Compiler Design. Fall Lexical Analysis. Sample Exercises and Solutions. Prof. Pedro C. Diniz

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1 Univerity of Southern Cliforni Computer Siene Deprtment Compiler Deign Fll 6 Lexil Anlyi Smple Exerie nd Solution Prof. Pedro C. Dini USC / Informtion Siene Intitute 4676 Admirlty Wy, Suite Mrin del Rey, Cliforni 99 pedro@ii.edu Lexil Anlyi Smple Exerie Fll 6

2 Univerity of Southern Cliforni Computer Siene Deprtment Prolem : Conidering the lphet Σ = {,} derive Non-Determiniti-Finite Automton (NFA) uing the Thompon ontrution tht i le to reognie the entene generted y the regulr expreion *( )*. Doe the entene w = elong to the lnguge generted y thi regulr expreion? Jutify. Solution: Conidering impliftion of the omintion of the NFA during the Thompon ontrution (e.g., equene of two -trnition n e onidered ingle -trnition) poile NFA i hown elow nd where the trt tte i the tte leled The word w = elong to the lnguge generted y thi RE eue there i pth from the trt tte to the epting tte tht pell out the word, repetively the pth,,4,6,8,,,,,,. Prolem : Strting from the NFA you derived in prolem, onvert it to DFA uing the u-et ontrution deried in l. Verify tht the DFA yield the me output the originl NFA for the me input tring w. Solution: Uing the -loure nd DFAedge omputtion deried in l, we hve the following mpping of et of tte of the NFA to et of tte of the DFA: I = -loure () = {,,, 4, } I = DFAedge(I,) = -loure ({,,, 4, }, ) = {6} I = DFAedge(I,) = -loure ({,,, 4, }, ) = {,,, 4,, 7} I = DFAedge(I,) = -loure ({6}, ) = Ierr I4 = DFAedge(I,) = -loure ({6}, ) = {8,,, } I = DFAedge(I,) = -loure ({,,, 4,, 7}, ) = {6, 9,,, } I6 = DFAedge(I,) = -loure ({,,, 4,, 7}, ) = {,,, 4,, 7} = I I7 = DFAedge(I4,) = -loure ({8,,, }, ) = Ierr I8 = DFAedge(I4,) = -loure ({8,,, }, ) = {,, } I9 = DFAedge(I,) = -loure ({6, 9,,, }, ) = Ierr I = DFAedge(I,) = -loure ({6, 9,,, }, ) = {8,,,, } I = DFAedge(I8,) = -loure ({,, }, ) = Ierr I = DFAedge(I8,) = -loure ({,, }, ) = {,, } = I8 I = DFAedge(I,) = -loure ({8,,,, }, ) = Ierr I4 = DFAedge(I,) = -loure ({8,,,, }, ) = I8 Effetively there re only 8 tte hown in the DFA elow nd thi i lerly not the miniml DFA tht n reognie thi regulr expreion. Lexil Anlyi Smple Exerie Fll 6

3 Univerity of Southern Cliforni Computer Siene Deprtment, Ierr I I I4 I8 I I I For the input entene w = in hi DFA we would reh the tte I8, through tte I, I4 nd I8 nd thu epting thi tring. Prolem : Strting from the DFA you ontruted in the previou prolem, onvert it to miniml DFA uing the minimition lgorithm deried in l. Verify tht the DFA yield the me output the originl NFA for the me input tring w. Solution: The firt prtition would hve the two et of tte P = {I, I, I, Ierr} nd = {I4, I, I8, I}. A eond prtition would mintin nd prtition P into {I, I, Ierr} nd generte P = {I} ine I goe to itelf on nd to on. Next the lgorithm would generte = {I) nd next generte P = {Ierr} leving P = {I}. The finl DFA would hve therefore tte h hown elow., P P P Agin on the input tring w = the DFA would trvere P,, nd loop in therefore epting the tring. Importnt Note: If you try to pply the me ontrution, i.e. prtitioning the et of tte trting with n inomplete DFA where the trp tte i miing you might e urpried to find tht you reh t n inorret miniml DFA. Lexil Anlyi Smple Exerie Fll 6

