Top-down vs Bottom-up. Bottom up parsing. Sentential form. Handles. Handles in expression example

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1 Bottom up prsing Generl e LR0) LR LR1) LLR o est exploit JvCUP, should understnd the theoreticl sis LR prsing); op-down vs Bottom-up Bottom-up more powerful thn top-down; Cn process more powerful grmmr thn LL, will explin lter. Bottom-up prsers re too hrd to write y hnd ut JvCUP nd ycc) genertes prser from spec; Bottom up prser uses right most derivtion op down uses left most derivtion; Less grmmr trnsltion is required, hence the grmmr looks more nturl; Intuition: ottom-up prse postpones decisions out which production rule to pply until it hs more dt thn ws ville to top-down. Will explin lter 1 2 Bottom up prsing trt with string of terminls; xmple: Build up from leves of prse Grmmr: tree; pply productions ckwrds; When rech strt symol & exhusted input, done; hift-reduce is common ottomup technique. Notice the lue d should not e reduced into B in step 2. rm Be rm de rm cde rm cde How to get the right reduction steps? Reduce cde to : cde cde c de d e d e B e Be Be c B d ententil form ententil Form ny string tht cn e derived from non-terminls. Cn consist of terminls nd non terminls. xmple: ententil forms:,,... Right sententil form: otined y right most derivtion entence ententil form with no non-terminls; is sentence. 3 4 Hndles rm Be rm de rm cde rm cde Be c B d Informlly, hndle of sententil form is sustring tht cn e reduced. c is hndle of the right sententil form cde, ecuse c, nd fter c is replced y, the resulting string de is still right sententil form. Is d hndle of cde? No. this is ecuse cbe is not right sententil form. Formlly, hndle of right sententil form γ is production β nd position in γ where the string β my e found nd replced y. If * rm αw rm αβw, then β in the position fter α is hndle of αβw. When the production β nd the position re cler, we simply sy the sustring β is hndle. 5 Hndles in expression exmple int * int ) Conser the string: int * int int he rightmost derivtion rm rm rm int rm int* int rm int*int int For unmiguous grmmr, there is exctly one hndle for ech right-sententil form. he question is, how to find the hndle? Oservtion: he sustring to the right of hndle contins only terminl symols. 6 1

2 hift-reduce prsing Brek the input string into two prts: un-digested prt nd semigested prt Left prt of input prtly processed; Right prt completely unprocessed. int foo doule n) { return int) n1 ; } hifted, prtly reduced o fr unprocessed Use stck to keep trck of tokens seen so fr nd the rules lredy pplied ckwrds reductions) hift next input token onto stck When stck top contins good right-hnd-se of production, reduce y rule; hift-reduce min loop hift: If cn t perform reduction nd there re tokens remining in the unprocessed input, then trnsfer token from the input onto the stck. Reduce: If we cn find rule α, nd the contents of the stck re βα for some β β my e empty), then reduce the stck to β. he α is clled hndle. Recognizing hndles is key! ccept: is t the top of the stck nd input now empty, done rror: other cses. Importnt fct: Hndle is lwys t the top of the stck. 7 8 xmple 1 Prse tck Remining input )$ Prser ction hift prenthesis onto stck )$ hift onto stck )$ Reduce: pop RH of production, push LH, input unchnged) )$ Reduce: )$ hift right prenthesis ) $ Reduce: ) $ Reduce: $ Reduce: $ Done: ccept Grmmr: > > > ) rm rm rm ) rm ) rm ) Input string: ) hift-reduce xmple 2 Prse tck Remining Input ction ) ) $ ) $ ) $ ) $ ) $ ) $ ) $ ) $ ) $ $ $ $ $ hift hift Reduce Reduce hift hift Reduce Reduce ; Ignore: ) hift ) Reduce ) Reduce Reduce ccept ) Input: ) ) ) ) ) ) Note tht it is the reverse of the following rightmost derivtion: rm rm rm ) rm ) rm ) rm ) rm ) 9 10 Conflicts during shift reduce prsing Reduce/reduce conflict stck input... ) Which rule we should use, or? hift/reduce conflict iftt if ) if ) else stck Input... if ) else... Both reduce nd shift re pplicle. Wht we should do next, reduce or shift? LRK) prsing Left-to-right, Rightmost derivtion with k-token lookhed. L - Left-to-right scnning of the input R - Constructing rightmost derivtion in reverse k - numer of input symols to select prser ction Most generl prsing technique for deterministic grmmrs. fficient, le-sed prsing Prses y shift-reduce Cn mssge grmmrs less thn LL1) Cn hndle lmost ll progrmming lnguge structures LL LR CFG In generl, not prcticl: tles too lrge 10^6 sttes for C, d). Common susets: LR, LLR 1)

