[PDF] - F all 1998 F ormal Language Theory Dr. R. Bo y er W eek PDF Document

SLIDE 1 F all 1998 F

rmal

Language Theory Dr. R. Bo y er W eek F

ur:

Equiv alence b et w een Finite Automata and Regular Expressions 1. Example

f

Con v ersion

f

NF A with e-mo v es with a DF A W e consider the NF A M = (Q; ; ; q ; fq 3 g); where Q = fq ; q 1 ; q 2 ; q 3 ; q 4 g; with t w

letter

alphab et

=

fa; bg; and with

ne

accepting state fq 3 g and

is

giv en b y the follo wing table: NF A T ransitions state a b e 1

2

1

3

2 4

1

3

4
4

1

3

Computation

f

e-closures states e-closure q fq ; q 1 ; q 2 ; q 3 g q 1 fq 1 ; q 3 g q 2 fq 1 ; q 2 ; q 3 g q 3 fq 3 g q 4 fq 3 ; q 4 g Computation

f

the states and transitions

f

the equiv alen t DF A: Since the

riginal

NF A has initial state f0g; the corresp

nding

DF A has initial state whic h is its e-closure: E (0) = f0; 1; 2; 3g 1

SLIDE 2 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions DF A T ransitions set

f

states a b f0; 1; 2; 3g E (1) [ E (2) = f1; 3; 4g E (4) = f3; 4g f1; 3; 4g E (1) = f1; 3g E (4) = f3; 4g f3; 4g E (1) = f1; 3g E (4) = f3; 4g f1; 3g ; E (4) = f3; 4g ; ; ; The accepting states are an y sets

f

states that con tain an accepting state

f

the

riginal

NF A. So, the accepting sets

f

the equiv alen t DF A are f0; 1; 2; 3g; f1; 3; 4g; f3; 4g and f1; 3g: 2. Prop

sition.

Let r b e a regular expression. Then there exists a NF A that accepts L(r ): The construction using induction

n

the n um b er

f
p

erators in r : It is con v enien t to normalize the t yp e

f

NF A used in the construction. W e require that the NF A has

ne

nal state and no transitions

ut
f

this unique nal state. It is easy to v erify that this do es not restrict the languages that will b e recognized. Base Case: r con tains zero

p

erators. Then either r = ;

r

r =

2

: F

r

r = ;; the NF A has t w

states

q and q f ; where q is the start state and q f 2

SLIDE 3 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions is the accepting state. There is no edge connecting them. F

r

r =

;

again there are t w

states

q and q f as b efore. There is

ne

edge connecting them lab eled with the letter

:

The induction h yp

thesis

is that the result holds for all regular expressions with k

p

erators

r

less. No w, assume r is a regular expression con taining k + 1

p

erators. There are three cases to consider. Case One: r = r 1 [ r 2 : W e rst state the construction informally . Then r 1 = L(M 1 ), where M 1 has initial state q 1 and unique accepting state f 1 ; and r 2 = L(M 2 ), where M 2 has initial state q 2 and unique accepting state f 2 : Then NF A that accepts the union r will ha v e initial state q and unique accepting state f ; further there are e-transitions from q to b

th

initial states q 1 and q 2 and e-transitions from the nal states f 1 and f 2 to f : In detail, let the NF A's b e giv en as M 1 = (Q 1 ;

1

;

1

; q 1 ; ff 1 g) and M 2 = (Q 2 ;

2

;

2

; q 2 ; ff 2 g) where Q 1 \ Q 2 = ;: W e create a new initial state q and a new nal state f : T ak e M = (Q 1 [ Q 2 [ fq ; f g;

1

[ S 2 ; ; q ; ff g); where (q ; e) = fq 1 ; q 2 g; (q ; a) =

1

(q ; a); q 2 Q 1 n ff 1 g; a 2

1

[ feg; (q ; a) =

2

(q ; a); q 2 Q 2 n ff 1 g; a 2

2

[ feg; (f 1 ; e) = (f 2 ; e) = ff g 3

SLIDE 4 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions Case Tw

:

r = r 1

r

2 : Let M 1 and M 2 b e as ab

v

e. Informally , the NF A that accepts r is

btained

b y connecting the nal state

f

M 1 to the initial state

f

M 2 : More precisely , M = (Q 1 [ Q 2 ;

1

[ S 2 ; ; q 1 ; ff 2 g); where (q ; a) =

1

(q ; a); q 2 Q 1 n ff 1 g; a 2

1

[ feg; (f 1 ; e) = fq 2 g; (q ; a) =

2

(q ; a); q 2 Q 2 n ff 2 g; a 2

2

[ feg; Case Three: r = (r 1 )

:

Let the NF A M = (Q 1 [ fq ; f g; ; ; q ; ff g); where (q ; e) = fq 1 ; f g; (f 1 ; e) = fq 1 ; f g; (q ; a) =

1

(q ; a); q 2 Q 1 ; a 2

1

3. Example. W e apply the ab

v

e construction to r =

1
[

1: 4. Prop

sition.

The class

f

languages accepted b y DF A's are closed under the follo wing

p

erations: (1) union, concatenation, Kleene star; (2) complemen tation; 4

SLIDE 5 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions (3) in tersection. Note: w e shall presen t t w

dieren

t constructions for in tersection. The rst will b e somewhat implicit and mak es use

f

(2) while the second construction will use the cross pro duct

f

automata. Let M 1 = (Q 1 ; ;

1

; s 1 ; F 1 ) and M 2 = (Q 2 ; ;

2

; s 2 ; F 2 ): Then w e let M = (Q 1

Q

2 ; ;

;

(s 1 ; s 2 ); F ) where

((q

; q ); a) = ( 1 (q ; a);

(q

; a)); q 2 Q 1 ; q 2 Q 2 ; a 2 ; F = F 1

F

2 = f(f ; f ) : f 2 F 1 ; f 2 F 2 g: The idea

f

the cross pro duct construction is that a computation in M actu- ally will trace

ut

the computation in the t w

mac

hines M 1 and M 2 sim ulta- neously . Note: if the collection

f

accepting states is c hanged to (F 1

Q

2 ) [ (Q 1

F

2 ); then M w

uld

accept the union L(M 1 ) [ L(M 2 ): So, this construction will pro duce a DF A that accepts the union rather than a NF A. 5. There are algorithms to answ er the follo wing questions ab

ut

DF A's: (1) giv en w 2

;

is w 2 L(M )? (2) Is L(M ) = ;? 5

SLIDE 6 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions (3) Is L(M ) =

?

(4) Giv en t w

DF

A's M 1 and M 2 ; is L(M 1 )

L(M

2 )? (5) Giv en t w

DF

A's M 1 and M 2 ; is L(M 1 ) = L(M 2 )? Pro

f.

6. Prop

sition.

If L = L(M ); where M is a DF A, then L is a regular language; that is, there is a regular expression r suc h that L = L(r ): W e shall study t w

constructions

for this corresp

ndence.

The rst is a graph based algorithm that builds up the regular expression as no des are deleted from the state diagram. The second is a metho d analogous to solving a linear system

f

equations. T

presen

t the graph

rien

ted algorithm, w e need to in tro duce an ev en more general notion

f

NF A. It will ha v e the expanded prop ert y that its edges ma y b e lab eled b y regular expressions, not simply b y a 2

r

e: W e will denote this new class

f

automata b y GNF A. Next, w e describ e ho w a string w is accepted. No w, in an usual NF A, the mac hine matc hes an input sym b

l

with an edge lab el in

rder

to mak e a mo v e. F

r

a GNF A, the automaton will consume, p erhaps, more than

ne

sym b

l

in

rder

to mak e the next mo v e. It will consume a substring that b elongs to the regular language whic h is denoted b y the edge lab el. F

r

a tec hnical normalization reason, w e shall assume that the GNF A M has a start state s that has NO edge coming in to it and that M has a unique 6

SLIDE 7 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions nal state ff g with NO edges lea ving it. F urther, assume f 6= s: F

r

completeness, w e giv e a detailed explanation

f

acceptance. Let M b e a GNF A, and let w 2

:

If w 2 L(M ); then w = w 1 w 2 : : : w k and there is a sequence

f

states q = s; q 1 ; : : : ; q k = f ; suc h that w i 2 L(R i ); where

(q

i1 ; q i ) = R i ; where R i is the regular expression lab el. Finally , w e can describ e the con v ersion pro cess

f

a DF A in to an equiv alen t regular expression. Step 1: Con v ert the giv en DF A M in to a GNF A M b y in tro ducing a new start state, new nal state, and the necessary transitions. Supp

se

M has k states. Step 2: If k = 2; then M has just the start state and the unique nal state. It is clear that L(M ) = L(R); where R =

(s;

f ): Step 3: "No de Reduction Step" If k > 2; w e remo v e a no de to pro duce an equiv alen t automaton M 00 : In particular, select a no de q 00 6= s; f : Then dene M 00 = (Q n fq 00 g; ;

00

; s; ff g) where

00

(q i ; q j ) =

(q

i ; q j ); if either

(q

i ; q 00 ) = ;

r
(q

00 ; q i ) = ;;

therwise,
00

(q i ; q j ) = R 1

(R

2 )

R

3 [ R 4 ; where R 1 =

(q

i ; q 00 ); R 2 =

(q

00 ; q 00 ); R 3 =

(q

00 ; q j ); R 4 =

(q

i ; q j ): 7

SLIDE 8 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions Step 4: Rep eat Step 3 un til M 00 has t w

states:

s and f : Note: This construction is also v alid for an y NF A. I w

rded

it for DF A's to emphasize the new automaton

f

t yp e GNF A. W e need to v erify that Step 3 do es indeed pro duce an equiv alen t automaton. W e can argue b y visualizing paths through the state diagram

f

the GNF A. W e next compare the ab

v

e graph algorithm with the approac h tak en in the textb

k.

As usual, the language L is accepted b y the DF A M = (fq 1 ; : : : ; q n g; ;

;

q 1 ; F ): W e let R k ij denote the set

f

all strings that tak e the automaton M from state q i to state q j without going through an y state n um b ered k

r

larger; that is,

nly

strings are allo w ed that start at q i and end at q j and

nly

use states q 1 ; : : : ; q k 1 as in termediate states. Suc h sets

f

strings satisfy an imp

rtan

t inductiv e iden tit y: R k +1 ij = R k ij [ R k ik (R k k k )

R

k k j : T

understand

this iden tit y , consider the follo wing. An y string w that is con tained in R k +1 ij either uses state q k

r

not. If w do es not use state q k ; then it m ust lie in R k ij : So, w e m ust examine ho w state q k is used b y the string w : Of course, q k m ust b e used as some in termediate state. So, w uses a path 8

SLIDE 9 We ek F

ur

: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions from state q i to q k and from q k to q j ; suc h that

nly

states q 1 ; : : : ; q k are used as in termediate states. So, it seems that w e m ust add the set R k ik R k k j : In fact, there is a further p

ssibilit

y . W e can also use paths that cycle through state q k : So, the set

f

strings that use state q k is: R k ik (R k k k )

R

k k j : One detail to consider is that (R k k k )

is

prop erly larger than R k ij itself, b ecause q k is not allo w ed as an in termediate state for strings from R k ij : W e can easily describ e the set

f

strings R 1 ij : F

r

i 6= j; R 1 ij = fa :

(q

i ; a) = q j g; while for i = j; R 1 ii = fa :

(q

i ; a) = q i g [ feg: W e sho w b y induction that R k ij is denoted b y a regular expression r k ij : F

r

k = 1; w e let, for i 6= j; r 1 ij = a 1 [ : : : [ a p ; for i = j; r 1 ii = e [ a 1 [ : : : [ a p ; where

(q

i ; a ` ) = q j ; 1

`
p:

If this set is empt y , then r 1 ij = ;; for i 6= j ; while for i = j; r 1 ii = e: W e conclude that R 1 ij = L(r 1 ij ): W e no w assume that the result holds for v alue k : That is, for an y set R k `m ; there exists a regular expression r k `m suc h that R k `m = L(r k `m : W e need to sho w that an y set R k +1 ij is giv en b y a regular expression. By the inductiv e prop ert y

f

the set R k +1 ij ; w e kno w that R k +1 ij = R k ij [ R k ik (R k k k )

R

k k j : By the induction h yp

thesis,

w e ha v e: R k ij [ R k ik (R k k k )

R

k k j = L(r k ij ) [ L(r k ik )L((r k k k )

L(r

k k j ) = L(r k ij [ r k ik ((r k k k )

r

k k j ): This nishes the induction. 9

SLIDE 10 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions Finally , w e

bserv

e that L(M ) = S fR n+1 1j : q j 2 F g: Example

f

con v erting a DF A in to a regular expression. Giv e a regular expression for the deterministic nite automaton M with states fq 1 ; q 2 ; q 3 g; with transitions: State a b 1 3 2 2 1 1 3 1 1 The accepting states are fq 2 ; q 3 g and the initial state is fq 1 g: W e compute the regular expressions r k ij ; for k = 1; 2; 3; 4: Recall: r k +1 ij = r k ij [ r k ik (r k k k )

r

k k j : F

r

k = 1; w e are in the base case. r 1 11 = e; r 1 12 = a; r 1 13 = b; r 1 21 = a; r 2 22 = e; r 2 23 = b; r 1 31 = ;; r 1 32 = a [ b; r 1 33 = e: 10

