Unit i

Introduction toAutomata Theory
Unit – I
By Dr.T.P.Latchoumi
BY DR.T.P.LATCHOUMI

UNIT 1
• Introduction to Automaton
• Mathematical concepts
• Formal Languages: Strings, Languages, Properties
• Finite Representation : Regular Expressions
• Problems related to regular expressions
• Finite Automata :Deterministic Finite Automata
• Nondeterministic Finite Automata
• Finite Automaton with €- moves
• Problems related to Deterministic and Nondeterministic Finite
Automata
• Problems related to Finite Automaton with €- moves
• Minimization of DFA
• Problems related to Minimization of DFA
• Regular Languages : Equivalence of Finite Automata and
Regular Languages
• Equivalence of Finite Automata and Regular Grammars
• Problems related to Equivalence of Finite Automata and
Regular Languages and Regular Grammars
• Variants of Finite Automata :Two-way Finite Automaton Mealy
Machines
• Properties of Regular Languages: Closure Properties
• Set Theoretic Properties & Other Properties
• Pumping Lemma
BY DR.T.P.LATCHOUMI

What is Automata Theory?
Study of abstract computing devices, or ―machines‖
Automaton = an abstract computing device
Note: A ―device‖ need not even be a physical hardware!
A fundamental question in computer science:
Find out what different models of machines can do and cannot do
The theory of computation
Computability vs. Complexity
BY DR.T.P.LATCHOUMI

(A pioneer of automata theory)
Alan Turing (1912-1954)
Father of Modern Computer Science
English mathematician
Studied abstract machines called Turing machines even before computers existed
Heard of the Turing test?
BY DR.T.P.LATCHOUMI

Theory of Computation: A Historical
Perspective
BY DR.T.P.LATCHOUMI

Why Automata Theory?
Automata theory is important because it allows scientists to understand how
machines solve problems.
An automaton is any machine that uses a specific, repeatable process to convert
information into different forms. Modern computers are a common example of
an automaton.
BY DR.T.P.LATCHOUMI

Introduction to formal proof
The Statement ―If A then B‖ means that B is deduced from A.
In this statement A is called hypothesis and B is called conclusion
statement.
BY DR.T.P.LATCHOUMI

Types of formal proof
Deductive proof
◦ It consists of a sequence of statements given with logical reasoning to prove
the first (or initial) statement.
◦ The first statement is called hypothesis.
Inductive proof
◦ It is a recursive kind of proof which consists of sequence of parameterized
statements that cure statement itself with lower values of its parameters
BY DR.T.P.LATCHOUMI

Inductive Proof (Example)
BY DR.T.P.LATCHOUMI

Cont.….
1. Show that the base case is true
2. Assume it is true for some n=k
3. Show that this implies that it is also true for n=k+1
BY DR.T.P.LATCHOUMI

Cont.…..
BY DR.T.P.LATCHOUMI

Example 2
Prove that1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 6
For all positive integers n.
Solution: Statement P (n) is defined by
1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 2
We first show that p (1) is true.
LHS = 1 2 = 1
RHS = 1 (1 + 1) (2*1 + 1)/ 6 = 1
Both sides of the statement are equal hence p (1) is true.
BY DR.T.P.LATCHOUMI

Cont.…
2. We now assume that p (k) is true
1 2 + 2 2 + 3 2 + ... + k 2 = k (k + 1) (2k + 1)/ 6 and show
that p (k + 1) is true by adding (k + 1) 2 to both sides of the
above statement
BY DR.T.P.LATCHOUMI

Additional forms of Proofs
I. Proofs about sets
II. Proofs by contradiction
III. Proofs by counter example
BY DR.T.P.LATCHOUMI

Proofs about sets
Set is a collection of elements or items. Let us consider two sets A and B.
Define the expression R as union of A and B (i.e. AUB) and S as union of B and
A (i.e. BUA), where U is the Union operation U on two sets A and B such that if
x ϵ AUB iff x is in A or x is in B.
Prove that R=S(i.e, x is in R if and only if it is in S)
BY DR.T.P.LATCHOUMI

Proof
To prove R=S, we must prove
AUB=BUA
S.No. Statement Justification
1 x is in AUB Given
2 x is in A or x is in B By definition of union & (1)
3 x is in B or x is in A By definition of union & (2)
4 x is in BUA By definition of union & (3)
i) x is in AUB then it is in BUA
ii) x is in BUA then it is in AUB
S.No. Statement Justification
1 x is in BUA Given
2 x is in B or x is in A By definition of union & (1)
3 x is in A or x is in B By definition of union & (2)
4 x is in AUB By definition of union & (3)
BY DR.T.P.LATCHOUMI

