Results on ResourceBounded Measure

Harry Buhrman and Stephen Fenner and Lance Fortnow

1

Centrum vo or Wiskunde en Informatica

2

University of Southern Maine

3

CWI The University of Chicago

Abstract We construct an oracle relative to which NP has pmeasure

p

but D has measure in EXP This gives a strong relativized negative

answer to a question p osed by Lutz Lut Secondly we give strong

evidence that BPP is small We show that BPP has pmeasure unless

EXP MA and thus the p olynomialtime hierarchy collapses This con

trasts with the work of Regan et al RSC where it is shown that

Ppol y do es not have pmeasure if exp onentially strong pseudorandom

generators exist

Intro duction

Since the intro duction of resourceb ounded measure by Lutz Lut many re

searchers investigated the size measure of complexity classes in exp onential

time EXP A particular p oint of interest is the hypothesis that NP do es not

have pmeasure Recent results have shown that many reasonable conjectures

in computational complexity theory follow from the hyp othesis that NP is not

small ie NP and hence it seems to b e a plausible scientic hyp oth

p

esis LM Lut

P

He shows In Lut Lutz shows that if NP then BPP is low for

p

P

that this even follows from the seemingly weaker hyp othesis that

p

He asks whether the latter assumption is weaker or equivalent to NP

p

In this pap er we show that relative to some oracle the two assumptions are not

equivalent

p

We show a relativized world where D EXP whereas NP has no Pbi

immune sets This immediately implies via a result of Mayordomo Maya

p P

that in this relativized world NP has pmeasure and D and hence has

measure in EXP and thus do es not have pmeasure or even p measure

URL httpwwwcwinlcwip eopleHarryBuhrmanhtml Email

buhrmancwinl Partially supp orted by the Dutch foundation for scientic research

NWO by SION pro ject and by the Europ ean Union through Neuro

COLT ESPRIT Working Group Nr and HCM grant nr ERBPL

URL httpwwwcsusmmaineedufenner Email fennercsusmmaineedu

Partially supp orted by NSF grant CCR

URL httpwwwcsuchicagoedufortnow Email fortnowcsuchicagoedu Sup

p orted in part by NSF grant CCR the Dutch Foundation for Scientic

Research NWO and a Fulbright Scholar award

NP

This shows in a very strong way that relativized measure for NP and P

NP NP

dier NP whereas P Here P is the class of sets

p p

recognized by p olynomial time Turing machines that are allowed two queries to

NP

an NP oracle We show that our results cannot b e improved to P

Secondly we investigate the p ossibility that BPP do es not have pmeasure

Intuitively BPP is a feasible complexity class close to P and therefore it

should b e the case that BPP is small We give very strong evidence supp orting

this intuition We show that BPP unless EXP MA and thus the

p

p olynomialtime hierarchy collapses

Since BPP Ppol y our result contrasts with the one by Regan Sivakumar

and Cai RSC where it is shown that Ppol y unless exp onentially

p

strong pseudorandom generators do not exist

Preliminaries



We let f g and identify strings in with natural numb ers via the

usual binary representation We x N N to b e a standard enumeration of

all nondeterministic p olynomialtime oracle Turing machines NOTMs where

i

for each i and input of length n N runs in time n for all oracles All our

i

machines run using symb ols and blanks Fix a deterministic oracle TM M

A

p 

which accepts some standard complete language for EXP for all A

m

n

We may assume that M runs in time We let h i b e the standard pairing

 p

function and we note that x y hx y i for all x y A set is in D if it can

b e expressed as the dierence of two sets in NP

The notations R Q R and Q denote the real numb ers the rational num

b ers the p ositive real numb ers and the p ositive rational numb ers resp ectively

