regular languages, regular expressions, and pumping lemma · 2014-02-14 · regular languages,...
TRANSCRIPT
2/14/14 Costas Busch - RPI 1
Regular Languages, Regular Expressions, and Pumping Lemma
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Non Turing-Acceptable
Turing-Acceptable
decidable
Context-sensitive
Context-free
Regular
The Chomsky Hierarchy
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Standard Representations of Regular Languages
Regular Languages
DFAs
NFAs Regular Expressions
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When we say: We are given a Regular Language
We mean:
L
Language is in a standard representation
L
(DFA, NFA, or Regular Expression)
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1L 2L
21LLConcatenation:
*1LStar:
21 LL ∪Union:
Are regular Languages
For regular languages and we will prove that:
1L
21 LL ∩
Complement:
Intersection:
RL1Reversal:
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We say: Regular languages are closed under
21LLConcatenation:
*1LStar:
21 LL ∪Union:
1L
21 LL ∩
Complement:
Intersection:
RL1Reversal:
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1LRegular language
( ) 11 LML =
1M
Single accepting state
NFA 2M
2L
Single accepting state
( ) 22 LML =
Regular language
NFA
Take two languages
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}{1 baL n=a
b
1M
{ }baL =2ab2M
0≥n
Example
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Union NFA for
1M
2M
21 LL ∪
λ
λ
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ab
ab
λ
λ
}{1 baL n=
}{2 baL =
}{}{21 babaLL n ∪=∪NFA for
Example
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Concatenation NFA for 21LL
1M 2Mλ
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NFA for
ab ab
}{1 baL n=}{2 baL =
}{}}{{21 bbaababaLL nn ==
λ
Example
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Star Operation NFA for *1L
1M
λ
λ
*1L∈λλ λ
1
21
Lwwwww
i
k
∈
=
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NFA for *}{*1 baL n=
ab
}{1 baL n=
λ
λ
λ
λ
Example
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Reverse RL1
1MNFA for
ʹ′1M
1. Reverse all transitions
2. Make initial state accepting state and vice versa
1L
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}{1 baL n=a
b
1M
}{1nR baL =
ab
ʹ′1M
Example
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Complement
1. Take the DFA that accepts 1L
1M1L ʹ′1M1L
2. Make accepting states non-final, and vice-versa
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}{1 baL n=
ab
1M
ba,
ba,
}{*},{1 babaL n−= a
b
ʹ′1M
ba,
ba,
Example
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Intersection
1L regular
2L regular We show 21 LL ∩
regular
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DeMorgan’s Law: 2121 LLLL ∪=∩
21 , LL regular
21 , LL regular
21 LL ∪ regular
21 LL ∪ regular
21 LL ∩ regular
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Example
}{1 baL n=
},{2 baabL =
regular
regular }{21 abLL =∩
regular
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1Lfor for 2LDFA 1M
DFA
2M
Construct a new DFA that accepts
Machine Machine
M 21 LL ∩
M simulates in parallel and 1M 2M
Another Proof for Intersection Closure
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States in M
ji pq ,
1M 2MState in State in
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1M 2M
1q 2qa
transition 1p 2pa
transition
DFA DFA
11, pq a
New transition
MDFA
22, pq
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0q
initial state 0p
initial state
New initial state
00, pq
1M 2MDFA DFA
MDFA
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iq
accept state
jp
accept states
New accept states
ji pq ,
kp
ki pq ,
1M 2MDFA DFA
MDFA
Both constituents must be accepting states
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Example:
}{1 baL n=
ab
1M
0≥n
}{2mabL =
b
b
2M
0q 1q 0p 1p
0≥m
2q 2pa
a
ba, ba,
ba,
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00, pq
Automaton for intersection
}{}{}{ ababbaL nn =∩=
10, pqa
21, pq
b
ab 11, pq
20, pq
a
12, pq
22, pq
b
ba,
a
b
ba,
b
a
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M simulates in parallel and 1M 2M
M accepts string w if and only if:
accepts string w1Mand accepts string w2M
)()()( 21 MLMLML ∩=
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Regular Languages and Regular Expressions
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Regular Expressions Regular expressions describe regular languages Example: describes the language
*)( cba ⋅+
{ } { },...,,,,,*, bcaabcaabcabca λ=
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Recursive Definition αλ,,∅
( )11
21
21
*rrrrrr
⋅
+
Are regular expressions
Primitive regular expressions:
2r1rGiven regular expressions and
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Examples
( ) )(* ∅+⋅⋅+ ccbaA regular expression:
( )++ baNot a regular expression:
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Languages of Regular Expressions : language of regular expression Example
( )rL r
( ) { },...,,,,,*)( bcaabcaabcacbaL λ=⋅+
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Definition For primitive regular expressions:
( )
( ) { }
( ) { }aaL
L
L
=
=
∅=∅
λλ
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Definition (continued) For regular expressions and
1r 2r
( ) ( ) ( )2121 rLrLrrL ∪=+
