theory of automata and formal languages unit 1

05/01/2023 Abhimanyu Mishra(CSE) JETGI

THEORY OF AUTOMATA AND FORMAL LANGUAGES

UNIT-I

ABHIMANYU MISHRAASSISTANT PROF.(CSE)

JETGI


What is Automata?AUTOMATAAn Automation derived from the Greek word which means "self-acting". An automaton (Automata in plural) is an theoretical self-propelledcomputing device which follows a predetermined sequence of operations automatically. “A set of string all of which are chosen form some ∑*,where ∑ is particular alphabet, is called a language". If ∑ is an alphabet, and L<= ∑*, then L is said to be language over alphabet ∑.

KLEENE CLOSURE“If S is a set of words then by S* we mean the set of all finite strings formed by concatenating words from S, where any word may be used as often we like, and where the null string is also included”

For example if ∑={0,1}

∑* ={ €,0,1,00,01,10,11,000,001,…………}


Finite automation is called “finite” because number of possible states and number of letter in the alphabet are both finite, and “automation” because the change of the state is totally governed by the input. it is deterministic, since, what state is next is automatic not will-full, just as the motion of the hands of the clock is automatic, while the motion of hands of a human is presumably the result of desire and thought.

THERE ARE TWO TYPES OF FINITE AUTOMATA

(A) DETERMINISTIC FINITE AUTOMATA(DFA)(B) NON DETERMINISTIC FINITE AUTOMATA(NDFA)

FINITE AUTOMATA


DETERMINISTIC FINITE AUTOMATA(DFA) IS A QUINTUPLE (Q,Σ, δ,q0,F)

1. where :Q is a finite set of states.2. Σ is a finite set of symbols, called the alphabet of the automaton.3. δ is the transition function, that is, δ: q × Σ → Q.4. q0 is the start state, that is, the state of the automaton before any input has been processed, where q0 Q.∈5. F is a set of states of q (i.e. F Q) called ⊆ accept states.

(A) DETERMINISTIC FINITE AUTOMATA(DFA)


A.1 PROCESSING OF STRINGS BY DFA

• Suppose a1,a2,………………… an is a sequence of input

symbols. We start out with deterministic finite automata having q1,q2,………………… qn states where q0 is the initial state and qn is the final state and transition function and processed as………..

δ(q0, a1 ) = q1

δ(q1, q2 ) = q2

. . . . δ(qn-1 , qn ) = qn Input a1,a2,………………… an is said to be “accepted” since qn is the member of the final states, and if not then it is “ rejected”


A.2 TRANSITION GRAPH

• A transition graph is a collection of three things:1. A finite set of states, at least one of which is designated as the start state. start state

final state

2. An alphabet Σ of possible input letters from which input strings are formed.

3. A finite set of transitions that show how to go from some states.


A.3 TRANSITION DIAGRAM

Transition Diagram Notations: A transition diagram for DFA, M=(Q,Σ, δ,q0,F) is a graph defined as follows:

(a) For each state in Q there is a node represented by the circle.(b) For each state q in Q and each input symbol a in Σ, let δ(q,a)=P. a a a,b or or b

(c) If any state q in Q is the starting state then it represented by the circle with arrow as

q P q P q P

q


A transition table is a conventional, tabular representation of a function like δ that takes two arguments and return a state. The rows of the table correspond to the states, and the columns correspond to the input.State δ(q,a). For example:

q0 is the starting state and q2 is the final state

A.4 TRANSITION TABLES

δ/ Σ a b

q0 q1 q2

q1 q2 q0

q2 q2 q2


And the transition diagram:

b a

a

a,b

Q={q0, q1, q2 }F={q2 }δ(q0 , a) = q1 , δ(q0 , b) = q2

δ(q0 , a) = q2 , δ(q0 , b) = q0

δ(q0 , a) = q2 , δ(q0 , b) = q2

q0

q0

q1

q2


Example 1:-• Design a FA that accepts set of strings such that every string end in 00, over

alphabet: {0,1} Σ = { 0,1}

Sol:- We have M= (Q,Σ, δ,q0,F) be the DFA q0 : initial state, Σ : {0,1} is given

0,1 Start 0 0

First we will decide the approach to design FA. It is not an easy task to think FA as a whole, So first we have to fulfill minimum condition i.e every string end in 00.

q0 q1q2


Facts in Designing Procedure of Finite Automata

(a) In the designing of FA, First of all we have to analyze the set strings language which is accepted by the FA.

(b) Make sure that every state is check for the output state for every symbol of Σ.

(c) Make sure, that there is one initial state and at least one final state in transition diagram of FA.

(d) Make sure that no state must have two different output state for a single input symbol.


Nondeterministic finite automaton with ε-moves (NFA-ε) is a further simplification to NFA. this automaton replaces the transition function with the one that allows the empty string ε as a possible input. the transitions without overwhelming an input symbol are called ε-transitions.

