Automata Theory Tutorial 1u622e

0

Formal Languages and Automata Theory

About the tutorial The theory of computation deals with the computation logic with respect to simple machines, termed as automata. This tutorial will help you in understanding the formal languages and automata theory along with appropriate references and examples.

Audience This tutorial has been designed to help beginners understand formal languages and automata theory.

Prerequisites Knowledge of mathematics and formal language basics is mandatory.

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Formal Languages and Automata Theory

Contents About the tutorial ............................................................................................................................................ i Audience .......................................................................................................................................................... i Prerequisites .................................................................................................................................................... i Copyright & Disclaimer Notice ......................................................................................................................... i Contents ......................................................................................................................................................... ii

1.

INTRODUCTION TO THEORY OF COMPUTATION....................................................................... 1 What is Automata? ......................................................................................................................................... 1 Related Terminologies .................................................................................................................................... 1 Deterministic and Nondeterministic Finite Automaton ................................................................................... 3 Non Deterministic Finite Automaton (NDFA) .................................................................................................. 4 Acceptors, Classifiers, and Transducers........................................................................................................... 5 Acceptability by DFA and NDFA ...................................................................................................................... 6 Converting a NDFA to an equivalent DFA ........................................................................................................ 6 DFA Minimization Using Myhill-Nerode Theorem ........................................................................................... 9 DFA Minimization Using Equivalence Theorem ............................................................................................. 11 Moore and Mealy Machines ......................................................................................................................... 13 Difference between Mealy Machine and Moore Machine ............................................................................ 14 Conversion of a Moore Machine to Its equivalent Mealy Machine ............................................................... 14 Conversion of a Mealy Machine to Equivalent Moore Machine .................................................................... 16 Summary ...................................................................................................................................................... 17

2.

FORMAL LANGUAGES AND CLASSIFICATION OF GRAMMARS ................................................. 18 Grammar ...................................................................................................................................................... 18 Derivations from a Grammar ........................................................................................................................ 19 Language Generated by a Grammar .............................................................................................................. 19 Construction of a Grammar generating a Language ...................................................................................... 20

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Formal Languages and Automata Theory

Chomsky Classification of Grammars ............................................................................................................ 21 Type-3 Grammar ........................................................................................................................................... 22 Type-2 Grammar ........................................................................................................................................... 22 Type-1 Grammar ........................................................................................................................................... 23 Type-0 Grammar ........................................................................................................................................... 23 Summary ...................................................................................................................................................... 24

3.

REGULAR LANGUAGES AND REULAR GRAMMAR .................................................................... 25 Regular Expressions ...................................................................................................................................... 25 Some RE Examples ........................................................................................................................................ 25 Regular Sets and Their Properties ................................................................................................................. 26 Identities Related to Regular Expressions ..................................................................................................... 28 Regular Expression Corresponding to Finite Automaton ............................................................................... 29 Constructing Finite Automaton from Regular Expression .............................................................................. 31 Finite Automata with Null Moves (NFA-ε) ..................................................................................................... 32 Removal of Null Moves from Finite Automata .............................................................................................. 33 Pumping Lemma for Regular Languages........................................................................................................ 34 Applications of Pumping Lemma ................................................................................................................... 35 Method to Prove That a Language L is Not Regular ....................................................................................... 35 Complement of a DFA ................................................................................................................................... 36

4.

CONTEXT FREE LANGUAGES AND GRAMMARS ....................................................................... 37 Context Free Grammar ................................................................................................................................. 37 Generation of Derivation Tree ...................................................................................................................... 37 Sentential Form and Partial Derivation Tree ................................................................................................. 38 Leftmost and Rightmost Derivation of a String ............................................................................................. 39 Ambiguity in Context Free Grammars ........................................................................................................... 39 Closure property of CFL ................................................................................................................................. 40 Simplification of CFGs ................................................................................................................................... 41

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Formal Languages and Automata Theory

Reduction of CFG .......................................................................................................................................... 41 Removal of Unit productions ........................................................................................................................ 42 Removal of Null productions......................................................................................................................... 43 Chomsky Normal Form ................................................................................................................................. 44 Algorithm to convert into Chomsky Normal Form: ........................................................................................ 44 Greibach Normal Form .................................................................................................................................. 46 Left and Right Recursive Grammars .............................................................................................................. 48

5.

PUSHDOWN AUTOMATA ........................................................................................................ 49 Basic Structure of Push Down Automata (PDA) ............................................................................................. 49 Terminologies Related to PDA ...................................................................................................................... 50 Acceptance by PDA ....................................................................................................................................... 51 Correspondence between PDA and CFL ........................................................................................................ 52 Algorithm to Find PDA Corresponding to a Given CFG................................................................................... 52 Algorithm to Find CFG Corresponding to a given PDA ................................................................................... 53 Parsing and PDA............................................................................................................................................ 53 Design of Top-Down Parser ........................................................................................................................... 54 Design of a Bottom-Up Parser ....................................................................................................................... 54

6.

