4. Consider the following languages. • ANFAFIN = {(N) |N is an NFA and L(N) is finite} • EREX = {(R) | R is a regular expression and L(R) = 0} • EQREX = {(R1, R2) | Ry and R2 are regular expressions and L(R) = L(R2)} • EQNFAFINSET = {(N,U)| N is an NFA, U is a finite set of words, and L(N) =U} • ALLREX = {(R) | R is a regular expression and L(R) = {0,1}"} (a) Answer the following questions based on the NFA N as shown below. Simply say yes or no. (i) Is (N) E ANFAFIN? (ii) Is ((0|1)*) € EREX?

4. Consider the following languages. • ANFAFIN = {(N) |N is an NFA and L(N) is finite} • EREX = {(R) | R is a regular expression and L(R) = 0} • EQREX = {(R1, R2) | Ry and R2 are regular expressions and L(R) = L(R2)} • EQNFAFINSET = {(N,U)| N is an NFA, U is a finite set of words, and L(N) =U} • ALLREX = {(R) | R is a regular expression and L(R) = {0,1}"} (a) Answer the following questions based on the NFA N as shown below. Simply say yes or no. (i) Is (N) E ANFAFIN? (ii) Is ((0|1)*) € EREX?

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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4. Consider the following languages. ANFAFIN = {{N) | N is an NFA and L(N) is finite} EREX = {(R) | R is a regular expression and L(R) = 0} • EQREX = {(R1, R2) | Ry and R2 are regular expressions and L(R;) = L(R2)} • EQ NFAFINSET = {(N,U) | N is an NFA, U is a finite set of words, and L(N) = U} • ALLREX = {{R) | R is a regular expression and L(R) = {0,1}*} %3D (a) Answer the following questions based on the NFA N as shown below. Simply say yes or no. 1 (i) Is (N) € ANFAFIN? (ii) Is (0|1)") € EREX? (iii) Is (00", 0*0) € EQREX?
Create a PDA that accepts the language L = {w ∈ {a.b}* : na(v) ≥ nb(v), where v is any prefix of w}
8. Build a DPDA that accepts the language L = {(ab)" (aab)b² n ≥0
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Decidable Languages Let R and S are regular expressions. Show that the following language L is decidable. L = {(R, S) | L(R) C L(S)}
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Construct regular expressions over Σ = {0,1} representing the following languages: 1. (w | |w|= 3 mod 4}
Design DPDA to accept language L= {xe [a, b]* N (x) > N,(x)). N(x) > N,(x) means number of a's are greater than number of b's in string x