lect22-PSPACE_COMPLETEPROBLEMSpnp4

# lect22-PSPACE_COMPLETEPROBLEMSpnp4 - 1 Polynomial Space The...

This preview shows pages 1–10. Sign up to view the full content.

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Polynomial Space The classes PS and NPS Relationship to Other Classes Equivalence PS = NPS A PS-Complete Problem 2 Polynomial-Space-Bounded TMs A TM M is said to be polyspace-bounded if there is a polynomial p(n) such that, given input of length n, M never uses more than p(n) cells of its tape. L(M) is in the class polynomial space , or PS . 3 Nondeterministic Polyspace If we allow a TM M to be nondeterministic but to use only p(n) tape cells in any sequence of IDs when given input of length n, we say M is a nondeterministic polyspace-bounded TM. And L(M) is in the class nondeterministic polyspace , or NPS . 4 Relationship to Other Classes Obviously, P PS and NP NPS . If you use polynomial time, you cannot reach more than a polynomial number of tape cells. Alas, it is not even known whether P = PS or NP = PS . On the other hand, we shall show PS = NPS . 5 Exponential Polytime Classes A DTM M runs in exponential polytime if it makes at most c p(n) steps on input of length n, for some constant c and polynomial p. Say L(M) is in the class EP . If M is an NTM instead, say L(M) is in the class NEP ( nondeterministic exponential polytime ). 6 More Class Relationships P NP PS EP , and at least one of these is proper. A diagonalization proof shows that P EP . PS EP requires proof. Key Point : A polyspace-bounded TM has only c p(n) different IDs. We can count to c p(n) in polyspace and stop it after it surely repeated an ID. 7 Proof PS EP Let M be a p(n)-space bounded DTM with s states and t tape symbols. Assume M has only one semi-infinite tape. The number of possible IDs of M is sp(n)t p(n) . States Positions of tape head Tape contents 8 Proof PS EP (2) Note that (t+1) p(n)+1 &gt; p(n)t p(n) . Use binomial expansion (t+1) p(n)+1 = t p(n)+1 + (p(n)+1)t p(n) + Also, s = (t+1) c , where c = log t+1 s. Thus, sp(n)t p(n) &lt; (t+1) p(n)+1+c . We can count to the maximum number of IDs on a separate tape using base t+1 and p(n)+1+c cells a polynomial. 9 Proof PS EP (2) Redesign M to have a second tape and to count on that tape to sp(n)t p(n) . The new TM M is polyspace bounded....
View Full Document

## lect22-PSPACE_COMPLETEPROBLEMSpnp4 - 1 Polynomial Space The...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online