Consider the following simple monotone circuit model: each gate is just a binary OR. What is the complexity of a function $f(x)=Ax$ where $A$ is a Boolean $n \times n$ matrix with $O(n)$ 0's? Can it be computed by linear size OR-circuits?

More formally, $f$ is a function from $n$ to $n$ bits. The $i$-th output of $f$ is $\bigvee_{j=1}^{n}(A_{ij} \land x_j)$ (i.e., an OR of the subset of input bits given by the $i$-th row of $A$).

Note that $O(n)$ 0's split the rows of $A$ into $O(n)$ ranges (subsets consisting of consecutive elements of $[n]$). This makes it possible to employ known range query data structures. E.g., a sparse table data structure can be turned into an OR-circuit of size $O(n\log n)$. Yao's algorithm for range semigroup operator queries can be turned into an almost linear circuit (of size $O(\alpha(n) \cdot n)$ where $\alpha(n)$ is inverse Ackermann)

In particular, I don't even know how to construct a linear size circuit for a special case where each row of $A$ contains exactly two zeros. While the case of exactly one zero in each row is easy. (Each output function can be computed by an OR of a prefix $[1..k-1]$ and a suffix $[k+1..n]$, which can be precomputed by $2n$ OR-gates.)

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    $\begingroup$ One upper bound is known: it is at most rk(A) times n divided by log n, where rk(A) is the OR-rank of a boolean matrix A (= minimum number of all-1 submatrices whose OR coincides with A). See Lemma 2.5 in this book. So, how large (at most) the boolean rank of an nxn matrix with O(n) zeroes can be? $\endgroup$
    – Stasys
    Dec 30, 2017 at 22:24
  • $\begingroup$ @Stasys Thank you, Stasys! Already for the matrix with zero diagonal the OR-rank is linear, right? $\endgroup$ Dec 31, 2017 at 9:58
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    $\begingroup$ The OR rank of your matrix (zero diagonal, and 1s elsewhere) is at most 2\log n: label rows/columns by binary strings of length \log n, and consider rectangles {(r,c) : r(i)=a, c(i)=1-a} for a=0,1. Note also that Lemma 2.5 is an upper bound. A lower bound in terms of OR rank is given in Thm. 3.20. Also, log of OR rank is exactly the nondeterministic communication complexity of matrices. $\endgroup$
    – Stasys
    Dec 31, 2017 at 13:15
  • $\begingroup$ @Stasys oh, yes, right! $\endgroup$ Jan 3, 2018 at 9:44

2 Answers 2


This is a partial (affirmative) answer in the case when we have an upper bound on the number of zeros in every row or in every column.

A rectangle is a boolean matrix consisting of one all-1 submatrix and having zeros elsewhere. An OR-rank $rk(A)$ of a boolean matrix is the smallest number $r$ of rectangles such that $A$ can be written as a (componentwise) OR of these rectangles. That is, every 1-entry of $A$ is a 1-entry in at least one of the rectangles, and every 0-entry of $A$ is a 0-entry in all rectangles. Note that $\log rk(A)$ is exactly the nondeterministic communication complexity of the matrix $A$ (where Alice gets rows, and Bob columns). As OP wrote, every boolean $m\times n$ matrix $A=(a_{i,j})$ defines a mapping $y=Ax$, where $y_i=\bigvee_{j=1}^na_{i,j}x_j$ for $i=1,\ldots,m$. That is, we take a matrix-vector product over the boolean semiring.

The following lemma is due to Pudlák and Rödl; see Proposition 10.1 in this paper or Lemma 2.5 in this book for a direct construction.

Lemma 1: For every boolean $n\times n$ matrix $A$, the mapping $y=Ax$ can be computed by an unbounded fanin OR-circuit of depth-3 using at most $O(rk(A)\cdot n/\log n)$ wires.

We also have the following upper bound on the OR-rank of dense matrices. The argument is a simple variation of that used by Alon in this paper.

Lemma 2: If every column or every row of a boolean matrix $A$ contains at most $d$ zeros, then $rk(A)=O(d\ln|A|)$, where $|A|$ is the number of $1$s in $A$.
Proof: Construct a random all-$1$ submatrix $R$ by picking each row independently with the same probability $p=1/(d+1)$. Let $I$ be the obtained random subset of rows. Then let $R=I\times J$, where $J$ is the set of all columns of $A$ that have no zeros in the rows in $I$.

A $1$-entry $(i,j)$ of $A$ is covered by $R$ if $i$ was chosen in $I$ and none of (at most $d$) rows with a $0$ in the $j$-th column was chosen in $I$. Hence, the entry $(i,j)$ is covered with probability at least $p(1-p)^{d}\geq pe^{-pd-p^2d}\geq p/e$. If we apply this procedure $r$ times to get $r$ rectangles, then the probability that $(i,j)$ is covered by none of these rectangles does not exceed $(1-p/e)^r\leq e^{-rp/e}$. By the union bound, the probability that some $1$-entry of $A$ remains uncovered is at most $|A|\cdot e^{-rp/e}$, 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.

  • $\begingroup$ Thanks a lot, Stasys! This is nice! In the meantime, Ivan Mihajlin came with another proof. I've posted it below. $\endgroup$ Jan 3, 2018 at 10:14

(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.

  • $\begingroup$ A nice argument. But it seems to be tailor made for the case of d=2 zeros per row. What about d>2 zeros? $\endgroup$
    – Stasys
    Jan 3, 2018 at 20:19
  • $\begingroup$ @Stasys, it is doable if I'm not mistaken. I've updated the answer. $\endgroup$ Jan 4, 2018 at 10:12

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