Compilers. Top-down analysis. Third course Spring term. Top-down analysis. Top-down analysis Overview. Introduction

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1 Complers Thrd course prng term Alfonso Orteg: nrque Alfonsec: Overvew Blnd serch, depth-frst serch nd wdth-frst serch. low cktrckng wth LL(1) grmmrs Procedures for modfyng grmmrs lmnton of left-recurson lmnton of lmd-rules Grech Norml Form Defnton Procedure for otnng the GNF of grmmr xmples LL(1) grmmrs Intl exmples Defnton of LL(1) grmmrs yntctc nlysers sed on LL(1) grmmrs Introducton We need to fnd dervton from the xom to the progrm nlysed. However, dependng on the grmmr, the numer of possle dervtons s extremely lrge. There re generl strteges to solve ny prolem wth lnd serch: Bredth-frst serch nd depth-frst serch. In redth-frst serch, t s gurnteed tht we shll fnd soluton. In depth-frst serch, there mght e rnches wth nfnte lengths. If tht s not the cse, sometmes t s possle to fnd soluton wth less effort thn n redth-frst serch. 3 4

2 Top-down wth cktrckng: concepts We cn consder top-down nlyss wth cktrckng prser s prtculr cse of depth-frst lnd serch. The crteron to decde whether rule s pplcle n gven poston s the concdence wth the non-termnl tht s n the left-hnd sde of the rule. We shll not contnue long n nlyss pth when we hve termnls tht do not concde wth the progrm. We cn fnsh the process n two stutons: If we hve een le to generte ll the termnls n the progrm sequence. In ths cse, we hve een le to construct syntctc tree for the progrm, whch s correct. If t ws not possle to fnsh wth the dervton tree, ut we hve tred ll the rules n ech poston nd there s no other opton vlle to try. In ths cse, the progrm wll e syntctclly ncorrect. Ths technque wll e llustrted wth the followng exmples: yntctc nlyss of the word

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5 L next pl ACCPTD yntctc nlyss of the word 19 20

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7 25 26 RJCTD 27 28

8 Anlyss of the word { - +}, > { - +}, > { - +}, > - - { - +}, >

9 { - +}, > - { - +}, > { - +}, > { - +}, >

10 { - +}, > { - +}, > { - +}, > { - +}, >

11 { - +}, > { - +}, > { - +}, > { - +}, >

12 Consder the nlyss of { - +}, > { - +}, > - - ACCPTD Conclusons If there re left-recursve rules, the nlyss of the word my provoke tht the lgorthm enters n nfnte loop. In the followng sldes, we shll study the possltes for vodng tht fct, nd we shll see other propertes of grmmrs whch re nterestng for uldng effcent top-down syntctc nlysers. Overvew Blnd serch, depth-frst serch nd wdth-frst serch. low cktrckng wth LL(1) grmmrs Procedures for modfyng grmmrs lmnton of left-recurson lmnton of lmd-rules Grech Norml Form Defnton Procedure for otnng the GNF of grmmr xmples LL(1) grmmrs Intl exmples Defnton of LL(1) grmmrs yntctc nlysers sed on LL(1) grmmrs 47 48

13 Propertes of grmmrs for top-down nlyss Propertes It s esy to uld LL(1) nlysers, n whch the nput s red from left to rght, nd the dervtons n the tree re lso mde from left to rght. The numer 1 mens tht t needs one look-hed symol from the nput strng. In order to uld n LL(1) nlyser, the grmmr hs to e expressed n LL(1) form. Possle steps to trnsform grmmr nto LL(1) form re: Removng ll the left-recursve rules. Removng nccessle symols. Removng rules tht generte the empty word: A λ xpressng the grmmr n Grech Norml Form. Left fctorston (whch s necessry condton for LL(1) grmmrs) { - +}, > + - Grech Norml Form Removng left-recursve rules Let us nlyse the followng exmple: { - + λ}, > λ λ - λ λ Grech Norml Form Removng left-recursve rules Grech Norml. Form Removng left-recursve rules: frst exmple It s dffcult to thnk of generl procedure from the prevous exmple. However, the technque cn e generlsed. G=<{,T,F}, {-,*,} { +T T T T*F F F }, > A Aα 1... Aα n β 1... β m A β 1 X... β m X X α 1 X... α n X λ G =<{,,T,T,F}, {-,*,} { T +T λ T FT F }, > 51 52