TURING MACHINE .................................................................................................................. 56 Definition ...................................................................................................................................................... 56 Comparison with the Previous Automaton: .................................................................................................. 56 Language Accepted and Decided by a Turing machine .................................................................................. 57 Deg a Turing Machine .......................................................................................................................... 57 Multi-tape Turing Machine ........................................................................................................................... 59 Multi-track Turing Machine .......................................................................................................................... 60 Non-Deterministic Turing machine ............................................................................................................... 60 Turing Machine with Semi-Infinite Tape ....................................................................................................... 61 Time and Space Complexity of a Turing Machine .......................................................................................... 61

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Formal Languages and Automata Theory

Linear Bounded Automata ............................................................................................................................ 62

7.

DECIDABILITY AND RECURSIVELY ENUMERABLE LANGUAGES ................................................ 63 Decidability and Decidable Languages .......................................................................................................... 63 Some More Decidable Problems ................................................................................................................... 64 TM Halting problem ...................................................................................................................................... 65 Rice Theorem ................................................................................................................................................ 66 Undecidability of Post Correspondence Problem .......................................................................................... 67

v

1. INTRODUCTION

The theory of computation deals with the computation logic with respect to simple machines, termed as automata. It is the study of abstract computer devices or intangible machines.

What is Automata? The term “Automata” is derived from the Greek word “αὐτόματα” which means "selfacting". 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 (q0 ∈ Q).



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

Related Terminologies 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 alphabets.

String 

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



Example:

‘cabcad’ is a valid string on the alphabet set Σ = {a, b, c, d} 1

Formal Languages and Automata Theory

Length of a String 

Definition: It is the number of symbols in it. (Denoted by |S|).



Examples:

If S=‘cabcad’ , |S|= 6 If |S|= 0, it is called empty string (Denoted by

λ or ε)

Kleene Star 

Definition: The set Σ* is the infinite set of all possible strings of all possible lengths over Σ including λ.



Representation:

Σ* = Σ0 U Σ1 U Σ2 U…… 

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+ = Σ0 U Σ1 U Σ2 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:

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Formal Languages and Automata Theory

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

Deterministic and Nondeterministic Finite Automaton Finite Automaton can be classified into two types: 1. Deterministic Finite Automaton (DFA) 2. 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. As it has finite number of states, 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 (q0 ∈ 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 Transition function δ as shown by the following table: 3

Formal Languages and Automata Theory

Present State

Next State for Next State for Input 0 Input 1

a

a

b

b

c

a

c

b

c

Graphical Representation: 1

0

a

b 1

0

c

1

0

Non Deterministic Finite Automaton (NDFA) In NDFA, for a particular input symbol, the machine can move to any combination of the states in the machine, or 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 a NDFA A 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 U ε}→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).

Graphical Representation of a NDFA: (Same as DFA) A 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. 4

Formal Languages and Automata Theory



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} Transition function  as shown below: Next State for Next State for Input 0 Input 1

Present State a

a, b

b

b

c

a, c

c

b, c

c

Graphical Representation: 1

0

a

b 0, 1

0, 1

c

0, 1

0

Acceptors, Classifiers, and Transducers 

Acceptor (Recognizer): An automation 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 terminates.



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

Mealy Machine (The output depends only on the current state.) 5

Formal Languages and Automata Theory

o

Moore Machine (The output depends both on the current input as well as the current state.)

Acceptability by DFA and NDFA A string is accepted by a DFA/NDFA if the DFA/NDFA starting at the initial state ends in an accepting state (any of final states) after reading the string wholly. A string S is accepted by a DFA/NDFA (Q, Σ, δ, q 0, F), if 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, q 0, F), if 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}. Example: Let us consider the DFA shown in the following diagram. From the DFA, the acceptable strings can be derived. 0 1

a

c

0 1 1 d 0

Strings accepted by the above DFA: {0, 00, 11, 010, 101, ...........} Strings not accepted by the above DFA: {1, 011, 111, ........}

Converting a NDFA to an equivalent DFA 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 1 Input: A NDFA 6

Formal Languages and Automata Theory

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 {Q0, 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 is the final states of the equivalent DFA.

Example: Let us consider the NDFA shown in the following diagram. Using algorithm 1, we find the equivalent DFA. The state table of the DFA and the state diagram is shown as follows:

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Formal Languages and Automata Theory

1 c

b 0

0

0

δ(q,0)

δ(q,1)

a

{a,b,c,d,e}

{d,e}

b

{c}

{e}

c

∅

{b}

d

{e}

∅

e

∅

∅

1 0, 1

e

a

0

q

0, 1

d

0

Conversion of the above NDFA to DFA using Algorithm 1: 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

∅

∅

D

e

∅

{c,e}

∅

b

B

c

e

C

∅

b

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Formal Languages and Automata Theory

DFA Minimization Using Myhill-Nerode Theorem Algorithm 2 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 minimize the DFA shown in the following diagram using Algorithm 2.

