Timeline for Diameter of m-fold cover
Current License: CC BY-SA 2.5
8 events
when toggle format | what | by | license | comment | |
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Mar 1 at 18:16 | comment | added | Anton Petrunin | Do you know other proofs that use the same type of argument? (I realized that it was used in the following puzzle from the mind-benders of Winkler: "You have two sets (one set red, one blue) of n n-sided dice, each die labeled with the numbers 1 through n. You roll all 2n dice simultaneously. Prove that there must be a nonempty subset of the red dice and a nonempty subset of the blue dice with the same sum!") | |
Dec 19, 2014 at 5:28 | comment | added | Włodzimierz Holsztyński | **** Nice! **** | |
Jun 25, 2013 at 15:29 | vote | accept | Anton Petrunin | ||
Mar 3, 2010 at 9:56 | comment | added | Sergei Ivanov | @Alon: I've edited the second paragraph to make it more clear. | |
Mar 3, 2010 at 9:54 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
clarified argument
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Mar 3, 2010 at 1:01 | comment | added | Anton Petrunin | @Alon. Otherwise the distance is $\le m\cdot \mathrm{diam}$ | |
Mar 3, 2010 at 0:36 | comment | added | Alon Amit | I must be missing something - why do you say that L(bi) < L(ai) in the second paragraph? Why is that a strict inequality? | |
Mar 3, 2010 at 0:08 | history | answered | Sergei Ivanov | CC BY-SA 2.5 |