密集线性算子的或电路复杂度

考虑以下简单的单调电路模型：每个门只是一个二进制OR。函数的复杂度是多少，其中是 0的布尔矩阵？可以通过线性或电路计算吗？f （x ）= A x A n × n O （n $f(x)=Ax$$A$$n \times n$$O(n)$

更正式地，是从到位的函数。第个输出是（即，由第行给出的输入位的子集的或。˚F Ñ Ñ 我˚F ⋁ Ñ Ĵ = 1（甲我Ĵ ∧ X Ĵ）我$f$$n$$n$$i$$f$$\bigvee_{j=1}^{n}(A_{ij} \land x_j)$$i$$A$

请注意， 0将的行划分为范围（由的连续元素组成的子集）。这使得可以采用已知的范围查询数据结构。例如，稀疏表数据结构可以变成大小为的OR电路。Yao的范围半组算子查询算法可以变成几乎线性的电路（大小为，其中是逆阿克曼倒数）O （n ）A O （n ）[ n ] O （n log n $O(n)$$A$$O(n)$$[n]$$O(n\log n)$ø （α （Ñ ）⋅ Ñ ）α （Ñ $O(\alpha(n) \cdot n)$$\alpha(n)$

特别是，我什至不知道如何为每一行都恰好包含两个零的特殊情况构造一个线性尺寸电路。每行恰好为零的情况很容易。（每个输出函数都可以由前缀和后缀进行“ 或”运算，可以通过或”门进行预计算。）$A$$[1..k-1]$$[k+1..n]$$2n$

当我们在每行或每列中的零个数有一个上限时，这是一个部分（肯定）答案。

甲矩形是由一个全1子矩阵和其它地方具有零的布尔矩阵。布尔矩阵的OR秩r k （A ）是矩形的最小数量r，因此A可以写为这些矩形的（按分量）OR。即，A的每个1个条目在至少一个矩形中是1个条目，并且A的每个0个条目在所有矩形中是0个条目。请注意，log r k （A ）恰好是矩阵A的不确定通信复杂度$rk(A)$$r$$A$$A$$A$$\log rk(A)$$A$（爱丽丝在其中获取行，鲍勃在其中列）。如OP写道，每布尔米× Ñ矩阵甲= （一个我，Ĵ）定义了映射Ŷ = 甲X，其中 ÿ 我 = ⋁ Ñ Ĵ = 1一个我，Ĵ X Ĵ用于我= 1 ，... ，米。也就是说，我们在布尔半环上采用矩阵向量乘积。 $m\times n$$A=(a_{i,j})$$y=Ax$$y_i=\bigvee_{j=1}^na_{i,j}x_j$$i=1,\ldots,m$

以下引理归功于Pudlák和Rödl；见命题10.1 本文 或引理2.5在这本书中的直接施工。

引理1：对于每一个布尔Ñ × Ñ矩阵甲，映射ÿ = 甲X可以通过深度-3的使用至多一个无限扇入OR电路来计算ø （ř ķ （甲）⋅ ñ /登录Ñ ）导线。 $n\times n$$A$$y=Ax$$O(rk(A)\cdot n/\log n)$

在稠密矩阵的OR列上，我们还有以下上限。该论点是Alon在本文中使用的论点的简单变体。

引理2：如果布尔矩阵A的每一列或每一行最多包含d个零，则r k （A ）= O （d ln | A |），其中| A | 是A中的1秒数。 $A$$d$$rk(A)=O(d\ln|A|)$$|A|$$1$$A$

证明： 构造一个随机清一色1子矩阵[R通过以相同的概率独立地采摘的每一行p = 1 /（d + 1 ）。让我成为获得的随机行子集。然后让R = I × J，其中J是A的所有列的集合，在I的行中没有零。 $1$$R$$p=1/(d+1)$$I$$R=I\times J$$J$$A$$I$

阿1个 -entry （我，Ĵ ）的甲由覆盖ř如果我被选择在 我和无的（至多d）行与0在Ĵ被选为在第柱予。因此，条目（我，Ĵ ）覆盖有概率至少p （1 - p ）d ≥ p ë - p d - p 2 d$1$$(i,j)$$A$$R$$i$$I$$d$$0$$j$$I$$(i,j)$p / ë。如果我们将这个过程 [R次获得 [R矩形，那么概率（我，Ĵ ）是由没有这些矩形不超过所覆盖（1 - p / ë ）[R ≤ é - [R p / ë。受联合约束，最多仍可以发现 A的1个条目的概率 | A | ⋅ ë - - [R p /$p(1-p)^{d}\geq pe^{-pd-p^2d}\geq p/e$$r$$r$$(i,j)$$(1-p/e)^r\leq e^{-rp/e}$$1$$A$$|A|\cdot e^{-rp/e}$, which is smaller than $1$ for $r=O(d\ln|A|)$. $\Box$

Corollary: If every column or every row of a boolean matrix $A$ contains at most $d$ zeros, then the mapping $y=Ax$ can be computed by an unbounded fanin OR-circuit of depth-3 using $O(dn)$ wires.

I guess that a similar upper bound as in Lemma 2 should also hold when $d$ is the average number of $1$s in a column (or in a row). It would be interesting to show this.

