From:           John Conway <>
Date:           Mon, 13 Dec 93 09:42:47 EST
To:   ,
Subject:        Re:  Aperiodic Space-filling tile?

    My tile is a very simple one.  It is aperiodic (only)
in the sense that no tiling with it admits a translational
symmetry.  This seems to be the standard sense now, and it
is a good one because in particular it implies that the
tiles can't be grouped into equal clumps that are all
embedded similarly in the tiling.

   Take a triangular prism:

           / | \
          /__|  \
          \  \  |
           \  \ |

Make it "generic" - in other words, avoid certain equalities
between the sides and angles.  In particular, don't make the
side faces be rectangles.  [The apparent right angles in the
triangles of my sketch are there only because of the limitations
of this keyboard.]

    Now take another such prism, again as generic as you can 
get it, subject to the condition that one of its parallelogram
faces has the same shape as one of the ones on the first prism,
oriented in such a way that you can stick the two prisms together
to make a "biprism" in which no two triangles share an edge.
Then this biprism is my tile.

    Which biprisms are sufficiently "generic", you may ask.  I
don't know a very easy way to find out, though almost all of
them are, of course.  Here is what's needed:

    1) It must be impossible to fit the pieces together in any
way except those in which the constituent prisms line up
in parallel, triangle-to-triangle:

            / | \
           /__|  \
           \  \  |
            \  \ | \
             \__\|  \
              \  \  |
               \  \ |    

    This sticks the biprisms together in two-dimensional "waffles"
which are then stacked to form the whole tiling.  

    2)  The angle of rotation between two adjacent "waffles" must
not be  60  or  90 degrees.

    I just worked out one case.  I took a rhombus made of two
triangles root2,root2,1  for the common face,  and took points
vertically above and below the midpoints of the edges to be
the other four vertices:

                /  \
               A    B
              /      \
             /        \
             \        /
              \      /
               D    C
                \  /

   A and C are above the plane,  B  and  D below it, so that the
"roof edges" of the prisms are  AC  and  BD.

    I made the angles of the triangular faces at these vertices
be right angles, but that gives a very flat biprism.  I suggest
to make them something like  30  degrees for a better shape.

[This will still be generic enough.]    

  I imagine you will be a bit disappointed, because the tiling
is fairly "regular".  But in fact the tiles are necessarily
embedded in infinitely many distinct ways, however you do it.