hw_08_sol - Problem Set 8 Spring 10 Due: Tuesday, April 20,...

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Unformatted text preview: Problem Set 8 Spring 10 Due: Tuesday, April 20, in class before the lecture. Please follow the homework format guidelines posted on the class web page: http://www.cs.uiuc.edu/class/sp10/cs373/ 1. E.T. [ Category : Puzzle, Points : 5] Show that the problem of deciding whether aliens exist in the universe is decidable. More precisely, show that there is a Turing machine that will print YES if there are aliens in the universe, and NO otherwise! Solution: A problem is decidable if there exists a Turing machine that decides it. However, we don't need to be abe to construct one machine . We simply have to prove the existence of a machine that decides it. Pick a TM M 1 that always prints YES and halts, and pick another TM M 2 that always prints NO and halts. We have two cases: If aliens exist then TM M 1 does the job, if aliens don't exist then TM M 2 does the job. So in both cases there is a TM that does the job and this problem is hence decidable (of course we don't know which of these two TMs actually decides it, but we know that such a TM exists!). In general, for any one question, or any nite number of questions, there is always a decider. Decidability makes sense only when the language is in nite. So, the set of all theorems mankind has proven is, trivially, decidable. This does not mean that we are over-trivializing unknown problems. It just means that the theory of computation considers any nite language trivial. The theory of com- putation is more about the structure of in nite sets (languages) that can be de ned using nite means (like a machine). What these sets mean is not, technically speaking, part of the study. This is often a source of great confusion in people who don't under- stand this di erence, but use Turing's undecidability arguments in their philosophical arguments (very similar to how quantum theory is misinterpreted in many arguments). 2. Reduction la Rice's Theorem [ Category : Proof, Points : 20] A language L * is closed under reversal if for every w L , w R L . Show that L rev = {h M i | M is a TM and L ( M ) is closed under reversal } is undecid- able. You may not simply appeal to Rice's theorem (however, you can adapt the proof of Rice's theorem to solve this problem). 1 Solution: First note that if | | = 1 , then L is decidable (because every string in * will be the reverse of itself and so every language will be the reverse of itself). So assume a,b and a 6 = b . We prove that L rev is not decidable which is enough to show that L rev is not decidable. We have L rev = {h M i | h M i is not a TM code or, it is a TM and L ( M ) is not closed under reversal } Consider a TM M built this way: Algorithm M ( x ) 1. v Simulate M ( w ) 2. if v = Yes and x = ab 3. then return Yes 4. else return No You can see that we have these two properties: h M,w i A TM = M ( w ) = Y es = L ( M ) = { ab } = h M i L rev h M,w i / A TM = M ( w ) 6 = Y es = L ( M ) = = h M i /...
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hw_08_sol - Problem Set 8 Spring 10 Due: Tuesday, April 20,...

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