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Formal Languages and Automata Theory

0, 1 1

1

b

f

d 0 0

0

1 1

a

c 1

e 0

0

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

b

C

d

e

f

c

d

e

f

a b c d e f Step 2: We mark the state pairs: a

b

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

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Formal Languages and Automata Theory

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}. So the final minimized DFA will contain three states {f}, {a, b} and {c, d, e}. The minimized DFA is as shown in the following diagram:

0, 1

0

(a, b)

1

(c,d,e) 1

( f

0,1

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 states are distinguishable.

Algorithm 3 

Step 1: All the states Q are divided in two partitions: final states and non-final states and is denoted by P0. All the states in partitions 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 so that δ(X, S) and δ(Y, S) are (k-1)-distinguishable.



Step 3: If Pk≠Pk-1, Then Repeat Step 2 otherwise Go To Step 4.

Step 4: Combine kth equivalent sets and make them the new states of the reduced DFA.Example: Let us considered the following DFA:

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Formal Languages and Automata Theory

0, 1 1

1

b

q

δ(q,1)

a

b

c

b

a

d

c

e

f

d

e

f

e

e

f

f

f

f

f

d 0 0

δ(q,0)

0

1 1

a

c 1

e 0

0

Let us apply algorithm 3 to this 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)

0

δ(q,1)

(a, b)

(a, b)

(c,d,e)

(c,d,e)

(c,d,e)

(f)

(f)

(f)

(f)

(a, b)

0, 1 1

(c,d,e) 1

0,1

( f

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Formal Languages and Automata Theory

Moore and Mealy Machines Finite automata may have outputs corresponding to each transition. There are two types of finite state machines those generate output: 3. Mealy Machine 4. Moore machine

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

Example: Refer the following state diagram of a Mealy Machine: 0 /x2 0 /x3, 1 /x2 b

1 /x3

0 /x1 a

0 /x3 1 /x1

d

c 1 /x1

Moore Machine Moore machine is a 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× Σ → O 13

Formal Languages and Automata Theory



q0 is the initial state from where any input is processed (q0 ∈ Q).

Example: Refer the state diagram of Moore Machine: 0 0, 1 0

1

b/x1

d /x3 a/ λ 1

c/x2

0 1

Difference between Mealy Machine and Moore Machine Here is the difference between the machines: Sr. No.

Mealy Machine

Moore Machine

1.

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

2.

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

3.

Output edges.

4.

In Moore machines, more logic is Mealy machines react faster to needed to decode the outputs since it inputs has more circuit delays.

changes

at

the

clock

only

upon

the

Input change can cause change in output change as soon as logic is done.

Conversion of a Moore Machine to Its equivalent 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. 14

Formal Languages and Automata Theory



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 under the output columns of Mealy Machine state table wherever Qi appears in the next state.

Example: Let us consider the following state table of Moore machine: Present State

Next State a=0

a=1

a

d

b

1

b

a

d

0

c

c

c

0

d

b

a

1

Output

Now we apply algorithm 4 to convert it to Mealy Machine. Step 1 & 2: State table after states 1 and 2: Next State Present State

a=0 State

a=1 Output

State

a

d

b

b

a

d

c

c

c

d

b

a

Output

Step 3: State table after Mealy Machine: 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 15

Formal Languages and Automata Theory

Conversion of a Mealy Machine to Equivalent 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 beginning which give 0 output.Example: Let us consider the following Mealy Machine: Next State Present a=0 State

a=1

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

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). Hence we divide b into b0, b1 and c into c0, c1. Present Next State State a=0 a=1

Output

a

d

b1

1

b0

a

d

0

b1

a

d

1

c0

c1

c0

0

c1

c1

c0

1

d

b0

a

0

16

Formal Languages and Automata Theory

Summary This chapter introduces finite state automata or simply finite automata, FA. The two types of finite automata are Deterministic Finite Automata or DFA and Nondeterministic Finite State Automata or NDFA. In DFA, the transition from a state to the next state can be exactly determined which is not possible in NDFA. For practical purposes, NDFA needs to be converted to DFA. Section 1.6 gives an algorithm for NDFA to DFA conversion. The chapter includes two algorithms to minimize a DFA. In the concluding sections, FA with outputs is discussed.

17

2. FORMAL LANGUAGES AND CLASSIFICATION OF GRAMMARS Formal Languages and Automata Theory

In the literary sense of the term, grammars denote syntactical rules for conversation in natural languages. Linguistics has 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 Non-terminal symbols.



T or  is a set of Terminal symbols.



