However if the goal is to convert <move> (0 2 5) into the 15 element
boolean expressing same, a simpler method may be:
(i.15) e. move NB. e. is “element of” being true
(1) or false (0), so avoids having to expand the 15x15 identity at all
Correct. I had already made that substitution, at another commenter's
suggestion, and it worked fine.
But the 15 element binary expansions would have to be redone for each new 3
element move. My thought was to generate them once in the id matrix and
then pull them out as needed with
move { id
Six of one...
On Mon, May 29, 2017 at 10:21 AM, Rob Hodgkinson <[email protected]> wrote:
> Not sure if this helps Michael, but I think that you are using the id to
> generate the steps shown here from one of your earlier emails …
>
> >> f =: ((1 0 0 0 0 0 0 0 0 0 0 0 0 0 0) |.!.0~ 3 1 $ -)
> >> f move
> >> 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
> >> 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
> >> 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
> >> +./f move
> >> 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 NB. Move 0 2 5 expressed as 15 element
>
> which you have tried to substitute with:
> (id {~)
>
> However if the goal is to convert <move> (0 2 5) into the 15 element
> boolean expressing same, a simpler method may be:
> (i.15) e. move NB. e. is “element of” being
> true (1) or false (0), so avoids having to expand the 15x15 identity at all
> 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0
>
> This might obviate the need for id in the first place and simplify your
> overall solution.
>
> HTH, Regards Rob
>
>
> > On 29 May 2017, at 9:22 am, Michael Rice <[email protected]> wrote:
> >
> > Thanks.
> >
> > I found that after I couldn't find it and wrote my own, but it seemed to
> be
> > changing one of my function parameters, the left one, so I moved back to
> my
> > definition.
> >
> > I'm again having a similar problem stringing these verbs, adverbs, etc.
> > together to form the function I want.
> >
> > I created an 15 by 15 identity matrix, call it id, and substituted this
> >
> > (1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 |.!.0~
> > 3 1 $ -)
> >
> > for this
> >
> > (id {~)
> >
> > Much cleaner and, no doubt, faster. Win. Win.
> >
> > Naively, I thought this simple substitution of the two would work
> >
> > make_move =: ((+. *. -.@:*.) +./) (id {~)
> >
> > Nope. Syntax error.
> >
> > Have already tried several other variations this AM with the same result.
> >
> > I'm using "Learning J" as a reference and NuVoc.
> >
> > http://www.jsoftware.com/help/learning/contents.htm
> >
> >
> >
> >
> >
> >
> >
> > On Sun, May 28, 2017 at 9:57 PM, David Lambert <[email protected]>
> wrote:
> >
> >> xor can be not equal ~:
> >>
> >> or bitwise 2b10110 b.
> >>
> >>
> >>
> >> On 05/28/2017 08:00 AM, [email protected] wrote:
> >>
> >>> Message: 5
> >>> Date: Sat, 27 May 2017 23:51:29 -0400
> >>> From: Michael Rice<[email protected]>
> >>> To: programming<[email protected]>
> >>> Subject: Re: [Jprogramming] Function creation, without creating
> >>> subfunctions
> >>> Message-ID:
> >>> <CAGw08vM_5CttMFY45F6zAAto-doycQYkXqRf8uRWC=0Ug9xcOw@mail.
> >>> gmail.com>
> >>> Content-Type: text/plain; charset="UTF-8"
> >>>
> >>> Yes, that worked. New board created, without changing the old board or
> the
> >>> move.
> >>>
> >>> board =: 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1
> >>> move =: 0 2 5
> >>> make_move =: ((+. *. -.@:*.) +./) (1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
> >>> |.!.0~
> >>> 3 1 $ -)
> >>> board make_move move
> >>> 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1
> >>> board
> >>> 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1
> >>> move
> >>> 0 2 5
> >>>
> >>
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