The shape of the arguments to the verb that's being inserted is 3x4, so when
that verb returns an atom, as dyadic + does, the shape of the result is 3x4.
But dyadic ? (Deal[1]) doesn't return an atom: it returns a list of the number
of items of its left argument. So for each of the 3x4 applications of dyadic ?
returns a list of 3 elements, so the shape of the result is 3x4x3. There's
probably a more official way to say that, but that's my model of J, so far.
It might be slightly less confusing to use arguments that aren't also the shapes of those arguments. E.g.,
The left argument is a list of length 2, and the right argument is a list of
length 4, so there are 2x4 pairs between each of which is inserted a Deal.
Each Deal chooses 3 items from i. 6 without replacement.
3 3 ?/ 6 6 6 6
5 0 2
3 5 0
1 4 5
1 0 3
0 3 2
0 5 2
1 2 4
3 0 1
so the shape of the result is 2x4x3.
$ 3 3 ?/ 6 6 6 6
2 4 3
Does that seem less odd?
... peter
[1] http://www.jsoftware.com/help/dictionary/d640.htm
Linda Alvord wrote:
f=: 13 :'0=?~ x#y'
f
0 = [: ?~ #
Maybe someday I'll just write expressions like yours easily..... The idea
seems so simple now.
However, I discovered this oddity:
3 3 3 +/ 4 4 4 4
7 7 7 7
7 7 7 7
7 7 7 7
3 3 3 ?/ 4 4 4 4
3 2 1
1 3 0
2 3 1
3 2 1
2 3 1
2 1 3
1 3 0
1 0 2
2 0 3
0 3 1
0 1 3
2 3 1
This seems odd:
Linda
-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Bo Jacoby
Sent: Thursday, November 08, 2012 4:46 AM
sTo: programm...@jsoftware.com
Subject: Re: [Jprogramming] Arc consistency in J
Linda, stick to integer arithmetic:
5 (0=[:?~#) 8
0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0
- Bo
________________________________
Fra: Linda Alvord <lindaalv...@verizon.net>
Til: programm...@jsoftware.com
Sendt: 10:10 torsdag den 8. november 2012
Emne: Re: [Jprogramming] Arc consistency in J
ee=:(]%2) > ?~@$
ee
0.5 > ?~@$
ff=: 13 :'0.5 > ?~x$y'
ff
0.5 > [: ?~ $
5 ff 8
0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1
0 0 0 0 1 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
J is so smart, it eliminate the need for *
hh=: 13 :' ?~x$y'
hh
[: ?~ $
]A=:5 hh 8
4 7 1 6 0 5 3 2
4 2 3 1 5 7 0 6
7 3 5 4 1 2 6 0
5 3 2 4 1 7 6 0
2 5 4 0 3 6 7 1
0.5 > A
0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0
Mind boggling!
Linda
-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Devon
McCormick
Sent: Wednesday, November 07, 2012 12:56 PM
To: J-programming forum
Subject: Re: [Jprogramming] Arc consistency in J
At first glance, I thought the right tine of this fork
(2 %~ ]) > [: (? ]) $
could be replaced by an idiom I frequently use
(?@$)
but then realized that what we need is
(?~@$)
so "dd" can be written as
(]%2) > ?~@$
On Wed, Nov 7, 2012 at 1:31 AM, Michal D.
<michal.dobrog...@gmail.com>wrote:
Thanks Roger, that makes sense now. The history of J is one of it`s
intriguing aspects for sure.
Re: Linda: I would call it a v(erb) as opposed to a N(oun). But what
do I know? ;-)
Mike
On Tue, Nov 6, 2012 at 8:53 AM, Roger Hui <rogerhui.can...@gmail.com>
wrote:
'noun verb verb' is a fork and is interpreted as 'noun"_ verb verb'
(noun"_
is a constant verb whose result is noun).
http://keiapl.org/anec/#nvv
'verb verb noun' can not be made into a fork because 'verb noun'
already has an interpretation (*viz*., apply verb to noun).
On Tue, Nov 6, 2012 at 8:47 AM, Michal D.
<michal.dobrog...@gmail.com
wrote:
Change from a Noun to a verb, view its tacit version and apply
it to
data:
dd=: 13 :'(y%2) > (?]) x$y'
dd
(2 %~ ]) > [: (? ]) $
That is quite cool. I'm surprised that you can automatically get
the
tacit
definition. Does this work for any explicitly defined verb?
I'm also surprised at the way %~ came out. Do left hand
arguments not require a & to bind the argument? It is strange to
me that
(1) works
but
(2) does not. It seems to me that (3) is the logical way to
phrase
either
of them (ie. a fork with a constant right / left side). To
reiterate,
why
does (1) work?
(1) (2 %~ ]) > [: (? ]) $
(2) (] % 2) > [: (? ]) $
(3a) (2: %~ ]) > [: (? ]) $
(3b) (] %~ 2:) > [: (? ]) $
(4a) (%&2 ]) > [: (? ]) $ NB. incorrect (hook caught me
out
again)!
(4b) ([: %&2 ]) > [: (? ]) $ NB. correct
Cheers,
Mike
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--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
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