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MathGroup Archive 2006

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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
> 


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