Two rare distances for orders five through fourteen except eleven

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This checks Section 2 and Appendix A of the independent input against the supplied source theorems and existing finite evidence. It proves

$$R(A)\ge2\qquad(5\le |A|\le10). \tag{1}$$

The later unique-diameter theorem also proves $R(A)\ge2$ when $|A|=12$, as shown below. The subsequently checked six-distance classification also settles orders 13 and 14. Thus the established range is $5\le n\le14$, except $n=11$.

All multiplicities count unordered pairs, with threshold the full cardinality $n=|A|$. The general assertions of Problem 132 remain unresolved.

Counting and the exact even-order boundary

If $R(A)=1$, the sole rare class is the diameter. With $q$ distinct distances, the pair budget gives

$$\binom n2\ge1+(q-1)(n+1),\qquad q\le\lfloor n/2\rfloor. \tag{2}$$

If $n=2t$ and $q=t$, equality holds: the diameter occurs once and all other distances occur exactly $n+1$ times. Deleting either endpoint of the diameter leaves exactly $t-1$ classes, each of multiplicity at most $n+1$. None disappears, since at most $n-1$ incident pairs are removed. This is the same forced spectrum as in distance_partition_algebra; that note's symmetry exclusions remain valid additional restrictions.

Classification inputs and exact spectra

Erdős--Fishburn (1996), Theorem 1 gives $g(2)=5,g(3)=7,g(4)=9,g(5)=12$, where $g(k)$ is the largest size of a planar set with exactly $k$ distances. It also classifies all extremizers for $k\le4$: one type at five points, two at seven, and four at nine. The following multiplicities are deductions from explicit models, not a table quoted from that theorem.

The five-point model is a regular pentagon. The seven-point models are a regular heptagon and a regular hexagon with its center. Every distance of a regular odd $m$-gon occurs $m$ times.

For completeness, the four nine-point types can be exhausted without recovering coordinates from the source's omitted figures. The first is a regular nonagon. For the next two use axial lattice coordinates

$$\phi(x,y)=(x+y/2,\sqrt3\,y/2),\qquad Q(u,v)=u^2+uv+v^2.$$

Define

$$H=\{(0,0),(1,0),(0,1),(-1,1),(-1,0),(0,-1),(1,-1)\},$$ $$T_1=H\cup\{(1,1),(-1,2)\},$$ $$T_2=\{(0,0),(1,0),(0,1),(-1,1),(-1,2),(1,1), (2,0),(1,-1),(0,-1)\}.$$

Their squared-distance lists are both $(1,3,4,7)$, with respective multiplicities $(16,9,7,4)$ and $(15,9,6,6)$. Here is an integer certificate: distance-degree vectors in that order occur as follows; sum each column and divide by two for the pair counts.

For the last model put $s=\sqrt3$, $\zeta=e^{2\pi i/3}$, and

$$C=\{a\zeta^j:a\in\{1,1+s,-(2+s)\},\ j=0,1,2\}.$$

Within each three-point orbit, squared distances are $3a^2$, occurring three times. Between distinct signed radii $a,b$, the three same-index pairs have square $(a-b)^2$, and the six other pairs have square $a^2+ab+b^2$. Thus the three within-orbit squares are $3,12+6s,21+12s$; the same-index squares for the three orbit pairs are these same three values; and all other squares equal $6+3s$. In increasing order the spectrum is therefore

$$3:6,\quad6+3s:18,\quad12+6s:6,\quad21+12s:6.$$

The four sorted multiplicity multisets differ, so the models are pairwise nonsimilar. The classification allows only four types, hence these exhaust them. Every nonregular nine-point type has a class of multiplicity at least fifteen.

Proof of (1)

Suppose $R(A)=1$. For $n=5,7,9$, (2) and the bounds on $g(k)$ force respectively $q=2,3,4$. The table rules out each case.

For $n=6,8,10$, the same bounds force respectively $q=3,4,5$. The even-order equality case applies. Each of the two endpoint deletions must therefore be a regular $(n-1)$-gon: for $n=6$ this is the unique five-point type; for $n=8$ the nonregular seven-point type has multiplicity $12>9$; for $n=10$ each nonregular nine-point type has multiplicity at least $15>11$.

These two regular polygons share $n-2\ge4$ vertices. Three common vertices determine their circumcircle, and one common vertex on that circle determines the regular polygon of a specified order. Hence the polygons coincide, contradicting deletion of distinct endpoints. This proves (1).

The introduction of the supplied Clemen--Dumitrescu--Liu paper mentions the earlier results for $n=5,6$ and describes the range $n\ge7$ as open. We do not interpret that informal wording as a theorem about each fixed order: (1) is a checked deduction from the complete classifications, not a claim attributed to that introduction or a resolution for all $n$.

Twelve-point uniqueness and the existing eleven-point motif

Let

$$U=\{(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3), (2,0),(2,1),(2,2),(3,0),(3,1)\}.$$

Its squared-distance spectrum is

$$1:24,\qquad3:15,\qquad4:12,\qquad7:12,\qquad9:3. \tag{3}$$

An exact certificate is the following grouped degree table in that order.

The column sums are $(48,30,24,24,6)$, yielding (3). Shinohara (2008), Theorem 1.2(b) states uniqueness up to similarity of the twelve-point five-distance set. Thus every such set has exactly three rare distances.

Deleting $(1,1)$ gives the input's eleven-point spectrum $(18,12,10,12,3)$. The existing motif note deletes $(2,1)$ instead; the isometry $(x,y)\mapsto(4-x-y,x)$ preserves $U$ and carries the first deletion to the second. Thus these are congruent examples, not different candidate families.