3 LR Prsing continued Dt structures: tck of sttes {s} ction tle ction[s,]; tle [s,x]; X N In LR prsing, push whole sttes on stck tck of sttes keeps trck of wht we ve seen so fr left context): wht we ve shifted & reduced & y wht rules. Use ction tles to dece shift vs reduce Use tle to move to new stte Min loop of LR prser Initil stte 0 strts on top of stck; Given stte t stte on top of stck nd the next input token : If ction[t, ] == shift i) Push new stte i onto stck Cll yylex to get next token If ction[t, ] == reduce y Y X1 Xn) Pop off n sttes to find u on top of stck Push new stte v = [u,y] onto stck If ction[t, ] == ccept), done! If ction[t, ] == error), cn t continue to successful prse xmple LR prse tle tte on O ction ) $ ccept 2 R2 R2 R2 R2 R R4 R4 R4 R4 R ) 2) 3) ) 4) 7 R3 R3 R3 R3 R3 8 R1 R1 R1 R1 R1 If ction[t, ] == shift), Push new stte ction[t, ] onto stck, Cll yylex to get next token If ction[t, ] == reduce y Y X1 Xn), Pop off n sttes to find u on top of stck, Push new stte v = [u,y] onto stck tte on ction O ) $ ccept 2 R2 R2 R2 R2 R R4 R4 R4 R4 R R3 R3 R3 R3 R3 8 R1 R1 R1 R1 R1 tte stck Remining Input Prser ction 01 $ ccept We explin how to construct this tle lter )$ hift 4 onto stte stck, move hed in input )$ Reduce 4), pop stte stck, goto 2, input unchnged )$ Reduce 2), goto 1 )$ hift 5 )$ hift 3 )$ hift 4 sw nother ) )$ Reduce 4), goto 2 )$ Reduce 2), goto 6 )$ hift 7 $ Reduce 3) ), goto $ Reduce 1), goto 1 * 1) 2) 3) ) 4) ypes of LR prsers LR k) LR k) -- imple LR LLR k) Lookhed LR k = # symols lookhed 0 or 1 in this clss Drgon ook hs generl cses trt with simplest: LR0) prser LR 0) prser dvntges: implest to understnd, mllest tles Disdvntges No lookhed, so too simple-minded for rel prsers Good cse to see how to uild tles, though. We ll use LR0) constructions in other LRk) prsers Key to LR prsing is recognizing hndles Hndle: sequence of symols encoded in top stck sttes representing right-hnd-se of rule we wnt to reduce y

