Automata Theory

Introduction to Formal Languages and Automata Theory

Automata Defintion:
The term "Automata" is derived from the Greek word "αὐτόματα" which means "self-
acting". An automaton (Automata in plural) is an abstract self-propelled computing device
which follows a predetermined sequence of operations automatically.

An automaton with a finite number of states is called a Finite Automaton (FA) or Finite
State Machine (FSM).

Formal definition of a Finite Automaton

An automaton can be represented by a 5-tuple (Q, Σ, δ, q0, F), where:

 Q is a finite set of states.

 Σ is a finite set of symbols, called the alphabet of the automaton.
 δ is the transition function.
 q0 is the initial state from where any input is processed (q 0 ∈ Q).
 F is a set of final state/states of Q (F ⊆ Q).

Terminology

Alphabet
 Definition: An alphabet is any finite set of symbols.

 Example: Σ = {a, b, c, d} is an alphabet set where ‘a’, ‘b’, ‘c’, and ‘d’ are
symbols.

String
 Definition: A string is a finite sequence of symbols taken from Σ.

 Example: ‘cabcad’ is a valid string on the alphabet set Σ = {a, b, c, d}

Length of a String
 Definition : It is the number of symbols present in a string. (Denoted by |S|).

 Examples:

o If S=‘cabcad’, |S|= 6

o If |S|= 0, it is called an empty string (Denoted by λ or ε)

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 Automata Theory

Kleene Star
 Definition: The Kleene star, Σ*, is a unary operator on a set of symbols or strings,
Σ, that gives the infinite set of all possible strings of all possible lengths over Σ
including λ.

 Representation: Σ* = Σ0 U Σ1 U Σ2 U……. where Σp is the set of all possible strings

of length p.

 Example: If Σ = {a, b}, Σ*= {λ, a, b, aa, ab, ba, bb,………..}

Kleene Closure / Plus

 Definition: The set Σ+ is the infinite set of all possible strings of all possible lengths
over Σ excluding λ.

 Representation: Σ+ = Σ1 U Σ2 U Σ3 U…….
Σ+ = Σ* − { λ }

 Example: If Σ = { a, b } , Σ+ ={ a, b, aa, ab, ba, bb,………..}

Language
 Definition : A language is a subset of Σ* for some alphabet Σ. It can be finite or
infinite.

 Example : If the language takes all possible strings of length 2 over Σ = {a, b},
then L = { ab, bb, ba, bb}

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 ta Theory
Types of Finite Automaton

Finite Automaton can be classified into two types:

 Deterministic Finite Automaton (DFA)

 Non-deterministic Finite Automaton (NDFA / NFA)

Deterministic Finite Automaton (DFA)

In DFA, for each input symbol, one can determine the state to which the machine will
move. Hence, it is called Deterministic Automaton
the machine is called Deterministic Finite Machine or Deterministic Finite
Automaton.

Formal Definition of a DFA

A DFA can be represented by a 5-tuple (Q, Σ, δ, q0, F) where:

 Q is a finite set of states.

 Σ is a finite set of symbols called the alphabet.
 δ is the transition function where δ: Q × Σ → Q
 q0 is the initial state from where any input is processed (q 0 ∈ Q).
 F is a set of final state/states of Q (F ⊆ Q).

Graphical Representation of a DFA

A DFA is represented by digraphs called state diagram.

 The vertices represent the states.

 The arcs labeled 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.

Example
Let a deterministic finite automaton be 

 Q = {a, b, c},
 Σ = {0, 1},
 q0={a},
 F={c}, and

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 Automata Theory

Transition function δ as shown by the following table:

Present State Next State for Next State for

Input 0 Input 1
a a b
b c a
c b c

Its graphical representation would be as follows:

1 0
a b
1 0 c
1
0

DFA – Graphical Representation

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 Non-deterministic Finite Automaton ry

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.

Formal Definition of an NDFA

An NDFA can be represented by a 5-tuple (Q, Σ, δ, q0, F) where:

 Q is a finite set of states.

 Σ is a finite set of symbols called the alphabets.

 δ is the transition function where δ: Q × Σ → 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 (q 0 ∈ Q).

 F is a set of final state/states of Q (F ⊆ Q).

Graphical Representation of an NDFA: (same as DFA)

An NDFA is represented by digraphs called state diagram.

 The vertices represent the states.

 The arcs labeled 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.

Example
Let a non-deterministic finite automaton be 

 Q = {a, b, c}
 Σ = {0, 1}
 q0 = {a}
 F={c}

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 Automata Theory

The transition function  as shown below:

Present State Next State for Next State for

Input 0 Input 1
a a, b b
b c a, c
c b, c c

Its graphical representation would be as follows:

1 0
a b
0, 1 0, 1 c
0 0, 1

NDFA – Graphical Representation

DFA vs NDFA
The following table lists the differences between DFA and NDFA.

DFA NDFA

The transition from a state is to a single The transition from a state can be to
particular next state for each input multiple next states for each input symbol.
symbol. Hence it is called deterministic. Hence it is called non-deterministic.

Empty string transitions are not seen in

NDFA permits empty string transitions.
DFA.

In NDFA, backtracking is not always

Backtracking is allowed in DFA
possible.

Requires more space. Requires less space.

A string is accepted by a NDFA, if at least

A string is accepted by a DFA, if it transits
one of all possible transitions ends in a
to a final state.
final state.

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Acceptors, Classifiers, and Transducers

Acceptor (Recognizer)
An automaton that computes a Boolean function is called an acceptor. All the states of
an acceptor is either accepting or rejecting the inputs given to it.

Classifier
A classifier has more than two final states and it gives a single output when it terminates.

Transducer
An automaton that produces outputs based on current input and/or previous state is called
a transducer. Transducers can be of two types:

 Mealy Machine The output depends both on the current state and the current
input.

 Moore Machine The output depends only on the current state.

Acceptability by DFA and NDFA

A string is accepted by a DFA/NDFA iff the DFA/NDFA starting at the initial state ends in
an accepting state (any of the final states) after reading the string wholly.

A string S is accepted by a DFA/NDFA (Q, Σ, δ, q0, F), iff

δ*(q0, S) ∈ F

The language L accepted by DFA/NDFA is

{S | S ∈ Σ* and δ*(q0, S) ∈ F}

A string S′ is not accepted by a DFA/NDFA (Q, Σ, δ, q0, F), iff

δ*(q0, S′) ∉ F

The language L′ not accepted by DFA/NDFA (Complement of accepted language L) is {S

| S ∈ Σ* and δ*(q0, S) ∉ F}

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Example
Let us consider the DFA shown in Figure 1.3. From the DFA, the acceptable strings can be
derived.

0
1
a c
0
1
1
d
0

Acceptability of strings by DFA

Strings accepted by the above DFA: {0, 00, 11, 010, 101, ...........}

Strings not accepted by the above DFA: {1, 011, 111, ........}

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 NDFA to DFA Conversion Automata Theory

Problem Statement
Let X = (Qx, Σ, δx, q0, Fx) be an NDFA which accepts the language L(X). We have to
design an equivalent DFA Y = (Qy, Σ, δy, q0, Fy) such that L(Y) = L(X). The following
procedure converts the NDFA to its equivalent DFA:

Algorithm
Input: An NDFA

Output: An equivalent DFA

Step 1 Create state table from the given NDFA.

Step 2 Create a blank state table under possible input alphabets for the equivalent
DFA.

Step 3 Mark the start state of the DFA by q0 (Same as the NDFA).

Step 4 Find out the combination of States {Q 0, Q1,... , Qn} for each possible input
alphabet.

Step 5 Each time we generate a new DFA state under the input alphabet columns,
we have to apply step 4 again, otherwise go to step 6.

Step 6 The states which contain any of the final states of the NDFA are the final
states of the equivalent DFA.

Example
Let us consider the NDFA shown in the figure below.

q δ(q,0) δ(q,1)

a {a,b,c,d,e} {d,e}

b {c} {e}

c ∅ {b}

d {e} ∅

e ∅ ∅

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Using the above algorithm, we find its equivalent DFA. The state table of the DFA is shown
in below.

q δ(q,0) δ(q,1)

[a] [a,b,c,d,e] [d,e]

[a,b,c,d,e] [a,b,c,d,e] [b,d,e]

[d,e] [e] ∅

[b,d,e] [c,e] [e]

[e] ∅ ∅

[c,e] ∅ [b]

[b] [c] [e]

[c] ∅ [b]

State table of DFA equivalent to NDFA

The state diagram of the DFA is as follows:

1 0
[a,b,c,d,e] [b,d,e] [c,e]

0 1

1 [b]
[a]

1 1 1
0
0
[d,e] [e] [c]

State diagram of DFA

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 Automata Theory
Minimization of Finite State Automata

DFA Minimization using Myphill-Nerode Theorem

Algorithm
Input DFA

Output Minimized DFA

Step 1 Draw a table for all pairs of states (Qi, Qj) not necessarily connected directly
[All are unmarked initially]

Step 2 Consider every state pair (Qi, Qj) in the DFA where Qi ∈ F and Qj ∉ F or vice
versa and mark them. [Here F is the set of final states]

Step 3 Repeat this step until we cannot mark anymore states:

If there is an unmarked pair (Qi, Qj), mark it if the pair {δ(Qi, A), δ (Qi, A)}
is marked for some input alphabet.

Step 4 Combine all the unmarked pair (Qi, Qj) and make them a single state in the
reduced DFA.

Example Let us use Algorithm 2 to minimize the DFA shown below.

0, 1
1 1
b f
0
0 0 1
1

a e
1 0
0

State Diagram of DFA

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 Automata Theory

Step 1 : We draw a table for all pair of states.

a b c d e f
a
b
c
d
e
f

Step 2 : We mark the state pairs:

a b c d e f
a
b
c ✓ ✓
d ✓ ✓
e ✓ ✓
f ✓ ✓ ✓

Step 3 : We will try to mark the state pairs, with green colored check mark, transitively.
If we input 1 to state ‘a’ and ‘f’, it will go to state ‘c’ and ‘f’ respectively. (c, f) is already
marked, hence we will mark pair (a, f). Now, we input 1 to state ‘b’ and ‘f’; it will go to
state ‘d’ and ‘f’ respectively. (d, f) is already marked, hence we will mark pair (b, f).

a b c d e f
a
b
c ✓ ✓
d ✓ ✓
e ✓ ✓
f ✓ ✓ ✓ ✓ ✓

After step 3, we have got state combinations {a, b} {c, d} {c, e} {d, e} that are
unmarked.

