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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