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