14 Grech Norml Form Lemm: removng left-recursve rules Grech Norml Form Removng lmd-rules very context-ndependent grmmr cn e trnsformed nto n equvlent grmmr wthout left-recursve rules. A left-recursve rule s rule of the followng form: A Ax,,A Σ N x (Σ T Σ N ) * The lemm s shown n constructve wy dong, for ech recursve rule, the followng tretment: Let < Σ T, Σ N,, P > e grmmr wth left-recursve rules: A Aα 1... Aα n β 1... β m Where {β } n =1 represent ll the non-left-recursve rules We cn susttute the prevous rules y the followng set of rules A β 1 X... β m X X α 1 X... α n X λ We cn modfy rule y susttutng non-termnl n ts rght-hnd sde y ll the rght-hnd sdes of the rules for tht non-termnl. The grmmr otned n ths wy genertes the sme lnguge s the orgnl one Grech Norml Form Grech Norml Form Removng lmd-rules Removng lmd-rules G =<{,,T,T,F}, {-,*,} { T +T λ T FT F }, > F ( cn e put n plce of F n) T FT G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > T T (T s set n the plce of T ) T G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > 55 56

15 Grech Norml Form Removng lmd-rules Grech Norml Form Removng lmd-rules We cn delete rule for non-termnl symol, provded tht we lso dd new rules y susttutng ll ppernces of the non-termnl y ll the rght-hnd sdes of the rules elmnted. A cse wth specl nterest s the rules tht generte the empty word, λ A lmd-rule s rule of the followng form: A λ,,a Σ N It s used to remove non-termnl symol from word. G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > G =<{,,T,T,F}, {-,*,} { T T +T +T T T F }, > Grech Norml Form Removng lmd-rules Grech Norml Form Removng lmd-rules Let us test some dervtons: G =<{,,T,T,F}, {-,*,} { T T +T +T T T F }, > G =<{,,T,T,F}, {-,*,} { T T +T +T T T T *FT *F F }, > G=<{,T,F}, {-,*,} { +T T T T*F F F }, > T + + T F T F ++ + T G =<{,,T,T,F}, {-,*,} { T T +T +T T T T *FT *F F }, > F + T 59 60

16 Grech Norml Form Removng lmd-rules Grech Norml Form Removng lmd-rules Let us test some dervtons: Let us test some dervtons: G=<{,T,F}, {-,*,} { +T T T T*F F F }, > T T * F ** * F T T * F * F T G =<{,,T,T,F}, {-,*,} { T T +T +T T T T *FT *F F }, > G=<{,T,F}, {-,*,} { +T T T T*F F F }, > T * + F T F F *+ T * F + T G =<{,,T,T,F}, {-,*,} { T T +T +T T T T *FT *F F }, > F Overvew Blnd serch, depth-frst serch nd wdth-frst serch. low cktrckng wth LL(1) grmmrs Procedures for modfyng grmmrs lmnton of left-recurson lmnton of lmd-rules Grech Norml Form Defnton Procedure for otnng the GNF of grmmr xmples LL(1) grmmrs Intl exmples Defnton of LL(1) grmmrs yntctc nlysers sed on LL(1) grmmrs Informlly, Formlly, Grech Norml Form Defnton A context-ndependent grmmr s n Grech Norml Form f nd only f the rghthnd-sde of ll the rules strts wth termnl symol, followed, optonlly, y nontermnls. < Σ T, Σ N,, P >, context-ndependent, s n Grech Norml Form s (def) r P r=a x,, Σ T x Σ * N 63 64

17 Grech Norml Form Theorem very context-ndependent grmmr whch does not generte the empty word cn e expressed n Grech Norml form. Grech Norml Form Proof of the theorem It wll e constructve proof, showng how the grmmr cn e trnsformed. The trnsformton s performed n the followng wy: 1. If the lnguge contns the empty word, dd the followng rule ( s the xom) λ 2. Remove ll the left-recursve rules pplyng the lemm seen efore. 3. The followng prtl orderng wll e estlshed etween the non-termnl symols, deduced from the producton rules: A <A j ( α Σ T* A A j α P) ( β Σ T * A j A β P) If ths orderng produces loop,.e., f α,β Σ T* A A j α P A j A β P we cn choose ny of the two followng optons: A <A j A j <A Grech Norml. Form xmple of step 3 G=<{,T,F}, {-,*,} { +T T, T T*F F, F }, > The followng prtl orderng cn e deduced from the rules n the grmmr: +T does not provde ny orderng. From T we cn deduce <T From T T*F we do not get ny orderng. From T F we cn deduce T<F From F we cnnot deduce nythng. In concluson, <T<F Grech Norml Form Proof of the theorem 4. We cn clssfy the rules n the followng three groups: Type 1: Rules of the form A x,, Σ T, x Σ * Type 2: Rules of the form A Bx,, A,B Σ N, x Σ * A<B Type 3: Rules of the form A Bx,, A,B Σ N, x Σ * B<A 67 68

