The language of the grammar G = ({S}, {a,b}, S, P) with rules S → aSa, S→ bSb, S →1, is context-free. A typical derivation in this grammar is S= aSa = aaSaa = %3D .aabSbaa = aabbaa a) L(G) = {ww: we {a,b}*} b) L(G) = {wwR: w e {a,b}} c) L(G) = {wwR: w e {a,b}*} O d) L(G) = {wwR: w e {a*,b*}} O

The language of the grammar G = ({S}, {a,b}, S, P) with rules S → aSa, S→ bSb, S →1, is context-free. A typical derivation in this grammar is S= aSa = aaSaa = %3D .aabSbaa = aabbaa a) L(G) = {ww: we {a,b}} b) L(G) = {wwR: w e {a,b}} c) L(G) = {wwR: w e {a,b}} O d) L(G) = {wwR: w e {a,b}} O

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Consider the grammar:S →aAbB | CCA →a | CB | εB →bB | dC →aC | d | ε where S, A, B, and C are non-terminals, S is the start variable, and a, b, c, and d are terminals. Does the grammar have a predictive recursive descent parser? If the grammar has a predictive recursive descent parser, show that the conditions of predictive parsing apply for every non-terminal. On the other hand, if the grammar does not have a predictive recursive descent parser, then show ALL the conditions of predictive parsing that fail.
Consider the context-free grammar G = (V, T, S, P), where V = {S, A}, T = {a, b}, and P consists of S → aSb | aAb, A → aA | λ. a. Show the derivation tree resulting from top-down parsing of the string aaaabb. b. Show the leftmost derivation of the string aaaabb.
A grammar is left recursive if the grammar has any non-terminal whose derivation contains itself as the left-most symbol as in the production. Consider the grammar G1: E → E + T|T T → T ∗ F|F F → (E)|id Show that G1 is left recursive and transform G1 into a non – left recursive grammar called G2. Construct a recursive – descent parser for G2.
Consider the grammar G below in the input alphabet {x, y, z, *}, with leading non-terminal S' in rule 1. S'→S $ S → E * F S → F E → x E → z F y F F → E a) Build the LR(1) automaton of grammar G. Is this grammar LR(1)? b) List the LR(1) items in the last automaton state reached (without performing reductions on the last character) when the input is: x* z zx zxy For this question, indicate the lookahead and the top of the stack, it is also necessary to list the states in the same order as the grammar rules. Also, list the lookaheads in the order {?, $, x, y, z, *} c) Check if G is LALR. If so, answer with the number of states eliminated in relation to the previous automaton. If not, answer with -1
Given the context-free grammarS -> aB | bAA -> aS | bAA | aB -> bS | aBB | b Construct a PDA that simulates the grammar, that is, the rules of the grammar must be part of the machine (left most derivation always done).
In Example 6.1, show a derivation tree for the string ababbac, using both the original and the modified grammar. EXAMPLE 6.1 Consider G = ({A, B}, {a, b, c}, A, P) with productions A→a|aaA|abBc,B→abbA|b.A→a|aaA|abBc,B→abbA|b. Using the suggested substitution for the variable B, we get the grammar Ĝ with productions A→a|aaA|ababbAc|abbc,B→abbA|b.A→a|aaA|ababbAc|abbc,B→abbA|b. The new grammar Ĝ is equivalent to G. The string aaabbc has the derivation A ⇒ aaA ⇒ aaabBc ⇒ aaabbc in G, and the corresponding derivation A ⇒ aaA ⇒ aaabbc in Ĝ. Notice that, in this case, the variable B and its associated productions are still in the grammar even though they can no longer play a part in any derivation. We will next show how such unnecessary productions can be removed from a grammar.
Theory :: 1. write a CFG for language L={ aibjck | k=i+j and i,j,k>0 } 2. write a CFG for language L = { w | w=wR } 3. write a CFG for language L = { w | number of (a) in w equals number of (b) in w } 4. show the derivation tree for string (a+a)*a in the following grammar G = ({E,T,F}, {a, +, *, (, )}, R, E}. The set of rules R is: E -> E + T | T T -> T * F | F F -> (E) | a
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