Remark: (added 04.01.2018) An analogue $rk(A)=O(d^2\log n)$ of Lemma 2 also holds when $d$ is the maximum average number of zeros in a submatrix of $A$, where the average number of zeros in an $r\times s$ matrix is the total number of zeros divided by $s+r$. This follows from Theorem 2 in N. Eaton and V. Rödl;, Graphs of small dimension, Combinatorica 16(1) (1996) 59-85. A slightly worse upper bound $rk(A)=O(d^2\ln^2 n)$ can be derived directly from Lemma 2 as follows.

Lemma 3: Let $d\geq 1$. If every spanning subgraph of a bipartite graph $G$ has average degree $\leq d$, then $G$ can be written as a union $G=G_1\cup G_2$, where the maximum left degree of $G_1$ and the maximum right degree of $G_2$ are $\leq d$.

Proof: Induction on the number $n$ of vertices. The base cases $n=1$ and $n=2$ are obvious. For the induction step, we will color the edges in blue and red so that the maximum degree in both blue and red subgraphs are $\leq d$. Take a vertex $u$ of degree $\leq d$; such a vertex must exists because also the average degree of the entire graph must be $\leq d$. If $u$ belongs to the left part, then color all edges incident to $u$ in blue, else color all these edges in red. If we remove the vertex $u$ then the average degree of the resulting graph $G$ is also at most $d$, and we can color the edges of this graph by the induction hypothesis. $\Box$

Lemma 4: Let $d\geq 1$. If the maximum average number of zeros in a boolean $n\times n$ matrix $A=(a_{i,j})$ is at most $d$, then $rk(A)=O(d^2\ln^2 n)$.

Proof: Consider the bipartite $n\times n$ graph $G$ with $(i,j)$ being an edge iff $a_{i,j}=0$. Then the maximum average degree of $G$ is at most $d$. By Lemma 3, we can write $G=G_1\cup G_2$, where the maximum degree of the vertices on the left part of $G_1$, and the maximum degree of the vertices on the right part of $G_2$ is $\leq d$. Let $A_1$ and $A_2$ be the complements of the adjacency matrices of $G_1$ and $G_2$. Hence, $A= A_1\land A_2$ is a componentwise AND of these matrices. The maximum number of zeros in every row of $A_1$ and in every column of $A_2$ is at most $d$. Since $rk(A)\leq rk(A_1)\cdot rk(A_2)$, Lemma 2 yields $rk(A)=O(d^2\ln^2 n)$. $\Box$

N.B. The following simple example (pointed by Igor Sergeev) shows that my "guess" at the end of the answer was totally wrong: if we take $d=d(A)$ to be the average number of zeros in the entire matrix $A$ (not the maximum of averages over all submatrices), then Lemma 2 can badly fail. Let $m=\sqrt{n}$, and put an identity $m\times m$ matrix in, say left upper corner of $A$, and fill the remaining entries by ones. Then $d(A)\leq m^2/2n < 1$ but $rk(A)\geq m$, which is exponentially larger than $\ln|A|$. Note, however, that the OR complexity of this matrix is very small, is $O(n)$. So, direct arguments (not via rank) can yield much better upper bounds on the OR complexity of dense matrices.

(I tried to post this as a comment to Stasys' answer above, but this text is too long for a comment, so posting it as an answer.) Ivan Mihajlin (@ivmihajlin) came up with the following construction. Similarly to Stasys' proof, it works for the case when the maximum (rather than average) number of 0’s in each row is bounded.

First, consider the case when every row contains exactly two zeros. Consider the following undirected graph: the set of vertices is $[n]$; two nodes $i$ and $j$ are joined by an edge, if there is a row having zeros in columns $i$ and $j$. The graph has $n$ edges and hence it contains a cut $(L,R)$ of size at least $n/2$. This cut splits the columns of the matrix into two parts ($L$ and $R$). Let now also split the rows into two parts: the top part $T$ contains all columns that have exactly one zero in both $L$ and $R$; the bottom part $B$ contains all the remaining rows. What is nice about the top part of the matrix ($T \times (L \cup R)$) is that it can be computed by $O(n)$ gates. For the bottom part, let’s cut all-1 columns out of it and make a recursive call. The corresponding recurrence relation is $C(n) \le an + C(n/2)$ implying $C(n)=O(n)$.

Now, generalize it to the case of at most $d$ zeros in every row. Let $C_d(n)$ be the complexity of an $n \times (\le dn)$ matrix with at most $d$ zeros per row (if there are more than $dn$ columns, then some of them are all-1). Partition the columns into two parts $L$ and $R$ such that at least $n(1-2^{-d})$ rows (call them $T$) satisfy the following property: if there are exactly $d$ zeroes in a row, then not all of them belong to the same part (denote the remaining rows by $B$). Then make three recursive calls: $T \times L$, $T \times R$, and $B \times (L \cup R)$. This gives a recurrence relation $C_d(n) \le an + 2\cdot C_{d-1}(n(1-2^{-d}))+C_d(2^{-d}n)$. This, in turn, implies that $C_d(n) \le f(d)\cdot n$. The function $f(d)$ is exponential, but still.