Re: subsets of a set
- To: mathgroup at smc.vnet.net
- Subject: [mg70033] Re: [mg70003] subsets of a set
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 1 Oct 2006 04:08:14 -0400 (EDT)
- Reply-to: hanlonr at cox.net
In version 5.1 or later use Subsets
n=4;
(Cases[Union[Union /@
Tuples[Range[n], n]], #1] & ) /@
Table[_, {i, n}, {j, i}]==
Rest@Split[Subsets@Range[n],
Length[#1]==Length[#2]&]==
Split[Subsets[Range[n],{1,n}],
Length[#1]==Length[#2]&]
True
Bob Hanlon
---- dimmechan at yahoo.com wrote:
> Actually it was more easily than I believed
>
> lst=Tuples[Range[7], 3];
> Cases[Union[Union /@ lst], {_, _, _}]
> {{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 2, 6}, {1, 2, 7}, {1, 3, 4},
> {1,3, 5}, {1, 3, 6},{1, 3, 7}, {1, 4, 5}, {1, 4, 6}, {1, 4, 7},
> {1, 5, 6}, {1, 5, 7}, {1,6, 7}, {2, 3, 4},{2, 3, 5}, {2, 3, 6},
> {2, 3, 7}, {2, 4, 5}, {2, 4, 6}, {2, 4, 7}, {2,5, 6}, {2, 5, 7},
> {2, 6, 7}, {3, 4, 5}, {3, 4, 6}, {3, 4, 7}, {3, 5, 6}, {3, 5, 7},
> {3,6, 7}, {4, 5, 6},{4, 5, 7}, {4, 6, 7}, {5, 6, 7}}
>
> Anyway, I really appreciate your guidance for other approaches.
>
> BTW I discovered that (not big deal for you...but quite big for me!)
>
> Flatten[Outer[List, Range[7], Range[7], Range[7]], 2] ==
> Tuples[Range[7], 3]
> True
>
> Also the following command gives all the sublsets of {1,2,3,4}
>
> (Cases[Union[Union /@ Tuples[Range[4], 4]], #1] & ) /@
> Table[_, {i, 1, 4}, {j, 1, i}]
> {{{1}, {2}, {3}, {4}}, {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4},
> {3, 4}}, {{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}},
> {{1, 2, 3, 4}}}
>
> Any other ideas to get the same output?
>
> Thanks in advance for any help.
>
> Dimitris Anagnostou
>