Theoretical Computer Science (Bridging Course) Regular Languages - - PowerPoint PPT Presentation

▶
theoretical computer
SMART_READER_LITE
LIVE PREVIEW

Theoretical Computer Science (Bridging Course) Regular Languages - - PowerPoint PPT Presentation

Jul 01, 2023 380 likes •1.45k views

Theoretical Computer Science (Bridging Course) Regular Languages Gian Diego Tipaldi Topics Covered Regular languages Deterministic finite automata Nondeterministic finite automata Closure Regular expressions Non-regular

slide-1
SLIDE 1

Theoretical Computer Science (Bridging Course)

Gian Diego Tipaldi

Regular Languages

slide-2
SLIDE 2

Topics Covered

  • Regular languages
  • Deterministic finite automata
  • Nondeterministic finite automata
  • Closure
  • Regular expressions
  • Non-regular languages
  • The pumping lemma
slide-3
SLIDE 3

Finite Automata

  • Supermarket door control
slide-4
SLIDE 4

Finite Automata

  • Supermarket door control

Front Pad Rear Pad

slide-5
SLIDE 5

Finite Automata

  • Supermarket door control

Open Closed

Rear Both Neither Rear Both Front Front Neither

slide-6
SLIDE 6

Finite Automata

  • Supermarket door control

Neither Front Rear Both Closed Closed Open Closed Closed Open Closed Open Open Open

slide-7
SLIDE 7

Finite Automata

  • Supermarket door control
  • Probabilistic counterparts exists
  • Markov chains
  • Bayesian networks

Neither Front Rear Both Closed Closed Open Closed Closed Open Closed Open Open Open

slide-8
SLIDE 8

Finite Automata

  • Supermarket door control
  • Probabilistic counterparts exists
  • Markov chains
  • Bayesian networks

Neither Front Rear Both Closed Closed Open Closed Closed Open Closed Open Open Open

slide-9
SLIDE 9

Finite Automata

slide-10
SLIDE 10

Finite Automata

  • States:
  • Alphabet:
  • Transition function: See edges
  • Start state:
  • Final states:
slide-11
SLIDE 11

Finite Automata

Which kind of input is accepted?

  • “aaaabbbbaaaa” ?
  • “000000” ?
  • An empty string?
  • “1000111” ?
slide-12
SLIDE 12

Finite Automata

Which language is accepted?

  • “aaaabbbbaaaa” ?
  • “000000” ?
  • An empty string?
  • “1000111” ?
slide-13
SLIDE 13

Finite Automata

Which language is accepted?

  • “M recognizes A”
  • “A is the language L(M)”
slide-14
SLIDE 14

Finite Automata – Example

  • Which language recognizes M?

q2 q1

1 1

slide-15
SLIDE 15

Finite Automata – Example

  • And in this case?

q2 q1

1 1

slide-16
SLIDE 16

Finite Automata – Example

  • What about this one?
slide-17
SLIDE 17

Finite Automata – Example

slide-18
SLIDE 18

Finite Automata – Example

  • Sums all numerical

symbols that reads, modulo 3.

  • Resets the count,

every time it receives <RESET>.

  • Accepts, if the sum

is a multiple of 3.

slide-19
SLIDE 19

Definition of Computation

slide-20
SLIDE 20

Designing Finite Automata

We want to accept binary strings with an odd number of 1s

slide-21
SLIDE 21

Designing Finite Automata

We want to accept binary strings with an odd number of 1s

  • 1. Design states

q2 q1

slide-22
SLIDE 22

Designing Finite Automata

We want to accept binary strings with an odd number of 1s

  • 1. Design states
  • 2. Design transitions

q2 q1

1 1

slide-23
SLIDE 23

Designing Finite Automata

We want to accept binary strings with an odd number of 1s

  • 1. Design states
  • 2. Design transitions
  • 3. Design start state and accept states

q2 q1

1 1

slide-24
SLIDE 24

Designing Finite Automata

We want to accept binary strings containing 001 as substring

slide-25
SLIDE 25

Designing Finite Automata

We want to accept binary strings containing 001 as substring

  • 1. No symbols of the string

q

1

slide-26
SLIDE 26

Designing Finite Automata

We want to accept binary strings containing 001 as substring

  • 1. No symbols of the string
  • 2. We have a 0

q0 q

1 1

slide-27
SLIDE 27

Designing Finite Automata

We want to accept binary strings containing 001 as substring

  • 1. No symbols of the string
  • 2. We have a 0
  • 3. We have a 00

q0 q

1 1

q00

1

slide-28
SLIDE 28

Designing Finite Automata

We want to accept binary strings containing 001 as substring

  • 1. No symbols of the string
  • 2. We have a 0
  • 3. We have a 00
  • 4. We have a 001

q0 q

1 1

q00

0,1 1

q001

slide-29
SLIDE 29

Regular Operations

Let A and B be languages, we have:

