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Unformatted text preview: Click to edit Master subtitle style 10/4/11 Reduction and Rice’s 10/4/11 Dec 2, © UCF (Charles E. 22 Two Interesting Sets • The sets Lr = { x | dom [x] is recursive } Lnr = { x | dom [x] is not recursive } • Lr is very easily confused with the set of indices of algorithms. It includes the indices of all algorithms, since their domains (all natural numbers) are clearly recursive. It also includes many indices of functions which diverge at some points where a corresponding algorithm might have produced a 0 output (rejection). • Our claim is that neither of these sets is re. 10/4/11 Dec 2, © UCF (Charles E. 33 Lr is Non-RE Let HALT(x,y) = i t STP(y, x, t) Consider again the set Lr = { x | dom [x] is recursive } Suppose Lr is re. We can show that this implies that the complement of Lu is also re, but then since Lu is re, we would have that Lu is recursive (decidable), an impossibility. We attack this by defining, for each function index x and input y, a function Fx,y( z ) = HALT(x, y) + HALT(<z>1 , <z>2) This function’s domain is Lu, if [x] (y) is defined, and is Ø, otherwise. Thus, Fx,y accepts a recursive language just in case (x, y) “ Lu (that is, [x] (y) is undefined). But (x, y) ) Lu just in case Fx,y’s index is in Lr. 10/4/11 Dec 2, © UCF (Charles E. 44 Lr Picture Proof x y x( y ) z1 ( z 2 ) z Given arbitrary x , y , define the function fx,y(z) = ¡ x(y) + ¤z1(z2) . The following illustrates fx,y,. Here, dom(fx,y,) = & if & x(y) & ; = K0 if & x(y)˛ Thus, ¤ x(y) ) iff fx,y is in Lr , and so ~ K0 ¢ 1 Lr. If Lr is re then so is ~K0 and hence K0 and its complement are both re, implying K0 is recursive, but that cannot be so, Hence Lr is not re....
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