Proximal short-edge anchors and the remote-anchor obstruction

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The short-edge inverse assumes linearly many edges of bounded smallest-distance rank touch a convex subset. The first result weakens incidence to bounded-distance proximity. The construction then explains why dropping that proximity cannot be justified by annular geometry alone.

A neighborhood version of the inverse

Theorem. Fix $B,\alpha,L>0$ and an integer $s\ge1$. Suppose $A$ has $n$ points, $q(A)\le Bn$, and at most $s-1$ distances smaller than an occurring $r$. Let $C_0\subseteq A$ be strictly convex. If at least $\alpha n$ unordered $r$-edges have an endpoint at distance at most $Lr$ from some point of $C_0$, then, for sufficiently large $n$, a circle contains $cn$ points of $A$, where $c>0$ depends only on the fixed parameters.

Orient each counted edge with one close endpoint $p$ first and assign it a point $c(p)\in C_0$ with $|p-c(p)|\le Lr$. A radius-$Lr$ disk contains at most

$$K=2s^2(\lceil6L\rceil+2)^2$$

points of $A$. To see this, cover it with at most $(\lceil6L\rceil+2)^2$ squares of side $r/3$. Each square has diameter less than $r$ and thus at most $s-1$ distance values. Two anchors and circle intersections bound its cardinality by $2(s-1)^2+2\le2s^2$ (a square with fewer than two points is immediate). The fixed-short-rank degree bound is $6s$, so each $c\in C_0$ receives at most $6sK$ ordered anchors.

Partition $C_0$ into $J\le12$ narrow convex chains, as in the original inverse proof. One chain $C$, of size $m$, receives $E\ge\alpha n/J$ anchors and hence has $m\ge E/(6sK)$. In particular $m=\Omega(n)$ and $q(A)=O(m)$.

The original chain lemma bounds the number of positive pinned distances, counted with their endpoints, from a chain vertex in any interval of length $r$ by $M=8(2s-1)$. If $p$ is assigned to $c\in C$ and $|p-x|$ lies in one such interval, then $|c-x|$ lies in an interval of length $(1+2L)r$. Thus the corresponding bound for the possibly off-chain anchor $p$ is

$$M'=(\lceil2L\rceil+2)M+1,$$

where the extra one allows $x=c$. Replace $M$ by $M'$ in the overloaded-bin proof. A retained apex $x\in C$ has both anchor distances in good neighboring bins, because the anchor edge has length $r$. The same linear triangle-type count supplies linearly many rich direct motions on $C$, and hence a positive circular fraction. No other step of that proof requires the anchor endpoint to lie on $C$.

Consequently, in a bounded-rarity sequence, for every fixed $L$ and fixed short rank, only $o(n)$ edges of that rank can have an endpoint within $Lr$ of any convex subset, uniformly over the subset. The neighborhood is of its vertices, not of the entire filled convex hull.

Remote anchors can have arbitrarily crowded annular bins

Let even $t\ge6$ tend to infinity, put $k=t^2/2$ and $H=t^{10}+1/2$, and take

$$P=\{0,\ldots,t-1\}^2,\qquad C=\{c_i=(i,H-i^2/(2H)):1\le i\le k\}.$$

Add $a=(-2H,-2H)$ and $b=(2H,-2H)$, and put $A=P\cup C\cup\{a,b\}$. The strictly concave parabola chain, together with the two lower corners, gives a strictly convex subset $C_0=C\cup\{a,b\}$ containing $P$ in its interior. Its consecutive upper-chain slopes strictly decrease; the two joining sides have slopes respectively greater and smaller than every chain slope. These also show directly that the grid lies below the upper boundary and above the lower side.

The minimum distance is one and has precisely the $2t(t-1)$ horizontal and vertical grid pairs. Chain pairs have distance strictly greater than one, and all other pairs outside the grid are much longer. Since $n=3t^2/2+2$, this minimum class is rich. It supplies linearly many translated anchors in each coordinate direction.

Nevertheless, for every $p=(u,v)\in P$, all $k$ chain vertices lie in one standard unit annular bin about $p$. Indeed,

$$|p-c_i|^2-(H-v)^2 =u^2-2ui+vi^2/H+i^4/(4H^2).$$

Its absolute value is less than $4t^3$, while $H-v>H/2$, so

$$\bigl||p-c_i|-(H-v)\bigr|<8t^3/H<1/4.$$

Since $H-v$ is a half-integer, every $|p-c_i|$ belongs to $[\lfloor H\rfloor-v,\lfloor H\rfloor-v+1)$.

No circle contains more than $2t+6=o(n)$ points of $A$: it meets each of the $t$ grid columns at most twice, the parabola at most four times (substitution gives a polynomial of degree four with nonzero leading coefficient), and contains at most the two added corners.

This is not an obstruction satisfying the small-palette hypothesis. In fact all distances within $C$ are distinct, since

$$|c_i-c_j|^2=(i-j)^2\left(1+\frac{(i+j)^2}{4H^2}\right).$$

Different positive values of $|i-j|$ give disjoint intervals: the correction to the integer square is less than $k^4/H^2<1$. For a fixed difference, the positive sum $i+j$ distinguishes the pairs. Thus $q(A)\ge\binom{k}{2}$ is quadratic.

The author checked the neighborhood packing and transfer of the triangle proof, and the exact parabola formulas, convexity, separation, and pair counts; the arguments are author-recorded. Remote anchors cannot be controlled by short rank, richness, translations, or annular geometry alone; a successful extension there must use the small global palette. No such extension is proved.