I thought that J did not represent negative zero?

Is it possible to trick J into revealing a negative zero?  If so, does
it involve foreigns or is there some native calculations that lead
here?

Thanks,

-- 
Raul

On Wed, Jan 16, 2013 at 7:26 AM, Henry Rich <henryhr...@nc.rr.com> wrote:
> On my awaking, there was a whiff of sulfur in the air, and a greenish
> haze... and somehow in my mind the idea that that last program won't work,
> because of the possibility of negative zero.  I'll stay relegated to imp
> status.
>
> Henry Rich
>
>
> On 1/15/2013 6:20 PM, Henry Rich wrote:
>>
>> Nah, that's not beyond impish.  The devilish solution is to take the
>> bitwise OR of the matrix with its conjugate transpose (but that's easier
>> in assembler language than in J:
>> (23 b.&.(a.&i.)&.(2&(3!:5))&.+. +@|:))
>> ).  And you need to be sure that the zeros on the lower diagonal and
>> below are true zeros!
>>
>> Henry Rich
>>
>> On 1/15/2013 6:03 PM, km wrote:
>>>
>>> Oh, boy!  (v1`v2) } y <--> (v1 y) } (v2 y)
>>>
>>> Brief and devilish, take care for your soul, Henry!
>>>
>>> --Kip
>>>
>>> Sent from my iPad
>>>
>>>
>>> On Jan 15, 2013, at 3:39 PM, Henry Rich <henryhr...@nc.rr.com> wrote:
>>>
>>>>    hft =: 0&=`(,: +@|:)}
>>>>
>>>> Henry Rich
>>>>
>>>> On 1/15/2013 5:25 AM, km wrote:
>>>>>
>>>>> This is an easy one.  A Hermitian matrix matches its conjugate
>>>>> transpose.  Write a verb hft that creates a Hermitian matrix from a
>>>>> triangular one that has a real diagonal.
>>>>>
>>>>>      ishermitian =: -: +@|:
>>>>>      ]A =: 2 2 $ 1 2j3 0 4
>>>>>   1 2j3
>>>>>   0   4
>>>>>      ]B =: hft A
>>>>>      1 2j3
>>>>>   2j_3   4
>>>>>      ishermitian A
>>>>>   0
>>>>>      ishermitian B
>>>>>   1
>>>>>
>>>>> Kip Murray
>>>>>
>>>>> Sent from my iPad
>>>>> ----------------------------------------------------------------------
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