Since the lexicographically least rotation necessarily begins with the
least element of a row,
llr=: {. @ (/:~) @ (([: I. ] = <./) |."0 1 ])"1
Less computation if the least element doesn't occur many times in a row.
On Sat, May 16, 2020 at 10:59 PM Roger Hui <[email protected]>
wrote:
> Find the lexicographically least of all the rotations of each row. Not
> very efficient, but works.
>
> llr=: {.@(/:~)@(i.@# |."0 1 ])"1
>
> $ odo 3#5
> 125 3
> $ llr odo 3#5
> 125 3
> $ ~. llr odo 3#5
> 45 3
>
>
>
> On Sat, May 16, 2020 at 10:50 PM Skip Cave <[email protected]>
> wrote:
>
>> The acid test is to find all the rotational duplicates of a full odometer
>> sequence:
>>
>> *odo=:#:i.@(*/)* NB. Odometer verb
>>
>> * $n=.odo 3#5 *
>>
>> *125 3*
>>
>> *rd=:3 :'(y i."_1 <./"1 y)|."_1 y'* NB. Roger's rotational duplication
>> verb
>>
>> *$rdn=.rd n* NB. Return all rows to key form.
>>
>> *125 3*
>>
>> * $rdn1=.~. rdn* NB. Remove all duplicate keys.
>>
>> *55 3*
>>
>> * 5 11${rdn1*
>>
>> *┌─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐*
>>
>> *│0 0 0│0 0 1│0 0 2│0 0 3│0 0 4│0 1 0│0 1 1│0 1 2│0 1 3│0 1 4│0 2 0│*
>>
>> *├─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤*
>>
>> *│0 2 1│0 2 2│0 2 3│0 2 4│0 3 0│0 3 1│0 3 2│0 3 3│0 3 4│0 4 0│0 4 1│*
>>
>> *├─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤*
>>
>> *│0 4 2│0 4 3│0 4 4│1 1 1│1 1 2│1 1 3│1 1 4│1 2 1│1 2 2│1 2 3│1 2 4│*
>>
>> *├─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤*
>>
>> *│1 3 1│1 3 2│1 3 3│1 3 4│1 4 1│1 4 2│1 4 3│1 4 4│2 2 2│2 2 3│2 2 4│*
>>
>> *├─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┼─────┤*
>>
>> *│2 3 2│2 3 3│2 3 4│2 4 2│2 4 3│2 4 4│3 3 3│3 3 4│3 4 3│3 4 4│4 4 4│*
>>
>> *└─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘*
>>
>>
>> We can see that some rotational dups were missed - 001-010, 002-020, etc.
>> 112-121, 113-131, etc.
>>
>> So there is still some work to do on Roger's rb verb.
>>
>> Skip
>>
>>
>>
>> On Sat, May 16, 2020 at 8:21 PM Roger Hui <[email protected]>
>> wrote:
>>
>> > > I could be wrong.
>> >
>> > Right. I was wrong.
>> >
>> >
>> > On Sat, May 16, 2020 at 6:18 PM Raul Miller <[email protected]>
>> wrote:
>> >
>> > > For example:
>> > >
>> > > A=: 2 3 1 2 1,: 2 1 2 3 1
>> > > 2 1 |.("_1) A
>> > > 1 2 1 2 3
>> > > 1 2 3 1 2
>> > > 2 4 |.("_1) A
>> > > 1 2 1 2 3
>> > > 1 2 1 2 3
>> > >
>> > > Thanks,
>> > >
>> > > --
>> > > Raul
>> > >
>> > > On Sat, May 16, 2020 at 9:11 PM Roger Hui <[email protected]>
>> > > wrote:
>> > > >
>> > > > The question is, do you get a unique key (signature) if you rotate a
>> > row
>> > > so
>> > > > that the first occurrence of the minimum value is first? I thought
>> the
>> > > > answer was yes after thinking about it for a minute. I could be
>> wrong.
>> > > >
>> > > >
>> > > > On Sat, May 16, 2020 at 5:53 PM Raul Miller <[email protected]>
>> > > wrote:
>> > > >
>> > > > > A critical question here is whether the minimum value can appear
>> more
>> > > > > than once in each row, or whether the examples (where each value
>> is
>> > > > > has a unique appearance in each row) are adequately complex.
