Timeline for Arguing that $L= \{a^n | n\geq 0\} \cup \{a^nb^n| n\geq 0\}$ is not $LL(k)$ for any $k$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2021 at 22:12 | vote | accept | Abhishek Ghosh | ||
| S Apr 28, 2021 at 21:09 | history | bounty ended | CommunityBot | ||
| S Apr 28, 2021 at 21:09 | history | notice removed | CommunityBot | ||
| Apr 22, 2021 at 23:23 | history | edited | John L. | CC BY-SA 4.0 |
Edited tags. This question is not " about general methods and techniques for proving multiple theorems".
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| Apr 21, 2021 at 17:23 | answer | added | kiv | timeline score: 3 | |
| Apr 21, 2021 at 5:47 | answer | added | Yuval Filmus | timeline score: 2 | |
| Apr 20, 2021 at 20:42 | comment | added | Abhishek Ghosh | Please help me out in understanding it.. | |
| Apr 20, 2021 at 20:37 | comment | added | Abhishek Ghosh | @Yuval Yes this is what Linz also says. But the example he talks about, he comes up with a non obvious transformation and proves that the problem was with the grammar and not with the language. Now trying to prove the LL(k) nature of an unknown language is not an easy task I guess, (from the non obvious transformations that Linz does in the example). But is there any easy way out for such problems in general, I mean the thinking approach. Well in CFG, we can at times argue whether or not the language is inherently ambiguous or not like in $L = \{a^nb^nc^m\} \cup \{ a^n b^m c^m\}$ | |
| Apr 20, 2021 at 20:02 | comment | added | Yuval Filmus | A language is $LL(k)$ if it has an $LL(k)$ grammar. | |
| S Apr 20, 2021 at 19:48 | history | bounty started | Abhishek Ghosh | ||
| S Apr 20, 2021 at 19:48 | history | notice added | Abhishek Ghosh | Draw attention | |
| Apr 9, 2021 at 19:47 | history | edited | Abhishek Ghosh | CC BY-SA 4.0 |
added 14 characters in body
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| Apr 3, 2021 at 17:29 | history | asked | Abhishek Ghosh | CC BY-SA 4.0 |