Midterm I Solutions CS164, Spring 2006

Transcription

1 Midterm I Solutions CS164, Spring 2006 Februry 23, 2006 Plese red ll instructions (including these) crefully. Write your nme, login, SID, nd circle the section time. There re 8 pges in this exm nd 4 questions, ech with multiple prts. Some questions spn multiple pges. All questions hve some esy prts nd some hrd prts. If you get stuck on question move on nd come bck to it lter. You hve 1 hour nd 20 minutes to work on the exm. The exm is closed book, but you my refer to your two pges of hndwritten notes. Plese write your nswers in the spce provided on the exm, nd clerly mrk your solutions. You my use the bcks of the exm pges s scrtch pper. Plese do not use ny dditionl scrtch pper. Solutions will be grded on correctness nd clrity. ch problem hs reltively simple nd strightforwrd solution. We might deduct pos if your solution is fr more complicted thn necessry. Prtil solutions will be grded for prtil credit. LOGIN: NAM: SID: Circle the time of your section: Fri 10:00 Fri 11:00 Fri 2:00 Problem Mx pos Pos TOTAL 100 1

2 1 Miscellneous (6 pos) () Fill the missing right-hnd side for the production of A in the grmmr below such tht the rules form LL(1) grmmr. S A A A b A A Solution: A ɛ Consider the following grmmr: S S b S b A A ε A b A (b) Is the bove grmmr LL(1)? Why or why not? Solution: It is not LL(1) becuse on non-terminl S nd lookhed there re two productions possible ( S nd b A). (c) Show n NFA tht ccepts the lnguge of this grmmr. Solution: The solution is shown below. You cn solve this problem without figuring out the lnguge before: ech non-terminl corresponds to n NFA stte, ɛ trnsitions mrk the ccepting sttes, ech production corresponds to trnsition. This works for ll grmmrs whose productions hve t most one non-terminl on the right-hnd-side, occurring t the end., b, b S b A 2

3 2 Finite Automt nd Regulr xpressions (30 pos) () Write regulr expression for the lnguge of strings over the lphbet Σ = {, b} tht hve n even number of occurrences of (this includes 0 occurrences). Solution: There re mny correct solutions. Perhps the simplest is: (b b ) A close miss is This one does not llow the string b. (b ) (b) [Long] Drw the DFA corresponding to the following NFA. For ech stte in the resulting DFA write to which NFA sttes it corresponds. Solution: The solution is shown below. Bi-directionl edges stnd for two edges going in opposite directions, with the sme lbel. ABK b CMK CL DBMK b DBL b 3

4 In the following questions the lphbet is {0, 1}. Write TRU if the lnguge is regulr nd FALS otherwise. xplin your nswers in one or two sentences. (c) The lnguge of strings with no occurrences of the substring 010. Solution: TRU. The complement of regulr lnguge is regulr. recognizes the 010 substring nd swp rejecting nd ccepting sttes. Tke the DFA tht (d) The lnguge of strings with more occurrences of 0 thn 1. Solution: FALS. This would require counting up for 0 nd down for 1. Unbounded counting cnnot be done with finite utomt. (e) The lnguge of strings whose length is multiple of 3. Solution: TRU. Counting modulo N requires only N sttes. (f) [Hrder] The lnguge of strings whose number of occurrences of 0 minus number of occurrences of 1 is multiple of 3. Solution: TRU. This is lso counting modulo 3 problem, with counting up for 0 nd down for 1. (g) [Hrder] The lnguge of strings of terminls nd non-terminls tht cn pper on the prsing stck during n LR(1) prse (stck is red from the bottom of the stck). Here the lphbet is the set of terminls nd non-terminls of the grmmr being prsed. Solution: TRU. After ll we use DFA to process the contents of the stck nd decide whether to shift or to reduce. Tke thn DFA nd mke every stte n ccepting stte. 4

5 3 LL Prsing (20 pos) Consider the following grmmr with terminls (, [, ), nd ]. S T S [S]S )S ɛ T (X) X T X [X]X ɛ () [Long] Fill in the tble with the First nd Follow sets for the non-terminls. First Follow S ( [ ) ɛ $ ] T ( ( [ ) ] $ X ( [ ɛ ) ] (b) [Depends on ()] Fill in the LL(1) prsing tble. ( [ ) ] $ S T S [ S ] S ) S ɛ ɛ T ( X ) X T X [ X ] X ɛ ɛ (c) [Depends on (b)] Is this grmmr LL(1)? Solution: Yes, becuse every cell in the tble bove hs t most one entry. 5

6 4 LR Prsing (44 pos) Consider the following grmmr over terminls, if, nd : S if () This grmmr is mbiguous. Show two prse trees for the string if. Solution: S S if if On the next pge there is prtilly completed DFA for this grmmr (stte 0 is the initil stte): (b) [Long] Complete the DFA bove. You need to do the following: Complete stte 0 by performing closure on the item listed. Fill in ll elements of sttes 2, 4, 5, 6, nd the lookhed items for stte 3. Fill in the missing trnsition lbels on edges. Write the necessry reduce-by nd ccept lbels on sttes. (c) [Depends on (b)] For ech stte with conflict, list the stte, the lookhed token, nd the type of conflict (i.e., shift-reduce conflict, or reduce-reduce conflict). Solution: Stte 5, two conflicts: shift-reduce conflict on if, shift-reduce conflict on. Stte 6, two conflicts: shift-reduce conflict on if, shift-reduce conflict on. 6

7 1 reduce by on $/if/,$/if/ 0 S,$,$/if/ if,$/if/,$/if/ 2 if, $/if/, $/if/ if, $/if/,$/if/ 3 S, $ ACCPT on $ if, $/if/ if, $/if/,$/if/, $/if/ if, $/if/,$/if/ 4 5 if if 6 if,$/if/ if,$/if/, $/if/, $/if/ if,$/if/, $/if/ reduce by if on $/if/ reduce by on $/if/ 7

8 (d) [Does not depend on (b)] Assume tht we dismbigute the grmmr by sying tht is left-ssocitive nd binds more tightly thn if, nd if is right ssocitive. Show the correct prse tree for the string if if Solution: S if if (e) [Depends on (b)] For ech of the conflicts tht you hve identified bove, show how it must be resolved to chieve the dismbigution described bove. For shift/reduce conflict sy whether we should shift or reduce. For reduce/reduce conflict sy which reduction we should choose. Solution: Stte 5: shift on if, shift on. Stte 6: reduce on if, reduce on. 8