1. Introduction and Preliminaries
A semiring is a set with addition (+) and multiplication (·) such that is an Abelian monoid with identity 0, is a monoid with identity 1, such that multiplication (·) distributes over addition (+) from both sides and such that for all . We use juxtaposition for · for convenience. If is an Abelian monoid, then is called a commutative semiring. If only 0 element of has an additive inverse then is called an antinegative semiring. Thus all rings with 1 are semirings, but no rings are antinegative. Let be the binary Boolean semiring such that multiplication and addition are defined as in the integer ring except that . The set, , of nonnegative integers and , the binary Boolean semiring, are examples of antinegative semirings. Zero-divisor is defined for semirings as if were a ring.
In the following, we assume that is an antinegative, commutative semiring and without zero divisors and is the binary Boolean semiring.
Let denote the set of all matrices with entries in with usual matrix addition and multiplication. Let if , let denote the m square identity matrix, denote the zero matrix in , and denote the matrix of all 1’s in . If the order is obvious, then we usually omit the subscripts, and we write .
The matrix is said to be of rankr if we have matrices and such that and r is the least integer with such a factorization. We denote this by . Let denote the set of all matrices with entries in . The Boolean rank of is the rank over and denoted . In addition, , and O is the unique matrix with rank 0.
The Boolean rank has been applied in various combinatorial theories and graph theory. As an example, if
is the adjacent matrix
G, the bipartite graph with bipartition
, then the Boolean rank of
B is the least integer of bicliques that cover the edges of
G, which is called the
biclique covering number [
1]. If
, then the
support of
B is the matrix
such that
if
and
if
We say that a matrix dominates a matrix if implies .
Given a matrix B, we let denote the row of B and let denote the column. Now, if and is a factorization of , then . Since each term is a matrix of rank 1, the rank of B, , is the minimum number of rank 1 matrices whose sum is B.
Given a matrix , a set of isolated entries is a set of locations, usually written as such that , no two entries in I are in the same column, no two entries in I are in the same row, and, if then or . That is, isolated entries are independent entries and any two isolated entries and do not lie in a submatrix of B of the form with all entries nonzero. The isolation number of B, , is the maximum cardinality of a set of isolated entries. Note that if and only if .
Let
with
. Then, we have
h rank 1 matrices
such that
Because each matrix of rank 1 can be permuted to the form
with
, we have that the matrix with two isolated entries of
B cannot be dominated by any one
among the rank 1 summand of
B in (1). Thus,
For , let be the set of the indices of the nonzero rows of and be the set of the indices of the nonzero columns of , . Let , the number of nonzero columns of and , the number of nonzero rows of .
Many sets, relations and functions related with matrices do not depend on the magnitude of the each entries of a matrix, but depend only on the fact of whose entry is nonzero or zero. Such combinatorially significant matrix properties have become more important in recent research. Of primary interest is the Boolean rank. Determining the rank of a Boolean matrix is known as an NP-Complete problem, (see [
2]), and consequently determining upper bounds on the rank of a Boolean matrix is of importance to many authors that use Boolean rank in their research. If the Boolean matrix is the reduced adjacent matrix of some bipartite graph, the isolation number of the Boolean matrix denotes the largest size of a non-competitive matching in its bipartite graph. This fact is related to the research of such combinatorial studies as the patient hospital problem, the stable marriage problem, and so on. An additional reason for research on the set of isolation entries and isolation number is that isolation number is a lower bound on the rank of a Boolean matrix [
1,
3]. While determining the isolation number and determining the Boolean rank of a Boolean matrix are NP-Complete problems [
4], for some matrices, determining the isolation number could be easier than determining the Boolean rank if the Boolean matrix is sparse:
Example 1. The Boolean rank of B is at most 6 from the first three rows and columns. However, to determine that the Boolean rank of B is not 5 requires much calculation if we do not consider the isolation number. However, it is easy to see that the isolation number is 6 from the bold ’s in B, which constitute a set of isolated entries. Hence, the Boolean rank of B is 6 from Equation (2). We note that if one of the non-bold 1’s were replaced by 0, the changed matrix still has isolation number 6 and Boolean rank 6.
