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May 24, 2012 at 13:02 vote accept Hernan_eche
May 24, 2012 at 13:02 history edited Hernan_eche CC BY-SA 3.0
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May 23, 2012 at 18:39 comment added Hernan_eche @Gilles you're right, and that's a source of misunderstanding in my question, I was trying to use the set and powerset cardinals to prove existence of undecidable language in an specific case, thanks
May 23, 2012 at 18:29 comment added Gilles 'SO- stop being evil' @Hernan_e If you write “undecidable” with no qualification, it means undecidable by a Turing machine (i.e. for the usual notion of computation). If you are studying decidability in other computation models, there may or may not be a finite decidable language, depending on the model.
May 23, 2012 at 18:18 comment added Hernan_eche @DavidLewis I still think it's not end of story, because I try to work with a finite model, and to see if there is a way to apply #set < #powerset to show there is a finite undecidable language, please read the comments on Ran G answer, thanks
May 23, 2012 at 18:18 comment added Hernan_eche @Gilles I agree, for a Turing Machine, but if you do not allow an unbounded program size, then I think there are bigger restrictions
May 23, 2012 at 18:02 vote accept Hernan_eche
May 23, 2012 at 18:05
May 23, 2012 at 5:30 answer added Ben timeline score: 2
May 23, 2012 at 4:55 answer added Ran G. timeline score: 15
May 22, 2012 at 19:42 comment added Gilles 'SO- stop being evil' @Hernan_e To explain why “there is no surjection from $\mathbb{N}$ onto $\mathscr{P}(\mathbb{N})$, thus there must exist an undecidable language” doesn't extend to finite languages: the reason is that a decidable language has to be enumerable, that is, at most as big as $\mathbb{N}$. If $L$ is finite then $\mathscr{P}(L)$ is bigger but still finite and hence smaller than $\mathbb{N}$.
May 22, 2012 at 17:38 history edited Kaveh
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May 22, 2012 at 17:16 history tweeted twitter.com/#!/StackCompSci/status/204984042817458176
May 22, 2012 at 14:10 comment added David Lewis Finite language are decidable, period, end of story. There are any number of algorithms for that. If you insist on the classical Turing Machine model, it can be done that way too, though less perspicaciously. No need to invoke finite-state automata or regular languages or any other automaton model, as they are in fact, overkill without any additional clarity vis-a-vis Turing Machines.
May 22, 2012 at 14:07 comment added Hernan_eche @Raphael That's this question intention, but I could create a new one if that's a more clear rewording
May 22, 2012 at 14:00 comment added Raphael As far as I know, the existence of undecidable languages does not follow immediately from the nonexistence of such a surjection; you need some tiny bits more. Why, that would make another wonderful question! Why don't you go ahead and ask it? From that one, you should see why the argument does not carry over to finite languages.
May 22, 2012 at 13:58 comment added Hernan_eche @Raphael It's ok, but I mention power set because sometimes I read "there is no surjection from $\mathbb{N}$ onto $\mathcal{P}(\mathbb{N})$, thus there must exist an undecidable language." I would like to understand why that didn't work with a finite set $L'$, with $|L'|\leq N$ with $N$ finite, instead of needing $\mathbb{N}$, that's why I put $\mathcal{P}(S)$
May 22, 2012 at 13:51 comment added Raphael I removed the references to the power set as it is not relevant here; $\mathcal{P}(S)$ is finite if and only if $S$ is finite.
May 22, 2012 at 13:50 history edited Raphael CC BY-SA 3.0
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May 22, 2012 at 13:47 answer added Raphael timeline score: 8
May 22, 2012 at 13:42 history edited Hernan_eche CC BY-SA 3.0
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May 22, 2012 at 13:36 history edited Hernan_eche CC BY-SA 3.0
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May 22, 2012 at 13:31 comment added Raphael There seem to be (at least) three questions here. Please concentrate on one and edit out the others.
May 22, 2012 at 13:26 history edited Hernan_eche CC BY-SA 3.0
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May 22, 2012 at 13:14 answer added Sam Jones timeline score: 10
May 22, 2012 at 13:11 history asked Hernan_eche CC BY-SA 3.0