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I am trying to think if we can have an injective mapping $\mathcal{f}$, say something like $\Sigma_{bool}= \{0,1\}^* $ to $\Sigma_{5} = \{0,1,2,3,4\}^* $ such that $|x| \leq 2.|\mathcal{f}(x)|$ The m …

There's a small mistake in your question. The FSM (DFA or NFA) are machines that accept regular languages. the CFL is also a language. They are accepted by a machine called PDA (PushDown Automata …

@Shaull's solution is perfect.. here's a slightly different version of it... try to see, that $L$ = $L_1 \cup L_2$ where $L_1 = \{ w \in \{0,1\}^* \mid |w|_0 \gt |w|_1 + 2\}$ $L_2 = \{ w \in …

For a language that is regular but not CF or CS, construct a DFA/NFA that accepts it convert it to a regular expression show that it can be represented as an union, intersection or concatenation of …

Given any Context-Free-Grammar, $G$, and another in Chomsky Normal Form, $G_c$, how can we check if both $G$ and $G_c$ generate the same language? One of the trivial ways I know of is to convert $G …

(a): Recall that regular languages are closed under complement, i.e if $L$ is regular, so is $L'$. ($F_{L'} = Q_L- F_L$) Assume that $L$ is regular. This follows: if $L = \{x \in \{0,1\}^* \mid |x|_0 …

It means this Language can 'generate' any string using the alphabets $a, b, c$, including the empty string $\epsilon$ There is no bound on the length of the string generated. Here's a NFA that acce …

$A \rightarrow ba \mid A \mid B \mid cA$ under this $bababaccc$ is also legal You could try something like this... $S \rightarrow cS|cA$ $A \rightarrow baA|B$ $B \rightarrow bb$

@Karolis Juodele already gave the answer, and your answer is also correct. Another thing to keep in mind is although infinite languages can be undecidable, some of them are regular.. ex. $\{0\}^* \su …