确定性通信复杂度与分区数

背景：

考虑其中Alice和Bob给出的通信复杂度通常的两方模型$n$比特串$x$和$y$和具有计算一些布尔函数$f(x,y)$，其中$f:\{0,1\}^n \times \{0,1\}^n \to \{0,1\}$。

我们定义以下数量：

$D(f)$（的确定性通信复杂$f$）：比特Alice和Bob需要对计算通信的最小数量$f(x,y)$确定性。

$Pn(f)$（的分区编号$f$）：对数的分区中的单色矩形的最小数目（或不相交的覆盖物）的（基数为2）$\{0,1\}^n \times \{0,1\}^n$。

在单色矩形$\{0,1\}^n \times \{0,1\}^n$是一个子集$R \times C$使得$f$取相同的值（即，单色）上的所有元素$R \times C$。

还要注意，分区号与“ protocol分区号”不同，后者是此问题的主题。

有关更多信息，请参见Kushilevitz和Nisan的文章。用它们的符号，我定义为$Pn(f)$是$\log_2 C^D(f)$。

注意：这些定义容易推广到非布尔函数$f$，其中的输出$f$是一些较大的集合。

已知结果：

已知的是，是一个下界d （˚F ），即，对于所有（布尔或非布尔）˚F，P Ñ （˚F ）≤ d （˚F ）。确实，D （f ）的大多数下界技术（或全部？）实际上是下界P n （f ）。（任何人都可以确认所有下限技术都是如此吗？）$Pn(f)$$D(f)$$f$$Pn(f) \leq D(f)$$D(f)$$Pn(f)$

还已知该边界至多是平方松散的（对于布尔或非布尔函数），即。总而言之，我们知道以下几点：$D(f) \leq (Pn(f))^2$

$Pn(f) \leq D(f) \leq (Pn(f))^2$

据推测，。（这是Kushilevitz和Nisan撰写的文本中的开放式问题2.10。）据我所知，这两个布尔函数之间最广为人知的分隔仅是乘数为2，如“通信复杂性中的线性阵列猜想是错误的”，作者：Eyal Kushilevitz，Nathan Linial和Rafail Ostrovsky。$Pn(f) = \Theta(D(f))$

更准确地说，它们表现出布尔函数的无限族，使得D （f ）≥ （2 - o （1 ））P n （f ）。$f$$D(f) \geq (2 - o(1)) Pn(f)$

题：

对于非布尔函数，和D （f ）之间最著名的分隔是什么？还是上面提到的因子2分离？$Pn(f)$$D(f)$

第2版​​中的新增功能：由于一周内没有收到答案，因此我也很高兴听到部分答案，猜想，传闻和传闻证据等。

这个问题刚刚解决！正如我提到的，众所周知

，$Pn(f) \leq D(f) \leq (Pn(f))^2$

但是证明或存在一个P n （f ）= o （D （f ））的函数是一个主要的开放问题。$Pn(f) = \Theta(D(f))$$Pn(f) = o(D(f))$

几天前，MikaGöös，Toniann Pitassi，Thomas Watson解决了该问题（http://eccc.hpi-web.de/report/2015/050/）。他们表明存在一个函数$f$其满足

。$Pn(f) = \tilde{O}((D(f))^{2/3})$

它们还显示了的单面版本的最佳结果，我将用P n 1（f ）表示，其中您只需要用矩形覆盖1输入。P n 1（f ）也满足 $Pn(f)$$Pn_1(f)$$Pn_1(f)$

，$Pn_1(f) \leq D(f) \leq (Pn_1(f))^2$

and they show that this is the best possible relation between the two measures, since they exhibit a function $f$ which satisfies

$Pn_1(f) = \tilde{O}((D(f))^{1/2})$.

您指出，的下限与所有现有的下限技术密切相关。对于布尔函数，只要对数秩猜想成立，这似乎是正确的。但是，P n （f ）可以指数地大于愚弄集的边界。$Pn(f)$$Pn(f)$

我不清楚在非布尔情况下和D （f ）可以相差多少。$Pn(f)$$D(f)$

在其余部分中，我将使这些评论更加精确。

KN (Kushilevitz and Nisan in their 1997 textbook) outline the three basic techniques for Boolean functions: size of a fooling set, size of a monochromatic rectangle, and rank of the communication matrix.

