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Thursday, September 02, 2021

The hierarchy and GapP

There is a great but little-known theorem from the early 90's by Seinosuke Toda and Mitsunori Ogihara (buried as Lemma 2.3 in their paper) that shows the polynomial-time hierarchy is randomly low for Gap-P.

Let M be a non-deterministic polynomial-time machine. #M(x) is the number of accepting paths of M on x. #P is the set of functions f such that f(x) = #M(x) for an NP machine M. 

Gap-M(x) is the difference between the number of accepting and rejecting paths. Unlike #M(x), Gap-M(x) could be negative. Gap-P is the set of functions such that f(x)=Gap-M(x) for an NP machine M. Equivalently Gap-P is the difference of two #P functions. Gap-P inherits all the nice closure properties of #P while being closed under negation and subtraction.

Theorem (Toda-Ogihara): Let q be a polynomial and f be a function in Gap-PPH. There is a function g in Gap-P and a polynomial p such that for all x, if you choose a binary string r randomly of length p(|x|), f(x) = g(x,r) with probability at least 1-2-q(|x|).

In other words, with randomness the PH as an oracle of Gap-P disappears.

Let me show you how the proof works for a specific f in GapPPH, namely the indicator function for SAT: f(φ) = 1 if φ is satisfiable and 0 otherwise.

Recall Valiant-Vazirani showed how to randomly reduce a formula φ to a set of formula ψ1,…,ψk such that

  1. If φ is not satisfiable then for all i, ψi will not be satisfiable
  2. If φ is satisfiable then with high probability for some i, ψi will have exactly one satisfying assignment.
Picking k > 4|x|q(|x|) will reduce the error in (2) to under 2-q(|x|). Let g(φ,r) use r to choose ψ1,…,ψk and output the Gap-P function 1-∏(1-#ψi) using the closure properties of Gap-P where #ψi is the number of satisfying assignments to ψi.
  • If φ is not satisfiable then for all i, #ψi=0 and g(φ,r) = 0.
  • If φ is satisfiable then with high probability for some i, #ψi=1 and thus g(φ,r) = 1.
Note that if r is a bad random sequence and φ is satisfiable, g(φ,r) might be an exponentially large integer, positive or negative. It's right most of the time but when it's wrong it's terribly wrong.

5 comments:

  1. Does derandomization of Valiant Vazirani derandomize the full result or only this example?

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    1. Great question. The proof of the full Toda-Ogihara result requires a version of Valiant-Vazirani relative to the polynomial-time hierarchy and thus a stronger derandomization assumption. An assumption like DTIME(2^n) not in DSPACE(2^{o(n)}) will give you a PRG sufficient to fully derandomize Toda-Ogihara.

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  2. Can there be a converse such as deterministic Toda-Ogihara implies the time space separation? Or is it possible results such as EXP in PSPACE abd deterministic Toda-Ogihara are both simultaneously possible as far as we know?

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    Replies
    1. Probably not a full converse but it would be interesting to see if there is some circuit lower bound one could obtain from a deterministic version of Toda-Ogihara. I'll leave that as an open question.

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  3. While on the question of derandomizing [VV85], here is a fine question: Is there a Pseudo-deterministic RP reduction from SAT to USAT, i.e., the reduction can use random bits but reruns of the reduction should produce the same instance of USAT with very high probability.

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