the derivation. The

leaves of the tree represent the terminal symbols that arise.

11.3.1 Definition

In the derivation process, if a left most (respectively, right most) variable is

replaced at every step, then the derivation is said to be leftmost derivation

(respectively, right most derivation).

11.3.2 Example


, , { }, ,T NG V a b V S S

where : , .S aSb S ab

The language to this grammar is 1n na b n

The sentence aaabbb is derived as follows



The derivation tree is:

11.3.3 Problem


{ , } { , , }, ,T NG V a b V S A B S

Where the productions are:

Construct the derivation trees for the strings.

i) abababba

ii) aababb



Solution:

Derivation tree: for (i)

Derivation tree for (ii):



11.4 Ambiguous Grammar

Let G = (VN, VT, , S) be a context free grammar. A grammar G is

ambiguous if and only if there exists at least one string w VT for which two

or more different derivation trees exist by applying either the left most

derivation or right most derivation. Note that after applying left most

derivation or right most derivation even though the derivations look different

and if the structure of derivation trees obtained is same, we cannot conclude

that the grammar is ambiguous. It is not the multiplicity of the derivations

that cause ambiguity. But, it is the existence of two or more derivation trees

for the same string w derived from the root labeled S.

Note:

1. Obtain a left most derivation and get a string w. Obtain the right most

derivation and get a string w. For both the derivations construct the

derivation trees. If there are two different derivation trees, then the

grammar is ambiguous.

2. Obtain the string w by applying left most derivation twice and construct

the derivation tree. If the two derivation trees are different, the grammar

is ambiguous.



3. Obtain the string w by applying right most derivation twice and construct

the derivation tree. If the two derivation trees are different, the grammar

is ambiguous.

4. Apply the left most derivation and get string. Apply the left most

derivation again and get a different string. The derivation trees obtained

will naturally be different and don not come to the conclusion that the

grammar is ambiguous.

11.4.1 Example

Verify whether or not the following grammar is ambiguous.

Solution: Consider the two leftmost derivations for the string aaaa.

Since there are two left most derivations for the same sentence aaaa, the

given grammar is ambiguous.



11.4.2 Example

Consider a grammar

G= Φ)S,},{a,({S},G With productions a.SS,SS:Φ

The sentence a + a + a can be derived in two different ways. The derivation

trees are distinct.

11.4.3 Example

Obtain the string aaabbabbba by applying left most derivation, the derivation

tree for the grammar is shown below.



Solution: Apply the set of productions

Observation: In this example, we have used the productions (different from

the application of production in the SAQ).

The derivation tree is given below.




1. Find the language for the grammar.

0,1 , , ,T NG V V S S

where the set of productions

: 11 , 0.S S S

2. Find the language L (G), generated by the grammar.

, , , , , ,T NG V x y z V S A S

where Z.AyA,AyA,SS,xS:Φ

3. Find the language L (G), generated by the grammar

, , , ,T NG V a b V S S

where

: , , .S aaS S a S b

4. Suppose G = ({S, A, B}, {a, b}, , S) where consists of the following

productions: . Then

show that w = abaaababaaab L(G).

5. The CFG to generate a language consisting of equal number of a’s and

b’s is ________

6. Obtain a CFG to generate unequal number of a’s and b’s.

7. Obtain a left most derivation for the string aaabbabbba using the

following grammar.

.

11.5 Summary

In this unit, we studied the formal languages and develop mathematical

expressions, phrase structure grammar. A simple device for the construction

of useful formal languages is also studied. We also examined several useful

methods like BNF, derivation trees, to represent productions and grammars.

Some types of grammars depending on their productions were discussed.

These are useful for generating algorithms.




1. Construct the grammar which generates the following language.

i) 1)/nnbn{aL

(Hint: Φ)S,{S},VNb},{a,(VTG , where Φ is: bSaSa,S


ii) 1}m1,/nba{aL mn

(Hint: ΦS,A},{S,Vb},{a,(VG NT

where baAab,AaAb,AbAa,AbAa,SaAb,S:Φ It

is a context – free – language).

2. Give the BNF notations for the productions of the grammar.

G = (V1 = {a, +, (,)}, Vn = {S,A, B} , S, ), where

: , , ,S a A A a B B a B B a

(Hint: ::S a A

::A a B

:: |B a B a ).

3. Draw the derivation tree for the sentence x2y2z for the grammar given in

problem 2.



Ans:

4. Explain the left most and right most derivation trees.

11.7 Answers


1. L(G)={0, 110, 11110, 1111110, – – – – –}

2. }1m,0n/ZYX{)G(L mn

3. }0n/ba{}0n/a{)G(L n21n2

4. G = (VN, VT, S, ) where

VN = {S, A, B}

VT = {a, b}

The set of productions: S aB bA

A aS bAA a

B bS aBB b


6. G = (VN, VT, S, ) where

VN = {S, A, B}

VT = {a, b}

The set of productions: S A B

A a aA bAA AbA AAb

B bbBaBB BaB BBa




7.

The derivation tree is given below.

References:

1. Hopcroft, J.E. and J.D. Ulman: “Formal Language and Their relations

to Automata”, Addison-Wesley Pub. Company, Inc. Reading 1969.

2. Nelson Raymond J: “Introduction to Automata”, John Wiley & Sons,

Inc New York, 1968.

3. J.A. Bonday and U.S.R. Murthy: Graph Theory with Applications,

Elsevier Science, New York, 1979.



4. Bernard Kolman, R.C. Busby, Sharon Ross, “Discrete Mathematical

Structures” PHI, 1999.

5. Graham R. L., Knuth, D. E., and Patashnik O., Concrete Mathematics,

A Foundation for Computer Science, 2nd Ed., Addison Wesley, 1994.

6. Harary F. "Graph Theory”, Narosa Publishing House, 1995.

7. Liu.CL, Elements of Discrete Mathematics, Mc. Graw Hill.

8. Narsing Deo “Graph Theory with Applications to Engineering and

Computer Sience”, Prentice – Hall of India Pvt. Ltd., 1997.

9. Richard Johnsonbaugh, Discrete Mathematics” Pearson Education

Asia, 2001

10. Rosen K. H., Hand Book of Discrete and Combinatorial Mathematics,

CRC Press, 1999.

11. Trembly, J.P., and Manohar, R. “Discrete Mathematical Structures with

Applications to Computer Science”, Mc-Graw Hill, 1975.

12. A. Salomma: “Computations and Automata”, in Encyclopedia of

Mathematics and Its Applications. Cambridge University Press (1985).

13. Peter Linz: An introduction to Formal languages and Automata,

Narosa Publ. company (2004).

_______________________

BC0052-Theory of Computer Science

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