1 cdt314 faber formal languages, automata and models of computation lecture 5 school of innovation,...
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CDT314
FABER
Formal Languages, Automata and Models of Computation
Lecture 5
School of Innovation, Design and Engineering Mälardalen University
2012
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Content
- More Properties of Regular Languages (RL)- Standard Representations of RL- Elementary Questions about RL- Non-Regular Languages- The Pigeonhole Principle- The Pumping Lemma- Applications of the Pumping Lemma- NFA-DFA repetition
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More Properties of
Regular Languages
Based on C Busch, RPI, Models of Computation
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We have shown
Regular languages are closed under
Union
Concatenation
Star operation
Reverse
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Namely, for regular languages and :1L 2L
21 LL
21LL
1L
Union
Concatenation
Star operation
Reverse RL1
Regular
Languages
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We will show that
Regular languages are also closed under
Complement
Intersection
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Namely, for regular languages and :1L 2L
1L
21 LL
Complement
Intersection
Regular
Languages
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Complement
Theorem For regular language
the complement is regular. LL
Proof
Take DFA that accepts and exchange:L
non-accepting states accepting states
LResulting DFA accepts
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Examplea
b ba,
ba,
0q 1q 2q
)*( baLL
a
b ba,
ba,
0q 1q 2q
)))((( bababaaLL
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Intersection
Theorem For regular languages and
the intersection is regular. 21 LL 1L 2L
Proof Apply DeMorgan’s Law:
2121 LLLL
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21 , LL regular
21 , LL regular
21 LL regular
21 LL regular
21 LL regular
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Standard Representations of
Regular Languages
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Regular Language Representations
DFA
NFA
Regular
Expression
Regular
Grammar
Regular
Language
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Elementary Questionsabout
Regular Languages
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Membership Question
Question: Given regular language
and string
how can we check if ?
L
Lw w
Answer: Take the DFA that accepts
and check if is accepted.
Lw
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Lw
DFAw
Lw
DFAw
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Take the DFA that accepts .
Check if there is a path from
the initial state to a final state.
L
Given regular language
how can we check
if is empty: ?
L
L )( L
Question:
Answer:
Empty Language Question
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DFA
L
L
DFA
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Given regular language
how can we check
if is finite?
L
L
Take the DFA that accepts .
Check if there is a walk with a cycle
from the initial state to a final state.
L
Question:
Answer:
Finiteness Question
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DFA
L is infinite
DFA
L is finite
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Given regular languages and
how can we check if ? 1L 2L
21 LL Question:
)()( 2121 LLLLFind ifAnswer:
Equality Question
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)()( 2121 LLLL
21 LL 21 LLand
21 LL
1L2L 1L2L
21 LL 12 LL 2L 1L
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)()( 2121 LLLL
21 LL 21 LLor
1L2L 1L2L
21 LL 12 LL
21 LL
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Non-Regular Languages
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Regular languages
ba * acb *
...etc
*)( bacb
Non-regular languages???
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How can we prove that a language
is not regular?
L
Prove that there is no DFA that accepts L
Problem: this is not easy to prove.
Solution: the Pumping Lemma !
a
b
b
a
a
b
}0:{ nbaL nn
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The Pigeonhole Principle
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The Pigeonhole Principle
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pigeons
pigeonholes
4
3
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A pigeonhole must
contain at least two pigeons
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...........
...........
pigeonsn
pigeonholesm mn
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The Pigeonhole Principle
...........
pigeons
pigeonholes
n
m
mn There is a pigeonhole
with at least 2 pigeons
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The Pigeonhole Principleand DFAs
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DFA with states 4
1q 2q 3qa
b
4q
b
b b
b
a a
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1q 2q 3qa
b
4q
b
b
b
a a
a
In walks of strings:
aab
aa
ano state
is repeated
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In walks of strings:
1q 2q 3qa
b
4q
b
b
b
a a
a
...abbbabbabb
abbabb
bbaa
aabb a state
is repeated
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If the walk of string has length
1q 2q 3qa
b
4q
b
b
b
a a
a
w 4|| w
then a state is repeated
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If in a walk of a string
transitions states of DFA
then a state is repeated
Pigeonhole principle for any DFA:
w
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In other words for a string
transitions are pigeons
states are pigeonholesq
a
w
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A string has length number of states w
A state must be repeated in the walk of wq
In general
...... ......
