Take the 2-minute tour ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.

For any language $L$ over $\{0,1\}^*$, a language $L'$ can be defined as $\{ a | ab \in L \text{ for some } b \in \{0,1\}^* \}$.

If $L$ is decidable, is $L'$ decidable?

I think that $L'$ should be decidable because we can create a Turing machine for $L'$ that will run the decider for $L$ on the input $w$ for $L'$, accept if it accepts, and otherwise will enumerate all the possible strings $b$ and run $wb$ on $L$. Does that make sense?

share|improve this question
    
How do you check if w is not in L'? There are infinitely many possible suffixes (in general case). –  Karolis Juodelė Feb 20 '13 at 19:10
    
Hmm, I guess I can't do that then. If that's the case, it should be undecidable and I should be able to reduce some undecidable problem to it right? –  Jay Feb 20 '13 at 20:33
add comment

1 Answer

Pick some self-terminating encoding of Turing machines. (That means that the encoding of one Turing machine is never a prefix of the encoding of another.) Consider the language $L$ consisting of strings $Mn$, where $M$ is an encoded Turing machine, $n$ is interpreted as a number, and $M$ halts within $n$ steps (on the empty input).

Does that answer your question?

share|improve this answer
    
I'm not sure I follow. L can be any language over {0,1}* so wouldn't this L you create only apply to one case? –  Jay Feb 20 '13 at 20:35
    
In my example $L$ is decidable but $L'$ is not. –  Yuval Filmus Feb 20 '13 at 21:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.