{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture7 - CSci 5403 Theorem(Cantor Let S P(S map elements...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 COMPLEXITY THEORY CSci 5403 LECTURE VII: DIAGONALIZATION Theorem (Cantor). Let ƒ : S ! P (S) map elements of the set S to subsets of S. Then ƒ is not onto. In particular, let D ƒ = { y | y ƒ(y) }. There is no s 2 S such that ƒ(s) = D ƒ : Suppose ƒ(s) = D ƒ . Then s 2 D ƒ , s ƒ(s) , s D ƒ . Example: Let S = {0,1,2} and let ƒ(0) = Ø, ƒ(1) = {1,2}, ƒ(2) = {0,1}. y 0 2 ƒ(y)? 1 2 ƒ(y)? 2 2 ƒ(y)? 0 N N N 1 N Y Y 2 Y Y N D ƒ Y N Y Theorem (Turing). The diagonal language D TM = { h M i | M does not accept on input h M i } is undecidable. Proof. Let ƒ TM : {0,1}* ! P ({0,1}*) map a TM’s code to the language it decides. If D TM is decidable, there must be a TM H such that ƒ TM ( h H i ) = D TM . But D TM is the diagonal set for ƒ TM . Cantor, Gödel, and Turing used it… must be good! (Insert cheesy 1970s theme music) TIME HEIRARCHY THEOREM Theorem. Let f and g be time constructible, and let g(n) = ω (f(n) 2 ). Then DTIME(f(n)) ( DTIME(g(n)). Corollary. P EXP. Proof. Let D f = { h M i | M does not accept h M i in f(| h M i |) steps. }. D f DTIME(f(n)), since it is the diagonal set for L f mapping a TM to the set it decides in time f(n). But recall that we can simulate a TM that runs in time t in time O(t 2 ). Thus D f 2 DTIME(f(n) 2 ).
Background image of page 1

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

View Full Document Right Arrow Icon
2 SPACE HEIRARCHY THEOREM Theorem. Let f and g be space constructible, and let g(n) = ω (f(n)). Then SPACE(f(n)) ( SPACE(g(n)). Corollary. L PSPACE. Proof. Analogous. We can simulate a machine using space s(n) in space O(s(n)).
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}