(In reply to
Solution-part 1 by Tristan)
I sneaked a peek at Charlie's tetrahedron answer (just the answer), and though it looks completely different, it simplifies to my answer. That's good, but the cube will be much harder, and much more likely to have an error.
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I'm encountering difficulties with the cube. I don't have any thorough method of counting all the rotational groups. Using the same method for solving part 2 will be much harder, and I'm searching for a more elegant method.
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Posted by Tristan
on 2004-09-25 17:26:47 |