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Mar 20, 2013 at 13:23 comment added e_noether @frafl thanks! that's neat and simple.
Mar 20, 2013 at 13:11 answer added e_noether timeline score: -1
Mar 20, 2013 at 13:07 comment added frafl @emmy: Ooops, typo: I meant of course $x_{\lfloor x/2\rfloor}$ i.e. the symbol at the $\lfloor x/2\rfloor$-th position.
Mar 20, 2013 at 12:52 comment added e_noether see this link
Mar 20, 2013 at 10:50 history edited Raphael
edited tags; edited tags
Mar 20, 2013 at 8:08 comment added Shaull @emmy: How would a (nondeterministic) 1CM decide $\overline{L}$?
Mar 20, 2013 at 5:38 comment added e_noether how about this language : $L = \{ a^ib^j \mid i=j \ or\ i=2j \} $ and $\bar{L} = \{a^ib^j \mid i \neq j \ and\ i\neq 2j\}$
Mar 20, 2013 at 4:20 comment added e_noether what do you mean by $x_{\lfloor|c|/2\rfloor}=a$ ?
Mar 19, 2013 at 16:44 history tweeted twitter.com/#!/StackCompSci/status/314054705372332033
Mar 19, 2013 at 16:27 comment added frafl A PDA with only one kind of symbols (aside from the bottom symbol) on its stack.
Mar 19, 2013 at 16:24 comment added Ran G. what is "one-counter"?
Mar 19, 2013 at 16:19 comment added frafl Something like $L=\{x\in\{a,b\}^*\mid x_{\lfloor|c|/2\rfloor}=a\}$?
S Mar 19, 2013 at 16:04 history edited Gilles 'SO- stop being evil' CC BY-SA 3.0
Cleaned up latex, notation and grammar.
S Mar 19, 2013 at 16:04 history suggested Sam Jones CC BY-SA 3.0
Cleaned up latex, notation and grammar.
Mar 19, 2013 at 15:56 review Suggested edits
Mar 19, 2013 at 16:04
Mar 19, 2013 at 15:44 history asked e_noether CC BY-SA 3.0