fixed a typo

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 $$\bigvee_{i=1}^{n}(A_{ij} \land x_j)$$$$\bigvee_{j=1}^{n}(A_{ij} \land x_j)$$ (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.)

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 $$\bigvee_{i=1}^{n}(A_{ij} \land x_j)$$ (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.)