Nonlinear analytic doubling and tripling increase rarity

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Subsequent extension: four_point_collisions proves the same cross-distance bound and rarity increment for $b=4$. The arguments below retain their original $b=2,3$ scope; the four-point gap is closed by that separate proof. General $b$ and an arbitrary-configuration reduction remain unresolved.

This approach permits exact distance coincidences throughout a nonlinear analytic deformation, unlike generic cluster replacement. The result is proved only for cluster sizes two and three. It does not resolve either unrestricted assertion of Problem 132.

Two separated collision limits

Let $b\in\{2,3\}$. Suppose the coordinates of

$$X(t)=\{X_1(t),\ldots,X_b(t)\},\qquad Y(t)=\{Y_1(t),\ldots,Y_b(t)\}$$

are real-analytic at $t=0$. Assume the points within each set are distinct for all sufficiently small $t>0$, and

$$X_i(t)\longrightarrow p,\qquad Y_j(t)\longrightarrow q, \qquad p\ne q.$$

Then, for all sufficiently small $t>0$, there are at least $b$ distinct distances between $X(t)$ and $Y(t)$.

Analyticity is used at the endpoint $0$, not merely on a punctured interval. A nonzero analytic germ has a first nonzero Taylor coefficient; thus its sign is fixed near zero. In particular, the equality pattern of the finitely many squared cross-distances is fixed for small positive $t$. Having too few values at arbitrarily small parameters would give the same number of analytic distance germs.

Two-point clusters

If all four cross-distances were equal, the two $Y$ points would be the two intersections of equal-radius circles centered at the $X$ points. Their midpoint would equal the midpoint of the $X$ points. Passing to the limit would give $q=p$, a contradiction.

Three-point clusters

After a fixed similarity, take $p=(0,0)$ and $q=(d,0)$ with $d>0$. Suppose there are only two squared cross-distance germs $a(t)<b(t)$. One value alone is impossible, since three distinct points cannot lie on two distinct circles of the same radius.

Call a pair within one cluster flat if its coordinate difference satisfies $\Delta x=o(\Delta y)$. If a pair is not flat, analyticity gives $\Delta y=O(\Delta x)$ and $\Delta x\not\equiv0$. For a nonflat pair $X_i,X_k$, its squared-distance row difference is

$$|X_i-Y_j|^2-|X_k-Y_j|^2 =-2d\,\Delta x+o(\Delta x), \tag{1}$$

uniformly over the three columns. Indeed, expand the difference as $\Delta x(x_i+x_k-2x(Y_j))+ \Delta y(y_i+y_k-2y(Y_j))$ and use the collision limits. Thus the two rows are strictly pointwise ordered for small $t$. With only the values $a,b$, one row must be constantly $a$ and the other constantly $b$. All three $Y$ points would lie on the intersection of two distinct circles, which has at most two points. This is impossible. The same argument for columns shows that every pair in either cluster is flat.

Their vertical coordinate germs are consequently distinct. Order both clusters increasingly by those coordinates, an order fixed for small $t>0$, and set $D_{ij}=|X_i-Y_j|^2$. For $i<k$, $j<\ell$,

$$D_{ij}+D_{k\ell}-D_{i\ell}-D_{kj} =-2(X_i-X_k)\cdot(Y_j-Y_\ell)<0. \tag{2}$$

The vertical product is positive and dominates the horizontal product. Thus $D$ is strictly Monge. For either adjacent column pair, its three row increments strictly decrease by (2). The possible increments are only $-(b-a),0,b-a$, so they must occur in the order $b-a,0,-(b-a)$. The first row therefore increases by $b-a$ across each of its two column steps. That requires three values, contradicting the assumed two-value palette. This proves the theorem.

Consequence for balanced collision families

Let $P$ be any $N$-point planar set, $N\ge2$, and let $A(t)$ consist of $b$ distinct analytic point curves converging to each point of $P$, where $b=2$ or $b=3$. Then $|A(t)|=n=Nb$ for sufficiently small $t>0$, and

$$R(A(t))\ge R(P)+1. \tag{3}$$

No generic choice of a scale parameter is required, and the branch shapes need not be constant or congruent.

To prove (3), fix a rare base distance $r$, with $m_P(r)\le N$. All pairs between its corresponding clusters form one limiting-distance group with exactly $b^2m_P(r)\le b^2N=bn$ unordered pairs. This group has at least $b$ actual distance classes by the preceding theorem, applied to any one of its cluster pairs. Hence at least one of these classes has at most $n$ pairs. Groups with different positive limiting distances do not merge for sufficiently small $t$, and none merges with an internal cluster distance, which tends to zero. Each rare base distance therefore supplies a distinct globally rare class.

The largest internal cluster distance supplies one more. It occurs only in clusters attaining that internal diameter, and in each such cluster it has at most $b$ pairs by the planar diameter bound. Its total multiplicity is at most $Nb=n$. This proves (3), with the full-set threshold throughout.

