Midterm 2 Solution Fall 2013

so intersect the complement of l with abc to get

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Unformatted text preview: gular, must symmetric(L) be regular? Explain. Yes. Let Lrev be the language {w : wrev ∈ L}. We have seen that regular languages are closed under reversal, so if L is regular so is Lrev. Now symmetric(L) = L ∩ Lrev, so because regular languages are closed under intersection, symmetric(L) is regular whenever L is. (b) [7 marks] If L is context- free, must symmetric(L) be context- free? Explain. No. Let L = {aibiaj : i,j ≥ 0}. Then L is a CFL. (Here’s a CFG: S → TA; T → aTb | ε; A → aA | ε.) But symmetric(L) = {aibiai : i ≥ 0}. By pumping, symmetric(L) is not a CFL even though L is. Question 3 [10 marks] Consider the language L = {aibjck: i≠j or j≠k}. Show that L is not a DCFL. You cannot show this by showing that L is not a CFL, because it is. (Here’s a CFG for it: S → RC | AT; A → aA | ε; B → bB...
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This note was uploaded on 01/16/2014 for the course CSE 355 taught by Professor Lee during the Fall '08 term at ASU.

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