We can recombine {c, d} {c, e} {d, e} into {c, d, e}

Hence we got two combined states as: {a, b} and {c, d, e}

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 Automata Theory

So the final minimized DFA will contain three states {f}, {a, b} and {c, d, e}

0 0, 1

1
(a, (c, d, e)
b)
1

(f
) 0, 1

State diagram of reduced DFA

DFA Minimization using Equivalence Theorem

If X and Y are two states in a DFA, we can combine these two states into {X, Y} if they
are not distinguishable. Two states are distinguishable, if there is at least one string S,
such that one of δ (X, S) and δ (Y, S) is accepting and another is not accepting. Hence, a
DFA is minimal if and only if all the states are distinguishable.

Algorithm 3
Step 1: All the states Q are divided in two partitions: final states and non-final
states and are denoted by P0. All the states in a partition are 0th equivalent.
Take a counter k and initialize it with 0.

Step 2: Increment k by 1. For each partition in Pk, divide the states in Pk into two
partitions if they are k-distinguishable. Two states within this partition X and
Y are k-distinguishable if there is an input S such that δ(X, S) and δ(Y, S)
are (k-1)-distinguishable.

Step 3: If Pk ≠ Pk-1, repeat Step 2, otherwise go to Step 4.

Step 4: Combine kth equivalent sets and make them the new states of the reduced
DFA.

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Example
Let us consider the following DFA:

0, 1
q δ(q,0) δ(q,1)
1 1
b f a b c
0
b a d
0 0 1
1 c e f

a d e f
e
1 0
0 e e f

f f f

Let us apply the above algorithm to the above DFA:

 P0 = {(c,d,e), (a,b,f)}
 P1 = {(c,d,e), (a,b),(f)}
 P2 = {(c,d,e), (a,b),(f)}

Hence, P1 = P2.

There are three states in the reduced DFA. The reduced DFA is as follows:

Q δ(q,0) δ(q,1) 0
0, 1 1
(a, b) (a, b) (c,d,e) (a, (c, d, e)
b)
(c,d,e) (c,d,e) (f)
1
(f) (f) (f) 0,1 (f
)
State Table and State Diagram of Reduced DFA

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 Moore and Mealy Machines omata Theory

Finite automata may have outputs corresponding to each transition. There are two types
of finite state machines that generate output:

 Mealy Machine
 Moore machine

Mealy Machine
A Mealy Machine is an FSM whose output depends on the present state as well as the
present input.

It can be described by a 6 tuple (Q, ∑, O, δ, X, q0) where −

 Q is a finite set of states.
 ∑ is a finite set of symbols called the input alphabet.
 O is a finite set of symbols called the output alphabet.
 δ is the input transition function where δ: Q × Σ → Q
 X is the output transition function where X: Q × Σ → O
 q0 is the initial state from where any input is processed (q0 ∈ Q).

The state table of a Mealy Machine is shown below –

Next state
Present
state input = 0 input = 1
State Output State Output
a b 𝑥1 c 𝑥1

b b 𝑥2 d 𝑥3

c d 𝑥3 c 𝑥1

d d 𝑥3 d 𝑥2

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 Automata Theory

The state diagram of the above Mealy Machine is:

0
/x2 0 /x3, 1
b /x2
0 1
/x1 /x3

0 d
a /x3
1
/x1 c 1
/x1

Moore Machine
Moore machine is an FSM whose outputs depend on only the present state.

A Moore machine can be described by a 6 tuple (Q, Σ, O, δ, X, q0) where:

 Q is a finite set of states.

 Σ is a finite set of symbols called the input alphabet.
 O is a finite set of symbols called the output alphabet.
 δ is the input transition function where δ: Q × Σ → Q
 X is the output transition function where X: Q → O
 q0 is the initial state from where any input is processed (q0 ∈ Q).

The state table of a Moore Machine is shown below –

Next State
Present State Output
Input = 0 Input = 1

a b c 𝑥2

b b d 𝑥1

c c d 𝑥2

d d d 𝑥3

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 Automata Theory

The state diagram of the above Moore Machine is:

0, 1
b/x1
0 1

d
a/x2 /x3

1 c/x2 0

Mealy Machine vs. Moore Machine

The following table highlights the points that differentiate a Mealy Machine from a Moore
Machine.

Mealy Machine Moore Machine

Output depends both upon present Output depends only upon the present
state and present input. state.

Generally, it has fewer states than Generally, it has more states than Mealy
Moore Machine. Machine.

Input change can cause change in output

Output changes at the clock edges.
change as soon as logic is done.

In Moore machines, more logic is needed to

Mealy machines react faster to inputs decode the outputs since it has more circuit
delays.

Moore Machine to Mealy Machine

Algorithm 4
Input: Moore Machine

Output: Mealy Machine

Step 1 Take a blank Mealy Machine transition table format.

Step 2 Copy all the Moore Machine transition states into this table format.

Step 3 Check the present states and their corresponding outputs in the Moore
Machine state table; if for a state Qi output is m, copy it into the output
columns of the Mealy Machine state table wherever Qi appears in the next
state.

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 Automata Theory

Example
Let us consider the following Moore machine:

Present Next State

Output
State a=0 a=1

a d b 1

b a d 0

c c c 0

d b a 1
State table of a Moore Machine

Now we apply Algorithm 4 to convert it to Mealy Machine.

Step 1 & 2:
Next State
Present State a=0 a=1
State Output State Output
a d b
b a d
c c c
d b a

The partial state table after steps 1 and 2

Step 3:
Next State
Present State a=0 a=1
State Output State Output
=> a d 1 b 0
b a 1 d 1
c c 0 c 0
d b 0 a 1

State table of an equivalent Mealy Machine

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 Automata Theory

Mealy Machine to Moore Machine

Algorithm 5:
Input: Mealy Machine

Output: Moore Machine

Step 1 Calculate the number of different outputs for each state (Qi) that are
available in the state table of the Mealy machine.

Step 2 If all the outputs of Qi are same, copy state Qi. If it has n distinct outputs,
break Qi into n states as Qin where n = 0, 1, 2.......

Step 3 If the output of the initial state is 1, insert a new initial state at the beginning
which gives 0 output.

Example
Let us consider the following Mealy Machine:

Next State

Present
a=0 a=1
State
Next Next
Output Output
State State
a d 0 b 1
b a 1 d 0
c c 1 c 0
d b 0 a 1
State table of a Mealy Machine

Here, states ‘a’ and ‘d’ give only 1 and 0 outputs respectively, so we retain states ‘a’ and
‘d’. But states ‘b’ and ‘c’ produce different outputs (1 and 0). So, we divide b into b0, b1
and c into c0, c1.

Present Next State

Output
State a=0 a=1
a d b1 1
b0 a d 0
b1 a d 1
c0 c1 c0 0
c1 c1 c0 1
d b0 a 0
State table of equivalent Moore Machine

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 Introduction to Grammars tomata Theory

In the literary sense of the term, grammars denote syntactical rules for conversation in
natural languages. Linguistics have attempted to define grammars since the inception of
natural languages like English, Sanskrit, Mandarin, etc.

The theory of formal languages finds its applicability extensively in the fields of Computer
Science. Noam Chomsky gave a mathematical model of grammar in 1956 which is
effective for writing computer languages.

Grammar
A grammar G can be formally written as a 4-tuple (N, T, S, P) where

 N or VN is a set of variables or non-terminal symbols

 T or  is a set of Terminal symbols
 S is a special variable called the Start symbol, S ∈ N
 P is Production rules for Terminals and Non-terminals. A production rule has the
form 𝛼 → 𝛽, where  and  are strings on 𝑉𝑁 ∪ Σ and least one symbol of  belongs
to VN.

Example
Grammar G1:

({S, A, B}, {a, b}, S, {S →AB, A →a, B →b})

Here,

S, A, and B are Non-terminal symbols;

a and b are Terminal symbols

S is the Start symbol, S ∈ N

Productions, P : S →AB, A →a, B →b

Example:
Grammar G2:

({S, A}, {a, b}, S,{S → aAb, aA →aaAb, A→ε } )

Here,

S and A are Non-terminal symbols.

a and b are Terminal symbols.

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 Automata Theory

ε is an empty string.

S is the Start symbol, S ∈ N

Production P : S → aAb, aA →aaAb, A→ε

Derivations from a Grammar

Strings may be derived from other strings using the productions in a grammar. If a
grammar G has a production α  β, we can say that x α y derives x β y in G. This
derivation is written as:
𝑮
𝒙𝜶𝒚 ⇒ 𝒙𝜷𝒚

Example
Let us consider the grammar:

G2 = ({S, A}, {a, b}, S, {S → aAb, aA →aaAb, A→ε } )

Some of the strings that can be derived are:

S  aAb using production S  aAb

 aaAbb using production aA  aAb

 aaaAbbb using production aA  aAb

 aaabbb using production A  ε

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 Language Generated by a Grammar ry

The set of all strings that can be derived from a grammar is said to be the language
generated from that grammar. A language generated by a grammar G is a subset formally
defined by
𝐺
𝐿(𝐺) = { 𝑊 | 𝑊 ∈ Σ∗ , 𝑆 ⇒ 𝑊}
If L(G1) = L(G2), the Grammar G1 is equivalent to the Grammar G2.

