Here's another power iterator approach:
{."2 ({. f"1 }.)^:(i.@#) !/~i.10
And here's a looping explicit approach:
orto=:3 :0
a=. 1 {. y
for. i.(#y)-1 do.
a=.a,{.y=.({:a)f"1}.y
end.
)
The explicit approach should be more efficient on very large arguments, but
has no advantage on a 10 by 10.
Thanks,
--
Raul
On Tue, Sep 16, 2014 at 4:46 AM, Aai <[email protected]> wrote:
> I'm not sure if this is less embarrassing, but here's an power iterator
> solution:
>
> 0{::([ ((,{.);]) {:@[ f"1 }.@])&>/^:(<:@#`(,:@{.;])) !/~i.10
>
>
>
> On 16-09-14 08:56, 'Bo Jacoby' via Programming wrote:
>
>> I managed to ortogonalize a 10 10 - matrix like this:
>>
>> f=.[:(%+./)(]*[:+/[*[)-[*[:+/*
>> orto =. 3 : 0
>> a=.,:{.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> a=.a,{.y=.({:a)f"1}.y
>> )
>> orto(!/~)i.10
>> 1 1 1 1 1 1 1 1 1 1
>> _9 _7 _5 _3 _1 1 3 5 7 9
>> 6 2 _1 _3 _4 _4 _3 _1 2 6
>> _42 14 35 31 12 _12 _31 _35 _14 42
>> 18 _22 _17 3 18 18 3 _17 _22 18
>> _6 14 _1 _11 _6 6 11 1 _14 6
>> 3 _11 10 6 _8 _8 6 10 _11 3
>> _9 47 _86 42 56 _56 _42 86 _47 9
>> 1 _7 20 _28 14 14 _28 20 _7 1
>> _1 9 _36 84 _126 126 _84 36 _9 1
>>
>>
>> How can the program orto be made less embarrassing?
>>
>> Thank you!
>>
>>
>> Bo.
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>>
>
> --
> Met vriendelijke groet,
> @@i = Arie Groeneveld
>
>
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