meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 6 months |
| seen | Apr 29 at 5:30 | |
| stats | profile views | 52 |
Delete Me
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Apr 16 |
awarded | Popular Question |
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Apr 15 |
awarded | Popular Question |
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Apr 3 |
accepted | Decidability Turing Machine Problem |
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Apr 3 |
comment |
Decidability Turing Machine Problem Wow thanks I didn't think of that. |
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Apr 3 |
asked | Decidability Turing Machine Problem |
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Mar 26 |
comment |
Proving non CFL with pumping lemma Got it, thanks. |
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Mar 26 |
accepted | Proving non CFL with pumping lemma |
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Mar 26 |
comment |
Proving non CFL with pumping lemma Thanks for the answer. I understand now why this would be a contradiction if $r$ is not 0 or $n$, but lets say $vxy$ is all $a$'s, how do we know that $v^{k}xy^{k}$ $k$>0 can never be a multiple of $p+1$ and thus be in $L$ and able to be pumped? |
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Mar 26 |
asked | Proving non CFL with pumping lemma |
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Mar 26 |
accepted | Push down automata problem |
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Mar 25 |
comment |
Push down automata problem No I do not but if its not too much trouble I can definitely try to learn to solve this. |
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Mar 25 |
asked | Push down automata problem |
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Mar 25 |
accepted | Context free language problem |
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Mar 25 |
comment |
Context free language problem Thanks for the help! |
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Mar 25 |
comment |
Context free language problem Ah ok so you can eliminate one of them because having the left side unbalanced and right side balanced is the same thing as having the right side imbalanced and left side balanced since both of them just contribute to the middle having one extra "a". Thus this eliminates the ambiguity in the language. Is this the right line of thought? |
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Mar 25 |
asked | Context free language problem |
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Mar 13 |
awarded | Nice Question |
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Mar 1 |
awarded | Popular Question |
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Feb 23 |
asked | stochastic dominance |
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Feb 22 |
accepted | Basic regular expressions problem |
