if m dfa 1 2 0 1 2 0 l m 10 2 1 q q q 1 2 0 1 0
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( ) ( ) if = M DFA 1 2 0 , 1 2 0 L ( M ) { 10 } - PDF document

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Review Languages and Grammars Alphabets, strings, languages CS 301 - Lecture 3 Regular Languages NFA DFA Equivalence Deterministic Finite Automata Regular Expressions Nondeterministic Finite Automata Today: Fall 2008

  1. Review • Languages and Grammars – Alphabets, strings, languages CS 301 - Lecture 3 • Regular Languages NFA DFA Equivalence – Deterministic Finite Automata Regular Expressions – Nondeterministic Finite Automata • Today: Fall 2008 – Equivalence of NFA and DFA – Regular Expressions – Equivalence to Regular Languages Example of equivalent Equivalence of Machines machines M NFA 1 0 L M { 10 } * ( ) 1 = q q 1 M 0 M 1 Machine is equivalent to machine 1 2 L M L M ( ) ( ) if = M DFA 1 2 0 , 1 2 0 L ( M ) { 10 } * 2 = 1 q q q 1 2 0 1 0 1

  2. Convert NFA to DFA Convert NFA to DFA a M NFA a a M λ q q q 0 1 2 a λ q q q NFA b 0 1 2 DFA M ′ b a q { } q 1 , q { } 0 2 Begin with a set M ′ DFA of start states From the NFA start states, where could we get on a? q List the set of all possible states you can reach using { } DFA start state = 0 just an a. Union of NFA Start States Convert NFA to DFA Convert NFA to DFA a M NFA NFA a M a λ q q q 0 1 2 a λ q q q 0 1 2 b b M ′ DFA a a DFA { } q M ′ { q 1 , q } 0 2 a b q { } q 1 , q { } 0 2 How did we get from q0 to q2 b λ using only a? (think ) ∅ From new set, where could we get on a? From the NFA start states, where could we get on b? ∅ The set of all possible states you can reach may be empty! 2

  3. Convert NFA to DFA Convert NFA to DFA NFA a a M NFA M a a q q λ q λ q q q 0 1 2 0 1 2 b b a a b b DFA M ′ DFA M ′ a a q q { } { } q 1 , q q 1 , q { } { } 0 0 2 2 b b From empty set, From new set, where could we we can only get a , b get on b? back to empyt set. ∅ ∅ Convert NFA to DFA Could This Produce Infinite States? a M NFA If the NFA has states a λ q q q q , q , q ,... 0 1 2 0 1 2 b There are a finite number of NFA states by definition a b DFA M ′ the DFA has states in the powerset a q { } q 1 , q { } 0 2 , q , q , q , q , q , q , q ,.... { }{ }{ }{ } ∅ 0 1 1 2 3 4 7 b If a node has any Powerset of finite set can be big, but it is not infinite! NFA Final state, a , b mark the node as ∅ Final in the DFA. 3

  4. Procedure NFA to DFA Procedure NFA to DFA { q , q ,..., q } 2. For every DFA’s state i j m 1. Initial state of NFA: q 0 Compute in the NFA * q , a , ( ) δ i { q , q ,..., q } = ′ ′ ′ i j m q ( ) { } * q , a , Initial state of DFA: δ 0 j Add transition to DFA ... ( ) { q , q ,..., q }, a { q , q ,..., q } ′ ′ ′ δ = i j m i j m Procedure NFA to DFA Procedure NFA to DFA 3. For any DFA state { q , q ,..., q } i j m Repeat Step 2 for all letters in alphabet, q until no more transitions can be added. If some is a final state in the NFA j { q , q ,..., q } Then, i j m is a final state in the DFA 4

  5. Proof Theorem L ( M ) L ( M ) = ′ Take NFA M Apply procedure to obtain DFA M ′ L M L M L M L M ( ) ( ) AND ( ) ( ) ⊆ ′ ⊇ ′ M Then and are equivalent : M ′ L ( M ) L ( M ) ′ = w ∈ L ( M ) L M L M First we show: ( ) ( ) ⊆ ′ w ∈ L ( M ) Take arbitrary: w q q M : f 0 We will prove: w L ( M ) ′ ∈ w  = σ σ σ 1 2 k q σ σ σ q M : 1 2 k f 0 5

