It seems to me (unless I've missed something?) that this method generates
the power set of the
original multiset (i.e. all subsets) rather than partitions.
Oops, sorry, wasn't concentrating.
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On 7/23/07, DavidA [EMAIL PROTECTED] wrote:
Here's the approach I would try.
1. Use Data.List.group to group your multiset, eg [1,1,2] - [[1,1],[2]]
2. Now apply your partitions function to each of the groups
[[1,1],[2]] - [ [([1,1],[]), ([1],[1]), ([],[1,1])], [([2],[]), ([],[2])]
]
(Actually
On 7/24/07, Brent Yorgey [EMAIL PROTECTED] wrote:
I'm not sure what a formal
mathematical definition would be off the top of my head; but in Haskell,
given a list L :: [a], I'm looking for all partitions P :: [[a]] where (sort
. concat $ P) == (sort L).
Here is quick attempt that requires Ord
On 7/24/07, Pekka Karjalainen [EMAIL PROTECTED] wrote:
On 7/24/07, Brent Yorgey [EMAIL PROTECTED] wrote:
given a list L :: [a], I'm looking for all partitions P :: [[a]] where
(sort
. concat $ P) == (sort L).
Here is quick attempt that requires Ord [a] and expects a sorted list.
It may very
Here's the approach I would try.
1. Use Data.List.group to group your multiset, eg [1,1,2] - [[1,1],[2]]
2. Now apply your partitions function to each of the groups
[[1,1],[2]] - [ [([1,1],[]), ([1],[1]), ([],[1,1])], [([2],[]), ([],[2])] ]
(Actually of course, you can probably write a simpler