Example
If there is a grammar

G: N = {S, A, B} T = {a, b} P = {S →AB, A →a, B →b}

Here S produces AB, and we can replace A by a, and B by b. Here, the only accepted
string is ab, i.e.,

L(G) = {ab}

Example
Suppose we have the following grammar:

G: N={S, A, B} T= {a, b} P= {S →AB, A →aA|a, B →bB|b}

The language generated by this grammar:

L(G) = {ab, a2b, ab2, a2b2, ………}

= { am bn | m ≥ 0 and n ≥ 0}

Construction of a Grammar Generating a Language

We’ll consider some languages and convert it into a grammar G which produces those
languages.

Example
Problem Suppose, L (G) = {am bn | m ≥ 0 and n > 0}. We have to find out the
grammar G which produces L(G).

Solution

Since L(G) = {am bn | m ≥ 0 and n > 0}

the set of strings accepted can be rewritten as:

L(G) = {b, ab,bb, aab, abb, …….}

Here, the start symbol has to take at least one ‘b’ preceded by any number of ‘a’ including
null.

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 Automata Theory

To accept the string set {b, ab, bb, aab, abb, …….}, we have taken the productions:

S →aS , S →B, B → b and B → bB

S →B→ b (Accepted)

S →B→ bB → bb (Accepted)

S →aS →aB→ab (Accepted)

S →aS →aaS →aaB → aab(Accepted)

S →aS →aB→abB→ abb (Accepted)

Thus, we can prove every single string in L(G) is accepted by the language generated by
the production set.

Hence the grammar:

G: ({S, A, B}, {a, b}, S, { S →aS | B , B → b | bB })

Example
Problem: Suppose, L (G) = {am bn | m> 0 and n ≥ 0}. We have to find out the
grammar G which produces L(G).

Solution:

Since L(G) = {am bn | m> 0 and n ≥ 0}, the set of strings accepted can be rewritten as:

L(G) = {a, aa, ab, aaa, aab ,abb, …….}

Here, the start symbol has to take at least one ‘a’ followed by any number of ‘b’ including
null.

To accept the string set {a, aa, ab, aaa, aab, abb, …….}, we have taken the productions:

S → aA, A → aA , A → B, B → bB ,B → λ

S → aA → aB→ aλ→a (Accepted)

S → aA → aaA→ aaB → aaλ→aa (Accepted)

S →aA →aB→abB→ abλ → ab (Accepted)

S → aA → aaA→ aaaA→aaaB → aaaλ→aaa (Accepted)

S → aA → aaA→ aaB→aabB → aabλ→aab (Accepted)

S → aA → aB→ abB→abbB → abbλ→abb (Accepted)

Thus, we can prove every single string in L(G) is accepted by the language generated by
the production set.

Hence the grammar:

G: ({S, A, B}, {a, b}, S, {S → aA, A → aA | B, B → λ | bB })

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 ry
Chomsky Classification of Grammars

According to Noam Chomosky, there are four types of grammars: Type 0, Type 1, Type 2,
and Type 3. The following table shows how they differ from each other:

Grammar Grammar Language

Automaton
Type Accepted Accepted

Recursively enumerable
Type 0 Unrestricted grammar Turing machine
language

Context-sensitive Linear-bounded
Type 1 Context-sensitive grammar
language automaton

Pushdown
Type 2 Context-free grammar Context-free language
automaton

Finite state
Type 3 Regular grammar Regular language
automaton

Take a look at the following illustration. It shows the scope of each type of grammar:

Recursively Enumerable

Context-Sensitive

Context - Free

Regular

Containment of Type 3 ⊆ Type 2 ⊆ Type 1 ⊆ Type 0

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 Automata Theory

Type - 3 Grammar
Type-3 grammars generate regular languages. Type-3 grammars must have a single
non-terminal on the left-hand side and a right-hand side consisting of a single terminal or
single terminal followed by a single non-terminal.

The productions must be in the form X → a or X → aY

where X, Y ∈ N (Non terminal)

and a ∈ T (Terminal)

The rule S → ε is allowed if S does not appear on the right side of any rule.

Example
X → ε
X → a | aY
Y →b

Type - 2 Grammar
Type-2 grammars generate context-free languages.

The productions must be in the form A → γ

where A ∈ N (Non terminal)

and γ ∈ (T∪N)* (String of terminals and non-terminals).

These languages generated by these grammars are be recognized by a non-deterministic

pushdown automaton.

Example
S → X a
X → a
X → aX
X → abc
X → ε

26
 Automata Theory

Type - 1 Grammar
Type-1 grammars generate context-sensitive languages. The productions must be in the
form

αAβ→αγβ

where A ∈ N (Non-terminal)

and α, β, γ ∈ (T ∪ N)* (Strings of terminals and non-terminals)

The strings α and β may be empty, but γ must be non-empty.

The rule S → ε is allowed if S does not appear on the right side of any rule. The languages
generated by these grammars are recognized by a linear bounded automaton.

Example
AB → AbBc
A → bcA
B → b

Type - 0 Grammar
Type-0 grammars generate recursively enumerable languages. The productions have no
restrictions. They are any phase structure grammar including all formal grammars.

They generate the languages that are recognized by a Turing machine.

The productions can be in the form of α→ β where α is a string of terminals and non-
terminals with at least one non-terminal and α cannot be null. β is a string of terminals
and non-terminals.

Example
S → ACaB
Bc → acB
CB → DB
aD → Db

27
 Automata Theory

Regular Grammar

28
 Regular Expressions Automata Theory

A Regular Expression can be recursively defined as follows:

1. ε is a Regular Expression indicates the language containing an empty string. (L (ε)

= {ε})

2. φ is a Regular Expression denoting an empty language. (L (φ) = { })

3. x is a Regular Expression where L={x}

4. If X is a Regular Expression denoting the language L(X) and Y is a Regular

Expression denoting the language L(Y), then

(a) X + Y is a Regular Expression corresponding to the language L(X) U L(Y)

where L(X+Y) = L(X) U L(Y).

(b) X . Y is a Regular Expression corresponding to the language L(X) . L(Y)

where L(X.Y)= L(X) . L(Y)

(c) R* is a Regular Expression corresponding to the language L(R*) where

L(R*) = (L(R))*
5. If we apply any of the rules several times from 1 to 5, they are Regular Expressions.

Some RE Examples
Regular
Regular Set
Expression

(0+10*) L= { 0, 1, 10, 100, 1000, 10000, … }

(0*10*) L={1, 01, 10, 010, 0010, …}

(0+ε)(1+ ε) L= {ε, 0, 1, 01}

Set of strings of a’s and b’s of any length including the null string.
(a+b)*
So L= { ε, a, b, aa , ab , bb , ba, aaa…….}

(a+b)*abb Set of strings of a’s and b’s ending with the string abb.
So L = {abb, aabb, babb, aaabb, ababb, …………..}

(11)* Set consisting of even number of 1’s including empty string,

29
 Automata Theory

So L= {ε, 11, 1111, 111111, ……….}

(aa)*(bb)*b Set of strings consisting of even number of a’s followed by odd

number of b’s , so L= {b, aab, aabbb, aabbbbb, aaaab, aaaabbb,
…………..}

(aa + ab + ba + String of a’s and b’s of even length can be obtained by

bb)* concatenating any combination of the strings aa, ab, ba and bb
including null, so L= {aa, ab, ba, bb, aaab, aaba, …………..}

30
 Regular Sets Automata Theory

Any set that represents the value of the Regular Expression is called a Regular Set.

Properties of Regular Sets

Property 1. The union of two regular set is regular.

Proof:

Let us take two regular expressions

RE1 = a(aa)* and RE2 = (aa)*

So, L1= {a, aaa, aaaaa,.....} (Strings of odd length excluding Null)

and L2={ ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null)

L1 ∪ L2 = { ε, a, aa, aaa, aaaa, aaaaa, aaaaaa,.......}

(Strings of all possible lengths including Null)

RE (L1 ∪ L2) = a* (which is a regular expression itself)

Hence, proved.

Property 2. The intersection of two regular set is regular.

Proof:

Let us take two regular expressions

RE1 = a(a*) and RE2 = (aa)*

So, L1 = { a,aa, aaa, aaaa, ....} (Strings of all possible lengths excluding Null)

L2 ={ ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null)

L1 ∩ L2 = { aa, aaaa, aaaaaa,.......} (Strings of even length excluding Null)

RE (L1 ∩ L2) = aa(aa)* which is a regular expression itself.

Hence, proved.

Property 3. The complement of a regular set is regular.

Proof:

Let us take a regular expression:

RE = (aa)*

31
 Automata Theory

So, L = {ε, aa, aaaa, aaaaaa, .......} (Strings of even length including Null)

Complement of L is all the strings that is not in L.

So, L’ = {a, aaa, aaaaa, .....} (Strings of odd length excluding Null)

RE (L’) = a(aa)* which is a regular expression itself.

Hence, proved.

Property 4. The difference of two regular set is regular.

Proof:

Let us take two regular expressions:

RE1 = a (a*) and RE2 = (aa)*

So, L1= {a,aa, aaa, aaaa, ....} (Strings of all possible lengths excluding Null)

L2 = { ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null)

L1 – L2 = {a, aaa, aaaaa, aaaaaaa, ....}

(Strings of all odd lengths excluding Null)

RE (L1 – L2) = a (aa)* which is a regular expression.

Hence, proved.

Property 5. The reversal of a regular set is regular.

Proof:

We have to prove LR is also regular if L is a regular set.

Let, L= {01, 10, 11, 10}

RE (L)= 01 + 10 + 11 + 10

LR= {10, 01, 11, 01}

RE (LR)= 01+ 10+ 11+10 which is regular

Hence, proved.

32
 Automata Theory

Property 6. The closure of a regular set is regular.

Proof:

If L = {a, aaa, aaaaa, .......} (Strings of odd length excluding Null)

i.e., RE (L) = a (aa)*

L*= {a, aa, aaa, aaaa , aaaaa,……………} (Strings of all lengths excluding Null)

RE (L*) = a (a)*

Hence, proved.

Property 7. The concatenation of two regular sets is regular.

Proof:

Let RE1 = (0+1)*0 and RE2 = 01(0+1)*

Here, L1 = {0, 00, 10, 000, 010, ......} (Set of strings ending in 0)

and L2 = {01, 010,011,.....} (Set of strings beginning with 01)

Then, L1 L2 = {001,0010,0011,0001,00010,00011,1001,10010,.............}

Set of strings containing 001 as a substring which can be represented by an RE:

(0+1)*001(0+1)*

Hence, proved.