4 Univerity of Southern Cliforni Computer Siene Deprtment Prolem 4: Conidering the lphet Σ = {,} ontrut Non-Determiniti-Finite Automton (NFA) uing the Thompon ontrution tht i le to reognie the entene generted y the regulr expreion (**)**. Doe the entene w = elong to the lnguge generted y thi regulr expreion? Jutify. Solution: The NFA i hown elow nd where the trt tte i tte leled S = -loure () = {,,, 4,,, } thi i finl tte eue of S = DFAedge(S,) = -loure (S, ) = {, 6, 8} S = DFAedge(S,) = -loure (S, ) = {,, 4,,, } finl tte DFAedge(S,) = -loure (S, ) = {, 6, 8} = S DFAedge(S,) = -loure (S, ) = {,, 4,,, } = S S = DFAedge(S,) = -loure (S, ) = {,, 4, 9,,, } finl tte S4 = DFAedge(S,) = -loure (S, ) = {6, 7, 8} DFAedge(S4,) = -loure (S4, ) = {,, 4, 9,,, } = S DFAedge(S4,) = -loure (S4, ) = {6, 7, 8} = S4 DFAedge(S,) = -loure (S, ) = {, 6, 8} = S DFAedge(S,) = -loure (S, ) = {,, 4,,, } = S Thi reult in the DFA hown elow with trting tte S. S S S S S S4 S4 Thi DFA n e further minimie y reogniing tht S, S nd S form elf-ontined prtition nd S nd S4 nother prtition reulting in the DFA on the right-hnd-ide. Lexil Anlyi Smple Exerie 4 Fll 6

5 Univerity of Southern Cliforni Computer Siene Deprtment Prolem : Conider the lphet Σ = {,}.. Contrut Non-Determiniti-Finite Automton (NFA) uing the Thompon ontrution tht i le to reognie the entene generted y the regulr expreion RE = ()*.()*.. Do the entene w = nd w = elong to the lnguge generted y thi regulr expreion? Jutify.. Convert the NFA in prt ) to DFA uing the uet ontrution. Show the mpping etween the tte in the NFA nd the reulting DFA. d. Minimie the DFA uing the itertive refinement lgorithm diued in l. Show your intermedite prtition reult nd doule hek the DFA uing the entene w nd w. Solution: ) A poile ontrution (lredy implified to hve only ingle -trnition etween tte) would reult in the NFA hown elow nd where the trt tte i leled. 4 6 ) Both word re reognied y thi NFA. Regrding the word there i pth from tte to the epting tte 6, nmely:,,,,4,,4,,6. Regrding the word the utomton my reh tte 6 following the pth,,,4,,4,,4,,6. ) Uing the uet ontrution we rrive t the following uet nd trnition. S = -loure () = {,,, 4,, 6} thi i finl tte eue of tte 6 S = DFAedge(S,) = -loure (goto(s, )) = {, 4,, 6} finl tte SE = DFAedge(S,) = -loure (goto(s, )) = { } S = DFAedge(S,) = -loure (goto(s, )) = {4,, 6} finl tte S = DFAedge(S,) = -loure (goto(s, )) = {, 4,, 6} finl tte DFAedge(S,) = -loure (goto(s, )) = {4,, 6} = S DFAedge(S,) = -loure (goto(s, )) = { } = SE DFAedge(S,) = -loure (goto(s, )) = {, 4,, 6} = S DFAedge(S,) = -loure (goto(s, )) = { } = SE Thi reult in the DFA hown elow with trting tte S. S SE S S S, d) Thi DFA n e further minimie y uing the itertive refinement prtitioning yeilding the equene of prtition indited elow. Lexil Anlyi Smple Exerie Fll 6

6 Univerity of Southern Cliforni Computer Siene Deprtment S SE S S S S P SE S S S P, P, Initil prtitioning on finl tte Prtitioning on S P SE S S S S SE S S P,, Prtitioning on finl miniml DFA Lexil Anlyi Smple Exerie 6 Fll 6

7 Univerity of Southern Cliforni Computer Siene Deprtment Prolem 6: Minimie the FA hown elow over the lphet {,}. Mke ure to onvert it firt into DFA nd them pply the itertive refinement lgorithm diued in l. trt 4 Solution: The firt oervtion i tht there re ome tte for whih one of the trnition on the lphet ymol i miing. Thi i the e with tte "" nd "". A uh need to ugment thi FA with "trp-tte" tht i non-epting hown elow in ). The reminder DFA reult from the pplition of the itertive prtitioning lgorithm deried in l. trt trt P 4, 4 P, ) Augmented DFA trt ) Prtion ed on epting nd non-epting tte trt P P 4 P, 4 P, P ) Prtion ed on '' trnition d) Prtion ed on '' trnition trt trt P 4 P e) Prtion ed on '' trnition P, f) Minimied DFA, Lexil Anlyi Smple Exerie 7 Fll 6

