Polynomial Space - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

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1

Polynomial Space

The classes

PS

and

NPS

Relationship to Other Classes

Equivalence

PS

NPS

A PS-Complete Problem

2

Polynomial-Space-Bounded TM’s

A TM M is said to be

polyspace-

bounded

if there is a polynomial p(n)

such that, given input of length n, Mnever uses more than p(n) cells of itstape.

L(M) is in the class

polynomial space, or

PS

4

Relationship to Other Classes

Obviously,

P

PS

and

NP

NPS



If you use polynomial time, you cannotreach more than a polynomial number oftape 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 andpolynomial 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 ).

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 ID’s of M is sp(n)t

p(n)

States

Positions oftape head

Tapecontents

8

Proof

PS

EP

• (2)

Note that (t+1)

p(n)+

p(n)t

p(n)



Use binomial expansion (t+1)

p(n)+

= t

p(n)+

• (p(n)+1)t

p(n)

• …

Also, s = (t+1)

c

, where c = log

t+

s.

Thus, sp(n)t

p(n)

< (t+1)

p(n)+1+c

We can count to the maximum number of ID’s on a separate tape using baset+1 and p(n)+1+c cells – a polynomial.

10

Savitch’s Theorem:

PS

NPS

Key Idea: a polyspace NTM has “only” c

p(n)

different ID’s it can enter.

Implement a deterministic, recursive function that decides, about the NTM,whether I

*J in at most m moves.

Assume m < c

p(n)

, since if the NTM

accepts, it does so without repeating anID.

11

Savitch’s Theorem – (2)

Recursive doubling

trick: to tell if I

*J in

< m moves, search for an ID K such thatI

*K and K

*J, both in < m/2 moves.

Complete algorithm: ask if I

0

*J in at

most c

p(n)

moves, where I

0

is the initial ID

with given input w of length n, and J isany of the ID’s with an accepting stateand length < p(n).

13

Stack Implementation of

f

I, J, m O(p(n))space

I, K, m/

O(p(n))space

L, K, m/

O(p(n))space

M, N, 1O(p(n))space

...

O(p

2

(n)) space

14

Space for Recursive Doubling

f(I, J, m) requires space O(p(n)) to store I, J, m, and the current K.



m need not be more than c

p(n)

, so it can be

stored in O(p(n)) space.

How many calls to f can be active at once?

Largest m is c

p(n)

16

PS-Complete Problems

A problem P in

PS

is said to be

PS-

complete

if there is a polytime reduction

from every problem in

PS

to P.



Note: it has to be polytime, not polyspace,because:

Polyspace can exponentiate the output size.

Without polytime, we could not deal with thequestion

P

=

PS

?

17

What PS-Completeness Buys

If some PS-complete problem is:

In

P

, then

P

=

PS

.

In

NP

, then

NP

=

PS

.

19

QBF’s – (2)

First-order predicate logic, withvariables restricted to true/false.

Basis:

Constants 0 (false) and 1 (true) areQBF’s.

A variable is a QBF, and that variableoccurrence is

free

in this QBF.

20

QBF’s – (3)

Induction: If E and F are QBF’s, so are:

E AND F, E OR F, and NOT F. 

Variables are bound or free as in E or F.

(



x)E and (



x)E for any variable x.



All free occurrences x are bound to thisquantifier, and other occurrences of variablesare free/bound as in E.

Use parentheses to group as needed.



Precedence: quantifiers, NOT, AND, OR.