Namely, for regular languages L1. Concatenation. Regular Languages. Star operation L 1. Complement. For regular language L the complement L is regular

Transcription

1 CD5560 FABER For guges, Autot d Modes of Coputtio eture 5 Märde Uiversit 2003 Cotet More Properties of Regur guges (R) Stdrd Represettios of R Eeetr Questios out R No-Regur guges The Pigeohoe Priipe The Pupig e Appitios of the Pupig e More Properties of Regur guges We hve prove Regur guges re osed uder Uio Cotetio Str opertio Ne, for regur guges 1 d 2 : Uio Cotetio Str opertio Regur guges We i prove Regur guges re so osed uder Copeet Itersetio Reverse Reverse R Ne, for regur guges 1 d 2 : Copeet Itersetio Regur guges Theore Copeet For regur guge the opeet is regur Te tht epts d hge: ofi sttes fi sttes (o-eptig sttes eptig sttes) Epe = ( * ),, q0 q1 q2 = ( + ( + )( + ), ) Resutig epts, q0 q1 q

2 Itersetio 1, 2 regur Theore For regur guges 1 d 2 the itersetio 1 2 is regur App DeMorg s : = , regur regur regur regur Stdrd Represettios of Regur guges Stdrd Represettios of Regur guges Meership Questio Regur guges Regur Epressios s NFAs Regur Grrs 13 Eeetr Questios out Regur guges 14 Questio: Aser: Give regur guge d strig ho e he if? Te the tht epts d he if is epted 15 Questio: Give regur guge ho e he if is ept:? ( = ) Aser: Te the tht epts Che if there is pth fro the iiti stte to fi stte =

3 Questio: Give regur guge ho e he if is fiite? is ifiite Questio: Give regur guges 1 d 2 ho e he if 1 = 2? Aser: Te the tht epts Che if there is ith e fro the iiti stte to fi stte is fiite Aser: Fid if ( ) ( ) = ( ) ( ) = ( ) ( ) = d 1 2 = or No-Regur guges 1 = No-regur guges Regur guges??? * * + + ( + )* et... Ho e prove tht guge is ot regur? Prove tht there is o tht epts Proe: this is ot es to prove Soutio: the Pupig e! The Pigeohoe Priipe

4 4 pigeos A pigeohoe ust oti t est to pigeos The Pigeohoe Priipe (Postfpriipe eer Brevduvepriipe?) 3 Pigeohoes (eer postf) The Pigeohoe Priipe pigeos pigeos pigeohoes... > pigeohoes > There is pigeohoe ith t est 2 pigeos The Pigeohoe Priipe d s ith 4 sttes I s of strigs: o stte is repeted I s of strigs:... stte is repeted q 1 q 2 q3 q 4 q 1 q 2 q3 q 4 q 1 q 2 q3 q

5 4 If the of strig hs egth the stte is repeted q 1 q 2 q 3 q 4 Pigeohoe priipe for : If i of strig trsitios sttes of the stte is repeted I other ords for strig trsitios re pigeos q sttes re pigeohoes I geer A strig hs egth uer of sttes A stte q ust e repeted i the of of The Pupig e for Regur guges Te ifiite regur guge tht epts sttes Te strig ith = + 1 If strig hs egth ( uer of sttes) the, fro the pigeohoe priipe: Write = There is ith e stte q is repeted i the

6 egths: (fro pigeo priipe, s q is the first repetitio i sequee) 1 (there is i the grph) Oservtio The strig is epted Oservtio The strig is epted Oservtio The strig is epted Geer The strig i = 0,1, 2,... i is epted The Pupig e Give ifiite regur guge there eists iteger for strig ith egth e rite = 1 ith d suh tht: i i = 0,1, 2, Appitios of the Pupig e Theore The guge = { : 0} is ot regur Use the Pupig e! = { : 0} Assue for otrditio tht is regur guge Sie is ifiite e pp the Pupig e

7 = { : 0} et e the iteger i the Pupig e Pi strig suh tht: e.g. pi egth = Write: = Fro the Pupig e it ust e tht: egth =, 1, 1 = We hve: = Fro the Pupig e: We hoose = i = 0 =, 1 i i = 0,1, 2, CONTRADICTION! 57 Our ssuptio tht is regur guge is ot true No-regur guges { : 0} Theore The guge R = { : Σ*} Σ = {, } Cousio is ot regur guge Regur guges * * + + ( + )* et... is ot regur Use the Pupig e! END OF PROOF R = { : Σ*} Assue for otrditio tht is regur guge Sie is ifiite e pp the Pupig e et e the iteger i the Pupig e Pi strig suh tht: egth pi = { : Σ*} = R Write it ust e tht egth Fro the Pupig e =, 1 =, 1 =

8 =, 1 = We hoose i = 0 So e get s o the eft, hie is o the right CONTRADICTION! 64 Our ssuptio tht is regur guge is ot true Cousio is ot regur guge No-regur guges { : 0} { R : Σ*} Regur guges * * + + ( + )* et... END OF PROOF Theore The guge + = { :, 0} + = { :, 0} + = { :, 0} is ot regur Use the Pupig e Assue for otrditio tht is regur guge Sie is ifiite e pp the Pupig e et e the iteger i the Pupig e Pi strig suh tht: pi egth = Write 2 it ust e tht egth 2 Fro the Pupig e =, 1 =, 1 2 = We hve: Fro the Pupig e Thus: 0 i i = 0,1, 2,... 0 = = = 2 =, 1 2 BUT: CONTRADICTION! = { :, 0}

9 Our ssuptio tht is regur guge is ot true Cousio is ot regur guge No-regur guges { R : Σ*} { : 0 } + { :, 0} Regur guges Theore The guge = {! : 0} is ot regur! = 1 2 ( 1) Use the Pupig e END OF PROOF = {! : 0} Assue for otrditio tht is regur guge Sie is ifiite e pp the Pupig e et e the iteger i the Pupig e Pi strig suh tht: pi = {! : 0} egth! = Write! = Fro the Pupig e it ust e tht egth, 1! = =, 1! We hve: =! =, 1 Fro the Pupig e: Ad sie:!+ 1 = {! : 0} Hoever for > 1! +!+!+! <! +! =! ( + 1) = ( +1)! i i = 0,1, 2,... Thus: 2 2 = =!+ There is p! + = p! 1! + < ( + 1)!! + p! for p

10 BUT:!+ = {! : 0} d 1 Our ssuptio tht is regur guge is ot true + No-regur guges { R : Σ*} { : 0} { :, 0} {! : 0}!+ Cousio is ot regur guge Regur guges CONTRADICTION! END OF PROOF Theore The guge i = { : i prie} is ot regur Use the Pupig e 85 i = { : i prie} Assue for otrditio tht is regur guge Sie is ifiite e pp the Pupig e 86 Fro the Pupig e it ust e tht egth, 1 Fro the Pupig e: i i = 0,1, 2,... The egth of egth + 1 = ust e prie for eh > strig of. 87 But egth( + 1 ) = egth( = egth( ) + egth( hih is ot prie! ) ) = = + ( egth( )) = (1 + egth( )) Our ssuptio tht is regur guge is ot true Cousio is ot regur guge + No-regur guges { R : Σ*} { : 0} { :, 0} {! : 0} Regur guges Thus: CONTRADICTION! 88 END OF PROOF 89 i = { : i prie} 90