8 Univerity of Southern Cliforni Computer Siene Deprtment Prolem 7: Conider the DFA elow with trting tte nd epting tte :, 4, ) Uing the Kleene ontrution lgorithm derive regulr expreion. ) Symplify the expreion found in ) ove. Solution: Before engging in very long equene of ontention nd ymoli opertion we mke the oervtion tht tte 4 i trp-tte. A uh we n fely ignore it ine it i not n epting tte. Wht thi men i tht we n ue the lgorithm imply with k = nd need to ue it to ompute R there re pth from the trt tte to the finl tte vi tte. Expreion for k = R = () R = ( ) R = () R = () R = ( ) Expreion for k = R = R (R )*R R = () R = R (R )*R R = ( ) R = R (R )*R R = () R = R (R )*R R = () R = R (R )*R R = ( ) Expreion for k = R = R (R )*R R = R = () R = R (R )*R R = ( ) ( )* ( ) ( ) = ( ) * R = R (R )*R R = ( ) ( )* () () = *() R = R (R )*R R = () ( )* ( ) () = () * R = R (R )*R R = ( ) ( )* ( ) ( ) = * R = R (R )*R R = ( ) ( )* () = ( ) * () R = R (R )*R R = () ( )* () = () *() Expreion for k = R = R (R )*R R = R = () R = R (R )*R R = ( ) * () (() *())* () * ( )* R = R (R )*R R = *() (() *())* () *() * () R = R (R )*R R = () *() (() *())* () * () * R = R (R )*R R = *() (() *())* () * * R = R (R )*R R = ( ) * () (() *())* () *() ( ) * () R = R (R )*R R = () *() (() *())* () *() () *() Lexil Anlyi Smple Exerie 8 Fll 6

9 Univerity of Southern Cliforni Computer Siene Deprtment Clerly thee expreion hve lot of redundnt term. In ft we re only intereted in R = ( ) * () (() *())* () * ( )* ine the trting tte i tte nd the only epting tte i tte. ) To implify the expreion found ove i very hrd, there re lot of prtil term in thi expreion. Inted you follow the edge in the DFA nd try to ompoe the regulr expreion (in ome ene ignoring wht you hve done in the previou tep). The implet wy to look t thi prolem i to ee tht there i prefix ( ) nd then you return to tte either y (*) or (), o ompt regulr expreion would e R = ( )(() ())* muh more ompt regulr expreion tht R ove. Prolem 8: Conider the lphet Σ = {; }. Define hort, poily the hortet, regulr expreion tht generte tring over Σ tht ontin extly one nd t let one. Solution: The tring mut either trt with one or n, o we n ue * preffix to ount for ny poile numer of leding efore the firt. The ingle mut hve t let one either preeeding it or jut efore it. After thi ingle there n e ny numer of triling. A poile (lthough not neerily unique) regulr expreion denoting the ought lnguge i R = *( )*. Prolem 9: Given n NFA with -trnition how do you derive n equivlent NFA without thoe - trnition? Jutify. Solution: For thoe tte with -trnition merge the edge of the tte t the end of thoe trnition into the urrent tte. Iterte until there re no -trnition left. You till hve NFA ut there will e t more multiple edge with the lel lel out of given tte. Prolem : Given n NFA with everl ept tte, how how to onvert it onto NFA with extly one trt tte nd one epting tte. Solution: The initl tte i the me in the originl NFA. A to the epting tte jut dd - trnition from the orignl epting tte to new epting tte nd lel tht epting tte the only epting tte of the new NFA. Lexil Anlyi Smple Exerie 9 Fll 6

10 Univerity of Southern Cliforni Computer Siene Deprtment Prolem : Given the Finite Automton elow with initil tte nd lphet {,} nwer the following quetion: 4, () Why i thi FA Non-Determiniti Finite Automton (NFA)? () Convert thi NFA to DFA uing the loure omputtion. () Minimie the reulting DFA. (d) Wht i the Regulr Expreion mthed y thi NFA? You re dvied not to ue the utomti Kleene ontrution or try to look t the input NFA ut rther the orretly minimied DFA. Solution: () Thi FA i lredy DFA, the only trnition tht re miing ll go to the trp tte. () The uet ontrution, whih in thi e yield the me originl DFA on the left where we hve noted the uet eh tte tnd for. {} {} {} {4} {},, () Uing the itertive prtition refinement we hve the following prtition (elow left) nd the reulting minimied DFA on the right gin with the tte releled. Notie tht thi i peil e where the refinement lgorithm nnot refine ny u-et of tte given tht the originl DFA w lredy miniml. {} P {} {} P {4} {} P,, (d) Bed on the lt DFA the regulr expreion identified y thi FA mut hve extly two oneutive ymol poily preeded y ingle ymol. For intne, the tring i in the lnguge ut the tring or re not. The regulr expreion n e denoted y *. Lexil Anlyi Smple Exerie Fll 6

11 Univerity of Southern Cliforni Computer Siene Deprtment Prolem : Drw the mllet DFA tht ept the lnguge derived y the regulr expreion (** **). Solution: A poile trnltion guided y the ide of the Thompon ontrution i hown elow. 6, 4 Uing the itertive refinement lgorithm for DFA minimition indeed onfirm tht thi i the mllet poile DFA. The equene of refinement i hown elow., 4 6, 4 6 4, 6, 4 6 4, 6 4, 6 Prolem : Given the regulr expreion RE = ( *)*(*) over the lphet Σ = {,} do the following trnformtion:. Derive NFA uing the Thompon ontrution ple of identifying the tring generted y thi regulr expreion.. Convert the NFA otined in ) to DFA.. Minimie the reulting DFA uing the itertive refinement lgorithm diued in l. d. Determine in how mny tep i the equene proeed. Jutify. Solution: Regrding ) the NFA i given elow. Here we mde ome very imple implifition tht two or more equene of -trnition n e onverted to ingle -trnition. Thi utntilly redue the numer of tte in the NFA nd thiu filitte the ontrution of the equivlent DFA. Lexil Anlyi Smple Exerie Fll 6

