Midterm 2 Solution Fall 2013

# Now symmetricl l lrev so because regular languages are

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Unformatted text preview: plain. The language is {aibjck : i ≥ j+k and i,j,k ≥ 0}. This is not regular. To explain, suppose to the contrary that it is regular, and let p be its pumping length. Then apcp is in the language, so we can write it as xyz with |xy| at most p, and |y| at least one so that xyiz is in the language for every non- negative integer i. But then y contains only a’s, so xy0z is NOT in the language. This is a contradiction, so the language is not regular. You cannot just write that “A DFA can’t do it” – that does not explain at all. Also, this question is NOT asking whether the grammar is regular – it is asking whether the language is regular. Question 2. [10 marks] Let L be a language. Then symmetric(L) = {w : w ∈ L and wrev ∈ L}. (a) [3 marks] If L is re...
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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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