Root signed-span coverage and its reset

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Standing

This is a proved refinement of the tensor coloring criterion. It does not yet iterate to unbounded coloring. The exact reset example below explains the missing induction hypothesis.

A necessary separation parameter

Let $B$ have $N$ nonzero elements and admit a proper $q$-coloring. For $C\subseteq B$ put $R(C)=B\setminus\Sigma(C)$, and define

$$R_q(B)=\max_f\sum_{c=1}^q|R(f^{-1}(c))|, \qquad \alpha_q(B)=\frac{R_q(B)}{qN}. \tag{1}$$

The maximum ranges over all proper $q$-colorings. Every root belongs to the signed span of its own color, so $0\leq\alpha_q\leq(q-1)/q$. Any family witnessing $\tau_q(B)$ must omit each root from each color's signed span in at least one layer. Counting these $qN$ requirements gives

$$\tau_q(B)\geq\lceil1/\alpha_q(B)\rceil \quad(\alpha_q(B)>0). \tag{2}$$

If $\alpha_q(B)=0$, every root is universal and $\tau_q(B)=\infty$. Positive $\alpha_q$ does not imply that $\tau_q$ is finite.

Nonjump recurrence — proved

Suppose $Y=B\otimes S_m$ admits a proper $q$-coloring, where $m\geq3$. Then

$$m\alpha_q(B)\geq1,\qquad \alpha_q(Y)\leq \min\left\{\alpha_q(B),\ 2\alpha_q(B)-\frac2m\right\}. \tag{3}$$

To prove this, fix a proper coloring of $Y$, and denote its color $c$ by $D_c$, with base layers $C_{j,c}$. For a root $b\in B$, exact membership is

$$b\otimes s_j\in\Sigma(D_c) \iff \exists u\in\bigcap_{i\ne j}\Sigma(C_{i,c}) \text{ with }u+b\in\Sigma(C_{j,c}). \tag{4}$$

Indeed, a representation of the left side has common layer sum $u$ off layer $j$, and sum $u+b$ at that layer, by the unique rational relation among the simplex directions.

Let $k_c(b)$ be the number of layers whose color-$c$ span omits $b$. Properness and the layer criterion force $k_c(b)\geq1$. A root over $b$ can be absent from $\Sigma(D_c)$ only at one of these omitted layers, since otherwise use $u=0$ in (4). If $k_c(b)=1$, it is not absent even there: use $u=-b$ in (4). Thus

$$\sum_c|Y\setminus\Sigma(D_c)| \leq\sum_{b,c}k_c(b)1_{k_c(b)\geq2}.$$

Put $K=\sum_{b,c}k_c(b)$ and $M=qN$. Each layer is a proper base coloring, so $M\leq K\leq mR_q(B)$. Since every $k_c(b)\geq1$,

$$\sum_{b,c}k_c(b)1_{k_c(b)\geq2} \leq\min\{K,2(K-M)\}.$$

Divide by $qmN$ and maximize over the coloring of $Y$ to get (3).

In particular, if $0<\alpha=\alpha_q(B)\leq1/2$ and $m=\lfloor1/\alpha\rfloor+1$, a nonjump yields

$$\alpha_q(Y)\leq\frac{2\alpha^2}{1+\alpha}. \tag{5}$$

If $1/2<\alpha<2/3$, repeated nonjump factors $S_3$ first lower it below $1/2$, since the gap below $2/3$ at least doubles at each step. These are genuine amplification statements. No assertion here bounds their total analytic loss through arbitrarily many chromatic levels.

Quantitative cheap amplification below a small threshold

Suppose a tensor base has $\chi(B)=q$, $\alpha_q(B)\leq a\leq1/4$, and rounding parameter $Q_0$. As long as the color number does not jump, set $a_0=a$, $m_n=\lfloor1/a_n\rfloor+1$, and $a_{n+1}=2a_n-2/m_n$. If $a_n=0$, every further simplex forces a jump, and arbitrarily large arity makes its analytic cost arbitrarily small. Otherwise (3) gives $\alpha_q\leq a_n$ at each stage, and $a_{n+1}\leq2a_n^2\leq a_n/2$.

The logarithmic growth of $Q$ is at most

$$\sum_n\frac{\log m_n}{m_n} \leq\sum_n a_n\log(2/a_n) \leq2a\log(4/a)\leq\log4.$$

Thus $Q\leq4Q_0$ and the total nonjump extension loss is at most $2Q_0\sum_na_n^2\leq(8/3)Q_0a^2$. After a sufficiently long prefix, $p=\lfloor1/(2a_n)\rfloor\geq q+1$ is below $\tau_q$, so it forces a jump with additional loss tending to zero as the prefix length grows.

