Covers the topic - Finite Automata, Conversion of NFA to DFA, ENFA to DFA

1. THEORY OF
COMPUTATION
Sub Code: CS8501
Credit : 3
Mrs.D.Jena Catherine Bel
Assistant Professor, CSE
Velammal Engineering College

2. Course Objective
The student should be made to:
• To understand the language hierarchy
• To construct automata for any given pattern and find its
equivalent regular expressions
• To design a context free grammar for any given language
• To understand Turing machines and their capability
• To understand undecidable problems and NP class problems
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
2

3. Theory of Computation
• What?
• The theory of computation is a branch of computer science and
mathematics combined
• deals with how efficiently problems can be solved on a model of
computation, using an algorithm.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
3

4. Application
• Where?
• Text processing
• Compilers
• Programming Languages
• Artificial Intelligence
• Genetic programming
• Neural Networks
• Robotics
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
4

5. Models
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
5

6. Models
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
6

7. UNIT I
AUTOMATA FUNDAMENTALS
Introduction to formal proof – Additional forms of Proof –
Inductive Proofs –Finite Automata – Deterministic Finite
Automata – Non-deterministic Finite Automata – Finite Automata
with Epsilon Transitions
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
7

8. UNIT II
REGULAR EXPRESSIONS AND LANGUAGES
Regular Expressions – FA and Regular Expressions – Proving
Languages not to be regular – Closure Properties of Regular
Languages – Equivalence and Minimization of Automata.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
8

9. UNIT III
CONTEXT FREE GRAMMAR AND LANGUAGES
CFG – Parse Trees – Ambiguity in Grammars and Languages –
Definition of the Pushdown Automata – Languages of a
Pushdown Automata – Equivalence of Pushdown Automata and
CFG, Deterministic Pushdown Automata.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
9

10. UNIT IV
PROPERTIES OF CONTEXT FREE LANGUAGES
Normal Forms for CFG – Pumping Lemma for CFL – Closure
Properties of CFL – Turing Machines – Programming Techniques
for TM.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
10

11. UNIT V
UNDECIDABILITY
Non Recursive Enumerable (RE) Language – Undecidable Problem
with RE – Undecidable Problems about TM – Post‘s
Correspondence Problem, The Class P and NP.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
11

12. Course Outcome
At the end of the course, the students will be able to:
1. Design of Finite State Machine for any pattern (Unit 1)
2. Design of regular expression for any pattern and analyze the
properties of regular languages (Unit 2)
3. Write Context free grammar for any construct and design of
Pushdown Automata (Unit 3)
4. Apply Normal forms of CFG and analyze the properties of CFG
(Unit 4)
5. Design Turing machines for any language and Propose
computation solutions using Turing machines(Unit 4)
6. Derive whether a problem is decidable or not.(Unit 5)
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
12

13. Text Books
1. J.E.Hopcroft, R.Motwani and J.D Ullman, ―Introduction to
Automata Theory, Languages and Computations, Second
Edition, Pearson Education, 2003.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
13

14. Reference Books
1. H.R.Lewis and C.H.Papadimitriou, ―Elements of the theory
of Computation, Second Edition, PHI, 2003.
2. J.Martin, ―Introduction to Languages and the Theory of
Computation, Third Edition, TMH, 2003.
3. Micheal Sipser, ―Introduction of the Theory and
Computation, Thomson Brokecole, 1997.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
14

15. Terminologies
• Alphabet
• Finite, non empty set of symbols
• Basic elements of a language
• Denoted by ∑
• String
• Finite sequence of symbols chosen from some alphabets
• Empty string
• Length of the string
• Power of an alphabet
• Language
• Set of all strings which are chosen from ∑*
15
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

16. Example
• English
• Alphabet – [a-z]
• String –{hi,hello,…}
• Binary number
• Alphabet – [0,1]
• String –{0,1,00,01,10,11…}
• Hexadecimal
• Alphabet –[0-9][a-e]
• String –[0,1,1A3,…]
16
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

17. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
17

18. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
18

19. UNIT I
INTRODUCTION TO AUTOMATA
Introduction to formal proof – Deductive proofs–Additional
forms of Proof – Contrapositive – Proof by contradictions –
Counterexamples –Inductive Proofs–Deterministic Finite
Automata – Non-deterministic Finite Automata – Finite Automata
with Epsilon Transitions – Formal Definition – Extended Transition
function – Language of Finite Automata – Interconversion of
Finite automata
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
19

20. Introduction to formal proof –
Additional forms of Proof –
Inductive Proofs

21. Introduction to Formal Proof
• Deductive Proof
• Reduction to definition
• Other theorem forms
• Theorem that appear not to be if-then statements
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
21

22. Deductive Proof
• A deductive proof consists of a sequence of statements whose
truth leads us from some initial statement called the
hypothesis or the given statement(s) to a conclusion
statement.
• Each step in the proof must follow by some accepted logical
principle from either the given facts or some of the previous
statement
• The hypothesis may be true or false, depending on the value
of its parameter
• If H then C . C is deducted from H
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
22

23. • Hypothesis : x ≥ 4
• Conclusion : 2x ≥ x2
• Parameter : x
• Proof:
• If x=3, then 23 ≥ 32  8 ≥ 9 which is false
• If x=4 then 24 ≥ 42  16 ≥ 16 which is true
• For each time x increments by 1, the LHS get incremented by 2
and the RHS
𝑥+1
𝑥
2
• If x ≥ 4, the
𝑥+1
𝑥
= 1.25, (1.25)2= 1.5625
• 1.5625 is less than 2
• Hence 2x ≥ x2 will be true for x ≥ 4
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
23

24. If x is the sum of the squares of four positive
integers, then 2x ≥ x2
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
24

25. Reduction to Definitions
• Convert all the terms in hypothesis to their definitions
• Deductive proof
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
25

26. Let S be a finite subset of some infinite
set U. Let T be the complement of S with
respect to U. Then T is infinite
• Let us assume T is finite, |T| = m
• As per the given statement, S is finite => |S|=n
• Then |S U T | = n+ m, which is finite which contradicts the
given statement, Hence T should be infinite.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
26

27. Other Theorem Forms
• Ways of saying If-Then
• H implies C
• H only if C
• C if H
• Whenever H holds, C follows
• If and only if statement
• A if and only if B
• If part : If B then A
• Only if part: If A then B
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
27

28. Theorem that appear not to be
If- Then Statement
• Example
𝑎
2
+ 𝑏
2
= 𝑐
2
sin 𝛼 ± sin 𝛽 = 2 sin
1
2
𝛼 ± 𝛽 cos
1
2
𝛼 ∓ 𝛽
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
28

29. Additional Forms of Proof
• Proofs by sets
• Proof by Contrapositive
• Proofs by contradiction
• Proofs by counterexample
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
29

30. Proofs by sets
R U (S ∩ T) = (R U S) ∩ (R U T)
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
30

31. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
31

32. Proof by Contrapositive
• The contrapositive of a statement negates the conclusion as
well as the hypothesis. It is logically equivalent to the original
statement asserted. Often it is easier to prove the
contrapositive than the original statement.
• If H then C is equivalent to If not C then not H
• Example:
• If x ≥ 4 then 2x ≥ x2
• If not 2x ≥ x2 then not x ≥ 4
• If 2x < x2 then x < 4
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
32

33. Proofs by contradiction
• The method of proof by contradiction is to assume that a
statement is not true and then to show that that assumption
leads to a contradiction.
• To prove if H then C is to prove If H then not C implies
falsehood.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
33

34. Proofs by counterexample
• A proof by counterexample is not technically a proof. It is
merely a way of showing that a given statement cannot
possibly be correct by showing an instance that contradicts a
universal statement.
• If integer x is a prime then x is odd
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
34

