ALG-B1-M05-P022 Periodic Marking

Combine suitable multiples of the two vectors to obtain \((D,0)\) and \((0,D)\).

Let \(u=(a,b-a)\), \(v=(b-a,-a)\). Then \(a u+(b-a)v=(a^2+(b-a)^2,0)=(D,0)\).

Also \((b-a)u-a v=(0,D)\). Thus the marking is periodic under shifts \((D,0)\) and \((0,D)\).

In a fundamental \(D\times D\) square, the number of marked cells equals the area of the square divided by the area of the fundamental parallelogram of the lattice. The determinant of \(u,v\) is \(-a^2-(b-a)^2=-D\), so that area is \(D\). Hence a square of area \(D^2\) contains exactly \(D\) marked cells.

Low-confidence source adapted as a lattice-periodicity sequence problem; diagram recommended.