laziness, impatience, and hubris | |
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Re: unordered sets of N elementsby dragonchild (Archbishop) |
on Jul 13, 2004 at 13:42 UTC ( [id://373979]=note: print w/replies, xml ) | Need Help?? |
You are looking for combinations. Just with additional constraints. Look at a simpler example - 2 dice. Your rules says that 1-3 and 3-1 are the same. So, we take the first number in the first die and look at the possibilities in the second - 6. Then we take the second number in the first die, leaving us 5 possibilities in the second die. (The sixth, 2-1, is disallowed). Following the pattern leaves us 6+5+4+3+2+1 = 21 possibilities.
Extending to three dice, we have the following - set to 1-1 and we have 6 possibilities. 1-2 and we have 5, etc. So, with the first die as 1, we have the above 21 possibilities. Setting the first die to 2 and we have 5 possibilities for the second and 4 for the third - leaving us 15 total possibilities. So, the total ends up being 21 + 15 + 10 + 6 + 3 + 1 = 56, which is borne out by your code. You can extend the pattern upwards. The actual formula involves a bunch of Sigmas.
I'm sure there's a straight formula, but my brain hurts. :-) ------
Then there are Damian modules.... *sigh* ... that's not about being less-lazy -- that's about being on some really good drugs -- you know, there is no spoon. - flyingmoose I shouldn't have to say this, but any code, unless otherwise stated, is untested
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