Timeline for Random planes separating points in $\mathbb{R}^3$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2021 at 19:13 | comment | added | Penelope Benenati | Thank you @LiviuNicolaescu ! | |
| Mar 15, 2021 at 18:56 | comment | added | Liviu Nicolaescu | @PenelopeBenenati The problem is equivalent to the following. Suppose that you have $n-1$ points on a circle. What is the probability that a uniformly chosen random diameter leaves $k$ of them on one side and $n-1-k$ on the other side. The special case $n-1=4$ and the points are the vertices of a square shows that for $k=2$ the probability is $1$ while for other $k$țs the answer is zero. The answer is dependent on the configuration of those $(n-1)$ points. At one extreme they can be clustered and at the other extrem uniformly distributed. | |
| Mar 15, 2021 at 15:25 | vote | accept | Penelope Benenati | ||
| Mar 15, 2021 at 14:44 | answer | added | Iosif Pinelis | timeline score: 4 | |
| Mar 15, 2021 at 12:16 | comment | added | Penelope Benenati | @M.Dus sorry, I did not understand why you wrote that I "didn't say wha is the law of $\mathbf{h}$", while I wrote "[...] Let $\mathbf{h}$ be a point selected uniformly at random from $S$ ". | |
| Mar 15, 2021 at 12:14 | comment | added | Penelope Benenati | @LiviuNicolaescu, thank you for your answer! How can we generalize it for the case in which we have $\mathbf{x}$ and $n-1$ other points when $n>3$? | |
| Mar 15, 2021 at 12:13 | comment | added | Penelope Benenati | Thank you @YaakovBaruch, there was a typo. | |
| Mar 15, 2021 at 12:13 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
edited body
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| Mar 15, 2021 at 11:35 | comment | added | Yaakov Baruch | You probably mean $\frac12\left(\arccos\left(\mathbf{x}^{\top}\mathbf{y}\right)+\arccos\left(\mathbf{x}^{\top}\mathbf{z}\right)-\arccos\left(\mathbf{y}^{\top}\mathbf{z}\right)\right)$ | |
| Mar 15, 2021 at 10:55 | comment | added | Liviu Nicolaescu | Might as well assume that $x$ is the point $(1,0,0)$. There is a half-circle of planes through $0$ and $x$. Denote by $\alpha$ the angle between the planes $oxy$ and $oxz$. The probability is $\alpha/\pi$. | |
| Mar 15, 2021 at 10:17 | comment | added | M. Dus | Note that $H$ separates $x$ from $y$ and $z$ if and only if $h^\perp x \cdot h^\perp y<0$ and $h^\perp x\cdot h^\perp z<0$. You didn't say what is the law of $h$, but if you know the marginal probability distributions, then you can compute everything. | |
| Mar 15, 2021 at 9:58 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
added 2 characters in body
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| Mar 15, 2021 at 9:27 | history | asked | Penelope Benenati | CC BY-SA 4.0 |