Problem Set 2 Fall 16 Due: Wednesday, September 21th, in class, before class begins.

Transcription

1 Problem Set 2 Fll 16 Due: Wednesdy, September 21th, in clss, before clss begins. 1. LL Prsing For the following sub-problems, consider the following context-free grmmr: S T$ (1) T A (2) T bbb (3) A T (4) A c (5) B T (6) B cbb (7) () Wht re the terminls nd non-terminls of this lnguge? (b) Show the derivtion of the string bcccb$ strting from S (specify which production you used t ech step), nd give the prse tree ccording to tht derivtion. (c) Give the first nd follow sets for ech of the non-terminls of the grmmr. (d) Wht re the predict sets for ech production? (e) Give the prse tble for the grmmr. Is this n LL(1) grmmr? Why or why not? (f) Show the steps your prser would tke to prse bcccb$. (g) Suppose we chnge rule 5 to A λ. Is the resulting grmmr LL(1)? Why or why not? (h) Is the grmmr LL(k) for ny k? Solution: () The non-terminls re ny of the symbols tht pper on the left hnd side of productions: S, T, A, B. The terminls re ny of the symbols tht pper in completely-rewritten strings (i.e., strings with no non-terminls in them). These re ll of the symbols tht don t pper on the right hnd side, except for λ, which is not symbol tht ppers in strings:, b, c, $. 1

2 (b) Here is the derivtion. I will use left derivtion for this (we will rewrite strting from the left-most non-terminl): S T$ using S T$ bbb$ bcbbb$ bctbb$ bcabb$ bccbb$ bcctb$ bccab$ bcccb$ using T bbb using B cbb using B T using T A using A c using B T using T A using A c (c) We begin by computing the first sets for ech non-terminl. To build up the first sets, we strt by exmining ech production to determine which constrints tht production cretes on the vrious first sets. In the first pss over ech production, we get: First(S) First(T) λ from production 1 F irst(t) F irst(a) = {} from production 2 F irst(t) F irst(bbb) = {b} from production 3 F irst(a) F irst(t) from production 4 F irst(a) F irst(c) = {c} from production 5 F irst(b) F irst(t) from production 6 F irst(b) F irst(cbb) = {c} from production 7 Solving these constrints to find the minimum ssignment to ech first set (using the method we discussed in clss), we get n initil guess for the first sets of: First(S) = {,b} First(T) = {,b} First(A) = {,b,c} First(B) = {,b,c} Becuse none of these first sets include λ, when we go over the productions gin, we don t hve to dd ny new constrints. (If the first set for non-terminl contined λ, then for ny production whose right-hnd-side strted with tht non-terminl, we would hve to look t the first set of the second symbol on the RHS). 2

3 We cn now compute the follow sets. Follow sets begin by looking t the nonterminls tht pper on the right hnd side of rules. F ollow(t) F irst($) = {$} from production 1 F ollow(t) F ollow(a) from production 4 F ollow(t) F ollow(b) from production 6 F ollow(a) F irst() = {} from production 2 F ollow(b) F irst(b) = {b} from production 3 F ollow(b) F irst(b) = {, b, c} from production 7 When we solve these set constrints, we get: Follow(T) = {,b,c,$} Follow(A) = {} Follow(B) = {,b,c} (d) We compute the predict sets for ech production using the method we discussed in clss: Predict(S T$) = {,b} Predict(T A) = {} Predict(T bbb) = {b} Predict(A T) = {,b} Predict(A c) = {c} Predict(B T) = {,b} Predict(B cbb) = {c} (e) The prse tble is just different representtion of the predict sets. Ech nonterminl gets row, nd ech terminl gets column. Ech entry is either production or n error. Entry T[V n ][V t ] = p if production p = V n α nd V n t Predict(p). In other words, if the non-terminl of production p is V n, then for every terminl V t in the predict set of p, T[V n ][V t ] = p. In this cse, we get the following tble: b c $ S 1 T 2 3 Becuse there re no conflicts in the predict tble (i.e., becuse A B ech production for given non-terminl hs disjoint predict set), this grmmr is LL(1). 3

