Abstract

The linear discrepancy problem is to round a given [0,1]-vector x to a binary vector y such that the rounding error with respect to a linear form is small, i.e., such that || A( x− y)|| ∞ is small for some given matrix A. The combinatorial discrepancy problem is the special case of x=( 1 2 ,…, 1 2 ) t . A famous result of Beck and Spencer [Math. Programming 30 (1984) 88] as well as Lovász et al. [European J. Combin. 7 (1986) 151] shows that the linear discrepancy problem is not much harder than this special case: Any linear discrepancy problem can be solved with at most twice the maximum rounding error among the discrepancy problems of the submatrices of A. In this paper, we strengthen this result for the common situation that the discrepancy of submatrices having n 0 columns is bounded by Cn 0 α for some C>0, α ∈ ]0,1] . In this case, we improve the constant by which the general problem is harder than the discrepancy one from 2 down to 2( 2 3 ) α . We also find that a random vector has expected linear discrepancy 2( 1 2 ) αCn α only. Hence in the typical situation that the discrepancy is decreasing for smaller matrices, the linear discrepancy problem is even less difficult compared to the discrepancy one than assured by previous results. We also obtain the bound lindisc( A, x)⩽2(2 α /(2 1− α −1)) C|| x|| 1 α . Our proofs use a reduction to Pusher–Chooser games.

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