S is the Start symbol, S ∈ N.



P is Production rules for Terminals and Non-terminals.

Example 1: 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 2: 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. ε is empty string.



S is the Start symbol, S ∈ N.

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Formal Languages and Automata Theory

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 xαy derives xβy in G. This derivation is written as: 𝐺

𝑥𝛼𝑦 ⇒ 𝑥𝛽𝑦 Example: Let us consider the grammar G2 given in example 2. 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

Language Generated by a Grammar 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 is formally defined by ∗

𝐿(𝐺) = { 𝑊 | 𝑊 ∈ Σ ,

𝐺

𝑆 ⇒ 𝑊}

If L (G1) = L (G2), the Grammar G1 is equivalent to the Grammar G2. Example: Consider a grammar G: →b}

N = {S, A, B}

T = {a, b}

P = {S →AB, A →a, 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: Consider a grammar G: →bB|b}

N={S, A, B}

T= {a, b}

P= {S →AB, A →aA|a, B 19

Formal Languages and Automata Theory

L (G) = {ab, a2b, ab2, a2b2, ………}

Construction of a Grammar generating a Language We consider some languages and convert it into a grammar G which produces those languages. Example: 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, start symbol have to take at least one ‘b’ preceded by any number of ‘a’ including null. To accept 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: 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:

20

Formal Languages and Automata Theory

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

…….}

Here, start symbol have to take at least one ‘a’ followed by any number of ‘b’ including null. To accept 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 })

Chomsky Classification of Grammars This hierarchy of grammars was described by Noam Chomosky. Four types of grammars: Grammar Type

Grammar Accepted

Language Accepted

Automaton

Type 0

Unrestricted grammar

Recursively enumerable language

Turing machine

Type 1

Context-sensitive grammar

Context-sensitive language

Linear-bounded automaton

Type 2

Context-free grammar

Context-free language

Pushdown automaton

Type 3

Regular grammar

Regular language

Finite state automaton

21

Formal Languages and Automata Theory

The following diagram shows Containment of Type 3 ⊆ Type 2 ⊆ Type 1 ⊆ Type 0:

Recursively Enumerable

Context - Sensitive

Context - Free

Regular

Type-3 Grammar 

They generate regular languages.



They 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 nonterminal.



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 X → aY

Type-2 Grammar 

They 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). 22

Formal Languages and Automata Theory



These languages generated by these grammars are be recognized by a nondeterministic pushdown automaton.

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

Type-1 Grammar 

They generate context-sensitive languages.



The productions must be in the form α A β → α γ β 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 nonempty.



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 

They 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 like 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. 23

Formal Languages and Automata Theory

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

Summary This chapter introduces the concepts of grammars and formal languages. Here, we have generated a language from the mathematical model of a grammar. Besides, we have constructed a grammar when the language is available. Grammars have been categorized into four types by Naom Chomsky, namely Type 3, Type 2, Type 1, and Type 0.

24

3. REGULAR LANGUAGES AND REULAR GRAMMAR Formal Languages and Automata Theory

Regular Expressions A Regular Expression (RE) can be recursively defined as follows: 1. 2. 3. 4. 5.

ε is a Regular Expression that indicates the language containing an empty string. (L (ε ) = {ε}) φ is a Regular Expression denoting an empty language. (L (φ) = { }) x is a Regular Expression where L={x} If X is a Regular Expression denoting the language L(X) and Y is a Regular Expression denoting the language L(Y), then o o o

X+Y is a Regular Expression corresponding to the language L(X) U L(Y) where L(X+Y)= L(X) U L(Y). X.Y is a Regular Expression corresponding to the language L(X). L(Y) where L(X.Y)= L(X) . L(Y) R* is a Regular Expression corresponding to the language L(R*) where L(R*) = (L(R))*

6. If we apply any of the rules several times from 1- 5 they are Regular Expressions.

Some RE Examples Here are some examples of REs: Regular Expression

Regular Set

(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}

(a+b)*

Set of strings of a’s and b’s of any length including the null string. So L= { ε, 0, 1,00,01,10,11,…….} Set of strings of a‟s and b‟s ending with the string abb , So L= {abb, aabb, babb, aaabb, ababb, …………..}

(a+b)*abb (11)*

Set consisting of even number of 1’s including empty string, So L= {ε, 11, 1111, 111111, ……….}

25

Formal Languages and Automata Theory

(aa)*(bb)*b

(aa + ab + ba + bb)*

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, …………..} String of a’s and b’s of even length can be obtained by concatenating any combination of the strings aa, ab, ba and bb including null, so L= {aa, ab, ba, bb, aaab, aaba, …………..}

Regular Sets and Their Properties 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 RE 1 = 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,.......} lengths excluding Null)

(Strings of all possible

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 RE 1 = 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 26

