Linear rarity under equality-preserving affine deformation

---

This verifies Section 3 and Appendix B of the independent input, and slightly strengthens its bound. All multiplicities count unordered pairs in the full $n$-point set. No assertion for arbitrary planar sets follows.

Equivalent affine and coordinate conditions

For a finite planar set $A$, the following are equivalent:

1. An invertible affine map whose linear part is not a similarity preserves every original equality of pair distances.
2. In some orthonormal coordinates, equal-length difference vectors have the same absolute horizontal displacement, and hence also the same absolute vertical displacement.
3. The matrices $dd^{\mathsf T}-ee^{\mathsf T}$, over all difference vectors $d,e$ of equal length in $A$, span a subspace $V$ of dimension at most one in the two-dimensional space of traceless symmetric $2\times2$ matrices.

For (1) to (2), put $Q=T^{\mathsf T}T$ for the linear part $T$ and use its orthonormal eigenbasis. Its positive eigenvalues $\lambda,\mu$ are distinct. Equal vectors in length satisfy both

$$d_x^2+d_y^2=e_x^2+e_y^2,\qquad \lambda d_x^2+\mu d_y^2=\lambda e_x^2+\mu e_y^2,$$

so $d_x^2=e_x^2$. Conversely, (2) permits any positive diagonal stretching with unequal scale factors. For (1) and (3), equality preservation is exactly orthogonality of $Q$ to $V$ under the trace inner product. The traceless part of a nonscalar $Q$ is nonzero. Conversely a nonzero traceless $S\perp V$ gives a positive definite nonscalar $Q=tI+S$ for sufficiently large $t$, and $Q=T^{\mathsf T}T$ for an invertible $T$.

No prohibition on mergers of originally unequal classes is needed.

A linear bound

Theorem. Under these equivalent conditions, for $n\ge2$,

$$R(A)\ge\left\lceil\frac n4\right\rceil. \tag{1}$$

In particular $R(A)\ge2$ for every $n\ge5$ in this family.

Work in coordinates from condition (2). Each distance class has fixed absolute displacement pair $(a,b)$. A vertical class ($a=0$) is a disjoint union of finite translation paths, so has at most $n-1$ edges. If all points lie on one vertical line, every class is rare and at least $n-1$ distances occur, proving (1).

Otherwise let $m=\min x$, $M=\max x$, $W=M-m>0$, and let $c$ points lie on the middle line $x=(m+M)/2$. A class with $a>W/2$ has maximum degree at most two: a vertex can have neighbors on at most one of $x-a,x+a$, and that line allows at most two ordinates. It is rare.

At least $(n-c)/2$ points lie strictly on one side of the middle line. Join those points to an anchor on the opposite extreme line. All these edges belong to rare classes with $a>W/2$. At most two have the same length, since both their absolute displacements are fixed by that length. This supplies at least $\lceil(n-c)/4\rceil$ rare classes. If $c\ge1$, distances from a lowest middle-line point to the other $c-1$ middle-line points give $c-1$ further distinct vertical classes. They cannot coincide with any class already counted.

For $c=0$ this proves (1). For $c\ge2$,

$$\left\lceil\frac{n-c}{4}\right\rceil+c-1 =\left\lceil\frac{n+3c-4}{4}\right\rceil \ge\left\lceil\frac n4\right\rceil.$$

It remains to handle $c=1$. Every class with $a=W/2$ is also rare in this case. On the three lines $x=m,(m+M)/2,M$, its graph is a star with the unique middle point as center, together with isolates. Every other point can have neighbors on only one other vertical line, so the remaining graph has maximum degree at most two. There are no edges between these two parts. Thus the total edge count is at most $n$. The distance from the unique middle point to an extreme anchor supplies one such class, distinct from all strictly long classes. Hence

$$R(A)\ge\left\lceil\frac{n-1}{4}\right\rceil+1 \ge\left\lceil\frac n4\right\rceil.$$

This completes the proof. The midpoint argument is a checked improvement over the input's $\lceil(n-1)/4\rceil$ bound. Without it, the long-class argument requires the strict inequality $a>W/2$; one cannot simply replace it by $a\ge W/2$ for an arbitrary middle fiber.

The input's separate five-point argument is also valid but unnecessary for (1). For $a,b>0$, a connected class component is a subgraph of the integer grid in the basis $(a,b),(a,-b)$, not in the original coordinate axes. A component of $u$ vertices occupying $r$ rows and $s$ columns has at most $2u-r-s\le2u-2\sqrt u$ edges. Summing bounds the full class by $2n-2\sqrt n$, whose floor is five at $n=5$.

Obstructions and comparison with earlier notes

An equilateral triangle forces $Q$ to be scalar: equality on the three unit directions $(1,0),(1/2,\sqrt3/2),(-1/2,\sqrt3/2)$ first forces the off-diagonal entry to vanish and then the diagonals to agree. A square also forces this, using its side equalities and then its diagonal equality. Thus this theorem is a restricted family result, not a general deformation reduction.

The safe direction of transfer was already proved in variational_reduction: if distinct points are deformed while preserving all original equalities, classes can merge but cannot split, so $R(\text{new})\le R(\text{old})$. Arbitrary perturbation goes in the wrong direction. A regular hexagon with its center has multiplicities $(12,6,3)$ and $R=2$; a generic small perturbation has 21 singleton classes. Those new rare classes give no lower bound on the original rarity.

Unlike the line-cover and dyadic criteria, this theorem requires no bound on covering lines and no arithmetic assumption. Nor does either of those criteria imply this one: a rational square already fails the affine test. These are different sufficient conditions, not an exhaustive list of planar configurations. A hypothetical bounded-$R$ sequence must eventually have $\dim V=2$ by (1), but that supplies no useful nonlinear deformation or equilibrium-stress signs.

The original argument, the matrix test, and the midpoint sharpening were checked by the author; the argument is author-recorded. Neither unrestricted assertion of Problem 132 is settled.