Short-length routes in low-cost networks via Poisson line patterns

Abstract

In designing a network to link n points in a square of area n, we might be guided by the following two desiderata. First, the total network length should not be much greater than the length of the shortest network connecting all points. Second, the average route length (taken over source-destination pairs) should not be much greater than the average straight-line distance. How small can we make these two excesses? Speaking loosely, for a nondegenerate configuration, the total network length must be at least of order n and the average straight-line distance must be at least of order n1/2, so it seems implausible that a single network might exist in which the excess over the first minimum is o(n) and the excess over the second minimum is o(n1/2). But in fact we can do better: for an arbitrary configuration, we can construct a network where the first excess is o(n) and the second excess is almost as small as O(log n). The construction is conceptually simple and uses stochastic methods: over the minimum-length connected network (Steiner tree) superimpose a sparse stationary and isotropic Poisson line process. Together with a few additions (required for technical reasons), the mean values of the excess for the resulting random network satisfy the above asymptotics; hence, a standard application of the probabilistic method guarantees the existence of deterministic networks as required (speaking constructively, such networks can be constructed using simple rejection sampling). The key ingredient is a new result about the Poisson line process. Consider two points a distance r apart, and delete from the line process all lines which separate these two points. The resulting pattern of lines partitions the plane into cells; the cell containing the two points has mean boundary length approximately equal to 2r + constant(log r). Turning to lower bounds, consider a sequence of networks in satisfying a weak equidistribution assumption. We show that if the first excess is O(n) then the second excess cannot be