12 Univerity of Southern Cliforni Computer Siene Deprtment Applying the uet ontrution we hve the following tte S = -loure () = {,, 4, 7, 8,,,,, 4, } thi i finl tte eue of S = DFAedge(S,) = -loure (goto(s, )) = {} S = DFAedge(S,) = -loure (goto(s, )) = {,,, 4, 7, 8, 9,,, 4, } finl S = DFAedge(S,) = -loure (goto(s, )) = {} S4 = DFAedge(S,) = -loure (goto(s, )) = {,,, 4, 6, 7, 8,,, 4, } finl S = DFAedge(S,) = -loure (goto(s, )) = {} = S S6 = DFAedge(S,) = -loure (goto(s, )) = {,,, 4, 7, 8, 9,,,, 4, } finl S7 = DFAedge(S,) = -loure (goto(s, )) = {} = S S8 = DFAedge(S,) = -loure (goto(s, )) = {} = S S9 = DFAedge(S4,) = -loure (goto(s4, )) = {} = S S = DFAedge(S4,) = -loure (goto(s4, )) = {,,, 4,, 7, 8, 8,,,, 4, } = S6 S = DFAedge(S6,) = -loure (goto(s6, )) = {,,, 4, } finl S = DFAedge(S6,) = -loure (goto(s6, )) = {,,, 4, 7, 8, 9,,, 4, } = S S = DFAedge(S,) = -loure (goto(s, )) = {} = S S4 = DFAedge(S,) = -loure (goto(s, )) = {,,, 4, 6, 7, 8,,, 4, }= S4 So there re totl of 7 tte, S, S, S, S, S4, S6, S nd the orreponding DFA i hown y the tle elow (old tte re finl tte) nd then in the figure elow the tle. Next Stte Current Stte S S S S S S4 S S S6 S S S S4 S S6 S6 S S S S S4, 4 6 Lexil Anlyi Smple Exerie Fll 6

13 Univerity of Southern Cliforni Computer Siene Deprtment, P 4 6 initil refinement on eptne riteri 6, 4 refinement of on P P, 4 6 P refinement of on P, P 4 6 refinement of P on P P P6 P P, 4 6 refinement of on P P6 P P, 4 6 finl refinement with 6 tte P Prolem 4: Given the DFA elow derie in Englih the et of tring epted y it., Solution: Any tring over the lphet tht ontin ny numer of oneutive, inluding none with preffix of ero or more oneutive. Lexil Anlyi Smple Exerie Fll 6

14 Univerity of Southern Cliforni Computer Siene Deprtment Prolem : Given regulr lnguge L. i.e., lnguge deried y regulr expreion, prove tht the revere of L i lo regulr lnguge (Note: the revere of lnguge L i L R where for eh word w in L, w R i in L R. Given word w over the given lphet, w R i ontruted y pelling w kwrd). Solution: If L i regulr lnguge then there exit DFA M tht reognie it. Now given M we n ontrut M tht reognie the revere of L with repet to the input lphet. We now derie how to ontrut M. M i repli of M ut revering ll the edge. The finl tte of M i the tte tht ued to e the trt tte of M. The trt tte of M i new tte with -trnition to the tte in M tht ued to e the finl tte of M. Now eue M might hve multiple finl tte M i y ontrution NFA. Given the equivlene of NFA nd regulr expreion we hve hown tht if L i regulr o i L R. Prolem 6: Drw the DFA ple of reogniing the et of ll tring eginning with whih interpreted the inry repreenttion of n integer (uming the lt digit to e proeed i the let ignifint) i ongruent to ero modulo i.e., the numeri vlue of thi inry repreenttion i multiple of. Solution: The hrd prt out thi prolem i tht you need to keep trk with the lredy oerved it wht the reminder of the diviion y i. Given tht you hve reminder you would need no more tht tte, one for eh of the reminder vlue through eing the tte tht repreent reminder of ero the epting tte, in thi e tte S. The DFA elow omplihe thi. You n verify thi DFA y trying the numer in inry or in inry. Notie tht in the lt tte S ny dditionl men you re hifting the it y one it, i.e., multiplying y, hene tying in the me tte. S S S S Prolem 7: Conider the lphet Σ = {,}.. Conider the Non-Determiniti Finite Automton (NFA) depited elow. Why i thi utomton non-determiniti? Explin the vriou oure on indeterminy.. Do the entene w = nd w = elong to the lnguge reognied y thi FA? Jutify.. Convert the NFA in prt ) to DFA uing the uet ontrution. Show the mpping etween the tte in the NFA nd the reulting DFA. d. Minimie the DFA uing the itertive refinement lgorithm deried in l. Show your intermedite prtition reult nd doule hek the DFA uing the entene w nd w. Lexil Anlyi Smple Exerie 4 Fll 6