walk of w
q
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The Pumping Lemmafor Regular Languages
see alsohttp://www.math.uu.se/~salling/Movies/Nonregularity.mov
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Take an infinite regular language L
DFA that accepts L
nstates
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Take string with w Lw
There is a walk with label w
.........
walk w
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If string has length w1|| nmw
then, from the pigeonhole principle:
a state is repeated in the walkq w
...... ......
walk w
( number of states)
q
n
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Write zyxw
...... ......
x
y
z
q
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myx ||
Lengths:
1|| y
...... ......
x
y
z
q
(from pigeon principle, as q is the first repetition in sequence)
(there is a walk in the graph)
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The string is accepted zxObservation:
...... ......
x
y
z
q
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The string
is accepted
zyyxObservation:
...... ......
x
y
z
q
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The string
is accepted
zyyyxObservation:
...... ......
x
y
z
q
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The string
zyx iGenerally:
...,2,1,0i
...... ......
x
y
z
q
is accepted
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The Pumping Lemma
Given an infinite regular language L
there exists an integer such that m
for any string with length Lw mw ||
we can write zyxw
with andmyx || 1|| y
such that: Lzyx i ...,2,1,0i
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Applications of
the Pumping Lemma
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Theorem
The language}0:{ nbaL nn
is not regular.
Proof
Use the Pumping Lemma!
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Assume to the contrary,
that is a regular language.L
Since is infinite
we can apply the Pumping Lemma.
L
}0:{ nbaL nn
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Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||with length
mmbaw e.g. pick
m
}0:{ nbaL nn
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Write: zyxba mm
it must be that length
From the Pumping Lemma
1||,|| ymyx
Therefore:
1, kay k
babaaaaba mm ............
x y z
m m
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From the Pumping Lemma: Lzyx i ...,2,1,0i
We can choose
mmbazyx
0i
We have:
1, kay k
Lbaw mkm
CONTRADICTION!
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Therefore: Our assumption that
is a regular language is not true.
L
Conclusion
L is not a regular language.
END OF PROOF
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Regular languages
ba * acb *
...etc
*)( bacb
Non-regular languages
}0:{ nba nn
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Theorem The language
Proof
Use the Pumping Lemma!
is not regular.
*}:{ wwwL R },{ ba
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*}:{ wwwL R
Assume to the contrary,
that is a regular language.
Since is infinite
we can apply the Pumping Lemma.
L
L
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mmmm abbaw pick
Pick a string such that: w Lw
mw ||length
Let be the integer in the Pumping Lemma.m
*}:{ wwwL R
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Write zyxabba mmmm
it must be that length
From the Pumping Lemma
ababbabaaaaabba mmmm ..................
x y z
m m m m
1||,|| ymyx
1, kay k
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ababbabaaaaabba mmmm ..................
x y z
m m m m
1, kay k
We can choose 0i
CONTRADICTION!
So we get a’s on the left,
while is on the right:
km m
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Conclusion
L is not a regular language.
Our assumption that
is a regular language is not true.
Therefore:
END OF PROOF
L
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Regular languages
ba * acb *
...etc
*)( bacb
Non-regular languages
}0:{ nba nn *}:{ wwwR
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Theorem The language
is not regular.
Proof
Use the Pumping Lemma.
}0,:{ lncbaL lnln
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Assume to the contrary
that is a regular language.L
Since is infinite
we can apply the Pumping Lemma.
L
}0,:{ lncbaL lnln
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mmm cbaw 2Pick
Let be the integer in the Pumping Lemma.
Pick a string such that: w Lw
mw ||length
m
}0,:{ lncbaL lnln
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Write zyxcba mmm 2
cccbcabaaaaacba mmm ..................2
x y z
m m m2
it must be that length
From the Pumping Lemma
1||,|| ymyx
1, kay k
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From the Pumping Lemma
Lzyx i ...,2,1,0i
Thus:
Lcbazxzyx mmkm 20
Lzyx 0
mmm cbazyx 2We have:
1, kay k
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Lcba mmkm 2Therefore:
Lcba mmkm 2
BUT:
CONTRADICTION!
