cd5560 faber formal languages, automata and models of computation exercise 2
DESCRIPTION
CD5560 FABER Formal Languages, Automata and Models of Computation Exercise 2 Mälardalen University 2007. NEXT WEEK! Midterm Exam 1 Regular Languages. Place: U2-114 Time: Tuesday 2007-04-24, 10:15-12:00 It is OPEN BOOK . - PowerPoint PPT PresentationTRANSCRIPT
1
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
8
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 |||
9
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
10
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.
11
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).
14
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
15
Minimizing DFA’s
By Partitioning(Delmängdskonstruktion)
16
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
17
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?
18
• 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
19
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
20
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
21
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
22
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
23
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
24
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
25
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
26
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
27
Chomsky Hierarchy
28
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
29
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
30
Regular Languages
}{ nnba }{ Rww
Context-Free Languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln