FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY. FLAC (15-453) - Spring L. Blum

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1 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY THE PUMPING LEMMA FOR REGULAR LANGUAGES nd REGULAR EXPRESSIONS TUESDAY Jn 21 WHICH OF THESE ARE REGULAR? B = {0 n 1 n n 0} C = { w w hs equl numer of occurrences of 01 nd 10 } D = { w w hs equl numer of 1s nd 0s} THE PUMPING LEMMA Let L e regulr lnguge with L = Then there is positive integer P s.t. if w L nd w P then cn write w = xyz, where: 1. y > 0 (y isn t ) 2. xy P 3. For every i 0, xy i z L Why is it clled the pumping lemm? The word w gets PUMPED into something longer Proof: Let M e DFA tht recognizes L Let P e the numer of sttes in M Assume w L is such tht w P We show: w = xyz 1. y > 0 2. xy P 3. xy i z L for ll i 0 r 0 r j r k r w There must e j nd k such tht j < k P, nd r j = r k (why?) (Note: k - j > 0) Proof: Let M e DFA tht recognizes L Let P e the numer of sttes in M Assume w L is such tht w P We show: w = xyz y 1. y > 0 2. xy P 3. xy i z L for ll i 0 x z r 0 r j = r k r w There must e j nd k such tht j < k P, nd r j = r k 1

2 USING THE PUMPING LEMMA Let s prove tht B = {0 n 1 n n 0} is not regulr Assume B is regulr. Let w = 0 P 1 P If B is regulr, cn write w = xyz, y > 0, xy P, nd for ny i 0, xy i zis lso in B USING THE PUMPING LEMMA D = { w w hs equl numer of 1s nd 0s} is not regulr Assume D is regulr. Let w = 0 P 1 P (w is in D!) If D is regulr, cn write w = xyz, y > 0, xy P, where for ny i 0, xy i zis lso in D y must e ll 0s: Why? xy P y must e ll 0s: Why? xy P xyyz hs more 0s thn 1s Contrdiction! xyyz hs more 0s thn 1s Contrdiction! WHAT DOES C LOOK LIKE? C = { w w hs equl numer of occurrences of 01 nd 10} = { w w = 1, w = 0, w = or w strts with 0 nd ends with 0 or w strts with 1 nd ends with 1 } 1 0 0(0 1)*0 1(0 1)*1 REGULAR EXPRESSIONS (expressions representing lnguges) is regexp representing { } is regexp representing {} is regexp representing If R 1 nd R 2 re regulr expressions representing L 1 nd L 2 then: (R 1 R 2 ) represents L 1 L 2 (R 1 R 2 ) represents L 1 L 2 (R 1 )* represents L 1 * PRECEDENCE * EXAMPLE R 1 *R 2 R 3 = ( ( R 1 * ) R 2 ) R 3 2

3 { w w hs exctly single 1 } Wht lnguge does * represent? { w w hs length 3 nd its 3rd symol is 0 } { w every odd position of w is 1 } 1. EQUIVALENCE L cn e represented y regexp L is regulr L cn e represented y regexp L is regulr 1. Given regulr expression R, we show there exists NFA N such tht R represents L(N) Induction on the length of R: Bse Cses (R hs length 1): R= 2. L cn e represented y regexp L is regulr lnguge R= R= 3

4 Inductive Step: Assume R hs length k > 1, nd tht every regulr expression of length < k represents regulr lnguge Three possiilities for R: R= R 1 R 2 R= R 1 R 2 R= (R 1 )* (Union Theorem!) (Conctention) (Str) Therefore: L cn e represented y regexp L is regulr Give n NFA tht ccepts the lnguge represented y (1(0 1))* 1 1,0 2. L cn e represented y regexp NFA L is regulr lnguge Proof ide: Trnsform n NFA for L into regulr expression y removing sttes nd re-leling rrows with regulr expressions Add unique nd distinct strt nd ccept sttes While mchine hs more thn 2 sttes: Pick n internl stte, rip it out nd re-lel the rrows with regexps, to ccount for the missing stte NFA Add unique nd distinct strt nd ccept sttes While mchine hs more thn 2 sttes: Pick n internl stte, rip it out nd re-lel the rrows with regexps, to ccount for the missing stte 01*0 GNFA While mchine hs more thn 2 sttes: More generlly: R(, ) R(, ) R(, ) R(, ) 4

5 GNFA While mchine hs more thn 2 sttes:, More generlly: q 0 R(, )R(, )*R(, ) R(, ) R(q 0, ) = represents L(N) Formlly: Add q strt nd q ccept to crete G (GNFA) Run CONVERT(G): (Outputs regexp) If #sttes = 2 return the expression on the rrow going from q strt to q ccept Formlly: Add q strt nd q ccept to crete G (GNFA) Run CONVERT(G): (Outputs regexp) If #sttes > 2 select q rip Q different from q strt nd q ccept define Q = Q {q rip } } Defines: G (GNFA) define R s: R (q i,q j ) = R(q i,q rip )R(q rip,q rip )*R(q rip,q j ) R(q i,q j ) (R = the regexps for edges in G ) We note tht G nd G re equivlent return CONVERT(G ) Clim: CONVERT(G) is equivlent to G Proof y induction on k (numer of sttes in G) Bse Cse: k = 2 Inductive Step: Assume clim is true for k-1 stte GNFAs Recll tht G nd G re equivlent But, y the induction hypothesis, G is equivlent to CONVERT(G ) Thus: CONVERT(G ) equivlent to CONVERT(G) QED 5

6 ( )* Convert the NFA to regulr expression, ( )* ( ( )*)*( ( )*) DEFINITION DFA Regulr Lnguge NFA Regulr Expression Finish Chpter 1 of the ook for next time 6

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