4 LR les Given grmmr G, entify possile sttes for prser. ttes encpsulte wht we ve seen nd shifted nd wht re reduced so fr teps to construct LR tle: Construct sttes using LR0) configurtions or items); Figure out trnsitions etween sttes Configurtion configurtion or item) is rule of G with dot in the right-hnd se. If rule XYZ in grmmr, then the configs re XYZ XY Z X YZ XYZ Dot represents wht prser hs gotten in stck in recognizing the production. XYZ mens XYZ on stck. Reduce! X YZ mens X hs een shifted. o continue prse, we must see token tht could egin string derivle from Y. Nottionl convention: X, Y, Z: symol, either terminl or non-terminl,, c : terminl α, β, γ: sequence of terminls or non-terminls et of configurtions X YZ mens X hs een shifted. o continue prse, we must see token tht could egin string derivle from Y. ht is, we need to see token in FirstY) or in FollowY) if Y ε) Formlly, need to see token t such tht Y * t β for some β uppose Y α β lso in G. hen these configs correspond to the sme prse stte: X YZ Y α Y β ince the ove configurtions represent the sme stte, we cn: Put them into set together. dd ll other equivlent configurtions to chieve closure. lgorithm lter) his set represents one prser stte: the stte the prser cn e in while prsing string. rnsitions etween sttes Prser goes from one stte to nother sed on symols processed X YZ Y α Y β Y XY Z Model prse s finite utomton! When stte configurtion set) hs dot t end of n item, tht is F ccept stte Build LR0) prser sed on this F Constructing item sets & closure trting Configurtion: ugment Grmmr with symol dd production to grmmr Initil item set I0 gets Perform Closure on ht completes prser strt stte.) Compute uccessor function to mke next stte next item set) Computing closure ClosureI) 1. Initilly every item in I is dded to closurei) 2. If α B β is in closurei) for ll productions B γ, dd B γ 3. Repet step 2 until set gets no more dditions. xmple Given the configurtion set: { } Wht is the closure of { }: y rule 1 y rule 2 ) y rule 2 nd 3 y rule 2 nd 3 1) 2) 3) ) 4)

5 Building stte trnsitions LR les need to know wht stte to goto fter shift or reduce. Given et C & symol X, we define the set C = uccessor C,X) s: For ech config in C of the form Y α X β, 1. dd Y α X β to C 2. Do closure on C Informlly, move y symol X from one item set to nother; move to the right of X in ll items where dot is efore X; remove ll other items; compute closure. C X C uccessor exmple Given I= {,, ), } Wht is successori, )? move the fter : ) compute the closure: ) ) 1) 2) 3) ) 4) Construct the LR0) tle Construct F={I0, I1, I2,..., In} tte i is determined y Ii. he prsing ctions for stte i re: if α is in Ii, then set ction[i, ] to reduce α for ll inputs if is not ) If is in Ii, then set ction[i, $] to ccept. if α β is in Ii nd successorii, )=Ij, then set ction[i,j] to shift j. is terminl) he goto trnsitions for stte i re constructed for ll non-terminls using the rule: if successorii,)=ij, then goto[i, ]=j. ll entries not defined y ove rules re errors. he initil stte I0 is the one constructed from. teps of constructing LR0) tle 1. ugment the grmmr; 2. Drw the trnsition digrm; 1. Compute the configurtion set item set/stte); 2. Compute the successor; 3. Fill in the ction tle nd tle. 0) 1) 2) 3) ) 4) Configurtion set uccessor I0: ' I1 I1 I2 ) I3 I4 I1: ' ccept dot t end of rule) I5 I2: Reduce 2 dot t end) I3: ) I6 I6 I2 ) I3 I4 I4: Reduce 4 dot t end) I5: I8 ) I3 I4 I6: ) I7 I5 I7: ) Reduce 3 dot t end) I8: Reduce 1 dot t end) Item sets exmple 29 I 2 I 0 rnsition digrm ' ) I 1 ' I 3 I 4 ) ) I 5 ) I 6 ) I 7 I 8 ) ) 30 5