18 Grech Norml. Form xmple of step 4 G =<{,,T,T,F}, {-,*,} { T, +T λ, T FT,, F }, > Type 1 rules: +T T *FT F Type 2 rules: T T FT There re no type 3 rules. The lmd-rules wll e seen lter. Grech Norml Form Proof of the theorem 5. All the rules of type 3 wll e deleted. Type 3: Rules of the form A Bx,, A,B Σ N, x Σ * B<A To do ths, 1. We replce B wth ll the rght-hnd sdes for B. 2. Ths process s repeted for ll the type-3 rules. 3. If, durng ths process, there pper new left-recursve rules, we trnsform them usng the procedure seen efore. 4. If there pper nccessle symols, we elmnte them Grech Norml. Form xmple of step 5 Grech Norml. Form xmple of step 5 G =<{A,B,C}, {,} {A BC, B CA, C AB }, Type 3 rules: C AB We susttute A wth ts rght-hnd sdes (BC): C BCB Now, there s new type-3 rule: C BCB We susttute B wth ts left-hnd sdes (CA ): C CACB CB G =<{A,B,C}, {,} {A BC, B CA, C CACB CB }, G =<{A,B,C}, {,} {A BC, B CA, C BCB }, 71 72

19 Grech Norml Form Proof of the theorem 6. At ths pont, there should not e more type-3 rules. The next step wll e the removl of 2-type rules, strtng wth the non-termnls whch re t the end of the prtl orderng. Type 2: Rules of the followng form: A Bx,, A,B Σ N, x Σ * A<B To do ths, 1. B wll e replced y ll the rght-hnd sdes of the rules for B. 2. Ths s repeted untl we do not hve ny more type-2 or type-3 rules. 3. If there pper new left-recursve rules, they wll lso e removed. 4. Inccessle symols wll lso e removed. Grech Norml Form xmple of step 6 G =<{,,T,T,F}, {-,*,} { T, +T λ, T FT,, F }, > There re two type-2 rules, wth the orderng <T<F T T FT Frstly, we susttute F n T FT wth ts rght-hnd sdes (), nd otn: T T G =<{,,T,T,F}, {-,*,} { T, +T λ, T T,, F }, > Grech Norml. Form xmple of step 6 G =<{,,T,T,F}, {-,*,} { T, +T λ, T T,, F }, > Next, we susttute T n T wth ts rght-hnd sdes, (T ). We otn the rule T Grech Norml Form Proof of the theorem 7. Now, ll the rules elong to type 1: they ll hve the followng form: A x,, Σ T, x Σ * The only dfferent wth respect to Grech Norml Form my e due to rules hvng more thn one termnl symol n the rght-hnd sde. Ths cn e solved wth trvl susttuton, ddng new non-termnl symol, s n the followng exmple. A C We cn replce tht rule wth: A BC, B Where B s new non-termnl symol G =<{,,T,T,F}, {-,*,} { T, +T λ, T T,, F }, > 75 76

20 G =<{A,X}, {,} {A X X, X X λ}, Grech Norml. Form xmple of step 7 There re two rules n whch the termnl symol hs to e replced y the new non-termnl symol A : We dd the new rule A We susttute A X y A A X We susttute X X y X A X G =<{A,X,A }, {,} {A A X X, X A X λ, A }, 77 Grech Norml Form Proof of the theorem 8.The lst thng to do s the tretment of lmd-rules. In Grech Norml Form, these rules re fordden. The only excepton s when the lnguge contns the empty word, n whch there hs to e, necessrly, lmd rule for the xom of the grmmr. They should e removed usng the procedure prevously studed. NOT: For the purpose of uldng LL(1) complers, the grmmr needs not e exctly n Grech Norml form: ome of the lmd rules wll not e wrong for n LL(1) grmmr. Otherwse, they wll e removed usng the procedure lredy seen. When these rules re removed, sometmes t s dffcult to comply wth ll the condtons for LL(1) grmmrs. In ths cse, t my e necessry to lter the grmmr mnully to otn n equvlent one whch cn e restted s n LL(1) grmmr. 78 Grech Norml. Form xmple 1 1. As the lnguge does not contn the empty word, there s nothng to do. 2. Remove ll the left-recursve rules: G=<{,T,F}, {-,*,} { +T T T T*F F F }, > A Aα 1... Aα n β 1... β m A β 1 X... β m X X α 1 X... α n X λ G =<{,,T,T,F}, {-,*,} { T +T λ T FT F }, > Grech Norml. Form xmple 1 3. We estlsh the prtl orderng for the non-termnl symols. In generl, we cn estlsh the orderng usng the non-termnl symols n the orgnl grmmr, nd the ones dded n step 2 my e dded to the orderng f necessry. G=<{,T,F}, {-,*,} { +T T, T T*F F, F }, > The followng orderng wll e deduced from the prevous rules: From +T we do not deduce nythng From T we deduce <T From T T*F we deduce nothng From T F we deduce T<F From F we cn t deduce nythng. In concluson, <T<F 79 80

