Timeline for Distributing values of an array into smaller arrays with bounded sums
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2014 at 10:58 | comment | added | Raphael | This problem does indeed seem to admit trivial solutions if there are not too many big elements, and none at all otherwise. Are there no other restrictions, such as minimising the number of pairs in the result? | |
| Sep 1, 2014 at 10:55 | history | edited | Raphael | CC BY-SA 3.0 |
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| Sep 1, 2014 at 7:25 | answer | added | FrankW | timeline score: 1 | |
| Sep 1, 2014 at 7:21 | comment | added | FrankW | I have integrated your edit into the actual question, since future visitors will not care about the previous wrong version. | |
| Sep 1, 2014 at 7:18 | history | edited | FrankW | CC BY-SA 3.0 |
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| Sep 1, 2014 at 5:51 | history | edited | Adam Colvin | CC BY-SA 3.0 |
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| Aug 31, 2014 at 21:35 | comment | added | Florian F | There must be a typo. An array of 180 plus an array of 240 sums to 420. The sum of your items is 785, not a multiple of 420. And all elements are multiple of 15, except for the 5 which is not. You can not use it in a subarray that sums to 180 or 240. | |
| Aug 31, 2014 at 21:26 | comment | added | Florian F | Are you looking for one solution, or a general method to solve this kind of problem? | |
| Aug 31, 2014 at 7:12 | comment | added | David Richerby | What did you try? Where did you get stuck? | |
| Aug 31, 2014 at 6:31 | review | First posts | |||
| Aug 31, 2014 at 7:12 | |||||
| Aug 31, 2014 at 6:28 | history | asked | Adam Colvin | CC BY-SA 3.0 |