基本对称多项式的单调算术电路复杂度？

第 $k$ 个基本对称多项式 $S_k^n(x_1,\ldots,x_n)$ 是所有不同变量的乘积。我对该多项式的单调算术电路复杂度感兴趣。一个简单的动态编程算法（以及下面的图1）给出了一个具有门的电路。 $\binom{n}{k}$ $k$ $(+,\times)$ $(+,\times)$ $O(kn)$

问题： 是否知道的下限？ $\Omega(kn)$

甲电路是歪斜如果每个产品门的两个输入端的至少一个是可变的。这种电路实际上与开关和整流网络相同（有向无环图，其中的某些边缘用变量标记；每个st路径给出其标记的乘积，输出是所有st路径的总和）。早在40年前，马尔可夫就证明了一个令人吃惊的严格结果：的最小单调算术偏斜电路恰好具有乘积门。的上界如下从图1： $(+,\times)$ $S_k^n$ $k(n-k+1)$

但是我没有看到任何尝试证明非偏斜电路的下限。这仅仅是我们的“自大”，还是一路上观察到一些固有的困难？

PS我知道门对同时计算所有是必要的。这是从对0-1输入进行排序的单调布尔电路的大小的下限开始的；请参阅Ingo Wegener的书的第158页。所述AKS排序网络也意味着门在此（布尔值）的情况下就足够了。实际上，鲍尔（Baur）和斯特拉森（Strassen）已经证明了紧约束 $\Omega(n\log n)$ $S_1^n,\ldots,S_n^n$ $O(n\log n)$ $\Theta(n\log n)$ 的非单调算术电路的大小。但是单调算术电路呢？ $S_{n/2}^n$

— Stasys
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一个挑战是，如果您取消了“单调”限制，我们确实知道如何有效地计算这些东西。您可以使用基于FFT的多项式乘法来计算时间内所有（评估所有基本对称多项式）。因此，在单调电路模型中证明下界将需要证明 $S_0^n,\dots,S_n^n$ $n+1$ $O(n \log^2 n)$ $\Omega(nk)$ $\Omega(n^2)$ 多项式乘法的下界。

就是这样。引入形式未知数，并考虑多项式 $y$

P (y) = \prod_{i = 1}^{n} (1 + x_{i} y) .

$P(y) = \prod_{i=1}^n (1 + x_i y).$

Note that since the $x_i$ 's are known constants, this is a univariate polynomial with unknown $y$ and with degree $n$ . Now you can note that the coefficient of $y^k$ in $P(y)$ is exactly $S_k^n$ , so to evaluate all the $S_0^n,\dots,S_n^n$ , it suffices to compute $P(y)$ .

This makes it possible to compute $P(y)$ in $O(n \lg^2 n)$ time: build a balanced binary tree of polynomials with the $(1+x_i y)$ 's at the leaves, and multiply the polynomials. Multiplying two polynomials of degree $d$ takes $O(d \lg d)$ time using FFT techniques, so we get the recurrence $T(n) = 2 T(n/2) + O(n \lg n)$ , which solves to $T(n) = O(n \lg^2 n)$ . For convenience, I am ignoring $\text{poly}(\lg \lg n)$ factors.

If you care about the case where $k$ is very small, you can compute $S_0^n,\dots,S_k^n$ in $O(n \lg^2 k)$ time using similar tricks, keeping in mind that you only care about $P(x) \bmod y^{k+1}$ (i.e., throwing away all terms of $y^{k+1}$ or higher powers of $y$ ).

Of course, the FFT uses subtraction, so naively it's not expressible in a monotone circuit. I don't know whether there's some other way to multiply polynomials efficiently with monotone arithmetic circuits, but any efficient monotone method for polynomial multiplication immediately leads to an algorithm for your problem as well. So, lower bounds on your problem require/imply lower bounds for polynomial multiplication.

— D.W.
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D.W., thanks for recalling this construction! It is usually attributed to Ben-Or, and I should have mentioned it. The construction also gives a <i>formula</i> of size

O (n^{2})

$O(n^2)$ and depth only

3

$3$ (!) computing the operator

S_{0}^{n}, \dots, S_{n}^{n}

$S_0^n,\ldots,S_n^n$ (by evaluating

P (y)

$P(y)$ at some

n + 1

$n+1$ points). This was used to separate homogeneous and non-homogeneous small-depth formulas. But, as you mention, the construction substantially uses subtraction. So, my question asks: how "substantial" this use actually is? This could be interesting also in restricted-depth scenario.

— Stasys

@Stasys: I think subtraction is pretty crucial. Viz. the Nisan-Wigderson lower bound on depth 3 homogeneous circuits; in homogeneous depth 3 circuits, the point is that it's useless to compute terms whose degrees differ from the degree of the output. So this limits the kinds of cancellations that can happen. Whereas in the Ben-Or construction, to compute

S_{k}^{n}

$S_k^n$ , one needs to compute a polynomial of degree

n

$n$ (even though the output has degree

k < n

$k < n$ ), and then crucially use cancellation to get rid of the terms of degrees

> k

$> k$ . This isn't a proof, just some intuition...

— Joshua Grochow

@Joshua: yes, we know that the coefficients of the variable

y

$y$ in the polynomial

P (y, x)

$P(y,x)$ are exactly the polynomials

S_{k}^{n} (x)

$S_k^n(x)$ . But we need Gauss (and so - subtractions) to extract these coefficients from

n + 1

$n+1$ values of

P (y)

$P(y)$ on

n + 1

$n+1$ distinct points. My question asks whether the "monotone word" has no Gauss indeed, in this case. (With a guessed answer - NO.) I am not sure that for this, it is enough to get rid of terms of degrees

> k

$> k$ . We have to find these

k

$k$ first coefficients.

— Stasys