New Nonuniform lower bounds for - PowerPoint Presentation

Slide1

New Nonuniform lower bounds for uniform classes

Lance Fortnow

Georgia Institute of Technology

Rahul Santhanam

University of OxfordSlide2

Spoilers

For constants a, b with 1≤a<b

NTIME(

n

b

) is not contained in NTIME(N

a

) with N

1/B

bits of advice

Previously known with only o(log n) bits of advice

Other results

NTIME(N

B

) is not contained in

i.o

.-NTIME(N

A

) with sublinear nondeterminism

NP does not have NP-uniform nondeterministic circuits of size

n

k

RTIME(2

n

) not in RTIME(

n

c

)/n

1/2cSlide3

The Power Of More

Hartmanis

and Stearns 1965

A computer can do more given more time

A computer can do more given more memorySlide4

Deterministic Time HIerarchy

For A>B

DTIME(

n

A

) is not contained in DTIME(N

B

)

On input 0

n

simulate

M

n

(0

N

) for N

B

steps and accept if reject and vice versaSlide5

An Existential crisis

For A>B

NTIME(

n

A

) is not contained in

NTIME(N

B

)

On input 0

n

simulate

M

n

(0

N

) for N

B

steps and accept

if

reject and vice versa

Doesn’t work of machine has both accepting and rejecting pathsSlide6

Nondeterministic Time Hierarchy

For any real r and s, 1 ≤ r < s

NTIME(n

r

) is strictly contained in

NTIME(n

s

)

Reduce to deterministic diagonalization

First proved by Steve Cook in 1972

Seiferas

-Fischer-Meyer 1978

if t(n+1) = o(u(n)) then NTIME(t(n)) is strictly contained in NTIME(u(n))

Simplified proof by

Zàk

1983

New proof by Fortnow and

Santhanam

2011Slide7

Fortnow-Santhanam

Define a NTM M that on input 1

j

01

m

0w

if |w| ≤ t(j+m+2) then accept if both

M

j

(1

j

01

m

0w0) and Mj(1j01m0w1) acceptif |w| > t(j+m+2) then accept ifMj(1j01m0) rejects on the path specified by wSuppose M and Mj accept the same languageM(1j01m0) accepts ifM(1j01m00) and M(1j01m01) accept

M(1j01m0)

M(1j01m00)

M(1j01m01)Slide8

Fortnow-Santhanam

Define a NTM M that on input 1

j

01

m

0w

if |w| ≤ t(j+m+2) then accept if both

M

j

(1

j

01

m

0w0) and Mj(1j01m0w1) acceptif |w| > t(j+m+2) then accept ifMj(1j01m0) rejects on the path specified by wSuppose M and Mj accept the same languageM(1j01m0) accepts ifM(1j01m000) M(1j01m001) M(1j

01m010) M(1j01m011) accept

M(1j01m0)M(1j

01m00)M(1j

01m01)

M(1

j

01

m

0

00

)

M(1

j

01m001)

M(1j01m010)

M(1j01m011)Slide9

Fortnow-Santhanam

Define a NTM M that on input 1

j

01

m

0w

if |w| ≤ t(j+m+2) then accept if both

M

j

(1

j

01

m

0w0) and Mj(1j01m0w1) acceptif |w| > t(j+m+2) then accept ifMj(1j01m0) rejects on the path specified by wSuppose M and Mj accept the same languageM(1j01m0) accepts ifM(1j01m0000) M(1j01m0001) M(1j

01m0010) M(1j01m0011) M(1

j01m0100) M(1j01m0101) M(1j01m0110) M(1

j01m0111) acceptM(1j01m0)

M(1j01

m

0

0

)

M(1

j

01

m

01)

M(1j

01m000)M(1j

01m001)M(1j

01m010)

M(1j01m

011)

000

010

001

1

00

011

110

101

111Slide10

Fortnow-Santhanam

Define a NTM M that on input 1

j

01

m

0w

if |w| ≤ t(j+m+2) then accept if both

M

j

(1

j

01

m

0w0) and Mj(1j01m0w1) acceptif |w| > t(j+m+2) then accept ifMj(1j01m0) rejects on the path specified by wSuppose M and Mj accept the same languageM(1j01m0) accepts ifM(1j01m0w) for all wM(1j01m

0) rejects on all paths wContradictionM(1j

01m0)M(1j01m0

0)M(1j01m0

1)

M(1

j

01

m

0

00

)

M(1

j

01m001)

M(1j01m010)

M(1j01m011)

000

010

001

1

00

011

110

101

111

00…00

11..11

…

M(1

j

01

m

0)Slide11

Depth is length of the witness.

