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Re: How to calculate a union

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  • Subject: [mg104533] Re: How to calculate a union
  • From: Ramiro <ramiro.barrantes at>
  • Date: Tue, 3 Nov 2009 02:54:57 -0500 (EST)
  • References: <hcgmv6$d16$> <hcjiph$ji9$>


Thanks so much for this, I was thinking how one could calculate it.
Yes, if n=273 you get some serious amounts of numbers that I need to
add.  I don't think that it is even possible to store such large
amounts.  The thing is, for each possible k-tuple, say {1,3,4,8,9,273}
I can calculate the corresponding probability very easily, and I then
need to add all the probabilities of all the k-tuples and add/subtract
as necessary for a grand number that would correspond to the
probability of the union, the entire calculation is trivial, but there
are just too many numbers!!!  For n=273, the number of 140-tuples are

Any suggestions?


On Nov 1, 4:01 am, ADL <alberto.dilu... at> wrote:
> I do not know why Export seems to behave as you say, but you can
> compute in advance the number of terms in each file with this formula:
> filelengths = Reverse@CoefficientList[Expand[(x + 1)^n], x]
> where the first value (always 1) is for index 0, and so on.
> The total Total[filelengths] is of course equal to 2^n.
> With this, you might compute in advance the memory required for your
> operation and establish when it is too much.
> On Oct 31, 7:54 am, Ramiro <ramiro.barran... at> wrote:
> > Hi,
> > I am trying to compute the probability of a union, the formula
> > requires to sum and subtract all possible combinations of
> > intersections for all pairs, triplets, etc.  When n is big, getting
> > all the terms is a challenge.
> > ...

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