>> > > > >
>> > > > > Thanks,
>> > > > >
>> > > > > --
>> > > > > Raul
>> > > > >
>> > > > > On Sat, May 16, 2020 at 8:15 PM Roger Hui <
>> [email protected]
>> > >
>> > > > > wrote:
>> > > > > >
>> > > > > > Hmm, you just want the keys: rotate each row so that the minimum
>> > > item is
>> > > > > > first.
>> > > > > >
>> > > > > > (n i."_1 <./"1 n)|."_1 n
>> > > > > > 1 3 2 4
>> > > > > > 1 2 3 4
>> > > > > > 1 2 3 4
>> > > > > > 1 3 2 4
>> > > > > > 1 3 2 4
>> > > > > > 1 2 3 4
>> > > > > > 1 2 3 4
>> > > > > > 1 3 2 4
>> > > > > >
>> > > > > >
>> > > > > > On Sat, May 16, 2020 at 5:11 PM Roger Hui <
>> > [email protected]
>> > > >
>> > > > > wrote:
>> > > > > >
>> > > > > > > ((n i."_1 <./"1 n)|."_1 n) </. n
>> > > > > > > ┌───────┬───────┐
>> > > > > > > │2 4 1 3│2 3 4 1│
>> > > > > > > │3 2 4 1│3 4 1 2│
>> > > > > > > │1 3 2 4│4 1 2 3│
>> > > > > > > │4 1 3 2│1 2 3 4│
>> > > > > > > └───────┴───────┘
>> > > > > > >
>> > > > > > > Rotate each row so that the minimum item is first, then use
>> those
>> > > > > rotated
>> > > > > > > rows as keys.
>> > > > > > >
>> > > > > > >
>> > > > > > > On Sat, May 16, 2020 at 4:44 PM Skip Cave <
>> > [email protected]
>> > > >
>> > > > > wrote:
>> > > > > > >
>> > > > > > >> I have run across this issue a few times in the past.
>> > > > > > >> The following 8x4 array has several rows that are 'rotational
>> > > > > duplicates'.
>> > > > > > >>
>> > > > > > >> ]n=.8 4$2 4 1 3 2 3 4 1 3 4 1 2 3 2 4 1 1 3 2 4 4 1 2 3 1 2
>> 3 4
>> > 4
>> > > 1 3
>> > > > > 2
>> > > > > > >>
>> > > > > > >> 2 4 1 3
>> > > > > > >>
>> > > > > > >> 2 3 4 1
>> > > > > > >>
>> > > > > > >> 3 4 1 2
>> > > > > > >>
>> > > > > > >> 3 2 4 1
>> > > > > > >>
>> > > > > > >> 1 3 2 4
>> > > > > > >>
>> > > > > > >> 4 1 2 3
>> > > > > > >>
>> > > > > > >> 1 2 3 4
>> > > > > > >>
>> > > > > > >> 4 1 3 2
>> > > > > > >>
>> > > > > > >>
>> > > > > > >> Is it possible to develop a verb that would find the rows
>> that
>> > are
>> > > > > > >> rotational duplicates of each other. That is, find all the
>> rows
>> > > that
>> > > > > would
>> > > > > > >> be the same, if each row was rotated some integer value in
>> the
>> > > first
>> > > > > > >> dimension. The output of the verb would be the same shape
>> array,
>> > > but
>> > > > > with
>> > > > > > >> each duplicate row rotated such that they show as identical.
>> > > Picking
>> > > > > the
>> > > > > > >> 'standard' rotation for a set of rotational duplicates is up
>> to
>> > > the
>> > > > > > >> implementer.
>> > > > > > >>
>> > > > > > >>
>> > > > > > >> Skip
>> > > > > > >>
>> > > ----------------------------------------------------------------------
>> > > > > > >> For information about J forums see
>> > > > > http://www.jsoftware.com/forums.htm
>> > > > > > >>
>> > > > > > >
>> > > > > >
>> > > ----------------------------------------------------------------------
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>> > > > >
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>> > > > >
>> > > >
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>>
>
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