Filling a Box with Translates of Two Bricks

We give a new proof of the following interesting fact recently proved by Bower and Michael: if a d-dimensional rectangular box can be tiled using translates of two types of rectangular bricks, then it can also be tiled in the following way. We can cut the box across one of its sides into two boxes, one of which can be tiled with the first brick only and the other one with the second brick. Our proof relies on the Fourier Transform. We also show that no such result is true for translates of more than two types of bricks. Suppose we have at our disposal two types of d-dimensional rectangles (bricks), type A with dimensions (a1, . . . , ad) and type B with dimensions (b1, . . . , bd). We want to use translates of such bricks to fill completely, and with no overlaps (except at the boundaries of the bricks), a given d-dimensional rectangular box. We then say that these two bricks tile our box by translations. All rectangles that appear in this note are axis-aligned. Bower and Michael [1] recently showed the following nice result. A hyperplane cut is a separation of an axis-aligned box in d dimensions using a hyperplane of the type xj = α, ∗Supported in part by European Commission IHP Network HARP (Harmonic Analysis and Related Problems), Contract Number: HPRN-CT-2001-00273 HARP. the electronic journal of combinatorics 11 (2004), #N16 1 for some j = 1, . . . , d and some α ∈ R. A hyperplane cut separates such a box into two rectangular boxes. Theorem 1 (Bower and Michael [1]) If two bricks, of types A and B, tile a box Q (in dimension d ≥ 1) by translations then we can split Q into two other boxes Qa and Qb using a hyperplane cut, such that Qa can be tiled using translates of type A bricks only and Qb can be tiled using translates of type B bricks only. (For d = 1 the result is obvious.) The purpose of this note is to give a short proof of this fact using the Fourier Transform, a very natural tool for this problem, as will become apparent. The reader could consult [2] for more applications of the Fourier Transform to tiling problems. Indeed, suppose that A = (−a1/2, a1/2)×· · ·×(−ad/2, ad/2) and B = (−b1/2, b1/2)× · · ·×(−bd/2, bd/2) are the two bricks and Λa, Λb are two finite subsets of R which represent the translations of A and B that make up our box Q = (−1/2, 1/2) (as we may clearly assume without loss of generality). In other words, the indicator functions χA and χB of the two bricks satisfy ∑ λ∈Λa χA(x − λ) + ∑ λ∈Λb χB(x − λ) = χQ(x), a.e. x ∈ R. (1) The definition of the Fourier Transform f̂ of a function f ∈ L(R) that we use is