Re: find subsets
- To: mathgroup at smc.vnet.net
- Subject: [mg104725] Re: [mg104694] find subsets
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 8 Nov 2009 06:43:38 -0500 (EST)
- Reply-to: hanlonr at cox.net
Ns = Subsets[{a, b, c, d, e}, {3}];
Needs["Combinatorica`"]
Ns == KSubsets[{a, b, c, d, e}, 3]
True
Cases[Ns, {___, b, ___}]
{{a, b, c}, {a, b, d}, {a, b, e}, {b, c, d}, {b, c, e}, {b, d, e}}
% == Cases[Ns, _?(! FreeQ[#, b] &)]
True
%% == Select[Ns, ! FreeQ[#, b] &]
True
Cases[Ns, {___, b, c, ___}]
{{a, b, c}, {b, c, d}, {b, c, e}}
% == Cases[Ns, {___, b, ___, c, ___}]
True
%% == Cases[Ns, _?(! FreeQ[#, b] && ! FreeQ[#, c] &)]
True
Bob Hanlon
---- r_poetic <radford.schantz at mms.gov> wrote:
=============
Hello,
here is a quick question from a novice.
First, generate a list of subsets using Subsets or KSubsets
functions.
Ns = KSubsets[{a,b,c,d,e},3] = {{a,b,c},{a,b,d},{a,b,e},{a,c,d}, etc.}
How can one infer from that list a list of its members that contain
given elements, e.g. a list of members that contain b, or those that
contain both b and c? For illustration:
Some_function[Ns, b and c] = { {a,b,c},{b,c,d},{b,c,e}}
I keep thinking there must be a direct way to do this using Select or
some other function?