Nondeterministic Finite Automata CSCI 3130 Formal Languages and - - PowerPoint PPT Presentation

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Feb 15, 2024 550 likes •807 views

1/23 Nondeterministic Finite Automata CSCI 3130 Formal Languages and Automata Theory Siu On CHAN Chinese University of Hong Kong Fall 2017 2/23 Example from last lecture with a simpler solution 1 0 0 1 1 0 or 0 1 0 1 1 0 1 0 0

SLIDE 1

1/23

Nondeterministic Finite Automata

CSCI 3130 Formal Languages and Automata Theory Siu On CHAN

Chinese University of Hong Kong

Fall 2017

SLIDE 2

2/23

Example from last lecture with a simpler solution

Construct a DFA over alphabet {0, 1} that accepts all strings ending in 01

qε q0 q1 q00 q01 q10 q11

1 1 1 1 1 1 1

q0 q1 q2

1 1 1

Three weeks later: DFA minimization

SLIDE 3

3/23

Another example from last lecture

Construct a DFA over alphabet {0, 1} that accepts all strings ending in 101

q0 q1 q2 q2

1 1 1 1

SLIDE 4

4/23

String matching DFAs

Ending in 01

q0 q1 q2

1 1 1 Ending in 101

q0 q1 q2 q2

1 1 1 1 Fast string matching algorithms to turn a pattern into a string matching DFA and execute the DFA: Boyer–Moore (BM) and Knuth–Morris–Pratt (KMP) (won’t cover in class)

SLIDE 5

5/23

Nondeterminism

SLIDE 6

6/23

In a few lectures

q0 q1 q2 q2

1 1 1 1 What problems can finite state machines solve? We’ll answer this question in the next few lectures Useful to consider hypothetical machines that are nondeterministic

SLIDE 7

7/23

Even easier with guesses

Suppose we could guess when the input string has only 3 symbols lefu Accept strings ending in 101: 3 symbols lefu 1 1 1 This is not a DFA!

SLIDE 8

8/23

Nondeterministic finite automata

A machine that allows us to make guesses

q0 q1 q2 q3

0,1 1 1 Each state can have zero, one, or more outgoing transitions labeled by the same symbol

SLIDE 9

9/23

Choosing where to go

q0 q1 q2 q3

0,1 1 1 State q0 has two transitions labeled 1 Upon reading 1, we have the choice of staying at q0 or moving to q1

SLIDE 10

10/23

Ability to choose

q0 q1 q2 q3

0,1 1 1 State q1 has no transition labeled 1 Upon reading 1 at q1, die; upon reading 0, continue to q2

SLIDE 11

11/23

Ability to choose

q0 q1 q2 q3

0,1 1 1 State q1 has no transition going out Upon reading 0 or 1 at q3, die

SLIDE 12

12/23

Meaning of NFA

q0 q1 q2 q3

Guess if we are 3 symbols away from end of input If so, guess we will see the pattern 101 Check that we are at the end of input 0,1 1 1

SLIDE 13

13/23

How to run an NFA

q0 q1 q2 q3

0,1 1 1 input: 01101 The NFA can have several active states at the same time NFA accepts if at the end, one of its active states is accepting

SLIDE 14

14/23

Example

Construct an NFA over alphabet {0, 1} that accepts all strings containing the pattern 001 somewhere 11001010, 001001, 111001 should be accepted

ε, 000, 010101

should not

SLIDE 15

15/23

Example

Construct an NFA over alphabet {0, 1} that accepts all strings containing the pattern 001 somewhere

q0 q1 q2 q3

0,1 1 0,1

SLIDE 16

16/23

Definition

A nondeterministic finite automaton (NFA) is a 5-tuple (Q, Σ, δ, q0, F) where

◮ Q is a finite set of states ◮ Σ is an alphabet ◮ δ : Q × (Σ ∪ {ε}) → subsets of Q is a transition function ◮ q0 ∈ Q is the initial state ◮ F ⊆ Q is a set of accepting states

Difgerences from DFA:

◮ transition function δ can go into several states ◮ allows ε-transitions

SLIDE 17

17/23

Language of an NFA

The NFA accepts string x if there is some path that, starting from q0, ends at an accepting state as x is read from lefu to right The language of an NFA is the set of all strings accepted by the NFA

