Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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- Solve problem 24 please by following the directions.arrow_forwardThis question is about context-free grammars and parsing, and Turing machines (TM’s) i) Consider the following context-free grammar G over the alphabet Σ = {a, b, c}: S → aSb | ε | acSba | abc where S is the start variable. Using the rules of the grammar G, construct parse trees for thewords aabcb and aacbab. ii) Give 2 words (from Σ∗) that are not in the language accepted by the grammar G of Question i.Justify your answer, explaining why are the words you give not acceptedarrow_forward24. Show that the grammars and are not equivalent. S→ aSbbSa|SS|a SaSbbSalaarrow_forward
- THEORY of Automata. grammar Garrow_forward1. Prove that the language L(G) is not regular where G is the following context-free grammar: G = ({S,A,B,C}, {a,b}, {S->bB|C, A>b|C, B->Sb, C->aAa }, S). Note: You must first determine L(G).arrow_forward2. Using Pumping Lemma (slides 30-35 of the notes 'Regular Languages & Finite Automata-IV') one can show the language L= {a"b" | n€N } is not regular (We need this property in the notes 'Context-free Languages and Pushdown Automata I). This is done by way of contradiction. We assume L is regular. Since L is infinite, Pumping Lemma applies. We then consider the string s = ambm where m is the number of states in the DFA that recognizes L. Since the length of s is bigger than m, by Pumping Lemma, there exists strings x,y and z such that s = xyz, y + A, |xy| < 2m and xykz eL for all k e N. If |xy| < 2m then the first repeated state on the acceptance path cannot be a final state. Is this statement true, why or why not?arrow_forward
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