*THEORY OF COMPUTATION

TOPIC:COMPLEXITY CLASSES

SPACE COMPLEXITY

SHESHAN SRIVATHSA

*INTRODUCTION

Definition:Let M be a deterministic Turing Machine that halts on all inputs.Space Complexity of M is the function f:NN, where f(n) is the maximum number of tape cells that M scans on any input of length n.

For any function f:NN, we define:

SPACE(f(n))={L | L is a language decided by an O(f(n)) space DTM}

NSPACE(f(n))={L | L is a language decided by an O(f(n)) space NTM}

*INTRODUCTION

Low Space Classes:

Definitions (logarithmic space classes):

L = SPACE(logn)NL = NSPACE(logn)

*Analyzing Space Complexity

If M is a deterministic Turing machine that halts on an inputs, then the value of

spacereq(M) is the function f(n) defined so that, for any natural number n,f(n) is

the maximum number of tape squares that M reads on any input of length n.

If M is a nondeterministic Turing machine all of whose computational paths halt on all

inputs, then the value of spacereq(M) is the function f(n) defined so that, for any natural

number n,f(n) is the maximum number of tape squares that M reads on

any path that it executes on any input of length n.

*EXAMPLE

SAT is NP-complete.

SAT can be solved in linear(O(m)) space.

M1=“On input <>, where is a Boolean formula:

1. For each truth assignment to the variables x1,…xm of .

2. Evaluate on that truth assignment.

3. If ever evaluate to 1, ACCEPT; if not, REJECT.”

0 0 0 … 0

x1 x2 x3 … xn

*Relating Time and Space Complexity

Theorem: Given a Turing machine M = (K , 2:., r, 8, s, H) and assuming that spacereq(M) ~

n, the following relationships hold between M's time and space requirements:

spacereq(M) < timereq(M) E o (cspacereq(M) )

Proof: Spacereq(M) is bounded by timereq(M) since M must use at least one time step for every

tape square it visits.

The upper bound on timereq(M) follows from the fact, since M halts, the number of steps that it

can execute is bounded by the number of distinct configurations that it can enter. That number is

given by the function MaxConfigs(M), as described above. Since MaxConjigs(M) E

O(cspacereq(M»), timereq(M) E O(cspacereq(M»).

*Savitch’s Theorem

Proof:

Suppose language A can be decided by an NTM in k f(n) space,

for some constant k.

We shall show that it can be decided by a DTM in O((f(n))2) space.

Let be an input string to N.t: integer;

c1,c2: configurations of N on .

Theorem

For any function f: NR+, where f(n)n, we have

NSPACE(f(n)) SPACE(f2(n)).

*Savitch’s Theorem

CANYIELD(c1,c2,t) accept if starting from c1 N has a branch

entering configuration c2 in t steps; o/w rejects.CANYIELD=“On input c1,c2 and t:

1. If t=1, test whether c1=c2 or c1├ c2, then accept; else reject.

2. If t>1, then for each configuration cm of N on using space f(n).

3. Run CANYIELD(c1,cm, ) 4. Run CANYIELD(cm,c2, ) 5. If 3 & 4 both accept, then accept; else reject.”

caccept: accept configuration.cstart: start configuration of N on .

Select a constant d so that N has no more than 2df(n) configurations

using f(n) tape space, n=| |.M=“On input output the result of CANYIELD(cstart, caccept, 2df(n)).”CANYIELD is a recursive procedure:•Recursive depth:log22df(n)=O(f(n))•Each recursive call of CANYIELD use O(f(n)) space.•Thus, it uses O(f2(n)) space.

*Example of Savitch’s algorithmboolean PATH(a,b,d) {

if there is an edge from a to b then

return TRUE else { if (d=1) return FALSE for every vertex v (not a,b) { if PATH(a,v, d/2) and PATH(v,b, d/2) then return TRUE } return FALSE }}

1

4

32

*PSPACE AND NSPACE

PSPACE is the class of languages (or problems) that are decidable in polynomial space on a det. TM.

Similarly we can define NPSPACE to be the class of languages that are decidable in polynomial space by a NTM.

So, what is the relationship between PSPACE and NPSPACE?