Formal Languages and Automata Theory

Property 3: The complement of a regular set is regular Proof: Let us take a RE = (aa)* 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 27

Formal Languages and 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 a 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. R+ = 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) 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) 28

Formal Languages and Automata Theory

15.ε + RR* = ε + R*R = R*

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

Arden’s Theorem 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 = Q + QP + RPP [After putting the value R = Q + RP ] 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 q 1.q1 = q1R11 + q2R21 + … + qnRn1 + є q2 = q1R12 + q2R22 + … + qnRn2 ..………………………… …………………………… …………………………… …………………………… qn = q1R1n + q2R2n + … + qnRnn Rij represents the set of labels of edges from qi to qj, if no such edge exists then Rij = Ø. 29

Formal Languages and Automata Theory

Step 2: Solve these equations to get the equation for the final state in of RijProblem 1: Construct a regular expression corresponding to the finite automata given below: b

a q2 q3

b b

a q1

a

Solution: Here initial and final state is q1. The equations for the three states, q1, q2 and q3 are: q1 = q1a + q3a + є (є move is because q1 is the initial state 0) 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 = q1a + q1b(b + ab*)aa + є = q1(a + b(b + ab)*aa) + є = є (a+ b(b + ab)*aa)* =(a + b(b + ab)*aa)* Hence, the regular expression is

(Substituting value of q3) (Substituting value of q2)

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

Problem 2: Construct a regular expression corresponding to the finite automata given below:

30

Formal Languages and Automata Theory

0

0, 1 q1 q3

1

1 q 2

0

Solution: Here initial state is q1 and 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*

Constructing Finite Automaton from Regular Expression 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 f

a

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

q1

q1 a

b

q f

Case 3: For a regular expression (a+b) we can construct the following FA: 31

Formal Languages and Automata Theory

b q1

q f

a

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

q f 

Method:

Step 1 Construct a NFA with Null moves from the given regular expression.

Step 2 Remove Null transition from the NFA and convert it into equivalent DFA. Problem: Convert the following RA into equivalent DFA: 1 (0 + 1)* 0 Solution: We will concatenate three expressions “ 1 ” , “ ( 0 + 1 )* “ and “ 0 ” 0, 1

q0 1

q 1

q2

є

є

q3 0

q f

Now we will remove the є transitions. After we remove the є transitions from the NDFA we get the following: 0, 1 q0

q2 1

0

q f

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

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 null move. A NFA-ε is represented formally by a 5-tuple, (Q, Σ, δ, q0, F), consisting of 

a finite set of states Q 32

Formal Languages and Automata Theory



a finite set of input symbols Σ



a transition function δ : Q × (Σ ∪ {ε}) → 2Q



an initial state q0 ∈ Q



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

The finite automata with null moves is as given:

b

ε

a

ε

c

1

0

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

Removal of Null Moves from Finite Automata If in a 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. Problem: Convert the following NFA- ε to NFA without Null move. Finite automata with null moves is as given: 0 ε q1

q 1

0, 1

0 q2 1

Solution: 

Step 1: Here the ε transition is between q1 and q2, so Let q1 is X and qf is Y. 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: 33

Formal Languages and Automata Theory

0 0, 1 q1

q 1

0, 1

0 q2 1



Step 3: Here, q1 is an initial state so we make qf also an initial state.

So the FA becomes as follows: 0 0, 1 q1

q 1

0, 1

0 q2 1



Step4: Here qf is an final state so we make q1 also an final state.

Hence the final FA becomes as follows:

0 0, 1

q

q 1

0, 1

0 q2 1

Pumping Lemma for Regular Languages Theorem: Let L be a regular language. Then there exists a constant ‘c’ such that for every string w in L such that |w| ≥ c, we can break w into three strings, w = xyz, such that: 1. |y| > 0 34

Formal Languages and Automata Theory

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 that some language is regular. 1. If L is regular it satisfies Pumping Lemma 2. If L is non-regular it does not satisfy Pumping Lemma

Method to Prove That a Language L is Not Regular At first we have to assume that L is regular. Hence, the pumping lemma should hold for L. Use the pumping lemma to obtain a contradiction: 

Select w such that |w| ≥ c



Select y such that |y| ≥ 1



Select x such that |xy| ≤ c



Assign remaining string to z



Select k such that the resulting string is not in L.

Hence L is not regular Problem 1: Prove that L = {aibi | i ≥ 0} is not regular.

