Introduction to the Theory of Computation Automatacs.uni-salzburg.at/~anas/teaching/Automata2015/Week1.pdf · •Lectures and Instructions • Books Introduction to the Theory of

Automata1VO + 1PS

Lecturer: Dr. Ana Sokolova

http://cs.uni-salzburg.at/~anas/

Introduction to the Theory of Computation

http://cs.uni-salzburg.at/~anas/

• Lectures and Instructions

• Books Introduction to the Theory of Computation by M. Sipser Introduction to Automata Theory, Languages, and Computation by J. E. Hopcroft, R. Motwani, and J.D. Ullman

3.11.23.10. 11.12.1.12.24.11.17.11. 18.12.

Setup and Dates



3.11.23.10. 11.12.1.12.24.11.17.11. 18.12.

Setup and Dates

automata



3.11.23.10. 11.12.1.12.24.11.17.11. 18.12.

Setup and Dates

automata grammars



3.11.23.10. 11.12.1.12.24.11.17.11. 18.12.

Setup and Dates

automata grammars push-down

automata



3.11.23.10. 11.12.1.12.24.11.17.11. 18.12.

Setup and Dates


automata

Turing machines



3.11.23.10. 11.12.1.12.24.11.17.11. 18.12.

Setup and Dates


automata

Turing machines

computability

The Rules... Instructions

• Instruction exercises on the web http://cs.uni-salzburg.at/~anas/Ana_Sokolova/Automata2015.html after class

• To be solved by you, the students, (in groups of at most 3 people) and handed in as homework at the next meeting.

• In class I will present a sample solution and you, the students, will be asked to present solutions/discuss the exercises

Starting in 2 weeks!

http://cs.uni-salzburg.at/~anas/Ana_Sokolova/Automata2015.html

The Rules... Instructions

• One randomly chosen exercise will be graded each week

• The graded exercise will be returned at the next meeting.

• Grade based on (1) exam (2) the grades of the corrected exercise and (3) activity in class (ability to present solutions)

• All information about the course / rules / exams / grading is / will be on the course webpage

The Rules... Grading

• Written exam on December 18, 10 am - 12:30 pm

• Grade based on the number of points on the written exam (80%), homework grades, and activity in class (20%)

• For better grade oral exam after the written one upon appointment

• 55% of the maximal points are needed to pass.

Finite Automata

Alphabets and Languages

if y is not free in P and Q

Alphabet and words



∑ - alphabet (finite set) !∑n = {a1a2..an | ai ∈ ∑} is the set of words of length n !∑* = {w | ∃n ∈ N. ∃ a1, a2, .. , an ∈ ∑. w = a1a2..an} is the set of all words over ∑ !

Alphabet and words




Alphabet and words∑0 = {ℇ} contains only the

empty word




Alphabet and words∑0 = {ℇ} contains only the

empty word

A language L over ∑ is a subset L ⊆ ∑*

Deterministic Automata (DFA)


∑ = {0,1}

Informal example

q0 q1

01 0

1

M1:

alphabet



∑ = {0,1}

Informal example

q0 q1

01 0

1

M1:

alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0

1

M1:

alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0

1q0 is initial

M1:

alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0

1q0 is initial q1 is final

M1:

alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0


transitions, labelled by alphabet symbols

M1:

alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0



M1:

Accepts the language L(M1) = {w ∈∑* | w ends with a 0} = ∑*0

alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0



M1:

regular language


alphabet



∑ = {0,1}

Informal example

q0, q1 are states

q0 q1

01 0



M1:

regular expressionregular language


DFA


A deterministic finite automaton M is a tuple M = (Q, ∑, δ, q0, F) where ! Q is a finite set of states ∑ is a finite alphabet δ: Q x ∑⟶ Q is the transition function q0 is the initial state, q0 ∈Q F is a set of final states, F⊆Q

Definition

DFA



Definition

In the example M

M1 = (Q, ∑, δ, q0, F) for

DFA



Definition

In the example M

M1 = (Q, ∑, δ, q0, F) for

DFA



Definition

In the example MQ = {q0, q1}

M1 = (Q, ∑, δ, q0, F) for

DFA



Definition


∑ = {0, 1}

M1 = (Q, ∑, δ, q0, F) for

DFA



Definition


∑ = {0, 1}

F = {q1}

M1 = (Q, ∑, δ, q0, F) for

DFA



Definition


∑ = {0, 1}

F = {q1} δ(q0, 0) = q1 ,δ(q0, 1) = q0

δ(q1, 0) = q1, δ(q1, 1) = q0

DFA


The extended transition function

DFA


Given M = (Q, ∑, δ, q0, F) we can extend δ: Q x ∑⟶ Q to ! δ*: Q x ∑*⟶ Q !inductively, by: ! δ*(q, ε) = q and δ*(q,wa) = δ(δ*(q,w), a)


DFA




In M1, δ*(q0,110010) = q1

DFA




DefinitionThe language recognised / accepted by a deterministic finite automaton M = (Q, ∑, δ, q0, F) is !

L(M) = {w ∈ ∑*| δ*(q0,w) ∈ F}

In M1, δ*(q0,110010) = q1

DFA




DefinitionThe language recognised / accepted by a deterministic finite automaton M = (Q, ∑, δ, q0, F) is !

