not everything is regular

2NICTA Researcher,Adjunct at the Australian National University and Griffith University

Not Everything is Regular

Charles Gretton2

4 March 2014

NICTA Funding and Supporting Members and Partners

Not Everything is Regular Copyright NICTA 2014 Charles Gretton2 1/41

Language Class

• We have spoken about the regular languages.

• Any language described using a regular expression.

• Any language described using a DFA, NFA and ε-NFA.


Properties of a Class

1 Closure properties.• The reversal of a regular language is regular.• The star-closure of a regular language is regular.• The union of two regular languages is regular.• The intersection of two regular languages is regular.• The complement of a regular language is regular.• The concatenation of two regular languages is regular.• The difference of two regular languages is regular.• A homomorphism of a regular language is regular.• An inverse homomorphism of a regular language is regular.

2 Decision properties.• Is the language described empty?• Is the language described finite?• Is the string X a member of the language described?• Take two descriptions, do both describe the same language?• Take two descriptions, does one contain the language of the other?


Motivating Decision Properties of a Class

• You are given a transaction protocol, does a transaction necessarilyterminate? – i.e. Is the language finite?

• You are given a transaction protocol, can it fail? – i.e. Is thelanguage non-empty?

• Shortest regular expression or minimum-state DFA? – i.e. Are twodescriptions, one smaller than the other, describing the samelanguage?


Is the string X a member of the languagedescribed?

• We know how to do this...• Take “ababb”

A B C

D

a

b

ǫa

b

a, b

1



• Take “ababb”

bb

epsa

aa,b



• Take “ababb”• Starting

bb

epsa

aa,b



• Take “ababb”• Reading: a

bb

epsa

aa,b



• Take “ababb”• Reading: ab

bb

epsa

aa,b



• Take “ababb”• Reading: aba

bb

epsa

aa,b



• Take “ababb”• Reading: abab

bb

epsa

aa,b



• Take “ababb”• Reading: ababb

bb

epsa

aa,b


Is a Language Empty?

• Regular Expressions↔ DFA↔ NFA↔ ε-NFA

• Assume a DFA representation.

• Compute the set of states R reachable from the starting state q0.

• If R ∩ F 6= ∅, then no, else yes.


Is a Language Finite? – DFA example

• If a DFA has N states, and a string with N + i , i ≥ 0, symbols isaccepted, then no.

• Otherwise, yes.


Is a Language Finite? – DFA example – proof

• If a DFA has N states, and a string with N + i , i ≥ 0, symbols isaccepted, then no.

• For the N + i length input, clearly one state is traversed twice – i.e.pigeonhole principle.

• Therefore there must be a loop.

• QED


Is a Language Finite? – DFA example – proof

• If a DFA has N states, and a string with N + i , i > 0, symbols isaccepted, then no.

• For the N + i length input, clearly one state is traversed twice – i.e.Pigeon hole argument.

• Therefore there must be a loop.

• Otherwise, yes.


Is a Language Finite? – A finite test.

• How do we get a computer to decide if a language is finite?

• What is the finite test to eliminate the infinite strings of length N + i ,i > 0?

• We established that there must be a cycle.

• But that cycle can only be so long.......

• Maximally we traverse all the states twice – i.e. 2N − 1.

• If an automaton accepts a string of length N + i , i > 0, then there is acycle of length at most 2N − 1.



• What is a better algorithm than testing all strings of lengthless-than-or-equal 2N − 1?



• What is a better algorithm than testing all strings of lengthless-than-or-equal 2N − 1?

• Find cycles between q0 and states in F .

1 Eliminate unreachable states

2 Eliminate states from which the accepting states cannot be reached.

3 If any self loops remain, then it is infinite.

4 Is there a strongly connected component of the directed graph of size2 or greater? If so, again the language is infinite.


Language Equivalence. – DFA example

• Remember the product of two automata from the first lecture?

• A1 = 〈Q1,Σ, δ1, q10 ,F

1〉.

• A2 = 〈Q2,Σ, δ2, q20 ,F

2〉.

• A2 × A1 = 〈Q2 × Q1,Σ, δ2×1, (q20 , q

10),F 2 × F 1〉.

• The product has the number of sates Q2 × Q1

• taking q2 ∈ Q2 and q1 ∈ Q1

δ2×1((q2, q1), a) = (δ2(q2, a), δ1(q1, a))

• There is a one-to-one correspondence between pairs of states fromthe source automata, and states in the product.



• Remember the product of two automata from the first lecture?

• Product example:

ab

a,b

ba

a

b

a a

a

b

bba

b

A B

CD

AC AD

BC BD



• Make accepting states in the product where exactly one of thesource machines is accepting.

• This machine accepts strings iff they are in one of the sourcelanguages.

• Product example:

ab

a,b

ba

a

b

a a

a

b

bb a

b

A B

CD

AC AD

BC BD



• Product automaton describes the empty language if the two sourceautomata are equivalent.

