Re: [SPOILER] Re: more euler problems

cac <cac@[EMAIL PROTECTED]
> writes:
>Hmmm. My approach was to divide the triangle into an upper half and a 
>lower half, and enumerate the boundary as you did. Where I get wedged
>is the rotation and reflection issue. I need to figure out how many
>non-distinct colourings there are, and subtract that from my answer;

As far as I understand the task, rotation and reflection is a
non-issue.  If you enumerate all colourings for one orientation,
that's the answer.

If you had a way to create only one of, say, the two mirror images
(along some axis), then you would have to count it twice, unless the
colouring is symmetric.  But with the usual enumerative approach we
get both mirror images anyway.

- anton
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