( M ) { abba } = M Take DFA M a , b Definition: L M ( ) - PDF document

Review • Languages and Grammars CS 301 - Lecture 2 – Alphabets, strings, languages • Deterministic Finite Automata (DFA) Regular Languages and – Definition and Use Nondeterministic Finite • Today: Automata – Regular Languages – Nondeterministic Finite State Automata Fall 2008 – Equivalence with DFA Example Languages Accepted by DFAs L ( M ) { abba } = M • Take DFA M a , b • Definition: L M ( ) – The language contains all input q M strings accepted by 5 a , b a b a b a a b b q q q q q L ( M ) M 0 1 2 3 4 – = { strings that drive to a final state} accept 1

Formally Another Example L M , ab , abba ( ) { } = λ M Q , , , q , F • For a DFA ( ) = Σ δ 0 M a , b M • Language accepted by : L ( M ) { w * : * ( q , w ) F } = ∈ Σ δ ∈ 0 q 5 a , b a b a b a b b a w q q q q q 0 1 2 3 4 q q ′ q ∈ F ′ 0 accept accept accept Observation More Examples n L ( M ) { a b : n 0 } = ≥ M • Language rejected by : L M w * : * q , w F ( ) { ( ) } = ∈ Σ δ ∉ a , b a 0 w q q ′ q ∉ F ′ 0 a , b b q q q 0 1 2 trap state accept 2

L M L M ( ) = { all strings with prefix } ab = { all strings without ( ) 001 substring } a , b 1 0 0 , 1 a b q q q 0 1 2 1 a accept b 0 1 0 00 001 λ q a , b 0 3 Examples of regular languages: Regular Languages { abba } , ab , abba { } λ L M • A language is regular if there is a DFA a n such that { b : n 0 } ≥ L = L ( M ) ab { all strings with prefix } 001 { all strings without substring } There exist automata that accept these • All regular languages form a language family Languages 3

There exist languages which are not Regular: Another Example L { awa : w { a , b } * } = ∈ n n Example: L { a b : n 0 } a = ≥ b b There is no DFA that accepts such a language a q q q 0 2 3 b a (we will prove this later in the semester) q L = L ( M ) 4 a , b Nondeterministic Finite Accepter (NFA) Nondeterministic Finite Accepter (NFA) Alphabet = { a } Alphabet = { a } a Two choices a q q q q 1 2 1 2 a a q q 0 0 a a q q 3 3 4

Nondeterministic Finite Accepter (NFA) First Choice Alphabet = { a } a a Two choices a a q q q q No transition 1 2 1 2 a a q q 0 0 a a q q No transition 3 3 First Choice First Choice a a a a a a q q q q 1 2 1 2 a a q q 0 0 a a q q 3 3 5

First Choice Second Choice a a a a All input is consumed a a q q q q “accept” 1 2 1 2 a a q q 0 0 a a q q 3 3 Second Choice Second Choice a a a a a a q q q q 1 2 1 2 a a q q 0 0 a a No transition: q q 3 3 the automaton hangs 6

Second Choice When To Accept a String a a An NFA accepts a string: when there is a computation of the NFA Input cannot be consumed that accepts the string a q q 1 2 AND a q 0 a all the input is consumed and the automaton q should we reject aa? is in a final state 3 Rejection example Example aa is accepted by the NFA: a “accept” a q q 1 2 a a q q a 1 2 q q 1 2 a q a 0 a q q 0 a 3 q 0 q a “reject??” 3 because this q But this only tells 3 computation us that choice accepts aa didn’t work…. 7

First Choice First Choice a a “reject??” a a q q q q 1 2 1 2 a a q q 0 0 a a q q 3 3 Second Choice Second Choice a a a a q q q q 1 2 1 2 a a q q 0 0 a a q q 3 3 8

Second Choice An NFA rejects a string: a when there is no computation of the NFA that accepts the string: a q q 1 2 • All the input is consumed and the a automaton is in a non final state q 0 a OR q “reject??” 3 • The input cannot be consumed Rejection example Example a is rejected by the NFA: a a a “reject??” a q q 1 2 a q q a a q q 1 2 1 2 a a q q q 0 a 0 0 a a q “reject??” q 3 3 q 3 All possible computations lead to rejection 9

First Choice First Choice a a a a a a a a q q q q 1 2 1 2 a a q q No transition: 0 0 a a the automaton hangs q q 3 3 First Choice Second Choice a a a a a a Input cannot be consumed a a q q q q “reject??” 1 2 1 2 a a q q 0 0 a a q q 3 3 10

