Given some unrecognizable language $L$, is it possible for its complement $\overline{L}$ to also be unrecognizable?
If some other language $S$ and its complement $\overline{S}$ are both recognizable, then $S$ and $\overline{S}$ are decidable. If $\overline{S}$ is unrecognizable, then then $S$ is undecidable but still recognizable. Why do we ignore the idea that $S$ and $\overline{S}$ may both be unrecognizable? This implies that $\exists! s \in S \cup \overline{S} = \Sigma^*$ on which no machine halts, otherwise I don't see why we cannot have $x,y \in \Sigma^*$ and $x \neq y$ such that no machine halts on $x$ or $y$, where $x \in S$ and $y \in \overline{S}$.
Perhaps I am making a false assumption somewhere?
Asked By : baffld
Answered By : sdcvvc
I'll write "corecognizable" as a shortcut for "complement of recognizable". There are countably many recognizable languages and countably many corecognizable languages. Therefore, there are uncountably many languages which are neither recognizable nor corecognizable.
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Question Source : http://cs.stackexchange.com/questions/22814
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