This proves a quantitatively cheap jump given a small input coverage bound. It supplies no restart bound after the jump; (6) below disproves a general restart bound even below $2/3$.

An exact maximal reset after a jump

Let $B=S_3\otimes S_4$. Then

$$\chi(B)=2,\quad\tau_2(B)=\infty, \qquad \chi(B\otimes S_m)=3, \quad\alpha_3(B\otimes S_m)=2/3\quad(m\geq12). \tag{6}$$

Identify $B$ with the edges of $K_{3,4}$. The elementary cographic criterion proved for $K_{3,5}$ in the layer evidence applies with the same proof: a set is dissociated exactly when its complement graph is connected; a root outside it is in its signed span exactly when that edge is a bridge of the complement.

The following two edge-disjoint spanning paths partition $K_{3,4}$:

$$R_1-L_1-R_2-L_2-R_3-L_3-R_4, \qquad R_3-L_1-R_4-L_2-R_1-L_3-R_2.$$

Their complements are spanning trees, giving a proper two-coloring. Every dissociated set has size at most six. Thus in every proper two-coloring both classes have size six, both complements are trees, and both classes signed-span every root. This proves the first two assertions in (6); the chromatic jump then follows from the layer criterion and the three-palette upper bound.

Here is a fully rational certificate for the last assertion. Color the four columns of the $3$-by-$4$ edge array by the permutations

$$(123),\quad(231),\quad(312),\quad(123). \tag{7}$$

Call the classes $C_0,C_1,C_2$. Each class has four edges. Its complement is connected and bridgeless: suppressing the four degree-two right vertices gives a triangle on the left vertices with edge multiplicities $(2,1,1)$. Consequently each class is independent and $\Sigma(C_i)\cap B=C_i$.

Order the six tensor coordinates as $(11,12,13,21,22,23)$. Directly from the classes in (7), their rational spans have common intersection

$$U=\{(0,a,b,0,a-b,a):a,b\in\mathbb Q\}. \tag{8}$$

For verification, defining equations for their spans respectively are

$$\begin{array}{ll} C_0:&x_{21}=0,\quad -x_{12}+x_{13}+x_{22}=0,\\ C_1:&x_{11}=0,\quad -x_{13}-x_{22}+x_{23}=0,\\ C_2:&x_{21}-x_{11}=0,\quad x_{23}-x_{12}=0. \end{array}$$

Swap right columns 3 and 4, with induced ambient map $T$. On each row triple it sends $(x_1,x_2,x_3)$ to $(x_1-x_3,x_2-x_3,-x_3)$, so

$$T(U)=\{(-b,a-b,-b,-a,-b,-a):a,b\in\mathbb Q\}, \qquad U\cap T(U)=\{0\}. \tag{9}$$

For the last equality, coordinates $11$ and $21$ force both parameters of the vector in $T(U)$ to vanish.

Use six layer types: the three cyclic color relabelings of (7), and the three relabelings of its $T$-image. For $m\geq12$, use every type at least twice, filling any extra layers by repetitions. For each color, the intersection of its layer rational spans remains zero even after deletion of any one layer, by (8)–(9). The layers give a proper three-coloring. In (4), the only possible $u$ is zero; hence a tensor root belongs to a color's signed span exactly when it actually has that color. It is omitted by exactly the other two colors, proving $\alpha_3=2/3$, its absolute maximum.

The same witness gives the maximal value for any nonnegative weights on the tensor roots. Thus simply weighting root omissions cannot prevent this reset. This does not exclude other witness sets or a stronger invariant.

The same seed also loses anchors at boundary arity three

For $Y=S_3\otimes S_4\otimes S_3$, an additional exact certificate proves $\chi(Y)=3$ and

$$\tau_3(L(Y))\leq36 \tag{10}$$

for every linear scalar encoding $L$ preserving all ternary relations exactly. In particular no universal three-color anchor survives.

Run uv run --no-sync python wiki/research/erdos_774/evidence/tensor_boundary/main.py from the repository root. The standalone standard-library checker has fixed finite input: nine four-point base layers with exactly $3^4=81$ ternary sums each, and twelve specified automorphisms. It verifies the colorwise intersection sizes $(81,7,1),(81,7,1),(81,17,1)$, so the displayed tensor coloring is proper. Each color has twelve points.

Its third class lies in a rational hyperplane. The twelve automorphic copies have hyperplane normals of rank twelve, and each copied coloring has a class disjoint from a class of the original coloring. All classes therefore lie in one connected disjoint-class graph. A scalar common to all their signed spans has a common vector lift, by the exactness argument in the Grow boundary certificate. The full-rank normals force this vector to be zero. Taking all three cyclic color relabelings of each copied coloring makes these twelve partitions into a 36-layer witness for (10). The certificate page records what the checker verifies and how the certificate was found.