NonDeterministic Finite Automata Fall 2005 Costas Busch RPI

Non-Deterministic Finite Automata Fall 2005 Costas Busch - RPI 1

Nondeterministic Finite Automaton (NFA) Alphabet = Fall 2005 Costas Busch - RPI 2

Alphabet = Two choices Fall 2005 Costas Busch - RPI 3

Alphabet = Two choices No transition Fall 2005 Costas Busch - RPI 4

First Choice Fall 2005 Costas Busch - RPI 5

First Choice Fall 2005 Costas Busch - RPI 6

First Choice Fall 2005 Costas Busch - RPI 7

First Choice All input is consumed “accept” Fall 2005 Costas Busch - RPI 8

Second Choice Fall 2005 Costas Busch - RPI 9

Second Choice Fall 2005 Costas Busch - RPI 10

Second Choice No transition: the automaton hangs Fall 2005 Costas Busch - RPI 11

Second Choice Input cannot be consumed “reject” Fall 2005 Costas Busch - RPI 12

An NFA accepts a string: if there is a computation of the NFA that accepts the string i. e. , all the input is consumed and the automaton is in an accepting state Fall 2005 Costas Busch - RPI 13

Example is accepted by the NFA: “accept” because this computation accepts Fall 2005 “reject” Costas Busch - RPI 14

Rejection example Fall 2005 Costas Busch - RPI 15

First Choice Fall 2005 Costas Busch - RPI 16

First Choice “reject” Fall 2005 Costas Busch - RPI 17

Second Choice Fall 2005 Costas Busch - RPI 18

Second Choice Fall 2005 Costas Busch - RPI 19

Second Choice “reject” Fall 2005 Costas Busch - RPI 20

An NFA rejects a string: if there is no computation of the NFA that accepts the string. For each computation: • All the input is consumed and the automaton is in a non final state OR • The input cannot be consumed Fall 2005 Costas Busch - RPI 21

Example is rejected by the NFA: “reject” All possible computations lead to rejection Fall 2005 Costas Busch - RPI 22

Rejection example Fall 2005 Costas Busch - RPI 23

First Choice Fall 2005 Costas Busch - RPI 24

First Choice No transition: the automaton hangs Fall 2005 Costas Busch - RPI 25

First Choice Input cannot be consumed “reject” Fall 2005 Costas Busch - RPI 26

Second Choice Fall 2005 Costas Busch - RPI 27

Second Choice Fall 2005 Costas Busch - RPI 28

Second Choice No transition: the automaton hangs Fall 2005 Costas Busch - RPI 29

Second Choice Input cannot be consumed “reject” Fall 2005 Costas Busch - RPI 30

is rejected by the NFA: “reject” All possible computations lead to rejection Fall 2005 Costas Busch - RPI 31

Language accepted: Fall 2005 Costas Busch - RPI 32

Lambda Transitions Fall 2005 Costas Busch - RPI 33

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(read head does not move) Fall 2005 Costas Busch - RPI 36

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all input is consumed “accept” String Fall 2005 is accepted Costas Busch - RPI 38

Rejection Example Fall 2005 Costas Busch - RPI 39

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(read head doesn’t move) Fall 2005 Costas Busch - RPI 41

No transition: the automaton hangs Fall 2005 Costas Busch - RPI 42

Input cannot be consumed “reject” String Fall 2005 is rejected Costas Busch - RPI 43

Language accepted: Fall 2005 Costas Busch - RPI 44

Another NFA Example Fall 2005 Costas Busch - RPI 45

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“accept” Fall 2005 Costas Busch - RPI 49

Another String Fall 2005 Costas Busch - RPI 50

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“accept” Fall 2005 Costas Busch - RPI 57

Language accepted Fall 2005 Costas Busch - RPI 58

Another NFA Example Fall 2005 Costas Busch - RPI 59

Language accepted (redundant state) Fall 2005 Costas Busch - RPI 60

Remarks: • The symbol never appears on the input tape • Simple automata: Fall 2005 Costas Busch - RPI 61