Identities Related to Regular Expressions

Given R, P, L, Q as regular expressions, the following identities hold:

1. Ø* = ε

2. ε* = ε

3. RR* = R*R

4. R*R* = R*

5. (R*)* = R*

6. RR* = R*R

7. (PQ)*P =P(QP)*

8. (a+b)* = (a*b*)* = (a*+b*)* = (a+b*)* = a*(ba*)*

9. R + Ø = Ø + R = R (The identity for union)

10. Rε = εR = R (The identity for concatenation)

33
 Automata Theory

11. ØL = LØ = Ø (The annihilator for concatenation)

12. R + R = R (Idempotent law)

13. L (M + N) = LM + LN (Left distributive law)

14. (M + N) L = LM + LN (Right distributive law)

15. ε + RR* = ε + R*R = R*

34
 Automata Theory

In order to find out a regular expression of a Finite Automaton, we use Arden’s Theorem
along with the properties of regular expressions.

Statement:

Let P and Q be two regular expressions.

If P does not contain null string, then R = Q + RP has a unique solution that is R
= QP*

Proof:

R = Q + (Q + RP)P [After putting the value R = Q + RP]

= Q + QP + RPP

When we put the value of R recursively again and again, we get the following equation:

R = Q + QP + QP2 + QP3…..

R = Q (є + P + P2 + P3 + …. )

R = QP* [As P* represents (є + P + P2 + P3 + ….) ]

Hence, proved.

Assumptions for Applying Arden’s Theorem:

1. The transition diagram must not have NULL transitions
2. It must have only one initial state

Method
Step 1: Create equations as the following form for all the states of the DFA having
n states with initial state q1.

q1 = q1R11 + q2R21 + … + qnRn1 + є

q2 = q1R12 + q2R22 + … + qnRn2
..…………………………
……………………………
……………………………

……………………………
qn = q1R1n + q2R2n + … + qnRnn

35
 Automata Theory

Rij represents the set of labels of edges from qi to qj, if no such edge exists, then Rij = Ø

Step 2: Solve these equations to get the equation for the final state in terms of Rij

Problem
Construct a regular expression corresponding to the automata given below:

b a
q2
b q3
b a

q1
a

Finite automata

Solution

Here the initial state is q2 and the final state is q1.

The equations for the three states q1, q2, and q3 are as follows:

q1 = q1a + q3a + є (є move is because q1 is the initial state0

q2 = q1b + q2b + q3b
q3 = q2a

Now, we will solve these three equations:

q2 = q1b + q2b + q3b

= q1b + q2b + (q2a)b (Substituting value of q3)

= q1b + q2(b + ab)

= q1b (b + ab)* (Applying Arden’s Theorem)

q1 = q1a + q3a + є

= q1a + q2aa + є (Substituting value of q3)

= q1a + q1b(b + ab*)aa + є (Substituting value of q2)
= q1(a + b(b + ab)*aa) + є

36
 Automata Theory

= є (a+ b(b + ab)*aa)*

= (a + b(b + ab)*aa)*

Hence, the regular expression is (a + b(b + ab)*aa)*.

Problem
Construct a regular expression corresponding to the automata given below:

0 0, 1

q1 q3
1 1

q
0 2

Finite automata

Solution:

Here the initial state is q1 and the final state is q2

Now we write down the equations:

q1 = q10 + є
q2 = q11 + q20
q3 = q21 + q30 + q31

Now, we will solve these three equations:

q1 = є0* [As, εR = R]

So, q1 = 0*

q2 = 0*1 + q20
So, q2 = 0*1(0)* [By Arden’s theorem]

Hence, the regular expression is 0*10*.

37
 Construction of an FA from a RE tomata Theory

We can use Thompson's Construction to find out a Finite Automaton from a Regular
Expression. We will reduce the regular expression into smallest regular expressions and
converting these to NFA and finally to DFA.

Some basic RA expressions are the following:

Case 1: For a regular expression ‘a’, we can construct the following FA:

q1 q
a

Finite automata for RE = a

Case 2: For a regular expression ‘ab’, we can construct the following FA:

q1 q1 q
a b

Finite automata for RE = ab

Case 3: For a regular expression (a+b), we can construct the following FA:

q1 q
a

Finite automata for RE= (a+b)

38
 Automata Theory

Case 4: For a regular expression (a+b)*, we can construct the following FA:

a,b

Finite automata for RE= (a+b)*

Method:
Step 1 Construct an NFA with Null moves from the given regular expression.

Step 2 Remove Null transition from the NFA and convert it into its equivalent DFA.

Problem
Convert the following RA into its equivalent DFA: 1 (0 + 1)* 0

Solution:

We will concatenate three expressions "1", "(0 + 1)*" and "0"

0, 1

q0 q1 q2 q3 q
1 є є 0

NDFA with NULL transition for RA: 1 (0 + 1)* 0

Now we will remove the є transitions. After we remove the є transitions from the NDFA,
we get the following:

0, 1

q0 q2 q
1 0

NDFA without NULL transition for RA: 1 (0 + 1)* 0

It is an NDFA corresponding to the RE: 1 (0 + 1)* 0. If you want to convert it into a DFA,
simply apply the method of converting NDFA to DFA discussed in Chapter 1.

39
 Automata Theory

Finite Automata with Null Moves (NFA-ε)

A Finite Automaton with null moves (FA-ε) does transit not only after giving input from
the alphabet set but also without any input symbol. This transition without input is called
a null move.

An NFA-ε is represented formally by a 5-tuple (Q, Σ, δ, q0, F), consisting of

 Q : a finite set of states

 Σ : a finite set of input symbols
 δ : a transition function δ : Q × (Σ ∪ {ε}) → 2Q
 q0 : an initial state q0 ∈ Q

 F: a set of final state/states of Q (F⊆Q).

a b
ε ε

0 1

Finite automata with Null Moves

The above (FA-ε) accepts a string set: {0, 1, 01}

Removal of Null Moves from Finite Automata

If in an NDFA, there is ϵ-move between vertex X to vertex Y, we can remove it using the
following steps:

1. Find all the outgoing edges from Y.

2. Copy all these edges starting from X without changing the edge labels.
3. If X is an initial state, make Y also an initial state.
4. If Y is a final state, make X also a final state.

40
 Automata Theory

Problem
Convert the following NFA-ε to NFA without Null move.

0
ε 0, 1
q q
f

1
0
q
1

Finite automata with Null Moves

Solution

Step 1:

Here the ε transition is between q1 and q2, so let q1 is X and qf is Y.

Here the outgoing edges from qf is to qf for inputs 0 and 1.

Step 2:

Now we will Copy all these edges from q1 without changing the edges from qf and
get the following FA:

0
0, 1
0, 1
q q

1
0
q
1

NDFA after step 2

41
 Automata Theory

Step 3:

Here q1 is an initial state, so we make qf also an initial state.

So the FA becomes -

0
0, 1
q q 0, 1

1
0
q
1

NDFA after Step 3

Step 4:

Here qf is a final state, so we make q1 also a final state.

So the FA becomes -

0
0, 1
q q 0, 1

1
0
q
1

Final NDFA without NULL moves

42
 Pumping Lemma for Regular Languages ry

Theorem
Let L be a regular language. Then there exists a constant ‘c’ such that for every string
w in L:

|w| ≥ c

We can break w into three strings, w = xyz, such that:

1. |y| > 0

2. |xy| ≤ c

3. For all k ≥ 0, the string xykz is also in L.

Applications of Pumping Lemma

Pumping Lemma is to be applied to show that certain languages are not regular. It should
never be used to show a language is regular.

1. If L is regular, it satisfies Pumping Lemma.

2. If L does not satisfy Pumping Lemma, it is non-regular.

Method to prove that a language L is not regular:

1. At first, we have to assume that L is regular.
2. So, the pumping lemma should hold for L.
3. Use the pumping lemma to obtain a contradiction:
(a) Select w such that |w| ≥ c
(b) Select y such that |y| ≥ 1
(c) Select x such that |xy| ≤ c
(d) Assign the remaining string to z.
(e) Select k such that the resulting string is not in L.

Hence L is not regular.

43
 Automata Theory

Problem
Prove that L = {aibi | i ≥ 0} is not regular.

Solution:

1. At first, we assume that L is regular and n is the number of states.

2. Let w = anbn. Thus |w| = 2n ≥ n.

3. By pumping lemma, let w = xyz, where |xy|≤ n.

4. Let x = ap, y = aq, and z = arbn, where p + q + r = n, p ≠ 0, q ≠ 0, r ≠ 0. Thus

|y|≠ 0.

5. Let k = 2. Then xy2z = apa2qarbn.

6. Number of as = (p + 2q + r) = (p + q + r) + q = n + q

7. Hence, xy2z = an+q bn. Since q ≠ 0, xy2z is not of the form anbn.

8. Thus, xy2z is not in L. Hence L is not regular.

44
 DFA Complement Automata Theory

If (Q, Σ, δ, q0, F) be a DFA that accepts a language L, then the complement of the DFA
can be obtained by swapping its accepting states with its non-accepting states and vice
versa.

We will take an example and elaborate this below:

a, b
b
X Y
a
b
Z
a

DFA accepting language L

This DFA accepts the language

L = {a, aa, aaa , ............. }

over the alphabet

Σ = {a, b}

So, RE = a+.

Now we will swap its accepting states with its non-accepting states and vice versa and will
get the following:

a, b
b
X Y
a
b
Z
a

DFA accepting complement of language L

45
 Automata Theory

This DFA accepts the language

Ľ = {ε, b, ab ,bb,ba, ............... }

over the alphabet

Σ = {a, b}

Note: If we want to complement an NFA, we have to first convert it to DFA and then have
to swap states as in the previous method.

