Esta é uma resposta parcial (afirmativa) no caso em que temos um limite superior no número de zeros em todas as linhas ou colunas.
Um retângulo é uma matriz booleana que consiste em uma submatriz all-1 e possui zeros em outro lugar. Um OR-rank r k ( A ) de uma matriz booleana é o menor número r de retângulos, de modo que A possa ser escrito como um OR (componente a componente) desses retângulos. Ou seja, cada entrada de A é uma entrada em pelo menos um dos retângulos e cada entrada de A é entrada de 0 em todos os retângulos. Observe que log r k ( A ) é exatamente a complexidade de comunicação não determinística da matriz Ark(A)rAAAlogrk(A)A (onde Alice obtém linhas e colunas Bob). Como OP escreveu, toda matriz × n m × n A = (m×n b a i , j ) define um mapeamento y = A x , onde y i = ⋁ n j = 1 a i , j x j para i = 1 , … , m . Ou seja, tomamos um produto de vetor de matriz sobre o semiamento booleano.
A=(ai,j)y=Axyi=⋁nj=1ai,jxji=1,…,m
O seguinte lema é devido a Pudlák e Rödl; veja a Proposição 10.1 neste artigo
ou o Lema 2.5 deste livro para uma construção direta.
Lema 1: Para cada matriz booleana n × n A , o mapeamento y = A x pode ser calculado por um circuito OR de profundidade ilimitada de ventilador 3, usando no máximo fios O ( r k ( A ) ⋅ n / log n ) .
n×nAy=AxO(rk(A)⋅n/logn)
Também temos o seguinte limite superior no ranking OR de matrizes densas. O argumento é uma variação simples da usada por Alon neste artigo .
Lema 2: Se todas as colunas ou linhas de uma matriz booleana A contiver no máximo d zeros, então r k ( A ) = O ( d ln | A | ) , em que | Um | é o número de 1 s emAdrk(A)=O(dln|A|)|A|1 um .
A
Prova:
Construa uma submatriz aleatória all- 1 R escolhendo cada linha independentemente com a mesma probabilidade p = 1 / ( d + 1 ) . Seja eu o subconjunto aleatório obtido de linhas. Então deixe- R = I × J , onde J é o conjunto de todas as colunas de A que não têm zeros nas linhas em que eu .
1Rp=1/(d+1)IR=I×JJAI
A 11-entry (i,j)(i,j) of AA is covered by RR if ii was chosen in
II and none of (at most dd) rows with a 00 in the jj-th column
was chosen in II. Hence, the entry (i,j)(i,j) is covered with probability at least p(1−p)d≥pe−pd−p2d≥p/ep(1−p)d≥pe−pd−p2d≥p/e.
If we apply this procedure rr times to get rr rectangles,
then the probability that (i,j)(i,j) is covered by none of these
rectangles does not exceed (1−p/e)r≤e−rp/e(1−p/e)r≤e−rp/e. By the union bound, the
probability that some 11-entry of AA remains uncovered is at most
|A|⋅e−rp/e|A|⋅e−rp/e, which is smaller than 11 for r=O(dln|A|)r=O(dln|A|).
◻□
Corollary: If every column or every row of a boolean matrix AA contains at most dd zeros, then the mapping y=Axy=Ax can be computed by an unbounded fanin OR-circuit of depth-3 using
O(dn)O(dn) wires.
I guess that a similar upper bound as in Lemma 2 should also hold when dd is the average number of 11s in a column (or in a row). It would be interesting to show this.
Remark: (added 04.01.2018) An analogue rk(A)=O(d2logn)rk(A)=O(d2logn) of Lemma 2 also holds when dd is the maximum average number of zeros in a submatrix of AA, where the average number of zeros in an r×sr×s matrix is the total number of zeros divided by s+rs+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(d2ln2n)rk(A)=O(d2ln2n) can be derived directly from Lemma 2 as follows.
Lemma 3: Let d≥1d≥1. If every spanning subgraph of a bipartite graph GG has average degree ≤d≤d, then GG can be written as a union G=G1∪G2G=G1∪G2, where the maximum left degree of G1G1 and the maximum right degree of G2G2 are ≤d≤d.
Proof: Induction on the number nn of vertices. The base cases n=1n=1 and n=2n=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 ≤d≤d. Take a vertex uu of degree ≤d≤d; such a vertex must exists because also the average degree of the entire graph must be ≤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. ◻
Lemma 4: Let d≥1. If the maximum average number of zeros in a boolean n×n matrix A=(ai,j) is at most d, then rk(A)=O(d2ln2n).
Proof: Consider the bipartite n×n graph G with (i,j) being an edge iff ai,j=0. Then the maximum average degree of G is at most d. By Lemma 3, we can write G=G1∪G2, where
the maximum degree of the vertices on the left part of G1, and the maximum degree of the vertices on the right part of G2 is ≤d.
Let A1 and A2 be the complements of the adjacency matrices of G1 and G2.
Hence, A=A1∧A2 is a componentwise AND of these matrices.
The maximum number of zeros in every row of A1 and in every column of A2 is at most d. Since rk(A)≤rk(A1)⋅rk(A2), Lemma 2 yields rk(A)=O(d2ln2n). ◻
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=√n, and put an identity m×m matrix in, say left upper corner of A, and fill the remaining entries by ones. Then d(A)≤m2/2n<1 but rk(A)≥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.