15 Univerity of Southern Cliforni Computer Siene Deprtment 4 6 trt Solution: ) Thi i indeed NFA for two reon. Firt, it inlude -trnition. Seond, in tte there re two trnition on the me terminl or lphet ymol, "". ) Regrding the word there i pth from tte to the epting tte 6, nmely:,4,,6. Regrding the word the utomton will never e le to reh the tte 6 in order to pell out the hrter will neerily e in tte nd to reh tht tte we nnot hve two oneutive "" hrter we we would e either in tte or tte 4. ) Uing the uet ontrution we rrive t the following uet nd trnition. S = -loure ({}) = {,, 4} thi i not finl tte. S = DFAedge(S,) = -loure (goto(s, )) = {,,, 6} finl tte S = DFAedge(S,) = -loure (goto(s, )) = {} S9 = DFAedge(S,) = -loure (goto(s, )) = {,, 6} finl tte S4 = DFAedge(S,) = -loure (goto(s, )) = {,, 6} finl tte. S = DFAedge(S,) = -loure (goto(s, )) = {,, 6} finl tte. DFAedge(S,) = -loure (goto(s, )) = {} = S DFAedge(S4,) = -loure (goto(s4, )) = {,, 6} = S finl tte. DFAedge(S4,) = -loure (goto(s4, )) = {,, 6} = S4 finl tte. S6 = DFAedge(S,) = -loure (goto(s, )) = {, 6} finl tte. S7 = DFAedge(S,) = -loure (goto(s, )) = {, 6} finl tte. DFAedge(S6,) = -loure (goto(s6, )) = {, 6} = S6 finl tte. S8 = DFAedge(S6,) = -loure (goto(s6, )) = { } DFAedge(S7,) = -loure (goto(s7, )) = { } = S8 DFAedge(S7,) = -loure (goto(s7, )) = {, 6} = S7 finl tte. DFAedge(S8,) = -loure (goto(s8, )) = { } = S8 DFAedge(S8,) = -loure (goto(s8, )) = { } = S8 DFAedge(S9,) = -loure (goto(s9, )) = {,, 6} = S9 DFAedge(S9,) = -loure (goto(s9, )) = {} = S Thi reult in the DFA hown elow with trting tte S. trt S S S9 S S4 S S7 S6 S8, Lexil Anlyi Smple Exerie Fll 6

16 Univerity of Southern Cliforni Computer Siene Deprtment d) We n try to minimie thi DFA uing the itertive refinement lgorithm deried in l. The figure elow depit poile equene of refinement. For eh tep we indite the riteri ued to diriminte etween tte in the previou prtition. A n e een, the lgorithm led to no refinement, o thi i prtiulr e where the u-et lgorithm tht help u derive the DFA from it originl NFA doe yield miniml DFA. P P P P trt S S S S9 S4 S S7 S6 S8, trt S S S9 S S4 S S7 S6, S8 trt S S S S4 S9 S S7 S6, S8 P P P () Prtition etween epting nd non-epting tte () Dirimintion ed on "" trnition () Dirimintion ed on "" trnition trt P S P S S P9 S9 S4 S S7 S6 P P S8, trt P S P S S P6 S S4 P9 S9 S7 S6 P P S8, trt P S P7 P S S P6 S S4 P9 S9 S7 S6 P P S8, (d) Dirimintion ed on "" trnition (e) Dirimintion ed on "" trnition (f) Diriminion ed on "" trnition Lexil Anlyi Smple Exerie 6 Fll 6

17 Univerity of Southern Cliforni Computer Siene Deprtment Prolem 8: Conider the DFA elow with trting tte nd epting tte : trt ) Derie in Englih the et of tring epted y thi DFA. ) Uing the Kleene ontrution lgorithm derive the regulr expreion reognied y thi utomton ymplifying it muh poile. Solution: ) Thi utomton reognie ll non-null tring over the {,} lphet tht end with "". ) The derivtion re hown elow with the oviou ymplifition. Expreion for k = R = ( ) R = () R = () R = ( ) Expreion for k = R = R (R )*R R = ( ). ( )*. ( ) ( ) = * R = R (R )*R R = ( ). ( )*. () () = * * R = R (R )*R R = (). ( )*. ( ) () = + + R = R (R )*R R = (). ( )*. ( ) ( ) = (* ) Expreion for k = R = R (R )*R R = ( * * ). (* )*. ( + + ) ( * ) R = R (R )*R R = ( * * ). (* )*. ( + + ) ( * * ) R = R (R )*R R = ( * * ). (* )*. ( + + ) ( + + ) R = R (R )*R R = ( * * ). (* )*. ( + + ) (* ) L = R = ( * * ). (* )*. ( + + ) ( * * ) A n e een the implifition of ny of thee regulr expreion eyond the expreion for k= i firly omplited. Thi method, lthough orret y deign led to regulr expreion tht re fr from eing miniml or even the mot ompt repreenttion of the regulr lnguge the DFA reognie. Lexil Anlyi Smple Exerie 7 Fll 6