( ) ( ) ( )2121 rLrLrrL =⋅
( ) ( )( )** 11 rLrL =
( )( ) ( )11 rLrL =
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Example Regular expression: ( ) *aba ⋅+
( )( )*abaL ⋅+ ( )( ) ( )*aLbaL +=
( ) ( )*aLbaL +=
( ) ( )( ) ( )( )*aLbLaL ∪=
{ } { }( ) { }( )*aba ∪=
{ }{ },...,,,, aaaaaaba λ=
{ },...,,,...,,, baababaaaaaa=
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Example Regular expression
( ) ( )bbabar ++= *
( ) { },...,,,,, bbbbaabbaabbarL =
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Example Regular expression ( ) ( ) bbbaar **=
( ) }0,:{ 22 ≥= mnbbarL mn
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Example Regular expression *)10(00*)10( ++=r
)(rL = { all strings containing substring 00 }
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Example Regular expression )0(*)011( λ++=r
)(rL = { all strings without substring 00 }
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Equivalent Regular Expressions Definition: Regular expressions and are equivalent if
1r 2r
)()( 21 rLrL =
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Example L = { all strings without substring 00 }
)0(*)011(1 λ++=r
)0(*1)0(**)011*1(2 λλ +++=r
LrLrL == )()( 211r 2rand
are equivalent regular expressions
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Regular Expressions and
Regular Languages
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Theorem
Languages Generated by Regular Expressions
Regular Languages =
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Languages Generated by Regular Expressions
Regular Languages
⊆
Languages Generated by Regular Expressions
Regular Languages
⊇
Proof:
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Proof - Part 1
r)(rL
For any regular expression the language is regular
Languages Generated by Regular Expressions
Regular Languages
⊆
Proof by induction on the size of r
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Induction Basis Primitive Regular Expressions: αλ,,∅Corresponding NFAs
)()( 1 ∅=∅= LML
)(}{)( 2 λλ LML ==
)(}{)( 3 aLaML ==
regular languages
a
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Inductive Hypothesis Suppose that for regular expressions and , and are regular languages
1r 2r)( 1rL )( 2rL
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Inductive Step We will prove: ( )
( )
( )
( )( )1
1
21
21
*
rL
rL
rrL
rrL
⋅
+
Are regular Languages
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By definition of regular expressions: ( ) ( ) ( )
( ) ( ) ( )
( ) ( )( )
( )( ) ( )11
11
2121
2121
**
rLrL
rLrL
rLrLrrL
rLrLrrL
=
=
=⋅
∪=+
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)( 1rL )( 2rLBy inductive hypothesis we know: and are regular languages
Regular languages are closed under: ( ) ( )( ) ( )( )( )*1
21
21
rLrLrLrLrL ∪Union
Concatenation
Star
We also know:
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Therefore:
( ) ( ) ( )
( ) ( ) ( )
( ) ( )( )** 11
2121
2121
rLrL
rLrLrrL
rLrLrrL
=
=⋅
∪=+
Are regular languages
)())(( 11 rLrL = is trivially a regular language (by induction hypothesis)
End of Proof-Part 1
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Using the regular closure of these operations, we can construct recursively the NFA that accepts
M)()( rLML =
Example: 21 rrr +=)()( 11 rLML =
)()( 22 rLML =
)()( rLML =
λ
λ
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For any regular language there is a regular expression with
Proof - Part 2
Languages Generated by Regular Expressions
Regular Languages
⊇
Lr LrL =)(
We will convert an NFA that accepts to a regular expression
L
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Since is regular, there is a NFA that accepts it
LM
LML =)(
Take it with a single final state
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From construct the equivalent Generalized Transition Graph in which transition labels are regular expressions
M
Example:
aba,
cM
a
ba +
c
Corresponding Generalized transition graph
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Another Example:
ba +a
b
b0q 1q 2q
ba,a
b
b0q 1q 2q
b
bTransition labels are regular expressions
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Reducing the states: ba +
ab
b0q 1q 2q
b
0q 2q
babb*
)(* babb +
Transition labels are regular expressions
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Resulting Regular Expression:
0q 2q
babb*
)(* babb +
*)(**)*( bbabbabbr +=
LMLrL == )()(
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In General Removing a state:
iq q jqa b
cde
iq jq
dae* bce*dce*
bae*
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0q fq
1r
2r
3r4r
*)*(* 213421 rrrrrrr +=
LMLrL == )()(
The resulting regular expression:
By repeating the process until two states are left, the resulting graph is
Initial graph Resulting graph
End of Proof-Part 2
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Non-regular languages
(Pumping Lemma)
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Regular languages ba* acb +*
...etc*)( bacb ++
Non-regular languages }0:{ ≥nba nn
}*},{:{ bavvvR ∈
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Non Turing-Acceptable
Turing-Acceptable
decidable
Context-sensitive
Context-free
Regular
The Chomsky Hierarchy
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How can we prove that a language is not regular?