Solution: At first we assume that L is regular and n is the number of states Let w = anbn. Thus |w| = 2n ≥ n By pumping lemma, let w = xyz, where |xy|≤ n Let x = ap, y = aq and z = arbn , where p + q + r = n. p ≠ 0, q ≠ 0, r ≠ 0. Thus |y|≠ 0 Let k = 2. Then xy2z = apa2qarbn. Number of as = ( p + 2q + r ) = ( p + q + r ) + q = n + q Hence, xy2z = an+qbn. Since q ≠ 0, xy2z is not of the form anbn. 35

Formal Languages and Automata Theory

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

Complement of a DFA 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. The following DFA accepts language L: a, b b

X

Y

a b Z a

This DFA accepts language L = {a, aa, aaa , ............. } over alphabet Σ = {a, b} So, RE = a+Now we will swap it’s accepting states with its non-accepting states and vice versa and will get the following DFA accepting complement of language L: a, b b

X

Y

a Z

b

a

This DFA accepts language Ľ = {ε, b, ab, bb, ba, ............} 0ver alphabet Σ = {a, b}. Note:

If we want to complement a NFA we need to first convert it to DFA and then have to swap states as the previous method.

36

4. CONTEXT FREE LANGUAGES AND GRAMMARS Formal Languages and Automata Theory

Context Free Grammar 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: 1) The grammar ({A}, {a, b, c}, P, A), P: A → aA, A → abc. 2) The grammar ({S, a, b}, {a, b}, P, S), P: S → aSa, S → bSb, S → ε 3) 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: S

x1

x2

xn

37

Formal Languages and Automata Theory

There are two different approaches to draw a derivation tree: 

Top-down Approach:

o Starts with the starting symbol S o Goes down to tree leaves using productions 

Bottom-up Approach: o Starts from tree leaves o 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

SS

S

a

S

ε

b

a

S

b

a

S

b

ε

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. 38

Formal Languages and Automata Theory

Example: If in any CFG the productions are: S → AB, A → aaA | ε, B →Bb| ε, the partial derivation tree can be the following: S

A

B

If a partial derivation tree contains the root S, it is called 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 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 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

Ambiguity in Context Free Grammars If a context free grammar G has more than one derivation tree for some string w ∈ L (G), it is called 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: 39

Formal Languages and Automata Theory

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

XS

+

X

a

*

X

a

a

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

X

X

a

+

* X

XS

a

a

As there are two parse trees for a single string ‘a+a*a’ the grammar G is ambiguous.

Closure property of CFL Context free languages are closed under: 

Union



Concatenation 40

Formal Languages and Automata Theory



Kleene Star operation

Context free languages are not closed under: 

Intersection



Intersection with Regular Language



Complement

Simplification of CFGs In a CFG, it may happen that all the production rules and symbols are not needed for 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 derives some terminal and initialize i=1.



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

41

Formal Languages and 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 42

Formal Languages and Automata Theory



Step1: To remove A→B, add production A→x to the grammar a rule whenever B→x occurs in the grammar. [x ∈ Terminal , x can be Null]



Step2: Delete A→B from the grammar.



Step3: Repeat from step1 until all unit productions are removed.

Problem: Remove null production from the following: S → XY, X → a, Y → Z | b, Z → M, M → N, N → aSolution: There are 3 unit productions in the grammar Y → Z, Z → M, M → N At first we will remove M → N. removed.

As, N → a, we add M → a and M → N is

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 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

Removal Procedure 43

Formal Languages and Automata Theory



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 step2 and remove ϵ-productions.

Problem: Remove null production from the following: S→ASA | aB | b, A → B, B → b | ϵSolution: 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.

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, C are Non-terminals and a is a terminal.

Algorithm to convert into Chomsky Normal Form: 

Step1: If the start symbol S occurs on some right side, create a new start symbol S´ and a new production S´ → S.



Step2: Remove Null productions. (Using the Null production removal algorithm discussed earlier) 44

Formal Languages and Automata Theory



Step3: Remove unit productions. (Using the Unit production removal algorithm discussed earlier)



Step4: Replace each production A → B1…Bn where n > 2 with A → B1C where C → B2 …Bn. Repeat this step for all production which has two or more symbols in right side.



Step5: If the right side of any production is in the form A→ aB where a is 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: 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 Now we will remove the null production B → ϵ and A → ϵ. 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 → bNow we will remove the null 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 → bAfter removing A→ B, the production set becomes:

S0→ ASA | aB | a | AS | SA, S→ ASA | aB | a | AS | SA A → S | b B → b 45

Formal Languages and Automata Theory

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 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 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

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 terminal. 46

Formal Languages and Automata Theory

Algorithm to convert a CFG into Greibach Normal Form 

Step1: If the start symbol S occurs on some right side, create a new start symbol S´ and a new production S´ → S.



Step2: Remove Null productions. (Using the Null production removal algorithm discussed earlier)



Step3: Remove unit productions. (Using the Unit production removal algorithm discussed earlier)



Step4: Remove all direct and indirect left-recursion.



Step5: 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 is also no unit and null production in the production rule set. So, we can skip step1 to step3.

Step4: Now after replacing X in S → XY | Xo | p with mX | m we obtain 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

47

Formal Languages and Automata Theory

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

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: Case1: 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. Case2: vwx has no 0s. Here contradiction arrives. Hence, L is not context free Language.

