No-three selections in bounded-width digit supports

1. Problem, model and main results

The no-three-in-line problem asks for large subsets of an integer square with no three distinct points on a Euclidean line. The present community target is an unbounded family of side lengths $n$ with more than $1.8n$ points [1]. We prove a restriction on digit constructions: a bounded-width layered digit support containing at most $O(n^{2-\delta})$ points for any fixed $\delta>0$ cannot contain a positive-linear-size no-three subset, uniformly over all transition schedules. The new main theorem permits nondeterministic programs and arbitrary accepting-path multiplicity. The selection from the support may be completely arbitrary.

Fix an integer $b\geq2$. A width-$q$ layered deterministic program has $k+1$ layers, each with at most $q\geq1$ states, one initial state, and a set of final states. A digit and current state determine at most one next state. Its transitions may depend on the position and on the total length $k$. It reads the alphabet

$$ \mathcal A=\{0,\ldots,b-1\}^2,\qquad A=|\mathcal A|=b^2. $$

A word $w=d_1\cdots d_k$ encodes the vector

$$ [w]=\sum_{i=1}^k b^{k-i}d_i\in\{0,\ldots,b^k-1\}^2. $$

Leading zero pairs are allowed. Encoding at each fixed length is injective. Let $R_k$ be the set of vectors of accepted length-$k$ words, and let $\alpha(X)$ be the largest cardinality of a subset of a finite set $X$ containing no three distinct collinear points. All collinearity here is over the integers, not a finite field. Translation identifies our square with the usual $\{1,\ldots,b^k\}^2$.

A fixed deterministic finite automaton (DFA), unrolled across the layers, is a special case. We call the resulting family a fixed regular support.

A layered nondeterministic program has the same states and layers, but a digit and current state may have several possible next states. A word is accepted when at least one path starts at the initial state and ends in a final state. Its support $R_k$ counts distinct accepted words, not accepting paths. Theorems 1, 2 and 4 allow this more general model. The results retained in Theorems 5–7 concern deterministic programs as specified there.

Theorem 1 (strictly subquadratic supports). Fix $b,q,C_0$ and $\delta>0$. Uniformly over width-$q$ layered nondeterministic programs with

$$|R_k|\leq C_0b^{(2-\delta)k},$$

there are $C_1,\varepsilon>0$ such that

$$\alpha(R_k)\leq C_1b^{(1-\varepsilon)k}.$$

All constants depend only on the displayed fixed parameters, not on the transition schedule or the selection. In particular, supports with arbitrary fixed growth exponent between one and two are excluded from positive-linear-size no-three constructions.

Theorem 2 (finite entropy and width inequality). Suppose a width-$q$ layered nondeterministic program has a nonempty no-three selection of size $M$. For any $1\leq L\leq k$ put

$$ \Delta=k\log b-\log M,\quad Q=b^L,\quad N=k-L+1, \quad\kappa=\frac4{3\pi^2}, $$

$$E=1+8\pi^2+\frac{4Q^2L}{N}\bigl(\Delta+\log(2Q)\bigr).$$

Then

$$\log q\geq2L\log b-\frac L N\log|R_k|+4\log\kappa-\log E.\tag{E1}$$

Theorem 3 (bounded missing-pattern positions). Fix $b$ and a word length $\ell\geq1$. For a no-three set $S\subseteq\{0,\ldots,b^k-1\}^2$, let $B_\ell(S)$ count complete length-$\ell$ positions at which some paired-digit word is absent from all points of $S$. Put $B_\ell(S)=0$ when $k<\ell$. There are explicit $C,\gamma>0$, depending only on $b,\ell$, such that

$$|S|\leq Cb^k\exp\bigl(-\gamma B_\ell(S)\bigr).$$

Consequently a selection of size at least $c_0b^k$ has only $O_{b,\ell,c_0}(1)$ missing-pattern positions. A positive fraction of missing positions gives a power saving. Every proper scalar digit subshift is excluded, and a width-one product support admitting dense selection must occupy a positive fraction of the full grid.

Theorem 4 (width for intermediate support growth). Fix $b,C_0,c>0$ and $0<\delta\leq1$. If a width-$q_k$ layered nondeterministic program satisfies $|R_k|\leq C_0b^{(2-\delta)k}$ and $\alpha(R_k)\geq cb^k$, then

$$q_k=\Omega_{b,C_0,c,\delta}\left(\left(\frac{k}{(\log k)^2}\right)^{\delta/2}\right).$$

For deterministic linear-size supports, Theorem 6 below retains the stronger $\sqrt{k/\log k}$ bound.

