Theory of Automata and formal languages unit 1

THEORY OF AUTOMATAAND
FORMAL LANGUAGES
UNIT-I
ABHIMANYU MISHRA
ASSISTANT PROF.(CSE)
JETGI
Abhimanyu Mishra(CSE) JETGI12/31/2016

What is Automata?
AUTOMATA
An Automation derived from the Greek word which means "self-acting". An
automaton (Automata in plural) is an theoretical self-propelled
computing 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 q1
q2

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 NFA
4) 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

Solution
Step 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

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 }

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 ∈ ∈
Abhimanyu Mishra(CSE) JETGI
q0 q1
q2
12/31/2016

Example:2
Consider the NFA with ∈-transition M =(Q,Σ, δ,q0,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 q1
q2
12/31/2016

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(q0) = {q0, q1, q2 } (New Initial state for NFA
without ∈ transition)
(ii) Rest of the states are:
∈-closure(q1) ={q1, q2 } New State
∈-closure(q2) ={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)
= {q1, 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
12/31/2016

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
12/31/2016

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
12/31/2016

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
12/31/2016

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
12/31/2016

q0 q5 q1
q1 q2 q6
q4 q5 q1
q5 q6 q2
q6 q4 q6
12/31/2016

Step 5 Repeat step 4 same process
Again 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
12/31/2016

Step 6 Combined Set minimized sets
Now 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
12/31/2016

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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