Transcript Regular Expressions and Non

Regular Expressions and
Non-regular Languages
http://cis.k.hosei.ac.jp/~yukita/
Expressions and their values
Expression Value
Arithmetic
expression
Regular
expression
(5  3)  4
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(0  1)  0*
T helanguage thatconsistsof
or
strings begining with 0 or 1
(0  1)0*
followed by an arbitrarynumber of 0s.
Subexpressions 0, 1, 0  1, and 0* have values{0}, {1}, and
{0,1}, and{0}* , respectively. T he value of the whole expression
is {0,1}  {0}* .
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Definition 1.26
R is a regular expressionif
1. R  a  ,
2. R   ,
3. R  ,
4. R  ( R1  R2 ), where R1 and R2 are regular expressions,
5. R  ( R1  R2 ), where R1 and R2 are regular expressions, or
6. R  ( R1 ), where R1 is a regular expression.
*
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The values of atomic expressions
expression value
a
{a}

{ }

meaning
T helanguage thatconsistsof
only one stringa  * .
T helanguage thatconsistsof
only theemptystring  * .
  {}
T heemptylanguage
In manyreal programming languages, we must distinguish
alphabeta   and stringa  * ; theformershould be
writtenas ' a' while thelatter" a". But we will not follow
thisconvention
.
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Example 1.27
Let   {0,1} throughout in thefollowingexamples.
1. 0*10*  {w | w has exactlya single 1}.
2. *1*  {w | w has at least one1}.
3. * 001*  {w | w containsthestring001as a substring}.
4. ()*  {w | w is a stringof even length}.
7. 0* 0  1*1  0  1  {w | w startsand ends with thesame symbol}.
8. (0   )1*  01*  1*.
10. 1*   .
11. *  { }.
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Units for the binary operations
R    R shows t hat is t heunit for t heunion operat ion.
R    R shows t hat is t heunit for t heconcat enation operat ion.
We denot eby L( R ) t helanguage of R.
R   may not equal R.
For example,if R  0, t henL( R )  {0} but L( R   )  {0,  }.
In general, R    R.
For example,if R  0, t hen L( R )  {0} but L( R  )  .
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Theorem 1.28
• A language is regular if and only if some
regular expression describes it.
We break down this theorem as follows.
• Lemma 1.29
– If a language is described by a regular
expression, then it is regular.
• Lemma 1.32
– If a langulage is regular, then it is described
by a regular expression.
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Proof of Lemma 1.29
We shall convertR intoan NFA N .
a
Case 1 : R  a.
Case 2 : R   .
Case 3 : R  .
Case 4 : R  R1  R2 .
Case 5 : R  R1  R2 .
Next three slides
Case 6 : R  R
*
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Proof of Lemma 1.29
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Case 4: Let N1, N2, and N correspond to R1,
R2, and R, respectively.
N
N1


N2
Proof of Lemma 1.29
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Case 5: Let N1, N2, and N correspond to R1,
R2, and R, respectively.
N2
N1

N

Proof of Lemma 1.29
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Case 6: Let N1 and N correspond to R1 and
R, respectively.
N

