3
$\begingroup$

A bit more detail: I am generating a list of all 3-color combination from a list of colors (say 10, but can increase/decrease). I am generating triangles with these combinations, where each side has one of the 3 colors.

I want to tile a flat surface with these 3-color triangles, but I also want sides that are adjacent (from two neighboring triangles) to share the same color. And, finally, each combination should be repeated just once over the surface. I am also happy to hear a solution that doesn't depend on colors/geometry but numbers of course. I just don't know how to phrase it that way.

  • 0
    Shall we include or exclude the triangles in which some of the edges (or even all three) are colored the same? Also, what's that with "just once"? Isn't our plane _infinite_?2017-01-13
  • 0
    What about choosing a triangulation, color its edges then check if the involved triangles match any other constraint?2017-01-13
  • 0
    What shape is the flat surface, and how are pieces at the edge of the surface handled?2017-01-13
  • 0
    Thanks for the comments! Each color should appear just once maximum per triangle. When I say each combination just once I mean: I make a list of all combinations if I have say 10 starting colors and then want to tile them next to each other. (I guess this answer also the infinite question).2017-01-13
  • 0
    Jack, do you have any tips on what you suggested? I think I'm getting the general idea but (sorry!) don't know enough to know where to start.2017-01-13
  • 0
    Joffan, I guess following from what I mentioned just now: I would like to tile them side by side, it doesn't really matter what shape the whole thing will have or what kind of edges. thanks!2017-01-13
  • 1
    It might help if you gave an example or two of what you want, with a smaller list of colors, say $4$ or $5$ instead of $10$. For that matter, if there are only $3$ colors on the list, is there just one combination or two -- i.e., are mirror images considered the same or different?2017-01-13

1 Answers 1

4

You're looking for MacMahon triangles.