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Poly Geeks Gone Wild

So, what happens when you get a bunch of polyamorous geeks together at a party? Someone starts charting relationship configurations, and someone else starts wondering how many possible relationship configurations there are in a particular group, and someone else gets out a calculator and a sheet of paper, and...

As it turns out, the equation that will tell you for any size group of people n how many possible relationship configurations (couples, triads, and so on) are possible within that group is pretty complicated. It took a lot of work and many sheets of paper, and the considerable brainpower of a couple people with degrees in mathematics, but the equation is:



This will tell you for any group of people n how many possible relationship configurations exist in that group.

And the number goes up fast. Scary fast. For n=9, there are 502 possible relationship configurations in that group. The number of people in the Squiggle I belong to is 15; I haven't calculated the number of possible relationship configurations exist in such a group.

I think I'm going to make this formula into a T-shirt.

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Comments

( 14 comments — Leave a comment )
james_the_evil1
Apr. 9th, 2006 01:32 am (UTC)
Jeez Franklin, quit being a geek & get back to the party :-p
janetmiles
Apr. 9th, 2006 01:46 am (UTC)
6,355 (assuming I didn't screw anything up).
sweh
Apr. 9th, 2006 02:17 am (UTC)
That's a standard combinational formala for "n choose r"
eg
http://www.mathsrevision.net/alevel/pure/binomial.php

That was a formula taught to me at age 16 or 17; before I went to university. Ouch, over 2 decades ago. The mathematicians in your group should have known it without thinking!
tacit
Apr. 9th, 2006 04:56 pm (UTC)
ep. Actually, early on in the conversation, one of the math geeks said something along the line of "This looks a lot like Pascal's Triangle" to me, and predicted that the final equation would be a Riemann sum of "n choose r." He turned out to be right, of course, but we had to do a lot of arguing and scribbling on sheets of paper before we believed him. :)
kiwitayro
Apr. 9th, 2006 02:23 am (UTC)
josh (who understood this equation and has your same "got DSS?" shirt) wants to know what k is.
quaryn_dk
Apr. 9th, 2006 08:11 am (UTC)
I'm thinking k is just a counter, and n is the number of people in the group.
tacit
Apr. 9th, 2006 04:56 pm (UTC)
Bingo. n is the size of the group; k is the counter for the Riemann sum.
quaryn_dk
Apr. 9th, 2006 08:09 am (UTC)
Oh, good gracious. That's kinda frightening. I'll have to show that one to my computer-scientist husband...
quaryn_dk
Apr. 9th, 2006 08:20 am (UTC)
Wait a sec... I don't think this formula will work for a group of three... you'd end up with a zero in the denominator, wouldn't you?
blaisepascal
Apr. 9th, 2006 09:18 am (UTC)
Nope, because you'd only have one term to sum: 3!/(2!1!) which is 6/(2*1) = 3. His formula doesn't count the everyone-relationship, singleton relationships (it takes at least two to tango in Tacit's world) or the trivial nil relationship.
tacit
Apr. 9th, 2006 07:16 pm (UTC)
The sum shouldn't have excluded the everyone-relationship (that means the original version of the formula was off by one), but does intentionally exclude singletons and the nil relationship.
blaisepascal
Apr. 9th, 2006 09:22 am (UTC)
You do realise that n!/(k!(n-k)!) are just entries in the nth row of Pascal's Triangle, and that the sum of any row in Pascal's Triangle is 2^n right? Take away the k=0, k=1, and k=n entries (which are 1, n, and 1 respectively) and you get as a closed form 2^n-n-2.

I get, by both closed form and summing the damn thing, 501 for n=9 instead of your 502, and I get 32768-15-2 = 32751 for n=15.

Of course, I recognised Pascal's Triangle instantly.
tacit
Apr. 9th, 2006 05:00 pm (UTC)
Actually, that brings up a good point; the Riemann sum given here doesn't consider the n-case (that is, there should be 4 possibilities in a triad, the three dyads plus the triad itself). There are 502 possibilities for n=9, if you count the 9-some case as well. Didn't catch that last night; we need to add one to the result.

We (well, not me, but the math geeks among us) recognized the solution once we closed in on it, but we took a very backward and circuitous route to get there.
pstscrpt
Apr. 10th, 2006 02:12 pm (UTC)
When I was taking Discrete Math, using k as a variable bugged the shit out of me because my science classes had conditioned me to think of k as a constant.

In this case, though the denominator almst spells out "kink!", which I think outweighs it.
( 14 comments — Leave a comment )