1983 is not on the list of possibles for a Lattice Labyrinth based on the square lattice. It is a number divisible by 3, which makes it impossible for a Trefoil Labyrinth but a candidate for a Labyrinth based on the triangular lattice for which e2 + ef + f2 must equal 1983 and indeed the separation parameter pair (e,f) with values (38,11) provides the solution, BUT this corresponds to a Honeycomb Labyrinth with a repeat unit of twice 1983, that is with 3966 triangles in each supertile. Crikey, that’s the highest-order Labyrinth I’ve ever tried for, and not a member of any sub-family I’ve investigated – neither the (e,11) or (e,e-17) sub-families has hitherto attracted me. A Star Labyrinth can be constructed by simply dividing the Honeycomb supertile into its six arms, but these Star supertiles will each have an area of 661 triangles. A Diamond Labyrinth should also exist, but the supertiles of this will each have an area of 1322 triangles. So it seems that no labyrinth with a supertile area of 1983 triangles can exist UNLESS there is a Honeycomb Labyrith with supertile are 6 x 1983 = 11,898 triangles, and for which the separation parameters must obey e2 + ef + f2 = 5949. I must investigate , but for now we can celebrate the birthyears of TWO Olivers (there must be another somewhere) with Honeycomb lattice Labyrinth (38,11) – and what a dynamic individual he is, reaching far out from a powerful centre. To my surprise (thanks, O Platonic God) the construction which led to it is wonderfully simple and elegant. Here is one central supertile with six neighbours sketched in faintly.
POSTPOSTCRIPT: there IS indeed a possible Honeycomb Lattice Labyrinth (60,27), with repeat unit 11,898 triangles, the supertile of which can be dissected into six arms each of 1983 triangles. I fear I must divert a day into searching for the construction.