• Do you recall the empty test?


Is a Language Empty?

• Assume a DFA representation.

• Compute the set of states R reachable from the starting state q0.

• If R ∩ F 6= ∅, then no, else yes.


Is one language contained in another? – DFAexample

• Two regular languages L1 and L2 described using two automata:

1 A1 = 〈Q1,Σ, δ1, q10 ,F

1〉.

2 A2 = 〈Q2,Σ, δ2, q20 ,F

2〉.

• We want to test L1 ⊆ L2

• What are the accepting states of the product, if we want to testcontainment?


Is one language contained in another? – DFAexample

• Two regular languages L1 and L2 described using two automata:

1 A1 = 〈Q1,Σ, δ1, q10 ,F

1〉.2 A2 = 〈Q2,Σ, δ2, q2

0 ,F2〉.

• We want to test L1 ⊆ L2

• What are the accepting states of the product, if we want to testcontainment?

• Accepting in product state: Q1 label is accepting and Q2 label is not.

• If the result describes an empty language, then we can never acceptan A1 string without that also being accepted by A2.


Language Containment. – DFA example

• In product state: Q1 label is accepting and Q2 label is not.

• Starting at state “AC” this product accepts no strings, hencecontainment holds.

ab

a,b

ba

a

b

a a

a

b

bb a

b

A B

CD

AC AD

BC BD


Language Class – Is it regular?

• Consider the following language, described using a setcomprehension:

{anbn|n > 0}

• Do you think it is regular?


Language Class – Is it regular?


{anbn|n > 0}

• It is not regular!

• And our job is to prove it.

• We will do this using the pumping lemma.


Pumping Lemma

• L is regular.

• There is a finite positive integer constant N(L), so that forall W ∈ LN(L) ≤ |W |W = XYZ satisfying:

1 Y 6= ε

2 |XY | ≤ N(L)

3 ∀k ≥ 0, we have XY k Z ∈ L


Pumping Lemma – Example

• Take |Σ| = 1 and L is the set of strings of prime length.

• Is this a regular language?



• Take |Σ| = 1 and L are sequences of prime length.

• Is this a regular language?

• No, the machine would need infinite states.

• But let us prove it using the pumping lemma.



• Take |Σ| = 1 and L are sequences of prime length.• Proof by contradiction that this is not a regular language.• N(L) is finite remember.• Take a prime p satisfying p ≥ N(L) + 2.• For |XY | ≤ N(L) take |XYZ | = p with |Y | = m so |XZ | = p −m• The pumping lemma (which is true) tells us XY p−mZ ∈ L

|XY p−mZ | = |XZ |+ (p −m)|Y | = p −m + (p −m)m= (m + 1)(p −m)

(m + 1) > 1m = |Y | ≤ |XY | ≤ N(L); ; therefore

(p −m) > 1

• Oh no, p is nonprime, as it has two factors greater than 1.• QED


Pumping Lemma Continued


L = {anbn|n > 0}


• XYZ = aN(L)bN(L), |XYZ | ≥ N(L), |Y | > 0, |XY | ≤ N(L)

• Hmmm.. XYYZ ∈ L and XYYZ 6∈ {anbn|n > 0}

• QED


Properties of a Language Class

1 Closure properties.• The reversal of a regular language is regular.• The star-closure of a regular language is regular.• The union of two regular languages is regular.• The intersection of two regular languages is regular.• The complement of a regular language is regular.• The concatenation of two regular languages is regular.• The difference of two regular languages is regular.• A homomorphism of a regular language is regular.• An inverse homomorphism of a regular language is regular.

2 Decision properties.• Is the language described empty?• Is the language described finite?• Is the string X a member of the language described?• Take two descriptions, do both describe the same language?• Take two descriptions, does one contain the language of the other?


Motivating Closure Properties of a LanguageClass

• e.g. Regular languages as closed under union, concatenation andKleene-star.

• We used those properties to create a representation, namely regularexpressions.

• Properties will also help us show that a language is not regular.



• Consider the following language, described in natural language:

• Any string comprising an equal number of a and b symbols of length≤ 1000.




• Consider the following language, described in natural language:• Any string comprising an equal number of a and b symbols of length≤ 1000.

• Yes, it is regular!!• A regular language is a language that can be given by a regular

expression.• A finite language is a finite set of strings:

{X1,X2, ..,Xn}• The regular expression that gives that finite language is:

X1 + X2 + ..+ Xn

• QEDNot Everything is Regular Copyright NICTA 2014 Charles Gretton2 38/41



• Any string comprising an equal number of a and b symbols.








• And our job is to prove it, using closure properties.







• Property: The intersection of two regular languages is regular.

• Let L1 be the above language.

• Take the language L2 ∩ L(a∗b∗)

• This is not regular, the RHS is regular, therefore L2 is not regular.


not everything is regular

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