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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