Regular Languages (14 points) Solution: Problem 1 (6 points) Minimize the following automaton M. Show that the resulting DFA is minimal.

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1 Regular Languages (14 points) Problem 1 (6 points) inimize the following automaton Show that the resulting DFA is minimal. Solution: We apply the State Reduction by Set Partitioning algorithm (särskiljandealgoritmen) (Lecture 4 page 71). inimal automaton is: The above DFA is minimal because pairs of states are distinguishable: q B 1 q C λ λ q A q B

2 Problem (4 points) a) Convert the following NFA into Regular Grammar: We apply the algorithm explained in F4, starting with page 151. S aa A abs b b) Does the expression ((0+1)(0+1)*)*00(0+1)* denote the language of strings w such that w has at least one pair of consecutive zeros? Justify your answer clearly! YES. Here is why: x x* represents one or more x s. Thus (0+1)(0+1)* represents strings, each one of which is either a 0 or a 1. So this is all strings of zeros and ones of length 1 or greater. But when you apply * to this expression, you are now including λ making the expression ((0+1)(0+1)*)* the same as (0+1)*. So the expression as a whole represents any string that consists of 0 s and 1 s with a 00 somewhere in it. Problem 3 (4 points) Prove by using Pumping lemma that the following language is not regular: n m k L = { a b c n = m + k or m = n + k} Using the pumping lemma to prove that the language is not regular: Let w = a k b k c k. Note that w L because it satisfies the first condition (n = m + k). (Do not get confused by the two different uses of the variable name k.) Then y is a p, for some nonzero p. Pump in once. The resulting string is a k+p b k c k. It is not in L because it satisfies neither of the conditions for membership in L, as k + p k + k, and k k + p + k. Thus, L is not regular.

3 Context Free Languages (14 points) Problem 4 (5 points) Construct context free grammars to accept the following languages. a. L = { w Σ = {a,b}and w is odd} b. n j j n L = { a b a b n, j 0} a. S Σ A Σ A Σ S λ where Σ ={a, b} b. S asb X X bxa λ Problem 5 (5 points) Describe the following PDA. Test run. What language is accepted by the automaton? i j k L = { a b c i, j, k 0 and i = j or i = k)} Problem 6 (4 points) Is the following language context free? Justify your answer. If it is CF, give a grammar or an automaton. Otherwise prove that the language is not CF using the pumping lemma. i j k L = { a b c i, j, k 0 and j > max ( i, k)}

4 The language is not context free. We prove it using the pumping lemma. Let m be the pumping length from the pumping lemma and we choose the string w = uvxyz = a m b m+1 c m. (According to pumping lemma also uv m xy m z is in L for m 0.) Let the part 1 of the string contain all the a s, part all the b s, and part 3 contain all the c s. If either v or y go across numbered parts, pump up once. The resulting string will not be in L because it will not follow the form of the language. We consider the remaining cases: (1, 1): Pump up once. This increases the number of a s and thus the max of the number of a s and c s. But the number of b s is unchanged so it is no longer greater than that maximum. (, ): Pump down once. The max number of a s and c s is unchanged. But the number of b s is decreased and so it is no longer greater than that maximum. (3, 3): Same argument as (1, 1) but increases the number of c s. (1, ).(, 3): Pump down once. The max number of a s and c s is unchanged. But the number of b s is decreased and so it is no longer greater than that maximum. (1, 3): Not possible since vxy must be less than or equal to m. This means that language does not satisfy pumping lemma Recursively Enumerable Languages(1 points) Problem 7 (6 points) How does the following T work? Does it accept the input string? Trace the execution of this Turing machine on the input string.

5 T starts from the first symbol in the input aabbcc. The string gets accepted and tape contains n n n xxyyzz##. T accepts the language L = { a b c n 1 } Problem 8 (6 points) Prove each of the following languages decidable or undecidable! a) {( < 1 >,( < > ) L( 1) L( ) = }? Undecidable. This is a non trivial property of languages, so Rice's Theorem applies. b) {( < 1 >,( < > ) L( 1) L( ) = }? Undecidable. We give a reduction from E T (empty language T). Suppose the contrary, that this language is decidable. Let T be a Turing machine that decides it. Then we can construct a Turing machine T deciding E T, that behaves as follows on input <>. (1) Simulate T on input <, >, where is a Turing machine recognizing the empty language. () Accept if and only if T accepts. Suppose that accepts the empty language. Then T accepts <, >, and so T accepts. On the other hand, suppose that accepts some string. Then T rejects <, >, and so T rejects. Thus, T decides E T. c) Is L() infinite if is an arbitrary DFA? Decidable. An arbitrary DFA is a finite machine. We can inspect the coding of the machine and see if there are loops which make the language infinite. References Linz Peter, An Introduction to Formal Languages and Automata, Jones & Bartlett, 006 Rich Elaine, Automata, Computability and Complexity: Theory and Applications, Prentice Hall, 007 Sudkamp, Languages and achines, Addison Wesley 1998 Sipser ichael, Introduction to the Theory of Computation, PWS 1997 Hopcroft, otwani, Ullman, Introduction to Automata Theory, Languages, and Computation, A. Wesley 001

Computer Sciences Department

Computer Sciences Department 1 Reference Book: INTRODUCTION TO THE THEORY OF COMPUTATION, SECOND EDITION, by: MICHAEL SIPSER 3 D E C I D A B I L I T Y 4 Objectives 5 Objectives investigate the power of algorithms to solve problems.