1. For the following sub-problems, consider the following context-free grammar: S AA$ (1) A xa (2) A B (3) B yb (4)

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1 ECE 468 & 573 Problem Set 2: Contet-free Grammars, Parsers 1. For the following sub-problems, consider the following contet-free grammar: S $ (1) (2) (3) (4) λ (5) (a) What are the terminals and non-terminals of this language? nswer: Terminals: {,, $}. Non-terminals: {S,, } (b) Describe the strings are generated b this language. Is this a regular language (i.e., could ou write a regular epression that generates this language)? nswer: Yes, this is a regular language. To see how, note that turns into. Thus, the whole language can be captured b the following regular epression: $ (c) Show the derivation of the string $ starting from S (specif which production ou used at each step), and give the parse tree according to that derivation. nswer: S $ Rule 1 $ Rule 2 $ Rule 3 $ Rule 4 $ Rule 5 $ Rule 2 $ Rule 2 $ Rule 3 $ Rule 4 $ Rule 5 (6) 1

2 S $ λ λ (d) Give the first and follow sets for each of the non-terminals of the grammar. nswer: F irst(s) = {,, $} F irst() = {,, λ} F irst() = {, λ} and: F ollow(s) = {} F ollow() = {,, $} 2

3 F ollow() = {,, $} (e) What are the predict sets for each production? nswer: P redict(1) = {,, $} P redict(2) = {} P redict(3) = {,, $} P redict(4) = {} P redict(5) = {,, $} Note that even though the first-set of does not contain $, $ is in the predict set for Rule 3 because can go to λ, and hence the predict set must include the follow-set of. (f) Give the parse table for the grammar. Is this an LL(1) grammar? Wh or wh not? nswer $ S , , 5 5 This is not LL(1) because there are conflicts in the parse table. 2. for the following sub-problems, consider the following grammar: (a) Describe the strings generated b this language. S $ (7) (8) w (9) (10) z (11) nswer: There are man was to describe this string. Here s one. Ever can turn into the following: () (z w) 3

4 nd ever can turn into the following: which is thus equivalent to: Putting it all together, we get: w () (z w) w (() (z w) w) () (z w) (b) Is this language LL(1)? Wh or wh not? nswer: To figure this out, we must build the predict sets for these rules. So first, we compute the first and follow sets for each non-terminal: and: F irst(s) = {} F irst() = {} F irst() = {, z} F ollow(s) = {} F ollow() = {, z, $} F ollow() = {, z, $} Note that rule 1 lets us know that F ollow() F irst() and F ollow() $, rule 2 lets us know that F ollow() F ollow() and rule 4 lets us know that F ollow() F ollow(). If we put these together, we can compute the predict sets: P redict(7) = {} P redict(8) = {} P redict(9) = {} P redict(10) = {} P redict(11) = {z} We can see that there will be a predict conflict between rules 8 and 9. If we are epanding an, and our lookahead token is, we do not know whether to predict or w. Thus, this is not an LL(1) grammar. 4

5 (c) uild the CFSM for this grammar. State 0 S $ w State 1 S $ z State 2 S $ $ z State S $ State 3 w z z State 4 z State 5 State 8 w State 6 w State 7 State 9 w (d) uild the goto and action tables for this grammar. Is it an LR(0) grammar? Wh or wh not? nswer: The goto table can be read directl off the CFSM. Note how each state in the CFSM has one outgoing transition for each smbol to the right of the read head. The following states are Shift states: 0, 1, 2, 3, 5, 6. The following states are Reduce states: 4, 7, 8, 9. State 10 () is the accept state. This is an LR(0) grammar because there are no Shift/Reduce or Reduce/Reduce conflicts. Ever state is either a shift state or a reduce state (the accept state is just a special reduce state). 5

6 (e) Show the steps taken b the parser when parsing the string: zz. Give the action and show the state stack and remaining input for each step of the parse. Shift actions should be of the form Shift X where X is the state ou are shifting to, and Reduce actions should be of the form Reduce R, goto X where R is the rule being used to reduce, and X is the state the parser winds up in. nswer: Parse stack Smbol stack Remaining input ction 0 zz Shift zz Shift z z Reduce 11, goto z Reduce 8, goto z Shift z Shift z Shift z Shift z Reduce 11, goto Reduce 8, goto Reduce 10, goto Shift $ ccept (f) Suppose we change the last production to: w Is the resulting grammar LR(0)? Wh or wh not? nswer: No. To see wh, consider the CFSM that gets built: 6

7 State 0 S $ w State 1 S $ w State 2 S $ $ z State S $ State 3 w w State 4 w State 5 State 8 w State 6 w State 7 State 9 w w State 9 has a Reduce/Reduce conflict. If the parser gets to that state, it does not know whether to reduce using w or w. 7