Re: Help on Partitions, Again!!! (fwd)

*To*: mathgroup at smc.vnet.net*Subject*: [mg24658] Re: [mg24636] Help on Partitions, Again!!! (fwd)*From*: Rob Pratt <rpratt at email.unc.edu>*Date*: Mon, 31 Jul 2000 09:23:26 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Addendum to my previous response: Of course, the procedure I outlined will return both {{A,B,C},{D,E,F}} and {{D,E,F},{A,B,C}} as two different objects. If you consider these to be the same object, you could then map Sort to the resulting pairs and then apply Union to eliminate duplicates. Or you could use Union with an appropriate SameTest. (The original procedure produces 20 partitions, but I believe you only wanted 10.) But a more efficient approach would be to remove A from the original list and run the procedure on {B,C,D,E,F}, with the convention that the element A belongs to the first set in each pair. You could explicitly add in A as a final step. Also, if the two parameters add up to something less than the length of the list (such as 2 + 1 < 6), you could just apply the procedure to each subset of the appropriate size (such as 2 + 1 = 3). This is not overkill, as there will be no duplicates. Perhaps the best approach would be to look at the code used to define KSubsets and doctor it to suit your needs. Steven Skiena's book Implementing Discrete Mathematics describes the algorithms and code for the Combinatorica package (which he wrote). Rob Pratt Department of Operations Research The University of North Carolina at Chapel Hill rpratt at email.unc.edu http://www.unc.edu/~rpratt/ ---------- Forwarded message ---------- From: Rob Pratt <rpratt at email.unc.edu> To: mathgroup at smc.vnet.net Subject: [mg24658] Re: [mg24636] Help on Partitions, Again!!! Jose, If the two parameters will always add up to the length of the list (as in your example with 3 + 3 = 6), you can use KSubsets in the Combinatorica package. Just ask for the complement along with each subset. More explicitly, use KSubsets[{A,B,C,D,E,F},3], then map Complement[{A,B,C,D,E,F},#]& to the result, then combine the two lists in pairs with Transpose[{lis1,lis2}]. Rob Pratt Department of Operations Research The University of North Carolina at Chapel Hill rpratt at email.unc.edu http://www.unc.edu/~rpratt/ On Fri, 28 Jul 2000, Jose Prado de Melo wrote: > > > Hello, MathGroup > First of all, thanks for your attention. > To be more specific: > It's not too dificult to calculate the solution of the problem: > How many ways, can the set {A,B,C,D,E,F} be separeted into two parts > with three elements in each? > Answer: x = 6!/(2!.3!.3!) = 10 > I'm looking for a function to generate all the partitions using > Mathematica 3.0 . > I'm not sure, but I think the package Combinatorica doesn't have a > function to do this. > For example, I'm trying to think up a function f like this one: > > In[ ] = f [ {A,B,C,D,E,F},{3,3}] > Out [ ] = { { {A,B,C},{D,E,F} }, { { > A,B,F},{C,D,E}},...................} and so on. > In [ ] = Length[%] > Out [ ] = 10 > > Please, help me. > Thanks! > > > >