Proofs by
contradiction
Example: Prove PUQ = QUP
Proof: Initially assume, PUQ = QUP is not true
i.e. PUQ ≠ QUP
Now consider, x is in PUQ
x is in P or x is in Q (By Definition of Union)
x is in Q or x is in P (By Definition of Union)
x is in QUP (By Definition of Union)
This contradicts our assumption PUQ ≠ QUP
Hence the assumption which we made initially is false.
Therefore PUQ = QUP is proved.
• In proof by contradiction, to
prove the statement of the form
―if A true then B true‖, We start
with statement A is not true by
assumption and try to get the
conclusion statement B.
• When it impossible to reach B
then we contradict our self and
accept that A is true.
BY DR.T.P.LATCHOUMI

Proofs by counter
Example 1: There is no pair of integers a and b such that a mod b
= b mod a
Proof: There are two possibilities. They are either a>b or a<b.
If a=2 and b=3 then 2 mod 3 ≠ 3 mod 2
If a=3 and b=2 then 3 mod 2 ≠ 2 mod 3
If a=b=2 then 2 mod 2 = 2 mod 2, so the given statement is true
only if a=b, so we have to change the statement slightly, Thus, a mod b
= b mod a true only when a=b.
We proved by counter example; this type of proof is called proof
by counter example. Such proof is true only at specific conditions.
• To prove certain statements, we
need to check all the possible
conditions in which that
statement remains true.
• There are some situations in
which the statement cannot be
true.
BY DR.T.P.LATCHOUMI

Example :2
Let n = 1, P(1) = 42(1) + 1 + 3(1) + 2= 43 + 33 = 64 + 27 = 91 = 13.t where t= 7
Let n = k, P(k) = 42(k) + 1 + 3(k) + 2= 42k + 1 + 3k + 2 = 13.t, true for some integer t
Let n = k + 1, P(k + 1) = 42(k + 1) + 1 + 3(k + 1) + 2
= 42(k + 1) + 1 + 3(k + 1) + 2
= 42k + 2 + 1 + 3k + 1 + 2
= 42k + 1 + 2 + 3k + 2 + 1
= 42k + 1.42 + 3k + 2.31
= 42.42k + 1 + 3.3k + 2
= 42.42k + 1 + 423k+2 - 423k+2 + 3.3k + 2 {Add & Sub 423k+2}
= 42(42k + 1 + 3k+2) + 3k+2 (-42+ 3)
= 42(13.t)- 3k+2 (-42+ 3) {42k + 1 + 3k + 2 = 13.t} = 16(13.t)- 3k+2 (-13)
= 13[16. t + 3k+2] = Multiple of 13
P(k + 1) is multiple of 13, Hence proved.
Prove that every integer n ≥ 0, the
number 42n + 1 + 3n + 2 is multiple of
13 by induction.
BY DR.T.P.LATCHOUMI

Formal Language
 Language is a set of valid strings from some alphabet. The set may be empty, finite, or infinite.
 L (M) is the language defined by machine M and L(G) is the language defined by context free grammar.
The two notations for specifying formal languages are:
 Grammar or regular expression (Generative approach).
 Automaton (Recognition approach).
BY DR.T.P.LATCHOUMI

Alphabet
An alphabet is a finite nonempty set of symbols. The symbol Σ is used to denote the alphabet.
1) Σ = {0, 1}
2) Σ = {a, b, c, …, z}
3) Σ = Set of ASII character
BY DR.T.P.LATCHOUMI

String
A string is a finite sequence of symbols chosen from some alphabet.
1) Σ = {0,1} 11001 [ Empty string is denoted by ε]
2) Σ = {a, b, c}  aabbcbbbaa
The total number of symbols in the string is called length of the string
Example: 1) |0001|=4
2) |100|=3
BY DR.T.P.LATCHOUMI

Empty string (ε)
Empty string is the string with zero occurrence of symbol. This empty string is denoted by ε.
Power of an alphabet
BY DR.T.P.LATCHOUMI

Power of an alphabet
If Σ is an alphabet, then the set of all string of certain length forms an alphabet. 𝑘 to be
the set of string of length k for each symbol in Σ.
Example
∑ = {0, 1}
0 ={ε}
1 = {0, 1}
2 = {00, 01, 10, 11}
3 = {000, 001, 010, 011, 100, 101, 110, 111}
+ = 1 𝑈 2 𝑈 3 …
∑+ = 𝑖
∗ = 0 𝑈 1 𝑈 2 …
∑∗ = 𝑖
Set of all strings over an alphabet is denoted by ∑∗
Set of all strings over an alphabet except null string is denoted by ∑+
BY DR.T.P.LATCHOUMI