• NFAs are interesting because we can express languages easier than DFAs NFA Fall 2005 DFA Costas Busch - RPI 62

Formal Definition of NFAs Set of states, i. e. Input aplhabet, i. e. Transition function Initial state Accepting states Fall 2005 Costas Busch - RPI 63

Transition Function Is the result of following exactly one transition Fall 2005 Costas Busch - RPI 64

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Extended Transition Function Fall 2005 Costas Busch - RPI 68

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Formally : there is a walk from with label Fall 2005 Costas Busch - RPI to 71

The Language of an NFA Fall 2005 Costas Busch - RPI 72

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Formally The language accepted by NFA is: where (accepting state) and there is some Fall 2005 Costas Busch - RPI 77

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NFAs accept the Regular Languages Fall 2005 Costas Busch - RPI 79

Equivalence of Machines Definition: Machine is equivalent to machine if Fall 2005 Costas Busch - RPI 80

Example of equivalent machines NFA DFA Fall 2005 Costas Busch - RPI 81

We will prove: Languages accepted by NFAs Regular Languages NFAs and DFAs have the same computation power Fall 2005 Costas Busch - RPI Languages accepted by DFAs 82

We will show: Languages accepted by NFAs Regular Languages Fall 2005 Costas Busch - RPI 83

Proof-Step 1 Languages accepted by NFAs Regular Languages Proof: Every DFA is trivially an NFA Any language accepted by a DFA is also accepted by an NFA Fall 2005 Costas Busch - RPI 84

Proof-Step 2 Languages accepted by NFAs Regular Languages Proof: Any NFA can be converted to an equivalent DFA Fall 2005 Any language accepted by an NFA is also accepted by a DFA Costas Busch - RPI 85

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 86

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 87

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 88

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 89

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 90

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 91

Convert NFA to DFA NFA DFA Fall 2005 Costas Busch - RPI 92

NFA to DFA: Remarks We are given an NFA We want to convert it to an equivalent DFA With Fall 2005 Costas Busch - RPI 93

If the NFA has states the DFA has states in the powerset Fall 2005 Costas Busch - RPI 94

Procedure NFA to DFA 1. Initial state of NFA: Initial state of DFA: Fall 2005 Costas Busch - RPI 95

Example NFA DFA Fall 2005 Costas Busch - RPI 96

Procedure NFA to DFA 2. For every DFA’s state Compute in the NFA Add transition to DFA Fall 2005 Costas Busch - RPI 97

Exampe NFA DFA Fall 2005 Costas Busch - RPI 98

Procedure NFA to DFA Repeat Step 2 for all letters in alphabet, until no more transitions can be added. Fall 2005 Costas Busch - RPI 99

Example NFA DFA Fall 2005 Costas Busch - RPI 100

Procedure NFA to DFA 3. For any DFA state If some is accepting state in NFA Then, is accepting state in DFA Fall 2005 Costas Busch - RPI 101

Example NFA DFA Fall 2005 Costas Busch - RPI 102

Theorem Take NFA Apply conversion procedure to obtain DFA Then Fall 2005 and are equivalent : Costas Busch - RPI 103

Proof AND Fall 2005 Costas Busch - RPI 104

First we show: Take arbitrary: We will prove: Fall 2005 Costas Busch - RPI 105

NFA Fall 2005 Costas Busch - RPI 106

denotes Fall 2005 Costas Busch - RPI 107

We will show that if NFA then DFA Fall 2005 Costas Busch - RPI 108

More generally, we will show that if in : (arbitrary string) NFA then DFA Fall 2005 Costas Busch - RPI 109

Proof by induction on Induction Basis: NFA DFA Is true by construction of Fall 2005 Costas Busch - RPI 110

Induction hypothesis: NFA DFA Fall 2005 Costas Busch - RPI 111

Induction Step: NFA DFA Fall 2005 Costas Busch - RPI 112

Induction Step: NFA DFA Fall 2005 Costas Busch - RPI 113

Therefore if NFA then DFA Fall 2005 Costas Busch - RPI 114

We have shown: We also need to show: (proof is similar and omitted) Fall 2005 Costas Busch - RPI 115