Resource Bounded Measure

Classical Leb esque measure is an unusable to ol in complexity classes As these

classes are all countable everything we dene in such a class has measure Yet

we might wish to have a notion of abundance and randomness in complexity

classes Lutz Lut Lut intro duced the notion of resource bounded measure

and gave a to ol to talk ab out these notions inside complexity classes



Denition A martingale d is a function from to R with the prop erty



that dw dw dw for every w



Denition A pmartingale is a martingale d Q that is p olynomial

time computable

Denition A martingale d succeeds on a language A if

lim sup d n

A

n7!1

1

We write S d fA j d succeeds on Ag

Denition Let X b e a class of languages

X has pmeasure X i there exists a pmartingale d such that

p

1

X S d

X has pmeasure X i X

p p

X has pmeasure in EXP X jEXP i X EXP

p p

X has pmeasure in EXP X jEXP i X EXP

p p

One often denes measure in EXP using p measure where the martingale

O (1)

log n

can use time All of our results also hold in this weaker mo del

NP

Measure of NP versus Measure of P

In this section we concentrate on the question p osed by Lutz Lut We show

NP

that relative to some oracle NP do es not imply that P We

p p

do this in a very strong way by constructing an oracle such that NP do es not

p

contain Pbiimmune sets and D EXP

Theorem There exists an oracle A such that relative to A NP has no

p

Pbiimmune sets and D EXP

p

Proof We will co de EXP into D on one side of the oracle and prevent Pbi

 

immunity on the other ie strings in fx j x g will b e used to co de

p  

EXP into D while strings in fx j x g will co de the information to

nd an innite subset of each NP set or its complement Some diagonalization

will also b e necessary to force certain NP computations

To mix co ding with diagonalization we employ a simplied version of the

NP

trick used to construct an oracle for P NEXP BT FF For each x we

A

reserve two p otential regionsleft and right in which to co de M x only one

of which will actually b e used To co de correctly in a region we must let exactly

one string in the region enter A We will co de in the left region unless we have

to diagonalize against some NP machine which may necessitate adding several

strings of the left region to A If this happ ens we scrap the left region and co de

in the right region but we can do this only if our diagonalization hasnt already

put strings of the right region into A



We now pro ceed with the formal treatment For every x with jxj n

and b we call s an x b left coding string resp ectively an x b right



coding string if s xy b resp ectively s xy b for some y of length

n We identify left and right with and resp ectively We build the oracle A

in stages each successive stage extending a nite p ortion of As characteristic



function If is some partial characteristic function N an oracle



machine and x then the computation N x is dened as usual except

that when N makes any query outside domain it is answered negatively As

is customary we regard as a set of ordered pairs If is another characteristic

function we write to mean that extends Finally dene the tower

of s function tn for n by

t

tn

tn

Stage

End Stage

Stage n

We are given Set

n n

Forcing an NP computation If n tk for any k then set

right if s for some x b left co ding string s with jxj n

d

n

left otherwise

and go to step Otherwise let n tk for some k hi j i If there exists

a minimal such that b oth

n

a N has an accepting path in which all queries are in domain and

i

b for no x with jxj n and no x b rightco ding string s do es s

i

n

then set f g and set d right note that is only dened on

n

i

i n

strings no longer than n Otherwise set f g and set d left

n

Preserving computations of M For all x of length n run M x and ex

tend with just enough s to cover all queries made by M x not in

domain

 

