Automata & Complexity - Jörg Endrullisjoerg.endrullis.de/automata/01_intro.pdf/01_intro.pdf ·...

Automata & Complexity

Jörg Endrullis

Vrije Universiteit Amsterdam

Information

Lecturer:Jörg Endrullis

room: NU-12A-65

email: j.endrullis@vu.nl

Teaching assistant:Geoffrey Frankhuizen

Johannes van der Meer

Alexander van Oostveen

Course information:all relevant course information is on Canvas

link to the slides on Canvas

Course Structure

2 lectures per week

2 exercise classes per week

Homework:80% of the homework in an online system

20% of the homework written assignments(one written assignment at the end of the course)

70% of homework points to qualify for the exam

90% of homework points for 0.5 points bonus

Written exam

Material

Peter LinzAn Introduction to Formal Languages and Automata (5th edition)

Jones and Bartlett

Course Motivation

Computer is based on a universal computation mechanism.

Different applications =⇒ different formalisms:

Pattern recognition: regular languagesregular grammars, regular expressions, finite automata

Parsing: context-free languagescontext-free grammars, pushdown automata

Computation: recursively enumerable languagesTuring machines

Course Motivation

Computer is based on a universal computation mechanism.

Different applications =⇒ different formalisms:

Pattern recognition: regular languagesregular grammars, regular expressions, finite automata

Parsing: context-free languagescontext-free grammars, pushdown automata

Computation: recursively enumerable languagesTuring machines

Course Motivation: Automata are Ubiquitous

Theory of computation

Pattern matching

Parsing

Hardware design

Software modelling

Compiler construction

Formal verification

Number theory

Course Motivation: What can a Computer do?

What can a computer do?

Some (at first glance simple) problems are undecidable.

For example:program termination,Post correspondence problem,validity in predicate logic.

Some problems (NP-complete problems)are (probably) not efficiently solvable bya computer.

For example:travelling salesman problem,satisfiability in propositional logic.

Course Motivation

Typical questions that we will answer

What is a (programming) language?

How can languages be recognised by computers?(automata, regular expressions, grammars)

What problems can be solved by what types of automata?

How much time/memory is needed for solving a problem?

Aspects of languages:syntax: the form of the words in the language

semantics: the meaning of the words in the language

We will focus on the syntax.

Course Motivation

Typical questions that we will answer

What is a (programming) language?

How can languages be recognised by computers?(automata, regular expressions, grammars)

What problems can be solved by what types of automata?

How much time/memory is needed for solving a problem?

Aspects of languages:syntax: the form of the words in the language

semantics: the meaning of the words in the language

We will focus on the syntax.

WordsWord = finite sequence of symbols from a finite alphabet Σ.

notation for symbols: a,b, c, . . .

symbol from the alphabet: a ∈ Σ

notation for words: u, v ,w , x , y , z

empty word: λ

Note that, in programming, words are often called strings.

A computer program:is itself a word (a finite sequence of bits/bytes)

takes input word

produces output word

empty word: λ

takes input word

empty word: λ

takes input word

Operations on Words (1)

ConcatenationIf v = a1 · · · an and w = b1 · · · bm, then

vw = a1 · · · anb1 · · · bm

LengthIf v = a1 · · · an, then |v | = n.

The length can be defined inductively:

|λ| = 0|va| = |v |+ 1

ConcatenationIf v = a1 · · · an and w = b1 · · · bm, then

vw = a1 · · · anb1 · · · bm

LengthIf v = a1 · · · an, then |v | = n.

The length can be defined inductively:

|λ| = 0|va| = |v |+ 1

PowerThe power vk consists of k concatenations of v ’s:

v0 = λ

vk+1 = vkv

ReverseThe reverse of a1 · · · an is

(a1 · · · an)R = an · · · a1

The reverse can be inductively defined

λR = λ

(va)R = a(vR)

PowerThe power vk consists of k concatenations of v ’s:

v0 = λ

vk+1 = vkv

ReverseThe reverse of a1 · · · an is

(a1 · · · an)R = an · · · a1

The reverse can be inductively defined

λR = λ

(va)R = a(vR)

Formal Languages

Formal language = set of words

ExampleAll parseable C programs form a language.

