Cyclic and Non-cyclic Permutations

Question

Mathematica has a built in function to generate all permutations of a given list of elements; Permutations

I can't find an equivalent function to generate cyclic permutations only in the documentation. Here is my function that achieves this goal:

CyclicPermutations[list_] := 
 RotateRight[list, #] & /@ (Range[Length[list]] - 1)

Is there an in-built function somewhere that I've not been able to find?

And then a similar question which I don't have my own answer to. I would like to also generate all noncyclic permutations, ie. the set of permutations minus the set of cyclic permutations. I'm not sure of a good way to do this, I can think up some methods which use Permutations and my CyclicPermutations and then maybe DeleteCases, but I think this will be comparatively very inefficient. Does anyone else have a better method?

@yode Please post an answer. I did not remember that Permute can work with a group. — Szabolcs, Jul 08 '16 at 14:53

yode · Accepted Answer · 2021-12-05T15:10:07.717

13

Per the request, I post my comment as an answer:

First question

cy := Permute[#, CyclicGroup[Length@#]] &
cy[Range@5]

{{1, 2, 3, 4, 5}, {2, 3, 4, 5, 1}, {3, 4, 5, 1, 2}, {4, 5, 1, 2,3}, {5, 1, 2, 3, 4}}

Second question

We can use the Complement mentioned by J.M. in his comment. I suppose that the order is $5$; then, you can use the following method to get noncyclic permutations:

Complement[Permutations[Range[5]], cy[Range@5]]

{{1,2,3,5,4},{1,2,4,3,5},{1,2,4,5,3},{1,2,5,3,4},{1,2,5,4,3},{1,3,2,4,5},{1,3,2,5,4},<<101>>,{5,3,4,2,1},{5,4,1,2,3},{5,4,1,3,2},{5,4,2,1,3},{5,4,2,3,1},{5,4,3,1,2},{5,4,3,2,1}}

edited Dec 05 '21 at 15:10

answered Jul 08 '16 at 15:16

yode

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Sadly Permute[#, CyclicGroup[Length@#]] & proves to be orders of magnitude slower than what the OP started with! :-( – Mr.Wizard Jul 11 '16 at 16:31
Complement is great, thanks for your help – Jojo Jul 12 '16 at 13:09
@Mr.Wizard The CyclicPermutations just do one calculation to get a list.But the cy do n times... – yode Jul 12 '16 at 13:30
CyclicPermutations[Range@5] returns {{1, 2, 3, 4, 5}, {5, 1, 2, 3, 4}, {4, 5, 1, 2, 3}, {3, 4, 5, 1, 2}, {2, 3, 4, 5, 1}}. How is Permute better here? – Mr.Wizard Jul 12 '16 at 22:41

score 9 · Answer 2 · answered Jul 12 '16 at 23:41

9

cp=HankelMatrix[#, RotateRight@#] &;

Should perform quite well and returns packed array...

answered Jul 12 '16 at 23:41

ciao

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Great! I never remember these specialized generators. This clearly should be the Accepted answer. – Mr.Wizard Jul 13 '16 at 04:35
I guess my memory does not extend three years back :-/ – Mr.Wizard Jul 13 '16 at 05:01
I think this is a winning answer. :) – yode Jul 13 '16 at 05:07
But note my cy can be used by cy[{2, 7, 9}],which make life ease in some case.Of course,it out of this topic. – yode Jul 13 '16 at 05:10
@yode cp[{2, 7, 9}] works too though, doesn't it? – Mr.Wizard Jul 13 '16 at 05:12
@Mr.Wizard Er..little mistake by me,thanks for point out. :) – yode Jul 13 '16 at 05:16
@Mr.Wizard - had not seen that, but I've a library of sneaky functions using Hankel/Toeplitz matrices to build "interesting" configurations... glad you like it. – ciao Jul 13 '16 at 05:16
@ciao If you really do would you consider sharing it? I like your imagination and knowledge and I am sure I would learn from it. – Mr.Wizard Jul 13 '16 at 05:21
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@Mr.Wizard - sure, I'll gather/clean and put as answer to the "elegant array operations" (or whatever it's called) question when I have time. Remind me if I forget. – ciao Jul 13 '16 at 06:22
@ciao That question is about basic/common transformations and operations. Would not expression generation be better in its own Q&A? – Mr.Wizard Jul 13 '16 at 06:24

score 4 · Answer 3 · answered Jul 12 '16 at 22:51

At least in version 10.1 under Windows there is a performance problem with yode's Permute solution. For comparison here is his code, Joe's original code, and a variation of my own:

fn1[list_] := RotateRight[list, #] & /@ (Range[Length[list]] - 1)

fn2 = Permute[#, CyclicGroup[Length@#]] &;

fn3[a_] := Array[RotateLeft[a, #]&, Length @ a]

The results are all equivalent under sorting:

Sort @ # @ Range @ 4 & /@ {fn1, fn2, fn3}

{{{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}},
 {{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}},
 {{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}}}

The performance however is not!

AbsoluteTiming @ Timing @ Do[#@Range@500, {50}] & /@ {fn1, fn2, fn3} // Column

 {0.046702, {0.0312002, Null}}
{2.48765, {2.44922, Null}}
{0.0456291, {0.0156001, Null}}

Permute on CyclicGroup is some fifty times slower than the other methods here.

My fn3 is just a hair faster than fn1 and IMHO somewhat cleaner, so it is my proposal.

kglr · Answer 4 · 2017-07-17T20:36:36.173

A few more ... The first one using Partition not too horribly slow

fn4[a_] := Partition[a, Length@a, 1, {1, 1}]
fn5[a_] := ListConvolve[{1}, a, #] & /@ a
fn6[a_] := ArrayPad[a, {1, -1} (# - 1), "Periodic"] & /@ a

fn4@Range@4

{{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}}

Equal @@ (Sort@#@Range@4 & /@ {fn1, fn2, fn3, fn4, fn5, fn6})

True

First@AbsoluteTiming@Do[#@Range@500, {50}] & /@ {fn0, fn1, fn2, fn3, fn4, fn5, fn6} // 
 TableForm[#, TableHeadings -> {{"fn0", "fn1", "fn2", "fn3", "fn4", "fn5", "fn6"}, None}] &

where fn0[a_] := HankelMatrix[a, RotateRight@a] is from ciao's answer.

Cyclic and Non-cyclic Permutations

4 Answers4

First question

Second question

Linked