35. Inductive Proof
• Induction on integers
• Structural induction
• Mutual induction
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
35

36. Induction on integers
• Mathematical Induction is a mathematical technique which is
used to prove a statement, a formula or a theorem is true for
every natural number.
• The technique involves two steps to prove a statement, as
stated below −
• Step 1(Base step) − It proves that a statement is true for the
initial value.
• Step 2(Inductive step) − It proves that if the statement is true
for the nth iteration (or number n), then it is also true for
(n+1)th iteration ( or number n+1).
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
36

37. 1 + 2 + ... + n = n(n+1)/2
Proof. (Proof by Mathematical Induction)
Let's let P(n) be the statement "1 + 2 + ... + n = (n (n+1)/2."
Basis Step.
P(1) asserts "1 = 1(2)/2", which is clearly true. So we are done
with the initial step.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
37

38. Inductive Step.
Induction Hypothesis/ Inductive assumption:
Assume, 1 + 2 + ... + k = k (k+1)/2 is true
Prove for k+1,
(i.e) 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2.
1 + 2 + ... + k + (k+1)
= k(k+1)/2 + (k+1)
= (k(k+1) + 2 (k+1))/2
= (k+1)(k+2)/2.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
38

39. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
39

40. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
40

41. Structural Induction
• Structural induction is a proof methodology similar to
mathematical induction, only instead of working in the
domain of positive integers (N) it works in the domain of such
recursively defined structures!
• It is terrifically useful for proving properties of such structures.
• Its structure is sometimes “looser” than that of mathematical
induction.
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
41

42. Everytree has one more nodethan its edges
• If T is a tree and T has n nodes and e edges then n=e+1
• Basis:
• T is single node tree, then n=1, e=0, so n=e+1 holds true
• Inductive Hypothesis:
• Assume the statement S(Ti ) hold for i=1,2,3,…,K and Ti have ni
nodes and ei edges then ni = ei + 1
• Induction
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
42

43. Mrs.D.Jena
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AP/CSE,
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44. Every expressionhasanequalnumberofleftand
right parentheses
• Let G is an expression
• Basis :
• G is a number or variable, so the number of left and right
parenthesis is 0
• Inductive Hypothesis:
• Assume E and F are two expressions which has equal number of
left and right parentheses.
• Induction :
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
44

45. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
45

46. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
46

47. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
47

48. Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
48

49. Finite Automata
• Finite automata are used to recognize patterns.
• It takes the string of symbol as input and changes its state
accordingly. When the desired symbol is found, then the
transition occurs.
• At the time of transition, the automata can either move to the
next state or stay in the same state.
• Finite automata have two states, Accept state or Reject state.
When the input string is processed successfully, and the
automata reached its final state, then it will accept.
49
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

50. • Input Tape
• It is a linear tape having some number of cells. Each input symbol is
placed in each cell.
• Finite control
• It decides the next state on receiving particular input from input
tape.
• Tape reader
• It reads the cells one by one from left to right, and at a time only
one input symbol is read.
50
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

51. • A finite automaton consists of:
• a finite set S of N states
• a special start state
• a set of final (or accepting) states
• a set of transitions T from one state to another, labelled with
chars in C
51
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

52. • Execution of FA on an input sequence as follows:
• Begin in the start state
• If the next input char matches the label on a transition from the
current state to a new state, go to that new state
• Continue making transitions on each input char
• If no move is possible, then stop
• If in accepting state, then accept
52
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

53. Deterministic Finite Automata
• Deterministic refers to the uniqueness of the computation.
• On each input there is one and only one state to which the
automaton can transition from its current state
• DFA does not accept the null move.
53
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

