@ R.E. Boss:
Combinations k mmd n that result in 0=z1 won't work
5 mmd 13
|domain error: combnu
| z2=:(z1 combnu>:x)(+/@:{)"1 _|.|:(>:/&i.>:)x
it seems to me that in this case the result is equal to z0:
z0
0 3 6 9 12
compared with (see hereafter):
5 mmd17 13
0 3 6 9 12
@ R. Miller:
if you're after a final result then 2-stage filtering seems to be an option:
f1=: (#~ [: (= >./) [: <./"(1) 2 -~/\"1 ])
f2=:(#~ [: (= >./) [: >./"(1) 2 -~/\"1 ])
f2 f1 3 comb 8
0 3 7
0 4 7
f2 f1 4 comb 8
0 2 4 7
0 2 5 7
0 3 5 7
compared with R.E. Boss's:
>{: 4 mmd 8
0 2 4 7
0 2 5 7
0 3 5 7
and J. Randall's:
k 4 comb 8
0 2 4 7
0 2 5 7
0 3 5 7
Looking @ the differences of the last example:
<==> all unique permutations of 2 2 3
apply the running sum to these perms with a leading 0:
0,.+/\"1 perms
0 2 4 7
0 2 5 7
0 3 5 7
-->
mmd17 =: 4 : 0
ds=. (,(y-1)-+/@:])(x-2) $ y <....@%&<: x
if. ({.={:) ds do.
+/\0,ds
else.
0,.+/\"1 (|."0 _~ i...@# ) ds
end.
)
3 mmd17 8
0 3 7
0 4 7
4 mmd17 8
0 2 4 7
0 2 5 7
0 3 5 7
different result for:
k 6 comb 8
0 1 2 3 5 7
0 1 2 4 5 7
0 1 2 4 6 7
0 1 3 4 5 7
0 1 3 4 6 7
0 1 3 5 6 7
0 2 3 4 5 7
0 2 3 4 6 7
0 2 3 5 6 7
0 2 4 5 6 7
compared with:
>{: 6 mmd 8
6 mmd17 8
f2 f1 6 comb 8
0 1 2 3 4 7
0 1 2 3 6 7
0 1 2 5 6 7
0 1 4 5 6 7
0 3 4 5 6 7
Another one:
k 5 comb 8
0 1 3 5 7
0 2 3 5 7
0 2 4 5 7
0 2 4 6 7
>{: 5 mmd 8
f2 f1 5 comb 8
5 mmd17 8
0 1 2 3 7
0 1 2 6 7
0 1 5 6 7
0 4 5 6 7
--
Met vriendelijke groet,
=@@i
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