1. For the following sub-problems, consider the following context-free grammar: S AB$ (1) A xax (2) A B (3) B yby (5) B A (6)

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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 AB$ (1) A A (2) A B (3) A λ (4) B B (5) B A (6) B λ (7) (a) What are the terminals and non-terminals of this language? Terminals: {,, $}. Non-terminals: {S, A, B} (b) Is this a regular language (i.e., could ou write a regular epression that generates this language)? No, this is not a regular language. To see this, consider starting with S, and rewriting the first B into λ, leaving us with just A. Then note that each added to the beginning of the string must be matched b the same number of s at the end of the string. We can then turn the A into a B to generate matched s, etc. What we are generating is a palindrome made up of s and s. To create a palindrome, we would have to remember what the first half of the palindrome looked like to make sure the second half corresponds. That s not possible to do with a finite amount of state, so we can t build a finite automaton to capture these strings. Hence, no regular epression can match these strings, so the language is not regular. (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. (8) 1

2 S AB$ Rule 1 AB$ Rule 2 BB$ Rule 3 BB$ Rule 5 B$ Rule 7 B$ Rule 5 A$ Rule 6 A$ Rule 2 $ Rule 4 S A B $ A B B A B A λ λ (d) Give the first and follow sets for each of the non-terminals of the grammar. We go through each of the productions in turn, generating new set constraints until we can t generate an more. The set constraints we generate are: F irst(s) F irst(a) λ From production 1 2

3 F irst(a) {} From production 2 F irst(a) F irst(b) λ From production 3 F irst(a) {λ} From production 4 F irst(b) {} From production 5 F irst(b) F irst(a) λ From production 6 F irst(b) {λ} From production 7 F irst(s) F irst(b) λ From production 1 because λ F irst(a) F irst(s) {$} From production 1 because λ F irst(b) Solving these set constraints gives us: F irst(s) = {,, $} F irst(a) = {,, λ} F irst(b) = {,, λ} We then turn to the follow sets. Here, we go through the rules and add constraints based on non-terminals that appear on the right hand side: F ollow(a) F irst(b$) = {,, $} From production 1 F ollow(b) F irst($) = {$} From production 1 F ollow(a) F irst() = {} From production 2 F ollow(b) F ollow(a) From production 3 F ollow(b) F irst() = {} From production 5 F ollow(a) F ollow(b) From production 6 Solving these set constraints gives us: F ollow(s) = {} F ollow(a) = {,, $} F ollow(b) = {,, $} (e) What are the predict sets for each production? 3

4 P redict(1) = F irst(ab$) = {,, $} P redict(2) = F irst(a) = {} P redict(3) = (F irst(b) λ) F ollow(a) = {,, $} P redict(4) = F ollow(a) = {,, $} P redict(5) = F irst(b) = {} P redict(6) = (F irst(a) = λ) F ollow(b) = {,, $} P redict(7) = F ollow(b) = {,, $} Note that the predict sets for productions 3 and 6 include the follow sets of their left hand sides. That s because the first sets of the right hand sides include λ. If the right hand side can be rewritten to nothing, then we also need to consider what could come after the non-terminal we are predicting. (f) Give the parse table for the grammar. Is this an LL(1) grammar? Wh or wh not? Answer $ S A 2, 3, 4 3, 4 3, 4 B 6, 7 5, 6, 7 6, 7 This is not LL(1) because there are conflicts in the parse table. 2. for the following sub-problems, consider the following grammar: (a) Eplain wh this grammar is not LL(k) for an k. S AB$ (9) A A (10) A z (11) A (12) B B (13) B w (14) The eas answer is that the grammar is left-recursive (note the production A A). Without knowing how man s there are in the string, we don t know how man times to appl the production before turning the A into a z or an. If ou tr 4

5 to build the predict sets for this grammar, ou ll see that no matter how man lookahead tokens ou add, ou won t be able to resolve the conflict between productions 10 and 11 (when ou see a z) and 10 and 12 (when ou see an ). (b) Build the CFSM for this grammar. (Note that our state numbering ma be different.) State 0 S AB$ A A A z A z A State 1 S A B$ A A B B B w B State 2 $ S AB $ w State 8 A A State 3 S AB$ State 4 A z State 6 B B B B B w w State 9 B w State 5 A B State 7 B B A couple of things to note. In State 1, we had to consider what the B could become from configuration S A B$, which caused us to add all the configurations for B. In state 6, the initial configuration was B B, which means we have to worr about what the second B could become. Even though we re alread in the middle of matching a B, we haven t seen an of the second B, so the read head is at the beginning of those productions. (c) Build the goto and action tables for this grammar. Is it an LR(0) grammar? Wh or wh not? The goto table can be read directl off the CFSM it s just the transition table for the finite state machine. 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, 6. The following states are Reduce states: 4, 5, 7, 8, 9. State 3 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 (d) Show the steps taken b the parser when parsing the string: zw$. Give the action and show the state stack and remaining input for each step of the parse. Shift actions should be of the form Shift X where X is the state ou are shifting to, and Reduce actions should be of the form Reduce R, goto X where R is the rule being used to reduce, and X is the state the parser winds up in. Parse stack Smbol stack Remaining input Action 0 zw$ Shift z w$ Reduce 11, goto A w$ Shift A w$ Reduce 10, goto A w$ Shift A w$ Shift Aw $ Reduce 14, goto AB $ Reduce 13, goto AB $ Shift AB$ Accept (e) Suppose we change the last production to: B Is the resulting grammar LR(0)? Wh or wh not? No. To see wh, consider the CFSM that gets built: State 0 S AB$ A A A z A z State 4 A z State 5 A A State 1 S A B$ A A B B B State 6 B B B B B B State 7 B B B State 2 $ S AB $ State 8 A A B State 9 B State 3 S AB$ 6

7 State 8 has a Reduce/Reduce conflict. If the parser gets to that state, it does not know whether to reduce using A A or B. 7