So the current state of the art is that the minimum possible c is between 0.707... and 0.92... ? I wonder if a computer search on a reasonably-sized piece of Z^2 could give any insights. Also: having settled on a bijection F, make the following visual aid: show the points of Z^2, with an edge drawn between points P and Q if F(P) and F(Q) are adjacent in *their* grid. On Thu, May 28, 2015 at 2:44 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Veit Elser Take a look at this: http://arxiv.org/pdf/1305.1798.pdf Figure 4 shows such a bijection-generating fundamental domain. The diameter is given in equation (1), which, after you divide by 2, gives the bound 0.92 ... on c (for the L_infinity norm). This is actually a very good bound, considering the fact that 1/sqrt(2) = 0.707 is a lower bound. Exercise: Prove that c >= 1/sqrt(2).
WDS: Wow!! Well, this is not an "octagon", it is a very strange fractalish set, but it does have 8-way symmetry, and it tiles the plane, and it has bounded diameter, so I think Veit is correct. That's amazing!!!
However, before you get too excited, note this solution is very special to Wilson's specific problem, and will not solve the general-rotation, general-translation version of it.
Re VE's exercise, the usual square grid with square-centers G1 will necessarily contain some squares containing zero members of G2. (Expected membership=1, but nonzero variance.) Hence c>1/2. In fact, a square with side s arbitrarily near sqrt(2) will be empty and will be translatable to anyplace mod G2 with arbitrarily small error, thus proving c>=sqrt(1/2).
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