More precisely:Σ∗ is the set of all words over Σ.

Formal languageA (formal) language L is a subset of Σ∗, that is, L ⊆ Σ∗.

Formal Languages

Examples of Formal Languages

Let Σ = {a,b}.

{ab, aab, bbaaabb } is a finite language

{abna | n ≥ 1 } is an infinite language:

{aba, abba, abbba, abbbba, . . . }

{anbn | n ≥ 0 } is an infinite language:

{ λ, ab, aabb, aaabbb, aaaabbbb, . . . }

Let Σ = {a,b}.

Operations on Languages (1)

A language is a set of words.So the usual set operations have meaning for languages:

∈, ⊆, ∩, ∪, \, . . .

Complement

The complement L = all words that are not in the language L:

L = Σ∗ \ L

ConcatenationThe concatenation of languages L1 and L2 is defined as

L1L2 = { xy | x ∈ L1 ∧ y ∈ L2 }

ReverseThe reverse of a language L is

LR = { xR | x ∈ L }

∈, ⊆, ∩, ∪, \, . . .

Complement

L = Σ∗ \ L

L1L2 = { xy | x ∈ L1 ∧ y ∈ L2 }

LR = { xR | x ∈ L }

∈, ⊆, ∩, ∪, \, . . .

Complement

L = Σ∗ \ L

L1L2 = { xy | x ∈ L1 ∧ y ∈ L2 }

LR = { xR | x ∈ L }

∈, ⊆, ∩, ∪, \, . . .

Complement

L = Σ∗ \ L

L1L2 = { xy | x ∈ L1 ∧ y ∈ L2 }

LR = { xR | x ∈ L }

PowerThe n-th power of a language L is defined by induction on n:

L0 = { λ }

Ln+1 = LnL (n ≥ 0)

Attention: L2 = {uv | u, v ∈ L} 6= {uu | u ∈ L}

Kleene star

L∗ =

∞⋃i=0

Li = L0 ∪ L1 ∪ L2 ∪ L3 ∪ · · ·

∞⋃i=1

Li = L1 ∪ L2 ∪ L3 ∪ · · ·

Thus L∗ = L+ ∪ {λ}.

L0 = { λ }

Ln+1 = LnL (n ≥ 0)

Kleene star

L∗ =

∞⋃i=0

Li = L0 ∪ L1 ∪ L2 ∪ L3 ∪ · · ·

∞⋃i=1

Li = L1 ∪ L2 ∪ L3 ∪ · · ·

Thus L∗ = L+ ∪ {λ}.

L0 = { λ }

Ln+1 = LnL (n ≥ 0)

Kleene star

L∗ =

∞⋃i=0

Li = L0 ∪ L1 ∪ L2 ∪ L3 ∪ · · ·

∞⋃i=1

Li = L1 ∪ L2 ∪ L3 ∪ · · ·

Thus L∗ = L+ ∪ {λ}.

Exercise

LetΣ = {a,b }

L1 = {a, bb }

L2 = {ab, ba }

(In groups of two, 2 minutes)

Describe the following languages as sets:L1L2 =

{aab, aba, bbab, bbba }

{aa, abb, bba, bbbb }

answered in the lecture

Exercise

LetΣ = {a,b }

L1 = {a, bb }

L2 = {ab, ba }

Describe the following languages as sets:L1L2 = {aab, aba, bbab, bbba }

{aa, abb, bba, bbbb }

Exercise

LetΣ = {a,b }

L1 = {a, bb }

L2 = {ab, ba }

L21 = {aa, abb, bba, bbbb }

Exercise

LetΣ = {a,b }

L1 = {a, bb }

L2 = {ab, ba }

L21 = {aa, abb, bba, bbbb }

L31 = answered in the lecture

Exercise

LetΣ = {a,b }

L = {abn | n ≥ 0 }

Describe the following languages as sets:LR =

{bna | n ≥ 0 }

{ λ } ∪ {bw | w ∈ Σ∗ } ∪ {awau | w ,u ∈ Σ∗ }

ConclusionSets are not ideal to describe languages.

