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The Easy Witness Lemma

50%

Overview

Easy witness for NEXP

If $\NTIME[2^n]\subseteq\SIZE[s(n)]$, then there is a constant $t\in\N$ such that for every $L\in\NTIME[2^n]$ and $x\in L$, there exists a witness $w$ for $x$ such that there exists a circuit of size $s(s(n^t)^t)^t$ whose truth table is $w$.

When $s(n)$ is polynomial (resp. quasi-polynomial), then the size of the witness circuit is also polynomial (resp. quasi-polynomial). In the following, we sketch the proof of the theorem. For the simplicity of presentation, we consider the case where $s(n)=n^k$ for some $k\in\N$.

The high-level idea of the proof is using unconditional circuit lower bound to derive contradiction. By assuming there is no easy witness for $\NTIME[2^n]$, we then know that the witness itself is a hard function since it cannot be efficiently encoded by a small circuit. As a result, by feeding this hard function into the Nisan-Wigderson generator, we can then derandomize the Merlin-Arthur protocol for a hard language (who has unconditional circuit lower bound) into a nondeterministic time algorithm. By applying the assumption that $\NTIME[2^n]$ has a small circuit again, we get a contradiction and thus the witness can be encoded into a small circuit.

The above explanation is a little bit sketchy and we explain the proof in more details as follows.

Unconditional circuit lower bound

Here, we are using the most naive unconditional circuit lower bound by diagonalization. That is, enumerating all the possible algorithms/machines and negate on one input length with each of them. Concretely, we have the following lemma.

For any $k\in\N$, there exists a language $L_{\text{hard}}\in\TIME[2^{n^{k+1}}]$ such that $L\not\in\SIZE[n^{O(k)}]$.

The proof is actually quite simple. It is easy to see that such language $L_{\text{hard}}$ exists in $\SPACE[n^{k+1}]$ as $\SIZE[n^{O(k)}]$ can be enumerated in $n^{k+1}$ space. The lemma then follows by $\SPACE[n^{k+1}]\subseteq\TIME[2^{n^{k+1}}]$.

As an important remark, note that this lower bound can be strengthen in the sense that there would be no size $n^{O(k)}$ circuit that computes $L_{\text{hard}}$ infinitely often.

Collapse theorem

Recall that the premise of the easy witness lemma is $\NTIME[2^n]\subseteq\SIZE[n^k]\subseteq\Ppoly$. By the collapse theorem of BFNW93 (see also this note), we then have $\EXP=\MA$. Moreover, if one dig into the proof carefully, the size of the Merlin-Arthur protocol only has a small blow-up. Specifically, for $L_{\text{hard}}$ described in the previous subsection, there exists a Merlin-Arthur protocol of size at most $n^{O(k^2)}.

Note that this is an exponential speedup since the $L_{\text{hard}}$ originally lies in $\TIME[n^{k+1}]$ and by the lemma it does not have circuit of size $n^{O(k)}$. If one can use the size $n^{O(k^2)}$ Merlin-Arthur protocol to construct a smaller circuit for $L_{\text{hard}}$, then we get the contradiction.

Derandomization

Now, we are going to derandomize the size $n^{O(k^2)}$ Merlin-Arthur protocol for $L_{\text{hard}}$ to circuit of size $n^{O(k)}$.

As we are proving the easy witness lemma by contradiction, we assume that there exists a language $L\in\NTIME[2^n]$ such that it does not have easy witness. Concretely, for any constant $c\geq1$, (i) there exists infinitely many input length $n$ such that (ii) there exists $x\in\bit^n$ and $x\in L$ such that (iii) for all witness $w$ for $x$, for any circuit computes $w$ has size at least $n^{ck^2}$. Let $n_1,n_2,\dots$, $x_1,x_2,\dots$, $w_1,w_2,\dots$ be the corresponding hard input lengths, inputs, and witnesses.