18 Univerity of Southern Cliforni Computer Siene Deprtment Prolem 9: For the regulr expreion depited elow nwer the following quetion: RE = ( - +)?. (e E). (-9)+ ) Uing the Thompon ontrution (or implifition thereof) derive NFA tht reognie the tring peified y thi RE. ) Convert the NFA in ) to n equivlent DFA uing the uet ontrution. ) Mke ure the DFA M found in ) ove i miniml (you do not need to how tht it i in ft miniml) nd ontrut the DFA M tht ept the omplement of the regulr lnguge the firt DFA ept. Show tht M doe ept the word w = "e" whih i not epted y the originl DFA. Solution: ) The figure elow depit the NFA with -trnition nd where we hve implified the trnition on numeri digit. 'e' '' trt '-' 4 6 'E' 7... '9' 8 9 '+' ) We ue the uet ontrution y tring for every et of tte the other poile tte the NFA ould e in if it were to trvere -edge only. The lit elow depit the reult of thi ontrution where we lo derie the omputtion of edge DFA edge uing the funtion DFA_edge nd then omputing the e-loure of the reulting intermedite et of tte. -loure({}) DFA_edge(S,'-') DFA_edge(S,'+') DFA_edge(S,'e') DFA_edge(S,'E') DFA_edge(S,'<digit>') DFA_edge(S,'-') DFA_edge(S,'+') DFA_edge(S,'e') DFA_edge(S,'E') DFA_edge(S,'<digit>') = {,,,, 4} = S = -loure({}) = {,, 4} = S = -loure({}) = {,, 4} = S = -loure({}) = {, 6, 7} = S = -loure({}) = {, 6, 7} = S = -loure( {} ) = SE; error tte = -loure( {} ) = SE; error tte = -loure( {} ) = SE; error tte = -loure({}) = {, 6, 7 } = S = -loure({}) = {, 6,7 } = S = -loure( {} ) = SE; error tte DFA_edge(S,'-') = -loure( {} ) = SE; error tte DFA_edge(S,'+') = -loure( {} ) = SE; error tte DFA_edge(S,'e') = -loure( {} ) = SE; error tte DFA_edge(S,'E') = -loure( {} ) = SE; error tte DFA_edge(S,'<digit>') = -loure( {8} ) = { 6, 7, 8, 9, } = S; DFA_edge(S,'-') = -loure( {} ) = SE; error tte DFA_edge(S,'+') = -loure( {} ) = SE; error tte DFA_edge(S,'e') = -loure( {} ) = SE; error tte DFA_edge(S,'E') = -loure( {} ) = SE; error tte DFA_edge(S,'<digit>') = -loure( {8} ) = { 6, 7, 8, 9, } = S; The figure elow depit the DFA reulting from the pplition of the uet ontrution. Notie tht thi i fully peified DFA where eh tte inlude trnition for ll the hrter in the onidered lphet. Lexil Anlyi Smple Exerie 8 Fll 6

19 Univerity of Southern Cliforni Computer Siene Deprtment trt '-' '+' S 'e' 'E' '' '9'... S... '' '9' '' '9' 'e' 'E' '-' '+'... 'e' 'E' '' S S 'e' 'E' '-' '+' SE... '9' 'e' 'E' '-' '+' ''... '9' ) Let M e the DFA found in ). Thi DGA i in ft miniml. Informl proof: All tte re required. S i the trt tte. S pture the ft tht we hve oerved ign.. Stte S pture to ft we need to oerve t let one digit. All the digit re then ptured y tte S. Stte SE i the trp error tte. Let M e thi firt DFA. We ontrut M y mking ll epting tte of M non-epting tte nd ll non-epting tte of M epting tte of M. The tte trt i unhnged. Thi i DFA we only hnge the tte of non-epting tte. The trt tte i till ingleton nd ll trnition re lerly identified in the newly reted DFA hown elow. '-' '+' S S 'e' 'E' 'e' 'E' ''... S ''... '9' trt '' '9'... '' '9' S... '' '9' 'e' 'E' '-' '+'... 'e' 'E' '-' '+' SE '9' 'e' 'E' '-' '+' In thi newly reted DFA the input tring w = "e" tke the DFA to tte SE whih i n epting tte n thi "omplement" mhine. Lexil Anlyi Smple Exerie 9 Fll 6

