Timeline for Show that if L is regular, then third(L) is also regular. Hint: Construct an ϵ-NFA from the DFA for L
Current License: CC BY-SA 4.0
11 events
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Jun 16, 2020 at 10:30 | history | edited | CommunityBot |
Commonmark migration
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Dec 7, 2018 at 23:37 | history | edited | Alfred Kaminski | CC BY-SA 4.0 |
I added a new (comparatively-minor) question.
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Dec 7, 2018 at 23:33 | comment | added | Alfred Kaminski | Thank you both (regarding this series of comments). :) | |
Dec 6, 2018 at 18:58 | answer | added | Hendrik Jan | timeline score: 1 | |
Dec 6, 2018 at 15:51 | answer | added | Uli Schlachter | timeline score: 2 | |
Dec 6, 2018 at 15:32 | comment | added | Uli Schlachter | In the third line of the transition relation, if you replace $\langle q,4\rangle$ with $\langle q,1\rangle$ in $(\langle p,3\rangle,a,\langle q,4\rangle)$, then you no longer need the fourth copy of $A$. Also, the initial state should be $\langle s_0,1\rangle$. | |
Dec 6, 2018 at 3:16 | comment | added | Yuval Filmus | You ask whether $(p,a,q) \in \delta$ is the same as $\delta(p,a) = q$. It isn't since for an NFA, $\delta(p,a)$ is a set of states rather than a single state. | |
Dec 6, 2018 at 3:15 | comment | added | Yuval Filmus | I'm not sure why we need four copies – it seems that three should be enough. Perhaps you also need an additional initial state to handle $\epsilon$. | |
Dec 6, 2018 at 3:14 | comment | added | Yuval Filmus | The solution is indeed missing an initial state. | |
Dec 6, 2018 at 3:14 | history | edited | Yuval Filmus | CC BY-SA 4.0 |
added 7 characters in body
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Dec 5, 2018 at 23:07 | history | asked | Alfred Kaminski | CC BY-SA 4.0 |