21 Grech Norml. Form xmple 1 4. The rules wll e clssfed n three groups. 5. The grmmr does not contn type-3 rules, so there s nothng to e done 6. Type-2 rules removl: Grech Norml. Form xmple 1 G =<{,,T,T,F}, {-,*,} { T, +T λ, T FT,, F }, > Type-1 rules +T T *FT F Type-2 rules T T FT No type-3 rules. Lmd rules wll e treted t the end. G =<{,,T,T,F}, {-,*,} { T, +T λ, T FT,, F }, > There re two type-2 rules, wth the orderng <T<F T T FT We frst solve F n T FT nd replce t y ts rght-hnd sdes (), nd otn T T G =<{,,T,T,F}, {-,*,} { T, +T λ, T T,, F }, > Grech Norml. Form xmple 1 G =<{,,T,T,F}, {-,*,} { T, +T λ, T T,, F }, > Grech Norml Form xmple 1 7. As there re no termnl symols ncorrectly set n the rght-hnd sdes of the rules, there s nothng to do t ths step. Next, we delete T n T nd replce t wth ts rght-hnd sdes (T ), to otn T G =<{,,T,T,F}, {-,*,} { T, +T λ, T T,, F }, > 83 84

22 Grech Norml Form xmple 1 Grech Norml Form xmple 1 8. In the cse tht we wnt to otn the grmmr n Grech Norml form, the lst step would e to remove ll the lmd-rules. RMMBR: ths step s not strctly necessry for LL(1) prsers G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > G =<{,,T,T,F}, {-,*,} { T T +T +T T T F }, > G =<{,,T,T,F}, {-,*,} { T T +T +T T T F }, > G =<{,,T,T,F}, {-,*,} { T T +T +T T T T *FT *F F }, > When there re no more lmd-rules, we hve the grmmr, fnlly, n GNF Overvew Blnd serch, depth-frst serch nd wdth-frst serch. low cktrckng wth LL(1) grmmrs Procedures for modfyng grmmrs lmnton of left-recurson lmnton of lmd-rules Grech Norml Form Defnton Procedure for otnng the GNF of grmmr xmples LL(1) grmmrs Intl exmples Defnton of LL(1) grmmrs yntctc nlysers sed on LL(1) grmmrs Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 ome lmd rules re not ncorrect n LL(1) grmmrs. Therefore, we re only gong to remove those tht re not rght. The followng procedure wll e followed: When non-termnl hs lmd rule, e.g. X λ The followng stuton my rse: tht the sme non-termnl hs other rules wth dfferent rght-hnd sde. Becuse we re n step 8, we know tht ll rules re now type-1 rules, strtng wth termnl symol. X α,, Σ T α Σ * 87 88

23 Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 Imgne syntctc dervton tree for word (the termnl symols re represented n red colour) Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 Imgne tht we re dong the top-down prsng of tht word, nd we re just efore the symol X XY 1...Y m t t j t l t k t t t s t n If X s the only non-termnl tht remns, there would e the followng two possltes: X λ X α X α t...t n Intutvely, the effcency of LL(1), whch s etter thn smple top-down prsng wth slow cktrckng, s due to the ndexton of the rght-hnd sdes, for ech non-termnl, usng the next termnl to e nlysed. Ths wll e possle only f ech non-termnl hs just one rght-hnd sde strtng wth ech termnl. If we do not hve lmd rules (X λ), the top-down nlyss cn e done completely wthout ny rnchng n the nlyss. t t j t l t k t t t s t n t t j t l t k t t t s t n Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 If we hd the followng rules for non-termnl X, X β 1 Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 X t -1 β -1 X α X t +1 β +1 X t k β k,,{,...,t -1,,t +1,...,t k } Σ T t p t q p,q {1,...,-1,+1,...k} {β 1,...,β -1,,β +1,...,β k } Σ N * There would e just one posslty to choose f the next symol n the nput strng were. Y 1...Y m α t...t n Lmd-rules wll e prolemtc whenever they produce severl possltes of choosng the next rule to pply

