CMI2011-A-01

Question

Let L $\subseteq \{0, 1\}^∗$ be a language accepted by a finite automaton. Let $F$ be some subset of $\{0, 1\}^∗$, containing 2011 strings. Which of the following statements is true? Justify your answer.

• A. $L \: \: \cup \:\: F$ is always regular.
• B. $L \: \: \cup \: \:F$ is regular only when $L\: \cap \: F =\not{0}$ —that is, when $L$ and $F$ do not have any common string.
• C. $L \: \cup \: F$ is never regular.
• D. $L \: \cup \: F$ is regular only if $L$ contains $\in$ (the empty string).

Answer 1

consider a L = {strings starting with '0' }

assume F = {01,0011,000111........0²⁰¹¹ 1²⁰¹¹} //// F can be any set of string as it is finite i.e ...2011

now take give options

B) FALSE

L ∩ F = {01,0011,000111........0²⁰¹¹ 1²⁰¹¹}  and it is regular and is finite...

C) false

L = {strings starting with '0' }

F = {01,0011,000111........0²⁰¹¹ 1²⁰¹¹} 

L U F= { strings starting with '0' } is regular 

D) false

L = {strings starting with '0' }

F = {01,0011,000111........0²⁰¹¹ 1²⁰¹¹} 

L U F= { strings starting with '0' } is regular ...even F doesn't contain empty string...

A) TRUE it will be true...for any F union with L...coz L is accepted by finite automata and F is already finite (given)

Comments

Dts true... L is a regular language and F is a finite subset of this language as 2011 strings are given and  'finite subset of regular language is regular '..regular languges are closed under union hence L UNION F is regular

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