[PDF] - Recall: Regular Languages The language recognized by a finite PDF Document

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13-06-03 CSE 2001, Summer 2013 1

CSE 2001: Introduction to Theory of Computation

Summer 2013

Yves Lesperance

lesperan@cse.yorku.ca Office: CSEB 3052A Phone: 416-736-2100 ext 70146 Course page: http://www.cse.yorku.ca/course/2001

Slides are mostly taken from Suprakash Datta’s for Winter 2013

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Recall: Regular Languages

The language recognized by a finite automaton M is denoted by L(M). A regular language is a language for which there exists a recognizing finite automaton.

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Terminology: closure

A set is defined to be closed under an
peration if that operation on members
f the set always produces a member of

the same set. (adapted from wikipedia)

E.g.:

The integers are closed under addition, multiplication.
The integers are not closed under division
Σ* is closed under concatenation
A set can be defined by closure -- Σ* is called the

(Kleene) closure of Σ under concatenation.

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Terminology: Regular Operations

Pages 44-47 (Sipser) The regular operations are:

1. Union
2. Concatenation
3. Star (Kleene Closure): For a language A,

A* = {w1w2w3…wk| k ≥ 0, and each wi ∈A}

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Closure Properties

Set of regular languages is closed

under

- Complementation

– Union – Concatenation – Star (Kleene Closure)

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Complement of a regular language

Swap the accepting and non-accept

states of M to get M’.

The complement of a regular language

is regular.

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Other closure properties

Union: Can be done with DFA, but using a complicated construction. Concatenation: We tried and failed Star: ???

We introduced non-determinism in FA

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Recall: NFA drawing conventions

Not all transitions are labeled
Unlabeled transitions are assumed to

go to a reject state from which the automaton cannot escape

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Closure under regular operations

Union (new proof):

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Closure under regular operations

Concatenation:

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Closure under regular operations

Star:

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Incorrect reasoning about RL

Since L1 = {w| w=an, n ∈ N},

L2 = {w| w = bn, n ∈ N} are regular, therefore L1 • L2 = {w| w=an bn, n ∈ N} is regular

If L1 is a regular language, then

L2 = {wR| w ∈ L1} is regular, and Therefore L1 • L2 = {w wR | w ∈ L1} is regular

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Are NFA more powerful than DFA?

NFA can solve every problem that DFA

can (DFA are also NFA)

Can DFA solve every problem that NFA

can?

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Equivalence of NFA, DFA

Pages 54-58 (Sipser, 2nd ed)
We will prove that every NFA is

equivalent to a DFA (with upto exponentially more states).

Non-determinism does not help FA’s to

recognize more languages!

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Epsilon Closure

Let N=(Q,Σ,δ,q0,F) be any NFA
Consider any set R ⊆ Q
E(R) = {q|q can be reached from a state

in R by following 0 or more ε-transitions}

E(R) is the epsilon closure of R under ε-

transitions

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Proving equivalence

* languages all For Σ ⊆ L M N M L L iff N L L DFA NFA some for some for ) ( ) ( = =

One direction is easy:

A DFA M is also a NFA N. So N does not have to be `constructed’ from M

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Proving equivalence – contd.

N = (Q,Σ,δ,q0,F)
Construct M= (Q’,Σ,δ’,q’0,F’) such

that,

– for any string w ∈ Σ*, – w is accepted by N iff w is accepted by M

The other direction:

Construct M from N

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Special case

Assume that ε is not used in the NFA N.
Need to keep track of each subset of N
So Q’ = P (Q), q’0 = {q0}
δ’(R,a) = ∪(δ(r,a)) over all r ∈ R, R ∈ Q’
F’ = {R ∈ Q’| R contains an accept state of

N}

Now let us assume that ε is used.

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Construction (general case)

1. Q’ = P(Q)
2. q’0 = E({q0})
3. for all R ∈ Q’ and a∈ Σ

δ’(R, a) = {q ∈ Q|q ∈ E(δ(r,a)) for some r∈R}

4. F’ = { R ∈ Q’| R contains an accept

state of N}

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Why the construction works

for any string w ∈ Σ*,
w is accepted by N iff w is accepted by

M

Can prove using induction on the

number of steps of computation…

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State minimization

It may be possible to design DFA’s without the exponential blowup in the number of states. Consider the NFA and DFA below. 0,1 0,1 1 1 1 0,1 1 1 1

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Characterizing FA languages

Regular expressions

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Regular Expressions (Def. 1.52)

Given an alphabet Σ, R is a regular expression if: (INDUCTIVE DEFINITION)

R = a, with a∈Σ
R = ε
R = ∅
R = (R1∪R2), with R1 and R2 regular expressions
R = (R1•R2), with R1 and R2 regular expressions
R = (R1*), with R1 a regular expression

Precedence order: *, •,∪

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Regular Expressions

Unix ‘grep’ command: Global Regular

Expression and Print

Lexical Analyzer Generators (part of

compilers)

