Mods: Please don't move this. This post is a puzzle related to finding a 16 color algorithm for any generic soccerball or truncated icosahedron shape, in which every color is used exactly twice and no color touches another color more than once:

I had a brainstorm the other week about how to sticker the Tuttminx puzzle. It seems that with 32 faces, the Tuttminx has well too many colors for the average person to easily distinguish in average lighting. This may eventually lead to extreme eye fatigue. So what about 16 colors, but with non-opposite faces? My understanding is that every color has enough stickers for two hexagons and two pentagons. I don't know if it is entirely possible or not, but the rules for a coloring scheme in which every part is unique, dictate that no two edges or corners can have the exact same color combination. This means that the color scheme would work equally well on a Fuutminx. The edges are a simple place to start: to be all unique, basically requires that no face of color "A" is permitted to contact any face of color "B" more than once. The sixteen colors are of no consequence and can be decided later. Two faces of every color, "A" through "P" are chosen. It matters not whether each pair is pentagon or hexagon. Any color can be any combination. This obviously means that any particular color cannot be edge adjacent to, or diagonally adjacent to itself. This leaves 30-1-6-6=17 possible faces remaining for hexagons, and 30-1-5-5=19 possible remaining faces for pentagons. So, mostly, each face's same-colored mate will be located somewhere within, or bordering the opposite hemisphere of the puzzle. Like I said before, I don't know for sure it if is possible with 16 colors, but if it is, it would make an ideal coloring scheme for the Tuttminx. I guess one could start by placing sticky notes for "A" through "P" on a soccer ball and seeing where that goes...

I do have a very simple proof that 16 colors would be the theoretical minimum for this type of arrangement. If less than 16 colors are chosen for the Tuttminx, than at least some colors must be present on three or more faces. If we place color "A" on three pentagons (the face type with the fewest neighbors), and one "A" pentagon shares a border with colors "B", "C", "D", "E", & "F", then the next "A" pentagon will share borders with "G", "H", "I", "J", & "K", and finally the third "A" pentagon must share borders with "L", "M", "N", "O", and "P". Since "P" is the 16th letter of the alphabet, this means tiling the Tuttminx with 15 or fewer colors (in which every piece is unique) is impossible.

If anyone comes up with a solution, could you please show it by using paint (or similar graphics program) by pasting the letters "A" through "P" or the numbers "1" through "16", or the hexadecimal digits "0" through "F" (if you're a computer geek) into the truncated icosahedronal net:
Or, if you can prove it impossible, please provide a graphical explanation why it can't be done. Thanks...

I loved the question and had played a bit with it. Programming could do the job here,
and it won't be too hard for someone to get a nice answer.

I will just provide a start (in some way) of two non-symmetric configurations, where
it kind of shows such a ten-coloring of the hexagons, and a four coloring of the pentagons.





The reason I provided this, was to show yet another coloring way, where the pentagons
should inherit the colors of their five neighboring hexagons, and each piece (on the pentagon
face) should have a "super-stickering"which clearly indicates the face they belong to.
It makes much more sense, and only ten colors are used in total.

Now... how about combining 16 colors for the whole pattern?




Pantazis