Torsten Sillke, 05. June 1997

The Janus2 puzzle of Frits G"obel: (f.gobel@math.utwente.nl)
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Shape of the pieces:

  piece = a equilateral triangle and halve a square.

  The regular dodecagon can be build by:
    6 squares and 12 equilateral triangles.

  This is the basic grid into which the pieces fit.


  The pieces are handed. In a tiling of the 
  dodecagon we have 6 right ones (the outer)
  and 6 left ones (the inner).

Colors of the pieces:

  Three colors are in use: R(ed), B(lue), and Y(ellow).

  We have the following combinations:

  Color Table:
  ------------
  Top, Down, Number (pieces right handed):

   R    Y      4
   R    B      2
   B    Y      3
   B    R      2
   Y    R      1
   Y    B      -


Problem 4:

  This is very simple. There are three case for the color
  on both sides. The two others are on opposite sides.

  Top, Down, Both sides
   R     B       Y
   R     Y       B
   B     Y       R

  Flip the pieces and count the number of left and rigth ones.
  If you count 6:6 you get a solution.


Problem 3: The main problem

  It is best to start with the piece (Y,R) as it is singular.
  This one breaks the symmetry. We have only two positions for
  it: first an inner one and second an outer one.
  The we have many restrictions between inner and outer pieces:


     R  _  R  _  B  _  B  _  Y
     B     Y     Y     R     R     (I-O adjacency graph plus loops)

     The loops in this graph can be rejected as a combination
     of an inner (a,b) with an outer (a,b) results in the common
     neightbour an inner (c,c).

  Case: (Y,R) in inner position
     The only possible neightbours in outer position are (B,R).
     There common neightbours are determined as (B,Y).
     The outer neightbours of (B,Y) must be (R,Y) (as the (B,R) 
     are all used already). 
     There common neightbours are determined as (R,B).
     The outer neightbours of (R,B) must be (R,Y).
     The last piece (B,Y) fits into the last place.

     This gives the solution: (Top, Down coloring of the dual graph)

                R
                Y
               / \
          B___Y___B___R
          R   B   R   Y
           \ /     \ /
            R       Y
            Y       B
           / \     / \
          B___Y___B___R
          R   B   R   Y
               \ /
                R
                Y

  Case: (Y,R) in outer position
     The only possible neightbours in inner position are (B,R).
     But this is impossible as both (B,R) have common edge.