A new central limit theorem and decomposition for Gaussian polynomials, with an application to deterministic approximate counting

Abstract

One of the main results of this paper is a new multidimensional central limit theorem (CLT) for multivariate polynomials under Gaussian inputs. Roughly speaking, the new CLT shows that any collection of Gaussian polynomials with small eigenvalues (suitably defined) must have a joint distribution which is close to a multidimensional Gaussian distribution. The new CLT is proved using tools from Malliavin calculus and Stein’s method. A second main result of the paper, which complements the new CLT, is a new decomposition theorem for low-degree multilinear polynomials over Gaussian inputs. Roughly speaking, this result shows that (up to some small error) any such polynomial is very close to a polynomial which can be decomposed into a bounded number of multilinear polynomials all of which have extremely small eigenvalues. An important feature of this decomposition theorem is the delicate control obtained between the number of polynomials in the decomposition versus their eigenvalues. As the main application of these results, we give a deterministic algorithm for approximately counting satisfying assignments of a degree-d polynomial threshold function (PTF) over the domain $\{-1,1\}^n$; this is a well-studied problem from theoretical computer science. More precisely, given as input a degree-d polynomial $p(x_1,\dots ,x_n)$ over ${{\mathbb {R}}}^n$ and a parameter $\epsilon > 0$, the algorithm approximates

$$\begin{aligned} \mathop {\mathbf{Pr}}_{x \sim \{-1,1\}^n}[p(x) \ge 0] \end{aligned}$$

to within an additive $\pm \epsilon$ in time $O_{d,\epsilon }(1)\cdot \mathrm {poly}(n^d)$. (Since it is NP-hard to determine whether the above probability is nonzero, any sort of efficient multiplicative approximation is almost certainly impossible even for randomized algorithms.) Note that the running time of the algorithm (as a function of $n^d$, the number of coefficients of a degree-d PTF) is a fixed polynomial. The fastest previous algorithm for this problem (Kane, CoRR, arXiv:1210.1280, 2012), based on constructions of unconditional pseudorandom generators for degree-d PTFs, runs in time $n^{O_{d,c}(1) \cdot \epsilon ^{-c}}$ for all $c > 0$.

Appendix: A dealing with non-multilinear polynomials

The decomposition procedure that we use relies heavily on the fact that the input polynomials $p_i$ are multilinear. To handle general (non-multilinear) degree-d polynomials, the first step of our algorithm is to transform them to (essentially) equivalent multilinear degree-d polynomials. This is accomplished by a simple procedure whose performance is described in the following theorem.⁴ Note that given Theorem 55, we assume that the polynomial p given as input in Theorem 3 is multilinear.

Theorem 55

There is a deterministic procedure Multilinearize with the following properties: The algorithm takes as input a (not necessarily multilinear) variance-1 degree-d polynomial p over ${{\mathbb {R}}}^n$ and an accuracy parameter $\delta > 0$. It runs in time $O_{d,\delta }(1)\cdot poly(n^d)$ and outputs a multilinear degree-d polynomial q over $R^{n'}$, with $n' \le O_{d,\delta }(1) \cdot n$, such that

$$\begin{aligned} \left| \mathbf{Pr}_{x \sim N(0,1)^n}[p(x) \ge 0] - \mathbf{Pr}_{x \sim N(0,1)^{n'}}[q(x) \ge 0] \right| \le O(\delta ). \end{aligned}$$

Proof

The procedure Multilinearize is given below. (Recall that a diagonal entry of a q-tensor $f = \sum _{i_1,\dots ,i_q=1} f(i_1, \dots , i_q)\cdot e_{i_1} \otimes \cdots \otimes e_{i_q}$ is a coefficient $f(i_1,\dots ,i_q)$ that has $i_a=i_b$ for some $a \ne b.$)

It is clear that all the tensors $g_j$ are multilinear, so by Remark 11 the polynomial q that the procedure Linearize outputs is multilinear. The main step in proving Theorem 55 is to bound the variance of $\tilde{q}-q$, so we will establish the following claim:

Claim 56

$\mathbf{Var}[\tilde{q}-q] \le {\frac{d}{K}} \cdot \mathbf{Var}[\tilde{q}].$

Proof

We first observe that

$$\begin{aligned} \mathbf{Var}[\tilde{q}-q] = \mathbf{Var}\left[ \sum _{j=1}^d (I_j(\tilde{f}_j) - I_j(g_j)) \right] = \sum _{j=1}^d \mathbf{E}\left[ (I_j(\tilde{f}_j) - I_j(g_j))^2\right] , \end{aligned}$$

