day27 - Click to edit Master subtitle style 10/4/11...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Click to edit Master subtitle style 10/4/11 Reduction and Rices 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 functions domain is Lu, if [x] (y) is defined, and is , 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,ys 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....
View Full Document

This note was uploaded on 10/03/2011 for the course COT 5310 taught by Professor Staff during the Spring '08 term at University of Central Florida.

Page1 / 10

day27 - Click to edit Master subtitle style 10/4/11...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online