On 8/28/07, Roger Hui <[EMAIL PROTECTED]> wrote:
> Algorithm to find the next boolean list with m 1s:
> Find the rightmost 0 which has k 1s on its right
> where k>0. Move a 1 into that position and move
> the other (k-1) 1s all the way to the right.
nv2=:3 :0
one=.y i: 1
zer=.(one{.y) i: 0
assert.zer < one
1 0 (zer,one)} y
)
Note that neither nv nor nv2 deals well with cases which
do not have a successor with the supplied number of bits.
--
Raul
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