Exact boundary-anchor exclusion for the Grow source

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Theorem and scope

Let $A=\{4,5,9,11,13,27,30,33,81,85,89,243,248,253\}$ and $G=A\otimes S_3$. Then $\chi(G)=3$ and $U_3(G)=\{0\}$. Every linear scalar encoding preserving all ternary relations exactly also has $U_3=\{0\}$. Here $U_q$ is the universal signed-span set in the tensor coloring page.

Together with the large-arity obstruction, this means that no arity $m\geq3$ supplies a reusable three-color anchor from this particular Grow deletion, even after exact scalar realization. This retires the repeated gluing of this source, not every anchored construction. The source is the deletion of the point $3$ from the Grow–Whicher example.

Exact finite verification

Run, from the repository root,

uv run --no-sync python wiki/research/erdos_774/evidence/boundary_anchor/main.py

The checker reads the fixed certificate. It uses the standard library and the shared root tools checker, performs no search, and finishes in a few seconds. Its named checks verify:

• all $2^{14}=16384$ source subsets and exactly 56 proper two-color classes;
• the complete $U_2(A)$, of size 501 including zero;
• 12,730 canonical nonzero candidate orbits;
• 83 coloring templates, all six ray permutations of each, and their 498 proper palette bridges;
• exclusion of every candidate, with an independent ordinary-integer-set recomputation of all colorings and one exclusion per candidate.

A failed clause exits nonzero. The author ran the checker on the retained certificate; that run is author-recorded, not an independent review.

Discovery of the certificate

The certificate was found by finite constructive sampling, which is not part of the proof or of the normal run. With all $2^{14}$ source subsets tabulated and the exact layer-intersection test, the searches were: 100,000 unstructured source assignments and 300,000 triples of retained proper partitions (seed 42; 28,410 retained partitions; no proper tensor coloring found); all 84 single-point recolorings of one palette (no valid move); all 128 splits of one seven-point class and then all 7,168 splits across the 56 source classes (the inherited anchor survived); all $56\cdot128\cdot56=401408$ mixed-middle splits (a coloring omitting $3t_1$ was found); an initial elimination over 30,000 split-family and 30,000 refinement-family samples (seed 813743), reducing the 12,730 candidate orbits to 87 and then 54; an expanded elimination of at most 200,000 outer-row proposals (seed 666013), each admitted pair tested against all 56 middle partitions, succeeding at proposal 42,927 after 4,087 admissions and 228,872 proper-coloring checks; and a reverse compression scan of at most 1,040 reducing witnesses, retaining the 83 templates.

The families were, for proper pairs $P\sqcup Q=R\sqcup T=V\sqcup W=A$: splits $P=a\sqcup b$ with rows $(a,b,Q),(R,T,0),(b,a,Q)$; refinements $(P\cap R,Q\cap R,T),(V,W,0),(P\cap T,Q\cap T,R)$; and expanded proposals choosing proper outer pairs $P,Q$ and $P',Q'$, splitting their first classes into $a,b$ and $a',b'$ oppositely on their intersection, admitting only when $a\cup a'$ and $b\cup b'$ are dissociated, and testing $(a,b,Q),(R,T,0),(a',b',Q')$ for every middle pair. A separate omission check tested the three scalar possibilities $-3,30,-36$. The fixed certificate and the checker replace all of these searches for verification.

Candidate reduction

Use $t_1=(1,0),t_2=(0,1),t_3=(-1,-1)$. A color with layers $C_1,C_2,C_3$ is dissociated exactly when the layers are dissociated and their signed spans intersect only at zero. Also

Palette colorings with any given source two-color class on any pair of rays force a universal vector to satisfy $x,y,x-y\in U_2(A)$. Ray permutations and overall sign reversal give a representative with $x>0$ and $0\leq2y\leq x$, unless the vector is zero. This gives the 12,730 candidates. For each, (1) fails on a class of a symmetry-expanded witness. The source anchor $3$ and the palette upper bound prove $\chi(G)=3$.

Why arbitrary exact scalar encodings are covered

Join color classes by an edge when their supports are disjoint. If an exact linear encoding $L$ had a universal nonzero scalar anchor $a$, choose $v_C\in\Sigma(C)$ with $L(v_C)=a$ in each class. For disjoint $C,D$, their representation difference is a ternary sum, so exactness forces $v_C=v_D$. All representatives therefore coincide on each connected component of this class graph.

All palettes with rows

where the displayed pairs are proper two-partitions of $A$, lie in one component: the three classes of each coloring are disjoint, and changing a single layer leaves its absent color class unchanged. This family includes all two-ray classes used to bound candidates.

Every certificate template shares its third class with a palette (2). Each ray-permuted template shares a class with a permuted palette; global color relabeling returns that palette to (2). All these bridges and relabelings are explicitly checked. Thus all bounding classes and all symmetry-expanded witnesses share one vector representative of $a$. The candidate exclusion forces that vector to be zero, a contradiction. Neither overlapping-support subtraction nor symmetry of $L$ is assumed.