6 he prsing tle tte on O ction ) $ ccept 2 R2 R2 R2 R2 R R4 R4 R4 R4 R R3 R3 R3 R3 R3 8 R1 R1 R1 R1 R1 31 Prsing n erroneous input tte stck Input Prser ction 0 $ hift 4 0) 1) 2) 3) ) 4) 0 4 $ Reduce 4), pop 4, $ Reduce 2), pop 2, $ Push $ No ction [5, ] rror! tte on ction O ) $ ccept 2 R2 R2 R2 R2 R R4 R4 R4 R4 R R3 R3 R3 R3 R3 8 R1 R1 R1 R1 R1 32 uset construction nd closure LR0) grmmr I 0 ' I 2 I 0 ' I 1 ' I 2 I 3 I 4 I 4 grmmr is LR0) if the following two conditions hold: 1. For ny configurtion set contining the item α β, there is no complete item B γ in tht set. No shift/reduce conflict in ny stte in tle, for ech stte, either shift or reduce 2. here is t most one complete item α in ech configurtion set. No reduce/reduce conflict in tle, for ech stte, use sme reduction rule for every input symol. Very few grmmrs meet the requirements to e LR0). I ) [] I 2 I 1 I 0 ' ' ) [] [] I 3 [ I 4 ) ) [] I 9 [ ] ) [] I 5 I 10 ) [] [ ] [] I 11 ] ) I 7 Incomplete digrm ) I 6 ) I 8 35 LR Prse tle incomplete) tte on O ction ) $ [ ] ccept 2 R2 R2 R2 R R3 R3 R3 9 R R5 R5 R5 R5 8 R1 R1 R1 R R5 R5 R5 R5 0) 1) 2) 3) 4) ) 5) [] 36 6

7 tte stck Input Prser ction 0 [ ] $ hift [ ] $ [ ] $ shift [ 4 ] $ Reduce 0 4 9[ 2 ] $ Reduce 0 4 9[ 10 ] $ [ 10 11] $ Reduce [] 0 2 $ Reduce 0 1 $ $ $ Reduce $ Reduce 0 1 $ ccept V= ) [] V I 0 ' V= ) [] V I 2 > I 4 I 1 [] V ' hift/reduce conflict: [] Reduce/reduce conflict: V LR Prse tle incomplete) tte on O ction ) $ [ ] ccept 2 R2 R2 R2 R R4 R4 R4 9 R R5 R5 R5 R5 8 R1 R1 R1 R ) 1) 2) 3) V= 4) 5) ) 6) [] 7) V 39 LR0) key points Construct LR tle trt with ugmented grmmr ) Generte items from productions. Insert the Dot into ll positions Generte item sets or configurtion sets) from items; they re our prser sttes. Generte stte trnsitions from function successor stte, symol). Build ction nd tles from sttes nd trnsitions. les implement shift-reduce prser. View [sttes nd trnsitions] s finite utomton. n Item represents how fr prser is in recognizing prt of one rule s RH. n Item set comines vrious pths the prser might hve tken so fr, to diverge s more input is prsed. LR0) grmmrs re esiest LR to understnd, ut too simple to use in rel life prsing. 40 imple LR1) prsing: LR Construction for LR tles LR0) One LR0) stte mustn t hve oth shift nd reduce items, or two reduce items. o ny complete item dot t end) must e in its own stte; prser will lwys reduce when in this stte. LR Peek hed t input to see if reduction is pproprite. Before reducing y rule XYZ, see if the next token is in Follow ). Reduce only in tht cse. Otherwise, shift. 1. Construct F = {I0, I1,... In }, the LR0) item sets. 2. tte i is Ii. he prsing ctions for the stte re: ) If α is in Ii then set ction[i,] to reduce > α for ll in Follow) is not '). ) If ' is in Ii then set ction[i,$] to ccept. c) If α β is in Ii nd successorii, ) = Ij, then set ction[i,] to shift j must e terminl). 3. he goto trnsitions for stte i re constructed for ll non-terminls using the rule: If successorii, ) = Ij, then [i, ] = j. 4. ll entries not defined y rules 2 nd 3 re errors. 5. he initil stte is closure of set with item

