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May 20, 2021 at 1:32 history closed D.W. Duplicate of Context-free grammar for $L = \{a^{2^k}, k \in\mathbb{N}\}$
S Apr 22, 2015 at 6:34 history suggested Alejandro Sazo CC BY-SA 3.0
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Apr 22, 2015 at 1:03 review Suggested edits
S Apr 22, 2015 at 6:34
Oct 27, 2014 at 13:45 comment added Raphael Our reference question may provide some clues.
Oct 27, 2014 at 13:44 history edited Raphael
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Oct 26, 2014 at 23:09 history tweeted twitter.com/#!/StackCompSci/status/526511008659931136
Oct 26, 2014 at 22:49 vote accept MannfromReno
Oct 26, 2014 at 21:18 answer added Rick Decker timeline score: 3
Oct 26, 2014 at 18:12 comment added MannfromReno Hmm, you're right, we wouldn't know what u,v,w,x and y are. Now I might be more lost, I'll keep reviewing. Thank you.
Oct 26, 2014 at 17:26 answer added Yuval Filmus timeline score: 2
Oct 26, 2014 at 17:22 history edited Juho CC BY-SA 3.0
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Oct 26, 2014 at 17:18 comment added Roi Divon It's not true. you don't know what $u,v,w,x$ and $y$ are. Try showing that $2^k < |uv^2wx^2y| < 2^{k+1}$
Oct 26, 2014 at 17:13 comment added MannfromReno Okay that makes sense, so with $uv^2wx^2y$ the length of this would be an odd length if u, v, and y are all raised to 1, so $u^1v^2x^1y^2z^1$ is an odd length.
Oct 26, 2014 at 17:10 history edited MannfromReno CC BY-SA 3.0
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Oct 26, 2014 at 17:03 comment added Roi Divon Asumme $L$ is CFL. Let $k$ be the constant of the pumping lemma. let $z=0^{2^k} \in L$.. from the lemma we get that $z=uvwxy$. Consider $uv^2wx^2$. by the lemma, it's in $L$. What can you say about the length of this word? Try prooving that the length isn't a power or 2
Oct 26, 2014 at 16:35 review First posts
Oct 26, 2014 at 16:41
Oct 26, 2014 at 16:32 history asked MannfromReno CC BY-SA 3.0