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Jun 16, 2020 at 10:30 history edited CommunityBot
Commonmark migration
Dec 7, 2018 at 23:37 history edited Alfred Kaminski CC BY-SA 4.0
I added a new (comparatively-minor) question.
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
Dec 5, 2018 at 23:07 history asked Alfred Kaminski CC BY-SA 4.0