Point selection problem
How to define "median" for point set on plane?
Is there always a point in many triangles?
The planar case
Given $n$ points on the plane in general position.
There exists a point in $\ge$$\frac{2}{9}$${n\choose 3}-O(n^2)$ triangles.
The constant $\frac{2}{9}$ is the best possible by stretched grid!
Higher dimensional case
Given $n$ points in $\mathbb{R}^d$ in general position.
A simplex is the convex hull of $d+1$ points.
There exists a point in $\ge c_d{n \choose d+1}$ simplices.
We cannot hope for $c_d > \frac{(d+1)!}{(d+1)^d}$ by stretched grid.
Timeline
Bárány: a point is in $\ge$ $c_d$$n\choose d+1$ simplices.
| 1982 | Bárány | $\frac{1}{(d+1)^d}$ | |
| 2003 | Wagner | $\frac{(d^2+1)}{(d+1)^{d+1}}$ | |
| 2010 | Gromov | $\frac{2d}{(d+1)(d+1)!}$ $< \frac{(d+1)!}{(d+1)^d}$ |
Plan of talk
| ▶ | Introduce a colored variant |
| Karasev's proof with a twist |
A colored variant
Given $n$ red points, $n$ green points and $n$ blue points.
There is a point in at least $pn^3$ colorful triangles.
Probabilistic equivalence
Given $n$ red points, $n$ green points and $n$ blue points.
A point in random colorful triangle with probability $\ge p$.
Given 3 (continuous) probability measures $m_0$, $m_1$ and $m_2$.
A point in random colorful triangle with probability $\ge$ $p$.
Timeline (cont.)
| 2010 | Gromov | $\frac{2}{9}$ $\to \frac{2d}{(d+1)(d+1)!}$ | $m_i = m_j$ | |
| 2012 | Karasev | $\frac{1}{6}$ $\to \frac{1}{(d+1)!}$ | ||
| $\frac{2}{9}$ $\to \frac{2d}{(d+1)(d+1)!}$ | $m_1$ $=$ $m_2$ | |||
| 2014 | J. | $\frac{2}{9}$ $\to \frac{2d}{(d+1)(d+1)!}$ |
Plan of talk
| ▶ | Introduce a colored variant |
| ▶ | Karasev's proof with a twist |
Karasev's proof
Given 3 (continuous) probability measures $m_0$, $m_1$ and $m_2$.
Proof by contradiction. Assume for every point $v$,
$\operatorname{Pr}\big(v \text{ is in} $ $ \big) < p := 2/9.$
Plane (plus a point at infinity) and sphere are the same.
Preparation
- Choose a sufficiently fine triangulation of sphere.
- Add "buttress" inside the sphere.
Auxiliary map
Gradually build a map from "sphere with buttress" to sphere:
such that:
- it is identity when restricted to the sphere;
- it is economical: e.g., image of each "spike" is small.
Build economical map gradually
Step 1: Vertex $O$ (sphere center) $\mapsto \infty$ (the north pole).
| Claim | $\operatorname{Pr}\big(\overline{P\infty} \text{ intersects }$ $\big)$ $< p$ for some $\overline{P\infty}$. |
| Obs | is in $\Longleftrightarrow$ intersects . |
$\operatorname{Pr}\big($ intersects $\big)=\operatorname{Pr}\big($ is in $\big) < p$
$\implies \operatorname{Pr}\big($ intersects $\;\big\vert\; $$\big) < p$ for some
Set $a := m_1(A)$, $b := m_2(A)$, $x := \overline{m}(A) = (a + b)/2$.
$2x(1-x)\le $ $a(1-b)+b(1-a) \le$ $\operatorname{Pr}\big($ int. $\big)$ $< 2p$
$\implies x(1-x) < p = 2/9$. WLOG assume $\overline{m}(A) = x < 1/3$.
Degree argument
We have found a (continous) map $f$ such that
$\overline{m}\big(f(OPQ)\mathrm{mod2}\big)<\frac{1}{3}$ $\implies \overline{m}\big(f(\text{spike})\mathrm{mod2}\big) < 1$.
| Def | Degree of map $f$ is $\left|f^{-1}(\text{generic point})\right|$. |
Note there are plenty of points not in $f(\text{spike})\mathrm{mod2}$.
| Obs | $\sum$deg of $f$ on each spike is even. |
| Cor | $\text{deg on sphere} + 2\sum \text{deg on }OPQ$ is even. |
| ! | Degree of identity map is even. Impossible! |