(54)

where the first equality is because $g_0=f_0$ and the second is by Claim 12 and the fact that each $I_j(\tilde{f}_j), I_j(g_j)$ has mean 0 for $j \ge 1.$ Now fix a value $1 \le j \le d.$ Since each $g_j$ is obtained from $\tilde{f}_j$ by zeroing out diagonal elements, again using Claim 12 we see that

$$\begin{aligned} \mathbf{E}\left[ (I_j(\tilde{f}_j) - I_j(g_j))^2\right] = j! \cdot \Vert \tilde{f}_j - g_j\Vert ^2_F, \end{aligned}$$

(55)

where the squared Frobenius norm $\Vert \tilde{f}_j - g_j\Vert ^2_F$ equals the sum of squared entries of the tensor $\tilde{f}_j - g_j$. Now,observe that the entry $\alpha _{i_1,\dots ,i_j}=f_j(i_1, \ldots , i_j)$ of the tensor $f_j$ maps to the entry

$$\begin{aligned}&\alpha _{i_1, \ldots , i_j} \frac{(e_{i_1,1} + \cdots + e_{i_1, K})}{\sqrt{K}} \otimes \ldots \otimes \frac{(e_{i_j,1} + \cdots +e_{i_j, K})}{\sqrt{K}}\\&\quad = \alpha _{i_1, \ldots , i_j} \cdot \sum _{(\ell _1, \ldots , \ell _{{j}}) \in [K]^{{j}}} \frac{1}{K^{{j}/2}} \otimes _{{{a=1}}}^{{j}} e_{i_{{a}},\ell _{{a}}} \end{aligned}$$

when $\tilde{q}$ is constructed from p. Further observe that all $K^j$ outcomes of $\otimes _{{{a=1}}}^{{j}} e_{i_{{a}},\ell _{{a}}}$ are distinct. Since $g_j$ is obtained by zeroing out the diagonal entries of $\tilde{f}_j$, we get that

$$\begin{aligned} \Vert \tilde{f}_j - g_j \Vert _F^2 = \sum _{{(i_1, \ldots , i_j) \in [n]^j}} (\alpha _{i_1, \ldots , i_j})^2 \cdot \frac{1}{K^j} \cdot |{\mathcal {S}}_{K, j}| \end{aligned}$$

where the set ${\mathcal {S}}_{K,j} = \{{(\ell _1, \ldots , \ell _j) \in [K]^j : \ell _1, \ldots , \ell _j} \text { are not all distinct}\}$. It is easy to see that $|{\mathcal {S}}_{K, j}| \le (j^2 \cdot K^j)/{K}$, so we get

$$\begin{aligned} \Vert \tilde{f}_j - g_j \Vert _F^2 \le \sum _{{(i_1, \ldots , i_j) \in [n]^j}} (\alpha _{i_1, \ldots , i_j})^2 \cdot \frac{j^2}{{K}}. \end{aligned}$$

Returning to (54) and (55), this yields

$$\begin{aligned} \mathbf{Var}[q-\tilde{q}] \le \sum _{j=1}^d j! \cdot \sum _{{(i_1, \ldots , i_j) \in [n]^j}} (\alpha _{i_1, \ldots , i_j})^2 \cdot \frac{j^2}{{K}} \le \frac{d^2}{K} \cdot \left( \sum _{j=1}^d j! \cdot \sum _{{(i_1, \ldots , i_j) \in [n]^j}} (\alpha _{i_1, \ldots , i_j})^2\right) . \end{aligned}$$

Using Fact 13 and Claim 12, we see that

$$\begin{aligned} \mathbf{Var}[p] = \sum _{j=1}^d \mathbf{Var}[I_j(f_j)] = \sum _{j=1}^d \mathbf{E}[I_j(f_j)^2] = \sum _{j=1}^d j! \cdot \sum _{{(i_1, \ldots , i_j) \in [n]^j}} (\alpha _{i_1, \ldots , i_j})^2. \end{aligned}$$

It is easy to see that $\mathbf{Var}[\tilde{q}]=\mathbf{Var}[p]$, which equals 1 by assumption, so we have that $\mathbf{Var}[q-\tilde{q}]\le {\frac{d^2}{K}} \cdot \mathbf{Var}[\tilde{q}]$ as desired. $\square$

To finish the proof of Theorem 55, observe that by our choice of K we have $\mathbf{Var}[q-\tilde{q}] \le (\delta /d)^{3d} \cdot \mathbf{Var}[ \tilde{q}]$. Since $q-\tilde{q}$ has mean 0 and $\mathbf{Var}[\tilde{q}]=1$ we may apply Lemma 6, and we get that $|\mathbf{Pr}_{x \sim N(0,1)^{n'}}[q(x) \ge 0] - \mathbf{Pr}_{x \sim N(0,1)^{n'}}[\tilde{q}(x) \ge 0]| \le O(\delta ).$ The theorem follows by observing that the two distributions $p(x)_{x \sim N(0,1)^n}$ and $\tilde{q}(x)_{x \sim N(0,1)^{n'}}$ are identical. $\square$