Timeline for Find all the possible ways of partitioning a list into a set of pairs of element
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 13, 2015 at 18:17 | comment | added | LLlAMnYP | @belisarius That's true, however all partitions that we seek are split into Length[list]/2 chunks, so this slightly narrows down the search space, no? @user2596320 I think I'm onto a non-bruteforce algorithm that may be much more efficient. | |
| Jul 13, 2015 at 18:15 | comment | added | sdnnds | belisarius's answer does the job as far as I can tell. | |
| Jul 13, 2015 at 18:13 | comment | added | sdnnds | @LLlAMnYP this does something different, it finds all partitions in a given number of sets but they can all be of different length. | |
| Jul 13, 2015 at 17:30 | comment | added | LLlAMnYP | Perhaps more efficient is KSetPartitions[Length[list],Length[list]/2] | |
| Jul 13, 2015 at 12:13 | comment | added | sdnnds | This seems to work! I'll test it later and get back here for feedback. | |
| Jul 13, 2015 at 11:04 | review | Low quality posts | |||
| Jul 13, 2015 at 11:05 | |||||
| Jul 13, 2015 at 10:45 | history | answered | Dr. belisarius | CC BY-SA 3.0 |