48

5. PUSHDOWN AUTOMATA

Formal Languages and Automata Theory

Basic Structure of Push Down Automata (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 a finite amount of information but a PDA can 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 have to read the top of the stack in every transition. Takes input

Finite control unit

Accept or reject

Push or Pop

Input tape Stack 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 49

Formal Languages and Automata Theory



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 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 : Input

Stack top

Push

Symbol

Symbol

Symbol

q

a, b→c

1

q 2

This means at state q1, if we encounter 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 (ID) The configuration 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) This implies that while taking a transition from state p to state q, the input symbol is ‘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 ⊢* symbol for it. 50

Formal Languages and Automata Theory

Acceptance by PDA There are two different ways to define PDA acceptability.

1)

Final state acceptability

In this 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.

2)

Empty stack acceptability

In this final state acceptability 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 empty stack is: L (PDA) = {w | (q0, w, I) ⊢* (q, ε, ε), q ∈ Q} Example 1: Construct PDA that accepts L= {0n 1n | n≥0} Solution: The following diagram shows PDA for L= {0n 1n | n≥0} 0, ε →0 1, 0→ ε

ε , ε→$ q1

1, 0→ ε

q2

ε, $→ ε q3

q4

This language accepts L = {ε, 01, 0011, 000111, ............................. }. Here, in this example we the number of ‘a’ and ‘b’ have to be same. Initially we put a special symbol ‘$’ into the empty stack. Then at state q2 if we encounter input 0 and top is Null, we push 0 into stack. This may iterate. And if we encounter input 1 and top is 0, we pop this 0. Then at state q3 if we encounter input 1 and top is 0, we pop this 0. This may also iterate. And if we encounter input 1 and top is 0, we pop the top element. Then if the special symbol ‘$’is encountered at top of the stack, it is popped out and it finally goes to accepting state q4 Example 2: 51

Formal Languages and Automata Theory

Construct PDA that accepts L= { wwR | w = (a+b)* } Solution: a,a → ε

a, ε →a b, ε →b ε, ε → ε

ε , ε→$ q1

b, b→ ε

q2

ε, $→ ε q3

q4

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

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

Correspondence between PDA and CFL 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 for can be built for the grammar G. Also, if P is a Push down automaton an equivalent context free grammar G can be constructed where L (G) = L (P). 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): 

Step1: Convert the productions of the CFG into GNF.



Step2: The PDA will have only one state {q}.



Step3: Start symbol of CFG will be the start symbol in the PDA.



Step4: 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.



Step5: 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: 52

Formal Languages and Automata Theory

G = ({S, X}, {a, b}, P, S) where 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,  )} 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. 

Step1: 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.



Step2: For every w, x, y, z ∈ Q, add the production rule Xwx → XwyXyx in grammar G.



Step3: For w ∈ Q add the production rule Xww→  in grammar G.

Parsing and PDA Parsing is used to derive a string using 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 input and builds a parse tree. There are two approaches of parser: 

Top-Down Parser: Top down parsing starts from the top with the startsymbol 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 parse tree.

53

Formal Languages and Automata Theory

Design of Top-Down Parser For the 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 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)

Design of a Bottom-Up Parser For the 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 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:

54

Formal Languages and Automata Theory

(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)

55

6. TURING MACHINE

Formal Languages and Automata Theory

A Turing Machine (TM), invented in 1936 by Alan Turing, is an accepting device which accepts the languages (Recursively enumerable set) generated by Type 0 grammars.

Definition Turing Machine is a mathematical model which consists of an infinite length tape divided into cells on which input is given. TM consists of a head which reads the input tape. A state 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 Turing machine can be formally described as 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: Machine

Stack Data Structure

Deterministic?

Finite Automaton

N.A

Yes

Push down 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} 56

Formal Languages and Automata Theory



q0= {q0}



B = blank symbol



F = {qf }

δ is given by: Tape alphabet symbol

Present State ‘q0’

Present State ‘q1’

Present State ‘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.

Language Accepted and Decided by a Turing machine A TM accepts a language if it enters into a final state for any input string w. A language is recursively enumerable (Generate 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 arise some cases where TM does not stop. Such TM accepts the language but it does not decide it.

Deg a Turing Machine 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 state q2 and writes B (blank). If M is in q2, on scanning α, it enters state q1 and writes B (blank).