Exercise

LetΣ = {a,b }

L = {abn | n ≥ 0 }

Describe the following languages as sets:LR = {bna | n ≥ 0 }

{ λ } ∪ {bw | w ∈ Σ∗ } ∪ {awau | w ,u ∈ Σ∗ }

Exercise

LetΣ = {a,b }

L = {abn | n ≥ 0 }

L = { λ } ∪ {bw | w ∈ Σ∗ } ∪ {awau | w ,u ∈ Σ∗ }

Exercise

LetΣ = {a,b }

L = {abn | n ≥ 0 }

L = { λ } ∪ {bw | w ∈ Σ∗ } ∪ {awau | w ,u ∈ Σ∗ }

LR= answered in the lecture

Exercise

LetΣ = {a,b }

L = {abn | n ≥ 0 }

L = { λ } ∪ {bw | w ∈ Σ∗ } ∪ {awau | w ,u ∈ Σ∗ }

LR = answered in the lecture

Exercise

LetΣ = {a,b }

L = {abn | n ≥ 0 }

L = { λ } ∪ {bw | w ∈ Σ∗ } ∪ {awau | w ,u ∈ Σ∗ }

LR = answered in the lecture

Deterministic Finite Automata

A deterministic finite automaton, short DFA, consists of:a finite set Q of statesa finite input alphabet Σa transition function δ : Q × Σ→ Qa starting state q0 ∈ Qa set F ⊆ Q of final states

Understanding the transition function δ : Q × Σ→ QIf the automaton in state q reads the symbol a, then theresulting state is δ(q,a).

Example DFALet M = (Q, Σ, δ, q0,F ) with Q = {q0,q1}, Σ = {a,b}, F = {q0},

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Deterministic Finite Automata

A deterministic finite automaton, short DFA, consists of:a finite set Q of statesa finite input alphabet Σa transition function δ : Q × Σ→ Qa starting state q0 ∈ Qa set F ⊆ Q of final states

Understanding the transition function δ : Q × Σ→ QIf the automaton in state q reads the symbol a, then theresulting state is δ(q,a).

Example DFALet M = (Q, Σ, δ, q0,F ) with Q = {q0,q1}, Σ = {a,b}, F = {q0},

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

DFA’s Reading Words

Let M = (Q, Σ, δ, q0,F ) be a DFA. We write

(q,aw) ` (q ′,w) if δ(q,a) = q ′

Here, (q,w) indicates that M is in state q and reads word w .

Let M = (Q, Σ, δ, q0,F ) with Q = {q0,q1}, Σ = {a,b}, F = {q0},

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) `

(q0,bba) ` (q1,ba) ` (q0,a) ` (q0, λ)

We define `∗ as the reflexive transitive closure of `.

Continuing the above example, we have

(q0,abba) `∗ (q0, λ)

(q,aw) ` (q ′,w) if δ(q,a) = q ′

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) `

(q0,bba) ` (q1,ba) ` (q0,a) ` (q0, λ)

(q0,abba) `∗ (q0, λ)

(q,aw) ` (q ′,w) if δ(q,a) = q ′

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) ` (q0,bba) `

(q1,ba) ` (q0,a) ` (q0, λ)

(q0,abba) `∗ (q0, λ)

(q,aw) ` (q ′,w) if δ(q,a) = q ′

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) ` (q0,bba) ` (q1,ba) `

(q0,a) ` (q0, λ)

(q0,abba) `∗ (q0, λ)

(q,aw) ` (q ′,w) if δ(q,a) = q ′

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) ` (q0,bba) ` (q1,ba) ` (q0,a) `

(q0, λ)

(q0,abba) `∗ (q0, λ)

(q,aw) ` (q ′,w) if δ(q,a) = q ′

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) ` (q0,bba) ` (q1,ba) ` (q0,a) ` (q0, λ)

(q0,abba) `∗ (q0, λ)

(q,aw) ` (q ′,w) if δ(q,a) = q ′

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

(q0,abba) `∗ (q0, λ)

Transition Function in Table Notation

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Hint: transition function δ can be written in the form of a table:

δ q0 q1a q0 q1b q1 q0

DFA’s as Transition Graphs

A DFA can be visualised as a transition graph, consisting of:states are the nodes of the graph

arrows with labels from Σ

starting state indicated by an extra incoming arrow

final states indicated by double circleWe have an arrow from q to q ′ with label a if δ(q,a) = q ′.