24 Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 Let us mgne tht we hve the followng three rules: X α, X λ y Y 1 γ 1 A termnl cn e derved drectly from X (X α), nd from Y 1, whch follows X (Y 1 γ 1 ) Borrdo de X Y 1 γ 1 Adjustng grmmr n Grech Norml Form to LL(1) Chnges n step 8 In order to dentfy these stutons, we need to know: Whch re the frst termnls derved y X (n our exmple, ) Whch s the frst termnl tht cn e derved y wht s fter X,.e. next(x) The followng fgures descre the two possle cses: XY 1...Y m λ Y 1...Y m Y 2...Y m γ 1 B σxy 1 B V τiwω W Y 1...Y m I ξx τ IW V ω t...t n t...t n t...t n σ X Y 1...Y m ξ X Y 1...Y m X λ X α λ λ XY 1...Y m Y 1...Y m t...t n t...t n t...t n α t...t n Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 8. If we wnt to generte top-down syntctc nlyser wth the LL(1) technque, we hve to study, n step 8, whch lmd-rules produce mgutes. 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > Who cn follow? next( ) We study ll the rght-hnd sdes tht contn G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > Who cn follow? next( ) We study ll the rght-hnd sdes tht contn In the two rules, ppers s the lst symol n the rght-hnd sde. cn e followed y nythng tht follows the lefthnd sde of those rules: Let s strt wth the lmd-rule λ

25 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > Therefore: next( ) s ncluded n next( ) ovous next() s ncluded n next( ). As s the xom, nd t does not pper n ny other rght-hnd sde. next() s the end-of-progrm symol, $. o next( ) s {$}. On the other hnd, frst( ) = {+} G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > Who cn follow T? next(t ) We study ll the rght-hnd sdes tht contn T We cn conclude tht the frst lmd rule s correct for LL(1), nd cn e left lke tht. Let us contnue wth the second lmd rule (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > Who cn follow T? next(t ) We study ll the rght-hnd sdes tht contn T There s one rule n whch T s followed y. In two other rules, t ppers s the lst symol of the rght-hnd sde. Therefore, next(t ) wll e: frst( ) next(t) next(t ) -- ovous G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > We cn focus now on next(t) Let us see ll the rules n whch T ppers n the rght-hnd sde

26 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) tep 8 for exmple 1 G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > We cn focus now on next(t) Let us see ll the rules n whch T ppers n the rght-hnd sde. It only ppers followed y. Therefore, next(t) = frst( ) In concluson, next(t ) wll e: frst( ) next(t) = frst( ) next(t ) -- ovous G =<{,,T,T,F}, {-,*,} { T +T λ T T F }, > Fnlly, we need to check whether the lmd rule produces ny mguty durng the nlyss. For nstnce, The frst termnl derved y T s *. The termnls n next(t ) re just +. Therefore, next(t ) = {+} Becuse the termnls re dfferent (* y +) we cn conclude tht the lmd rule s pproprte for n LL(1) grmmr Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2 Otn, f possle, the LL(1) equvlent grmmr from the followng one: =<{A,B}, {,} {A B B A }, Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2 1. The lnguge does not contn the empty word, so there s nothng to do. 2. There re no left-recursve rules, so there s gn nothng to do here. 3. We shll estlsh the prtl orderng etween the non-termnl symols. =<{A,B}, {,} {A B B A }, From the prevous rules, we cn deduce two dfferent prtl orderngs: From A B we cn deduce A<B From B A we cn deduce B<A In summry, there re two optons: Opton 1: B<A Opton 2: A<B We shll study the two possltes seprtely, to see how the choce tken ffects the result

27 Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 1 B<A 4. The rules wll e clssfed n three groups Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 1 B<A 5. The type-3 rule wll e elmnted y susttutng B Type 1 rules A B Type 2 rules B A Type 3 rules A B =<{A,B}, {,} {A B B A }, Type-3 rules: A B The B wll e replced y ts rght-hnd sdes (A nd ), nd we otn: A A =<{A,B}, {,} {A B B A }, =<{A,B}, {,} {A A B A }, Adjustng grmmr n Grech. Norml Form to LL(1) Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 1 B<A In ll the steps, we need to check whether there s ny nccessle symol. xmple 2, opton 1 B<A We lso need to elmnte ll the left-recursve rules tht hve turned up: =<{A,B}, {,} {A A B A }, =<{A}, {,} {A A }, It cn e esly seen tht, n ths grmmr, B cnnot e generted from the xom. It s nccessle, so we cn elmnte t nd ll the rules where t s t the left-hnd sde: B A =<{A}, {,} {A A }, 107 The rule A A s left-recursve, so we pply the lemm whch sd: If we hve grmmr < Σ T, Σ N,, P >, the left-recursve rules A Aα 1... Aα n β 1... β m wll e susttuted y A β 1 X... β m X X α 1 X... α n X λ In our exmple, A X X X X λ =<{A,X}, {,} {A X X X X λ}, 108