SLIDE 18

18/23

ε-transitions

ε-transitions can be taken for free: q0 q1 q2 ε

a b a accepts a, b, aab, bab, aabab, … rejects

ε, aa, ba, bb, …

SLIDE 19

19/23

Example

q0 q1 q2 ε,1 ε

alphabet Σ = {0, 1} states Q = {q0, q1, q2} initial state q0 accepting states F = {q2} table of transition function δ

inputs

1 ε

states

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

SLIDE 20

20/23

Running NFA

ε

q0 q1 q2 ε,1

00

q0 q1 q2 ε,1

1 2 3

q0 q1 q2 ε,1

1 4 3 2

q0 q1 q2 ε,1

1 3 2 4

SLIDE 21

21/23

Running NFA

001

q0 q1 q2 ε

, 1

4 3 2

101

q0 q1 q2 ε,1

1 3 2

11

q0 q1 q2 ε,1

ε q0 q1 q2 ε,1

SLIDE 22

22/23

Language of this NFA

q0 q1 q2 ε,1 ε

What is the language of this NFA?

SLIDE 23

23/23

Example of ε-transitions

Construct an NFA that accepts all strings with an even number of 0s or an

dd number of 1s

q r r s s

even number of 0s

dd number of 1s

1 1 1 1

Nondeterministic Finite Automata

CSCI 3130 Formal Languages and Automata Theory Siu On CHAN

Fall 2017

Example from last lecture with a simpler solution

Construct a DFA over alphabet {0, 1} that accepts all strings ending in 01

Three weeks later: DFA minimization

Another example from last lecture

Construct a DFA over alphabet {0, 1} that accepts all strings ending in 101

String matching DFAs

Ending in 01

q0 q1 q2

1 1 1 Ending in 101

q0 q1 q2 q2

1 1 1 1 Fast string matching algorithms to turn a pattern into a string matching DFA and execute the DFA: Boyer–Moore (BM) and Knuth–Morris–Pratt (KMP) (won’t cover in class)

Nondeterminism

In a few lectures

q0 q1 q2 q2

1 1 1 1 What problems can finite state machines solve? We’ll answer this question in the next few lectures Useful to consider hypothetical machines that are nondeterministic

Even easier with guesses

Suppose we could guess when the input string has only 3 symbols lefu Accept strings ending in 101: 3 symbols lefu 1 1 1 This is not a DFA!

Nondeterministic finite automata

A machine that allows us to make guesses

q0 q1 q2 q3

0,1 1 1 Each state can have zero, one, or more outgoing transitions labeled by the same symbol

Choosing where to go

q0 q1 q2 q3

0,1 1 1 State q0 has two transitions labeled 1 Upon reading 1, we have the choice of staying at q0 or moving to q1

Ability to choose

q0 q1 q2 q3

0,1 1 1 State q1 has no transition labeled 1 Upon reading 1 at q1, die; upon reading 0, continue to q2

Ability to choose

q0 q1 q2 q3

0,1 1 1 State q1 has no transition going out Upon reading 0 or 1 at q3, die

Meaning of NFA

q0 q1 q2 q3

Guess if we are 3 symbols away from end of input If so, guess we will see the pattern 101 Check that we are at the end of input 0,1 1 1

How to run an NFA

q0 q1 q2 q3

0,1 1 1 input: 01101 The NFA can have several active states at the same time NFA accepts if at the end, one of its active states is accepting

Example

Construct an NFA over alphabet {0, 1} that accepts all strings containing the pattern 001 somewhere 11001010, 001001, 111001 should be accepted

ε, 000, 010101

should not

Example

Construct an NFA over alphabet {0, 1} that accepts all strings containing the pattern 001 somewhere

q0 q1 q2 q3

0,1 1 0,1

Definition

A nondeterministic finite automaton (NFA) is a 5-tuple (Q, Σ, δ, q0, F) where

Difgerences from DFA:

Language of an NFA

The NFA accepts string x if there is some path that, starting from q0, ends at an accepting state as x is read from lefu to right The language of an NFA is the set of all strings accepted by the NFA

ε-transitions

ε-transitions can be taken for free: q0 q1 q2 ε

a b a accepts a, b, aab, bab, aabab, … rejects

ε, aa, ba, bb, …

Example

q0 q1 q2 ε,1 ε

alphabet Σ = {0, 1} states Q = {q0, q1, q2} initial state q0 accepting states F = {q2} table of transition function δ

1

ε

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

Running NFA

ε

00

Running NFA

001

101

11

Language of this NFA

q0 q1 q2 ε,1 ε

What is the language of this NFA?

Example of ε-transitions

Construct an NFA that accepts all strings with an even number of 0s or an

q r r s s

even number of 0s

1 1 1 1

Example of ε-transitions

Construct an NFA that accepts all strings with an even number of 0s or an

q0 r0 r1 s0 s1