Reread the dictionary and I learned the correct interpretation of a negative
rank.
A negative rank is complementary: u"(-r) y is equivalent to u"(0>.(#$y)-r)"_
y .
I had the misconception that it should be u"(0>.(#$y)-r) y .
> On Aug 7, 2017, at 4:02 PM, Louis de Forcrand <ol...@bluewin.ch> wrote:
>
> The same observations go for all operators which depend on their argument’s
> rank, such as @, &, or &. :
>
> <@(,"_1)~ i.3
> ┌───┐
> │0 0│
> │1 1│
> │2 2│
> └───┘
>
> Louis
>
>> On 07 Aug 2017, at 16:55, Louis de Forcrand <ol...@bluewin.ch> wrote:
>>
>> Yes I guess it could be replaced by
>>
>> u”_1”_1 _
>>
>> I would guess it is the intended behavior as b.0 reports
>> infinite rank, but that is what I find strange.
>>
>> Louis
>>
>>> On 07 Aug 2017, at 16:47, Xiao-Yong Jin <jinxiaoy...@gmail.com> wrote:
>>>
>>> I remembered I had problem with this and ended up specifying both left and
>>> right rank.
>>>
>>> JVERSION
>>> Engine: j806/j64avx/darwin
>>> Beta-4: commercial/2017-06-27T12:55:06
>>> Library: 8.06.03
>>> Platform: Darwin 64
>>> Installer: J806 install
>>> InstallPath: /applications/j64-806
>>> Contact: www.jsoftware.com
>>> +"_1 _ b.0
>>> _ _ _
>>> +"_1 _/~i.3
>>> 0 1 2
>>> 1 2 3
>>> 2 3 4
>>>
>>> I'm not sure if it is the intended behavior.
>>>
>>>
>>>> On Aug 7, 2017, at 3:40 PM, Louis de Forcrand <ol...@bluewin.ch> wrote:
>>>>
>>>> +"0/~ i.3
>>>> 0 1 2
>>>> 1 2 3
>>>> 2 3 4
>>>> +"_1/~ i.3
>>>> 0 2 4
>>>> +"0 b.0
>>>> 0 0 0
>>>> +"_1 b.0
>>>> _ _ _
>>>>
>>>> I understand that this is dictionary compliant:
>>>>
>>>> "In general, each cell of x is applied to the entire of y . Thus x u/ y is
>>>> equivalent to x u"(lu,_) y where lu is the left rank of u ."
>>>>
>>>> +"_1 b.0
>>>> _ _ _
>>>>
>>>> So u"_1/ -: u"_ _ . Wouldn’t it be better though if u"_1/ -: u"_1 _ ,
>>>> or if (u”_1 b.0) -: _1 _1 _1 (or any other negative rank)?
>>>> I ran into this while trying to use ,"_1/ , which I can replace by >@{@,&<
>>>> ,
>>>> but I still find this strange.
>>>>
>>>> Louis
>>>>
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