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title: MDS count
---
Following Larry's book 17.8.3, $d\geq 8r+1$

1) Disks of radius $4r$ around vertices do not overlap
$$D_{4r}(v_{i_j})=\{(i_1,i_2,i_3,i_4)|~~ (i_1+i_2+i_3+i_4=d) ~~\wedge~~ (i_j\geq d-4r)\}$$

WLOG, we want to show that $D:=D_{4r}(v_{1})\cap D_{4r}(v_{2})=\emptyset$.
**Proof**
\begin{align*}
(i_1,i_2,i_3,i_4)&\in D \Leftrightarrow \\&(i_1+i_2+i_3+i_4=d) \wedge (i_1\geq d-4r) \wedge (i_2\geq d-4r)\\
&\Rightarrow i_1+i_2\geq 2d-8r\geq 2d-(d-1)=d+1~~\text{since}~~ d\geq 8r+1
\end{align*}
Contradiction, since $i_1+i_2+i_3+i_4=d$.
2. Tubes of radius $2r$ around edges overlap only inside the disks of radius $4r$ around common vertices
$$D_{2r}(v_i,v_j)=\{(i_1,i_2,i_3,i_4)|~(i_1+i_2+i_3+i_4=d)~ \wedge ~(i_j+v_j\geq d-2r).\}$$
WLOG we want to show that
$$E_1:=[D_{2r}(v_1,v_2)\cap D_{2r}(v_1,v_3)]\subset D_{4r}(v_{1}),$$
and
$$E:=D_{2r}(v_1,v_2)\cap D_{2r}(v_3,v_4)=\emptyset.$$
**Proof**
\begin{align*}
(i_1,i_2,i_3,i_4)&\in E_1 \Leftrightarrow \\&(i_1+i_2+i_3+i_4=d) \wedge (i_1+i_2\geq d-2r) \wedge (i_1+i_3\geq d-2r)\\
&\Rightarrow 2i_1+i_2+i_3\geq 2d-4r\\
&\Rightarrow i_1+d-i_4\geq 2d-4r~~\Rightarrow i_1-i_4\geq d-4r\\
&\Rightarrow i_1\geq i_1-i_4\geq d-4r,
\end{align*}
and
\begin{align*}
(i_1,i_2,i_3,i_4)&\in E \Leftrightarrow \\&(i_1+i_2+i_3+i_4=d) \wedge (i_1+i_2\geq d-2r) \wedge (i_3+i_4\geq d-2r)\\
&\Rightarrow i_1+i_2+i_3+i_4\geq 2d-4r\\
&\Rightarrow d\geq 2d-4r~~\Rightarrow d\leq 4r,
\end{align*}
contradiction.

Then Peter's projection idea follows here. . `` *The notation is a bit messy but the idea is simple. Suppose, without loss of generality, that the edge in question is in the z-axis and we project into the x-y plane. Suppose also all points in the 4r balls at the ends of the edge have been determined. We want to determine the remaining points in the 2r-tube.
So we start at one vertex ball and work our way through the layers in the tube parallel to that ball. Then consider the 3D smoothness conditions.
All the coefficients in them, except those going with points in the new layer are zero. Looking closely we see that those 3D conditions involving the remaining points are exactly the 2D conditions in the projection of the orange in question. The formula follows from there.*""

3. Layers of radius $r$ around triangular faces overlap only inside the tubes of radius $2r$ of the common edge.

\begin{align*}
D_r(v_i,v_i,v_k):=&\{(i_1,i_2,i_3,i_4)|~(i_1+i_2+i_3+i_4=d)~ \wedge ~(i_j+v_j+v_k\geq d-r)\}\\
=&\{(i_1,i_2,i_3,i_4)|~(i_1+i_2+i_3+i_4=d)~ \wedge ~(i_l\leq r)\}
\end{align*}
WLOG we want to prove that
$$L:=[D_r(v_1,v_2,v_3)\cap D_r(v_1,v_2,v_4)]\subset D_{2r}(v_1,v_2))$$
**Proof**
\begin{align*}
(i_1,i_2,i_3,i_4)&\in L \Leftrightarrow \\&(i_1+i_2+i_3+i_4=d) \wedge (i_3\leq r) \wedge (i_4\leq r)\\
&\Rightarrow i_3+i_4+i_3\leq 2r\\
&\Rightarrow i_1+i_2\geq d-2r.
\end{align*}
4. How many points are left inside of each tetra?
\begin{align*}
T:=&\{(i_1,i_2,i_3,i_4)|~(i_1+i_2+i_3+i_4=d)~ \wedge ~(i_j< d-4r, ~~\forall j)\\
&\wedge ~(i_j> r,~~ \forall j)~~\wedge ~~(i_j+i_k<d-2r,~~\forall j\neq k\}
\end{align*}
We want cardinality of $T$
According to Peter's note
$$card(T)=\binom{d-4r-1}{3}-4\binom{r}{3}$$
**Proof**

TBD