A long while back I did a writeup on ` (tie) and its relationship to gerunds, 
and tried to briefly define what a “gerund” is:

http://code.jsoftware.com/wiki/User:Dan_Bron/backtick 
<http://code.jsoftware.com/wiki/User:Dan_Bron/backtick>

-Dan



> On Oct 5, 2015, at 7:10 AM, Raul Miller <[email protected]> wrote:
> 
> But tie (`) is a conjunction which is used to produce gerunds.
> 
> Thanks,
> 
> -- 
> Raul
> 
> 
> On Mon, Oct 5, 2015 at 4:14 AM, Linda A Alvord <[email protected]> 
> wrote:
>> Trace is great!  It seems to think tie is a conjunction rather than a 
>> gerund. Also, the little '`' symbol for tie has disappeared completely.  
>> Somehow I never discovered trace, but it should come in handy!
>> 
>> Thanks,  Linda
>> 
>> -----Original Message-----
>> From: [email protected] 
>> [mailto:[email protected]] On Behalf Of Raul Miller
>> Sent: Monday, October 5, 2015 3:08 AM
>> To: Programming forum
>> Subject: Re: [Jprogramming] problem using ,:`]@.
>> 
>> Does this get you closer to where you want to be?
>> 
>>   require'trace'
>>   trace'(2+3)&*'
>>   trace '[: (*: d. 1) %:'
>> 
>> Love,
>> 
>> --
>> Raul
>> 
>> 
>> On Sun, Oct 4, 2015 at 1:31 PM, Jose Mario Quintana
>> <[email protected]> wrote:
>>> This is a similar question: Where is the + in (2+3)&* ?
>>> 
>>> (9!:3) 5 4 1  NB. Linear, tree and atomic representations
>>> 
>>> (2+3)&*
>>> 
>>>   (2+3)&*
>>> 5&*
>>>      ┌─ 5
>>> ── & ─┴─ *
>>> ┌─────────────┐
>>> │┌─┬─────────┐│
>>> ││&│┌─────┬─┐││
>>> ││ ││┌─┬─┐│*│││
>>> ││ │││0│5││ │││
>>> ││ ││└─┴─┘│ │││
>>> ││ │└─────┴─┘││
>>> │└─┴─────────┘│
>>> └─────────────┘
>>> 
>>> Plus does not appear because 2+3 is resolved as 5 which is apparent in the
>>> atomic representation.  Likewise, ,:`]  in the verb expression
>>> ,:`]@.(<:@#@$) is resolved as a gerund,
>>> 
>>> ,:`]@.(<:@#@$)
>>> ,:`]@.(<:@#@$)
>>>             ┌─ ,:
>>>       ┌─────┴─ ]
>>>       │           ┌─ <:
>>> ── @. ─┤     ┌─ @ ─┴─ #
>>>       └─ @ ─┴─ $
>>> ┌────────────────────────────────────┐
>>> │┌──┬───────────────────────────────┐│
>>> ││@.│┌──────────┬──────────────────┐││
>>> ││  ││┌─┬──────┐│┌─┬──────────────┐│││
>>> ││  │││0│┌──┬─┐│││@│┌──────────┬─┐││││
>>> ││  │││ ││,:│]││││ ││┌─┬──────┐│$│││││
>>> ││  │││ │└──┴─┘│││ │││@│┌──┬─┐││ │││││
>>> ││  ││└─┴──────┘││ │││ ││<:│#│││ │││││
>>> ││  ││          ││ │││ │└──┴─┘││ │││││
>>> ││  ││          ││ ││└─┴──────┘│ │││││
>>> ││  ││          ││ │└──────────┴─┘││││
>>> ││  ││          │└─┴──────────────┘│││
>>> ││  │└──────────┴──────────────────┘││
>>> │└──┴───────────────────────────────┘│
>>> └────────────────────────────────────┘
>>> 
>>> There are other conjunction expressions that are also resolved, for
>>> example, d. in
>>> 
>>>   (*: d. 1) @: %:
>>> +:@:%:
>>>       ┌─ +:
>>> ── @: ─┴─ %:
>>> ┌────────────┐
>>> │┌──┬───────┐│
>>> ││@:│┌──┬──┐││
>>> ││  ││+:│%:│││
>>> ││  │└──┴──┘││
>>> │└──┴───────┘│
>>> └────────────┘
>>> 