20 Univerity of Southern Cliforni Computer Siene Deprtment Prolem Conider the lphet Σ = {,, }.. Conider the Non-Determiniti Finite Automton (NFA) depited elow. Why i thi utomton non-determiniti? Explin the vriou oure on indeterminy.. Do the entene w = nd w = elong to the lnguge generted y thi FA? Jutify.. Convert the NFA in prt ) to DFA uing the uet ontrution. Show the mpping etween the tte in the NFA nd the reulting DFA. d. By inpetion n you derive regulr expreion (RE) thi NFA detet?,, 4,, trt 6, Solution: ) Thi FA i non-determiniti for t let one of it tte there re two trnition on the me lel to two ditint tte. Thi i the e with tte where on '' the FA goe to tte nd 6. Similrly, tte nd exhiit the me ehvior. ) The entene w = elong to the lnguge epted y thi FA there i pth through tte, nd tht peel out thi ord nd tte i n epting tte. A to the word w = ny word tht egin with '' led to tte 4 whih i trp nd non-epting tte. A uh w doe not elong to the lnguge reognied y thi FA. ) A implified u-et ontrution i hown elow: S = -loure ({}) = {} thi i not finl tte. DFAedge(S, ) = -loure (goto(s, )) = {,6} = S DFAedge(S, ) = -loure (goto(s, )) = {4} DFAedge(S, ) = -loure (goto(s, )) = {4} DFAedge(S, ) = -loure (goto(s, )) = {4} = S = S = S DFAedge(S, ) = -loure (goto(s, )) = {,4} = S4 DFAedge(S, ) = -loure (goto(s, )) = {4,} = S DFAedge(S, ) = -loure (goto(s, )) = {4} DFAedge(S, ) = -loure (goto(s, )) = {4} DFAedge(S, ) = -loure (goto(s, )) = {4} = S = S = S DFAedge(S4, ) = -loure (goto(s4, )) = {4} = S DFAedge(S4, ) = -loure (goto(s4, )) = {4} = S DFAedge(S4, ) = -loure (goto(s4, )) = {,4} = S4 DFAedge(S, ) = -loure (goto(s, )) = {4} = S DFAedge(S, ) = -loure (goto(s, )) = {4,} = S DFAedge(S, ) = -loure (goto(s, )) = {4} = S Lexil Anlyi Smple Exerie Fll 6

21 Univerity of Southern Cliforni Computer Siene Deprtment In thi ontrution we hve implified the reulting DFA y mking the trnition etween tte {,6} on '' direlty move to tte 4 whih i trp or error tte. A more mehnil proedure would hve een the retion of n dditionl trp tte tht i equivlent to tte 4.,4,,6, 4,, trt 4,, d) By inpetion of the two pth from tte to either tte or tte we n immeditely ee tht the lnguge epted y thi FA n e deried y the regulr expreion RE = * *. Prolem Conider the lphet Σ = {,, }. Contrut DFA tht ept the following lnguge. Argue tht your DFA i miniml (with repet to the numer of tte) or ue the DFA minimition lgorithm to how tht tht i the e.. {w {,} w = n, w = m,m,n N}. {w {, } w = n, n N} nd w doen t ontin the utring Solution:. trt. 6, 7,, trt 4 Lexil Anlyi Smple Exerie Fll 6

22 Univerity of Southern Cliforni Computer Siene Deprtment Prolem Develop regulr expreion (RE) tht detet the longet tring over the lphet {-} with the following propertie:. The tring egin with n hrter nd end with hrter;. After the firt, the tring n inlude n lterntion of two uequene, nmely equene tht egin with hrter followed y ero or more hrter ending with n e hrter nd uequene tht egin with hrter ut doe not hve ny hrter until terminting t hrter.. If t ny point in the uequene there i x hrter, tht peifi equene i onidered terminted, i.e., the x t the e hrter in the firt type of equene nd the t hrter in the eond type of equene. 4. The two uequene nnot e neted ut n e repeted in ny lternting order. A n exmple the tring efghte i to e epted well the tring xfft Quetion: e. Develop regulr expreion tht pture the truture of the eptle tring deried ove. Ue hort-hnd to pture uet of hrter in the lphet o tht your deritpion nd the orreponding FA re kept hort. f. Uing the regulr expreion define in etion ) devie the orreponding Non-Determiniti Finite Automton (NFA) uing the Thompon ontrution deried in l. g. Convert the NFA in etion ) to DFA uing the uet ontrution. Show the mpping etween the tte in the NFA nd the reulting DFA. h. Minimie the DFA derived in etion ) (or how it i lredy miniml) uing the itertive refinement lgorithm deried in l. Solution: ) Among the mny poile olution for regulr expreion the one elow eem to e the mot intuitive. Here we ue the hort-hnd notstx for ll hrter in the lphet exept nd t i.e., notstx = {-r, u, v, y, }. RE =. [ (. ( * ). (x e) ) (. ( notstx * ). (x t) ) ] *. ) The NFA elow reflet the truture of the RE preented ove where we hve lo ompreed ome -trnition for revity, in prtiulr the one tht reflet Kleene loure nd lterntion trt 4 /{tx} ) The uet ontrution of thi NFA i hown elow where gin we hve merged trnition on hrter with nlogou ehvior. Lexil Anlyi Smple Exerie Fll 6

