Not a worry. I can supply an upper bound. Run vertical lines through the points of G1. Points of G1 with the same x coordinate lie on each vertical line, separated by 1 unit. Call this figure L1. Now rotate L1 by pi/4, giving parallel lines through the points of G2, call this figure L2. Now consider a third set of parallel lines at angle pi/8 to the x-axis, each line of L3 passing through an integer point (0, k). Adjoin each line of L3 to the region strictly between it and the next line above it, you get a set S3 of disjoint half-open strips that cover the plane. Note that each line of L1 intersects each strip of S3 in a half-open segment of length 1, which must include exactly one point of G1. Likewise, each line of L2 intersects each strip of S3 in a half-open segment of length 1, which include exactly one point of G2. Let the intersection of L1 with S be the set of segments A1. Let the intersection of L2 with S be the set of segments A2. Pair each segment of A1 with the segment of A2 by closest midpoint. This induces a bijection between A1 and A2. With a little analytic geometry, we can bound the distance between the midpoints of A1 and A2, and hence the distance between any point P1 on A1 and any point P2 on A2, that is, |P1-P2| <= D for some finite D (definitely D < 2). Each segment of A1 includes a unique point P1 of G1, each segment of A2 includes a unique point P2 of G2. The map f_S(P1) = P2 is a bijection between all points of G1 in S and all points of G2 in S, with |P-f_S(P)| <= D. The union f of f_S for all S is a bijection between all points of G1 and G2 with |P-f(P)| <= D. D is therefore the sought-after upper bound. Could anybody follow this?
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Friday, May 29, 2015 3:13 PM To: math-fun Subject: Re: [math-fun] Wilson's grid problem
The thing that is so challenging about the problem with Z^2(exp(pi/4)) is that the pair of superposed grids is not periodic, so no periodic matching can work. The creeping fear is that no matter how clever your matching procedure is, it will eventually maze itself and leave a point with no available partner closer than (per my earlier joke) 17,000.