  • Union:
  • Concatenation:
  • Star:
  • Example
slide-30
SLIDE 30

Closure of Regular Languages

slide-31
SLIDE 31

Proof by Construction

slide-32
SLIDE 32

Proof by Construction

slide-33
SLIDE 33

Example

  • L(M1) = {w|w contains a 1}
  • L(M2) = {w|w contains at least two 0s}

q1 q2 0,1 1 p2 p1 p3 1 0,1 1

slide-34
SLIDE 34

Example

  • L(M1) = {w|w contains a 1}
  • L(M2) = {w|w contains at least two 0s}

q1 q2 0,1 1 p2 p1 p3 1 0,1 1

q1 p1 q1 p2 q1 p3 q2 p1 q2 p2 q2 p3

slide-35
SLIDE 35

Example

  • L(M1) = {w|w contains a 1}
  • L(M2) = {w|w contains at least two 0s}

q1 q2 0,1 1 p2 p1 p3 1 0,1 1

q1 p1 q1 p2 q1 p3 q2 p1 q2 p2 q2 p3

1 1 1 0,1 1 1

slide-36
SLIDE 36

Example

  • L(M1) = {w|w contains a 1}
  • L(M2) = {w|w contains at least two 0s}

q1 q2 0,1 1 p2 p1 p3 1 0,1 1

q1 p1 q1 p2 q1 p3 q2 p1 q2 p2

1 1 1 0,1 1 1

q2 p3

slide-37
SLIDE 37

Closure of Regular Languages

slide-38
SLIDE 38

Closure of Regular Languages

  • Non deterministic

finite automata

slide-39
SLIDE 39

Nondeterministic Automata

  • Deterministic (DFA)
  • One successor state
  • 𝜁 transitions not allowed
  • Nondeterministic (NFA)
  • Several successor states possible
  • 𝜁 transitions possible

q2 q1 q3 q4

0,1 0,1 1 0,ε 1

slide-40
SLIDE 40

Nondeterministic Computation

slide-41
SLIDE 41

Example Run

q1 q1 q3 q2 q1 q3 q1 q2 q1 q3 q4 q4 q4 q2 q1 q3 q3 q1 q4 q4

1 1 1

q2 q1 q3 q4

0,1 0,1 1 0,ε 1

Input: w = 010110

NFA 𝑂1

slide-42
SLIDE 42

Which language is accepted?

slide-43
SLIDE 43

Nondeterministic Automata

slide-44
SLIDE 44

Nondeterministic Automata

slide-45
SLIDE 45

Definition of computation

slide-46
SLIDE 46

A NFA has an equivalent DFA

NFA recognizing language 𝐵 DFA recognizing language 𝐵

slide-47
SLIDE 47

Equivalence NFA and DFA

Theorem 1.39: Every nondeterministic finite automaton has an equivalent deterministic finite automaton. Corollary 1.40: A language is regular if and only if some nondeterministic finite automaton recognizes it.

slide-48
SLIDE 48

Proof: Theorem 1.39

slide-49
SLIDE 49

Proof: Theorem 1.39

slide-50
SLIDE 50

Proof: Theorem 1.39

slide-51
SLIDE 51

Proof: Theorem 1.39 (ctd.)

slide-52
SLIDE 52

Example

  • Consider the following NFA
  • What is the corresponding DFA?
slide-53
SLIDE 53

Example

  • Resulting DFA for the example before
slide-54
SLIDE 54

Example

  • Simplified DFA for the example before
slide-55
SLIDE 55

Closure of Regular Operations

slide-56
SLIDE 56

Closure of Regular Operations

  • Regular languages are closed under

the union operation

slide-57
SLIDE 57

Proof

slide-58
SLIDE 58

Closure of Regular Operations

  • Regular languages are closed under

the concatenation operation

slide-59
SLIDE 59

Proof

slide-60
SLIDE 60

Closure of Regular Operations

  • Regular languages are closed under

the star operation

slide-61
SLIDE 61

Proof

slide-62
SLIDE 62

Regular Expressions

slide-63
SLIDE 63

Regular Expressions – Examples

slide-64
SLIDE 64

Regular Expressions – Examples

slide-65
SLIDE 65

Applications of Regular Expressions

  • Design of compilers
  • Search for strings (awk, grep, …)
  • Programming languages
  • Bioinformatics (repetitive patterns)
slide-66
SLIDE 66

Equivalence of RE and NFA

Theorem 1.54 (page 66): A language is regular if and only if some regular expression describes it.

slide-67
SLIDE 67

Equivalence of RE and NFA

Theorem 1.54 (page 66): A language is regular if and only if some regular expression describes it. Two directions to consider RE <-> NFA

slide-68
SLIDE 68

Equivalence of RE and NFA

Lemma 1.55 (page 67): If a language is described by some regular expression, then it is regular. Lemma 1.60 (page 69): If a language is regular, then it can be described by some regular expression.

