On the Discrepancy of Strongly Unimodular Matrices

A (0, 1) matrix A is strongly unimodular if A is totally unimodular and every matrix obtained from A by setting a nonzero entry to 0 is also totally unimodular. Here we consider the linear discrepancy of strongly unimodular matrices. It was proved by Lovaz, et.al. [5] that for any matrix A, lindisc(A) # herdisc(A). (1) When A is the incidence matrix of a set-system, a stronger inequality holds: For any family H of subsets of {1, 2, . . . , n}, lindisc(H) # (1 - t n )herdisc(H). where t n # 2 -2 n (J. Spencer, [6]). In this paper we prove that the constant t n can be improved to 3 -(n+1)/2 for strongly unimodular matrices. # The first author is supported by NSF Grant DMS-9304580. + The second author is supported by Courant Instructorship, New York University. 1 1 Introduction and results A matrix A is said to be totally unimodular if the determinant of each square submatrix of A is 0 or 1. Clearly the entries of a totally unimodular matrix must be 0 or 1. A matr..