46
 Automata Theory

Context-Free Grammars

47
 Context-Free Grammars ry

Definition: A context-free grammar (CFG) consisting of a finite set of grammar rules is

a quadruple (N, T, P, S) where

 N is a set of non-terminal symbols.

 T is a set of terminals where N ∩ T = NULL.

 P is a set of rules, P: N → (N U T)*, i.e., the left-hand side of the production rule
P does have any right context or left context.

 S is the start symbol.

Example
 The grammar ({A}, {a, b, c}, P, A), P : A → aA, A → abc.
 The grammar ({S, a, b}, {a, b}, P, S), P: S → aSa, S → bSb, S → ε
 The grammar ({S, F}, {0, 1}, P, S), P: S → 00S | 11F, F → 00F | ε

Generation of Derivation Tree

A derivation tree or parse tree is an ordered rooted tree that graphically represents the
semantic information a string derived from a context-free grammar.

Representation Technique:
1. Root vertex: Must be labeled by the start symbol.
2. Vertex: Labeled by a non-terminal symbol.
3. Leaves: Labeled by a terminal symbol or ε.

If S → x1x2 …… xn is a production rule in a CFG, then the parse tree / derivation tree will
be as follows:

x1 x2 xn

48
 Automata Theory

There are two different approaches to draw a derivation tree:

1. Top-down Approach:
(a) Starts with the starting symbol S
(b) Goes down to tree leaves using productions
2. Bottom-up Approach:
(a) Starts from tree leaves
(b) Proceeds upward to the root which is the starting symbol S

Derivation or Yield of a Tree

The derivation or the yield of a parse tree is the final string obtained by concatenating the
labels of the leaves of the tree from left to right, ignoring the Nulls. However, if all the
leaves are Null, derivation is Null.

Example
Let a CFG {N,T,P,S} be

N = {S}, T = {a, b}, Starting symbol = S, P = S → SS | aSb | ε

One derivation from the above CFG is “abaabb”

S → SS → aSbS →abS → abaSb → abaaSbb → abaabb

S S

a S b a S b

ε a S b

49
 Automata Theory

Sentential Form and Partial Derivation Tree

A partial derivation tree is a sub-tree of a derivation tree/parse tree such that either all of
its children are in the sub-tree or none of them are in the sub-tree.

Example
If in any CFG the productions are:

S → AB, A → aaA | ε, B →Bb| ε

the partial derivation tree can be the following:

A B

If a partial derivation tree contains the root S, it is called a sentential form. The above
sub-tree is also in sentential form.

Leftmost and Rightmost Derivation of a String

 Leftmost derivation - A leftmost derivation is obtained by applying production to
the leftmost variable in each step.

 Rightmost derivation - A rightmost derivation is obtained by applying production

to the rightmost variable in each step.

Example
Let any set of production rules in a CFG be

X → X+X | X*X |X| a

over an alphabet {a}.

The leftmost derivation for the string "a+a*a" may be:

X → X+X→ a+X→ a+ X*X →a+a*X→ a+a*a

The stepwise derivation of the above string is shown as below:

50
 Automata Theory

Step 1:
Step 2:
X X

X + X X + X

a
Step 3:

X
Step 4:
X
X + X

X + X
X * X
a
X * X
a
Step 5:
X a

X + X

X * X
a

a a

The rightmost derivation for the above string "a+a*a" may be:

X → X*X→ X*a → X+X*a →X+a*a→ a+a*a

51
 Automata Theory

The stepwise derivation of the above string is shown as below:

Step 1:
Step 2:
X X

X * X X * X

X Step 4:
Step 3:
X
X * X

X * X
X + X
a
X + X
a

Step 5: a
X

X * X

X + X
a

a a

Left and Right Recursive Grammars

In a context-free grammar G, if there is a production in the form X → Xa where X is a
non-terminal and ‘a’ is a string of terminals, it is called a left recursive production. The
grammar having a left recursive production is called a left recursive grammar.

And if in a context-free grammar G, if there is a production is in the form X →

aX where X is a non-terminal and ‘a’ is a string of terminals, it is called a right recursive
production. The grammar having a right recursive production is called a right recursive
grammar.

52
 Ambiguity in Context-Free Grammars ry

If a context free grammar G has more than one derivation tree for some string w ∈ L(G),
it is called an ambiguous grammar. There exist multiple right-most or left-most
derivations for some string generated from that grammar.

Problem
Check whether the grammar G with production rules:

X → X+X | X*X |X| a

is ambiguous or not.

Solution

Let’s find out the derivation tree for the string "a+a*a". It has two leftmost derivations.

Derivation 1: X → X+X→ a +X→ a+ X*X →a+a*X→ a+a*a

Parse tree 1:

X + X

X * X
a

a a

53
 Automata Theory

Derivation 2: X → X*X→X+X*X→ a+ X*X →a+a*X→ a+a*a

Parse tree 2:

X * X

X + X
a

a a

Since there are two parse trees for a single string "a+a*a", the grammar G is ambiguous.

54
 Automata Theory
CFL Closure Property

Context-free languages are closed under:

 Union
 Concatenation
 Kleene Star operation

Union
Let L1 and L2 be two context free languages. Then L1  L2 is also context free.

Example:

Let L1 = { anbn , n>0}. Corresponding grammar G1 will have P: S1  aAb|ab

Let L2 = { cmdm , n≥0}. Corresponding grammar G2 will have P: S2  cBb| ε

Union of L1 and L2, L = L1  L2 = { anbn }  { cmdm }

The corresponding grammar G will have the additional production S  S1 | S2

Concatenation
If L1 and L2 are context free languages, then L1L2 is also context free.

Example:

Union of the languages L1 and L2, L = L1L2 = { anbncmdm }

The corresponding grammar G will have the additional production S  S1 S2

Kleene Star
If L is a context free language, then L* is also context free.

Example:

Let L = { anbn , n≥0}. Corresponding grammar G will have P: S  aAb| ε

Kleene Star L1 = { anbn }*

The corresponding grammar G1 will have additional productions S1  SS1 | ε

Context-free languages are not closed under:

 Intersection : If L1 and L2 are context free languages, then L1  L2 is not

necessarily context free.

 Intersection with Regular Language : If L1 is a regular language and L2 is a

context free language, then L1  L2 is a context free language.

 Complement : If L1 is a context free language, then L1’ may not be context free.

55
 Automata Theory
CFG Simplification

In a CFG, it may happen that all the production rules and symbols are not needed for the
derivation of strings. Besides, there may be some null productions and unit productions.
Elimination of these productions and symbols is called simplification of CFGs.
Simplification essentially comprises of the following steps:

 Reduction of CFG
 Removal of Unit Productions
 Removal of Null Productions

Reduction of CFG
CFGs are reduced in two phases:

Phase 1: Derivation of an equivalent grammar, G’, from the CFG, G, such that each
variable derives some terminal string.

Derivation Procedure:

Step 1: Include all symbols, W1, that derive some terminal and initialize i=1.

Step 2: Include all symbols, Wi+1, that derive Wi.

Step 3: Increment i and repeat Step 2, until Wi+1 = Wi.

Step 4: Include all production rules that have Wi in it.

Phase 2: Derivation of an equivalent grammar, G”, from the CFG, G’, such that each
symbol appears in a sentential form.

Derivation Procedure:

Step 1: Include the start symbol in Y1 and initialize i = 1.

Step 2: Include all symbols, Yi+1, that can be derived from Yi and include all
production rules that have been applied.

Step 3: Increment i and repeat Step 2, until Yi+1 = Yi.

56
 Automata Theory

Problem
Find a reduced grammar equivalent to the grammar G, having production rules, P: S 
AC | B, A  a, C  c | BC, E  aA | e

Solution

Phase 1:

T = { a, c, e }

W1 = { A, C, E } from rules A  a, C  c and E  aA

W2 = { A, C, E } U { S } from rule S  AC

W3 = { A, C, E, S } U 

Since W2 = W3, we can derive G’ as:

G’ = { { A, C, E, S }, { a, c, e }, P, {S}}

where P: S  AC, A  a, C  c , E  aA | e

Phase 2:

Y1 = { S }

Y2 = { S, A, C } from rule S  AC

Y3 = { S, A, C, a, c } from rules A  a and C  c

Y4 = { S, A, C, a, c }

Since Y3 = Y4, we can derive G” as:

G” = { { A, C, S }, { a, c }, P, {S}}

where P: S  AC, A  a, C  c

Removal of Unit productions

Any production rule in the form A → B where A, B ∈ Non-terminal is called unit
production.

Removal Procedure:

Step 1: To remove A→B, add production A→x to the grammar rule whenever B→x
occurs in the grammar. [x ∈ Terminal, x can be Null]

Step 2: Delete A→B from the grammar.

Step 3: Repeat from step 1 until all unit productions are removed.

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 Automata Theory

Problem
Remove unit production from the following:

S → XY, X → a, Y → Z | b, Z → M, M → N, N → a

Solution:

There are 3 unit productions in the grammar:

Y → Z, Z → M, and M→N

At first, we will remove M → N.

As N → a, we add M → a, and M → N is removed.

The production set becomes

S → XY, X → a, Y → Z | b, Z → M, M → a, N → a

Now we will remove Z → M.

As M → a, we add Z→ a, and Z → M is removed.

The production set becomes

S → XY, X → a, Y → Z | b, Z → a, M → a, N → a

Now we will remove Y → Z.

As Z → a, we add Y→ a, and Y → Z is removed.

The production set becomes

S → XY, X → a, Y → a | b, Z → a, M → a, N → a

Now Z, M, and N are unreachable, hence we can remove those.

The final CFG is unit production free:

S → XY, X → a, Y→a|b

Removal of Null Productions

In a CFG, a non-terminal symbol ‘A’ is a nullable variable if there is a production A → ϵ or
there is a derivation that starts at A and finally ends up with

ϵ: A → .......… → ϵ

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 Automata Theory

Removal Procedure:
Step1 Find out nullable non-terminal variables which derive ϵ.

Step2 For each production A → a, construct all productions A → x where x is obtained

from ‘a’ by removing one or multiple non-terminals from Step 1.

