Timeline for Balancing all but one weight
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2015 at 12:30 | history | edited | Gamow |
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| Jan 24, 2015 at 7:01 | vote | accept | xnor | ||
| Jan 15, 2015 at 10:12 | answer | added | Gamow | timeline score: 6 | |
| Jan 15, 2015 at 9:35 | comment | added | JiK | Essentially the same problem (with real number weights) in Math.SE, answered: math.stackexchange.com/questions/1002267/… Comments include perhaps a simpler answer to the integer case. | |
| Jan 15, 2015 at 6:11 | comment | added | xnor | @EFrog That's fine, the claim is there will be at least one such choice of token. | |
| Jan 15, 2015 at 6:07 | comment | added | EFrog | I'm confused. If I have tokens that weigh 2, 3, 4 and I remove any one of them, the remaining two tokens can't be split into equal-sized groups of the same weight... | |
| Jan 15, 2015 at 0:45 | comment | added | xnor | @Lopsy Whoops, I forgot the condition that you must partition into two equal-size sets. Thanks for catching that. | |
| Jan 15, 2015 at 0:44 | history | edited | xnor | CC BY-SA 3.0 |
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| Jan 15, 2015 at 0:38 | comment | added | Lopsy | 1, 1, 1, 1, 3? | |
| Jan 14, 2015 at 23:58 | answer | added | Florian F | timeline score: 6 | |
| Jan 14, 2015 at 23:26 | history | asked | xnor | CC BY-SA 3.0 |