54. Formal Definition of DFA
• A deterministic finite automaton (DFA) is a 5-tuple
(Q,Σ,δ,q0,F),where
• Q is a finite set called the states,
• Σ is a finite set called the alphabet,
• δ:Q×Σ→Q is the transition function,
• q0 ∈ Q is the start state, and
• F ⊆ Q is the set of accepting states.
54
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

55. Transition Table
• A transition table is a tabular representation of the transition
function that takes two arguments and returns a state.
• The column contains the state in which the automaton will be
on the input represented by that column.
• The row corresponds to the state the finite control unit can be
in.
• The entry for one row corresponding to state q and the
column corresponds to input a is the state δ(q, a).
55
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

56. Transition Diagram
• Transition graph can be interpreted as a flowchart for an
algorithm recognizing a language.
• A transition graph consists of three things:
• A finite set of states, at least one of which is designated the start
state and some of which are designated as final states.
• An alphabet Σ of possible input symbols from which the input
strings are formed.
• A finite set of transitions that show the change of state from the
given state on a given input.
56
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

57. Example DFA
57
states A b
q0 q1 q2
q1 q1 q3
q2 q2 q3
*q3 q3 q3
A=({q0,q1,q2,q3},{a,b}δ,q0,{q3})
δ is given by
δ(q0,a)=q1
δ(q0,b)=q2
δ(q1,a)=q1
δ(q2,b)=q2
δ(q1,b)=q3
δ(q2,a)=q3
δ(q3,a)=q3
δ(q3,b)=q3
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

58. • Design a DFA with ∑ = {0, 1} accepts those string which starts
with 1 and ends with 0.
• Design a DFA with ∑ = {0, 1} accepts the only input 101.
58
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

59. • Design a DFA with ∑ = {0, 1} accepts the strings with an even
number of 0's end by single 1.
59
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

60. Extended transition function δ
• The DFA define a language: the set of all strings that result in a
sequence of state transitions from the start state to an accepting
state
• Extended Transition Function
• Describes what happens when we start in any state and follow any
sequence of inputs
• If δ is our transition function, then the extended transition function is
denoted by δ
• The extended transition function is a function that takes a state q and
a string w and returns a state p (the state that the automaton
reaches when starting in state q and processing the sequence of
inputs w)
• Let w=va then
δ(q, va) = δ(δ (q, v), a). 60
Mrs.D.Jena
Catherine
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AP/CSE,
VEC

61. Language accepted by DFA
• The language of a DFA A = (Q, Σ, δ, q0, F), denoted L(A) is
defined by
L(A) = { w | δ(q0, w) is in F }
• The language of A is the set of strings w that take the start
state q0 to one of the accepting states
• If L is a L(A) from some DFA, then L is a regular language
61
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

62. Nondeterministic Finite
Automata
• An NFA is like a DFA, except that it can be in several states at
once.
• This can be seen as the ability to guess something about the
input.
• Useful for searching texts
62
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

63. Formal Definition of NFA
• A nondeterministic finite automaton (NFA) is a 5-tuple
(Q,Σ,δ,q0,F),where
• Q is a finite set called the states,
• Σ is a finite set called the alphabet,
• δ:Q×Σ→P(Q) is the transition function,
• q0 ∈ Q is the start state, and
• F ⊆ Q is the set of accepting states.
63
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

64. Extended transition function δ
• The extended transition function is a function that takes a
state q and a string w and returns a set of states P (The set of
possible state that the automaton reaches when starting in
state q and processing the sequence of inputs w)
• Let w=va then
δ 𝑞0, 𝑣𝑎 = 𝑞′∈ δ 𝑞0,𝑣
δ( 𝑞′,a)
64
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

65. Language accepted by NFA
• The language L(A) accepted by the NFA A is defined as follows:
L(A) = {w | δ(q0, w) ∩ F ≠ ∅}
65
Mrs.D.Jena
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AP/CSE,
VEC

66. Example NFA
66
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67. • Design an NFA with ∑ = {0, 1} in which double '1' is followed by
double '0'.
• Design an NFA with ∑ = {0, 1} accepts all string in which the
third symbol from the right end is always 0.
67
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

