What projection order and crossing inequalities omit

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These are checks of a new sweep/order approach, not a solution of Problem 132.

An affine sweep can have all distances rare

Every finite planar configuration admits an orientation-preserving affine image with the same order type and cyclic projection-sweep sequence, up to reparametrizing the direction, in which all distances are rare.

Let $v,w$ range over its nonzero difference vectors with $v\ne\pm w$. Choose a positive-definite symmetric matrix $H$ avoiding all finitely many hyperplanes

$$v^{\mathsf T}Hv=w^{\mathsf T}Hw.$$

Each hyperplane is proper because $vv^{\mathsf T}\ne ww^{\mathsf T}$. With $T=H^{1/2}$, equal distances in $T(A)$ therefore correspond only to a single displacement up to sign. The graph for such a displacement is a disjoint union of finite translation paths, so has at most $n-1$ edges.

The determinant of $T$ is positive, preserving orientation signs. Projection of $T(A)$ in direction $u$ is projection of $A$ in direction $T^{\mathsf T}u$. The induced map on the direction circle preserves cyclic order, including simultaneous crossings. Thus the sweep ordering does not retain the equal-length amplitudes. This does not prove that all order-type methods must fail; it identifies information that cannot be discarded.

Unlike the equality-preserving affine criterion, this deformation can split the original distance classes. It gives no lower bound for $R(A)$ from the transformed configuration.

Few-valued strict Monge matrices need not be Euclidean

Crossing diagonals in a convex quadrilateral have greater total length than either pair of opposite sides. Opposite-chain distance blocks consequently satisfy Monge inequalities, with the sign depending on column orientation. These inequalities and a linear number of distinct entries alone do not impose Euclidean rank.

For $1\le i,j\le h$ and $L>0$, set

$$D_{ij}=L+(i-j)^2+\tfrac14\delta_{ij}.$$

This matrix has exactly $h$ distinct entries. For $i<i'$ and $j<j'$,

$$D_{ij}+D_{i'j'}-D_{ij'}-D_{i'j} =-2(i'-i)(j'-j)+E,\qquad |E|\le\tfrac12,$$

so it is strictly Monge. Reversing the columns reverses the sign. Nevertheless its entrywise squared matrix satisfies

$$\operatorname{rank}(D^{\circ2})\ge h-5.$$

Indeed, the matrix $P_{ij}=(L+(i-j)^2)^2$ has rank at most five, by expansion in the row powers $1,i,\ldots,i^4$, while

$$D^{\circ2}=P+(L/2+1/16)I.$$

The latter scalar is nonzero, and the rank inequality gives the claim. A planar bipartite squared-distance matrix has rank at most four, since

$$|p_i-q_j|^2=|p_i|^2+|q_j|^2-2p_i\cdot q_j.$$

For $h\ge10$ the displayed Monge matrix cannot be such a distance matrix. Taking $L\ge h^2$ even puts all its entries within a factor of two. This is an obstruction to the scalar relaxation, not a nonexistence result for actual planar configurations with few distances.

Limits of distinct-entry inductions for Monge matrices

The nonlinear collision approach raised the separate question whether a strictly Monge $b\times b$ matrix must have at least $b$ distinct entries. This was neither proved nor refuted. The following abstract-matrix checks delimit proposed proofs; they are not planar configurations or counterexamples to Problem 132.

First, neither a row with $b$ distinct values nor a monochromatic-forest argument is available. The matrix

$$\begin{pmatrix}-1&0&0\\0&1/2&0\\0&0&-1\end{pmatrix}$$

is strictly Monge (all four adjacent defects are $-1/2$), but every row and column has only two values. Its zero entries form a bipartite six-cycle. A three-point analytic collision example also rules out the corresponding fully distinct pinned-distance shortcut for actual Euclidean clusters.

Second, an induction deleting one row and one column to remove an extreme entry fails. Consider

$$\begin{pmatrix} 0&5&4&11&9&13\\ 5&9&7&13&10&13\\ 0&3&0&5&1&3\\ 11&13&9&13&8&9\\ 7&8&3&6&0&0\\ 13&13&6&8&1&0 \end{pmatrix}.$$

Its adjacent rectangle defects are all $-1$, except the defect with upper-left position $(5,2)$, which is $-2$. Every general rectangle defect is a sum of adjacent defects, proving strict Monge. Minimum positions include $(1,1),(3,3),(5,5)$, and maximum positions include $(2,6),(4,4),(6,2)$. Neither support can be covered by one row and one column. The author checked this finite integer certificate directly.

There is an exact alternative formulation of the unresolved entry-count question. For a strictly Monge matrix $M$, the matrix $E(t)=(e^{-tM_{ij}})$ is totally positive for all sufficiently large $t$. In each minor, uncrossing an inversion strictly decreases the sum of the selected $M$ entries. The diagonal permutation is therefore the unique minimum-cost permutation. Its positive determinant term dominates the finite exponential sum for large $t$. Conversely, minus the entrywise logarithm of a totally positive matrix is strictly Monge by its positive $2\times2$ minors. These transformations preserve the number of distinct entries. Thus the question is equivalent to asking whether a totally positive $b\times b$ matrix must have at least $b$ distinct entries. This equivalence does not supply that lower bound.

Small numerical optimizations conducted during the exploration are not being treated as exact classifications or as proofs of additional orders.

A necessary isosceles surplus under R equal to one

Let $d_r(p)$ be the distance-$r$ degree of $p$, and put

$$I(A)=\sum_r\sum_{p\in A}\binom{d_r(p)}2.$$

This counts isosceles triples with a distinguished apex and an unordered base; an equilateral triangle contributes three. If $R(A)=1$, then

$$I(A)\ge \begin{cases} \binom n2+n-3,&n\text{ odd},\\ \binom n2+3n/2-4,&n\text{ even}. \end{cases} \tag{1}$$

Write $M=\binom n2$, $q=q(A)$, and let the diameter have $e$ pairs. Every other class is rich. The elementary integer inequalities $\binom d2\ge2d-3$ and $\binom d2\ge d-1$ give

$$I(A)\ge4(M-e)-3n(q-1)+2e-n =4M-2e-3nq+2n. \tag{2}$$

The pair budget gives $q\le\lfloor n/2\rfloor$. For odd $n$ at $q=(n-1)/2$, it also gives $e\le(n+3)/2$; substituting in (2) proves (1). If $q\le(n-3)/2$, use $e\le n$ in (2) to obtain the stronger $I(A)\ge M+3n$.

For even $n$ at $q=n/2$, the exact spectrum is one diameter pair and $n+1$ pairs in each of the other $n/2-1$ classes. Each rich class contributes at least $4(n+1)-3n=n+4$ to $I(A)$; the diameter contributes zero. This gives $(n/2-1)(n+4)=M+3n/2-4$. For smaller $q$, (2) gives the stronger bound $M+3n/2$.

If $k_{ab}$ is the number of points of $A$ on the perpendicular bisector of $ab$, then $I(A)=\sum_{\{a,b\}}k_{ab}$ and

$$\sum_{\{a,b\}}\binom{k_{ab}}2\ge I(A)-M.$$

Thus a counterexample requires linearly many choices of a base and two equidistant apices. This does not force a line containing many points: the count could be spread over many bisectors with exactly two apices. No geometric argument concentrating that surplus was established.

The author checked the affine hyperplanes, sweep reparametrization, Monge and rank calculations, and both parity cases in (1); the argument is author-recorded.