Outline

• Time-Space Tradeoffs
• Quiz
• An Oracle approach to PH
• Circuits

Time-Space Tradeoffs for SAT

• Recall last Wednesday we were trying to prove: Theorem (Fortnow 1998). For every two functions `S, T: NN -> NN`, define `TISP(T(n), S(n))` to be the set of languages decided by a TM `M` that on every input `x` takes at most `O(T(|x|))` steps and uses at most `O(S(|x|))` cells of its read-write tapes. Then, `SAT !in TISP(n^(1.1), n^(0.1))`.
• We first showed it sufficed to prove `NTIME(n)` is not contained in `TISP(n^(1.2), n^(0.2))` via looking at Cook-Levin.
• Next we proved the following variant of Nepomnjasci's Theorem: `TISP(n^(12), n^(2)) subseteq Sigma_2TIME(n^8)`
• To finish this proof we need one more claim.

Speeding-up Nondeterminism

Claim. Suppose that `NTIME(n) subseteq DTIME(n^(1.2))`. Then `Sigma_2TIME(n^8) subseteq NTIME(n^(9.6))`.

Proof. Using the equivalence between alternating time and PH, `L` is in `Sigma_2TIME(n^8)` iff there is a TM `M` such that
`x in L iff exists u in {0, 1}^(c|x|^8)forall v in {0,1}^(d|x|^8) M(x, u, v) = 1`
for some constants c,d where `M` runs in time `O(|x|^8)`. Yet if `NTIME(n)subseteq DTIME(n^(1.2))`, then by a padding argument, we have a deterministic algorithm `D` that on inputs `x, u` with `|x| =n` and `|u| = cn^8` runs in `O((n^8)^(1.2)) = O(n^(9.6))`-time and returns 1 iff there exists some `v in {0,1}^(dn^8)` such that `M(x, u, v) = 0`. Thus,
`x in L iff exists u in {0,1}^(c|x|^8)D(x,u) = 0`. Q.E.D.

The two claims now show the theorem for suppose `NTIME(n) subseteq TISP(n^(1.2), n^(0.2))`. This assumption together with a padding argument shows `NTIME(n^(10)) subseteq TISP(n^(12), n^2)` which by Nepomnjascii's Theorem implies `NTIME(n^(10)) subseteq Sigma_2TIME(n^8)`. But the last claim then implies this is contained in `NTIME(n^(10)) subseteq NTIME(n^(9.6))` contradicting the nondeterministic time hierarchy.

Quiz

Which of the following is known to be true?

1. `TISP(n^(14), n^(2)) subseteq Sigma_2TIME(n^9)`.
2. It is possible that `NL=Sigma_i^p` and `PSPACE = Pi_i^p`.
3. `PSPACE stackrel(subset)(ne) AP`.

PH via Oracle Machines

Theorem. For every `i ge 2`, `Sigma_i^p = NP^(Sigma_(i-1)SAT)`, where the latter class denotes the set of languages decided by polynomial time NDTMs with access to an oracle for `Sigma_(i-1) SAT`.

Proof. We show the idea by showing `Sigma_2^p = NP^(SAT)`. Suppose `L in Sigma_2^p`. So there is some `M` and polynomial `q` such that
`x in L iff exists u_1 in {0,1}^(q(|x|))forall u_2 in {0,1}^(q(|x|))M(x, u_1, u_2) = 1`.
For every fixed `u_1` and `x`, the statement "for every `u_2, M(x, u_1, u_2) = 1`" is the negation of an NP-statement and hence its truth could be determined using an oracle for SAT. Thus an NDTM N given oracle access for SAT could decide `L`: On input `x`, it nondeterministiscally guesses `u_1` and then uses the oracle to decide if `forall u_2 in {0,1}^(q(|x|))M(x, u_1, u_2) = 1`. We see that `x in L` iff there exists some choice `u_1` that makes `N` accept.

On the other hand, suppose that `L` is decided by a p-time NDTM `N` with access to SAT. `N` makes at most polynomially many queries to SAT. Let `c_1, ... c_m` denote the transition function choices `N` makes, let `u_1`, .., `u_k` represent guesses of satisfying assignments queries that `N` makes, and let `a_1`, ... `a_m` be the answers. Then `x in L` iff
`exists c_1, ... c_m, a_1, ...a_k, u_1, ... u_k forall v_1, ..., v_k` such that
`N` accepts `x` using choices `c_1`...`c_m` and answers `a_1`, ..., `a_k` AND
`forall i in [k]` if `a_i = 1` then on query `phi_i(u_i) = 1` AND `forall i in [k]` if `a_i =0` then `phi_i(v_i) = 0`
So is `Sigma_p^2`.

This shows we can define `Sigma_i^p` as `NP^(Sigma_(i-1)^p)`.