The Origami Forum

is it possible to make this model...[http://picasaweb.google.com/lh/photo/I0 ... zHx-QeN1vA]...with only 3 colors and no two adjacent units are the same color? I'm too lazy to try and figure it out on my own, so a picture showing the pattern would be great

Check http://books.google.com/books?id=_Nw4Hg ... q=&f=false
It includes a description and an image of the graph that you need to colour.

The basic idea is: first draw a graph, then find a Hamiltonian circuit in it (travel to all vertices without visiting any of the edges more than once, start and end in the same vertex). Next colour the Hamiltonian circle by alternating between two colours. Then determine the colours of the remaining edges. This is where you get to start using your third colour.
While this is not a complete solution, it is a very good starting point. I'm not sure how hard it is to manually find a Hamiltonian circuit, but you can read up on some theory / good algorithms. Maybe someone has posted a solution somewhere online, too.

-- Sara

Yes, it's possible!
But it's very difficult to do with 90 edges because it's an NP-complete problem.
If you PM to me your e-mail address I'll send you a colored graph

I just did a search for a Hamiltonian circuit, and found this video:
http://www.youtube.com/watch?v=eIjcTZGTqQ8
Looks like the remaining edges (the ones not on the Hamiltonian circuit) will all be in the third colour.

This should solve the problem (and is more readily available than an image sent privately). I hope others will profit from it.

-- Sara

Sara wrote: This should solve the problem (and is more readily available than an image sent privately). I hope others will profit from it.

The problem is not so simple as it seems (http://en.wikipedia.org/wiki/NP-complete_problem).
You have to know at least the fundamentals of graph theory.

Anyway, if someone else is interested this is the graph (of course there are other solutions):


[img]http://img16.imageshack.us/img16/6301/bucky90.jpg[/img]

Thanks for posting that image. I noticed that your colouring doesn't actually imply a Hamiltonian circuit, so I decided to give it a go. Below you'll see an image of an alternate colouring.
[fi-ori has circuits of length 6, 10, 10, 34. Check for example red and blue as the colours that imply the circuits.]



The interesting thing about this colouring is that every pentagon and hexagon has at least one module of each colour. It will give you a second option apart from the one fi-ori suggested. It'd be nice to see a picture of two truncated icosahedrons, each using a different colouring scheme. Anyone up for it?

By the way, I found it quite hard to extract the Hamiltonian circuit from the video, so I searched for one myself. So it might be a different one than the one that's shown in the video I linked to above. Not sure.

More: http://www.happyfolding.com/gallery-tru ... _colouring

thank you to everyone for their help. I have a 4 day old baby and have NO energy (or time) to figure this out. Folding is really helping me keep my sanity

Nice coloration Sara! I'll give it a try!

This is an example of my 3-coloration of a truncated icosahedron made with PHiZZ units.
The colors are different from the graph but the pattern is the same



This is the same 3-coloration of another truncated icosahedron

Here's mine from PHiZZ units:



More: http://www.happyfolding.com/gallery-hul ... cosahedron

I managed to pull off the 3 colouration too!!

I know this may come as a shock to you, but ...
I fear you folded too many white modules, and not enough from the other two colours...

-- Sara

3 shades of white, no?

here it is... http://www.facebook.com/photo.php?pid=1 ... 1435370357

Nicely done!

What kind of unit you used?

Here I used Tom Hull's Bouncy units:



instead of pentagon and hexagons there are five- and six-pointed stars

I used Francis Ow's 135 degree module.