I like “normalize” personally.

The ideas I associate with “signature” are not robust enough for a reliable
nub (and so would require additional care elsewhere).

If that matters...

Thanks,

—
Raul

On Wednesday, February 13, 2019, Roger Hui <[email protected]>
wrote:

> In this context, I prefer words like signature or representative rather
> than canonical form.  Shorter and less scary and puts one's mind on the
> right track in multiple problems.
>
> e.g. What's a good representative for identity problems?  ~. i. ]  (index
> in nub).  What's a good representative for ordering problems?  (Seen a
> couple of weeks ago.)  /:~@, i.!.0 ]  (ordinals).
>
>
>
>
>
> On Wed, Feb 13, 2019 at 5:19 PM Henry Rich <[email protected]> wrote:
>
> > This is what I was looking for.  It works on REB's testcase but has less
> > than quadratic run time I think.
> >
> > NB. Get # left-shifts to canonicalize y
> >
> > canonshift =: 3 : 0
> >
> > NB. Try each atom of y until we find one that works
> >
> > for_t. /:~ ~. y do.
> >
> >    NB. get spacing between positions of t, including the wraparound
> >
> >    cyclt =. 2 -~/\ tpos =. (, (#y) + {.) t I.@:= y
> >
> >    NB. if there is only 1 value, use its position
> >
> >    if. 1 = # cyclt do. {. tpos end.
> >
> >    NB. If all spacings are the same, try the next value
> >
> >    if. (}. -: }:) cyclt do. continue. end.
> >
> >    NB. Canonicalize cyclt.  Use its result to canonicalize y
> >
> >    (canonshift cyclt) { tpos return.
> >
> > end.
> >
> > NB. No atom worked; must be abcabc...; canonize by moving smallest to
> front
> >
> > (i. <./) y
> >
> > )
> >
> >     ~: (|.~ canonshift)"1 a
> > 1 1 1 1 1 0 1 1 1 1 1 0
> >
> > The canonical form used here does not always put the smallest atom at the
> > front, but I think it causes vector that differ only by a rotation to
> > canonicalize identically.
> >
> > Henry Rich
> >
> >
> >
> > On 2/13/2019 7:29 PM, Roger Hui wrote:
> > > Idea k: a minimum vector necessarily begins with a minimum sub-sequence
> > in
> > > x,(k-1){.x of length k , itself necessarily begins with the minimal
> item.
> > >
> > >
> > > On Wed, Feb 13, 2019 at 9:52 AM Roger Hui <[email protected]>
> > wrote:
> > >
> > >> Yes, well, left as an exercise for the reader. :-)
> > >>
> > >> Idea: the minimum rotation of a vector necessarily begins with its
> > minimal
> > >> item.
> > >>
> > >> On Wed, Feb 13, 2019 at 9:34 AM Henry Rich <[email protected]>
> > wrote:
> > >>
> > >>> Yes; but now suppose the lines are very long.  Is there a way to find
> > >>> the signature (I would call it a canonical form) that doesn't require
> > >>> enumerating rotations?  (I haven't found a good way yet).
> > >>>
> > >>> Henry Rich
> > >>>
> > >>> On 2/13/2019 12:16 PM, Roger Hui wrote:
> > >>>> For each row, find a "signature", then find the nub sieve of the
> > >>>> signatures.  The signature I use here is the minimum of all possible
> > >>>> rotations.
> > >>>>
> > >>>>      signature=: {. @ (/:~) @ (i.@# |."0 1 ])
> > >>>>
> > >>>>      ~: signature"1 a
> > >>>> 1 1 1 1 1 0 1 1 1 1 1 0
> > >>>>
> > >>>>
> > >>>>
> > >>>>
> > >>>> On Wed, Feb 13, 2019 at 8:55 AM R.E. Boss <[email protected]>
> > wrote:
> > >>>>
> > >>>>> Let the 12 x 20 matrix be defined by
> > >>>>> a=: 0 : 0
> > >>>>>    1  4  4  1 _4 _4  1  1 _4 _1 _1 _4 _4 _1  4  4 _1 _1  4  1
> > >>>>>    1  4  4  1 _4 _4  1  1 _4 _1 _1 _4 _4 _1  4  1  4 _1 _1  4
> > >>>>>    1  4  4  1 _4 _1 _4  1  1 _4 _1 _4 _4 _1  4  1  4 _1 _1  4
> > >>>>>    4  1  1  4 _1  4  1 _4 _4  1 _4 _1 _1 _4  1 _4 _1  4  4 _1
> > >>>>>    4  1  1  4 _1  4  1 _4 _4  1  1 _4 _1 _1 _4 _4 _1  4  4 _1
> > >>>>> _1  4  1  1  4  4  1 _4 _4  1  1 _4 _1 _1 _4 _4 _1  4  4 _1
> > >>>>> _1  4  4 _1 _4 _4 _1 _1 _4  1  1 _4 _4  1  4  4  1  1  4 _1
> > >>>>> _1  4  4 _1 _4 _4 _1 _1 _4  1  1 _4 _4  1  4 _1  4  1  1  4
> > >>>>> _1  4  4 _1 _4  1 _4 _1 _1 _4  1 _4 _4  1  4 _1  4  1  1  4
> > >>>>>    4 _1 _1  4  1  4 _1 _4 _4 _1 _4  1  1 _4 _1 _4  1  4  4  1
> > >>>>>    4 _1 _1  4  1  4 _1 _4 _4 _1 _1 _4  1  1 _4 _4  1  4  4  1
> > >>>>>    1  4 _1 _1  4  4 _1 _4 _4 _1 _1 _4  1  1 _4 _4  1  4  4  1
> > >>>>> )
> > >>>>>
> > >>>>> Required is the nubsieve for the items modulo rotation.
> > >>>>> So two arrays are considered to be equal if one is a rotation of
> the
> > >>> other.
> > >>>>> The answer I found is
> > >>>>> 1 1 1 1 1 0 1 1 1 1 1 0
> > >>>>>
> > >>>>>
> > >>>>> R.E. Boss
> > >>>>>
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