Timeline for Proving that recursively enumerable languages are closed against taking prefixes
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 13, 2021 at 12:07 | comment | added | Dennis | @vonbrand, did you mean that if $L \in \mathsf{RE}$, so there exists an enumerator $T$. We can simulate $T$, and on every output "print out" the all the possible prefixes of each output of $T$? Thank you. | |
| Jan 19, 2013 at 9:08 | comment | added | Ran G. | @vonbrand I agree that the method you suggest is actually simpler (and way more elegant! +1 for that). However your answer is somewhat cryptic: although an expert will be able to understand it, a non-expert will not be able to figure it out. Could you please explain your answer in greater details? | |
| Jan 18, 2013 at 13:35 | comment | added | vonbrand | @Raphael: It feels simpler than the above. Just my opinion. | |
| Jan 18, 2013 at 13:05 | comment | added | Raphael | The question is old and there are already two upvoted answers here, one of which has been accepted. So the question stands as to why you think your pointer adds something new here, and why it is not "just" a comment (which I can concert it to). | |
| Jan 18, 2013 at 12:58 | comment | added | vonbrand | It was meant just as a pointer in the right direction. Building the relevant TMs is tedious, and adds nothing to understanding what is going on. | |
| Jan 18, 2013 at 12:15 | comment | added | Raphael | You should elaborate on that; if you compare yours to the other answers, you notice a lack of detail (and, arguably, effort). | |
| Jan 18, 2013 at 10:14 | review | Low quality posts | |||
| Jan 18, 2013 at 20:10 | |||||
| Jan 18, 2013 at 9:56 | history | answered | vonbrand | CC BY-SA 3.0 |