To clarify,
A conjunction F.. in this case, is documented according to the same explicit
defintion: either 2 : 'u F.. v' or {{u F.. v}}
In the above explicit definitions u is left argument to F.. v is right
argument.
lfold is an adverb. Has just one left argument u. That adverb will place its u
argument in "the v position" of F..
a tacit versions of lfold
lfold =: ~(] F..) NB. Adverb train
for easier display
- ~(+: @:)
+:@:(-~)
2 - ~(+: @:)3
2
+: 2 -~ 3
2
On Tuesday, December 6, 2022 at 06:31:21 p.m. EST, Henry Rich
<henryhr...@gmail.com> wrote:
No, u is the left argument and v is the right. u@v for example.
Henry Rich
On 12/6/2022 6:29 PM, Brian Schott wrote:
> Raul,
>
> I'm beginning to understand your explanation of lfold=: {{ ]F..(u~) }} .
> You're saying that j conjunctions have a left and right argument we call v
> and u, respectively
> (from the perspective of the interpreter).
> In that context, u is on the right of the conjunction, and that's why you
> used u here,
> not because it's the u in Folds. That is confusing, but I can accept that.
> Thank you, again,
>
> On Tue, Dec 6, 2022 at 5:36 PM Raul Miller <rauldmil...@gmail.com> wrote:
>
>> Context is important here.
>>
>> In
>> lfold=: {{ ]F..(u~) }}
>>
>> The u for lfold is the v for F..
>>
>> The J parser is not going to interpret u or v according to some
>> referenced definition -- it's going to interpret u or v according to
>> the definition it's currently interpreting.
>>
>> The same holds for x and y.
>>
>> Consider this example:
>>
>> thing1=: {{ y thing2 x }}
>> thing2=: ,&<
>> thing1/i.3
>> +-----+-+
>> |+-+-+|0|
>> ||2|1|| |
>> |+-+-+| |
>> +-----+-+
>> thing2/i.3
>> +-+-----+
>> |0|+-+-+|
>> | ||1|2||
>> | |+-+-+|
>> +-+-----+
>>
>> I hope this makes sense...
>>
>> --
>> Raul
>>
>>
>>
>>
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