  6. More generally, we will show that if in : M We will show that if w ∈ L ( M ) v a a  a (arbitrary string) = 1 2 n w  = σ σ σ 1 2 k a a a q n q 1 q 2 q q M : 0 m i j l σ q σ σ q 1 2 k M : f 0 σ σ σ M ′ : 1 2 k a a a M ′ : n 1 2 { 0 q } { q , … } f w L ( M ) … { 0 q } … { q , } { q , … } { q , … } ′ { q , } ∈ j i l m Proof by induction on | v | Induction hypothesis: 1 | v ≤ | k ≤ v = a Induction Basis: 1 v a a  a = 1 2 k a a a q q k q 1 2 q q M : d 0 c a i j q 1 q M : 0 i a M ′ : 1 a a a M ′ : 1 2 k { 0 q } { q , … } i { 0 q } { q , … } { q , … } { q , … } { q , … } j c d i 6

  7. Induction Step: | v | = k 1 Induction Step: | v | = k 1 + + v a   a a a v a v a   a a a v a ′ ′ = = = = 1 2 k k 1 k 1 1 2 k k 1 k 1     + +     + + v v ′ ′ a a a a a a a q q k q q q k q q 1 q 2 q 1 q 2 q k 1 M : M : + d d 0 c 0 c e i j i j v ′ v ′ a a a a a a a M ′ : k M ′ : k k 1 1 2 1 2 + … … { 0 q } … { q , } { q , … } { q , … } { 0 q } … { q , } { q , … } { q , … } { q , … } { q , } { q , } j j i c d i c d e v ′ v ′ w ∈ L ( M ) Therefore if L ( M ) L ( M ) We have shown: ′ ⊆ w  = σ σ σ 1 2 k q σ σ σ q k M : 1 2 f 0 L ( M ) L ( M ) We also need to show: ′ ⊇ (proof is similar) σ σ σ M ′ : 1 2 k { 0 q } { q , … } f w L ( M ) ′ ∈ 7

  8. We will prove: Languages = Regular accepted NFAs Accept Regular Languages by NFAs Languages Languages accepted by DFAs NFAs and DFAs have the same computation power Step 1 Step 2 Languages Languages Regular Regular accepted accepted Languages Languages by NFAs by NFAs Proof: Any NFA can be converted to an Proof: Every DFA is trivially an NFA equivalent DFA L Any language accepted by a DFA L Any language accepted by an NFA is also accepted by an NFA is also accepted by a DFA 8

  9. Recursive Definition Regular Expressions Primitive regular expressions: , λ , ∅ α • Regular expressions describe regular r r Given regular expressions and 1 2 languages r r + 1 2 ( a b c ) * + ⋅ • Example: r r ⋅ 1 2 Are regular expressions describes the language r * 1 a , bc * , a , bc , aa , abc , bca ,... { } { } = λ r ( ) 1 Languages of Regular Examples Expressions A regular expression: L r r ( ) • : language of regular expression ( a b c ) * ( c ) + ⋅ ⋅ + ∅ • Example ( a + b ) Not a regular expression: + L ( ( a b c ) * ) { , a , bc , aa , abc , bca ,... } + ⋅ = λ 9

  10. Definition (continued) Formal Definition r r • For regular expressions and 1 2 • For primitive regular expressions: L r r L r L r ( ) ( ) ( ) + = ∪ 1 2 1 2 L ( ) ∅ = ∅ L ( r r ) L ( ) ( ) r L r ⋅ = 1 2 1 2 L ( ) { } λ = λ L r * L r ) * ( ) ( ( ) = 1 1 L ( ) { } a a = L r L r ( ( ) ) ( ) = 1 1 Example Example L a b L a * ( ( ) ) ( ) L ( ( a b ) a * ) = + + ⋅ L a b L a * ( ) ( ) = + • Regular expression ( L ( ) a L ( ) b ) ( L ( ) a ) * = ∪ r a b * a bb ( ) ( ) = + + a ∪ b a ) * ( { } { } ) { } ( = L ( ) { r a , bb , aa , abb , ba , bbb ,... } { a , b }{ , a , aa , aaa ,... } = = λ { a , aa , aaa ,..., b , ba , baa ,... } = 10

  11. What’s Next • Read – Linz Chapter 1, 2.1, 2.2, 2.3, (skip 2.4), Chapter 3 – JFLAP Startup, Chapter 1, 2.1, (skip 2.2) 3, 4 • Next Lecture Topics from Chapter 3.2 and 3.3 – Regular Expressions and Regular Languages – Regular Grammars and Regular Languages • Quiz 1 in Recitation on Wednesday 9/17 – Covers Linz 1.1, 1.2, 2.1, 2.2, 2.3, and JFLAP 1, 2.1 – Closed book, but you may bring one sheet of 8.5 x 11 inch paper with any notes you like. – Quiz will take the full hour • Homework – Homework 2 is Due Thursday 11

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