4 (f) The steps the prser would tke: Prse Tble Remining Input Action S bcccb$ Predict 1 T$ bcccb$ Predict 3 bbb$ bcccb$ Mtch(b) Bb$ cccb$ Predict 7 cbbb$ cccb$ Mtch(c) BBb$ ccb$ Predict 6 TBb$ ccb$ Predict 2 ABb$ ccb$ Mtch() ABb$ ccb$ Predict 5 cbb$ ccb$ Mtch(c) Bb$ cb$ Predict 6 Tb$ cb$ Predict 2 Ab$ cb$ Mtch() Ab$ cb$ Predict 5 cb$ cb$ Mtch(cb$) (ccept) (g) Rebuilding the first nd follow sets for the non-terminls, we get: First(S) = {,b} First(T) = {,b} First(A) = {,b,λ} First(B) = {,b,c} Follow(T) = {,b,c,$} Follow(A) = {} Follow(B) = {,b,c} Putting them together, we get the following predict sets (note tht now we sometimes hve to worry bout follow sets!): Predict(S T$) = {,b} Predict(T A) = {} Predict(T bbb) = {b} Predict(A T) = {,b} Predict(A λ) = {} Predict(B T) = {,b} Predict(B cbb) = {c} 4

5 Now we hve some cler prediction conflicts. If we re trying to predict wht to rewrite n A into, nd we see n, we don t know whether to rewrite it into T or into λ. Therefore, the resulting grmmr is not LL(1). (h) The grmmr is not LL(k) for ny k. Note tht the letter s re lwys generted in pirs using production 1. Therefore, to determine whether to rewrite A into T or into λ, the prser needs to compre the number of s it hs red nd the number of s remining. If there re more s in the remining input, A should be rewritten into T; otherwise rewritten to λ. Nonetheless, the length of the input could be rbitrrily long. For ny integer k, consider the input of length 2k +1: 2k $. After reding the first k s nd rewriting A to T for k times, the prser won t be ble to determine how to rewrite the next A. 2. LR Prsing for the following sub-problems, consider the following grmmr: () Build the CFSM for this grmmr. S P$ (8) E P (9) E Eb (10) E b (11) P E (12) (b) Build the goto nd ction tbles for this grmmr. Is it n LR(0) grmmr? Why or why not? (c) Show the steps tken by the prser when prsing the string: b$. Give the ction nd show the stte stck nd remining input for ech step of the prse. Shift ctions should be of the form Shift X where X is the stte you re shifting to, nd Reduce ctions should be of the form Reduce R, goto X where R is the rule being used to reduce, nd X is the stte the prser winds up in. Solution: () Here is the CFSM: 5

6 Stte 2 S P.$ $ Stte 3 S P$. P Stte 0 S.AP$ Stte 1 S.P$ P.E E.P E.Eb E.b b Stte 6 E b. E b Stte 7 P E. E E.b Stte 4 E.P P.E E.P E.Eb E.b E b P Stte 5 E P. Stte 8 E Eb. Stte 9 P E. (b) Note tht in this CFSM, there re no sttes where there re both Shift configurtions nd Reduce configurtions. Every stte either hs ll Shift configurtions, or exctly one Reduce configurtion. Hence, this mchine hs no Shift/Reduce or Reduce/Reduce conflicts. As result, the grmmr is n LR(0) grmmr. The ction tble is just the tbulr representtion of this mchine. The goto nd ction tbles is s follows: Stte b $ E P Stte Action 0 Shift 1 Shift 2 Shift 3 Accept 4 Shift 5 Reduce 9 6 Reduce 11 7 Shift 8 Reduce 10 9 Reduce 12 6

7 (c) Let s see how the following string is prsed by this mchine. b$ Recll tht we prse by keeping stck of sttes (strting in stte 0). The stte t the top of the stck is the stte we re currently in. When we shift, we consume the next token off the input, nd use the goto tble to decide which tble to go to; tht stte is pushed onto the stck. When we reduce, we bck up s mny steps s there re symbols on the right hnd side of the rule we re reducing we pop tht mny symbols off the stck. Then, from the stte we end up in, we look t the goto tble to decide which stte to go to bsed on the symbol on the left hnd side of the rule: we Reduce (ccording to rule) nd goto (the next stte). Stte stck Remining input Action Explntion 0 b$ Shift 1 Consume from the input nd shift 01 b$ Shift 4 Consume from the input nd shift 014 b$ Shift 6 Consume b from the input nd shift 0146 $ Reduce 9, goto 7 Bck up one (b) nd replce with E 0147 $ Shift 9 Consume from the input nd shift $ Reduce 12, goto 5 Bck up two (E) nd replce with P 0145 $ Reduce 9, goto 7 Bck up one (P) nd replce with E 017 $ Shift 9 Consume from the input nd shift 0179 $ Reduce 12, goto 2 Bck up two (E) nd replce with P 012 $ Shift 3 Consume $ from the input nd shift 0123 Accept We hve mtched the string 7