Solutions to Exercises, Set 3

Transcription

1 Shool of Computer Siene, University of Nottinghm G5MAL Mhines nd their Lnguges, Spring 1 Thorsten Altenkirh Solutions to Exerises, Set 3 Fridy 3rd Mrh 1 1. () () L(+ +ǫ) = {L(E +F) = L(E) L(F)} L() L( ) L(ǫ) = {L(EF) = L(E)L(F)} L()L() L( )L( ) L(ǫ)L() = {L( ) = } L()L() L( ) L(ǫ)L() = {S = } L()L() L(ǫ)L() = {L(ǫ) = {ǫ}} L()L() {ǫ}l() = {{ǫ}s = S} L()L() L() = {L(x) = {x}} {}{} {} = {Contention of Lnguges} {} {} = {Set Union} {, } L((+)+( +)ǫ) = {L(E +F) = L(E) L(F)} L((+)) L(( +)ǫ) = {L(EF) = L(E)L(F)} L()L(+)L() L( +)L(ǫ) = {L(ǫ) = {ǫ}} L()L(+)L() L( +){ǫ} = {S{ǫ} = S} L()L(+)L() L( +) = {L(E +F) = L(E) L(F)} L()(L() L())L() L( ) L() = {L( ) = } L()(L() L())L() L() = {L(x) = {x}} {}({} {}){} {} = {Set Union} {}{,}{} {} = {Contention of Lnguges} {,} {} = {Set Union} {,, }. () ǫ+(+)(ǫ+) () 1

2 3. () (++) () (+) () (++) (++) (d) (+) (+) (++) (e) (+) (+) (f) (++(+) ) (g) (ǫ+)((+)(ǫ+)) (h) (ǫ++)((+)(ǫ++)). () Let us strt y onstruting NFAs for nd. I hve nmed the sttes to mke it esier for you to see how they orrespond to the finl NFA, ut normlly you wouldn t nme them until lter. 1 3 An NFA for is then onstruted y omposing the two NFAs sequentilly: 1 3 Stte 1 is now ded-end stte nd n e removed: 3 An NFA for () is then onstruted y dupliting ll rrows tht point to n epting stte to point to ll initil sttes, nd then dding n initil epting stte: 3

3 Finlly, n NFA for +() is onstruted y dding n NFA for in prllel: 3 () Let us strt with n NFA for +: 1 3 An NFA for (+) is then s follows: 1 3 3

4 To onstrut n NFA for (+), we first ple the previous NFA side y side with n NFA for : 1 3 The two re then omposed sequentilly: 1 3 Note tht the initil rrow ws removed from Stte 5, then dded gin s the initil rrow for Stte ws duplited to point to Stte 5.

5 At this point, it s good ide to remove ded-end sttes. Next let s onsider. First we ple the NFAs for nd side y side: We now onstrut n NFA for y omposing them in sequene. As N( ) hs no epting sttes, this just requires setting ll sttes from N() to not e initil The NFA for +(+) +ǫ onsists of N( ), N((+) ) nd N(ǫ) in prllel. 5

6 Finlly, Stte 7 n e eliminted s it is ded-end stte, nd sttes 8 nd 9 n e eliminted s they re unrehle () The empty word is just the empty list: ε :: Word σ ε = [] () Contention for finite lnguges n e defined using list omprehension: lngcont :: Lnguge σ Lnguge σ Lnguge σ lngcont l 1 l = [w 1 + w w 1 l 1,w l ] 6

7 It gets it trikier for infinite lnguges, s we wnt to ensure tht ll words get enumerted. For exmple, if l is infinite, then, using the ove definition, only the first word from l 1 will ever pper in the resultnt lnguge. As eh word in the resultnt lnguge is pir of words (one from eh rgument lnguge), the prolem is to ssign ountle ordering to ll suh pirs. This n e hieved y using the Cntor Piring Funtion to ssign n order of enumertion (see, for exmple, However, for this prolem, iterting through the pirs is ll tht is required, so defining n iterting funtion is suffiient (rther thn diretly defining the inverse piring funtion). nextpir :: (Int,Int) (Int,Int) nextpir (,n) = (n +1,) nextpir (m,n) = (m 1,n +1) Contention n now e defined reursively s follows: lngcont :: Lnguge σ Lnguge σ Lnguge σ lngcont l 1 l = laux (,) where laux (m,n) 1 = ((l 1!! m) + (l!! n)) : laux (nextpir (m,n)) 1 = laux (nextpir (m,n)) otherwise = [] where 1 = inbound m l 1 = inbound n l inbound :: Int [] Bool inbound n = not null drop n Note tht s the lnguges my e finite, we need to use inbound to hek tht we do not index list (lnguge in this se) out of ounds. If the ounds of oth lnguges re exeeded, then ll words hve een enumerted nd the reursion termintes. Further note tht this funtion ould e mde muh more effiient. For exmple, speilised uxiliry funtions ould e introdued for the ses when one of the two lnguges is disovered to e finite. () Exponentition of lnguges n e defined y primitive reursion: lngexp :: Lnguge σ Int Lnguge σ lngexp l = [ε] lngexp l n = lngcont l (lngexp l (n 1)) (d) The definition L = n= L n n e enoded firly diretly: kleenestr :: Eq σ Lnguge σ Lnguge σ kleenestr l = unions [lngexp l n n [..]] 7

8 (e) Finlly, enumerting the lnguge of regulr expression is strightforwrd enoding of its semntis using the funtions lredy defined: relng :: Eq σ RE σ Lnguge σ relng Empty = [] relng Epsilon = [ε] relng (Symol x) = [[x]] relng (Plus r1 r) = relng r1 union relng r relng (Sequene r1 r) = lngcont (relng r1) (relng r) relng (Str r) = kleenestr (relng r) relng (Pren r) = relng r 8