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Revision #4 to TR18-003 | 10th January 2019 17:33

Proving that prBPP = prP is as hard as proving that “almost NP” is not contained in P/poly

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Revision #4
Authors: Roei Tell
Accepted on: 10th January 2019 17:33
Downloads: 66
Keywords: 


Abstract:

What circuit lower bounds are necessary in order to prove that $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$? We show that the recent breakthrough result of Murray and Williams (STOC 2018) can be used to show a dramatic strengthening of the previously-known answer to this question. Specifically, we show that if $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$, then $NTIME[n^{f(n)}]\not\subseteq\mathcal{P}/\mathrm{poly}$, for essentially any $f(n)=\omega(1)$.

We also prove a technical strengthening of this result. Specifically, we show that if $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$, then for essentially any $s:\mathbb{N}\rightarrow\mathbb{N}$ it holds that $NTIME[s^{O(1)}\circ s^{O(1)}]\not\subseteq SIZE[s]$. Moreover, we show that size-$s$ circuits fail to compute the ``hard'' function in any interval of length $\poly(s(\poly(n)))$. The proof of this result uses tools of Murray and Williams, but relies on a different proof strategy. (Their proof strategy yields three compositions of $s$ instead of two, and does not yield the guarantee of failure in any small interval.)

Lastly, we present an alternative proof of the main result, which only relies on a generalization of the well-known lower bound of Santhanam (SICOMP, 2009).



Changes to previous version:

Replaced Thm 4 with a reference to a stronger result of Kinne, van Melkebeek, and Shaltiel ('10).


Revision #3 to TR18-003 | 5th January 2019 11:45

Proving that prBPP=prP is as hard as "almost" proving that P \ne NP





Revision #3
Authors: Roei Tell
Accepted on: 5th January 2019 11:45
Downloads: 53
Keywords: 


Abstract:

What circuit lower bounds are necessary in order to prove that $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$? We show that the recent breakthrough result of Murray and Williams (STOC 2018) can be used to show a dramatic strengthening of the previously-known answer to this question. Specifically, we show that if $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$, then $NTIME[n^{f(n)}]\not\subseteq\mathcal{P}/\mathrm{poly}$, for essentially any $f(n)=\omega(1)$.

We also prove several technical strengthenings of this result, which use different proof strategies than the one employed by Murray and Williams. In particular:

1. We prove that if $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$, then for essentially any $s:\mathbb{N}\rightarrow\mathbb{N}$ it holds that $NTIME[s^{O(1)}\circ s^{O(1)}]\not\subseteq SIZE[s]$. Moreover, we show that the failure of size-$s$ circuits to compute the ``hard'' functions happens in any interval of length $\poly(s(\poly(n)))$. (Directly invoking the Murray-Williams tools yields three compositions of $s$ instead of two, and does not yield the guarantee of failure in any small interval.)

2. We prove that under the weaker hypothesis $\mathcal{BPP}=\mathcal{P}$ (i.e., without the promise), one of the following holds: Either for essentially any $s:\mathbb{N}\rightarrow\mathbb{N}$ it holds that $NTIME[s^{O(1)}\circ s^{O(1)}]\not\subseteq SIZE[s]$, or the permanent of $\{0,1\}$-matrices over $\mathbb{Z}$ does not have polynomial-sized arithmetic circuits. This uses a well-known idea of Kabanets and Impagliazzo (2004).

Lastly, we present an alternative proof of the main result, which only relies on a generalization of the well-known lower bound of Santhanam (SICOMP, 2009).



Changes to previous version:

A significant revision of the paper, which includes several additional results, a revised exposition, and a minor change in the title.


Revision #2 to TR18-003 | 28th January 2018 15:50

Proving that prBPP=prP is as hard as "almost" proving that P \ne NP





Revision #2
Authors: Roei Tell
Accepted on: 28th January 2018 15:50
Downloads: 258
Keywords: 


Abstract:

What circuit lower bounds are necessary in order to prove that $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$? The main result in this paper is that if $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$, then polynomial-sized circuits cannot simulate non-deterministic machines that run in arbitrarily small super-polynomial time (i.e., $NTIME[n^{f(n)}]\not\subseteq\mathcal{P}/\mathrm{poly}$, for essentially any $f(n)=\omega(1)$). The super-polynomial time bound in the conclusion of the foregoing conditional statement cannot be improved (to conclude that $\mathcal{NP}\not\subseteq\mathcal{P}/\mathrm{poly}$) without unconditionally proving that $\mathcal{P}\ne\mathcal{NP}$.