28 Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 1 B<A 6. There no type-2 rules to elmnte 7. We elmnte ll the termnl symols tht re not n the frst poston of the rules: =<{A,X}, {,} {A X X X X λ}, We defne new non-termnl Z to derve the (Z ) And we susttute the n ll the rght-hnd sdes y Z : (A ZX y X ZX) =<{A,X},{,} {A ZX X X ZX λ Z }, Adjustng grmmr n Grech Norml Form to LL(1) xmple 2, opton 1 B<A 8. tudy of the λ-rules =<{A,X},{,} {A ZX X X ZX λ Z }, There s just one lmd-rule, for X. Let us study the set next(x), to see whether t s dfferent from frst(x) (cont.) Adjustng grmmr n Grech Norml Form to LL(1) xmple 2, opton 1 B<A Adjustng grmmr n Grech. Norml Form to LL(1) 4. Let us see now xmple 2, opton 2 A<B =<{A,X},{,} {A ZX X X ZX λ Z }, Concernng next(x), t ppers n three rules. From tht, we cn conclude tht next(x) contns: next(a) next(x) - ovous Becuse A s the xom, nd t does not pper n ny rght-hnd sde, we cn conclude tht next(a) wll e the end-of-progrm symol {$}, nd hence next(x) wll lso e {$}. On the other hnd, frst(x) = {} Type-1 rules: A B Type-2 rules: A B Type-2 rules: B A =<{A,B}, {,} {A B B A }, Ths lmd-rule s not wrong for n LL(1) grmmr

29 Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 2 A<B 5. We strt y elmntng the type-3 rules from the grmmr: =<{A,B}, {,} {A B B A }, Type-3 rules: B A We replce A wth ts rght-hnd sdes (B nd ) to get: B B =<{A,B}, {,} {A B B B }, Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 2 A<B Now, we need to elmnte left-recursve rules. The rule B B s left-recursve, so we pply the lemm. In the exmple, we get: B X X X X λ =<{A,B}, {,} {A B B B }, =<{A,B,X}, {,} {A B B X X X X λ}, Adjustng grmmr n Grech. Norml Form to LL(1) 6. Remove type-2 rules: There s one type-2 rule: A B xmple 2, opton 2 A<B =<{A,B,X}, {,} {A B B X X X X λ}, We susttute B wth ll ts rght-hnd sdres (X y X) to get: =<{A,B,X}, {,} {A X X B X X X X λ}, Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 2 A<B As efore, we need to check whether there s ny nccessle symol: =<{A,B,X}, {,} {A X X B X X X X λ}, Now, t s not possle to rrve to B from the xom, so we cn delete t nd ll the rules tht hve t n the left-hnd sde: B X X =<{A,X}, {,} {A X X X X λ},

30 Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 2 A<B 7. Removl of ll the termnls tht re not n the frst poston. =<{A,X}, {,} {A X X X X λ}, Frst, we cn elmnte the symol : we defne new non-termnl B to derve t (B ), nd we put the non-termnl n the fght-hnd sdes (A BX y X BX) =<{A,B,X}, {,} {A BX X X BX λ B }, 7. (cont.) Adjustng grmmr n Grech. Norml Form to LL(1) xmple 2, opton 2 A<B =<{A,B,X}, {,} {A BX X X BX λ B }, We lso hve msplced termnl : we defne new non-termnl Z to derve t, (Z ), nd we susttute t n the prolemtc rght-hnd sdes (A BXZ XZ) =<{A,B,X,Z}, {,} {A BXZ XZ X BX λ B Z }, Adjustng grmmr n Grech Norml Form to LL(1) xmple 2, opton 2 A<B 8. tudy of the λ-rules. 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) xmple 2, opton 2 A<B =<{A,B,X,Z}, {,} {A BXZ XZ X BX λ B Z }, We need to check whether next(x) nd frst(x) hve ny termnl symol n common. =<{A,B,X,Z}, {,} {A BXZ XZ X BX λ B Z }, Let us clculte next(x): X ppers followed y Z. X lso ppers t the end of rule. Therefore, next(x) wll contn: frst(z) = next(x) - ovous There s just one lmd-rule

31 8. (cont.) Adjustng grmmr n Grech Norml Form to LL(1) xmple 2, opton 2 A<B LL(1) lnguges re LL(1) Concept =<{A,B,X,Z}, {,} {A BXZ XZ X BX λ B Z }, But t s the cse tht: frst(x)s the symol next(x)=frst(z)= Let us clculte next(x): X ppers followed y Z. X lso ppers t the end of rule. Therefore, next(x) wll contn: frst(z) = next(x) ovous Those whose grmmrs pper n Grech Norml Form, nd there re not two rules for the sme non-termnl, strtng wth the sme termnl symol n ther rght-hnd sde. o we cn conclude tht ths lmd-rule produces mgutes durng the prsng LL(1) Convertng from GNF to LL(1) A grmmr n Grech Norml Form my not e LL(1) For nstnce, =<{U,V,W,X,Y,Z,T}, {,,c,d,e} { U V W V X cy W dz et }, U> Two of the rght-hnd sdes for U strt wth the sme termnl symol,. LL(1) Convertng from GNF to LL(1) ometmes, ths cm e solved wth common fctor. teps Otn the longest common prefx of the two rght-hnd sdes (n the exmple the symol ). Leve sngle rght-hnd sde whch strts wth tht common prefx, nd whch ends wth new non-termnl, e.g. K =<{U,V,W,X,Y,Z,T}, {,,c,d,e} { U K V X cy W dz et }, U>

