Re: Re: How to calculate a union

*To*: mathgroup at smc.vnet.net*Subject*: [mg104607] Re: [mg104533] Re: How to calculate a union*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Wed, 4 Nov 2009 01:41:42 -0500 (EST)*References*: <hcgmv6$d16$1@smc.vnet.net> <hcjiph$ji9$1@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

I haven't found the whole thread, or maybe you're not being specific enough. Could the sum you're talking about be StirlingS2? (look it up in Help) {1, 3, 4, 8, 9, 273} // Total 298 If you want to know how many ways 298 objects divide into 6 non-empty sets, the answer is six = StirlingS2[298, 6] 1075821248827096561510600533061021312147292178470709061394244250242240\ 4204344618445256417817622689352182506360801499420973972117501295752112\ 7858761331082536332833271473290414868861248314974063061884865879515098\ 26099869741728596859 or... the number of ways 298 divides into UP TO 6 non-empty sets: upTo6 = Total@StirlingS2[298, Range@6] 1075821248827096561510616896705905637902469164377225682313054167627282\ 0852663455496893658655311541729640462357823813462486418083469381491114\ 7144460089469673867410642929568451368797860156877490757740974231983345\ 77877398619316408371 Both numbers agree in the first 23 digits: -1 + Min@Position[IntegerDigits@upTo6 - IntegerDigits@six, _?Positive] 23 That's from a total 230 digits: Ceiling@Log[10., {upTo6, six, upTo6 - six}] {230, 230, 207} If that's off-topic... never mind! Bobby On Tue, 03 Nov 2009 01:54:57 -0600, Ramiro <ramiro.barrantes at gmail.com> wrote: > Hi, > > 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 > 699577983584109258531043307610450240451701311502431378368296040472451355165796880 > > Any suggestions? > > Ramiro > > On Nov 1, 4:01 am, ADL <alberto.dilu... at tiscali.it> 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. >> >> ADL >> >> On Oct 31, 7:54 am, Ramiro <ramiro.barran... at gmail.com> 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. >> > ... >> >> > > -- DrMajorBob at yahoo.com