}0,:{ lncbaL lnln
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Conclusion
L is not a regular language.
LTherefore: Our assumption that
is a regular language is not true.
END OF PROOF
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Regular languages
Non-regular languages
}0,:{ lncba lnln
*}:{ wwwR
}0:{ nba nn
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Theorem
The language
is not regular.
ProofUse the Pumping Lemma.
}0:{ ! naL n
nnn )1(21!
Factorial of n, (n!) is the product of all positive integers less than or equal to n. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120 The value of 0! is 1.
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Assume to the contrary
that is a regular language.L
Since is infinite
we can apply the Pumping Lemma.
L
}0:{ ! naL n
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!maw Pick
Pick a string such that: w
Lw mw ||length
Let be the integer in the Pumping Lemma.m
}0:{ ! naL n
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Write zyxam !
From the Pumping Lemma
aaaaaaaaaaam ...............!
x y z
m mm !
it must be that length 1||,|| ymyx
mkay k 1,
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From the Pumping Lemma:
Lzyx i ...,2,1,0i
Thus:
!mazyx
Lazyyxzyx km !2
Lzyx 2
We have:
mkay k 1,
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La km !Therefore:
!! pkm
}0:{ ! naL nSince:
mk 1
mk 1There is p
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However
)!1( += m)1(! += mm!! +< mmm
!!+£ mm!+£ mmkm !
for 1m
! ! ( 1)!m m k m
!! pkm for any p
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La km !
Therefore: La km !
BUT:
CONTRADICTION!
}0:{ ! naL n mk 1and
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Conclusion
L is not a regular language.
Our assumption that
is a regular language is not true.LTherefore:
END OF PROOF
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Regular languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln
*}:{ wwwR}0:{ nba nn
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Theorem
The language
is not regular.
ProofUse the Pumping Lemma.
}:{ primeiaL i
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Assume to the contrary,
that is a regular language.L
Since is infinite
we can apply the Pumping Lemma.
L
}:{ primeiaL i
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it must be that length 1||,|| ymyx
Lw mw ||length
From the Pumping Lemma:
Lzyx i ...,2,1,0i
The length of zxyw k 1
must be prime for each string of .
mk
L
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Thus:
But, choosing
which is not prime!
CONTRADICTION!
))(1(
))((
)()(
)()( 1
ylengthk
ylengthkk
ylengthxyzlength
zxyylengthzxylengthk
kk
)( mk
Lw
( )length xyz k
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Conclusion
L is not a regular language.
Our assumption that
is a regular language is not true.LTherefore:
END OF PROOF
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Regular languages
Non-regular languages
}0:{ ! nan}0,:{ lncba lnln
*}:{ wwwR}0:{ nba nn
}:{ primeiaL i
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Theorem
The language}0:{ nbaL nn
is not regular.
Proof
Use the Pumping Lemma!
For your exercise:An alternative variant of proof from slide 53.
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Assume to the contrary
that is a regular language.L
Since is infinite
we can apply the Pumping Lemma.
L
}0:{ nbaL nn
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Pick a string such that: w Lw
mw ||length
mmbaw Pick
Let be the integer in the Pumping Lemma.m
}0:{ nbaL nn
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Write: zyxba mm
it must be that: length
From the Pumping Lemma
1||,|| ymyx
Therefore: babaaaaba mm ............
1, kay kx y z
m m
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From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmbazyx
Lbazyyxzyx mkm 2
Lzyx 2
We have: 1, kay k
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Lba mkm Therefore:
}0:{ nbaL nnBUT:
Lba mkm
CONTRADICTION
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Our assumption that
is a regular language is not true.
L
Conclusion: L is not a regular language.
Therefore:
END OF PROOF
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Pumping Lemma, in short
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Observation:
Every language of finite size has to be regular.
