## A Lattice Labyrinth for two Olivers

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.

There’s going to be no stopping you two 1983 Olivers.

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.