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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