Previously exponential in witness size.

M(1

j

01

m

0)

M(1

j

01

m

0

0

)M(1j01m01)

M(1j01m0

00)M(1j01m0

01)M(1j01m0

10)

M(1

j

01

m

0

11

)

000

010

001

1

00

011

110

101

111

00…00

11..11

…

M(1

j

01

m

0)Slide12

Old Nonuniform lower bounds for uniform classes

Goal:

For

contants

a,b

with 1≤

a<b,

NTIME(

N

b

) is not contained in NTIME(N

a) with N1/B bits of adviceHow do we do this for DTIME?DTIME(Nb) is not contained in DTIME(Na) with N bits of adviceUse input as adviceM(x) simulates M|x|(x) using advice x for |x|A steps and do the oppositeSlide13

Theorem:

NTIME(

N

b

) is not contained in NTIME(N

a

) with N

1/B

bits of

advice for b>a

N is |1

j

01

m0| = J+M+2Need to encode advice for all lengths up to N+W (W = witness size < NA)Do the math: can handle N1/b bits of advice for b>aM(1j01m0)M(1j01m00)M(1j01

m01)

M(1

j01m000)

M(1j01m0

01

)

M(1

j

01

m

0

10

)M(1j01m0

11)

000

010

001

1

00

011

110

101

111

00…00

11..11

…

M(1

j

01

m

0)Slide14

Infinitely often hierarchies

For B>A there are languages in DTIME(N

B

) that differ from every language in DTIME(N

A

) for

all

but finitely

many input lengths.

There is an oracle relative to which NEXP is in

i.o

.-NP

Buhrman

-Fortnow-Santhanam 2009Slide15

Proof requires collapse at all these lengths

get

i.o

. if we could embed the entire diagram in one input length

Size of diagram is roughly 2

W

for W = witness size.

We can get NTIME

i.o

. hierarchy if we limit witness size to sublinear

M(1

j

01

m0)M(1j01m00)M(1j01m01)

M(1

j01m000)

M(1j01m001)

M(1j01

m

0

10

)

M(1

j

01

m

011)

000

010

001

1

00

011

110

101

111

00…00

11..11

…

M(1

j

01

m

0)Slide16

RE not in RTIME(Nc)/n

1/2c

Case 1: SAT in BPP/N

1/2c

By trying all advice and self-reduction, SAT in RTIME(2

N

1/2c

poly(n))

NTIME(n

3c/2

) is contained in RE

NTIME(n

3c/2) is not contained in NTIME(nC)/n1/2c by main theoremNTIME(nC)/n1/2c contains RTIME(nC)/n1/2c Slide17

RE not in RTIME(Nc)/n

1/2c

Case 2: SAT not in BPP/N

1/2c

SAT is in E is in RE

BPP/N

1/2c

contains RTIME(N

C

)/N

1/2cSlide18

NP does not have NP-uniform nondeterministic circuits of size nk

Suppose

NP does

have

NP-uniform nondeterministic circuits of size

n

k

We show every L in NP can be simulated in NTIME(N

2K+2

)/N

1/4k

Proof uses padded version of L with advice describing circuit and census

Contradicts main theorem for NTIME(N

4k)Slide19

Conclusions

Tighter bounds, perhaps sublinear advice

Bounds for advice in

i.o

. setting

One in a series of paper showing separations for nondeterministic computation

Could SAT not in

l

ogspace

or SAT not in deterministic linear time be far away?