Concatenation of strings
Let X and Y be the two strings over Σ = {a, b, c}. Then XY denotes the concatenation of X and Y.
X = aabb
Y = cccc
XY = aabbcccc
YX = ccccaabb
BY DR.T.P.LATCHOUMI

Language
It is a collection of all strings.
A set of strings of all which are chosen from ∑∗ is called a language, where ∑ is a particular alphabet set.
BY DR.T.P.LATCHOUMI

Substring, suffix and prefix
Substring: For a string w, the string x is a substring of w if the string x appears contiguously in w.
As such, for w =abcdef
we have that bcd is a substring of w,
but ace is not a substring of w.
Suffix and prefix: A string x is a suffix of w if its a substring of w appearing in the end of w.
Similarly, y is a prefix of w if y is a substring of w appearing in the beginning of w.
As such, for w =abcdef
we have that abc is a prefix of w,
and def is a suffix of w.
BY DR.T.P.LATCHOUMI

Binary relations on strings
1. Prefix - u is a prefix of v if there is a w such that v = uw.
◦ Examples:
◦  is a prefix of 0 since 0 = 0
◦ apple is a prefix of appleton since appleton = apple.ton
2. Suffix - u is a suffix of v if there is a w such that v = wu.
◦ Examples:
◦  is a suffix of 0 since 0 =0 
◦ ton is a suffix of appleton since= apple.ton
BY DR.T.P.LATCHOUMI

3. Substring - u is a substring of v if there are x and y such that v = xuy.
◦ Examples:
◦ let is a substring of appleton since appleton = app.let.on
◦ 0 is a substring of 0 since 0 = epsilon.0.epsilon
Observe that prefix and suffix are special cases
of substring.
BY DR.T.P.LATCHOUMI

Finite Automata
BY DR.T.P.LATCHOUMI

Types of finite automata (FA)
 Deterministic Finite Automata (DFA)
 Nondeterministic Finite Automata (NFA)
BY DR.T.P.LATCHOUMI

Transition
diagram
BY DR.T.P.LATCHOUMI

Transition
table
• A transition table is a conventional table
represent of a function like in FA.
• That takes two argument and return a
value.
• The rows of the table correspond to the
state and column corresponds to the input
symbols.
• Row corresponds to states of automata
and columns correspond to input symbols.
• The entry for row correspond to state q
and column corresponding to the input
symbol ‗a‘ is the state δ(q,a).
BY DR.T.P.LATCHOUMI

Deterministic Finite Automata
(DFA)
◦ Definition
◦ Finite automata are called Deterministic automata if there is only one path for its specified input from current
state to next state.
◦ A DFA can be represented by a 5-tuple (Q, ∑, δ, q0, F) where −
◦ Q is a finite set of states.
◦ ∑ is a finite set of symbols called alphabet.
◦ δ is the transition function where δ: Q × ∑ → Q
◦ q0 is the initial state from where any input is processed (q0 ∈ Q).
◦ F is a set of final state/states of Q (F ⊆ Q).
BY DR.T.P.LATCHOUMI

Graphical Representation of a
DFA
A DFA is represented by digraphs called state diagram.
 The vertices represent the states.
 The arcs labelled with an input alphabet show the transitions.
 The initial state is denoted by an empty single incoming arc.
 The final state is indicated by double circles.
BY DR.T.P.LATCHOUMI

Example 2
Design FA which accepts only those
strings which start with 0 and end
with 1 over the input set Σ = {0, 1}
BY DR.T.P.LATCHOUMI

Example 3
Design FA & State Table which
accepts odd number of 1‘s and any
number of 0‘s.
BY DR.T.P.LATCHOUMI

Nondeterministic Finite Automata
(NFA)
 FA is also called as NFA when many paths are specified in code from the current state to next state.
 In NDFA, for a particular input symbol, the machine can move to any combination of the states in the machine.
 In other words, the exact state to which the machine moves cannot be determined.
 Hence, it is called Non-deterministic Automaton.
 As it has finite number of states, the machine is called Non-deterministic Finite Machine or Non-deterministic Finite
Automaton.
BY DR.T.P.LATCHOUMI