N1


Proof of Lemma 1.29
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Generalized Nondeterministic
Finite Automaton
• is roughly a NFA in which the transition arrows may have
regular expressions as labels.
• We assume the following standard form for convenience,
which can always be attained with an easy modification.
– There is only one accept state and different from the start state.
– The start state has transition arrows going to every other state
but no arrows coming in from any other state.
– There is only a single accept state, and it has arrows coming in
from any other state but no arrows going to any other state.
– Except for the start and accept states, one arrow goes from
every state to every other state and also from each state to itself.
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Standard Form of GNFA
...
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Standard Form of GNFA
Q  {qstart }  Q  {qaccept} is a disjont union.
For each pair (qi , q j ) in ({qstart }  Q)  (Q  Q)  Q  {qaccept}
thereis one and only one directedarrow going from qi to q j .
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Equivalent GNFA with one fewer state
qi
R4
R1
qj
qi
( R1 )(R2 ) * ( R3 )  ( R4 )
qj
R3
qrip
R2
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Definition 1.33
A generalized nondeterministicfiniteautomatonis 5 - tuple
(Q, ,  , qstart , qaccept ), where R is theset of regular expressions,
1. Q is thefiniteset of states,
2.  is theinput alphabet,
3.  : (Q  {qaccept})  (Q  {qstart })  R,
4. qstart is thestart state,and
5. qaccept is theacceptstate.
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Computation with GNFA
A GNFA acceptsa stringw  w1w2  wk  * with wi  * ,
and a sequence of statesq0 , q1 ,, qk existssuch that
1. q0  qstart ,
2. qk  qaccept,
3. for each i, we havewi  L( Ri ), where Ri   (qi 1 , qi ).
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Converting GNFA
Convert(G ) :
1. Let k be thenumber of statesof G.
2. If k  2, thenreturn theregular expressionappearingon the
only arrow.
3. If k  2, we select any stateqrip  Q  {qstart , qaccept} and let
G be theGNFA (Q, ,  , qstart , qaccept ), where Q  Q  {qrip },
and for any qi  Q  {qaccept} and q j  Q  {qstart } let
 (qi , q j )  ( R1 )(R2 ) * ( R3 )  ( R4 ),
for R1   (qi , qrip ), R2   (qrip , qrip ), R3   (qrip , q j ), and R4   (qi , q j ).
4. ComputeConvert(G) and return this value.
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Claim 1.34 For any GNFA G, Convert(G) is
equivalent to G.
P r oof. B as is (k 2) : Obvious.
In du cti onste p: Assume t hat t heclaim is t ruefor k  1 st at es.
Suppose t hatG accept san input w. T hen,in an accept ingbranch
of t hecomput at ion, G ent ersa sequence of st at es
qstart , q1 , q2 , q3 ,  , qaccept.
If qrip  {qstart , q1 , q2 , q3 ,  , qaccept}, clearlyG also accept sw.
If qrip  {qstart , q1 , q2 , q3 ,  , qaccept}, removingeach run of
consecut ive qrip st at esformsan accept ingcomput at ion for G.
T hest at esqi and q j bracket inga run havea new regular expression
on t hearrow bet ween t hem t hatdescribes all st rings t akingqi t o q j
in via qrip on G . So G accept sw.
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Proof continued
For theotherdirection,suppose thatG acceptsan input w.
As each arrow between any twostatesqi and q j in G describes
thecollectionof strings takingqi and q j in G, eitherdirectly
or viaqrip , G must also accept w. T husG and G are equivalent.
T heinduction hypothesisstatesthat whenthealgorithmcalls
itself recursively on input G, theresult is a regular expression
thatis equivalent to G because G has k  1 states.Hence the
regular expressionalso is equivalent to G, and thealgorithm
is provedcorrect.
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Non-regularity
B, C , and D seem to require machineswith infinitenumber of states
to recognizethem.
B  {0 n1n | n  0}
C  {w | w has an equal number of 0s and1s}
D  {w | w has an equal number of occurrences of
01and10 as substrings}.
B and C will turn out tobe nonregularwhile D regular.
See, problem1.41.
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Theorem 1.37 Pumping Lemma
e is a number p
If A is a regular language, then ther
(thepumpinglength)where,if s  A with | s | p,
then s can be divided into threepieces,
s  xyz, satisfyingthefollowingconditions:
1. for each i  0, xy z  A,
2. | y | 0, and
i
3. | xy | p.
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Proof of Th 1.37
s  s1 s2  sk  sl  sn
  


q1 q2  qk  qk  qa
 ( s1s2 )(sk  sl )( sn )  xyz
where qa is an acceptingstate.
By thepigeonholeprinciple,
we can takethepumpinglength
as thenumber of statesplus one.
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Example 1.38
C lai m: B  {0 n1n | n  0} is not regular.
t B is regular.
Proof. Assume thecontrary hat
Let p be thepumpinglengt h.T hen,s  0 p1p can be decomposed
as s  xyz wit h | y | 1, and xy n z  B for any n  0.
T herecan be threecases y  0 k , y  0 k1l , and y  1l , for some
nonzerok , l. In each case, we can eazily see that xy n z  B, which
leads to contradiction.(n  2)
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Example 1.39
C laim: C  {w | w has an equal number of 0s and1s}
is not regular.
Proof. Assume thecontrary hat
t C is regular.
Let p be thepumpinglength.T hen,s  0 p1p can be decomposed
as s  xyz with | y | 1 and | xy | p, and xy n z  C for any n  0.
T hen wemust have y  0 k for some nonzerok .
We can eazily see that xy n z  C , which contradicts theassumption.
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Alternative proof of 1.39
T heclass of regular languages is closed under theitnersection
operation.T hisis eazy t oproveif we run t wo DFAs parallelyand
acceptonlystrings which are acceptedby bot h of theDFAs.
Now, assume C is regular.T hen, C  0*1*  B is also regular.
T hiscontradicts what we provedin Example1.38.
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Example 1.40
C laim: F  {ww | w {0,1}*} is not regular.
Proof. Assume thecontrary hat
t F is regular.
Let p be thepumpinglengthand let s  0 p10 p1 F . T hiss can be split
into pieceslike s  xyz with | y | 1 and | xy | p, and xy n z  F for any n  0.
T hen wemust have y  0 k for some nonzerok .
We can eazily see that xy n z  F , which contradicts theassumption.
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Example 1.41 Unary Language
C laim: D  {1 | n  0} is not regular.
n2
Proof. Assume thecontrary hat
t D is regular.
Let p be thepumpinglength and let s  1  D. T hiss can be split
p2
into pieceslike s  xyz with | y | 1 and | xy | p, and xy n z  D
for any n  0.
T helength of xy n z　grows linearly with n, while thelengths
of stringsin D grows as 0,1,4,9,16,25,36,49, 
T hese two factsare incompatible as can be easily seen.
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Example 1.42 Pumping Down
C laim: E  {0i1 j | i  j} is not regular.
Proof. Assume thecontrary hat
t E is regular.
Let p be thepumpinglength and let s  0 p 11p. T hiss can be split
into s  xyz with | y | 1 and | xy | p, and xy n z  E for any n  0.
T hen wemust have y  0 k for some nonzerok .
We can eazily see that xy0 z  xz  E , which contradicts theassumption.
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Problem 1.41 Differential Encoding
C laim: D  {w | w containsequal number of occurrences of the
substrings 01and10} is regular.
0
1
1
0
0xx0
0xx1
0
qstart
0
1
1xx1
1xx0
1
1
0
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