32 LL(1) Convertng from GNF to LL(1) The new non-termnl wll generte the remnng of the rght-hnd sdes, K V W. LL(1) Convertng from GNF to LL(1) We hve to tke cre to leve the rules gn n GNF... We cn derve the ntl non-termnls n the rules for K, so they strt wth on-termnl. In ths cse, we cn pply the rules for V nd W: =<{U,V,W,X,Y,Z,T}, {,,c,d,e} { U K K V W V X cy W dz et }, U> =<{U,V,W,X,Y,Z,T}, {,,c,d,e} { U K K X cy dz et V X cy W dz et }, U> LL(1) Convertng from GNF to LL(1)...nd we hve to remove the nccessle symols (V y W) Formlston of LL(1) grmmrs Concept In the remnng prt of ths lesson, we re gong to descre formlly LL(1) nlysers, whch we hve lredy ntroduced nformlly wth exmples. =<{U,V,W,X,Y,Z,T}, {,,c,d,e} { U K K X cy dz et }, U>

33 Formlston of LL(1) grmmrs xmple Consder the followng grmmr, nd the frst nd next sets for ech non-termnl: G=< {,,T,T,F}, {+,*,(,),d} { T +T λ T FT T *FT λ F () d }, > frst()=frst(t)=frst(f)={(,d} frst( )={+,λ} frst(t )={*,λ} next()=next( )={),$} next(t)=next(t )={+,),$} next(f)={+,*,),$} Constructng LL(1) nlysers Algorthm The followng lgorthm clcultes the nlyss tle T M ΣN ΣT +1 In ths mtrx, there wll e row for ech non-termnl, nd column for ech termnl, ncludng the end-of-progrm symol $. : 1. A α P repet: 1. frst(α) Σ T A α T[A,]. 2. If λ frst(α) then, next(a) A α T[A,] (note tht cn lso e $) Constructng LL(1) nlysers xmples In the grmmr for the prevous exmple Gven the followng grmmr, Constructng LL(1) nlysers xmples Σ N T T F d T FT F d + +T T λ Σ T {$} * ( T FT T *FT F () ) λ T λ $ λ T λ G=< {P,P,}, {,t,,e,} { P tpp P ep λ P>

34 We cn otn the next tle: Constructng LL(1) nlysers xmples LL(1), frst nd next sets Defnton We cn defne LL(1) grmmrs s those tht comply wth the followng condton: Σ N e Σ T {$} t $ The nlyss tle constructed wth the prevous procedure s determnstc,.e., ll the rows contn t most one rule. P P P tpp P P λ P ep P λ xmples The grmmr n the frst exmple s LL(1) The grmmr n the second exmple s not LL(1) electve top-down nlysers Concept LL(1) nlysers re lso clled wth no cktrckng, ecuse they re determnstc nd t wll never e necessry to cktrck durng the nlyss of progrm. They re lso clled recursve-descent prsers, ecuse of the knd of nlyss progrms tht re produced from LL(1) grmmrs. The reson of the effcency of these prsers s ecuse the rght-hnd sdes of the rules for ech non-termnl symol cn e consdered to e ndexed y the next termnl. electve top-down nlysers Automtclly uldng prser for n LL(1) grmmr Accordng to the procedure descred, n n LL(1) grmmr we re gong to hve two knds of rules: Rules for non-termnls tht hve λ-rule: U xx 1 X 2...X n yy 1 Y 2...Y m... zz 1...Z p λ Rules for non-termnls tht do not hve λ-rule: U xx 1 X 2...X n yy 1 Y 2...Y m... zz 1...Z p

