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CD5560

FABER

Formal Languages, Automata and Models of Computation

Exercise 2

Mälardalen University

2007

2

NEXT WEEK!Midterm Exam 1

Regular Languages

Place: U2-114

Time: Tuesday 2007-04-24, 10:15-12:00

It is OPEN BOOK.

(This means you are allowed to bring in one book of your choice.)

It will cover lectures 1 through 5 (Regular Languages).

3

Tenta 29 okt 1999; uppgift 2 (L Salling)

Construct (and explain)

},,{ cbawich strings contain all three symbols!

),( LabbcbaLaba

a) A regular expression over

4

Solution bcacba

abcbac

acbcab

baccacabba )()(

abccbcbaab )()( acbbcbcaac )()(

or

5

Construct (and explain)

},,{ cbawich strings contain all three symbols!

b) A minimal DFA

for a language L over

6

ca,

1

43

7

5

2

6

8

a

b

ba,

c

cba ,,

b

c

bc

b c

bc,

c

b

a

a

a

a

7

ca,

1

43

7

5

2

6

8

a

b

ba,

c

cba ,,

b

c

bc

b c

bc,

c

b

a

a

a

a

}7,6,5,4,3,2,1{ }8{

}6,5,4,3,2,1{a

}7{ }8{

}1{ }2{ }4{ }6{ }3{ }5{ }7{ }8{

c

}6,4,2,1{b

}3{ }5{ }7{ }8{

Särskiljningsalgoritm

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c) En reguljär grammatik för L

ca,

S

BC

F

D

A

E

G

a

b

ba,

c

cba ,,

b

c

bc

b c

bc,

c

b

a

a

a

a

cCbBaAS ||

cDbEaAA ||

aEcFbBB ||

bFaDcCC ||

bGcDaDD ||

cGbEaEE ||

aGcFbFF ||

cGbGaGG |||

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Tenta 24 okt 1994; uppgift 2 (L Salling)

Reguljära?

},{ ba

vars strängar innehåller ett jämnt antal a:n!

a) Språket över

Ja, språket är reguljärt och beskrivs med ett reguljärt uttryck:

)( ababb

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Tenta 24 okt 1994; uppgift 2 (L Salling)

}(,),,{ a

b) De välformade aritmetiska uttrycken formade i alfabetet

Nej, språket är inte reguljärt: Ta följande sträng:

))))(((( aaaaa N stycken a adderas

Om språket vore reguljärt skulle det kunna pumpas.

Men de N avslutande tecknen består enbart av höger- parenteser och kan

inte ändras utan att balansen med vänsterparenteserna förstörs.

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Tenta 15 mars 1995; uppgift 3 (L Salling)

Reguljära?

}

3|},{{

trängpalindromsenärsom

längdavprefixettharwbaw a)

Ja, språket är reguljärt och beskrivs med ett reguljärt uttryck:

))(( babbbbababaaaa

12

Tenta 15 mars 1995; uppgift 3 (L Salling)

Reguljära?

b)

}

3|},{{

trängpalindromsenärsom

MINSTlängdavprefixettharwbaw

Nej. Strängen22 aabbaw NN

vars enda palindromprefix längre än 2 är strängen själv, kan inte pumpas någonstans inuti b-block utan att falla ur språket.

13

Tenta 15 mars 1995; uppgift 3 (L Salling)

Reguljära?

c)

}

3|},{{

trängpalindromsenärsom

MINSTlängdavprefixingetwbaw

Nej.

Om det vore reguljärt skulle även föregående språk vara det (eftersom det är komplementspråk, och regulariteten bevaras under komplementbildning).

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Pumping Lemma is necessary but not sufficient for RL

OBS! The pumping lemma does not give a sufficient condition for a language to be regular! You can not use it to show that language is regular.

For example, the language

(strings over the alphabet {0,1} consisting of a nonempty even palindrome followed by another nonempty string) is not regular but can still be "pumped" with m = 4:

Suppose w=uuRv has length at least 4. If u has length 1, then |v| ≥ 2 and we can take y to be the first character in v. Otherwise, take y to be the first character of u and note that yk for k ≥ 2 starts with the nonempty palindrome yy. For a practical test that exactly characterizes regular languages, see the Myhill-Nerode theorem. The typical method for proving that a language is regular is to construct either a Finite State Machine or a Regular Expression for the language.