8 Properties of LR Pickier rule out setting ction tle is the only difference from LR0) tles; If G is LR it is unmiguous, ut not vice vers; tte cn hve oth shift nd reduce items, if Follow sets re disjoint. LR xmple Item sets I0 nd successor I0, ): ' ) [] [] ' ) [] LR0) prser sees oth shift nd reduce, ut LR prser consults Follow set: Follow) = {, ), ], $ } so mens reduce on or ) or ] or $ [] mens shift otherwise e.g. on [ ) LR xmple 2 ' V = ) V V ' V = ) V wo complete LR0) items, so reduce-reduce conflict in LR0) grmmr, ut: Follow) = {, ), $ } FollowV) = { = } Disjoint, so no conflict. eprte ction entries in tle. LR grmmr grmmr is LR if the following two conditions hold: If items α β nd B γ re in stte, then terminl FollowB). no shift-reduce conflict on ny stte. his mens the successor function for x from tht set either shifts to new stte or reduces, ut not oth. For ny two complete items α nd B β in stte, the Follow sets must e disjoint. Follow) FollowB) is empty.) no reduce-reduce conflict on ny stte. If more thn one non-terminl could e reduced from this set, it must e possile to uniquely determine which using only one token of lookhed. Compre with LR0) grmmr: 1. For ny configurtion set contining the item α β, there is no complete item B γ in tht set. 2. here is t most one complete item α in ech configurtion set. Note tht LR0) LR LR dc 3. d 4. c In 3 there is reduce/shift conflict: It cn e R4 or shift. By looking t the Follow set of, the conflict is removed. 0: ' dc d 1: d 2: d c d c c ction 3: dc c 4: d c d $ R4 6 R2 R3 5: dc 6: d Prse trce tte stck Input Prser ction 0 dc$ hift 2 0 2d c$ hift 3 0 2d 3c $ shift 5 0 2d 3c 5 $ Reduce $ ccept

9 Non-LR exmple dc 3. d c 0: ' dc d c c 9: c d 1: 2: d c d c 7: c 3: dc c 4: d 8: 3 hs shift/reduce conflict. By looking t Follow), oth nd re in the follow set. o under column we still don t know whether to reduce or shift. 5: dc 6: d 49 he conflict LR prsing tle ction c d $ /R5 R R2 6 R R4 9 R5 R5 Follow) = {, } 50 LR1) prsing Mke items crry more informtion. LR1) item: X1...Xi Xi1...Xj, tok erminl tok is the lookhed. Mening: hve sttes for X1...Xi on stck lredy expect to put sttes for Xi1...Xj onto stck nd then reduce, ut only if token following Xj is tok tok cn e $ plit Follow) into seprte cses Cn cluster items nottionlly: [ α, //c] mens the three items: [ α, ] [ α, ] [ α, c] Reduce α to if next token is or or c {,, c } Follow) LR1) item sets More items nd more item sets thn LR Closure: For ech item [ α Bβ, ] in I, for ech production B γ in G, nd for ech terminl in Firstβ), dd [B γ, ] to I dd only items with the correct lookhed) Once we hve closed item set, use LR1) successor function to compute trnsitions nd next items. xmple: Initil item: [, $] Wht is the closure? [ dc, $] [ d, $] [, $] [ c, ] dc d c LR1) successor function Given I n item set with [ α Xβ, ], dd [ α X β, ] to item set J. successori,x) is the closure of set J. imilr to successor function to LR0), ut we propgte the lookhed token for ech item. xmple 0: ', $ dc, $ d, $, $ c, c d 1:, $ 2: d c, $ d, $ c, LR1) tles ction tle entries: If [ α, ] Ii, then set ction[i,] to reduce y rule α is not '). If [, $] Ii then set ction[i,$] to ccept. If [ α β, ] is in Ii nd succii, ) = Ij, then set ction[i,] to shift j. Here is terminl. entries: For ech stte I & ech non-terminl : If succii, ) = Ij, then [i, ] = j. 9: c,