L
Prove that there is no DFA or NFA or RE that accepts L
Difficulty: this is not easy to prove (since there is an infinite number of them)
Solution: use the Pumping Lemma !!!
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The Pumping Lemma
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Take an infinite regular language L
There exists a DFA that accepts L
mstates
(contains an infinite number of strings)
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mw ≥||(number of states of DFA)
then, at least one state is repeated in the walk of w
q...... ...... 1σ 2σ kσ
Take string with Lw∈
kw σσσ 21=Walk in DFA of
Repeated state in DFA
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Take to be the first state repeated q
q....
w
There could be many states repeated
q.... ....
Second
occurrence First
occurrence
Unique states
One dimensional projection of walk :
1σ 2σ kσiσ jσ1+iσ 1+jσ
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q.... q.... ....
Second
occurrence First
occurrence 1σ 2σ kσiσ jσ1+iσ 1+jσ
wOne dimensional projection of walk :
ix σσ 1= jiy σσ 1+= kjz σσ 1+=
xyzw =We can write
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zyxw =
q... ...
x
y
z
In DFA:
...
...
1σ kσiσ1+iσjσ
1+jσ
contains only first occurrence of q
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Observation: myx ≤||length number of states of DFA
Since, in no state is repeated (except q)
xy
Unique States
q...
x
y
...
1σ iσ1+iσjσ
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Observation: 1|| ≥ylength
Since there is at least one transition in loop
q
y
...
1+iσjσ
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We do not care about the form of string z
q...
x
y
z
...
z may actually overlap with the paths of and x y
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The string is accepted
zxAdditional string:
q... ...
x z
...
Do not follow loop y
...
1σ kσiσ1+iσjσ
1+jσ
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The string is accepted
zyyx
q... ... ...
x z
Follow loop 2 times
Additional string:
y
...
1σ kσiσ1+iσjσ
1+jσ
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The string is accepted
zyyyx
q... ... ...
x z
Follow loop 3 times
Additional string:
y
...
1σ kσiσ1+iσjσ
1+jσ
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The string is accepted
zyx iIn General:
...,2,1,0=i
q... ... ...
x z
Follow loop times i
y
...
1σ kσiσ1+iσjσ
1+jσ
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Lzyx i ∈Therefore: ...,2,1,0=i
Language accepted by the DFA
q... ... ...
x z
y
...
1σ kσiσ1+iσjσ
1+jσ
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In other words, we described:
The Pumping Lemma !!!
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The Pumping Lemma: • Given a infinite regular language L
• there exists an integer m
• for any string with length Lw∈ mw ≥||
• we can write zyxw =
• with and myx ≤|| 1|| ≥y
• such that: Lzyx i ∈ ...,2,1,0=i
(critical length)
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In the book:
Critical length = Pumping length m p
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Applications of
the Pumping Lemma
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Observation: Every language of finite size has to be regular
Therefore, every non-regular language has to be of infinite size (contains an infinite number of strings)
(we can easily construct an NFA that accepts every string in the language)
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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
2. Choose a particular string which satisfies the length condition
Lw∈
3. Write xyzw =4. Show that Lzxyw i ∉=ʹ′ for some 1≠i
5. This gives a contradiction, since from pumping lemma Lzxyw i ∈=ʹ′
mw ≥||
1. Let be the critical length for m L
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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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Theorem: The language }0:{ ≥= nbaL nn
is not regular
Proof: Use the Pumping Lemma
Example of Pumping Lemma application
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Assume for contradiction 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 critical length for
Pick a string such that: w Lw ∈
mw ≥||and length
mmbaw =We pick
m
}0:{ ≥= nbaL nn
L
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with lengths
From the Pumping Lemma:
1||,|| ≥≤ ymyx
babaaaaabaxyz mm ............==
mkay k ≤≤= 1,
x y z
m m
we can write zyxbaw mm ==
Thus:
=w
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From the Pumping Lemma: Lzyx i ∈
...,2,1,0=i
Thus:
mmbazyx =
Lzyx ∈2
mkay k ≤≤= 1,
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From the Pumping Lemma:
Lbabaaaaaaazxy ∈= ...............2
x y z
km + m
Thus:
Lzyx ∈2
mmbazyx =
y
Lba mkm ∈+
mkay k ≤≤= 1,
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Lba mkm ∈+
}0:{ ≥= nbaL nnBUT:
Lba mkm ∉+
CONTRADICTION!!!
1≥k
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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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Regular languages
Non-regular language }0:{ ≥nba nn
)( **baL