Every eleven-point deletion of $U$ retains exactly ten pairs at square four, two or three at square nine, and at least eighteen at square one. All such deletions therefore have at least two rare distances. The motif note now gives a direct five-candidate certificate for its one-point extension obstruction, removing its former dependence on the exhaustive grid enumeration.

Twelve points and the remaining order eleven

The conditions below remain only necessary conditions at order eleven. Orders twelve through fourteen are excluded completely.

For $n=11$, (2) and $g(4)=9$ force exactly five distances. The four nondiameter classes have multiplicity at least twelve, and the diameter has at most seven pairs. No one vertex covers the diameter graph: deleting it would leave ten points with at most four distances. The twelve-point uniqueness theorem does not classify arbitrary eleven-point five-distance sets.

For $n=12$, $g(4)=9$ and (2) leave only five or six distances. The five-distance case is excluded by (3) and Shinohara's uniqueness theorem. In the six-distance case the forced multiplicities are $(13,13,13,13,13,1)$, with unique diameter. The new unique-diameter theorem excludes that case. Thus every twelve-point planar set has at least two rare distances. This uses no classification of eleven-point five-distance sets. The source uniqueness statement and the integer pair spectrum (3) were rechecked for this deduction.

Thirteen and fourteen points via the six-distance classification

Szöllősi--Östergård (2020), Theorem 16 proves $g(6)=13$ and classifies the three maximum types: a regular 13-gon, a regular 12-gon with its center, and a regular hexagram with its center. This is a published computer-assisted classification, not a new exhaustion performed here. The author checked the full primary statement and its continuation on pp. 15--16. It concerns arbitrary planar sets, not just spherical sets. The theorem's normalized isometry statement is used up to similarity.

For $n=13$, (2) and $g(5)=12$ force exactly six distances. The regular 13-gon has six classes of multiplicity 13. For the unit-circumradius 12-gon with its center, the squared-distance spectrum is

$$2-\sqrt3:12,\quad1:24,\quad2:12,\quad3:12,\quad 2+\sqrt3:12,\quad4:6.$$

For the hexagram use the axial norm $Q(x,y)=x^2+xy+y^2$ and the set

$$\begin{split} H={}&\{(0,0),(1,0),(0,1),(-1,1),(-1,0),(0,-1),(1,-1),\\ & (1,1),(-1,2),(-2,1),(-1,-1),(1,-2),(2,-1)\}. \end{split}$$

Its exact spectrum is

$$1:24,\quad3:18,\quad4:15,\quad7:12,\quad9:6,\quad12:3.$$

The degree vectors in that order are $(6,6,0,0,0,0)$ at the center, $(5,2,3,2,0,0)$ at each of the six unit-radius points, and $(2,3,2,2,2,1)$ at each of the six other points. Summing and halving gives the displayed unordered multiplicities. Thus the three types have respectively 6, 5, and 3 rare distances, contradicting $R=1$.

For $n=14$, (2) gives $q\le7$, while the known maxima through six distances force $q\ge7$. The even counting boundary therefore gives a unique diameter, ruled out by the unique-diameter theorem. This proves $R(A)\ge2$ for every fourteen-point planar set.

The finite certificate recomputes all 78 hexagram pairs exactly. Classification completeness is an external theorem; the finite pair check alone would not establish exhaustion. The archived rotational thirteen-point report is not needed to exclude order thirteen and is unreviewed.

Verification scope

The author checked the pair-budget and endpoint-deletion arguments, the model formulas, and the source theorem statements, read the relevant full-source passages after the digests, and verified all finite spectra and grouped degree tables with exact integer or $\mathbb Q(\sqrt3)$ arithmetic; the arguments are author-recorded. Those computations check explicit models; the exhaustion of types relies on the stated published classifications. No new grid enumeration, eleven-point classification, or unrestricted proof is claimed.

Comparison with the second independent input

Sections 4 and 7 of the second independent input agree with the checked classifications, pair budgets, and spectra above. Its displayed $T_2$ is carried to this note's $T_2$ by the axial-coordinate isometry $(a,b)\mapsto(b,a)$; it is not an additional nine-point type. The lattice counts and the twelve-point degree and diameter-pair claims were rechecked exactly. Its regular odd-polygon extension lemma is valid, but the two endpoint deletions above already avoid that extra lemma.

The input leaves the twelve-point six-distance spectrum $(1,13,13,13,13,13)$ unresolved. That is a limitation of the independent argument, not of the current research state: the unique-diameter theorem excludes it without classifying eleven-point five-distance sets. The warning that arbitrary eleven-point five-distance sets need not extend to $U$ remains essential; no such extension theorem is established.

Comparison with the third independent input

Sections 2 and 7.1 of the third independent report repeat the valid finite proof and exact spectra. Its lattice model $P$ is taken to $T_1$ by $(x,y)\mapsto(y,x)$, and its $Q$ is taken to $T_2$ by $(x,y)\mapsto(y-1,x-1)$. These are isometries of the axial norm $x^2+xy+y^2$, not additional classified types. The signed-radius exceptional model is identical to the one above. Its eight-point example, scaled by $1/(1+\sqrt3)$, is exactly the two-square model in circular configurations. The displayed multiplicities were rechecked by bounded exact pair counts.

Its twelve-point spectrum and eleven-point deletion spectrum agree with (3) and the deletion of $(1,1)$ from $U$. The report's reference to deleting $(0,-2)$ cannot itself be checked without its absent companion coordinate file; the explicit $U$-model supplies the needed certificate instead. Its endpoint-deletion approach stopping at ten does not reopen the twelve-point case settled above. No classification of arbitrary eleven-point five-distance sets is inferred.