Let M be NFA then M is Defined as: M = (Q,Σ, δ,q0,F)(1) a finite set of states q0

(2) a finite set of input symbol Σ(3) a transition function δ : q × Σ → P(q).(4) an initial (or start) state q0 Q∈(5) a set of states f distinguished as accepting (or final) states F Q⊆

(B) NON DETERMINISTIC FINITE AUTOMATA(NDFA)


0,1

0,1 0 0

START 0,1

1 1

according to analysis it is clear that NFA will accepts the strings of the pattern 00,11,101000,101100,……………………..

Example 1: Design a NFA for the language L= all strings over {0,1} that have at least two consecutive 0 and 1.

q0

q1

q2

q2

q4


Procedure for converting NFA to equivalent DFA.1) intially Q=ɸ.2) Put {q0 } into Q.3) Then for each state q in Q do the following add this new state, add δ(q,a) = ᴜp€q δ(p,a) to δ on the right hand side is that of NFA4) Repeat step 3 till new states.

TRANSFORMATION OF NFA TO DFA


Example:1 Convert the following NFA in to DFA

a,b b• Start a a a,b b

q0

q3

q2q1


SolutionStep 1 ( We get new state)δ(q0, a ) = {q0, q1 } New single state δ(q0, b ) = {q0 }Step 2 In step 1 we are getting a new state {q0, q1 }. Now repat step 1 for this

new state only, so all transition of a and b from {q0, q1 } as δ({q0 , q1 },a) = δ(q0, a ) U δ(q1, a ) = {q0, q1 } U {q0 } = {q0, q1, q2 } New single state δ({q0 , q1 },b) = δ(q0, b ) U δ(q1, b ) = {q0, q1 } Step3 Repeat step 2 δ({q0 , q1, q2 },a) = δ(q0, a ) U δ(q1, a ) U δ(q2, a) = {q0, q1 } U {q2 }U {q3 } = {q0 , q1, q2, q3 } New single state

δ({q0 , q1, q2 },b) = δ(q0, b ) U δ(q1, b ) U δ(q2, b) = {q0 , q1, q3 } New single state

05/01/2023

Step 4: Repeat Step 3 δ({q0 , q1, q2, q3 },a) = {q0 , q1, q2, q3 } OLD STATE δ({q0 , q1, q2, q3 },b) = {q0 , q1, q2, q3 } OLD STATE Step 5: Repeat Step 4 δ ({q0 , q1, q3 },a) = {q0 , q1, q2, q3 } OLD STATE δ ({q0 , q1, q3 },b) = {q0 , q1, q2, q3 } OLD STATE Now Transition table

δ/ Σ a b

q0 {q0, q1 } {q0 }

{q0, q1 } {q0, q1, q2 } {q0, q1 }

{q0, q1, q2 }* {q0 , q1, q2, q3 } {q0 , q1, q3 }

{q0 , q1, q2, q3 }* {q0 , q1, q2, q3 } {q0 , q1, q2, q3 }

{q0 , q1, q3 }* {q0 , q1, q2, q3 } {q0 , q1, q2, q3 }

Abhimanyu Mishra(CSE) JETGI


Lets Mark the whole structure in Alphabets

q0 A

{q0, q1 } B

{q0, q1, q2 } C

{q0 , q1, q2, q3 } D

{q0 , q1, q3 } E


A is initial state and C,D and E are final states since they contain q2, q3 as member which are final states of NFA

• Transition Table

δ/ Σ a b A B A

B C B

*C D E

*D D D

*E D D


Final Transition Diagram

• b b• Start a

• a

• b

• a a• b• a,b

A

D

E

C

B


NFA with -Transitions ∈

• IF a FA is modified to permit transition without input symbols, along with zero, one or more transition on input symbols, then we get NFA with -∈transition, because the transition made without symbols are called as -∈transitions.

0 1 0 Start ∈ ∈

q0 q1q2


Example:2 Consider the NFA with -transition M =(Q,Σ, δ,q∈ 0,F)Q={q0, q1, q2 }Σ={a,b,c} and moves∈Initial state ={q0 }F= { q2 }

δ/ Σ a b c ∈

q0 {q0} {ɸ} {ɸ} {q1}

q1 {ɸ} {q1} {ɸ} {q2}

*q2 {ɸ} {ɸ} {q2} {ɸ}


And the transition Diagram: NFA with -∈ transition

a b c

Start ∈ ∈ q0 q1q2


SOLUTION:- Step 1- Find the states of NFA without -transition including initial state and ∈ final states as follows:(i) Initial state will be -closure of initial state of NFA with -transitions in ∈ ∈ -closure(q∈ 0) = {q0, q1, q2 } (New Initial state for NFA without transition)∈(ii) Rest of the states are: -closure(q∈ 1) ={q1, q2 } New State -closure(q∈ 2) ={q2} New State(iii) The final states of NFA without -transition are all those new states which ∈ contains final state of NFA with - transition as members.∈ M’ =(Q’,Σ’, δ’,q0’,F’) Q’ =({q0, q1, q2 } ,{q1, q2 }, {q2} )Step 2-Now we have to decide δ’ to find out the transition as follows

