We can do an angelic solution using "larger absolute value" in place of bitwise
OR.
lav =: [`]@.(|@[ < |@])"0 NB. larger absolute value
A
1 2j3
0 4
(lav +@|:) A
1 2j3
2j_3 4
Kip Murray
Sent from my iPad
On Jan 15, 2013, at 5:20 PM, Henry Rich <[email protected]> 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 <[email protected]> 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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