Timeline for Is there a strictly non-deterministic one-counter language whose complement is one-counter?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2013 at 13:40 | comment | added | Raphael | Unfortunately, there is no proof here. | |
| Mar 20, 2013 at 13:26 | comment | added | frafl | It's OK to elaborate a bit especially if it's a nice exercise for you, but please add a short summary. Additionally you should not use an answer to reply to a comment, but in this case this reply may become an actual answer to your question, so I think it's OK, too. | |
| Mar 20, 2013 at 13:20 | comment | added | e_noether | yeah :) just wanted to prove it can be done by 1-counter automata | |
| Mar 20, 2013 at 13:18 | comment | added | frafl | Quite a long answer which is essentially just "Because its the union of three languages, each recognizing one interval of $i$s relative to $j$". | |
| Mar 20, 2013 at 13:11 | history | answered | e_noether | CC BY-SA 3.0 |