Hi all!
The idea here is to take all combinations of the possible number of
elements in a bucket and the number of buckets, and then select the
bucket combinations with the correct number of elements.
par=: 4 : '(1,.2</\"1(i.x)#/~(y=+/"1
o)#o=.((x$v)#:i.v^x){1+i.v=.1+y-x)<;.1[1+i.y'
2 par 3
┌───┬───┐
│1 │2 3│
├───┼───┤
│1 2│3 │
└───┴───┘
2 par 4
┌─────┬─────┐
│1 │2 3 4│
├─────┼─────┤
│1 2 │3 4 │
├─────┼─────┤
│1 2 3│4 │
└─────┴─────┘
2 par 5
┌───────┬───────┐
│1 │2 3 4 5│
├───────┼───────┤
│1 2 │3 4 5 │
├───────┼───────┤
│1 2 3 │4 5 │
├───────┼───────┤
│1 2 3 4│5 │
└───────┴───────┘
3 par 4
┌───┬───┬───┐
│1 │2 │3 4│
├───┼───┼───┤
│1 │2 3│4 │
├───┼───┼───┤
│1 2│3 │4 │
└───┴───┴───┘
3 par 5
┌─────┬─────┬─────┐
│1 │2 │3 4 5│
├─────┼─────┼─────┤
│1 │2 3 │4 5 │
├─────┼─────┼─────┤
│1 │2 3 4│5 │
├─────┼─────┼─────┤
│1 2 │3 │4 5 │
├─────┼─────┼─────┤
│1 2 │3 4 │5 │
├─────┼─────┼─────┤
│1 2 3│4 │5 │
└─────┴─────┴─────┘
3 par 6
┌───────┬───────┬───────┐
│1 │2 │3 4 5 6│
├───────┼───────┼───────┤
│1 │2 3 │4 5 6 │
├───────┼───────┼───────┤
│1 │2 3 4 │5 6 │
├───────┼───────┼───────┤
│1 │2 3 4 5│6 │
├───────┼───────┼───────┤
│1 2 │3 │4 5 6 │
├───────┼───────┼───────┤
│1 2 │3 4 │5 6 │
├───────┼───────┼───────┤
│1 2 │3 4 5 │6 │
├───────┼───────┼───────┤
│1 2 3 │4 │5 6 │
├───────┼───────┼───────┤
│1 2 3 │4 5 │6 │
├───────┼───────┼───────┤
│1 2 3 4│5 │6 │
└───────┴───────┴───────┘
Cheers,
Erling Hellenäs
On 2017-10-19 20:54, Erling Hellenäs wrote:
Yes. Sorry. Thanks. /Erling
On 2017-10-19 19:47, Raul Miller wrote:
Most people will not have a C:/Users/erling/ directory.
Presumably, though, you are referring to x!y which counts x
combinations of y.
http://www.jsoftware.com/help/dictionary/d410.htm
That's not quite what Skip was asking for, but it can be relevant in
an implementation.
Thanks,
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