Yoni Hassin
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 Jan15 comment Proving a language is not a CFL although it can be pumped $a^{n-t}b^{n-s}c^n$ ? And using the fact that $t+s>0$ proves the whole thing? Actually you claim that $L$ can not be pumped for any partition of $w=uv^ixy^iz$ ? Jan14 comment Proving a language is not a CFL although it can be pumped Are we speaking about the pumping lemma for CFL's? Because $vxy$ can be $a^tb^s$,$t+s \le n$ so $w$ can be pumped and nothing can be proved here, i am wrong? Jan14 comment Proving a language is not a CFL although it can be pumped After further reading i realized it has to proven with Ogden's lemma. Still working on that.. Jan14 comment Proving a language is not a CFL although it can be pumped I know it will work, that is the tricky part because the easy way to prove a language is not a CFL is contradict the pumping lemma which can't be done it that case. Jan14 comment Are these languages context free or not? What about $L_8$ and $L_9$ ? :) Jan14 comment Are these languages context free or not? LOL!! can't believe someone from my class just copied paste the whole exercise :)) I currently stuck at 8&9 Jan13 comment Finding an appropriate value to contradict the pumping lemma. Thank you again..! Jan7 comment Formula for an equation Can you please write the main idea behind the prove? Dec31 comment Decide if a formal language is a Context-free language Isn't it enough to show that $L=h^{-1}(L_2)$ ? So $L$ is closed under inverse homomorphism? Why is the second part needed? Please be a bit more specific I am having bad time over here :) Dec29 comment Question about the meaning of expression in cryptography So why $1^n$ and $0^n$ ? Why not other combinations? Dec29 comment Question about the meaning of expression in cryptography The special role of $0^k$ and $1^k$ in that question Dec28 comment Question about the meaning of expression in cryptography i know it represent a string of zeros or ones but why always in cryptography $1^k$ and $0^k$ are used? do they have a symbolic value? that the part that confuses me.