# P/Poly vs Uniform Complexity Classes

It is not known whether NEXP is contained in P/poly. Indeed proving that NEXP is not in P/poly would have some applications in derandomization.

1. What is the smallest uniform class C for which one can prove that C is not contained in P/poly?

2. Would showing that co-NEXP is not contained in P/poly have some other complexity theoretic consequences as in the case NEXP vs P/poly?

Note: I'm aware that $SP_2$ is known not to be contained in $Size[n^k]$ for each fixed constant $k$ (This was also shown for MA with 1 bit of advice). But in this question I'm not interested in results for fixed $k$. I'm really interested in classes which are different from P/Poly, even if these classes are very large.

• You are essentially asking for a problem with superpolynomial size lower bounds for general circuits. – Kaveh Jun 19 '16 at 22:39
• $\mathsf{MA}_\mathrm{exp}$ is known to not be in $\mathsf{P/poly}$. See the Wikipedia article for a short proof. – Robin Kothari Jun 19 '16 at 22:46
• P/poly is closed under complement, so it contains NEXP if and only if it contains coNEXP. – Emil Jeřábek Jun 20 '16 at 8:11
• Emil, Robin and Andrew, thanks for your answers. I think my question can be considered to be answered now. Would somebody write it in an answer so that I can accept it? – Springberg Jun 20 '16 at 15:35
• I believe that $MA_{exp}$ is the smallest uniform class with known superpolynomial lower bounds (people.cs.uchicago.edu/~fortnow/papers/nonrel.pdf), and that $O^P_2$ is the smallest one with arbitrary polynomial lower bounds (citeseerx.ist.psu.edu/viewdoc/…). – Alex Golovnev Jun 21 '16 at 18:33

$\let\mr\mathrm$There are several results in the literature stating that a certain class $C$ satisfies $C\nsubseteq\mr{SIZE}(n^k)$ for any $k$, and usually it is straightforward to pad them to show that any barely superpolynomially expanded version of $C$ is not in $\mr{P/poly}$.

Let me say that $f\colon\mathbb N\to\mathbb N$ is a superpolynomial bound if it is time-constructible, and $f(n)=n^{\omega(1)}$. For example, $n^{\log\log\log\log n}$ is a superpolynomial bound. In fact, an instructive exercise shows that if $g(n)$ is any unbounded monotone computable function, there is a superpolynomial bound $f$ such that $f(n)\le n^{g(n)}$.

First, direct diagonalization shows that $\Sigma_4^P\nsubseteq\mr{SIZE}(n^k)$ for any $k$. The same argument gives:

• If $f$ is any superpolynomial bound, then $\Sigma_4\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.

Proof sketch: For any $n$, let $C_n$ be the lexicographically first circuit of size $2f(n)$ that computes a Boolean function in $n$ variables not computable by a circuit of size $<f(n)$. Then, the language $L$ defined by $x\in L\iff C_{|x|}(x)=1$ works.

A well-known improvement states that $S_2\mr P\nsubseteq\mr{SIZE}(n^k)$ for any $k$. Likewise,

• If $f$ is any superpolynomial bound, then $S_2\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.

Proof sketch: If not, then in particular $\mr{NP}\subseteq S_2\mr P\subseteq\mr{P/poly}$, hence $\mr{PH}=S_2\mr P$. By a padding argument, $\Sigma_4\text-\mr{TIME}(f(n))\subseteq S_2\text-\mr{TIME}(f(n))\subseteq\mr{P/poly}$, quod non.

Oblivious classes do even better. Taking into account the objection raised by Apoorva Bhagwat, let $\mr{NLin=NTIME}(n)$. Then $\mr{NLin}\cup O_2\mr P\nsubseteq\mr{SIZE}(n^k)$ for any $k$, and the same argument yields:

• If $f$ is any superpolynomial bound, then $\mr{NLin}\cup O_2\text-\mr{TIME}(f(n))\nsubseteq\mr{P/poly}$.