The proof starts with bounded fibers of every fixed integer projection modulo $b^k$. Projection entropy then forces high paired-digit block entropy, quantified by a positive Fejér kernel and Parseval. A recombined hidden Markov law on the same accepted support converts local block entropy into a lower bound for the support cardinality. This permits arbitrary correlations in the selected no-three set and counts words rather than paths. Charging the projection entropy only at missing-pattern positions gives the stronger finite count bound in Theorem 3; a sliding-window program gives an alternative proof of its positive-fraction consequence.

The following complementary relative-capacity and stationary results use geometric line covers and exact collinear substitutions. They also retain the stronger width bound at linear support growth.

Theorem 5 (arbitrary selections). Fix $q$ and $b$. Uniformly over all width-$q$ layered deterministic programs, the following hold.

(i) For every $1<\lambda\leq A$, there are explicit constants $C,\gamma>0$ such that, whenever $|R_k|\geq\lambda^k$,

$$ \alpha(R_k)\leq C e^{-\gamma k}|R_k|. \tag{1} $$

(ii) If $|R_k|=O(b^k)$, then for some $\varepsilon>0$,

$$ \alpha(R_k)=O\bigl(b^{(1-\varepsilon)k}\bigr) =O(n^{1-\varepsilon}),\qquad n=b^k. \tag{2} $$

For a fixed regular support we also have the following all-growth completion.

(iii) There are constants $B_0,C_0$ such that, for all $k\geq1$,

$$ \alpha(R_k)\leq B_0+C_0\frac{|R_k|}{k}. \tag{3} $$

In particular, $\alpha(R_k)/|R_k|\to0$ along every sequence for which $|R_k|\to\infty$.

The proof of (1) partitions accepted words into Cartesian products of equal-length path alphabets. For each direction, partition every block alphabet into its parallel lines. Every product cell lies on one physical line. A geometric inequality bounds the sum of logarithmic line-cover losses over directions below by half the logarithm of the alphabet size, forcing exponential loss in one common direction. The proof of (3) supplements it with a residue-by-residue decomposition of paths through simple cycles.

Two bounds quantify the state complexity needed to escape this obstruction. The program or automaton may be different at every depth in both statements.

Theorem 6 (growing width). Fix $b\geq2$ and $M\geq c>0$. Suppose a width-$q_k$ layered deterministic program has a length-$k$ support satisfying $|R_k|\leq M b^k$ and $\alpha(R_k)\geq c b^k$. Then, along any unbounded sequence of such depths,

$$ q_k=\Omega_{b,c,M}\left(\sqrt{\frac{k}{\log k}}\right). \tag{4} $$

Theorem 7 (stationary direct output at one depth). For a $q$-state DFA put

$$ T_q=4q^2+7q,\qquad C_q=\sum_{h=0}^{q-1}b^{2h}. $$

If $k>T_q$ and $R_k$ itself is no-three, then $|R_k|\leq C_q$. Consequently, for no-three outputs with $|R_k|\geq c b^k$,

$$ \liminf\frac{q_k}{\sqrt{k}}\geq\frac12. \tag{5} $$

Only the target slice is assumed no-three in Theorem 7. Theorem 6 permits an unrestricted selection from a larger support and therefore addresses a different, more permissive model.

2. Related work and distinction

Polynomial growth and bounded numbers of words per length are classical topics in regular-language theory. Szilard, Yu, Zhang and Shallit studied regular languages with polynomial densities [2]; Păun and Salomaa studied thin and slender languages [3]. We use this established structural perspective, but prove the graph and counting facts needed below directly. We do not claim the polynomial-growth classification itself as new. The contribution is its interaction with scalar positional encoding: parallel-line product cells, directional loss and exchanges between different cyclic components produce actual integer lines, controlling arbitrary subsets rather than just accepted-language counts.

Previous work in this community proves sublinear no-three selection bounds for randomly relabeled bounded-degree grid supports [4]. Those are high-probability statements about sampling ensembles. Our bounds hold uniformly for every synchronous bounded-width layered support of strictly subquadratic size and give a power saving, without a distributional hypothesis. Earlier algebraic selection barriers [5] concern periodic rational graphs and bounded-complexity arithmetic or geometric selection rules. Here no complexity restriction is placed on the selection; the restriction is instead on the layered digit program defining its support. The two settings do not subsume one another.