L(M) = {w ∈ ∑*| δ*(q0,w) ∈ F}

In M1, δ*(q0,110010) = q1

L(M1) = {w0|w ∈ {0,1}*}

Regular languages and operations

Let ∑ be an alphabet. A language L over ∑ (L ⊆ ∑*) is regular iff it is recognised by a DFA.

Definition



Definition

L(M1) = {w0|w ∈ {0,1}*} is regular



Definition

Regular operations

L(M1) = {w0|w ∈ {0,1}*} is regular



Definition

Regular operations

L(M1) = {w0|w ∈ {0,1}*} is regular

Let L, L1, L2 be languages over ∑. Then L1 ∪ L2, L1·L2, and L* are languages, where ! L1·L2 = {w1·w2 | w1 ∈ L1,w2 ∈ L2} ! L* = {w | ∃n ∈ N. ∃ w1, w2, .. , wn ∈ L. w = w1w2..wn}



Definition

Regular operations

L(M1) = {w0|w ∈ {0,1}*} is regular

Let L, L1, L2 be languages over ∑. Then L1 ∪ L2, L1·L2, and L* are languages, where ! L1·L2 = {w1·w2 | w1 ∈ L1,w2 ∈ L2} ! L* = {w | ∃n ∈ N. ∃ w1, w2, .. , wn ∈ L. w = w1w2..wn}

ℇ ∈ L* always

Regular expressions

Definition

Regular expressions

Definition

finite representation of infinite languages

Regular expressions

Let ∑ be an alphabet. The following are regular expressions !!1. a for a ∈ ∑ 2. ε 3. ∅ 4. (R1 ∪ R2) for R1, R2 regular expressions 5. (R1·R2) for R1, R2 regular expressions 6. (R1)* for R1 regular expression

Definition


Regular expressions


Definition

inductive


Regular expressions


Definition

inductiveexample: (ab ∪ a)*


Regular expressions


Definition

inductiveexample: (ab ∪ a)*

corresponding languages

L(a) = {a} L(ε) = {ε} L(∅) = ∅

L(R1 ∪ R2) = L(R1) ∪ L(R2) L(R1·R2) = L(R1)·L(R2)

L(R1*) = L(R1)*


Equivalence of regular expressions and regular languages




Theorem (Kleene)A language is regular (i.e., recognised by a finite automaton) iff it is the language of a regular expression.



Theorem (Kleene)A language is regular (i.e., recognised by a finite automaton) iff it is the language of a regular expression.

Proof ⇐ easy, via the closure

properties discussed next, ⇒ not so easy, we’ll skip it for now…

Closure under regular operations


Theorem C1

The class of regular languages is closed under union


Theorem C1


also under intersection


Theorem C1



Theorem C2

The class of regular languages is closed under complement


Theorem C1


Theorem C3

The class of regular languages is closed under concatenation


Theorem C2



Theorem C1


Theorem C3


Theorem C4

The class of regular languages is closed under Kleene star


Theorem C2



Theorem C1


Theorem C3


Theorem C4



We can already prove these!

Theorem C2



Theorem C1


Theorem C3


Theorem C4


But not yet these two…


We can already prove these!

Theorem C2


Nondeterministic Automata (NFA)

∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:

q2 q3

0,11


∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:

sources of nondeterminism

q2 q3

0,11


∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:


q2 q3

0,11


∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:


q2 q3

0,11

no 1 transition


∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:


q2 q3

0,11

no 1 transition


∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:


q2 q3

0,11

no 0 transition

no 1 transition


∑ = {0,1}

Informal example

q0 q10,ε

0,1

1

M2:


q2 q3

0,11

no 0 transition

Accepts a word iff there exists an accepting run

NFA


A nondeterministic automaton M is a tuple M = (Q, ∑, δ, q0, F) where ! Q is a finite set of states ∑ is a finite alphabet δ: Q x ∑ε⟶ P(Q) is the transition function q0 is the initial state, q0 ∈Q F is a set of final states, F⊆Q

Definition

NFA



Definition

∑ε = ∑ ∪ {ε}

In the example M

NFA



Definition

∑ε = ∑ ∪ {ε}

In the example M M2 = (Q, ∑, δ, q0, F) for

NFA



Definition

∑ε = ∑ ∪ {ε}


NFA



Definition

Q = {q0, q1, q2, q3}

∑ε = ∑ ∪ {ε}


NFA



Definition

Q = {q0, q1, q2, q3}

∑ = {0, 1}

∑ε = ∑ ∪ {ε}


NFA



Definition

Q = {q0, q1, q2, q3}

∑ = {0, 1} F = {q3}

∑ε = ∑ ∪ {ε}


NFA



Definition

Q = {q0, q1, q2, q3}

∑ = {0, 1} F = {q3}

δ(q0, 0) = {q0} δ(q0, 1) = {q0,q1} δ(q0, ε) = ∅ …..

∑ε = ∑ ∪ {ε}

Documents

Introduction to the Theory of Computation Automatacs.uni-salzburg.at/~anas/teaching/Automata2015/Week1.pdf · •Lectures and Instructions • Books Introduction to the Theory of