First, fooling sets. A fooling set $S$ is monochromatic: there is some $z \in \{0,1\}$ such that $f(x,y) = z$ for every $(x,y)\in S$. Some final patching is then needed to take into account the other colour. This extra step can be avoided. Let $f \colon X \times Y \to \{0,1\}$ be a function. A pair of distinct elements $(x_1,y_1),(x_2,y_2) \in X \times Y$ is weakly fooling for $f$ if $f(x_1,y_1) = f(x_2,y_2)$ implies that either $f(x_1,y_2) \ne f(x_1,y_1)$ or $f(x_2,y_1) \ne f(x_1,y_1)$. A set $S \subseteq X \times Y$ is a weak fooling set for $f$ if every distinct pair of elements of $S$ is weakly fooling. KN implicitly state after the proof of 1.20 that the log-size of a weak fooling set is a lower bound for the communication complexity.

A largest weak fooling set picks a representative element from each monochromatic rectangle in a smallest disjoint set cover. The size of a largest weak fooling set is therefore at most as large as (the exponent of) the partition number. Unfortunately the bound provided by fooling sets is often weak. The proof of KN 1.20 shows that any function mapping each element $s$ of a weak fooling set $S$ to a monochromatic rectangle $R_s$ containing that element is injective. However, there can be many monochromatic rectangles $R$ in a smallest disjoint cover that do not appear in the image of $S$, with every element of $R$ weakly fooling with some but not all of the elements of $S$$S$$n$$1/4$ of all Boolean functions on $n$ variables have $Pn(f) = n$ yet have (weak) fooling sets of log-size $O(\log n)$. So the log of the size of a largest (weak) fooling set can be exponentially smaller than the communication complexity.

For rank, establishing a close correspondence between the rank of the matrix of the function and its partition number would establish a form of the log-rank conjecture (depending on the tightness of the correspondence). For instance, if there is a constant $a> 0$ such that $Pn(f) \le a\log rk(f)$ for every Boolean function $f$, then $D(f) \le (2a\log rk(f))^2$, and a kind of log-rank conjecture then holds for families of functions for which $rk(f)$ ultimately increases with $|X|+|Y|$, with exponent $2+\epsilon$ for any $\epsilon > 0$ achievable for sufficiently large $|X|+|Y|$. (Recall that the Lovász-Saks log-rank conjecture says that there is a constant $c>0$ such that $D(f) \le (\log rk(f))^c$ for every Boolean function $f$; here $rk(f)$ is the rank of the communication matrix of $f$ over the reals.)

Similarly, if there is only one quite large monochromatic rectangle together with many small ones, then the partition number gives a stronger bound than the log-size of a largest monochromatic rectangle. However, the log-rank conjecture is also equivalent to a conjecture about the size of a largest monochromatic rectangle (Nisan and Wigderson 1995, doi:10.1007/BF01192527, Theorem 2). So using monochromatic rectangles is not currently known to be "the same as" using the partition number, but they are closely related if the log-rank conjecture holds.

In summary, the log-size of a largest weak fooling set may be exponentially smaller than the partition number. There may be gaps between the other lower bound techniques and the partition number, but if the log-rank conjecture holds then these gaps are small.

By using notions of size that extend the usual one (of cardinality), the size of any monochromatic rectangle can be used to generalise fooling sets, and to lower bound the communication complexity (see KN 1.24). I am not sure how close the generalised largest "size" of any monochromatic rectangle must be to the communication complexity.

In contrast to the above discussion for Boolean functions, for non-Boolean functions the gap between $D(f)$ and $\log rk(f)$ may be exponential. KN 2.23 gives an example: let $f$ be the function that returns the size of the intersections of the sets represented by the two input characteristic vectors. For this function, the log-rank is $\log n$. Now the set of all pairs of non-intersecting sets has $3^n$ elements. As far as I can tell, there can be no monochromatic rectangles larger than this set. If this is correct, then $D(f) \ge Pn(f) \ge (2 - \log 3)n > 0.4n$, so for this function, $D(f)$, $Pn(f)$, and the log-size of a largest monochromatic rectangle are all within a factor of at most $2.5$ of each other, while being exponentially far from the log rank. Hence small separations between $Pn(f)$ and $D(f)$ may be possible in the non-Boolean case, but they are not related in an obvious way to the log-rank of the matrix of $f$. I am not aware of any published work discussing how these measures are related in the non-Boolean case.

Finally, Dietzfelbinger et al. also defined an extended fooling set bound, generalising the fooling condition from pairs ("order 1" subsets) to larger subsets of monochromatic elements; the extended fooling condition requires that the submatrix spanned by the monochromatic elements is not monochromatic. It is not clear how this behaves as the order of the monochromatic subsets increases, as one has to divide the size of the extended fooling set by the order, and consider the largest value over all orders. However, this notion ends up being a close lower bound to $Pn(f)$.