The layered model allows nonuniform position-dependent transition tables, a standard feature of branching-program models; the restriction used here is the prescribed synchronous most-significant-first digit order. The exponential-slice argument uses no density Ramsey theorem, spectral asymptotics or external automaton growth theorem. Its quantitative constants are conservative. The exact-depth bound complements the selection theorem by giving a stronger state requirement when the accepted set itself is the desired output.

Krieger, Rampersad and Shallit [6] give classical regular-language growth methods, including equal-length loop constructions; Byszewski, Konieczny and Müllner [7, Theorem D] study arithmetic-progression counts in one-dimensional automatic sets. Here the chosen subset need not itself be automatic. Payne and Wood [8] study general-position subsets with controlled maximum line occupancy. Glasscock, Moreira and Richter [9, Theorems A–C and Proposition 3.10] study digit-invariant integer sets, symbolic languages and sums/intersections associated with multiplicatively independent bases. Our projection lemma uses a single common scalar base and bounds arbitrary subsets avoiding integer collinear triples. Their results supply useful context but are not invoked to deduce our capacity bounds. We do not claim new general entropy, Fejér-kernel, hidden Markov or regular-language growth principles; the contribution is their quantitative interaction with this integer geometric condition.

3. Modular projections and their digit entropy

Throughout Sections 3–6, $S\subseteq\{0,\ldots,b^k-1\}^2$ is nonempty and no-three, $M=|S|$, and $X$ is uniform on $S/b^k$ in the two-dimensional torus. Put

$$ \Delta=k\log b-\log M,\qquad T(x)=bx\pmod1, \qquad\nu_j=\mathcal L(T^jX). $$

All logarithms are natural. For a nonzero integer vector $d=(a,c)$ define

$$ Y_d=(ab^kX_1+cb^kX_2)\bmod b^k, \qquad F_d=2(|a|+|c|+1). $$

The unmodded integer projection has range width at most $(|a|+|c|)(b^k-1)$. At most $|a|+|c|+1$ intercepts have one given residue, and each intercept line meets $S$ at most twice. Thus every fiber of $Y_d$ has at most $F_d$ points, giving

$$ C_d:=k\log b-H(Y_d)\leq\Delta+\log F_d. \tag{E2} $$

For $1\leq L\leq k$, let $D_{d,j}$ be the entropy deficit of the length-$L$ block at position $j$ in the $k$-digit expansion of $Y_d$, including leading zeros. Partition the digit positions into disjoint full $L$-blocks and short remainders, once at each offset modulo $L$. Entropy subadditivity bounds the sum of the full-block deficits at each offset by $C_d$. Every complete sliding block is counted once, hence

$$ \sum_{j=0}^{k-L}D_{d,j}\leq L C_d. \tag{E3} $$

Write $Q=b^L$ and $N=k-L+1$. Integer torus projections commute with $T$. The first $L$ digits of $d\cdot T^jX\pmod1$ therefore have exactly the block distribution used above. Replacing a circle point by the left endpoint of its $Q$-adic interval changes its first Fourier character by at most $2\pi/Q$. The character averages to zero under the uniform block law. Pinsker's inequality gives

$$ |\widehat\nu_j(d)|\leq\sqrt{2D_{d,j}}+\frac{2\pi}{Q}. \tag{E4} $$

For completeness, Pinsker here follows by grouping outcomes according to where one probability exceeds the other. The log-sum inequality reduces the claim to Bernoulli relative entropy, whose second derivative with respect to its first parameter is $1/(p(1-p))\geq4$; integration gives relative entropy at least twice squared total variation.

Two useful consequences of (E3)–(E4) are

$$ \frac1N\sum_{j<N}|\widehat\nu_j(d)| \leq\sqrt{\frac{2L(\Delta+\log F_d)}N}+\frac{2\pi}Q, \tag{E5} $$

and, using $(u+v)^2\leq2u^2+2v^2$,

$$ \frac1N\sum_{j<N}|\widehat\nu_j(d)|^2 \leq\frac{4L(\Delta+\log F_d)}N+\frac{8\pi^2}{Q^2}. \tag{E6} $$

Absolute values are taken before averaging, so cancellation between digit positions is unavailable. If $M=b^{k-o(k)}$, (E5), first with a fixed block length and then with that length tending to infinity, shows that most individual digit-shift laws approach Haar measure in every fixed finite-frequency weak test. The finite entropy estimate below avoids any appeal to an invariant-measure theorem.

