Outline

• Time-Space Tradeoffs
• In-Class Exercise

Time-Space Tradeoffs for SAT

• As an application of our notions of alternating time, we next prove lower bounds on the simultaneous space time requirements for any algorithm for SAT.

Theorem. 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))`.

Remark. Stronger results are known.

Proof.

We will show that `NTIME(n)` is not contained in `TISP(n^(1.2), n^(0.2))`. This implies the result for `SAT` by following the ideas of the proof of Cook-Levin, since an analysis of that proof yields a reduction from the task of deciding membership in an `NTIME(T(n))`-language to the task of deciding whether an `O(T(n) log T(n))`-sized formula is satisfiable, where output bit of this reduction can be computed in quasilinear time and polylogarithmic time and space. Hence, if `SAT in TISP(n^(1.1), n^(0.1))` then `NTIME(n) subseteq TISP(n^(1.1)polylog(n), n^(0.1)polylog(n))`.

More Proof of Time-Space Tradeoffs

Towards our proof, the next claim is a particular case of a general tradeoff result known as Nepomnjasci's Theorem.

Claim. `TISP(n^(12), n^(2)) subseteq Sigma_2TIME(n^8)`

Proof. Suppose that `L` is decided by a machine `M` using `n^(12)` times and `n^2` space. For every `x in {0,1}^star`, consider the configuration `G_(M,x)` of `M` on input `x`. Each configuration in this graph can be described by a string of length `O(n^2)` and `x` is in `L` iff then is a path of length `n^(12)` in this graph starting at `C_(s\t\a\r\t)` to an accepting configuration. There is such a path iff there exists `n^6` configurations `C_1 ... C_(n^6)` such that if we let `C_0 = C_(s\t\a\r\t)` then `C_(n^6)` is accepting and for every `i in [n^6]` the configuration `C_i` is computed from `C_(i-1)` within `n^6` steps. This later condition can be verified in `O(n^7)` time, giving an `O(n^8)`-time `Sigma_2`-TM for deciding membership in `L`.

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.

In-Class Exercise

Welcome to Non-Ad-Hoc-Use-the-Prerequisite Day^TM!

• Recall from CS154 the definition of DFA and PDA.
• By the way, the last time I taught PDAs, the inventor and patentor of the stack was still alive.
• This exercise involves working out the following three items:
  1. Show that any language recognized by a DFA is in logspace, `L`.
  2. Show the language of correctly matched parentheses is in `L`. i.e., strings like (()())(), but not like ()).
  3. Explain why it might be harder to determine if arbitrary languages recognized by PDAs are in `NL`. Give a condition on a PDA to that ensure that its language is in `NL`.

Post your write up to the Apr 12 In-Class Exercise Thread.

PH via Oracle Machines

Theorem. For every `i ge 2`, `Sigma_i^p = NP^(Sigma_(i-1)^p)` where `NP^(Sigma_(i-1)^p)` denote the class of languages recognized by NDTMs with access to a set in `Sigma_(i-1)^p`.

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)`.