Proof sketch: If $\mr{NLin\subseteq P/poly}$, then by padding, $\mr{NP\subseteq P/poly}$, which implies $\mr{PH}=O_2\mr P$. Then we proceed as before.

There are also results involving MA. The often mentioned result that $\mr{MA\text-EXP\nsubseteq P/poly}$ is an overkill. Santhanam proved $$\mr{promise\text-MA\cap promise\text-coMA\nsubseteq SIZE}(n^k)$$ for any $k$, and a similar argument gives:

• If $f$ is any superpolynomial bound, then $$\mr{promise\text-MA\text-TIME}(f(n))\cap\mr{promise\text-coMA\text-TIME}(f(n))\nsubseteq\mr{P/poly}.$$

Proof sketch: By Santhanam’s Lemma 11 (which is a sharpened version of the standard fact that $\mr{PSPACE=IP}$ with a PSPACE prover), there is a PSPACE-complete language $L$ and a randomized poly-time oracle TM $M$ such that on input $x$, $M$ only asks oracle queries of length $|x|$; if $x\in L$, then $M^L(x)$ accepts with probability $1$; and if $x\notin L$, then for any oracle $A$, $M^A(x)$ accepts with probability $\le1/2$.

For a suitable monotone polynomial $p$, let $A=(A_{\mr{YES}},A_{\mr{NO}})$ be the promise problem defined by \begin{align} (x,s)\in A_{\mr{YES}}&\iff\exists\text{circuit }C\,\bigl(p(|C|+|x|)\le f(|s|)\land\Pr[M^C(x)\text{ accepts}]=1\bigr),\\ (x,s)\in A_{\rlap{\mr{NO}}\phantom{YES}}&\iff\forall\text{circuit }C\,\bigl(p(|C|+|x|)\le f(|s|)\to\Pr[M^C(x)\text{ accepts}]\le1/2\bigr). \end{align} Let $h(x)$ be a polynomial reduction of $L$ to its complement, and let $B=(B_{\mr{YES}},B_{\mr{NO}})$ be the promise problem \begin{align} (x,s)\in B_{\mr{YES}}&\iff(x,s)\in A_{\mr{YES}}\land(h(x),s)\in A_{\mr{NO}},\\ (x,s)\in B_{\rlap{\mr{NO}}\phantom{YES}}&\iff(x,s)\in A_{\mr{NO}}\land(h(x),s)\in A_{\mr{YES}}. \end{align} If $p(n)$ is chosen suitably large, $$B\in\mr{promise\text-MA\text-TIME}(f(n))\cap\mr{promise\text-coMA\text-TIME}(f(n)).$$ So, let us assume for contradiction that $B$ has polynomial-size circuits, say, $B\in\mr{SIZE}(n^k)$. Let $s(n)$ denote the size of the smallest circuit computing $L$ on inputs of length $n$, and put $t(n)=f^{-1}(p(s(n)))$; more precisely, $$t(n)=\min\{m:p(s(n))\le f(m)\}.$$ Then $x\mapsto(x,1^{t(n)})$ is a reduction of $L$ to $B$, thus $L\in\mr{SIZE}(t(n)^k)$, which means $$s(n)\le t(n)^k.$$ But since $f$ is superpolynomial, we have $t(n)=s(n)^{o(1)}$. This gives a contradiction for $n$ sufficiently large.