This paper is a direct follow-up to the very recent breakthrough of Murray and Williams (ECCC, 2017), in which they proved a new ``easy witness lemma'' for $NTIME[o(2^n)]$. Our main contribution is in highlighting the strong ``barriers'' for proving $pr\mathcal{BPP}=pr\mathcal{P}$ that can be demonstrated using their results (and, as it turns out, also using previous results). We include three proofs of the main theorem: Two proofs that rely on various results from the work of Murray and Williams, and yield stronger forms of the main theorem (i.e., either use a weaker hypothesis or deduce a stronger conclusion); and a third proof that only relies on a generalization of the well-known lower bound of Santhanam (SICOMP, 2009).



Changes to previous version:

* Added a third proof of the main theorem, which actually deduces a stronger conclusion.
* The section that discusses the meaning of the new barrier for proving prBPP=prP now appears as Sec 1.3.


Revision #1 to TR18-003 | 15th January 2018 19:32

Proving that prBPP=prP is as hard as "almost" proving that P \ne NP





Revision #1
Authors: Roei Tell
Accepted on: 15th January 2018 19:32
Downloads: 217
Keywords: 


Abstract:

The main result in this paper is that any proof that $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$ necessitates proving that polynomial-sized circuits cannot simulate non-deterministic machines that run in arbitrarily small super-polynomial time (i.e., $NTIME[n^{f(n)}]\not\subseteq\mathcal{P}/\mathrm{poly}$, for essentially any $f(n)=\omega(1)$). The super-polynomial time bound in the conclusion of the foregoing conditional statement cannot be improved (to conclude that $\mathcal{NP}\not\subseteq\mathcal{P}/\mathrm{poly}$) without unconditionally proving that $\mathcal{P}\ne\mathcal{NP}$.

This paper is a direct follow-up to the very recent breakthrough of Murray and Williams (ECCC, 2017), in which they proved a new ``easy witness lemma'' for $NTIME[o(2^n)]$. Our main contribution is conceptual, in highlighting the strong ``barriers'' for proving $pr\mathcal{BPP}=pr\mathcal{P}$ (and even for proving much weaker statements) that can be demonstrated using their techniques. Following their approach, we apply the new lemma within the celebrated proof strategy of Williams (SICOMP, 2013), and derive the main result by using a parameter setting that is different than the ones they considered. We also include an alternative proof of the main theorem, which does not rely on the work of Murray and Williams, but rather uses a modification of the well-known lower bound of Santhanam (SICOMP, 2009).



Changes to previous version:

* Added Sec 1.1 to discuss the meaning of the new barrier for proving prBPP=prP.
* Added a new and alternative proof of the main theorem (i.e., Thm 1).
* Minor revisions to the high-level exposition.


Paper:

TR18-003 | 2nd January 2018 18:50

Proving that prBPP=prP is as hard as "almost" proving that P \ne NP





TR18-003
Authors: Roei Tell
Publication: 2nd January 2018 18:55
Downloads: 699
Keywords: 


Abstract:

We show that any proof that $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$ necessitates proving circuit lower bounds that almost yield that $\mathcal{P}\ne\mathcal{NP}$. More accurately, we show that if $promise\textrm{-}\mathcal{BPP}=promise\textrm{-}\mathcal{P}$, then for essentially any super-constant function $f(n)=\omega(1)$ it holds that $NTIME[n^{f(n)}]\not\subseteq\mathcal{P}/\mathrm{poly}$. The conclusion of the foregoing conditional statement cannot be improved (to conclude that $\mathcal{NP}\not\subseteq\mathcal{P}/\mathrm{poly}$) without \emph{unconditionally} proving that $\mathcal{P}\ne\mathcal{NP}$.

This paper is a direct follow-up to the very recent breakthrough of Murray and Williams (ECCC, 2017), in which they proved a new ``easy witness lemma'' for $NTIME[o(2^n)]$. Following their approach, we apply the new lemma within the celebrated proof strategy of Williams (SICOMP, 2013), and derive our result by using a parameter setting that is different than the ones they considered.



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