>>> or, if you prefer, in
>>> 
>>>   [: (*: d. 1) %:
>>> [: +: %:
>>>  ┌─ [:
>>> ──┼─ +:
>>>  └─ %:
>>> ┌──────────────┐
>>> │┌─┬──────────┐│
>>> ││3│┌──┬──┬──┐││
>>> ││ ││[:│+:│%:│││
>>> ││ │└──┴──┴──┘││
>>> │└─┴──────────┘│
>>> └──────────────┘
>>> 
>>> However, d. does not appear in the above linear representations but ` does
>>> appear in the linear representation of ,:`]@.(<:@#@$).  How come?
>>> Apparently the linear representation facility is smart in that sense,
>>> 
>>>   G=. ;:',: ]'
>>> 
>>>   G @.
>>> ,:`]@.
>>>       ┌─ ,:
>>>  ┌────┴─ ]
>>> ──┴─ @.
>>> ┌───────────────────┐
>>> │┌─┬───────────────┐│
>>> ││4│┌──────────┬──┐││
>>> ││ ││┌─┬──────┐│@.│││
>>> ││ │││0│┌──┬─┐││  │││
>>> ││ │││ ││,:│]│││  │││
>>> ││ │││ │└──┴─┘││  │││
>>> ││ ││└─┴──────┘│  │││
>>> ││ │└──────────┴──┘││
>>> │└─┴───────────────┘│
>>> └───────────────────┘
>>> 
>>> What I have never been able to figure out, completely, is the criteria used
>>> for resolving some conjunctions (and adverbs) but not resolving other ones.
>>> 
>>> 
>>> On Sun, Oct 4, 2015 at 12:16 AM, Linda A Alvord <[email protected]>
>>> wrote:
>>> 
>>>> Two thoughts about this thread:
>>>> 
>>>>  Where is the tie in the tree?
>>>>  A little shorter mat.
>>>> 
>>>>   bill=:,:`]@.(<:@#@$)
>>>>   bill 'ab'
>>>> ab
>>>>   5!:5 <'bill'
>>>> ,:`]@.(<:@#@$)
>>>>   5!:4 <'bill'
>>>>             ┌─ ,:
>>>>       ┌─────┴─ ]
>>>>       │           ┌─ <:
>>>> ── @. ─┤     ┌─ @ ─┴─ #
>>>>       └─ @ ─┴─ $
>>>> 
>>>>   mat
>>>> (_2 {. 1 1 , $) $ ]
>>>>   mat2
>>>> ] $~ _2 {. 1 1 , $
>>>> 
>>>> 
>>>> Linda
>>>> -----Original Message-----
>>>> From: [email protected] [mailto:
>>>> [email protected]] On Behalf Of 'Pascal Jasmin'
>>>> via Programming
>>>> Sent: Saturday, October 3, 2015 12:29 PM
>>>> To: [email protected]
>>>> Subject: Re: [Jprogramming] problem using ,:`]@.
>>>> 
>>>> g^:f  as if in J is one of the cooler patterns of the language.
>>>> if. 1 = f(y) do. g(y) else. y end.
>>>>      From: Nollaig MacKenzie <[email protected]>
>>>> To: [email protected]
>>>> Sent: Saturday, October 3, 2015 1:06 AM
>>>> Subject: Re: [Jprogramming] problem using ,:`]@.
>>>> 
>>>> fie is my preference, I think - I wouldn't have thought of getting a no-op
>>>> by doing an op 0 times :-)
>>>> 
>>>> fum looked perverse at first, but quickly became intriguing. It would have
>>>> other uses:
>>>> 
>>>> fum (2 3$'abcdef'),:2 3$'UVWXYZ'
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> On 2015.10.02 21:48:39, you,
>>>> the extraordinary Raul Miller, spake thus:
>>>> 
>>>>> It sounds like you are asking for this:
>>>>> 
>>>>>   fie=: ,:^:(2>#@$)
>>>>> 
>>>>> but you might want to consider alternatives such as:
>>>>> 
>>>>>   foe=: , $~ _2{.1,$
>>>>> 
>>>>> or
>>>>> 
>>>>>   fum=: ,.&.|:
>>>>> 
>>>>> Good luck,
>>>>> 
>>>>> --
>>>>> Raul
>>>> 
>>>> --
>>>> Nollaig MacKenzie
>>>> http://www.yorku.ca/nollaig
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