If we prefer a result with a non-promise version of MA, Miltersen, Vinodchandran, and Watanabe proved $$\mr{MA\text-TIME}(f(n))\cap\mr{coMA\text-TIME}(f(n))\nsubseteq\mr{P/poly}$$ for a half-exponential function $f$. We can improve it in two ways: first, it holds for $\tfrac1k$-exponential bounds for any constant $k$, and second, it holds for oblivious classes. Here, a $\tfrac1k$-exponential function is, roughly speaking, a function $f$ such that $\underbrace{f\circ\dots\circ f}_k=\exp$. See the Miltersen–Vinodchandran–Watanabe paper and references therein for the precise definition; it involves a well-behaved family of well-behaved functions $e_\alpha(x)$, $\alpha\in\mathbb R_+$, such that $e_0(x)=x$, $e_1(x)=e^x-1$, and $e_{\alpha+\beta}=e_\alpha\circ e_\beta$. Also, if $f(n)\le e_\alpha(\mr{poly}(n))$ and $g(n)\le e_\beta(\mr{poly}(n))$, then $f(g(n))\le e_{\alpha+\beta}(\mr{poly}(n))$. Then we have:

• $\mr{OMA\text-TIME}(e_\alpha)\cap\mr{coOMA\text-TIME}(e_\alpha)\nsubseteq\mr{P/poly}$ for any $\alpha>0$.

Proof sketch: Assume otherwise. Fix an integer $k$ such that $1/k<\alpha$. Let me abbreviate $$\mr{OcOMT}(f)=\mr{OMA\text-TIME}(\mr{poly}(f(\mr{poly}(n)))\cap\mr{coOMA\text-TIME}(\mr{poly}(f(\mr{poly}(n))).$$ By padding, we have $$\tag{1}\mr{OcOMT}(e_{\beta+1/k})\subseteq\mr{SIZE}(e_\beta(\mr{poly}(n)))$$ for any $\beta\ge0$. Moreover, using e.g. Santhanam’s Lemma 11 above, we have the implication $$\tag{2}\mr{PSPACE\subseteq SIZE}(e_\beta(\mr{poly}(n)))\implies\mr{PSPACE\subseteq OcOMT}(e_\beta).$$ Since trivially $\mr{PSPACE\subseteq OcOMT}(e_1)$, a repeated application of (1) and (2) shows $\mr{PSPACE\subseteq SIZE}(e_{(k-1)/k}(\mr{poly}(n)))$, $\mr{PSPACE\subseteq OcOMT}(e_{(k-1)/k})$, $\mr{PSPACE\subseteq SIZE}(e_{(k-2)/k}(\mr{poly}(n)))$, $\mr{PSPACE\subseteq OcOMT}(e_{(k-2)/k})$, and so on. After $k$ steps, we reach $$\mr{PSPACE\subseteq P/poly}\qquad\text{and}\qquad\mr{PSPACE=OMA\cap coOMA}.$$ Using padding once more, we get $$\mr{DSPACE}(e_{1/k})\subseteq\mr{OcOMT}(e_{1/k})\subseteq\mr{P/poly},$$ which contradicts the results above, as $e_{1/k}$ is a superpolynomial bound.

Since nobody posted an answer, I will answer the question myself with the comments posted in the original question. Thanks to Robin Kothari, Emil Jerabek, Andrew Morgan and Alex Golovnev.

$MA_{exp}$ seems to be the smallest uniform class with known superpolynomial lower bounds.

$O_2^P$ seems to be the smallest known class not having circuits of size $n^k$ for each fixed $k$.

By diagonalization, it follows that for any super-polynomial (and space-constructible) function $s$, $DSPACE[s(n)]$ doesn't have polynomial-size circuits. $PSPACE$ versus $P/poly$ is still open.

$P/poly$ is closed under complement, so it contains $NEXP$ if and only if it contains $coNEXP$.

Please correct me if I'm wrong, but as far as I can tell, we actually don't know a fixed-polynomial size lower bound for $O_2^P$. This is because the usual Karp-Lipton argument doesn't go through for $O_2^P$, since we don't know whether $\textsf{NP}\subseteq O_2^P$ (in fact, this is equivalent to asking whether $\textsf{NP}\subseteq \textsf{P/poly}$). However, we do know that $\textsf{NP}^{O_2^P}$ isn't contained in $\textsf{SIZE}(n^k)$ for any $k$, as shown by Chakaravarthy and Roy.