(We can easily construct an NFA that accepts every string in the language, or a union of regular expressions)
Therefore, every non-regular language
has to be of infinite size. (contains an infinite number of strings)
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Suppose you want to prove that
An infinite language is not regular
1. Assume the opposite: is regular
2. The pumping lemma should hold for
3. Use the pumping lemma to obtain a contradiction
L
L
L
4. Therefore, is not regular L
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Explanation of Step 3: How to get a contradiction
with pumping lemma
i. Find a particular string which satisfies
the conditions of the pumping lemma
Lw
ii. Write xyzw
iii. Show that Lzxyw i for some 1i
iv. This gives a contradiction, since from
pumping lemma Lzxyw i
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Note: It suffices to show that
only one string
gives a contradiction Lw
You don’t need to obtain
contradiction for every Lw
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Extra Examples on Regular Languages
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FA RE
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FA RE
or
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NFA DFA
Subset Construction
Delmängdkonstruktion
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Convert an NFA to DFAusing the subset construction
108
Each state of the DFA is a set of states of the NFA.
Start by the initial state of the DFA which is the -closure of the initial state of the NFA (all the states you reach by -transitions from the initial state) .
CLOSE{0}={0, 1, 3}=S0
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Determine the transition function of the DFA on all inputs. Begin with the initial state S0, and determine the transition on input a.
CLOSE{0} = {0, 1, 3} = S0
(S0, a) = CLOSE{1, 2}
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The e-closure of the set {1, 2} is {1, 2, 3}This is a new state in the DFA, call it S1.
CLOSE{0} = {0, 1, 3} = S0
(S0, a) = CLOSE{1, 2} = {1, 2, 3} = S1
111
With the initial state S0, determine the transition on input b. CLOSE{0} = {0, 1, 3} = S0
(S0, a) = CLOSE{1, 2} = {1, 2, 3} = S1
(S0, b) = CLOSE{3}
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The -closure of the set {3} is {1, 3}. This is a new state in the DFA, call it S2.
CLOSE{0} = {0, 1, 3} = S0
(S0, a) = CLOSE{1, 2} = {1, 2, 3} = S1
(S0, b) = CLOSE{3} = {1, 3} = S2
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Determine the transition function of the DFA from state S1 oninputs a and b. On a there is no where to go in the NFA, so wecreate a “sink“(“trap”) state for DFA.
(S0, a) = CLOSE{1, 2} = {1, 2, 3} = S1
(S0, b) = CLOSE{3} = {1, 3} = S2
(S1, a) = CLOSE{} = = S3
114
Determine the transition function of the DFA from state S1 oninputs a and b. On b there is a transition to an existing state S2.
(S0, a) = CLOSE{1, 2} = {1, 2, 3} = S1
(S0, b) = CLOSE{3} = {1, 3} = S2
(S1, a) = CLOSE{} = = S3
(S1, b) = CLOSE{3} = {1, 3} = S2
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Determine the transition from state S2 on inputs a and b.
(S0, a) = CLOSE{1, 2} = {1, 2, 3} = S1
(S0, b) = CLOSE{3} = {1, 3} = S2
(S1, a) = CLOSE{} = = S3
(S1, b) = CLOSE{3} = {1, 3} = S2
(S2, a) = CLOSE{} = = S3
(S2, b) = CLOSE{3} = {1, 3} = S2
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Determining the transition from state S2 on inputs a and b is easy;from the empty set of states there are no transitions in the NFA. Inthe DFA this is represented by a transition from the empty setback to itself.
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The final states of the DFA are determined from the final states of the NFA. State 3 was the only final state in the NFA. (Any set of NFA states containing a final state is a final state in the DFA.)
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NFA DFA RESULT
State elimination
l+10* 0*11
0*1
q0 q1 q2
01
q0 q2
01
(l+10*)(0*1)*0*11
q0 q2
(l+10*)(0*1)*0*11 + 01
Example 7
Convert the following NFA to DFA.
s
p
0
ε0 ε
0
1
1
ε
0
r
srp
0 qrp
1
s
q, r, p
0
10
1
q
s
p
0
0
0
1
10
r
q
0q
1
q r p
0
1
0
1
Converting the NFA to DFA
p r sp r s
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Application: Lex - lexical analyzer
Lex
RE NFA DFAMinimalDFA
The final states of the DFA are associated with actions.