{ | , {0,1} } Ruu v u v

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Minimizing DFA’s

By Partitioning(Delmängdskonstruktion)

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Minimizing DFA’s

Different methods

All involve finding equivalent states:

States that go to equivalent states under all inputs

We will use the Partitioning Method

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Minimizing DFA’s by Partitioning

Consider the following DFA (from Forbes Louis):

• Accepting states are yellow• Non-accepting states are blue• Are any states really the same?

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• S2 and S7 are really the same: Both Final states Both go to S6 under input b Both go to S3 under an a • S0 and S5 really the same. Why?• We say each pair is equivalent

Are there any other equivalent states?We can merge equivalent states into 1 state

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Partitioning AlgorithmFirst

Divide the set of states into

Final and Non-final states

Partition I

Partition II

a b

S0 S1 S4

S1 S5 S2

S3 S3 S3

S4 S1 S4

S5 S1 S4

S6 S3 S7

*S2 S3 S6

*S7 S3 S6

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Partitioning AlgorithmNow

See if states in each partition each go to the same partition

S1 & S6 are different from the rest of the states in Partition I (but like each other)

We will move them to their own partition

a b

S0 S1 I S4 I

S1 S5 I S2 II

S3 S3 I S3 I

S4 S1 I S4 I

S5 S1 I S4 I

S6 S3 I S7 II

*S2 S3 I S6 I

*S7 S3 I S6 I

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Partitioning Algorithm

a b

S0 S1 S4

S5 S1 S4

S3 S3 S3

S4 S1 S4

S1 S5 S2

S6 S3 S7

*S2 S3 S6

*S7 S3 S6

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Partitioning AlgorithmNow again

See if states in each partition each go to the same partition

In Partition I, S3 goes to a different partition from S0, S5 and S4

We’ll move S3 to its own partition

a b

S0 S1 III S4 I

S5 S1 III S4 I

S3 S3 I S3 I

S4 S1 III S4 I

S1 S5 I S2 II

S6 S3 I S7 II

*S2 S3 I S6 III

*S7 S3 I S6 III

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Partitioning Algorithm Note changes in S6,

S2 and S7

a b

S0 S1 III S4 I

S5 S1 III S4 I

S4 S1 III S4 I

S3 S3 IV S3 IV

S1 S5 I S2 II

S6 S3 IV S7 II

*S2 S3 IV S6 III

*S7 S3 IV S6 III

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Partitioning AlgorithmNow S6 goes to a different partition on an a from S1

S6 gets its own partition.

We now have 5 partitions

Note changes in S2 and S7

a b

S0 S1 III S4 I

S5 S1 III S4 I

S4 S1 III S4 I

S3 S3 IV S3 IV

S1 S5 I S2 II

S6 S3 IV S7 II

*S2 S3 IV S6 V

*S7 S3 IV S6 V

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Partitioning AlgorithmAll states within each of the 5 partitions are identical.

We might as well call the

states I, II III, IV and V.

a b

S0 S1 III S4 I

S5 S1 III S4 I

S4 S1 III S4 I

S3 S3 IV S3 IV

S1 S5 I S2 II

S6 S3 IV S7 II

*S2 S3 IV S6 V

*S7 S3 IV S6 V

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Partitioning Algorithm

a b

I III I

*II IV V

III I II

IV IV IV

V IV II

Here they are:

Va

a

aa ab

b

b

bb

b

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Chomsky Hierarchy

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Automata theory: formal languages and formal grammars

Chomsky hierarchy

Grammars Languages Minimal automaton

Type-0 Unrestricted Recursively enumerable Turing machine

Type-1 Context-sensitive Context-sensitive Linear-bounded

Type-2 Context-free Context-free Pushdown

Type-3 Regular Regular Finite

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Grammar Languages Automaton Production

rules

Type-0 Recursively enumerable

Turing machine No restrictions

Type-1 Context-sensitive Linear-bounded non-deterministic Turing machine

Type-2 Context-free Non-deterministic pushdown automaton

Type-3 Regular Finite state automaton and

Automata theory: formal languages and formal grammars

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Regular Languages

}{ nnba }{ Rww

Context-Free Languages

Non-regular languages

}0:{ ! nan}0,:{ lncba lnln