35 electve top-down nlysers Automtclly uldng prser for n LL(1) grmmr: λ rules The next C functon wll e generted: electve top-down nlysers Automtclly uldng prser for n LL(1) grmmr: rules wthout λ The next C functon wll e generted: nt U(chr * strng, nt ) { f ( < 0) return ; /* ths propgtes errors */ swtch (strng[]) { cse x; ++; =X1(strng, ); =X2(strng, ); =Xn(strng, ); rek; cse y: ++; =Y1(strng, ); =Y2(strng, ); =Ym(strng, ); rek; cse z: ++; =Z1(strng, ); =Z2(strng, ); =Zp(strng, ); rek; } return ; } nt U(chr * strng, nt ) { f ( < 0) return ; /* Ths propgtes errors */ swtch (strng[]) { cse x; ++; =X1(strng, ); =X2(strng, ); =Xn(strng, ); rek; cse y: ++; =Y1(strng, ); =Y2(strng, ); =Ym(strng, ); rek; cse z: ++; =Z1(strng, ); =Z2(strng, ); =Zp(strng, ); rek; /* nd-of-strng goes to the defult cse */ defult: return n; /* The error wll e dfferent for ech rule */ } return ; } electve top-down nlysers Complete exmple Let us strt wth the followng context-ndependent grmmr electve top-down nlysers Complete exmple The followng s the grmmr n Grech Norml Form G 3 =<{,T,F},{,+,-,*,/,(,)} { T+ T- T T F*T F/T F F () }, > G 3 =<{,T,M,,P,D,C}, {,+,-,*,/,(,)} { PTM (CPTM DTM (CDTM M (CM PT (CPT DT (CDT (C PT (CPT DT (CDT (C T PT (CPT DT (CDT (C M + - P * D / C )}, >

36 electve top-down nlysers Complete exmple The followng s possle LL(1) grmmr, otned from the GNF grmmr y tkng common fctor n the rght-hnd sdes of the rules. G 3 =<{,T,M,,P,D,C}, {,+,-,*,/,(,)} { V (CV V *TX /TX + - λ X + - λ T U (CU U *T /T λ C )}, > electve top-down nlysers Complete exmple The followng would e the LL(1) nlyser (contnued) nt (chr * strng, nt ) { f ( < 0) return ; /* Ths propgtes prevous errors */ swtch (strng[]) { cse : ++; =V(strng, ); rek; cse ( : ++; =(strng, ); =C(strng, ); =V(strng, ); rek; defult: return 1;/*no λ*/ } return ; } nt V(chr * strng, nt ) { f ( < 0) return ; /* Ths propgtes prevous errors */ swtch (strng[]) { cse * : cse / : ++; =T(strng, ); =X(strng, ); rek; cse + : cse - : ++; =(strng, ); rek; /* λ */ } return ; } electve top-down nlysers Complete exmple The followng would e the LL(1) nlyser (contnued) electve top-down nlysers Complete exmple The followng would e the LL(1) nlyser (contnued) nt X(chr * strng, nt ) { f ( < 0) return ; /* Ths propgtes prevous errors */ swtch (strng[]) { cse + : cse - : ++; =(strng, ); rek; /* λ */ } return ; } nt T(chr * strng, nt ) { f ( < 0) return ; /* Ths propgtes prevous errors */ swtch (strng[]) { cse : ++; =U(strng, ); rek; cse ( : ++; =(strng, ); =C(strng, ); =U(strng, ); rek; defult: return 2;/*no λ*/ } return ; } nt U(chr * strng, nt ) { f ( < 0) return ; /* Ths propgtes prevous errors */ swtch (strng[]) { cse * : cse / : ++; =T(strng, ); rek; /* λ */ } return ; } nt C(chr * strng, nt ) { f ( < 0) return ; /* Propgte prevous errors */ swtch (strng[]) { cse ) : ++; rek; defult: return 4;/*no λ*/ } return ; }

37 electve top-down nlysers electve top-down nlysers Complete exmple Complete exmple A strng x would e nlysed wth the followng cll: The followng strng wll not e recognsed xom( x, 0); If the vlue returned equls the length of the strng, t ws correct. If t s negtve numer, t ws ncorrect. The followng executon exmple llustrtes the LL(1) nlyss: ( + *, 0) 0[1]: ( I + I * I, 0) 1[1]: V( I + I * I, 1) 2[1]: ( I + I * I, 2) 3[1]: V( I + I * I, 3) 4[1]: T( I + I * I, 4) 5[1]: U( I + I * I, 5) 5[1]: returned 5 4[1]: returned 5 4[1]: X( I + I * I, 5) 4[1]: returned 5 3[1]: returned 5 2[1]: returned 5 1[1]: returned 5 0[1]: returned ( + *, 0) 0[1]: ( I + I *, 0) 1[1]: V( I + I *, 1) 2[1]: ( I + I *, 2) 3[1]: V( I + I *, 3) 4[1]: T( I + I *, 4) 4[1]: returned -2 4[1]: X( I + I *, -2) 4[1]: returned -2 3[1]: returned -2 2[1]: returned -2 1[1]: returned -2 0[1]: returned yntctc nlyss Blogrphy [Alf] Teorí de Autómts y lengujes formles M. Alfonsec y otros [Hop] Introduccón l teorí de utómts, lengujes y computcón Hopcroft, J.; Motwn, R.; Ullmn, J. [Aho] Compldores. Prncpos, técncs y herrments A. V. Aho; R. efth; J. D. Ullmn 147