Mike asked: "Bo: is it true that (3$3) is not a list of integers but (3#3) 
is?".  No, I don't think so.  My main point is that evaluating the expression   
5(0.5>[:?~$)8    involves noninteger arithmetic while evaluating the equivalent 
simpler expression     5(0=[:?~$)8    does not.involve noninteger arithmetic. 
The matrix elements of   5([:?~$)8    are nonnegative integers, and a 
nonnegative integer less than a half equals zero. 
- Bo





>________________________________
> Fra: km <k...@math.uh.edu>
>Til: "programm...@jsoftware.com" <programm...@jsoftware.com> 
>Sendt: 6:55 lørdag den 10. november 2012
>Emne: Re: [Jprogramming] Arc consistency in J
> 
>About 3$3 and 3#3 I suggest you experiment.  Type 3$3 then type 3#3 .  Next 
>type $ 3$3 and $ 3#3 .  What do you learn?  The following will help you learn 
>about $ .
>
>http://www.jsoftware.com/docs/help701/dictionary/d210.htm
>
>Kip Murray
>
>Sent from my iPad
>
>
>On Nov 9, 2012, at 11:21 PM, "Michal D." <michal.dobrog...@gmail.com> wrote:
>
>> Some nice things learned: ?~ instead of (?]).
>> 
>> Bo: is it true that (3$3) is not a list of integers but (3#3) is?
>> 
>> Fascinating discussion on ?/... so much intricatness.
>> 
>> Mike
>> 
>> On Thu, Nov 8, 2012 at 11:12 PM, Linda Alvord <lindaalv...@verizon.net>wrote:
>> 
>>> The dawn finally broke.  It's a deal!
>>> 
>>>    3 10 4 ? 10 10 10
>>> 8 2 5 0 0 0 0 0 0 0
>>> 4 1 0 9 3 7 6 8 5 2
>>> 9 0 1 8 0 0 0 0 0 0
>>> 
>>>    3 10 4 ?/ 10 10 10
>>> 6 0 5 0 0 0 0 0 0 0
>>> 9 5 0 0 0 0 0 0 0 0
>>> 2 6 1 0 0 0 0 0 0 0
>>> 
>>> 8 4 9 5 6 1 3 0 2 7
>>> 9 3 5 8 0 6 4 7 1 2
>>> 1 6 8 0 4 2 5 3 9 7
>>> 
>>> 1 9 5 7 0 0 0 0 0 0
>>> 3 9 8 0 0 0 0 0 0 0
>>> 1 3 7 0 0 0 0 0 0 0
>>> 
>>> So:
>>> 
>>>   8 8 8 8 8 ? 8 8 8 8 8
>>> 0 1 7 6 5 3 4 2
>>> 0 7 3 2 5 1 4 6
>>> 5 0 7 6 4 1 3 2
>>> 6 1 5 7 3 0 2 4
>>> 4 7 3 5 6 0 2 1
>>> 
>>> There is only one zero in each row.
>>> 
>>> It is nice when the fog lifts.  Thanks to everyone who helped me.
>>> 
>>>   f=:0 = # ? #
>>>   5 f 8
>>> 0 0 0 0 0 1 0 0
>>> 0 0 0 0 0 0 1 0
>>> 0 0 1 0 0 0 0 0
>>> 0 0 0 0 0 0 0 1
>>> 0 0 0 0 0 1 0 0
>>> 
>>> 5!:4 <'f'
>>>  -- 0
>>>  +- =
>>> --+   -- #
>>>  L---+- ?
>>>      L- #
>>> 
>>> 
>>> Quite a pretty picture.
>>> 
>>> Linda
>>> 
>>> 
>>> 
>>> -----Origineal Message-----
>>> From: programming-boun...@forums.jsoftware.com
>>> [mailto:programming-boune...@forums.jsoftware.com] On Behalf Of Peter B.
>>> Kessler
>>> Sent: Thursday, November 08, 2012 9:24 PM
>>> To: programm...@jsoftware.comn
>>> Subject: Re: [Jprogramming] Arc consistency in J
>>> 
>>> 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
>>>>>>>> -----------------------------------------------------------------
>>>>>>>> -
>>>>>>>> ---- For information about J forums see
>>>>>>>> http://www.jsoftware.com/forums.htm
>>>>>>> -------------------------------------------------------------------
>>>>>>> -
>>>>>>> -- For information about J forums see
>>>>>>> http://www.jsoftware.com/forums.htm
>>>>>> --------------------------------------------------------------------
>>>>>> -
>>>>>> - For information about J forums see
>>>>>> http://www.jsoftware.com/forums.htm
>>>>> 
>>>>> 
>>>>> --
>>>>> Devon McCormick, CFA
>>>>> ^me^ at acm.
>>>>> org is my
>>>>> preferred e-mail
>>>>> ---------------------------------------------------------------------
>>>>> - For information about J forums see
>>>>> http://www.jsoftware.com/forums.htm
>>>>> 
>>>>> ---------------------------------------------------------------------
>>>>> - For information about J forums see
>>>>> http://www.jsoftware.com/forums.htm
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>> 
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
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>
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