On Wednesday, December 11, 2013 10:31:03 AM UTC-5, Ivan Andrus wrote:
>
> On Dec 11, 2013, at 8:22 AM, geo909 <[email protected] <javascript:>>
> wrote:
>
> Hi all,
>
> From wikipedia:
>
> *In mathematics, in the areas of combinatorics and computer
> science, a Lyndon word is a string that is strictly smaller in
> lexicographic order than all of its rotations. Lyndon words are named after
> mathematician Roger Lyndon, who introduced them in 1954, calling them
> standard lexicographic sequences*
>
> For example [1,1,2,1,3] is a Lyndon word, but [1,3,1,1,2] is not.
>
> I need to check as efficiently as possible if a given integer sequence is
> a Lyndon word or not. Is there such an option in sage?
> I check the section for Lyndon words in the
> manual<http://www.sagemath.org/doc/reference/combinat/sage/combinat/lyndon_word.html>but
> apparently the only way one can check something like that, is in a
> situation like so:
>
> sage: LyndonWord([2,1,2,3])Traceback (most recent call last):...ValueError:
> Not a Lyndon word
>>
>>
> Is this my only option for checking if something is a Lyndon word? This
> suggests that there is inherently some check in this
> function, but I do not want to use a ValueError for that..
>
> Any advice?
>
>
> You can use two question marks to see the source code. You can also edit
> the source code, so either of the two below should show you what's
> happening:
>
> LyndonWord??
> edit(LyndonWord)
>
> In this case the code of interest is the following:
>
> super(LyndonWord,self).__init__(parent=LyndonWords(),data=data)
> if check and not self.is_lyndon():
> raise ValueError("Not a Lyndon word")
>
> so it looks like
>
> LyndonWord([2,1,2,3],check=false).is_lyndon()
>
> should do what you want without having to catch a ValueError.
>
Thanks, that's it! Actually it's even a bit shorter:
a=[2,1,1,3,1,2]
Word(a).is_lyndon()
False
Thanks a lot.
>
> -Ivan
>
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