On 8/28/07, Roger Hui <[EMAIL PROTECTED]> wrote:
> How do you run nv2 so that you get all possible
> boolean vectors of length n with m 1s?
I misunderstood.
Correcting that fault makes this approach more complex:
nv3=:3 :0
one=.y i: 1
zer=.y i: 0
if. one < zer do.
zer=. (one {. y) i: 0
1 0 (zer+0 1)} y /: y+2*(2+zer)<.i.#y
else.
1 0 (zer+0 1)} y
end.
)
Thanks,
--
Raul
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