In this paper, we investigate the question: given a fixed
h, if the isolation number of
B is
h what can be the upper bound of the rank of
B and the rank of support of
B? Some terms not defined here can be found in [
5] or in [
6].
2. Upper Bounds for Isolation Numbers of a Matrix in
Lemma 1. For , we have in
Proof. It follows from the fact that is antinegative and in □
Lemma 2. For , we have
Proof. If then B has a rank-one factorization such that with and Therefore,
from Lemma 1.
Thus, □
We may have for some .
Example 2. Consider and in Then, but . Hence, However,
Lemma 3. For , we have
Proof. If and are any isolated entries in B, then and , and that or Hence, and are isolated entries in so we have
Conversely, if and are any isolated entries in then and and that or Hence, and are isolated entries in B, so we have □
Theorem 1. If , then we have if and only if .
Proof. Let . If , then so that and since by , we have .
Conversely, suppose on the contrary that there exists a matrix such that , . Then, there exists two non-equal and nonzero rows of , say ith and jth. Hence, without loss of generality, there exists a k such that and . Then, and any unit entry in jth row of constitute a set of two isolated entries. Thus, , a contradiction. □
Thus, the subset of of matrices with isolation number 1 is the same set of matrices whose support has Boolean rank 1.
Lemma 4. Let . Then, if , then , and if , then .
Proof. If , then we have by Theorem 1. Since from Lemma 3 and , we have that .
Next, suppose that
and
. Then, we have a factorization of
as
with
and
. Then, the three rows of
D generate all the rows of
. Since
,
D cannot have Boolean rank 2 or less. Thus, we have
Therefore, we have a factorization of
D as
with
and
Then, the three column of
E generate all the columns of
D and
. Therefore, it is sufficient to consider
matrices of rank 3. However, there are only 10 following
matrices of rank 3 up to permutations:
Since can be permuted to and can be permuted to , and can be permuted to with transposing. Therefore, there are only seven non-equivalent matrices of rank 3. However, these matrices have three isolation entries on the main diagonal. Thus, we have a contradiction to the conditions that and . Thus, if then . □
Theorem 2. Let . Then, we have that if and only if .
Proof. From Lemma 4, we have proved the sufficiency. Thus, we will prove the necessity.
Assume that there is such that and . By Lemma 4, , and hence . Thus, we choose B such that if then . Suppose that for with . That is, h is the minimum integer such that and . Suppose that the number of nonzero rows of is the minimum among the ’s. Let us permute the rows of so that the first rows of are nonzero. For , let be the matrix such that its the first j row entries are the first j row entries of and whose other row entries are all zero. Let be the matrix whose last row entries are the last row entries of and whose other row entries are all zero. Then, . In addition, any set of isolated entries of constitutes a set of isolated entries for . Now, from , and the fact that or , there is some t such that . Since by the choice of , for this t, we have that since . Thus, , which is impossible since . Therefore, . □
Now, as we see in the next example, we can find some such that and
Example 3. For , let . Then, we can easily show that but , where [7]. Hence, but . We define a tournament matrix as the adjacent matrix of a directed graph that is called a tournament graph, T. This tournament matrix is characterized by both and .
Now, we can ask the following: How much difference can there be between the isolation number of a matrix in and the Boolean rank of its support matrix? For each , can we determine the matrix forms in for which ? Of course, we determined those matrices if or only in the above theorems. For , we obtain new characterization of the matrix form in the following:
Theorem 3. Let with . Then, if and only if there are permutation matrices and such that , where and is dominated by a tournament matrix.
Proof. Suppose that . Then, we permute B by permutation matrices and so that the set of isolated entries are in the positions, . That is, if , then is the set of isolated entries in X. Therefore, , with and for every i and from the definition of the isolated entries. Thus, where is an m-square matrix which is dominated by a tournament matrix. Thus, where and we have no restrictions on D.
Conversely, if and where is an m-square matrix which is dominated by a tournament matrix, then the entries on the diagonal of C constitute a set of isolated entries for . Therefore, B has a set of m isolated entries. Thus, . □
Corollary 1. Let with . If there exist permutations and such that where is a diagonal matrix or a triangular matrix with nonzero diagonal entries, then .