1. New Nonuniform lower bounds for uniform classes
Lance Fortnow
Georgia Institute of Technology
Rahul Santhanam
University of Oxford

2. Spoilers For constants a, b with 1≤a<b Other results
NTIME(nb) is not contained in NTIME(Na) with N1/B bits of advice
Previously known with only o(log n) bits of advice
Other results
NTIME(NB) is not contained in i.o.-NTIME(NA) with sublinear nondeterminism
NP does not have NP-uniform nondeterministic circuits of size nk
RTIME(2n) not in RTIME(nc)/n1/2c

3. The Power Of More Hartmanis and Stearns 1965
A computer can do more given more time
A computer can do more given more memory

4. Deterministic Time HIerarchy
For A>B
DTIME(nA) is not contained in DTIME(NB)
On input 0n simulate Mn(0N) for NB steps and accept if reject and vice versa

5. An Existential crisis For A>B
NTIME(nA) is not contained in NTIME(NB)
On input 0n simulate Mn(0N) for NB steps and accept if reject and vice versa
Doesn’t work of machine has both accepting and rejecting paths

6. Nondeterministic Time Hierarchy
For any real r and s, 1 ≤ r < s
NTIME(nr) is strictly contained in NTIME(ns)
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 2011

7. Fortnow-Santhanam Define a NTM M that on input 1j01m0w
M(1j01m0)
M(1j01m00)
M(1j01m01)
Define a NTM M that on input 1j01m0w
if |w| ≤ t(j+m+2) then accept if both
Mj(1j01m0w0) and Mj(1j01m0w1) accept
if |w| > t(j+m+2) then accept if Mj(1j01m0) rejects on the path specified by w
Suppose M and Mj accept the same language
M(1j01m0) accepts if
M(1j01m00) and M(1j01m01) accept

8. Fortnow-Santhanam Define a NTM M that on input 1j01m0w
M(1j01m0)
M(1j01m00)
M(1j01m01)
M(1j01m000)
M(1j01m001)
M(1j01m010)
M(1j01m011)
Define a NTM M that on input 1j01m0w
if |w| ≤ t(j+m+2) then accept if both
Mj(1j01m0w0) and Mj(1j01m0w1) accept
if |w| > t(j+m+2) then accept if Mj(1j01m0) rejects on the path specified by w
Suppose M and Mj accept the same language
M(1j01m0) accepts if
M(1j01m000) M(1j01m001) M(1j01m010) M(1j01m011) accept

9. Fortnow-Santhanam Define a NTM M that on input 1j01m0w
M(1j01m0)
M(1j01m00)
M(1j01m01)
Define a NTM M that on input 1j01m0w
if |w| ≤ t(j+m+2) then accept if both
Mj(1j01m0w0) and Mj(1j01m0w1) accept
if |w| > t(j+m+2) then accept if Mj(1j01m0) rejects on the path specified by w
Suppose M and Mj accept the same language
M(1j01m0) accepts if
M(1j01m0000) M(1j01m0001) M(1j01m0010) M(1j01m0011)
M(1j01m0100) M(1j01m0101) M(1j01m0110) M(1j01m0111) accept
M(1j01m000)
M(1j01m001)
M(1j01m010)
M(1j01m011)
000
001
010
011
100
101
110
111

10. Fortnow-Santhanam Define a NTM M that on input 1j01m0w
M(1j01m0)
M(1j01m00)
M(1j01m01)
M(1j01m000)
M(1j01m001)
M(1j01m010)
Define a NTM M that on input 1j01m0w
if |w| ≤ t(j+m+2) then accept if both
Mj(1j01m0w0) and Mj(1j01m0w1) accept
if |w| > t(j+m+2) then accept if Mj(1j01m0) rejects on the path specified by w
Suppose M and Mj accept the same language
M(1j01m0) accepts if
M(1j01m0w) for all w
M(1j01m0) rejects on all paths w
Contradiction
M(1j01m011)
000
001
010
011
100
101
110
111 …
00…00
11..11
M(1j01m0)

11. Depth is length of the witness.
M(1j01m0)
M(1j01m00)
M(1j01m01)
M(1j01m000)
M(1j01m001)
M(1j01m010)
M(1j01m011)
000
001
010
011
100
101
110
111
Depth is length of the witness.
Previously exponential in witness size. …
00…00
11..11
M(1j01m0)

12. Old Nonuniform lower bounds for uniform classes
Goal: For contants a,b with 1≤a<b,
NTIME(Nb) is not contained in NTIME(Na) with N1/B bits of advice
How do we do this for DTIME?
DTIME(Nb) is not contained in DTIME(Na) with N bits of advice
Use input as advice
M(x) simulates M|x|(x) using advice x for |x|A steps and do the opposite

13. Do the math: can handle N1/b bits of advice for b>a
M(1j01m0)
M(1j01m00)
M(1j01m01)
M(1j01m000)
M(1j01m001)
M(1j01m010)
M(1j01m011)
000
001
010
011
100
101
110
111 …
00…00
11..11
M(1j01m0)
Theorem: NTIME(Nb) is not contained in NTIME(Na) with N1/B bits of advice for b>a
N is |1j01m0| = J+M+2
Need 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>a

14. Infinitely often hierarchies
For B>A there are languages in DTIME(NB) that differ from every language in DTIME(NA) for all but finitely many input lengths.
There is an oracle relative to which NEXP is in i.o.-NP
Buhrman-Fortnow-Santhanam 2009

15. Proof requires collapse at all these lengths
M(1j01m0)
M(1j01m00)
M(1j01m01)
M(1j01m000)
M(1j01m001)
M(1j01m010)
M(1j01m011)
000
001
010
011
100
101
110
111 …
00…00
11..11
M(1j01m0)
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 2W for W = witness size.
We can get NTIME i.o. hierarchy if we limit witness size to sublinear

16. RE not in RTIME(Nc)/n1/2c
Case 1: SAT in BPP/N1/2c
By trying all advice and self-reduction, SAT in RTIME(2N1/2c poly(n))
NTIME(n3c/2) is contained in RE
NTIME(n3c/2) is not contained in NTIME(nC)/n1/2c by main theorem
NTIME(nC)/n1/2c contains RTIME(nC)/n1/2c

17. RE not in RTIME(Nc)/n1/2c
Case 2: SAT not in BPP/N1/2c
SAT is in E is in RE
BPP/N1/2c contains RTIME(NC)/N1/2c

18. NP does not have NP-uniform nondeterministic circuits of size nk
Suppose NP does have NP-uniform nondeterministic circuits of size nk
We show every L in NP can be simulated in NTIME(N2K+2)/N1/4k
Proof uses padded version of L with advice describing circuit and census
Contradicts main theorem for NTIME(N4k)

19. 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 logspace or SAT not in deterministic linear time be far away?