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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 overtrivializing 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 à la Rice's Theorem [ Category : Proof, Points : 20] A language L ⊆ Σ * 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  Σ  = 1 , then L is decidable (because every string in Σ * will be the reverse of itself and so every language will be the reverse of itself). So assume a,b ∈ Σ 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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 Spring '08
 Viswanathan,M
 Halting problem, TM, Mq, Lall, queue automaton

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