23 Univerity of Southern Cliforni Computer Siene Deprtment 7,9 trt,,,4,7,9 6,8,,,,4,,,,6,,,4,,4,,6 7 /tx 8, /tx d) The minimition of thi DFA uing the itertive refinement lgorithm i ummried elow. We egin y imple prtitionetween the epting tte nd the non-epting tte. Thi immeditely low u to iolte the epting tte 7. The eond prtition ued the hrter the dirimintig token howing tht tte,,,4,,,,6 nd,,,4,,4,,6 my e equivlent. The lt prtition digrm reflet the ft tht ll tte re dijoint. Lexil Anlyi Smple Exerie Fll 6

24 Univerity of Southern Cliforni Computer Siene Deprtment Lexil Anlyi Smple Exerie 4 Fll 6 P P P P6 P7 trt,,,4,7,9,,,4,,,,6 6,8, 7 7,9 /tx /tx 8,,,,4,,4,,6 P P P6 P7 trt,,,4,7,9,,,4,,,,6 6,8, 7 7,9 /tx /tx 8,,,,4,,4,,6 P8 P trt /tx /tx P P trt,,,4,7,9,,,4,,,,6 6,8, 7 7,9 /tx /tx 8,,,,4,,4,,6 P P trt,,,4,7,9,,,4,,,,6 6,8, 7 7,9 /tx /tx 8,,,,4,,4,,6 P trt,,,4,7,9,,,4,,,,6 6,8, 7 7,9 /tx /tx 8,,,,4,,4,,6

25 Univerity of Southern Cliforni Computer Siene Deprtment Prolem Given the DFA elow over the lphet {,} determine the following:. Ue the dynmi-progrmming Kleene lgorithm to derive the regulr expreion tht denote the lnguge epted y it. Mke ure your lelling of the DFA tte i orret nd tht the DFA i ompletely peified, i.e., eh tte h trnition on ll the lphet hrter.. Derie uintely wht re the word epted y thi DFA?, trt Solution:. Ue the dynmi-progrmming Kleene lgorithm to derive the regulr expreion tht denote the lnguge epted y it. Mke ure your lelling of the DFA tte i orret nd tht the DFA i ompletely peified, i.e., eh tte h trnition on ll the lphet hrter. Expreion for k = R = () R = () R = () R = Ø R = ( ) R = Ø R = Ø R = () R = ( ) Expreion for k = R = R (R )*R R = ()() * () () = () R = R (R )*R R = ()() * () () = () R = R (R )*R R = ()() * () () = () R = R (R )*R R = R = Ø R = R (R )*R R = R = ( ) R = R (R )*R R = R = Ø R = R (R )*R R = R = Ø R = R (R )*R R = R = () R = R (R )*R R = R = ( ) Lexil Anlyi Smple Exerie Fll 6

26 Univerity of Southern Cliforni Computer Siene Deprtment Expreion for k = R = R (R )*R R = (). ( ) *. Ø () = () R = R (R )*R R = (). ( ) *. ( ) () = ( + ) R = R (R )*R R = (). ( ) *. Ø () = () R = R (R )*R R = ( ) ( )* Ø Ø = Ø R = R (R )*R R = ( ) ( )* ( ) ( ) = ( * ) R = R (R )*R R = ( ) ( )* Ø Ø = Ø R = R (R )*R R = (). ( )*. Ø Ø = Ø R = R (R )*R R = (). ( )*. ( ) () = (). ()* R = R (R )*R R = (). ( )*. Ø ( ) = ( ) Expreion for k = R = R (R )*R R = (). ()*. Ø () = Ø R = R (R )*R R = (). ()*. (). ()* ( + ) R = R (R )*R R = (). ()*. ( ) () = (). ()* R = R (R )*R R = Ø. ()*. Ø Ø = Ø R = R (R )*R R = Ø. ()*. (). ()* ( * ) = ( * ) R = R (R )*R R = Ø. ()*. ( ) Ø = Ø R = R (R )*R R = ( ). ()*. Ø Ø = Ø R = R (R )*R R = ( ). ()*. (). ()* (). ()* = ()*. (). ()* R = R (R )*R R = ( ). ()*. ( ) ( ) = ()* The lnguge reognied y thi DFA n e expreed y the regulr expreion R. L = R = (). ( )*. ( ) = (). ( )* =.*. Derie uintely wht re the word epted y thi DFA? Thi utomton reognie ll inrie tring tht denote integer tht re non-negtive power of two nd uh n e deried y the ompt regulr expreion: *. Lexil Anlyi Smple Exerie 6 Fll 6