Σ = {a,b }

q0 q2 q4

Exercise (1)

We define a DFA M = (Q, Σ, δ, q0,F ) with

Q = {q0,q1} Σ = {a,b} F = {q0}

and the transition function given by

(In groups of two, 1 minute)

Visualise M as a transition graph.

q0 q1b

Exercise (1)

We define a DFA M = (Q, Σ, δ, q0,F ) with

Q = {q0,q1} Σ = {a,b} F = {q0}

and the transition function given by

Visualise M as a transition graph.

q0 q1b

Exercise (2)

Note that the transition with label a,bis shorthand for two transitions: onewith label a and one with label b.

Translate the graph to a deterministic finite automaton:states Q = ?alphabet Σ = ?transition function δ : Q × Σ→ Q is given by

δ z0 z1 z2ab

starting state ?final states F = ?

DFA’s Reading Words continued...

If (q,w) `∗ (q ′, λ), we also write

q w−→ q ′

In other words,

q w−→ q ′

means that there is a path from q to q ′ in the automaton withtransitions labelled a1, . . . ,an such that w = a1 · · · an.

DFA’s Reading Words continued...

If (q,w) `∗ (q ′, λ), we also write

q w−→ q ′

In other words,

q w−→ q ′

means that there is a path from q to q ′ in the automaton withtransitions labelled a1, . . . ,an such that w = a1 · · · an.

Regular Languages

A DFA defines (accepts) a language!

The language accepted by DFA M = (Q, Σ, δ, q0,F ) is

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

The word abba is accepted by M, that is, abba ∈ L(M).

A language L is regular if there exists a DFA M with L(M) = L.

Regular Languages

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

Regular Languages

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

Regular Languages

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

Regular Languages

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

δ(q0,a) = q0 δ(q1,a) = q1

δ(q0,b) = q1 δ(q1,b) = q0

Then we have

Exercise

Let M be the following DFA:

q0 q1b

What is the language accepted by M? (describe in words)

L(M) consists of all words that contain an even number of b’s.

Exercise

q0 q1b

Exercise

q0 q1b

Determinism

DFAs are deterministicFor every state q ∈ Q and every symbol a ∈ Σ, the state q hasprecisely one outgoing arrow with label a.

Recall that δ is a function from Q × Σ to Q.

Hence, for every input word, there is precisely one path fromthe starting state through the transition graph.

The following picture shows the path for aaba:

q0 q2 q4

Determinism

DFAs are deterministicFor every state q ∈ Q and every symbol a ∈ Σ, the state q hasprecisely one outgoing arrow with label a.

Recall that δ is a function from Q × Σ to Q.

Hence, for every input word, there is precisely one path fromthe starting state through the transition graph.

The following picture shows the path for aaba:

q0 q2 q4

Exercise (1)

(Individually, 10 seconds)

Show that the following language is regular.

Construct a deterministic finite automaton for the language:

{ λ }

Exercise (2)

(Individually, 1 minutes)

{anb | n ≥ 0 }

Exercise (3)

{a2n+1 | n ≥ 0 } ∪ {b2n | n ≥ 0 }

Exercise (4)

Show that the following languages are regular.

Construct deterministic finite automata for the languages:

{w ∈ {a,b}∗ | w contains the subword bab }

{w ∈ {a,b}∗ | w does not contain the subword bab }

Theorems about Regular Languages (1)

TheoremIf L is a regular language, then L is also regular.

Proof.Let L be regular.

Then there exists a DFA M = (Q, Σ, δ, q0,F ) with L(M) = L.

Then N = (Q, Σ, δ, q0,Q \ F ) is a DFA with L(N) = L.

