Re: Help on Partitions, Again!!!
- To: mathgroup at smc.vnet.net
- Subject: [mg24651] Re: Help on Partitions, Again!!!
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Mon, 31 Jul 2000 09:23:21 -0400 (EDT)
- References: <8lt1mt$2m2@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
We can use the Combinatorica function KSubsets (but a custom function should
be quicker)
<< DiscreteMath`Combinatorica`
lst = {A, B, C, D, E, F};
KSubsets[lst, 3]
{#, Complement[lst, #]} & /@ %
Sort /@ %
Union[%]
{{{A, B, C}, {D, E, F}}, {{A, B, D}, {C, E, F}},
{{A, B, E}, {C, D, F}}, {{A, B, F}, {C, D, E}},
{{A, C, D}, {B, E, F}}, {{A, C, E}, {B, D, F}},
{{A, C, F}, {B, D, E}}, {{A, D, E}, {B, C, F}},
{{A, D, F}, {B, C, E}}, {{A, E, F}, {B, C, D}}}
Length[%]
10
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Jose Prado de Melo" <jpmelo at iconet.com.br> wrote in message
news:8lt1mt$2m2 at smc.vnet.net...
>
>
> 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!
>
>
>