10 LR1) digrm 0: ', $ dc, $ d, $, $ c, c 9: c, d 1:, $ 2: d c, $ d, $ c, 7:, $ c 3: dc, $ c, 4: d, $ 8:, $ dc 3. d c 5: dc, $ 6: d, $ Crete the LR1) prse tle ction c d $ R R2 6 R R4 9 R nother LR1) exmple Crete the trnsition digrm 0: ', $, $, /, / 9:, / 1:, $ 2:, $, $, $ 7:, /, /, / 3:, $ 4:, $, $, $ 5:, $ 8:, / 0) ' 1) 2) 3) 6:, $ 57 Prse tle stte ction $ ccept R R3 6 R R2 R2 9 R3 R3 58 Prse trce stck remining input prse ction 0 $ 9 09 $ R $ R $ R $ R2 023 $ R1 01 $ ccept LR1) grmmr grmmr is LR1) if the following 2 conditions re stisfied for ech configurtion set: For ech item [ α β, ] in the set, there is no item in the set of the form [B γ, ] In the ction tle, this trnsltes to no shift/reduce conflict. If there re two complete items [ α, ] nd [B β, ] in the set, then nd should e different. In the ction tle, this trnsltes to no reduce/reduce conflict Compre with the LR grmmr For ny item α β in the set, with terminl, there is no complete item B γ in tht set with in FollowB). For ny two complete items α nd B β in the set, the Follow sets must e disjoint. Note tht LR1) LR1) LR0) LR1) LR1)

11 LR1) tles continued LR1) tles cn get ig exponentil in size of rules Cn we keep the dditionl power we got from going LR LR without tle explosion? LLR! We split LR1) sttes to get LR1) sttes, mye too ggressively. ry to merge item sets tht re lmost enticl. ricky it: Don t introduce shift-reduce or reduce-reduce conflicts. LLR pproch Just sy LLR it s lwys 1 in prctice) Given the numerous LR1) sttes for grmmr G, conser merging similr sttes, to get fewer sttes. Cndtes for merging: sme core LR0) item) only differences in lookheds xmple: 1: X α β, //c 2: X α β, c/d 12: X α β, //c/d ttes with sme core items 0: ', $, $, /, / 9:, / 1:, $ 2:, $, $, $ 7:, /, /, / 3:, $ 4:, $, $, $ 5:, $ 8:, / 0) ' 1) 2) 3) 6:, $ Merge the sttes 0: ', $, $, /, / 59:, //$ 1:, $ 2:, $, $, $ 47:, //$, //$, //$ 3:, $ 4:, $, $, $ 5:, $ 68:, //$ 0) ' 1) 2) 3) 6:, $ Merge the sttes 0: ', $, $, /, / 59:, //$ 1:, $ 2:, $, $, $ 47:, //$, //$, //$ 3:, $ 68:, //$ 0) ' 1) 2) 3) Follow)={ $ } fter the merge Wht hppened when we merged? hree fewer sttes Lookhed on items merged. In this cse, lookhed in merged sets constitutes entire Follow set. o, we mde LR1) grmmr y merging. Result of merge usully not LR1)

12 conflict fter merging 0: ', $ Bc, $ Cd, $ Bd, $ Cc,$ 1:, $ 2: Bc, $ Cd, $ B e, c C e,d 4: Bd, $ Cc, $ B e, d C e, c e e 1) Bc Cd Bd Cc 2) B e 3) C e 3: B e, c C e, d 5: B e, d C e, c Prcticl consertion miguity in LR grmmrs G: G produces multiple rightmost derivtions. i.e. cn uild two different prse trees for one input string.) Rememer: * ) We dded terms nd fctors to force unmiguous prse with correct precedence nd ssocitivity Wht if we threw the grmmr into n LR-grmmr tle-construction mchine nywy? Conflicts = multiple ction entries for one cell We choose which entry to keep, toss others Precedence nd ssocitivity in JvCUP 0: ' * ) 1: * 3: 2: ) * ) * 4: * ) 5: * * ) 6: ) * * 8: * * ) 9: ) * ) * 7: * 69 JvCup grmmr terminl PLU, IM; precedence left PLU; precedence left IM; ::= PLU IM ID Wht if the input is xyz? When shifting conflicts with reducing production contining, choose reduce Wht if the input is xy*z? Wht if the input is x*yz? 70 rnsition digrm for ssignment expr 0: ', $, $ V=, $ V, = V 1:, $ 3: V =, $ 2:, $ V, = = 4: V=, $ V, $ n, $ V, $ V n 5: V=, $ 6: V, $ 7: n, $ 8: V, $ V= V V n 71 Why re there conflicts in some rules in ssignments? P 0: ' P, $ P m, $ P Pm, $ P, $ P, m P m, m P Pm, m 2: P, $ m 1: P m, $/m *** hift/reduce conflict found in stte #0 etween P ::= *) nd P ::= *) m under symol m Non LR1) grmmr P m Pm It is n miguous grmmr. here re two rightmost/leftmost derivtions for sentence m: P Pm m P m 72 12