Here it is important that for every input word w :There is precisely one path starting at q0 labelled with w .

There is precisely one state q with q0w−→ q, and

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

There is precisely one state q with q0w−→ q,

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ F

w ∈ L ⇐⇒ w ∈ L(M) ⇐⇒ q ∈ (Q \ F ) ⇐⇒ w ∈ L(N)

TheoremIf L1 and L2 are regular, then L1 ∪ L2 is regular.

Construction (Product)There exists a DFAs

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

such that L(M1) = L1 and L(M2) = L2.

Idea: We run M1 and M2 in parallel.

We define a DFA N = (Q, Σ, δ, q0,F ) whereQ = Q1 ×Q2 = { (q1,q2) | q1 ∈ Q1, q2 ∈ Q2 }

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

Then it follows that L(N) = L(M1) ∪ L(M2) = L1 ∪ L2.

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

We define a DFA N = (Q, Σ, δ, q0,F ) where

Q = Q1 ×Q2 = { (q1,q2) | q1 ∈ Q1, q2 ∈ Q2 }

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

We define a DFA N = (Q, Σ, δ, q0,F ) whereQ =

Q1 ×Q2 = { (q1,q2) | q1 ∈ Q1, q2 ∈ Q2 }

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) =

(δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))

q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 =

(q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

{(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

M1 = (Q1, Σ, δ1,q0,1,F1) M2 = (Q2, Σ, δ2,q0,2,F2)

δ((q1,q2),a) = (δ1(q1,a), δ2(q2,a))q0 = (q0,1, q0,2)

F = {(q1,q2) ∈ Q | q1 ∈ F1 or q2 ∈ F2}

QuestionLet L1 and L2 be regular languages.

How to change the product construction to show thatL1 ∩ L2 is regular, and

L1 \ L2 is regular ?

TheoremIf L is regular, then LR is also regular.

This proof will be an exercise.

Exercise

Is the following language regular?

{anbn | n ≥ 0 }

This language is not regular!

Intuition: a DFA has only a finite memory (the states).

(Groups of two, 2 minutes)

Give a deterministic automaton with infinitely many states for

{anbn | n ≥ 0 }

Answer in the lecture.

QuestionIs every finite language regular ?

Exercise

{anbn | n ≥ 0 }

Exercise

{anbn | n ≥ 0 }

Exercise

{anbn | n ≥ 0 }

Exercise

{anbn | n ≥ 0 }

Exercise

{anbn | n ≥ 0 }

Nondeterministic Finite Automata

Nondeterministic finite automaton (NFA)

A state can have zero or more outgoing arrows with thesame label.

Allows for empty steps: arrows with label λ that do not‘eat’ a symbol from the input word.

DFA’s and NFA’s are used everywhere:in software engineering,

for modelling of hardware circuits,

in compilers, and

in network protocols.

Nondeterministic finite automaton (NFA)

A state can have zero or more outgoing arrows with thesame label.

Allows for empty steps: arrows with label λ that do not‘eat’ a symbol from the input word.

DFA’s and NFA’s are used everywhere:in software engineering,

for modelling of hardware circuits,

in compilers, and

in network protocols.

NFA’s are defined like DFA’s, except for the transition function!

A nondeterministic finite automaton, short NFA, consists of:a finite set Q of states

a finite input alphabet Σ

a transition function δ : Q × (Σ ∪ {λ}) → 2Q

a starting state q0 ∈ Q

a set F ⊆ Q of final states

Here 2Q is the set of all subsets of Q:

2Q = {S | S ⊆ Q }

NFA’s are defined like DFA’s, except for the transition function!

A nondeterministic finite automaton, short NFA, consists of:a finite set Q of states

a finite input alphabet Σ

a transition function δ : Q × (Σ ∪ {λ}) → 2Q

a starting state q0 ∈ Q

a set F ⊆ Q of final states

Here 2Q is the set of all subsets of Q:

2Q = {S | S ⊆ Q }

NFA’s Reading Words

Let M = (Q, Σ, δ, q0,F ) be a NFA. We write

(q, αw) ` (q ′,w) if q ′ ∈ δ(q, α) with α ∈ Σ ∪ {λ}

Note that if α = λ, thenthe state changes (q to q ′), but

the input word stays the same (λw = w).