Second Choice Second Choice a a a a a a a a q q q q 1 2 1 2 a a q q 0 0 a a No transition: q q 3 3 the automaton hangs Second Choice aaa is rejected by the NFA: a a a Input cannot be consumed “reject??” a q q 1 2 a a a q q q q 1 2 1 2 a a q q q 0 0 a a 0 a q q 3 3 q “reject??” “reject??” 3 All possible computations lead to rejection 11

Lambda Transitions L = { aa } Language accepted: a q q 1 2 a q a a λ q q q q 0 a 0 1 2 3 q 3 a a a a a a a a λ λ q q q q q q q q 0 1 2 3 0 1 2 3 12

(read head does not move) a a a a a a a a λ q λ q q q q q q q 0 1 2 3 0 1 2 3 Rejection Example all input is consumed a a a a a “accept” a a a a λ λ q q q q q q q q 0 1 2 3 0 1 2 3 String is accepted aa 13

(read head doesn’t move) a a a a a a a a a a λ q λ q q q q q q q 0 1 2 3 0 1 2 3 Input cannot be consumed a a a a a a “reject??” a a a a λ λ q q q q q q q q 0 1 2 3 0 1 2 3 No transition: aaa the automaton hangs String is rejected 14

Another NFA Example L = { aa } Language accepted: a a a b λ q q q q q q λ q q 0 1 2 3 1 2 3 0 λ a b a b a a b b q q λ q q q λ q q q 1 2 3 1 2 3 0 0 λ λ 15

a b a b “accept” a a b b q q λ q q q λ q q q 1 2 3 1 2 3 0 0 λ λ Another String a b a b a b a b a a b b λ λ q q q q q q q q 0 1 2 3 0 1 2 3 λ λ 16

a b a b a b a b a a b b λ λ q q q q q q q q 0 1 2 3 0 1 2 3 λ λ a b a b a b a b a a b b λ λ q q q q q q q q 0 1 2 3 0 1 2 3 λ λ 17

a b a b a b a b “accept” a a b b λ λ q q q q q q q q 0 1 2 3 0 1 2 3 λ λ Language accepted Another NFA Example L ab , abab , ababab , ... { } = + ab { } = 0 0 , 1 q q a q b q q λ q 1 2 q 0 1 2 3 1 0 λ λ 18

Remarks: Language accepted • The symbol never appears on the λ input tape • Simple automata: 0 M M 2 1 0 , 1 q q q 1 2 q q 0 1 (redundant 0 0 state) L ( M ) = {} λ 1 • NFAs are interesting because we can Formal Definition of NFAs express languages easier than DFAs M Q , , , q , F ( ) = Σ δ 0 • Q : Set of states, i.e. { q , q , q } 0 1 2 NFA a M M DFA 1 2 Input aplhabet, i.e. { a , b } : Σ q a 2 q q Transition function 0 1 : δ a a q : q q Initial state 0 0 1 F : Final states L ( M ) = { a } L ( M ) = { a } 1 2 19

Transition Function δ q 0 1 , q ( ) { } ( q , 0 ) { q , q } δ = δ = 1 1 0 2 Note the function returns a set…. What did a DFA transition funciton return? 0 0 0 , 1 0 , 1 q q q q q q 0 1 2 0 1 2 1 1 λ λ ( q , ) { q , q } ( 2 q , 1 ) δ λ = δ = ∅ 0 0 2 0 0 0 , 1 0 , 1 q q q q q q 0 1 2 0 1 2 1 1 λ λ 20

Extended Transition Function * q 0 , a q * q , aa q , q ( ) { } ( ) { } δ = δ = 1 0 4 5 q q q q 4 5 4 5 a a a a a a b b λ λ q q q q q q q q 0 3 0 3 1 2 1 2 λ λ Formally q * q , w q q ( ) : there is a walk from to ∈ δ j i i j with label w * q , ab q , q , q ( ) { } δ = 0 2 3 0 w q q i j q q 4 5 a a w  = σ σ σ 1 2 k a b λ q q q q 0 1 2 3 q σ σ σ q 1 2 k i j λ 21

The Language of an NFA F = q 0 , q { } 5 q q q q 4 5 4 5 F = q 0 , q { } 5 a a a a a a b b λ λ q q q q q q q q 0 1 2 3 0 1 2 3 λ λ aa ∈ L ( M ) * q , aa ) { q , q } * q , ab ) { q , q , q } ab ∈ L M ( ( ( ) δ = δ = 0 4 5 0 2 3 0 F F ∈ ∈ F = q 0 , q F = q 0 , q { } { } 5 5 q q q q 4 5 4 5 a a a a a a b b q q q λ q q q q λ q 0 1 2 3 0 1 2 3 λ λ ) { } aba ∉ L M * ( q , abaa q , q * ( q 0 , aba ) { } q ( ) δ = δ = 0 4 5 1 F F ∉ ∈ 22