Q6) Consider the alphabet Σ = {a,b}. Consider the Non-Deterministic Finite Automaton (NFA). start O a = a 4 & a 2 a LO 5 a a & & 6 a. Why is this automaton non-deterministic? Explain the various source on indeterminacy. "aba" and w2 = "aab" belong to the language recognized by this FA? b. Do the sentences w1 Justify.

Q6) Consider the alphabet Σ = {a,b}. Consider the Non-Deterministic Finite Automaton (NFA). start O a = a 4 & a 2 a LO 5 a a & & 6 a. Why is this automaton non-deterministic? Explain the various source on indeterminacy. "aba" and w2 = "aab" belong to the language recognized by this FA? b. Do the sentences w1 Justify.

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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6. Give a nondeterministic finite automaton that accepts the language generated by the regular expression (a | ba | baaa)* ww 7. Convert the previous into a DFA
1. Give a deterministic finite automaton (DFA) which accepts the follow- ing language: L₁ = {ue {a,b,c}* | u contains the subwords ab and cc} -
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Draw the state diagram of a FINITE AUTOMATON (not DFA) that accepts the language of all strings over {a, b} that begin with aa AND contain the substring aba. Your finite automaton should not be overly complicated.
How would you prove that a formal language is not regular? Select one: O By using the Pumping lemma for regular languages. O By constructing a finite automaton that accepts that language. O By using Induction on number of steps of the derivation. O By constructing a context-free grammar that accepts that language.
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Hello I need help in theoretical computer science Let G be a grammar with the start variable S that describes a language using the alphabet Σ = {a, b, c} and whose rule set contains the following rules: • S ⇒ ABC, • A⇒aa and A⇒AA, • B ⇒ b, • C⇒c and C⇒CC. Show that the language L (G) is right-linear.
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2. 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?
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