13 slightly chnged grmmr, still not LR P 0: ' P, $ P m, $ P mp, $ P, $ 2: P, $ m It is n miguous grmmr. here re two prse trees for sentence m: P mp m P m 1: P m, $ P m P, $ P m, $ P mp, $ P, $ P 3: P mp, $ Non LR1) grmmr P m m P Reduce/Reduce conflict found in stte #1 etween P ::= m *) nd P ::= *) under symols: {OF} Produced from jvcup 73 Modified LR1) grmmr 0: ' P, $ P, $ P mp, $ P m 1: P, $ 2: P m P, $ P mp, $ P, $ Note tht there re no conflicts he derivtion: P mp mmp mm P LR1) grmmr P m P 3: P mp, $ 74 nother wy of chnging to LR1) grmmr 0: ' P, $ P Q, $ P, $ Q m, $ Q mq, $ P m 1: P, $ 2: Q m, $ Q m Q, $ Q m, $ Q mq, $ Q LR1) grmmr P Q Q m m Q 3: Q mq, $ LR grmmrs: comprison dvntges Disdvntges LR0) mllest tles, esiest to uild Indequte for mny PL structures LR1) LLR1) LR1) LR0) LR1) LLR LR1) CFG More inclusive, more inform?on thn LR0) me size tles s LR, more lngs, efficient to uild Most precise use of lookhed, most PL structures we wnt Mny useful grmmrs re not LR 1) empiricl, not mthem?cl les order of mgnitude > LR1) he spce of grmmrs CFG he spce of grmmrs CFG Wht re used in prctice Unmiguous CFG Unmiguous CFG LL1) LR1) LL1) LR1) LLR1) LLR1) LR1) LR1) LR0) LR0)

14 Verifying the lnguge generted y grmmr o verify grmmr: every string generted y G is in L every string in L cn e generted y G xmple: ) ε the lnguge is ll the strings of lnced prenthesis, such s ), )), ) ))) Proof prt 1: every sentence derived from is lnced. sis: empty string is lnced. induction: suppose tht ll derivtions fewer thn n steps produce lnced sentences, nd conser leftmost derivtion of n steps. such derivtion must e of the form: ) *x) *x)y Proof prt 2: every lnced string cn e derived from Bsis: the empty string cn e derived from. Induction: suppose tht every lnced string of length less thn 2n cn e derived from. Conser lnced string w of length 2n. w must strt with. w cn e written s x)y, where x, y re lnced. Hierrchy of grmmrs CFG is more powerful thn R ype n grmmr is more powerful thn type n1 grmmr xmple: Σ={, } he lnguge of ny string consists of nd CFG ε Lnguge 5 Cn e descrie y R he lnguge of plindromes consist of nd ε Cn e descried y CFG, ut not R Lnguge 1: ny string of nd Lnguge 2: plindromes Lnguge 3 Lnguge 2 Lnguge 4 RG Lnguge 1 When grmmr is more powerful, it is not tht it cn descrie lrger lnguge. Insted, the power mens the ility to restrict the set. More powerful grmmr cn define more complicted oundry etween correct nd incorrect sentences. herefore, more different lnguges Metphoric comprison of grmmrs R drw the rose use stright lines ruler nd -squre suffice) CFG pproximte the outline y stright lines nd circle segments ruler nd compsses) 81 14