Undecidability. We are not so much concerned if you are slow as when you come to a halt. (Chinese Proverb)
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1 We are not so much concerned if you are slow as when you come to a halt. (Chinese Proverb) CS /55 Theory of Computation
2 The is A TM = { M,w M is a TM and w L(M)} A TM is Turing-recognizable. Proof Sketch: Use the universal Turing machine. U = Input M,w, M is a TM and w is a string:. Simulate M on input w.. If M ever enters its accept state, accept. If M ever enters its reject state, reject. If M loops on w, so does U. If A TM were decidable, this implies being able to decide when A TM does not halt. CS /55 Theory of Computation
3 Proving is It turns out A TM is undecidable. Elements of the proof: TMs can be represented as strings. TMs can be input to TMs. TMs that are input can be simulated. leads to a contradiction. CS /55 Theory of Computation
4 is A set A is countable if there is an onto function f from N to A. f is onto if for all a A, there exists an n N such that f(n) = a. If there is also an onto function g from A to N, then A is said to have the same size as N. Consider E = {,,6,8,...}. Then f(n) = n is an onto function from N to E. And g(e) = e/ is an onto function from E to N. Therefore, E has the same size as N. Σ is countable and has the same size as N (homework problem). CS /55 Theory of Computation
5 of Rational Numbers is The set of positive rational numbers is countable. Proof: Consider the infinite matrix: We enumerate by traversing each lower left to upper right diagonal. CS /55 Theory of Computation 5
6 is can be used to show that some sets are uncountable. The set of real numbers R is uncountable. Proof: Suppose R is countable. Then, there is an onto function f from N to R. Let d(x,i) be the ith digit in x s fraction. Construct y such that d(y,i) d(f(i),i). y differs from every f(i), thus f is not onto. A contradiction! Therefore, R is uncountable. CS /55 Theory of Computation 6
7 is Suppose we start listing our favorite real numbers. n f(n) Then we can create a real number different from any number in the list such as CS /55 Theory of Computation 7
8 Non-Turing-Recognizable Languages Non-TR Languages is Some languages are not Turing-recognizable. Proof Sketch: The set of TMs is countable. The set of all languages is uncountable. Because TMs are countable and languages are not, some languages must be missed. CS /55 Theory of Computation 8
9 Characteristic Functions is A language can be represented by its characteristic sequence, an infinite sequence of bits. E.g., if A contains strings with an odd number of s, then A s characteristic sequence χ A is: Σ = { ǫ, 0,, 00, 0, 0,, 000, 00,...} A = {, 0, 0, 00,...} χ A = The set of characteristic sequences is uncountable (homework problem). CS /55 Theory of Computation 9
10 is is Proof: Suppose TM H decides A TM, that is: { accept if M accepts w H( M,w ) = reject if M does not accept w Let D be the following TM: D(M) = { accept if H rejects M,M reject if H accepts M,M Note: If H is a decider (doesn t loop forever), that implies that D is also a decider. CS /55 Theory of Computation 0
11 The is What happens with D(D)? There are two cases: Case : Suppose D accepts D. Then the definition of D implies that H rejects D,D. But by the definition of H, it follows that D rejects D. Case : Suppose D rejects D. Then the definition of D implies that H accepts D,D. But by the definition of H, it follows that D accepts D. A contradiction! Thus, it is not possible that H decides A TM. CS /55 Theory of Computation
12 Turing-Unrecognizable Languages is A language is co-turing-recognizable if it is the complement of a Turing-recognizable language. A language is decidable if and only if it is both Turing-recognizable and co-turing-recognizable. Proof Sketch: If A is decidable, both A and A are Turing-recogizable. If M recognizes A and M recognizes A, then A can be decided by running M and M in parallel. A TM is not Turing-recognizable. Proof Sketch: If A TM were Turing-recognizable, then A TM would be decidable. CS /55 Theory of Computation
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