slide-69
SLIDE 69

Proof RE -> NFA

slide-70
SLIDE 70

Proof RE -> NFA: Case 1

  • a
slide-71
SLIDE 71

Proof RE -> NFA: Cases 2 & 3

slide-72
SLIDE 72

Proof RE -> NFA: Case 4, 5 & 6

slide-73
SLIDE 73

Example

Let consider the expression (ab U a)*

  • Convert the expression into a NFA
  • Start from the smallest subexpression
  • Include the other subexpressions
  • Note: The states might be redundant!
slide-74
SLIDE 74

Example: (ab U a)*

  • a
  • b
  • ab
  • ab U a
  • (ab U a)*

a b a b ε a b ε a ε ε a b ε a ε ε ε ε ε

slide-75
SLIDE 75

Exercise: (ab U a)*

  • Let’s do it together!
slide-76
SLIDE 76

Exercise: (a U b)*aba

a b ε ε ε a b ε a ε a b ε a ε ε ε ε

slide-77
SLIDE 77

Proof NFA -> RE

Lemma 1.60 (page 69): If a language is regular, then it can be described by a regular expression. Two steps:

  • Convert DFA into GNFA
  • Convert GNFA into regular expression
slide-78
SLIDE 78

Generalized NFA

  • Labels are regular expressions
  • States connected in both directions
  • Start state only exit transitions
  • Accept state only incoming transitions
  • Only one accept state
slide-79
SLIDE 79

Generalized NFA

qstart qaccept b ab Ø b* ab* ab∪ba a* (aa)* aa

slide-80
SLIDE 80

Generalized NFA

slide-81
SLIDE 81

Generalized NFA

slide-82
SLIDE 82

Proof DFA -> GNFA

  • Add a new start state
  • Connect it with 𝜁 transitions
  • Add a new accept state
  • Connect it with 𝜁 transitions
  • Replace multiple labels with unions
  • Add transitions with ∅ when not present

in the original DFA

slide-83
SLIDE 83

Proof DFA -> GNFA

  • DFA
  • GNFA
slide-84
SLIDE 84

Convert GNFA into RE

3 state DFA 5 state GNFA 4 state GNFA 2 state GNFA 3 state GNFA Regular Expression

slide-85
SLIDE 85

Convert GNFA into RE

slide-86
SLIDE 86

Ripping of States

Replace one state with the corresponding regular expression

q2 q1 q2 qrip q1 R4 R1 R2 R3 (R1)(R2)* (R3) ∪ R4

slide-87
SLIDE 87

Example: From DFA to GNFA

slide-88
SLIDE 88

Example: Rip State 2

slide-89
SLIDE 89

Example: Rip State 1

slide-90
SLIDE 90

Another Example

1 2

a a

3

b b a b

s 1 2

a a

3

b b a b

ε

a

ε ε

s 2 3 a

ε ε

a aa ∪b ab b ba ∪a bb

Rip 1: Rip 2: GNFA: DFA: s 3 a

a(aa ∪b)* a(aa ∪b)*ab ∪b (ba ∪a) (aa ∪b)* ∪ε (ba ∪a) (aa ∪b)*ab ∪ bb

s a Rip 3:

(a(aa ∪b)*ab ∪b)((ba ∪a) (aa ∪b)*ab ∪ bb)*((ba ∪a) (aa ∪b)* ∪ε) ∪a(aa ∪b)*

slide-91
SLIDE 91

Equivalence Proof

slide-92
SLIDE 92

Equivalence Proof

slide-93
SLIDE 93

Equivalence Proof

  • q2

q1 q2 qrip q1 R4 R1 R2 R3 (R1)(R2)* (R3) ∪ R4

slide-94
SLIDE 94

Nonregular Languages

  • Finite automata have finite memory
  • Are the following language regular?
  • How can we prove it mathematically?

{0 1 | 0} { | h as an eq u al n u m b er o f 0 s an d 1 s} { | h as an eq u al n u m b er o f o ccu ren ces o f 0 1 an d 1 0 }

n n

B n C w w D w w    

slide-95
SLIDE 95

The Pumping Lemma

slide-96
SLIDE 96

Proof Idea

  • Let M be a DFA recognizing A
  • Let p be the numbers of states in M
  • Show that s can be broken into xyz
  • Prove the conditions holds
slide-97
SLIDE 97

Proof Idea

  • Let M be a DFA recognizing A
  • Let p be the numbers of states in M
  • Show that s can be broken into xyz
  • Prove the conditions holds
slide-98
SLIDE 98

Proof Idea

  • Let M be a DFA recognizing A
  • Let p be the numbers of states in M
  • Show that s can be broken into xyz
  • Prove the conditions holds
slide-99
SLIDE 99

Proof of the Pumping Lemma

slide-100
SLIDE 100

Use of the Pumping Lemma

slide-101
SLIDE 101

Nonregular Languages

slide-102
SLIDE 102

Nonregular Languages

slide-103
SLIDE 103

Nonregular Languages

slide-104
SLIDE 104

Nonregular Languages

slide-105
SLIDE 105

Example Exam Question

slide-106
SLIDE 106

Summary

  • Deterministic finite automata
  • Regular languages
  • Nondeterministic finite automata
  • Closure operations
  • Regular expressions
  • Nonregular languages
  • The pumping lemma