Timeline for Fraction of the sets receive each color
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2017 at 6:35 | history | bounty ended | Karo | ||
| Nov 24, 2017 at 6:35 | vote | accept | Karo | ||
| Nov 23, 2017 at 14:12 | comment | added | fedja | @Karo Also, if you do not like the continuous version, it is enough to use the discrete mesh with step $1/5$, say, because $\frac 25+\frac 32<2$ | |
| Nov 23, 2017 at 10:05 | comment | added | domotorp | @Karo I think fedja is using some continuous version of it, like en.wikipedia.org/wiki/…. | |
| Nov 23, 2017 at 9:47 | comment | added | Karo | @fedja Your fix is not very clear to me - how do you choose the triple real point $(x,y,z)$? Are you still using Sperner's lemma? | |
| Nov 23, 2017 at 9:44 | history | edited | domotorp | CC BY-SA 3.0 |
added update
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| Nov 23, 2017 at 6:50 | comment | added | fedja | @Dap That is a valid objection, of course, but the fix is easy. Let us just allow ourselves real coordinates (some balls can be shared when the dividing bars move through them). Then we have a triple real point $(x,y,z)$ with the sum $n$. Notice that we can move the bars to the nearest integers changing two numbers by $\le\frac 12$ and the third number by at most $1$, so the sum of any two changes will be below $3/2$. Thus, we will have the inequalities $r_i\ge m_i-3/2$, which automatically improve to $-1$ since everything is an integer now. Beautiful solution, Domotorp! | |
| Nov 23, 2017 at 6:06 | comment | added | Dap | In the last step, it's not clear to me why any vertex of the small triangle satisfies the conditions. Suppose we have a triangle with vertices $(r+1,g,b),(r,g+1,b),(r,g,b+1)$ (ignoring the $1/n$ scaling), colored red, green and blue respectively. Why should the red vertex have at least a third of sets receiving green? Compared to the green vertex, one green element has changed to red, which can decrease $g_i$ and increase $m_i,$ so sets that had $g_i\geq m_i$ are only guaranteed to satisfy $g_i\geq m_i-2.$ | |
| Nov 23, 2017 at 5:00 | history | edited | domotorp | CC BY-SA 3.0 |
replaced improper terminology barycentric subdivision
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| Nov 22, 2017 at 22:19 | history | answered | domotorp | CC BY-SA 3.0 |