SLIDE 11 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions F

r

the remaining sets, w e need to use the inductiv e set iden tit y . r 2 11 = r 1 11 [ r 1 11 (r 1 11 )

r

1 11 = e r 2 12 = r 1 12 [ r 1 11 (r 1 11 )

r

1 12 = a [ ae

e

= a r 2 13 = r 1 13 [ r 1 11 (r 1 11 )

r

1 13 = b [ ee

b

= b r 2 21 = r 1 21 [ r 1 21 (r 1 11 )

r

1 11 = a [ ae

e

= a r 2 22 = r 1 22 [ r 1 21 (r 1 11 )

r

1 12 = e [ ae

a

= e [ aa r 2 23 = r 1 23 [ r 1 21 (r 1 11 )

r

1 13 = b [ ae

b

= b [ ab r 2 31 = r 1 31 [ r 1 31 (r 1 11 )

r

1 11 = ; [ ;e

e

= ; r 2 32 = r 1 32 [ r 1 31 (r 1 11 )

r

1 12 = (a [ b) [ ;e

e

= ; r 2 33 = r 1 33 [ r 1 31 (r 1 11 )

r

1 13 = (a [ b) [ ;e

a

= a [ b r 3 11 = r 2 11 [ r 2 12 (r 2 22 )

r

2 21 = e [ a(e [ aa)

a

= (aa)

r

3 12 = r 2 12 [ r 2 12 (r 2 22 )

r

2 22 = a [ a(e [ aa)

(e

[ aa) = a(aa)

r

3 13 = r 2 13 [ r 2 12 (r 2 22 )

r

2 23 = (b [ ab) [ a(e [ aa)

(b

[ ab) = a

b

r 3 21 = r 2 21 [ r 2 22 (r 2 22 )

r

2 21 = a [ (e [ aa)(e [ aa)

a

= a(aa)

r

3 22 = r 2 22 [ r 2 22 (r 2 22 )

r

2 22 = (e [ aa) [ (e [ aa)(e [ aa)

(e

[ aa) = (aa)

r

3 23 = r 2 23 [ r 2 22 (r 2 22 )

r

2 23 = (b [ ab) [ (e [ aa)(e [ aa)

(b

[ ab) = a

11

SLIDE 12 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions r 3 31 = r 2 31 [ r 2 32 (r 2 22 )

r

2 21 = ; [ (a [ b)(e [ aa)

a

= (a [ b)(aa)

a

r 3 32 = r 2 32 [ r 2 32 (r 2 22 )

r

2 22 = (a [ b) [ (a [ b)(e [ aa)

(e

[ aa) = (a [ b)(aa)

r

3 33 = r 2 33 [ r 2 32 (r 2 22 )

r

2 23 = e [ (a [ b)(e [ aa)

(b

[ ab) = e [ (a [ b)a

b

Finally , w e

nly

need to compute r 4 12 and r 4 13 : r 4 12 = r 3 12 [ r 3 13 (r 3 33 )

r

3 32 = a(aa)

[

a

[e

[ (a [ b)a

b]
(a

[ b)(aa)

=

a(aa)

[

a

b[(a

[ b)a

b]
(a

[ b)(aa)

r

4 13 = r 3 13 [ r 3 13 (r 3 33 )

r

3 33 = a

b

[ a

b[(a

[ b)a

b]
a
b

= a

b[(a

[ b)a

b]
F
r

the second metho d, w e rst presen t a result that is kno wn as Arden's Lemma: Arden's Lemma: Let A; B

with

e = 2 A: Then the equation: X = A

X

[ B has the unique solution X = A

B

: Step 1: If X is a solution, then A

B
X

: 12

SLIDE 13 We ek F

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: : : : : Equivalenc e b etwe en A utomata and R e gular Expr essions T

see

this, note that A

B

= (A + [ e)B = A + B [ B = A(A

B

) [ B : Step 2: X

A
B

: By Step 1, X = A

B

[ C ; since A

B
X

with C \ A

B

= ;: W e w an t to sho w that C = ;: No w X = AX [ B ; so A

B

[ C = A(A

B

[ C ) [ B = A + B [ AC [ B = A + B [ B [ AC = (A + [ e)B [ AC = A

B

[ AC : Next, consider the relation: (A

B

[ C ) \ C = (A

B

[ AC ) \ C : Then C = AC \ C ; so C

AC

: Since e 6= A; the shortest string in AC m ust b e longer than the shortest string in C : Hence, AC = C = ;: W e conclude: A

B

is the unique solution. Note: If e 2 A; then the solution A

B

is no longer unique but it is the smallest solution. 13