δ’ ({q0, q1, q2},a) = -closure(δ(∈ q0, q1, q2 ),a) = -closure(δ(∈ q0, a)U(δ(q1, a)U(δ(q2, a)) = -closure(δ(∈ q0U ɸU ɸ) = -closure(∈ q0) = {q0, q1, q2 }


Similarly, δ’ ({q0, q1, q2},b) = -closure(δ(∈ q0, q1, q2 ),b) = -closure(δ(∈ q0, b)U(δ(q1, b)U(δ(q2, b)) = -closure(δ(∈ ɸ U q1U ɸ) = -closure(∈ q1) = {q 1, q2 } δ’ ({q0, q1, q2},c) = -closure(δ(∈ q0, q1, q2 ),c) = -closure(δ(∈ q0, c)U(δ(q1, c)U(δ(q2, c)) = -closure(δ(∈ ɸ U ɸU q2) = -closure(∈ q2) = { q2}

So we write same for others, δ’ ({q1, q2},a) = ɸ δ’ ({q1, q2},b) = {q1, q2} δ’ ({q1, q2},c) = {q2} δ’ ({q2},a) = ɸ δ’ ({q2},b) = ɸ δ’ ({q2},c) = {q2}


So, transition table for NFA -transition will be as:∈

δ/ Σ a b c

{q0, q1, q2}* {q0, q1, q2} {q1, q2 } { q2}

{q1, q2 }* ɸ {q1, q2 } { q2}

{ q2}* ɸ ɸ { q2}


{q0, q1, q2} as qx

{q1, q2 } as qy

{ q2} as qz

• a b start b

c c

c

NFA without -transition∈

qx

qz

qy


MINIMIZATION ALGORITHM FOR DFA

(i) all states are unreachable from the initial state via any set of transition of DFA M are removed(ii) draw the transition tables for rest states, after removing the unreachable states(iii) split the transition table in two tables(iv) find the similar rows(v) Repeat the same process(vi) Now combined the Reduced tables, and removing non reachable states, dead states.

Minimization of DFA’S


Example 1 : Minimize the DFA given below

• Start b a b

b a a b a a b a

a b a b b

q0

q4

q7

q6

q1

q0

q3

q2


STEP 1: Remove the unreachable states from the given DFA. the DFA given in diagram q3 is unreachable state so remove it.STEP 2: Draw the Transition table for rest states.

Solution :

δ/ Σ a b

q0 q5 q1

q1 q2 q6

*q2 q2 q0

q4 q5 q7

q5 q6 q2

q6 q7 q6

q7 q2 q6


Step 3: Now divide rows of transition table in two sets as:• (i)

• (ii)

q0 q5 q1

q1 q2 q6

q4 q5 q7

q5 q6 q2

q6 q7 q6

q7 q2 q6

*q2 q2 q0


row 2 and row 6,are similar since q1 and q7 transit to same states on a and b, so skip of them row 6 then q7 is replaced by q1 in the rest

Step 4: Apply step 4 on both the sets individually. Now consider set (i)

q0 q5 q1

q1 q2 q6

q4 q5 q7

q5 q6 q2

q6 q7 q6

q7 q2 q6


q0 q5 q1

q1 q2 q6

q4 q5 q1

q5 q6 q2

q6 q4 q6


Step 5 Repeat step 4 same processAgain Row 1 and Row 3 Similar

q0 q5 q1

q1 q2 q6

q4 q5 q1

q5 q6 q2

q6 q4 q6

q0 q5 q1

q1 q2 q6

q5 q6 q2

q6 q0 q6


Step 6 Combined Set minimized setsNow below minimized Transition table.

δ/ Σ a b

q0 q5 q1

q1 q2 q6

q5 q6 q2

q6 q0 q6

*q2 q2 q0


Transition Diagram of Minimized DFA

Start b b a b a a a b

b a

q6

q0

q2q5

q1


Myhill-Nerode theorem is used to eliminate useless states from a DFA.

Myhill-Nerode Theorem says the three statements are equivalent:

(i) The language L, a subset of Σ*, is accepted by a DFA (ii) L is the union of some of the equivalence classes of right invariant equivalence relation of finite index.(iii) Let equivalence relation RL be defined by: xRLy if and only if for all Z in Σ*, xz is in L exactly when yz is in L. Then RL is of finite index

THANKS

MYHILL-NERODE THEOREM

theory of automata and formal languages unit 1

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