Step3 Combine the original productions with the result of step 2 and remove ϵ-
productions.

Problem
Remove null production from the following:

S→ASA | aB | b, A → B, B → b | ϵ

Solution:

There are two nullable variables: A and B

At first, we will remove B → ϵ.

After removing B → ϵ, the production set becomes:

S→ASA | aB | b | a, A → B| b | ϵ, B → b

Now we will remove A → ϵ.

After removing A → ϵ, the production set becomes:

S→ASA | aB | b | a | SA | AS | S, A → B| b, B→b

This is the final production set without null transition.

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 Automata Theory
Chomsky Normal Form

A CFG is in Chomsky Normal Form if the Productions are in the following forms:

 A→a
 A → BC
 S→ϵ

where A, B, and C are non-terminals and a is terminal.

Algorithm to Convert into Chomsky Normal Form:

Step 1 If the start symbol S occurs on some right side, create a new start symbol S’
and a new production S’ → S.

Step 2 Remove Null productions. (Using the Null production removal algorithm
discussed earlier)

Step 3 Remove unit productions. (Using the Unit production removal algorithm
discussed earlier)

Step 4 Replace each production A → B1…Bn where n > 2 with A → B1C where C → B2
…Bn. Repeat this step for all productions having two or more symbols in the
right side.

Step 5 If the right side of any production is in the form A→ aB where a is a terminal
and A, B are non-terminal, then the production is replaced by A→ XB and X→
a. Repeat this step for every production which is in the form A→ aB.

Problem:
Convert the following CFG into CNF

S→ ASA | aB, A → B | S, B→b|ϵ

Solution:

(1) Since S appears in R.H.S, we add a new state S0 and S0→S is added to the production
set and it becomes:

S0→S, S→ ASA | aB, A → B | S, B→b|ϵ

(2) Now we will remove the null productions:

B→ϵ and A→ϵ

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 Automata Theory

After removing B → ϵ, the production set becomes:

S0→S, S→ ASA | aB | a, A → B | S | ϵ, B → b

After removing A → ϵ, the production set becomes:

S0→S, S→ ASA | aB | a | AS | SA | S, A → B | S, B → b

(3) Now we will remove the unit productions.

After removing S→ S, the production set becomes:

S0→S, S→ ASA | aB | a | AS | SA, A → B | S, B → b

After removing S0→ S, the production set becomes:

S0→ ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA

A → B | S, B→b

After removing A→ B, the production set becomes:

S0 → ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA

A→S|b

B→b

After removing A→ S, the production set becomes:

S0 → ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA

A → b |ASA | aB | a | AS | SA, B → b

(4) Now we will find out more than two variables in the R.H.S

Here, S0→ ASA, S → ASA, A→ ASA violates two Non-terminals in R.H.S.

Hence we will apply step 4 and step 5 to get the following final production set which is in
CNF:

S0→ AX | aB | a | AS | SA

S→ AX | aB | a | AS | SA

A → b |AX | aB | a | AS | SA

B→b

X→ SA

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 Automata Theory

(5) We have to change the productions S0→ aB, S→ aB, A→ aB

And the final production set becomes:

S0→ AX | YB | a | AS | SA

S→ AX | YB | a | AS | SA

A → b |AX | YB | a | AS | SA

B→b

X→ SA

Y→a

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 Automata Theory
Greibach Normal Form

A CFG is in Greibach Normal Form if the Productions are in the following forms:

A→b

A → bD1…Dn

S→ϵ

where A, D1,....,Dn are non-terminals and b is a terminal.

Algorithm to Convert a CFG into Greibach Normal Form:

Step 1 If the start symbol S occurs on some right side, create a new start symbol
S’ and a new production S’ → S.

Step 2 Remove Null productions. (Using the Null production removal algorithm
discussed earlier)

Step 3 Remove unit productions. (Using the Unit production removal algorithm
discussed earlier)

Step 4 Remove all direct and indirect left-recursion.

Step 5 Do proper substitutions of productions to convert it into the proper form of

GNF.

Problem:
Convert the following CFG into CNF

S→ XY | Xn | p

X → mX | m

Y → Xn | o

Solution:

Here, S does not appear on the right side of any production and there are no unit or null
productions in the production rule set. So, we can skip Step 1 to Step 3.

Step 4:

Now after replacing

X in S → XY | Xo | p

with

mX | m

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 Automata Theory

we obtain

S → mXY | mY | mXo | mo | p.

And after replacing

X in Y→ Xn | o

with the right side of

X → mX | m

we obtain

Y→ mXn | mn | o.

Two new productions O→ o and P → p are added to the production set and then we came
to the final GNF as the following:

S → mXY | mY | mXC | mC | p

X→ mX | m

Y→ mXD | mD | o

O→o

P→p

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 Automata Theory
Pumping Lemma for Context Free Grammars

Lemma:
If L is a context-free language, there is a pumping length p such that any string
w ∈ L of length ≥ p can be written as w = uvxyz, where vy ≠ ε, |vxy| ≤ p,
and for all i ≥ 0, uvixyiz ∈ L.

Applications of Pumping Lemma

Pumping lemma is used to check whether a grammar is context free or not. Let us take
an example and show how it is checked.

Problem:
Find out whether the language L= {xnynzn | n ≥1} is context free or not.

Solution:

Let L is context free. Then, L must satisfy pumping lemma.

At first, choose a number n of the pumping lemma. Then, take z as 0n1n2n.

Break z into uvwxy, where

|vwx| ≤ n and vx ≠ ε.

Hence vwx cannot involve both 0s and 2s, since the last 0 and the first 2 are at least
(n+1) positions apart. There are two cases:

Case 1: vwx has no 2s. Then vx has only 0s and 1s. Then uwy, which would have to be
in L, has n 2s, but fewer than n 0s or 1s.

Case 2: vwx has no 0s.

Here contradiction occurs.

Hence, L is not a context-free language.

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Pushdown Automata

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 ry
Pushdown Automata

Basic Structure of PDA

A pushdown automaton is a way to implement a context-free grammar in a similar way
we design DFA for a regular grammar. A DFA can remember a finite amount of information,
but a PDA can remember an infinite amount of information.

Basically a pushdown automaton is:

"Finite state machine" + "a stack"

A pushdown automaton has three components:

 an input tape,
 a control unit, and
 a stack with infinite size.

The stack head scans the top symbol of the stack.

A stack does two operations:

 Push: a new symbol is added at the top.

 Pop: the top symbol is read and removed.

A PDA may or may not read an input symbol, but it has to read the top of the stack in
every transition.

Takes
input Finite control
Accept or reject
unit

Push or Pop

Input tape

Stack

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 Automata Theory

A PDA can be formally described as a 7-tuple (Q, Σ, S, δ, q0, I, F):

 Q is the finite number of states

 Σ is input alphabet
 S is stack symbols
 δ is the transition function: Q × (Σ∪{ε}) × S × Q × S*
 q0 is the initial state (q0 ∈ Q)
 I is the initial stack top symbol (I ∈ S)
 F is a set of accepting states (F ∈ Q)

The following diagram shows a transition in a PDA from a state q1 to state q2, labeled as
a,b → c :

Stack top Push

Input
Symbol Symbol
Symbol

a,b→c
q q
1 2

This means at state q1, if we encounter an input string ‘a’ and top symbol of the stack is
‘b’, then we pop ‘b’, push ‘c’ on top of the stack and move to state q2.

Terminologies Related to PDA

Instantaneous Description
The instantaneous description (ID) of a PDA is represented by a triplet (q, w, s) where

 q is the state
 w is unconsumed input
 s is the stack contents

Turnstile Notation
The "turnstile" notation is used for connecting pairs of ID's that represent one or many
moves of a PDA. The process of transition is denoted by the turnstile symbol "⊢".

Consider a PDA (Q, Σ, S, δ, q0, I, F). A transition can be mathematically represented by

the following turnstile notation:

(p, aw, Tβ) ⊢ (q, w, αb)

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 Automata Theory

This implies that while taking a transition from state p to state q, the input symbol ‘a’ is
consumed, and the top of the stack ‘T’ is replaced by a new string ‘α’.

Note: If we want zero or more moves of a PDA, we have to use the symbol (⊢*) for it.

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 Acceptance by Pushdown Automata a Theory

There are two different ways to define PDA acceptability.

Final State Acceptability

In final state acceptability, a PDA accepts a string when, after reading the entire string,
the PDA is in a final state. From the starting state, we can make moves that end up in a
final state with any stack values. The stack values are irrelevant as long as we end up in
a final state.

For a PDA (Q, Σ, S, δ, q0, I, F), the language accepted by the set of final states F is:

L(PDA) = {w | (q0, w, I) ⊢* (q, ε, x), q ∈ F}

for any input stack string x.

Empty Stack Acceptability

Here a PDA accepts a string when, after reading the entire string, the PDA has emptied its
stack.

For a PDA (Q, Σ, S, δ, q0, I, F), the language accepted by the empty stack is:

L(PDA) = {w | (q0, w, I) ⊢* (q, ε, ε), q ∈ Q}

Example
Construct a PDA that accepts L= {0n 1n | n ≥ 0}

Solution

0, ε →0 1, 0→ ε

ε , ε→$ 1, 0→ ε ε, $→ ε
q1 q2 q3 q4

PDA for L= {0n 1n | n≥0}

This language accepts L = {ε, 01, 0011, 000111, ............................. }

Here, in this example, the number of ‘a’ and ‘b’ have to be same.

 Initially we put a special symbol ‘$’ into the empty stack.

 Then at state q2, if we encounter input 0 and top is Null, we push 0 into stack. This
may iterate. And if we encounter input 1 and top is 0, we pop this 0.

 Then at state q3, if we encounter input 1 and top is 0, we pop this 0. This may also
iterate. And if we encounter input 1 and top is 0, we pop the top element.

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 Automata Theory

 If the special symbol ‘$’ is encountered at top of the stack, it is popped out and it
finally goes to the accepting state q4.