A sharp example excluding a pinned-distance shortcut

The three-point lower bound cannot be strengthened to the assertion that some point sees three different cross-distances. Put $s=\sqrt{1+3t^2}$ and take

$$X(t)=\{(0,-t),(1-s,0),(0,t)\},$$ $$Y(t)=\{(1,-t),(s,0),(1,t)\}.$$

These converge to $(0,0)$ and $(1,0)$, respectively. Their squared cross-distance matrix is

$$\begin{pmatrix} 1&1+4t^2&1+4t^2\\ 1+4t^2&(2s-1)^2&1+4t^2\\ 1+4t^2&1+4t^2&1 \end{pmatrix}.$$

For $t>0$ its three displayed values are distinct: the middle diagonal exceeds $1+4t^2$ because $1+2t^2>\sqrt{1+3t^2}$. Every row and column nevertheless uses only two values. Thus a fully distinct pinned spectrum is not available even in this collision setting.

Precise gap for larger clusters

The analogous assertion for arbitrary fixed $b$ was not proved or disproved. With three or more palette values, pointwise ordered rows need not be constant, so the two-circle argument in (1) no longer applies. The initial four-point attempt stopped at this obstacle, but the subsequent four-point proof resolves it by reflection pairs and a three-value Monge argument. The remaining cluster-size question concerns $b\ge5$.

Even when every pair is flat, the argument would need a general distinct-entry bound for strictly Monge matrices. The proposed bound of at least $b$ values in a $b\times b$ matrix is not established here. The weaker proved estimate is

$$b\le k(k-1)+1$$

when there are $k$ values: differences between two fixed rows strictly increase across the columns and lie in a difference set of size at most $k(k-1)+1$.

Replacing the strict Monge inequalities by their ordinal consequence is insufficient. The matrix

$$\begin{pmatrix}0&0&1\\1&0&1\\1&0&0\end{pmatrix}$$

has two values but satisfies $\max(M_{i\ell},M_{kj})>\min(M_{ij},M_{k\ell})$ for every $i<k$, $j<\ell$: the maximum on the left is one and the minimum on the right is zero in every case.

First nonzero Taylor coefficients alone also lose information. A parabolic coefficient matrix $a_i+b_j-2u_i v_j$ can have $m^3$ distinct row parameter pairs and $m^3$ distinct column parameter pairs, but only $O(m^2)$ values: take $a_i,b_j\in\{1,\ldots,m^2\}$ and $u_i,v_j\in\{1,\ldots,m\}$ independently. Higher-order Euclidean terms must enter any argument that handles these repeated transverse coordinates. This example concerns coefficient matrices, not exact planar cross-distances.

The author checked the two- and three-point arguments, the full-set pair budget in (3), and the exact sharp example; they are author-recorded. The general collision problem and its matrix reductions remain provisional directions. No reduction of arbitrary planar configurations to such analytic collision families has been established.

An exact monic-quadratic model for the remaining cluster question

Choose distinct real numbers $u_1,\ldots,u_b$ and distinct real pairs $(v_j,\beta_j)$, and set

$$X_i(t)=(0,tu_i),\qquad Y_j(t)=\left(\sqrt{1+t^2(\beta_j-v_j^2)},\,tv_j\right).$$

For sufficiently small $t$, the positive square root is analytic, all branches are distinct within their clusters, and the two limits are $(0,0)$ and $(1,0)$. Direct expansion gives the exact identity

$$|X_i(t)-Y_j(t)|^2 =1+t^2\bigl(u_i^2-2u_iv_j+\beta_j\bigr).$$

Consequently, $b$ distinct monic quadratic polynomials whose collective values on a $b$-element real set number fewer than $b$ would disprove the proposed general cluster theorem. No such polynomials have been found, and no theorem excluding them has been established here. This is a concrete exact subproblem, not an equivalent reformulation of all analytic cluster families or a reduction of arbitrary planar sets.

A per-class forest argument also fails. Take

$$X_1=(0,-t),\quad X_2=(0,0),\quad X_3=(0,t),$$ $$Y_1=(\sqrt{1-t^2/4},-t/2),\quad Y_2=(\sqrt{1-t^2/4},t/2),\quad Y_3=(\sqrt{1-t^2},0).$$

Their squared cross-distance matrix is

$$\begin{pmatrix} 1&1+2t^2&1\\ 1&1&1-t^2\\ 1+2t^2&1&1 \end{pmatrix}.$$

The unit-distance class is a six-cycle, although both clusters shrink to distinct points. This disproves acyclicity of individual cross classes, not the already proved three-cluster bound or the proposed general bound. The author checked both displayed identities and branch distinctness; they are author-recorded. The general strict-Monge palette bound and the nonflat-pair analysis for $b\ge5$ remain unresolved. No computation from this continuation is being promoted to an all-size Monge theorem.