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peter linz
sushmita
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Theory of Computation
Sep 9, 2018
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The language L1={a^n b^n} union {b}
The language L2={a^n b^n} union {a}
They both are deterministic CFL.
Am i right?
theory-of-computation
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sushmita
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Sep 9, 2018
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Shaik Masthan
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Sep 9, 2018
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yes mam...
either you have 1 a ===> accepted ( due to final state )
or n a's then check n b's ===> accepted
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Shaik Masthan
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Sep 9, 2018
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you can't have extra b....
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srestha
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Sep 9, 2018
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ok yes
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Peter Linz Edition 4 Exercise 8.1 Question 8 (Page No. 212)
Determine whether or not the following languages are context-free. (a) $L=$ {$a^nww^Ra^n : n ≥ 0, w ∈$ {$a,b$}*} (b) $L=$ {$a^nb^ja^nb^j : n ≥ 0, j ≥ 0$}. (C) $L=$ {$a^nb^ja^jb^n : n ≥ 0, j ≥ 0$}. (d) $L=$ {$a^nb^ja^kb^l : n + j ≤ k + l$ ... $ L=$ {$a^nb^nc^j : n ≤j$}. (g) $L=$ {$w ∈$ {$a, b, c$}* $: n_a(w)= n_b (w)=2n_c(w)$}.
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Peter Linz Edition 4 Exercise 8.1 Question 5 (Page No. 212)
Is the language L = {$a^nb^m : n = 2^m$} context-free?
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Peter Linz Edition 4 Exercise 8.1 Question 1 (Page No. 212)
Show that the language $L=${$a^nb^nc^m,n\neq m$} is not context-free.
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Peter Linz Edition 4 Exercise 7.4 Question 9 (Page No. 204)
Give LL grammars for the following languages, assuming $Σ =$ {$a,b, c$}. (i) $L=$ {$a^nb^mc^{n+m}:n\geq0,m\geq0$} . (ii) $L=$ {$a^{n+2}b^mc^{n+m}:n\geq0,m\geq0$} . (iii) $L=$ {$a^nb^{n+2}c^{m}:n\geq0,m\gt1$} . (iv) $L=$ {$w:n_a(w)\lt n_b(w)$} . (v) $L=$ {$w:n_a(w)+n_b(w)\neq n_c(w)$} .
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Jun 25, 2019
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