[SPOILER] Re: more euler problems

Albert van der Horst <albert@[EMAIL PROTECTED]
> writes:
>I have other things to do, but cannot help to tackle the colored
>triangle

I guess you mean Problem 189.  I looked at this for half an hour or
so, and I think I have an approach to solve it, but am too lazy to
program it.  Here's my approach:

Divide the triange into 4 equally sized subtriangles; each of the
subtriangles has 4 element triangles on each edge.  There's a central
subtriangle and 3 corner subtriangles; they interface with the center
triangle along one edge, and not at all with each other.

Now, enumerate the possible colourings of one subtriangle, and for
each of the 81 colour combinations on one edge, remember how many
colourings have that edge.

Next, look at the interface of two subtriangles: for all compatible
combinations of edges, multiply the number of colourings represented
by the combinations and add up the products.  That gives you the
number of colourings for the combination of two subtriangles.

Ideally, the three edges of the central subtriangle would be
independent enough that we can just apply the factor we determined
from adding the first corner triangle two times more to get the number
for the combination of all subtriangles.

But I suspect it won't be so easy.  Then we have to deal with all the
egdes of the central subtriangle at once, i.e., 19683 potential
combinations.  Should not be too bad, either.

- anton
-- 
M. Anton Ertl  http://www.complang.tuwien.ac.at/anton/home.html
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