True or False Statements and provide proof with your answer. a. Suppose L is the empty language, then L only includes "epsilon", Furthermore, "epsilon" is the "identity" for concatenation since when concatenated with any string it yields the other string as a result. That is, for any string w, (epsilon)w = w(epsilon) = w. b. Suppose L us a regular language. Then, L is a set of stings satisfying a certain condition or conditions. So, L is unique. On the other hand, there can be more than one DFA (e,g,, two DFAs) accepting L. c. Consider finite antomata and regular expressions. The regular
True or False Statements and provide proof with your answer.
a. Suppose L is the empty language, then L only includes "epsilon", Furthermore, "epsilon" is the "identity" for concatenation since when concatenated with any string it yields the other string as a result. That is, for any string w, (epsilon)w = w(epsilon) = w.
b. Suppose L us a regular language. Then, L is a set of stings satisfying a certain condition or conditions. So, L is unique. On the other hand, there can be more than one DFA (e,g,, two DFAs) accepting L.
c. Consider finite antomata and regular expressions. The regular expression " 1{}*(0+1) " can be simplified to " 1(0+!) ".
d. Consider the following language L, L = {ab, aabb, aaabbb, aaaabbbb, ......}. Note that L includes all strings of a's followed by the same number of b's. (Assume Sigma = {a,b}). Then, we can design the DFA for accepting L.
e. If E and F are regular expressions, then E + F is a regular expression denoting the union of L(E) and L(F). That is, L(E+F) = L(E) U L(F).
f. If L U M is NOT a regular language, then it is NOT true that L and M are regular languages.'
g. It is possible to convert any NFA to a DFA that accepts the same language.
h. The following four are equivalent as different notations for the same class of languages known as a regular language: Regular expressions, DFA, NFA, epsilon-NFA.
i. Consider the following language L: L = { 0^n 1^n 2^n | n >= 1}. Examples of string in L include {012, 001122, 000111222, ...}. Then, we can use "pumping lemma for regular languages" to prove that L is not a regular language.
j. {epsilon, 01, 0011, 000111, ...} is the language of all strings of n's 0's followed by n 1's, for some n > 0.
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