68. Epsilon Nondeterministic Finite
Automata
• Formal Definition
A ε-NFA is a quintuple A=(Q,Σ,δ,q0,F) where
• Q is a set of states
• Σ is the alphabet of input symbols
• ε is never a member of Σ. Σε is defined to be (Σ ∪ ε)
• δ: Q × Σε → P(Q) is the transition function
• q0 ∈ Q is the initial state
• F ⊆ Q is the set of final states
68
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

69. Example ε-NFA
ε-NFA for a language which contain Os followed by 0 or more 1s.
69
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VEC

70. ε- closure
• Epsilon means present state can goto other state without any
input. This can happen only if the present state have epsilon
transition to other state.
• Epsilon closure is finding all the states which can be reached
from the present state on one or more epsilon transitions.
ε- closure (0)={0,1,2}
ε- closure(1)={1,2}
ε- closure(2)={2}
70
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

71. ε-closure of state a
71
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72. ε-closure of state 0,1,2,3,4
72
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73. Mrs.D.Jena
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73

74. δ - ε-NFA
δ 𝑞0, 𝑣𝑎 = 𝑖=1
𝑘
δ(𝑝𝑖, 𝑎) = 𝑗=1
𝑚
𝐸𝑐𝑙𝑜𝑠𝑒(rj)
Where δ (q0,v)= {p1,p2,…,pk } &
𝑖=1
𝑘
δ(𝑝𝑖, 𝑎)= {r1,r2,…,rm }
74
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

75. Language accepted by ε-NFA
The language of an ε-NFA E = (Q, Σ, δ, q0, F) is
L(E) = {w | ˆ δ(q0, w) ∩ F ≠ ∅}
75
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

76. Relationship between FAs
76
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77. • Every DFA is NFA but not vice versa.
• Both NFA and DFA have same power and each NFA can be
translated into a DFA.
• There can be multiple final states in both DFA and NFA.
• NFA is more of a theoretical concept.
• DFA is used in Lexical Analysis in Compiler.
• Every NFA is ε-NFA but not vice-versa
77
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

78. Conversion of NFA to DFA
• Subset Construction Method
1. Create state table from the given NFA.
2. Create a blank state table under possible input alphabets for the
equivalent DFA.
3. Mark the start state of the DFA by q0 (Same as the NFA).
4. Find out the combination of States {Q0, Q1,... , Qn} for each
possible input alphabet.
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.
6. The states which contain any of the final states of the NFA are
the final states of the equivalent DFA.
78
Mrs.D.Jena
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Bel,
AP/CSE,
VEC

79. NFA
79
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VEC

80. Subset construction
80
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VEC

81. Transition Diagram for the
subset
81
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VEC

82. Converted DFA
82
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83. NFA  DFA
State 0 1
→q0 q0 q1
q1 {q1, q2} q1
*q2 q2 {q1, q2}
83
Mrs.D.Jena
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84. Now we will obtain δ' transition for state q0.
δ'([q0], 0) = [q0]
δ'([q0], 1) = [q1]
The δ' transition for state q1 is obtained as:
δ'([q1], 0) = [q1, q2] (new state generated)
δ'([q1], 1) = [q1]
84
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

85. Now we will obtain δ' transition on [q1, q2].
δ'([q1, q2], 0) = δ(q1, 0) ∪ δ(q2, 0)
= {q1, q2} ∪ {q2}
= [q1, q2]
δ'([q1, q2], 1) = δ(q1, 1) ∪ δ(q2, 1)
= {q1} ∪ {q1, q2}
= {q1, q2}
= [q1, q2]
The state [q1, q2] is the final state as well because
it contains a final state q2. 85
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

86. State 0 1
→[q0] [q0] [q1]
[q1] [q1, q2] [q1]
*[q1, q2] [q1, q2] [q1, q2]
The transition table for the constructed DFA will
be:
86
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

