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THEORY OF COMPUTATION TOPIC:COMPLEXITY CLASSES SPACE COMPLEXITY SHESHAN SRIVATHSAPowerPoint Presentation

THEORY OF COMPUTATION TOPIC:COMPLEXITY CLASSES SPACE COMPLEXITY SHESHAN SRIVATHSA

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THEORY OF COMPUTATION TOPIC:COMPLEXITY CLASSES SPACE COMPLEXITY SHESHAN SRIVATHSA

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THEORY OF COMPUTATION TOPIC:COMPLEXITY CLASSES SPACE COMPLEXITY SHESHAN SRIVATHSA

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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}

- Low Space Classes:
Definitions (logarithmic space classes):

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

- If M is a deterministic Turing machine that halts on an inputs, then the value ofspacereq(M) is the function f(n) defined so that, for any natural number n,f(n) isthe 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 onany path that it executes on any input of length n.

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SAT is NP-complete.

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

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

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

Evaluate on that truth assignment.

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

x1 x2 x3… xn

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

Theorem

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

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

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 .

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:

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

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

Run CANYIELD(c1,cm, )

Run CANYIELD(cm,c2, )

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.

boolean 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

}

}

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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.

Definition:

A language B is PSPACE-Completeif

- BPSPACE
- For every ASPACE, APB.
- If B is just satisfies Condition 2, we say B is PSPACE-hard

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

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

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.

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. TQBFis PSPACE-hard.

Since TQBFPSPACE, it’s PSPACE-Complete.

PSPACE-complete:

The generalized version of some common games we play are PSPACE-complete:

Tic-Tac-Toe

Game of the Amazons

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

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 if

- BNL
- For every ANL, ALB.

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 the

halting states y and n are reversed decides ~L. spacereq(M' ) = spacereq(M) .

So, if L E dspace(f( n)), so is ~L.

REFERENCES

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/.