ReductionKey - 1. Consider the set of indices DEFINED =...

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1. Consider the set of indices DEFINED = { f | 5 x ϕ f (x) }. Use Rice’s Theorem to show that DEFINED is not decidable. Hint: There are two properties that must be demonstrated. Defined is not trivial as the index of S(x) = x+1 is in, but (x) = μ y [ y == y+1] is not. Let f and g be indices of two arbitrary effective procedures such that the dom(f) = dom(g). f DEFINED 5 x ϕ f (x) by definition of DEFINED dom(f) ≠ φ since f converges somewhere dom(g) ≠ φ since dom(g) = dom(f) 5 x ϕ g (x) since domain is not empty g DEFINED by definition of DEFINED 2. Let P = { f | 5 x ϕ f (x) converges in at most x steps }. Why does Rice’s theorem not tell us anything about the undecidability of P? Because P is not an I/O behavior; it is a performance behavior. To see this, consider the two functions F(x) = 0 and G(x) = μy [y>x] – μy [y>x] . 2200 x F(x) = G(x), but F is in P and G is not. 3. Show that DEFINED is not decidable by reducing K 0 to this set. Let f be the index of an arbitrary function, F, and x be an arbitrary input. Define G fx (y) = F(x)-F(x). G fx (y) is defined everywhere and thus in DEFINED if <f,x> is in K 0 . G fx (y) is undefined everywhere and thus not in DEFINED if <f,x> is not in K 0 .
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This note was uploaded on 07/14/2011 for the course COT 4610 taught by Professor Dutton during the Fall '10 term at University of Central Florida.

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ReductionKey - 1. Consider the set of indices DEFINED =...

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