Example
Construct a PDA that accepts L= { wwR | w = (a+b)* }

Solution

a, ε →a a,a → ε
b, ε →b b, b→ ε

ε , ε→$ ε, ε → ε ε, $→ ε
q1 q2 q3 q4

PDA for L= {wwR | w = (a+b)*}

Initially we put a special symbol ‘$’ into the empty stack. At state q2, the w is being read.
In state q3, each 0 or 1 is popped when it matches the input. If any other input is given,
the PDA will go to a dead state. When we reach that special symbol ‘$’, we go to the
accepting state q4.

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 mata Theory
PDA & Context-Free Grammar

If a grammar G is context-free, we can build an equivalent nondeterministic PDA which

accepts the language that is produced by the context-free grammar G. A parser can be
built for the grammar G.

Also, if P is a pushdown automaton, an equivalent context-free grammar G can be

constructed where

L(G) = L(P)

In the next two topics, we will discuss how to convert from PDA to CFG and vice versa.

Algorithm to find PDA corresponding to a given CFG

Input: A CFG, G= (V, T, P, S)

Output: Equivalent PDA, P= (Q, Σ, S, δ, q0, I, F)

Step 1 Convert the productions of the CFG into GNF.

Step 2 The PDA will have only one state {q}.

Step 3 The start symbol of CFG will be the start symbol in the PDA.

Step 4 All non-terminals of the CFG will be the stack symbols of the PDA and all
the terminals of the CFG will be the input symbols of the PDA.

Step 5 For each production in the form A→ aX where a is terminal and A, X are
combination of terminal and non-terminals, make a transition δ (q, a, A).

Problem
Construct a PDA from the following CFG.

G = ({S, X}, {a, b}, P, S)

where the productions are:

S → XS |  , A → aXb | Ab | ab

Solution

Let the equivalent PDA,

P = ({q}, {a, b}, {a, b, X, S}, δ, q, S)

where δ:

δ (q,  , S) = {(q, XS), (q,  )}

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 Automata Theory

δ(q,  , X) = {(q, aXb), (q, Xb), (q, ab)}

δ(q, a, a) = {(q,  )}

δ(q, 1, 1) = {(q,  )}

Algorithm to find CFG corresponding to a given PDA

Input: A CFG, G= (V, T, P, S)

Output: Equivalent PDA, P = (Q, Σ, S, δ, q0, I, F) such that the non- terminals of the
grammar G will be {Xwx | w,x ∈ Q} and the start state will be Aq0,F.

Step 1 For every w, x, y, z ∈ Q, m ∈ S and a, b ∈ Σ, if δ (w, a, ) contains (y, m)

and (z, b, m) contains (x, ), add the production rule Xwx → a Xyzb in
grammar G.

Step 2 For every w, x, y, z ∈ Q, add the production rule Xwx → XwyXyx in grammar
G.

Step 3 For w ∈ Q, add the production rule Xww→  in grammar G.

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 a Theory
Pushdown Automata and Parsing

Parsing is used to derive a string using the production rules of a grammar. It is used to
check the acceptability of a string. Compiler is used to check whether or not a string is
syntactically correct. A parser takes the inputs and builds a parse tree.

A parser can be of two types:

 Top-Down Parser: Top-down parsing starts from the top with the start-symbol
and derives a string using a parse tree.

 Bottom-Up Parser: Bottom-up parsing starts from the bottom with the string and
comes to the start symbol using a parse tree.

Design of Top-Down Parser

For top-down parsing, a PDA has the following four types of transitions:

 Pop the non-terminal on the left hand side of the production at the top of the stack
and push its right-hand side string.

 If the top symbol of the stack matches with the input symbol being read, pop it.

 Push the start symbol ‘S’ into the stack.

 If the input string is fully read and the stack is empty, go to the final state ‘F’.

Example
Design a top-down parser for the expression "x+y*z" for the grammar G with the following
production rules:

P: S → S+X | X, X → X*Y | Y, Y → (S) | id

Solution

If the PDA is (Q, Σ, S, δ, q0, I, F), then the top-down parsing is:

(x+y*z, I) ⊢(x +y*z, SI) ⊢ (x+y*z, S+XI) ⊢(x+y*z, X+XI)

⊢(x+y*z, Y+X I) ⊢(x+y*z, x+XI) ⊢(+y*z, +XI) ⊢ (y*z, XI)

⊢(y*z, X*YI) ⊢(y*z, y*YI) ⊢(*z,*YI) ⊢(z, YI) ⊢(z, zI) ⊢(ε, I)

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 Automata Theory

Design of a Bottom-Up Parser

For bottom-up parsing, a PDA has the following four types of transitions:

 Push the current input symbol into the stack.

 Replace the right-hand side of a production at the top of the stack with its left-hand
side.

 If the top of the stack element matches with the current input symbol, pop it.

 If the input string is fully read and only if the start symbol ‘S’ remains in the stack,
pop it and go to the final state ‘F’.

Example
Design a top-down parser for the expression "x+y*z" for the grammar G with the following
production rules:

P: S → S+X | X, X → X*Y | Y, Y → (S) | id

Solution

If the PDA is (Q, Σ, S, δ, q0, I, F), then the bottom-up parsing is:

(x+y*z, I) ⊢ (+y*z, xI) ⊢ (+y*z, YI) ⊢ (+y*z, XI) ⊢ (+y*z, SI)

⊢ (y*z, +SI) ⊢ (*z, y+SI) ⊢ (*z, Y+SI) ⊢ (*z, X+SI) ⊢ (z, *X+SI)

⊢ (ε, z*X+SI) ⊢ (ε, Y*X+SI) ⊢ (ε, X+SI) ⊢ (ε, SI)

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 Automata Theory

Turing Machine

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 omata Theory
Turing Machine

A Turing Machine is an accepting device which accepts the languages (recursively

enumerable set) generated by type 0 grammars. It was invented in 1936 by Alan Turing.

Definition
A Turing Machine (TM) is a mathematical model which consists of an infinite length tape
divided into cells on which input is given. It consists of a head which reads the input tape.
A state register stores the state of the Turing machine. After reading an input symbol, it
is replaced with another symbol, its internal state is changed, and it moves from one cell
to the right or left. If the TM reaches the final state, the input string is accepted, otherwise
rejected.

A TM can be formally described as a 7-tuple (Q, X, Σ, δ, q0, B, F) where:

 Q is a finite set of states

 X is the tape alphabet
 Σ is the input alphabet
 δ is a transition function; δ : Q × X → Q × X × {Left_shift, Right_shift}.
 q0 is the initial state
 B is the blank symbol
 F is the set of final states

Comparison with the previous automaton:

The following table shows a comparison of how a Turing machine differs from Finite
Automaton and Pushdown Automaton.

Machine Stack Data Structure Deterministic?

Finite Automaton N.A Yes

Pushdown Automaton Last In First Out(LIFO) No

Turing Machine Infinite tape Yes

Example of Turing machine:

Turing machine M = (Q, X, Σ, δ, q0, B, F) with

 Q = {q0, q1, q2, qf}

 X = {a, b}
 Σ = {1}

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 Automata Theory

 q0= {q0}
 B = blank symbol
 F = {qf }

δ is given by:

Tape alphabet Present State Present State Present State

symbol ‘q0’ ‘q1’ ‘q2’

a 1Rq1 1Lq0 1Lqf

b 1Lq2 1Rq1 1Rqf

Here the transition 1Rq1 implies that the write symbol is 1, the tape moves right, and the
next state is q1. Similarly, the transition 1Lq2 implies that the write symbol is 1, the tape
moves left, and the next state is q2.

Time and Space Complexity of a Turing Machine

For a Turing machine, the time complexity refers to the measure of the number of times
the tape moves when the machine is initialized for some input symbols and the space
complexity is the number of cells of the tape written.

Time complexity all reasonable functions:

T(n) = O(n log n)

TM's space complexity:

S(n) = O(n)

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 y
Accepted Language & Decided Language

A TM accepts a language if it enters into a final state for any input string w. A language is
recursively enumerable (generated by Type-0 grammar) if it is accepted by a Turing
machine.

A TM decides a language if it accepts it and enters into a rejecting state for any input not
in the language. A language is recursive if it is decided by a Turing machine.

There may be some cases where a TM does not stop. Such TM accepts the language, but
it does not decide it.

Designing a Turing Machine

The basic guidelines of designing a Turing machine have been explained below with the
help of a couple of examples.

Example 1
Design a TM to recognize all strings consisting of an odd number of α’s.

Solution

The Turing machine M can be constructed by the following moves:

 Let q1 be the initial state.

 If M is in q1; on scanning α, it enters the state q2 and writes B (blank).

 If M is in q2; on scanning α, it enters the state q1 and writes B (blank).

 From the above moves, we can see that M enters the state q1 if it scans an even
number of α’s, and it enters the state q2 if it scans an odd number of α’s. Hence q2
is the only accepting state.

Hence,

M = {{q1, q2}, {1}, {1, B}, δ, q1, B, {q2}}

where δ is given by:

Tape alphabet
Present State ‘q1’ Present State ‘q2’
symbol

α BRq2 BRq1

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 Automata Theory

Example 2
Design a Turing Machine that reads a string representing a binary number and erases all
leading 0’s in the string. However, if the string comprises of only 0’s, it keeps one 0.

Solution

Let us assume that the input string is terminated by a blank symbol, B, at each end of the
string.

The Turing Machine, M, can be constructed by the following moves:

 Let q0 be the initial state.

 If M is in q0, on reading 0, it moves right, enters the state q1 and erases 0. On

reading 1, it enters the state q2 and moves right.

 If M is in q1, on reading 0, it moves right and erases 0, i.e., it replaces 0’s by B’s.
On reaching the leftmost 1, it enters q2 and moves right. If it reaches B, i.e., the
string comprises of only 0’s, it moves left and enters the state q3.

 If M is in q2, on reading either 0 or 1, it moves right. On reaching B, it moves left

and enters the state q4. This validates that the string comprises only of 0’s and 1’s.

 If M is in q3, it replaces B by 0, moves left and reaches the final state qf.

 If M is in q4, on reading either 0 or 1, it moves left. On reaching the beginning of

the string, i.e., when it reads B, it reaches the final state qf.