Next, apply the Nisan-Wigderson generator using $w_i$, we get a PRG with seed length $O(ck^2\log n)$ and running time $2^{O(n)}$ that fools circuir of size at most $n_i^{ck^2/g}$ for some $g\geq1$ independent to $c$. By picking $c$ large enough, we can have $n_i^{ck^2/g}$ larger than the size of the Merlin-Arthur protocol for $L_{\text{hard}}$. That is, the PRG fools the Arthur in the protocol on input length $n_i$. Note that the crucial part here is that the PRG only runs in time $2^{O(n)}$ instead of time $2^{n^{O(k)}}$.

Finally, we are going to design a nondeterministic algorithm running runs in time $2^{O(n)}$ for $L_{\text{hard}}$ infinitely often by using this PRG and simulating the Merlin-Arthur protocol. This can be easily done by first nondeterministically guessing $w_i$ for each $n_i$ and running the Merlin-Arthur protocol with the PRG. Note that the totoal running time is $2^{O(n)}\times2^{O(ck^2n)}\times n^{O(k^2)}=2^{O(n)}$ and the correctness follows from the correctness of the Merlin-Arthur protocol and the property of PRG.

Now that $L_{\text{hard}}\in\NTIME[2^{O(n)}]$ infinitely often, by the premise of the easy witness lemma, $L_{\text{hard}}\in\SIZE[O(n)^k]$ infinitely often. This contradicts to the unconditional circuit lower bound for $L_{\text{hard}}$ and thus we conclude that the lemma holds.

Easy witness for NP and NQP

Murray and Williams improved the easy witness lemma for $\NEXP$ to smaller nondeterministic time classes such as $\NP$ and $\NQP$. As a consequence, they got $\class{ACC}$ circuit lower bound against $\NP$ and $\NQP$. The proof structure of this fine-grained easy witness lemma is similar to the IKW one but several non-trivial ideas are required. In the following, we focus on $\NP$ for the simplicity of the presentation.

Recall that the high-level strategy of the proof of easy witness lemma is to contradict an unconditional circuit lower bound by derandomizing a Merlin-Arthur protocol using the non-existence of easy witness as hardness. The first issue that comes up would be small Merlin-Arthur protcol for languages in $\NP$. Note that the Merlin-Arthur protocol for $\NEXP$ is due to the collapse theorem while there is no known collapse theorem for lower classes such as $\NP$.

As a result, the contradiction of MW easy witness lemma is no longer from the unconditional lower bound for exponential time classes, instead, they directly used the $\MA/1$ lower bound by Santhanam (see also this note). However, one caveat here is that Santhanam’s lower bound does not hold for the infinitely many setting and thus the IKW type contradiction would not hold. (Recall that in the end of IKW’s proof, they got a circuit for $L_{\text{hard}}$ infinitely often).

To overcome this infinitely many versus all but finitely many issue, Murray and Williams came up with a stronger lower bound for $\MA$ with slightly more bits of advice. It turns out that such lower bound suffices for their purpose. In the following, we will first state this ‘Almost’ an almost-everywhere circuit lower bound for $\MA$ and then see how to use it to get easy witness lemma for $\NP$.

‘Almost’ an almost-everywhere circuit lower bound for $\MA$

For any $k\in\N$, there exists a language $L_k$ such that there is a Merlin-Arthur protocol with $O(\log n)$ bits of advice for $L_k$ that runs in $n^{O(k^2)}$ time and for all but finitely many $n$, either (i) there is no size $n^{O(k)}$ circuit for $L_k$ on inputs of length $n$ or (ii) there is no size $n^{O(k^2)}$ circuit for $L_k$ on inputs of length $n^{O(k)}$.

Proof of the easy witness lemma for NP

Now that we have an almost almost-everywhere lower bound for $\MA/O(\log n)$, let us prove the easy witness lemma for $\NP$. Similarly, for any $c\geq1$ large enough, let us assume there exists a language $L$ in $\NTIME[n^k]$ such that (i) there exists infinitely many input lengths $n_1,n_2,\dots$ such that (ii) there exists $x_i\in\bit^{n_i}$ and $x_i\in L$ such that (iii) there exists a witness $w_i$ for $x_i$ having circuit complexity at least $n^{ck^3}$.

With the same derandomization argument as IKW, we can then design a nondeterministic algorithm for $L_k$ that runs in time For each input length $n_i$, one can then derandomize a Merlin-Arthur protocol running in time $n^{O(k^2)}$