57

Formal Languages and Automata Theory

From the above moves, we can see that M enters state q1 if it scans an even number of α’s and enters 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’ symbol α

BRq2

Present State ‘q2’ BRq1

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 keep one 0. Solution: Let us assume that the input string is 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 state q1 and erases 0. On reading 1, it enters state q2 and moves right. If M is in q1, on reading 0, it moves right and erases 0, i.e. 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 state q3. If M is in q2, on reading either 0 or 1, it moves right. On reaching B, it moves left and enters 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 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 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:

58

Formal Languages and Automata Theory

Tape alphabet symbol

Present State ‘q0’

Present State ‘q1’

Present State ‘q2’

Present State ‘q3’

Present State ‘q4’

0

BRq1

BRq1

0Rq2

-

0Lq4

1

1Rq2

1Rq2

1Rq2

-

1Lq4

B

BRq1

BLq3

BLq4

0Lqf

BRqf

Multi-tape Turing Machine 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 6-tuple (Q, X, B, δ, q0, 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. 59

Formal Languages and Automata Theory

Multi-track Turing Machine Multi-track TM, a specific type of Multi-tape Turing machine, contains multiple tracks but just one tape head reads and writes on all tracks. Here, a single tape head read n symbols from the 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 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).

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 con-figuration. 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 60

Formal Languages and Automata Theory



F is the set of final states

Turing Machine with Semi-Infinite Tape A Turing Machine with a semi-infinite tape has left end but no right end. The left end is limited with an end marker. En d Head It is a two-track tape1) Upper track: It represents the cells right of the initial head position. 2) Lower track: It represents the cells 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.

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(nlogn). TM's space complexity S(n)= O(n).

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Formal Languages and Automata 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 LBA 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 left end marker



MR is 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

Left End Marker

End

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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7. DECIDABILITY AND RECURSIVELY ENUMERABLE LANGUAGES Formal Languages and Automata Theory

Decidability and Decidable Languages 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 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, …………..} 63

Formal Languages and Automata Theory

Divide the number ‘m’ by all the numbers between ‘2’ and ‘√m’ starting from ‘2’. If any of these numbers give remainder zero, then it goes to “Rejected state” otherwise to “Accepted state”. So, here the answer could be made by ‘Yes’ or ‘No’. Hence, it is a decidable problem. Example 2: Given 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.

Input string w

Qr

w∉ L

Qf

w∈ L

Qi

DFA

Some More Decidable Problems 1) Does DFA accept the empty language? 2) Is L1∩ L2=Ø for regular sets? Etc. 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

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).

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Formal Languages and Automata Theory

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 it may be recursively enumerable languages. Non-Turing acceptable languages

Undecidable languages Decidable languages

Example: 1) The halting problem of Turing machine 2) The mortality problem. 3) The mortal matrix problem 4) The Post correspondence problem etc.

TM 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 will show it is contradicting to itself. We will call this Turing machine as Halting machine that produce a ‘yes’ or ‘no’ in finite amount of time. If the HM finishes in finite amount of time the output comes as ‘yes’ otherwise as ‘no’. The following is the block diagram of a Halting machine:

Input string

Halting Machine

Yes

(HM halts on input w)

No (HM does not halt on input w) Now we will design an inverted halting machine will be designed as (HM)’ as: 

If H returns YES then loop forever 65

Formal Languages and Automata Theory



If H returns NO then halt

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

Input string

Yes Qi

Qj

Halting Machine No

Further a machine (HM)2 with input itself is constructed as follows: 

If (HM)2 halts on input, loop forever

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

Rice Theorem Theorem: L= { <M> | L (M) ∈P } is undecidable where p is non-trivial property of the Turing machine is undecidable.

If the following two properties hold, it is proved as undecidable: Property 1: If M1 and M2 recognize the same language then either <M1><M2>∈L or <M1><M2>∉ L Property 2: For some M1 and M2 such that <M1> ∈L and <M2>∉ L Proof: Let there are two Turing machines X1 and X2. Let us assume <X1> ∈L such that L (X1) = φ and <X2> ∉ L. For an input ‘w’ in a particular instant, perform the following steps: 1. If X accepts w, then simulate X2 on x. 2. Run Z on input <W>. 3. If Z accepts <W>, Reject it and if Z rejects <W>, accept it. 66

Formal Languages and Automata Theory

If X accepts w, then L (W) = L (X2), and so <W> ∉ P. If M does not accept w, then L (W) = L(X1) = φ, and so <W> ∈ P Here the contradiction arises. Hence, it is undecidable.

Undecidability of Post Correspondence Problem The Post Correspondence Problem (P), introduced by Emil Post in 1946, is an undecidable decision problem. The P 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: M N Here, x2 x1x3=’aaabbaaa’

x1 abb bba

x2 aa aaa

x3 aaa aa

and y2 y1y3=’aaabbaaa’

We can see that x2 x1x3=y2y1 y3. Hence, the solution is i=2, j =1 and k=3. Example 2: Find whether the lists M = ( ab, bab, bbaaa ) and N = ( a, ba, bab ) have a Post Correspondence Solution? Solution: M

x1 ab

x2 bab

x3 bbaaa 67

Formal Languages and Automata Theory

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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