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Treebeard's Stumper Answer
26 October 2001

Tricky Tetrahedra

Here's a tricky visualization stumper. Imagine a perfect tetrahedron. That's a triangular pyramid where every side including the base is an identical triangle. Now imagine lots of tetrahedra and start to color their sides. If you have four colors to work with, then how many different pyramids can you paint so that a duplicate cannot be found by rotating or turning the pyramids? For help, there's a nifty way to make a tetrahedron from a small envelope at http://tremor.nmt.edu/tetra.html. How about coloring a cube with six colors?

Four different ways of visualizing a tetrahedron: as a wire frame, or an animation (by Rüdiger Appel), or a colored model, or a paper plan ready to fold up. How many unique colorings are possible, with just four colors to work with, if every face has a single color?


It's hard to visualize coloring the faces of a triangular pyramid because of the strong symmetry. With two colors, each can cover 1, 2, or 3 sides. There are 6 ways of picking 2 different colors from the 4 available, so that makes 3 x 6 = 18 colorings. Similarly,
with 1 color:1 wayx  4 color choices=4
with 2 colors:3 waysx6 color choices=18
with 3 colors:3 waysx4 color choices=12
with 4 colors:2 waysx1 color choice=2
That makes a total of 36 unique colorings. There are details and colored pictures below.

Notes:

Here's how I figure it. The colors are RGBY (Red, Green, Blue, and Yellow), and the flag on the pyramid shows the color of the hidden side. What makes this such a hard visuaization stumper is that each of these tricky tetrahedra is identical when rotated due to the overwhelming symmetry. Try to visualize how these are all really the same!

  
    
 
 
With one color:
   This is easy. Each face is the same color, and there are four colors available, so there's 1 x 4 = 4 possible colorings.
    
 
 
With two colors:
   With two colors, each can cover 1, 2, or 3 sides. There are 6 ways of picking 2 different colors from the 4 available, so that makes 3 x 6 = 18 colorings.

The colorings are:
  RG(RGGG RRGG RRRG)
  RB(RBBB RRBB RRRB)
  RY(RYYY RRYY RRRY)
  GB(GBBB GGBB GBBB)
  GY(GYYY GGYY GGGY)
  BY(BYYY BBYY BBBY)

    
 
 
With three colors:
   With three colors, each can cover 2 sides in turn for 3 combinations. There are 4 ways of picking 3 different colors from the 4 available, so that makes 3 x 4 = 12 colorings.

The colorings are:
  RGB(RRGB RGGB RGBB)
  RGY(RRGY RGGY RGYY)
  GBY(GGBY GBBY GBYY)
  RBY(RRBY RBBY RBYY)

    
 
 
With four colors:
   This is the tricky one since there are two mirror images that cannot be matched by rotating, even though there's only one way of picking 4 colors from the 4 available. That makes 2 x 1 = 2 colorings.

That's 4 + 18 + 12 + 2 = 36 colorings total with 4 available colors.

Coloring a cube with six colors is much trickier than this! Symmetry and mirror images abound. I haven't had time to work it out, but Graybear sent this analysis in betwen figuring grades at school:

When I say 'triples adjacent', I mean that three sides of the same color are arranged like the 1, 2, and 3 on a standard die. 'Triples inline', would imply the 1, 2, and 6 would be the same color, for example. 'Doubles adjacent', likewise, would be two adjacent sides of the same color (e.g. 1 and 2), while 'doubles opposite would mean the 1 and 6 are the same color. You can probably deduce the rest.

I. One color: 6 possibilities
II. Two colors
  A. 5 (sides of one color) + 1 (side of another color): 30 poss.
  B. 4 + 2
    1) doubles adjacent: 30 poss.
    2) doubles opposite: 30 poss.
  C. 3 + 3
    1) triples adjacent: 15 poss.
    2) triples inline: 15 poss.
III. Three colors
  A. 4 + 1 + 1
    1) singles adjacent: 60 poss.
    2) singles opposite: 60 poss.
  B. 3 + 2 + 1
    1) triples adjacent: 120 poss.
    2) triples inline
      a) doubles adjacent: 120 poss.
      b) doubles opposite: 120 poss.
  C. 2 + 2 + 2
    1) three doubles adjacent: 40 poss. ?
    2) two doubles adj, one opp: 60 poss.
    3) three doubles opposite: 20 poss.
IV. Four colors
  A. 3 + 1 + 1 + 1
    1) triples adjacent: 120 poss.
    2) triples inline: 180 poss.
  B. 2 + 2 + 1 + 1
    1) both doubles adj, singles opp: 90 poss.
    2) both doubles adj, singles adj: 180 poss.
    3) one double opp, one double adj: 90 poss.
    4) both doubles opposite: 90 poss.
V. Five colors: 450 poss.
VI. Six colors: 30 poss.

TOTAL: 6 + 120 + 600 + 750 + 450 + 30 = 1956
I've found two other answers on the Web at Ken Duisenberg's Puzzle of the Week that don't agree with Graybear's: either 2226 or 1866. Graybear is giving the total numbers for all six colors, so I divided his results by the number of ways to select n colors out of 6 to find how many ways there are of coloring the cube with just 1,2,3,...,6 colors, using each color at least once in each case. These divisors come from Pascal's Triangle:

                             1
                           1   1
                         1   2   1
                       1   3   3   1
                     1   4   6   4   1
                   1   5  10   10  5   1
   This row -->  1   6  15  20   15  6   1
So using each of n colors (n=1,2,3,4,5,6) at least once, in how many ways can you color a cube, such that a duplicate cannot be found simply by rotating the cube? Here are the candidates:

number of colors1   2   3   4   5   6   total

Graybear18305075301956
Graybear (2)18305675302046
Al Zimmermann & Gerd Baron18306875302226
Joseph DeVincentis18305645301866

ways to select that
many colors from 6
615201561 

Graybear sent this update (07nov01):

I found 90 more combinations for four colors. They fall into the category of 1 double opposite, 1 double adjacent, singles adjacent, so there is a total of 180 in that category instead of 90. This brings my 4-color total of 840 (56 x 15) in line with one of the other solutions. I'm now standing by my total of 2046 solutions.

The disagreements are with 4 and 5 colors. This should make it much easier to think about and settle.

What about the other Platonic Solids: octahedrons, dodecahedrons, and icosahedrons? And there's many other polyhedra to color if we relax the requirement that all faces be congruent. Is there a general solution for how many unique ways there are to color a n-faced polyhedron with n-colors?

Here are a few Web links for more research:

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Copyright © 2001 by Marc Kummel / mkummel@rain.org