87. Conversion of ε-NFA to DFA
• Modified subset construction
1. Find the ε-closure for the starting state of ε - NFA as a starting
state of DFA.
2. Find the states for each input symbol that can be traversed from
the present. That means the union of transition value and their
closures for each state of NFA present in the current state of
DFA.
3. If a new state is found, take it as current state and repeat step
2.
4. Repeat Step 2 and Step 3 until there is no new state present in
the transition table of DFA.
5. Mark the states of DFA as a final state which contains the final
state of ε-NFA. 87
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

88. Example
• Let us obtain ε-closure of each state.
ε-closure {q0} = {q0, q1, q2}
ε-closure {q1} = {q1}
ε-closure {q2} = {q2}
ε-closure {q3} = {q3}
ε-closure {q4} = {q4}
88
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

89. • let ε-closure {q0} = {q0, q1, q2} be state A
δ'(A, 0) = ε-closure {δ((q0, q1, q2), 0) }
= ε-closure {δ(q0, 0) ∪ δ(q1, 0) ∪ δ(q2, 0) }
= ε-closure {q3}
= {q3} call it as state B.
δ'(A, 1) = ε-closure {δ((q0, q1, q2), 1) }
= ε-closure {δ((q0, 1) ∪ δ(q1, 1) ∪ δ(q2, 1)}
= ε-closure {q3}
= {q3} = B.
89
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

90. δ'(B, 0) = ε-closure {δ(q3, 0) }
= ϕ
δ'(B, 1) = ε-closure {δ(q3, 1) }
= ε-closure {q4}
= {q4} i.e. state C
δ'(C, 0) = ε-closure {δ(q4, 0) }
= ϕ
δ'(C, 1) = ε-closure {δ(q4, 1) }
= ϕ
90
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

91. ε-closure(q0) = {q0, q1, q2}
ε-closure(q1) = {q1, q2}
ε-closure(q2) = {q2}
91
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

92. δ'(A, 0) = ε-closure{δ((q0, q1, q2), 0)}
= ε-closure{δ(q0, 0) ∪ δ(q1, 0) ∪ δ(q2, 0)}
= ε-closure{q0}
= {q0, q1, q2}
δ'(A, 1) = ε-closure{δ((q0, q1, q2), 1)}
= ε-closure{δ(q0, 1) ∪ δ(q1, 1) ∪ δ(q2, 1)}
= ε-closure{q1}
= {q1, q2} call it as state B
δ'(A, 2) = ε-closure{δ((q0, q1, q2), 2)}
= ε-closure{δ(q0, 2) ∪ δ(q1, 2) ∪ δ(q2, 2)}
= ε-closure{q2}
= {q2} call it state C
92
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

93. δ'(B, 0) = ε-closure{δ((q1, q2), 0)}
= ε-closure{δ(q1, 0) ∪ δ(q2, 0)}
= ε-closure{ϕ}
= ϕ
δ'(B, 1) = ε-closure{δ((q1, q2), 1)}
= ε-closure{δ(q1, 1) ∪ δ(q2, 1)}
= ε-closure{q1}
= {q1, q2} i.e. state B itself
δ'(B, 2) = ε-closure{δ((q1, q2), 2)}
= ε-closure{δ(q1, 2) ∪ δ(q2, 2)}
= ε-closure{q2}
= {q2} i.e. state C itself
93
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

94. δ'(C, 0) = ε-closure{δ(q2, 0)}
= ε-closure{ϕ}
= ϕ
δ'(C, 1) = ε-closure{δ(q2, 1)}
= ε-closure{ϕ}
= ϕ
δ'(C, 2) = ε-closure{δ(q2, 2)}
= {q2}
94
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC

95. ENFA to DFA
Mrs.D.Jena
Catherine
Bel,
AP/CSE,
VEC
95