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midterm-sol - CS 154 Introduction to Automata and...

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CS 154 - Introduction to Automata and Complexity Theory Solutions to Sample Midterm The questions below are from previous exams in CS154. They should give you some idea of what kinds of questions to expect and what we want for solutions. The actual midterm won’t be this long and will have a little more emphasize on decidability. Problem 1. [20 points] Decide if the following statements about languages over { 0 , 1 } are TRUE or FALSE, and circle the right answer using the boxes provided on the side. You must also give a brief explanation of your answer to receive full credit. F (a). If a DFA M has a loop then the language L ( M ) is infinite. FALSE – The loop might not be reachable from the initial state, or might not be possible to reach a final state from it. F (b). There is a regular language L for which there is exactly one regular expression R with L ( R ) = L . FALSE – For every regular expression R , R + R accepts the same language. F (c). Let L be a language and h a homomorphism. If h ( L ) is regular, then L must be regular. FALSE – Consider L = { 0 n 1 n | n 0 } and the homomorphism h (0) = ǫ, h (1) = ǫ . T (d). Let L be a regular language, and L R its reverse. The language L · L R is regular. TRUE – If L is regular then so is L R (to see this consider a DFA for L , reverse all transitions, add a new initial state with an ǫ -transition to all previous final states, and make the previous intial state final.) The concatenation of two regular languages is another regular.
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Problem 2. [30 points] a). [15 points] Consider the following language over the alphabet Σ = { a, b, c } . L = { a n b p ( c + b ) n - p | 1 n and 1 p n } . Here, ( c + b ) n - p means a sequence of n p symbols from the set { c, b } . Show that L is non-regular using closure properties. Do not use the pumping lemma, but you can use any language proven in the book to be non-regular (or regular). Solution: Consider the following homomorphism: h ( a ) = a, h ( b ) = b, h ( c ) = b . The image of L is h ( L ) = { a n b n | n 1 } which is non-regular. Therefore L has to be non-regular.
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