>From page two of the notes I reconstruct the same output ` >>> (Permutation(1,5)(2,6,4)).list() [0, 5, 6, 3, 2, 1, 4] ` Could it be that the semantics (which are the same when you start with an ordered list and do only one cycle) are the confusing issue in that you thought it meant elements but it really refers to position?
/c On Friday, March 12, 2021 at 12:15:35 PM UTC-6 [email protected] wrote: > Hi Chris, > > The convention that I'm familiar with is that the notation (both cycle > notation and the two-line notation) represents the exchange of *elements*, > not positions. See for example > > > http://www.math.caltech.edu/~2015-16/1term/ma006a/17.%20More%20permutations.pdf > > pp. 4--5. > > So my interpretation of t*s where t=(1,2) and s=(2,3) (multiplying from R > to L) would be: > > 1 -s-> 1 -t-> 2 > 2 -s-> 3 -t-> 3 > 3 -s-> 2 -t-> 1, > > hence (1,2,3) ("1 goes to 2, 2 goes to 3, 3 goes to 1"), or (in list form) > [0, 2, 3, 1]. > > Is transposition of elements from right to left (my interpretation) > equivalent to transposition of positions from left to right (your > interpretation)? I can't think of any counterexamples, but I'll chew on it. > > At the very least, I think that just specifying a multiplication direction > without specifying what's being permuted (elements or positions) is > ambiguous. > > Thanks again, > Alex > > On Fri, Mar 12, 2021 at 8:29 AM Chris Smith <[email protected]> wrote: > >> My thinking is expression in the transformations of the original list of >> items, [0,1,2,3]. If you first transpose the 2nd and third position you get >> [0,1,3,2] and then if you transpose 1st and 2nd position you get [0,3,1,2]. >> You'll see my name all over the docs for that module so if you can find the >> error in my thinking here, you are close to the source ;-) >> >> /c >> >> On Thursday, March 11, 2021 at 9:47:05 PM UTC-6 [email protected] wrote: >> >>> Hi Chris, >>> >>> Thanks for your response. When you write, >>> >>> > If you let `p = Permutation(1,2)(2,3)` then `p.list()` gives `[0, 3, >>> 1, 2]` which is consistent with R to L interpretation >>> >>> I think this is incorrect (and I contend that the docs are incorrect on >>> this point as well). >>> Multiplying the transpositions (1,2)(2,3) from R to L, we end up with >>> the cycle (1,2,3), >>> which in list form is [0, 2, 3, 1] (if `p.list()` is the second line of >>> 2-line permutation notation). >>> >>> What do you think? >>> >>> On Thu, Mar 11, 2021 at 7:33 PM Chris Smith <[email protected]> wrote: >>> >>>> So documentation here, "The composite of two permutations p*q means >>>> first apply p, then q" should read "...apply q, then p", right? This would >>>> be an easy issue to open and fix if there is consensus that it is wrong as >>>> written. But note that using the composition of function syntax reverses >>>> the order, "One can use also the notation p(i) = i^p, but then the >>>> composition rule is (p*q)(i) = q(p(i)), not p(q(i)):" >>>> >>>> /c >>>> >>>> On Thursday, March 11, 2021 at 8:37:25 PM UTC-6 Chris Smith wrote: >>>> >>>>> Given elements `0,1,2,3`, `Permutation(1,2)(2,3)` interpreting R to L >>>>> gives `0123->0132->0312`; interpreting L to R gives `0123->0213->0231` >>>>> >>>>> If you let `p = Permutation(1,2)(2,3)` then `p.list()` gives `[0, 3, >>>>> 1, 2]` which is consistent with R to L interpretation. So the assumption >>>>> that spelling it `Permutation(1,2)*Permutation(2,3)` means left to right >>>>> must be wrong? >>>>> >>>>> /c >>>>> >>>>> On Monday, February 22, 2021 at 3:51:02 PM UTC-6 [email protected] >>>>> wrote: >>>>> >>>>>> Hi everyone, >>>>>> >>>>>> I've been experimenting with the "Permutations" module, trying to >>>>>> follow the examples in the documentation here: >>>>>> >>>>>> https://docs.sympy.org/latest/modules/combinatorics/permutations.html >>>>>> >>>>>> As expected, >>>>>> >>>>>> Permutation(1, 2)(2, 3) == Permutation(1, 2) * Permutation(2, 3) >>>>>> >>>>>> But doesn't this mean that the permutations are applied from left to >>>>>> right, since (as described in the docs) left-to-right permutation >>>>>> multiplication p*q is equivalent to composition q o p? >>>>>> >>>>>> If so, this contradicts the documentation's claim that "The >>>>>> convention is that the permutations are applied from *right to left* >>>>>> ". >>>>>> >>>>>> If not, I must be confused about something, and would appreciate any >>>>>> corrections. >>>>>> >>>>>> Thanks for your help, >>>>>> Alex >>>>>> >>>>>> -- >>>> You received this message because you are subscribed to a topic in the >>>> Google Groups "sympy" group. >>>> To unsubscribe from this topic, visit >>>> https://groups.google.com/d/topic/sympy/5MTQFwB7xIo/unsubscribe. >>>> To unsubscribe from this group and all its topics, send an email to >>>> [email protected]. >>>> To view this discussion on the web visit >>>> https://groups.google.com/d/msgid/sympy/7556df78-eb14-408c-bf38-326dafaa1318n%40googlegroups.com >>>> >>>> <https://groups.google.com/d/msgid/sympy/7556df78-eb14-408c-bf38-326dafaa1318n%40googlegroups.com?utm_medium=email&utm_source=footer> >>>> . >>>> >>> -- >> You received this message because you are subscribed to a topic in the >> Google Groups "sympy" group. >> To unsubscribe from this topic, visit >> https://groups.google.com/d/topic/sympy/5MTQFwB7xIo/unsubscribe. >> To unsubscribe from this group and all its topics, send an email to >> [email protected]. >> > To view this discussion on the web visit >> https://groups.google.com/d/msgid/sympy/b0302a20-7afa-48e7-ac63-2f467c0b164cn%40googlegroups.com >> >> <https://groups.google.com/d/msgid/sympy/b0302a20-7afa-48e7-ac63-2f467c0b164cn%40googlegroups.com?utm_medium=email&utm_source=footer> >> . >> > -- You received this message because you are subscribed to the Google Groups "sympy" group. 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