381hw13solns

# 381hw13solns - CS 381 Homework#13 Problem 1 Question 9.3.7...

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CS 381 Homework #13 Problem 1 Question 9.3.7 Since we are trying to prove the following to be non-RE, we reduce a known problem which is non- RE to the given problems. a) We reduce the complement of L u to this problem, which is the complement of the halting problem for Turing Machines. The crux of the argument is that we can convert any TM M into another TM M' , such that M' halts on input w if and only if M accepts w . The construction of M' from M is as follows: 1. Make sure that M' does not halt unless M accepts. Thus, add to the states of M a new state p , in which M' runs right, forever; i.e., ° (p,X) = (p,X,R) for all tape symbols X . If M would halt without accepting, say ° (q,Y) is undefined for some nonaccepting state q , then in M' , make ° (q,Y) = (p,Y,R) ; i.e., enter the right-moving state and make sure M' does not halt. 2. If M accepts, then M' must halt. Thus, if q is an accepting state of M , then in M' , ° (q,X) is made undefined for all tape symbols X . 3. Otherwise, the moves of M' are the same as those of M . The above construction reduces the complement of L u to the complement of the halting problem. That is, if M accepts w , then M' halts on w , and if not, then not. Since the complement of L u is non- RE, so is the complement of the halting problem. b) Solution is similar in spirit to 9.3.4(d), proving L(M) = L(M) R is non-RE We reduce the problem L e (does a TM accept the empty language?) to the problem L(M 1 ) ° L(M 2 ) = 0. Given any TM M , we convert it into 2 other TMs M 1 & M 2 (with M 1 being nondeterministic), such that L(M 1 ) ° L(M 2 ) = 0 if and only if M accepts the empty language. If M accepts the empty language, we make M 1 & M 2 accept the empty language as well (giving L(M 1 ) ° L(M 2 ) = 0). Else if L(M) ± 0, then we make L(M 1 ) = L(M 2 ) = {01}

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• Fall '05
• HOPCROFT
• Halting problem, Alan Turing

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