Coding computations of M For all x of length n let y b e the

lexicographically least string if one exists such that jy j n and neither

the x d co ding string nor the x d co ding string corresp onding to

n n

y is in domain If M accepts set fxy d g otherwise set

n

fxy d g

n

Set to b e extended with just enough s to cover all remaining x b d

n

co ding strings for all b d fleft rightg and x of length n

End Stage

S

Let A b e such that extends for all n x for any x

n A n A

n

 B

For any B dene the language L by

if either B contains an x rightco ding string or

B

B contains no x dco ding strings for any d fleft rightg

L x

otherwise

B

A A  B p

We now show that L x M x for all x and Clearly L coD

A A A

p p

EXP D hence coD

Pick an n large enough and x an input x of length n In Step of Stage

n n

n such a y must exist there are at most x b dco ding strings

queried by M on inputs of length n b ecause of the running time of M and

 2

log n log n

less than n n total strings queried by the N in Step of Stages

i

n

through n Thus there are less than x b dco ding strings in domain

at Step of Stage n

The fact that

A A

M x L x

is now easily seen rst we observe that no x b rightco ding string for any

b gets into A in Steps or of any stage Thus we have two cases

d left For any b and d fleft rightg the only x b dco ding string

n

that ever enters A do es so in Step of Stage n This unique string is an

A

x left co ding string if M x accepts and is otherwise an x left

co ding string thus is satised

d right Exactly one x b rightco ding string enters A It is an x right

n

A

co ding string i M x accepts Again is satised

A

A

It remains to show that NP has no P biimmune sets This will b e done if

A

A

we can show that for any L NP there exist P sets Q and R with Q innite

such that L Q R or at least the symmetric dierence of L Q and R is

A

nite Let L LN for some xed i Let

i

n

Q f j j n thi j i g

i

n n

R Q f j Ag

A

The sets Q and R are clearly in P Pick n thi j i for j large enough so that

n i A n

thi j i n and consider Step of Stage n If exists then N

i

i

n n n

accepts and A so R If no such exists then R To see that

A n

i i

N rejects we simply observe that d d d d left so

n n

n n

i

i

no x b rightco ding strings enter A in any of the stages n through n Therefore

n A

rejects A preserves our conditions on the nonexistence of and so N

i

Corollary There exists an oracle relative to which NP has pmeasure and

p

D EXP and thus has pmeasure in E and in EXP

We actually get something more from the construction ab ove relative to A

we have EXP NP coNP That is EXP can b e computed in NP coNP

with one bit of advice for strings of length n namely d On input x of length n

n

A

A

A

an NP machine accepting LM resp ectively LM simply checks if there

is some x d co ding string resp ectively some x d co ding string in A

n n

A natural question is whether Theorem and Corollary are tight It could

NP

still happ en that NP and P The next theorem discards

p p

this p ossibility

NP

Theorem If P then NP

p p

p

NP

Proof P implies that SAT is weakly complete for EXP Amb os

p

tt

p

Spies Mayordomo and Zheng ASMZ have shown that the weakly

tt

p

completeness notion coincides with weakly completeness for EXP Hence

m

p

SAT is weakly complete for EXP and thus NP

p

m

p

Corollary Relative to the oracle constructed in Theorem it holds that D

p NP

coD P

BPP likely has measure

In this section we investigate the consequences of BPP not having pmeasure

We will see that this is unlikely since it would collapse the p olynomialtime

hierarchy Hence we provide strong evidence that BPP

p

Theorem If BPP then EXP MA

p

p p p

Since MA BM EXP MA implies that PH

We use the following Theorem from Babai Fortnow Nisan and Wigder

son BFNW stating that if EXP MA then BPP can b e simulated in sub ex

p onential time for innitely many input lengths

Theorem BFNW If EXP MA then for al l L BPP and for al l

0 n n

there exists a set L DTIME such that for innitely many n L

0 n

L

We will see that if BPP can b e simulated in sub exp onential time for in

nitely many input lengths then it has pmeasure Taking this together with

Theorem yields that EXP MA implies that BPP which proves

p

Theorem

Theorem If for al l languages L BPP there exists an and a set

n 0 n 0 n

such that for innitely many n L L then L DTIME

BPP

p

Proof Sketch We will construct a martingale that succeeds on all sets in BPP

k

0

b e the machine that runs in time n for some xed k Let L BPP and let M

L

0

that runs in sub exp onential time and accepts L If we are b etting on strings of

n 0 n

0

to predict exactly length n such that L L then we can use M

L

n

the next bit and hence we win times The problem however is that we do

0

is going to b e correct We overcome this problem by not know for which n M

L

the following strategy

n

Assume that our initial capital is We reserve to b et against the strings

0

of length n using M to predict the next bit ie whether the next string of

L

0

length n is in L We b et everything won so far on the strings of length n to

0

At the last string of length n we set aside what if any we the outcome of M

L

have won b etting on the strings of length n

n

n 0 n n

Observe that if n is a length such that L L then we win

and this is greater than n So for innitely many n we add n to our capital and

hence the liminf of this martingale go es to innity

To make the construction work uniformly for all L BPP we simulate all

n n i

the DTIME machines with a single DTIME machine allo cating of

our initial capital to machine i see Lut Mayb

Acknowledgment

We thank Leen Torenvliet for comments on an earlier version and Dieter van

Melkeb eek for helpful discussions on the writeup of the pro of of Theorem

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P

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a

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