Formal Definition of an NDFA
An NDFA can be represented by a 5-tuple (Q, ∑, δ, q0, F) where −
Q is a finite set of states.
∑ is a finite set of symbols called alphabets.
δ is the transition function where δ: Q × ∑ → 2Q (Here the power set of Q (2Q) has been taken because in case of
NDFA, from a state, transition can occur to any combination of Q states)
q0 is the initial state from where any input is processed (q0 ∈ Q).
F is a set of final state/states of Q (F ⊆ Q).
BY DR.T.P.LATCHOUMI

Graphical Representation of an
NDFA: (same as DFA)
An NDFA is represented by digraphs called state diagram.
 The vertices represent the states.
 The arcs labelled with an input alphabet show the transitions.
 The initial state is denoted by an empty single incoming arc.
 The final state is indicated by double circles.
BY DR.T.P.LATCHOUMI

Example 1
Let a non-deterministic finite automaton be
Q = {a, b, c}
∑ = {0, 1}
q0 = {a}
F = {c}
BY DR.T.P.LATCHOUMI

Example 2
Construct NFA
where L= {0101* + 0100 n ≥ 0}
over the string ∑= {0,1}
BY DR.T.P.LATCHOUMI

Differences between DFA and NFA
DFA NFA
The transition from a state is to a single next state for each input symbol.
Hence it is called deterministic.
The transition from a state can be to multiple next states for each input
symbol. Hence it is called non-deterministic.
The transition mapping for DFA is Q x ∑ —> Q. The transition mapping for NFA is Q x (∑ U {ε}) —> 2Q
Empty string transitions are not seen in DFA. NDFA permits empty string transitions.
Backtracking is allowed in DFA In NDFA, backtracking is not always possible.
Requires more space. Requires less space.
A string is accepted by a DFA if it transits to a final state. A string is accepted by a NDFA, if at least one of all possible transitions
ends in a final state.
BY DR.T.P.LATCHOUMI

FA with Epsilon transition
The ε transition in NFA is given to move from one state to another without having any
symbol from input.
The language accepted by NFA with epsilon transition is represented 5 tuples (Q, ∑, δ,
q0, F) Where, M is the NFA with ε moves.
Q is the finite set of non-empty states.
∑ is the finite set of non-empty symbols.
δ is the finite set of transitions.
q0 is the initial state.
F is the final state.
BY DR.T.P.LATCHOUMI

Epsilon closure (p)
It is set of all state which are reachable from state p or epsilon transition such that
1) ε closure (p) ={p}, Where p € Q
2) If there exist ε -closure (p) ={q} and δ (q, ε) = r, then ε closure (p) = {q, r}.
BY DR.T.P.LATCHOUMI

Example 1
Find the ε closure for the following
NFA with ε
BY DR.T.P.LATCHOUMI

Example 2
Construct NFA with ɛ which accepts a
language consisting g the strings of any
number of a‘s followed by any number of
b‘s followed by any number of c‘s.
Solution
Here any number of a‘s or b‘s or c‘s means
zero or more in number.
That means there can be zero or more a‘s
followed by zero or more b‘s followed by
more c‘s.
Hence the NFA with ɛ moves can be
Transition Table: a B c ɛ
q0 q0 Φ ϕ q1
q1 ϕ q1 ϕ q2
q2 ϕ Φ q2 ϕ
We can parse the string aabbcc as follows.
 (q0,aabbcc) -  (q0,abbcc)
 (q0,bbcc)
 (q1, bbcc)
 (q1, bcc)
 (q2, cc)
 (q2, c)
 (q2, ɛ )
Hence, we reached the accept state after scanning the complete input string.
BY DR.T.P.LATCHOUMI

Conversion from NFA with ε to
NFA without ε
1) Find all the ε transition for each state from Q that will be called as ε closure {qi} Where {qi} is the
element of Q
2) δ‘ transition can be obtained from δ transition. i.e., an ε closure of δ.
3) Step 2 is repeated for each input symbol and for each state of given NFA.
4) Using the resultant state, the transition table for equivalent NFA without ε can be built.
BY DR.T.P.LATCHOUMI

Example
Convert the given NFA with ε
transition to NFA without ε
transition
BY DR.T.P.LATCHOUMI

Theorem
Equivalence of
NFA with ε
without ε
moves
BY DR.T.P.LATCHOUMI

Conversion of
NFA to DFA
BY DR.T.P.LATCHOUMI

Theorem
Equivalence of NFA and DFA
BY DR.T.P.LATCHOUMI

Method for conversion NFA with ε to
DFA
BY DR.T.P.LATCHOUMI

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