Again we write

q w−→ q ′

whenever (q,w) `∗ (q ′, λ).

It means that there is a path from q to q ′ with transitionslabelled α1, . . . , αn ∈ (Σ ∪ {λ}) such that w = α1 · · ·αn.

Again we write

q w−→ q ′

Again we write

q w−→ q ′

The language accepted by NFA M = (Q, Σ, δ, q0,F ) is

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

But now paths are not unique!

q0 q1a

This automaton can accept the word ab in 2 different ways!

(q0,ab) ` (q1,b) ` (q1, λ)

(q0,ab) ` (q1,b) ` (q0, λ) ` (q1, λ)

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

q0 q1a

(q0,ab) ` (q1,b) ` (q1, λ)

(q0,ab) ` (q1,b) ` (q0, λ) ` (q1, λ)

L(M) = {w ∈ Σ∗ | (q0,w) `∗ (q, λ) with q ∈ F }

= {w ∈ Σ∗ | q0w−→ q with q ∈ F }

q0 q1a

(q0,ab) ` (q1,b) ` (q1, λ)

(q0,ab) ` (q1,b) ` (q0, λ) ` (q1, λ)

DFA’s and NFA’s are Equally Expressive

TheoremA language L is accepted by a NFA ⇐⇒ L is regular.

Construction (Powerset)Let M = (Q, Σ, δ, q0,F ) be a NFA.Idea: state of DFA = set of all states the NFA can be inWe construct a DFA ND = (QD, Σ, δD,q0D,FD) where

QD = 2Q

δD(X ,a) =

{q ′ ∈ Q | (q,a) `∗ (q ′, λ) for some q ∈ X }

{q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }

{X ⊆ Q | X ∩ F 6= ∅ }

For every w ∈ Σ∗ and q ∈ Q it holds that

(q0,w) `∗ (q, λ) ⇐⇒ (q0D,w) `∗ (X , λ) with q ∈ X .

From this property it follows that L(ND) = L(M).

Construction (Powerset)Let M = (Q, Σ, δ, q0,F ) be a NFA.

Idea: state of DFA = set of all states the NFA can be inWe construct a DFA ND = (QD, Σ, δD,q0D,FD) where

QD = 2Q

δD(X ,a) =

{q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }

{X ⊆ Q | X ∩ F 6= ∅ }

Construction (Powerset)Let M = (Q, Σ, δ, q0,F ) be a NFA.Idea: state of DFA = set of all states the NFA can be in

We construct a DFA ND = (QD, Σ, δD,q0D,FD) where

QD = 2Q

δD(X ,a) =

{q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }

{X ⊆ Q | X ∩ F 6= ∅ }

QD = 2Q

δD(X ,a) =

{q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }

{X ⊆ Q | X ∩ F 6= ∅ }

QD = 2Q

δD(X ,a) = {q ′ ∈ Q | (q,a) `∗ (q ′, λ) for some q ∈ X }

{q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }

{X ⊆ Q | X ∩ F 6= ∅ }

QD = 2Q

q0D = {q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }FD =

{X ⊆ Q | X ∩ F 6= ∅ }

QD = 2Q

q0D = {q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }FD = {X ⊆ Q | X ∩ F 6= ∅ }

QD = 2Q

q0D = {q ′ ∈ Q | (q0, λ) `∗ (q ′, λ) }FD = {X ⊆ Q | X ∩ F 6= ∅ }

Exercise

Given is the following NFA:

q0 q1a

Construct a DFA that accepts the same language.

Looking Forward

Read:Linz 1.2–1.2, 2.1–2.3

Do the following exercises:Linz 1.2: 2, 4, 8, 10

Linz 2.1: 1, 2d, 3, 7b, 9b,f, 11

Linz 2.2: 12

Linz 2.3: 2, 3, 6, 12

Following lecture:Alternative descriptions of regular languagse:

regular expressions

grammars

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