PSPACE = NPSPACE

Theorem: PSPACE = NPSPACE

Proof : If L is in PSPACE, then it must also be in NPSPACE because

the deterministic Turing machine that decides it in polynomial time is

also a nondeterministic Turing machine that decides it in polynomial

time.

It is an important corollary of Savitch's Theorem.

*PSPACE Completeness

Definition:

A language B is PSPACE-Complete if

1. BPSPACE

2. For every ASPACE, APB.

If B is just satisfies Condition 2, we say B is PSPACE-hard

*P, NP, and PSPACE

Theorem: P PSPACE⊆

Proof: If a language is decided by some DTM M

in f(n) time, M cannot see more than f(n) cells.

Thus, TIME(f(n)) ⊆ SPACE(f(n)),

So that P ⊆ PSPACE

*PSPACE and EXPTIME

Proof: If a language is decided by some DTM M in f(n)

space (where f(n) >=n), M can visit at most f(n) 2 O(f(n))

configurations (why?) Thus, M must run in f(n) 2 O(f(n))

time.

In other words, SPACE(f(n)) ⊆ TIME(2O(f(n))),

so that PSPACE ⊆ EXPTIME.

P NP P SPACE EXPTIME⊆ ⊆ ⊆

*A Hierarchy ofComplexity

*The Language QBF

Mathematical statements usually involve quantifiers: 8 (for all) and 9

(there exists)

• E.g., 8x F(x) means for every value of x, the statement F(x) is TRUE

• E.g., 9x F(x) means there exists some value of x such that F(x) is

TRUE

•Boolean formulas with quantifiers are called quantified Boolean

formulas

• E.g., 9y ( y = x+1) and 8x (9y ( y x)) are quantified Boolean

formulas

All of the following are quantified Boolean expressions.

• (P 1\ ,R)~S

• 3P «P 1\ ,R) ~ S)

• \iR (3P «P 1\ ,R) ~ S»

• \iS (\iR (3P «P 1\ ,R) ~ S»).

All quantified Boolean expressions are in prenex normal form.

*TQBF is PSPACE Complete

Theorem: TQBF is PSPACE-Complete

Proof:

(s,t,|V|) is TRUE iff there is a path from s to t.

is constructible in poly-time.

Thus, any PSPACE language is poly-time reducible to TQBF,

i.e. TQBF is PSPACE-hard.

Since TQBFPSPACE, it’s PSPACE-Complete.

*EXAMPLE(1)PSPACE-complete:The generalized version of some common games we play are PSPACE-complete:

Game of the Amazons Tic-Tac-Toe

*EXAMPLE(2)

Generalized Geography : Each city chosen must begin with the same letter that ended the previous city name.

*The Classes L and NL

L=SPACE(log n)

NL=NSPACE(log n)

E.G : PATH={<G,s,t> | G is a directed graph that has a directed path

from s to t}.

NL Completeness :

A language B is NL-Complete if1. BNL2. For every ANL, ALB.

*The Closure of Space Complexity ClassesUnder Complement

Deterministic Space-Complexity Classes are Closed Under Complement

Theorem: For every function f(n) , dspace(f( n)) = co-dspace(f(n)).

Proof: If L is a language that is decided by some deterministic Turing machine M,th en the deterministic Turing machine M' that is identical to M except that thehalting states y and n are reversed decides ~L. spacereq(M' ) = spacereq(M) .So, if L E dspace(f( n)), so is ~L.

• www.cs.tau.ac.il/~safra/Complexity/Space.ppt space complexity

• people.cs.nctu.edu.tw/~sctsai/fc/notes/SpaceComplexity.ppt

• http://homepage.cs.uri.edu/faculty/hamel/courses/2013/spring2013/csc544/lecture-notes/18-space-complexity.pdf

• 2008 Elaine A Rich, Automata, Computability And Complexity, Theory And Application (CHAPTER 29).

• http://www.cs.elte.hu/~lovasz/kurzusok/complexity.pdf.

• http://en.wikipedia.org/wiki/True_quantified_Boolean_formula

• http://www.cs.nthu.edu.tw/~wkhon/lectures/.

*REFERENCES