Hence,

M = {{q0, q1, q2, q3, q4, qf}, {0,1, B}, {1, B}, δ, q0, B, {qf}}

where δ is given by:

Tape
Present Present Present Present Present
alphabet
State ‘q0’ State ‘q1’ State ‘q2’ State ‘q3’ State ‘q4’
symbol

0 BRq1 BRq1 0Rq2 - 0Lq4

1 1Rq2 1Rq2 1Rq2 - 1Lq4

B BRq1 BLq3 BLq4 0Lqf BRqf

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 tomata Theory
Types of Turing Machines

Multi-tape Turing Machines have multiple tapes where each tape is accessed with a
separate head. Each head can move independently of the other heads. Initially the input
is on tape 1 and others are blank. At first, the first tape is occupied by the input and the
other tapes are kept blank. Next, the machine reads consecutive symbols under its heads
and the TM prints a symbol on each tape and moves its heads.

En
d

En
d

En
d

Head

A Multi-tape Turing machine can be formally described as a 6-tuple (Q, X, B, δ, q 0, F)

where:

 Q is a finite set of states

 X is the tape alphabet

 B is the blank symbol

 δ is a relation on states and symbols where

δ: Q ×Xk →Q× (X× {Left_shift, Right_shift, No_shift })k

where there is k number of tapes

 q0 is the initial state

 F is the set of final states

Note: Every Multi-tape Turing machine has an equivalent single-tape Turing machine.

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 tomata Theory

Multi-track Turing Machine

Multi-track Turing machines, a specific type of Multi-tape Turing machine, contain multiple
tracks but just one tape head reads and writes on all tracks. Here, a single tape head
reads n symbols from n tracks at one step. It accepts recursively enumerable languages
like a normal single-track single-tape Turing Machine accepts.

A Multi-track Turing machine can be formally described as a 6-tuple (Q, X, Σ, δ, q0, F)

where:

 Q is a finite set of states

 X is the tape alphabet

 Σ is the input alphabet

 δ is a relation on states and symbols where

δ(Qi, [a1, a2, a3,....]) = (Qj, [b1, b2, b3,....], Left_shift or Right_shift)

 q0 is the initial state

 F is the set of final states

Note: For every single-track Turing Machine S, there is an equivalent multi-track Turing
Machine M such that L(S) = L(M).

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 ry

Non-Deterministic Turing machine

In a Non-Deterministic Turing Machine, for every state and symbol, there are a group of
actions the TM can have. So, here the transitions are not deterministic. The computation
of a non-deterministic Turing Machine is a tree of configurations that can be reached from
the start configuration.

An input is accepted if there is at least one node of the tree which is an accept
configuration, otherwise it is not accepted. If all branches of the computational tree halt
on all inputs, the non-deterministic Turing Machine is called a Decider and if for some
input, all branches are rejected, the input is also rejected.

A non-deterministic Turing machine can be formally defined as a 6-tuple (Q, X, Σ, δ, q0,

B, F) where:

 Q is a finite set of states

 X is the tape alphabet

 Σ is the input alphabet

 δ is a transition function;

δ : Q × X → P(Q × X × {Left_shift, Right_shift}).

 q0 is the initial state

 B is the blank symbol

 F is the set of final states

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 ry

Semi-infinite Tape Turing Machine

A Turing Machine with a semi-infinite tape has a left end but no right end. The left end is
limited with an end marker.

En
d

Head

It is a two-track tape:

1. Upper track: It represents the cells to the right of the initial head position.

2. Lower track: It represents the cells to the left of the initial head position in reverse
order.

The infinite length input string is initially written on the tape in contiguous tape cells.

The machine starts from the initial state q0 and the head scans from the left end marker
‘End’. In each step, it reads the symbol on the tape under its head. It writes a new symbol
on that tape cell and then it moves the head either into left or right one tape cell. A
transition function determines the actions to be taken.

It has two special states called accept state and reject state. If at any point of time it
enters into the accepted state, the input is accepted and if it enters into the reject state,
the input is rejected by the TM. In some cases, it continues to run infinitely without being
accepted or rejected for some certain input symbols.

Note: Turing machines with semi-infinite tape are equivalent to standard Turing
machines.

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 tomata Theory
Linear Bounded Automata

A linear bounded automaton is a multi-track non-deterministic Turing machine with a tape

of some bounded finite length.

Length = function (Length of the initial input string, constant c)

Here,

Memory information ≤ c × Input information

The computation is restricted to the constant bounded area. The input alphabet contains
two special symbols which serve as left end markers and right end markers which mean
the transitions neither move to the left of the left end marker nor to the right of the right
end marker of the tape.

A linear bounded automaton can be defined as an 8-tuple (Q, X, Σ, q0, ML, MR, δ, F) where:

 Q is a finite set of states

 X is the tape alphabet

 Σ is the input alphabet

 q0 is the initial state

 ML is the left end marker

 MR is the right end marker where MR≠ ML

 δ is a transition function which maps each pair (state, tape symbol) to (state, tape
symbol, Constant ‘c’) where c can be 0 or +1 or -1

 F is the set of final states

End End

Left End Marker Right End Marker

A deterministic linear bounded automaton is always context-sensitive and the linear

bounded automaton with empty language is undecidable.

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 Automata Theory

Decidability

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 Automata Theory
Language Decidability

A language is called Decidable or Recursive if there is a Turing machine which accepts

and halts on every input string w. Every decidable language is Turing-Acceptable.

Non-Turing acceptable languages

Turing acceptable
languages
Decidable
languages

A decision problem P is decidable if the language L of all yes instances to P is decidable.

For a decidable language, for each input string, the TM halts either at the accept or the
reject state as depicted in the following diagram:

Decision on Halt

Rejected

Input

Accepted

Turing Machine

Example 1
Find out whether the following problem is decidable or not:

Is a number ‘m’ prime?

Solution

Prime numbers = {2, 3, 5, 7, 11, 13, …………..}

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 Automata Theory

Divide the number ‘m’ by all the numbers between ‘2’ and ‘√m’ starting from ‘2’.

If any of these numbers produce a remainder zero, then it goes to the “Rejected
state”, otherwise it goes to the “Accepted state”. So, here the answer could be
made by ‘Yes’ or ‘No’.

Hence, it is a decidable problem.

Example 2
Given a regular language L and string w, how can we check if w∈ L?

Solution

Take the DFA that accepts L and check if w is accepted

Qr w∉ L
Input
string w
Qi

Qf w∈ L

DFA

Some more decidable problems are:

1. Does DFA accept the empty language?

2. Is L1∩ L2=Ø for regular sets?

Note:

1. If a language L is decidable, then its complement L' is also decidable.

2. If a language is decidable, then there is an enumerator for it.

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 Automata Theory
Undecidable Languages

For an undecidable language, there is no Turing Machine which accepts the language and
makes a decision for every input string w (TM can make decision for some input string
though). A decision problem P is called “undecidable” if the language L of all yes instances
to P is not decidable. Undecidable languages are not recursive languages, but sometimes,
they may be recursively enumerable languages.

Non-Turing acceptable languages

Undecidable languages

Decidable
languages

Example:

 The halting problem of Turing machine

 The mortality problem
 The mortal matrix problem
 The Post correspondence problem, etc.

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 ta Theory
Turing Machine Halting Problem

Input: A Turing machine and an input string w.

Problem: Does the Turing machine finish computing of the string w in a finite number of
steps? The answer must be either yes or no.

Proof: At first, we will assume that such a Turing machine exists to solve this problem
and then we will show it is contradicting itself. We will call this Turing machine as a Halting
machine that produces a ‘yes’ or ‘no’ in a finite amount of time. If the halting machine
finishes in a finite amount of time, the output comes as ‘yes’, otherwise as ‘no’. The
following is the block diagram of a Halting machine:

Input Yes (HM halts on input w)

string Halting
Machine
No (HM does not halt on input w)

Now we will design an inverted halting machine (HM)’ as:

 If H returns YES, then loop forever.

 If H returns NO, then halt.

The following is the block diagram of an ‘Inverted halting machine’:

Infinite loop

Yes
Qi Qj
Input Halting
string Machine
No

Further, a machine (HM)2 which input itself is constructed as follows:

 If (HM)2 halts on input, loop forever.

 Else, halt.

Here, we have got a contradiction. Hence, the halting problem is undecidable.

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 mata Theory
Post Correspondence Problem

The Post Correspondence Problem (PCP), introduced by Emil Post in 1946, is an

undecidable decision problem. The PCP problem over an alphabet  is stated as follows:

Given the following two lists, M and N of non-empty strings over :

𝑀 = (𝑥1 , 𝑥2 , 𝑥3 , … … … , 𝑥𝑛 )

𝑁 = (𝑦1 , 𝑦2 , 𝑦3 , … … … , 𝑦𝑛 )

We can say that there is a Post Correspondence Solution, if for some 𝑖1 , 𝑖2 , … … … … 𝑖𝑘 , where
1 ≤ 𝑖𝑗 ≤ 𝑛, the condition 𝑥𝑖1 … … . 𝑥𝑖𝑘 = 𝑦𝑖1 … … . 𝑦𝑖𝑘 satisfies.

Example 1
Find whether the lists

M = (abb, aa, aaa) and N = (bba, aaa, aa)

have a Post Correspondence Solution?

Solution

x1 x2 x3

M Abb aa aaa

N Bba aaa aa

Here,

x2 x1x3 = ‘aaabbaaa’

and y2 y1y3 = ‘aaabbaaa’

We can see that

x2 x1x3 = y2y1 y3

Hence, the solution is i=2, j =1, and k=3.

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 Automata Theory

Example 2
Find whether the lists M = (ab, bab, bbaaa) and N = (a, ba, bab) have a Post
Correspondence Solution?

Solution

x1 x2 x3

M ab bab bbaaa

N a ba bab

In this case, there is no solution because:

| x2 x1x3 | ≠ | y2y1 y3 | (Lengths are not same)

Hence, it can be said that this Post Correspondence Problem is undecidable.

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