Both use regular expression to DFA

conversion

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Examples

e1 = a ∪ b, L(e1) = {a,b}
e2 = ab ∪ ba, L(e2) = {ab,ba}
e3 = a*, L(e3) = {a}*
e4 = (a ∪ b)*, L(e4) = {a,b}*
e5 = (em . en), L(e5) = L(em) • L(en)
e6 = a*b ∪ a*bb,

L(e6) = {w| w ∈ {a,b}* and w has 0 or more a’s followed by 1 or 2 b’s}

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Characterizing Regular Expressions

We prove that languages described by Regular expressions (RE) and Regular Languages are the same set, i.e., RE = RL

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Thm 1.54: RL ~ RE

We need to prove both ways:

If a language is described by a regular expression,

then it is regular (Lemma 1.55) (We will show we can convert a regular expression R into an NFA M such that L(R)=L(M))

The second part:

If a language is regular, then it can be described by a regular expression (Lemma 1.60)

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Regular expression to NFA

Claim: If L = L(e) for some RE e, then L = L(M) for some NFA M Construction: Use inductive definition 1. R = a, with a∈Σ 2. R = ε 3. R = ∅ 4. R = (R1∪R2), with R1 and R2 regular expressions 5. R = (R1•R2), with R1 and R2 regular expressions 6. R = (R1*), with R1 a regular expression a

1. 2. 3. 4,5,6: similar to

closure of RL under regular operations.

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Examples of RE to NFA conv.

L = {ab,ba} L = {ab,abab,ababab,……} L = {w | w = ambn, m<10, n>10}

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Back to RL ~ RE

The second part (Lemma 1.60):

If a language is regular, then it can be described by a regular expression.

Proof strategy:

§ regular implies equivalent DFA. § convert DFA to GNFA (generalized NFA) § convert GNFA to NFA.

GNFA: NFA that have regular expressions as transition labels

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Example GNFA

qS qA

01* 0* ∪ 11 0110 ∅ ε

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Generalized NFA - defn

Generalized non-deterministic finite automaton M=(Q, Σ, δ, qstart, qaccept) with

Q finite set of states
Σ the input alphabet
qstart the start state
qaccept the (unique) accept state
δ:(Q - {qaccept})×(Q - {qstart}) → R is the transition

function (R is the set of regular expressions over Σ) (NOTE THE NEW DEFN OF δ)

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Characteristics of GNFA’s δ

δ:(Q\{qaccept})×(Q\{qstart}) → R

The interior Q\{qaccept,qstart} is fully connected by δ From qstart only ‘outgoing transitions’ To qaccept only ‘ingoing transitions’ Impossible qi→qj transitions are labeled “δ(qi,qj) = ∅” qS qA R∈R Observation: This GNFA recognizes the language L(R)

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Proof Idea of Lemma 1.60

Proof idea (given a DFA M): Construct an equivalent GNFA M’ with k≥2 states Reduce one-by-one the internal states until k=2 This GNFA will be of the form This regular expression R will be such that L(R) = L(M) qS qA R

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DFA M → Equivalent GNFA M’

Let M have k states Q={q1,…,qk} with start state q1

Add two states qaccept and qstart

qS q1

ε

Connect qstart to original q1:

qi

∅

qj

Complete missing transitions by

qA qj

ε

Connect old accepting states to qaccept
Join multiple transitions:

qi qj

1

becomes qi

0∪1

qj

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Remove Internal state of GNFA

If the GNFA M has more than 2 states, ‘rip’ internal qrip to get equivalent GNFA M’ by:

Removing state qrip: Q’=Q\{qrip}
Changing the transition function δ by

δ’(qi,qj) = δ(qi,qj) ∪ (δ(qi,qrip)(δ(qrip,qrip))*δ(qrip,qj))

for every qi∈Q’\{qaccept} and qj∈Q’\{qstart} qi

R4∪(R1R2*R3)

qj qi R2 qj

R4

qrip R1 R3 =

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Proof Lemma 1.60

Let M be DFA with k states Create equivalent GNFA M’ with k+2 states Reduce in k steps M’ to M’’ with 2 states The resulting GNFA describes a single regular expressions R The regular language L(M) equals the language L(R) of the regular expression R

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Proof Lemma 1.60 - continued

Use induction (on number of states of

GNFA) to prove correctness of the conversion procedure.

Base case: k=2.
Inductive step: 2 cases – qrip is/is not on

accepting path.

qi

R4∪(R1R2*R3)

qj qi R2 qj

R4

qrip R1 R3 =

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Recap RL = RE

Let R be a regular expression, then there exists an NFA M such that L(R) = L(M) The language L(M) of a DFA M is equivalent to a language L(M’) of a GNFA = M’, which can be converted to a two-state M’’ The transition qstart ⎯R→ qacceptof M’’