Re: combinatorics problem
- To: mathgroup at smc.vnet.net
- Subject: [mg7551] Re: [mg7514] combinatorics problem
- From: "w.meeussen" <w.meeussen.vdmcc at vandemoortele.be>
- Date: Fri, 13 Jun 1997 19:37:51 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Gee, i must be thick today,
i don't understand why
In[5]:=
Prepend[#,1]&/@(Permutations[Range[4]]+1)
isn't enough to get what you want.
tell me where i'm off track.
wouter.
At 10:48 10-06-97 -0400, Xah Lee wrote:
>A combinatorics related programing problem.
>
>I have a list Permutations[Range[n]]. I want to reduce the list by the
>following criteria:
>
>(1) Two lists are considered the same if one is the other shifted.
>(RotateRight).
>
>(2) Two lists are considered the same if one is the other reversed.
>
>The following code captures my criteria.
>
>Clear[sameTestQ];
>sameTestQ[li1_?PermutationQ,li2_?PermutationQ]:=
> Or@@(SameQ[li1,#]&/@
> Flatten[{#,Reverse at #}&/@Table[RotateRight[li2,i],{i,Length at li2}],
> 1])/;(Length at li1==Length@li2)
>
>Ideally, I should get what I want by
>
>Union[Permutation[Range at n], SameTest->sameTestQ]
>
>but that doesn't work. The reason can be seen in the example:
>
>Union[{{2,1,3,4},{3,4,1,2},{3,4,2,1}},SameTest->sameTestQ]
>
>It returns the argument unchanged. The first and last elements are the same,
>but wasn't filtered out probably because Union only compare adjacent pairs
>after sorting.
>
>I'm working my way using DeleteCases or Select and I'll get it sooner or
>later. I'm wondering if you had similar problems and solutions before. It
>does looks like the best way is simply try to generate my list directly,
>perhaps by writing my own fuction. The problem is then a mathematical one.
>
>The solution answers the question: Given n points in space, in how many ways
>can one connect them together to form a loop.
>
> Xah
> xah at best.com
> http://www.best.com/~xah/SpecialPlaneCurves_dir/specialPlaneCurves.html
> Mountain View, CA, USA
>
>
>
>
Dr. Wouter L. J. MEEUSSEN
eu000949 at pophost.eunet.be
w.meeussen.vdmcc at vandemoortele.be