4. Fejér smoothing gives an explicit paired-block entropy bound

Let $W_j$ be the length-$L$ paired-digit block of $X$ at position $j$. It takes values in the $Q^2$ squares of the $Q$-grid. Set

$$ \kappa=\frac4{3\pi^2},\qquad E=1+8\pi^2+\frac{4Q^2L}{N}\bigl(\Delta+\log(2Q)\bigr). $$

We prove

$$ \frac1N\sum_{j<N}H(W_j) \geq2\log Q+4\log\kappa-\log E. \tag{E7} $$

Put $R=\lfloor Q/2\rfloor\geq1$. The normalized Fejér kernel is

$$ F_R(t)=\frac1R\left(\frac{\sin(\pi Rt)}{\sin(\pi t)}\right)^2 =\sum_{|h|<R}\left(1-\frac{|h|}R\right)e^{2\pi iht}, $$

with its continuous value at integers. It is nonnegative and has integral one. For $|t|\leq1/Q$, the inequalities $\sin(\pi u)\geq2u$ on $0\leq u\leq1/2$ and $|\sin(\pi t)|\leq\pi|t|$ imply

$$ F_R(t)\geq\frac{4R}{\pi^2}\geq\kappa Q, $$

since $R\geq Q/3$. Let $\Phi(x,y)=F_R(x)F_R(y)$ and $f_j=\Phi*\nu_j$. For any two points in the same half-open $Q$-grid cell, the coordinate differences have representatives of magnitude at most $1/Q$. Thus, throughout a cell $C$,

$$ f_j(z)\geq\kappa^2Q^2\nu_j(C). $$

Integrating the square over the grid gives

$$ \int f_j^2\geq\kappa^4Q^2\sum_C\nu_j(C)^2. $$

Shannon entropy is at least collision entropy, by Jensen's inequality, so

$$ H(W_j)\geq2\log Q+4\log\kappa-\log\int f_j^2. $$

Parseval for the finite trigonometric polynomial $f_j$ gives

$$ \int f_j^2=1+ \sum_{\substack{d\ne0\\|d_1|,|d_2|<R}} |\widehat\Phi(d)|^2|\widehat\nu_j(d)|^2. $$

There are at most $Q^2$ nonzero summands, their weights are at most one, and $F_d\leq4R-2\leq2Q$ throughout this frequency set. Equation (E6) therefore bounds the average of these energies by $E$. Averaging the entropy inequality and using concavity of the logarithm proves (E7). If $R=1$ there are no nonzero frequencies and the same estimate holds. The kernel lower bound includes cell boundaries, so no assumption about boundary mass is needed.

5. Recombining selected paths without preserving their correlations

Here the layered program may be nondeterministic. Start with any probability law on selected accepted words, and choose one accepting path per word. Denote this path law by $P$. Partition the digit positions into consecutive blocks, with possible short initial and final blocks. Write $Z_i$ for the state before block $i$, $B_i$ for its label word, and $Z_{i+1}$ for its terminal state.

Generate a new law $\widetilde P$ by starting at the prescribed initial state and successively using the kernels

$$ P(B_i,Z_{i+1}\mid Z_i). $$

Induction preserves every marginal $(Z_i,B_i,Z_{i+1})$. States of zero original probability are never reached. Each generated block is a valid path segment, and the terminal state remains accepted. Therefore the generated word law is supported on $R_k$, even if several generated paths spell the same word.

The chain rule and conditional independence under the generated law imply

$$ \begin{aligned} \log|R_k|&\geq H_{\widetilde P}(B_1,\ldots,B_s)\\ &=\sum_iH_{\widetilde P}(B_i\mid B_1,\ldots,B_{i-1})\\ &\geq\sum_iH_{\widetilde P}(B_i\mid B_1,\ldots,B_{i-1},Z_i)\\ &=\sum_iH_{\widetilde P}(B_i\mid Z_i) =\sum_iH_P(B_i\mid Z_i). \end{aligned} $$

The conditional-independence equality is a property of $\widetilde P$, not of the original selection $P$. This distinction permits arbitrary correlations in the selected set. In a deterministic program the state is determined by previous labels and the first entropy inequality becomes equality with the sum of conditional block entropies.

Discard the nonnegative conditional entropies of short blocks. For a full block, $H_P(B_i\mid Z_i)\geq H_P(B_i)-\log q$. Apply this argument at all $L$ offsets. Each complete sliding block appears once, and hence

$$ L\log|R_k|\geq\sum_{j=0}^{k-L}H_P(W_j)-N\log q. \tag{E8} $$

Combine (E8) with (E7), using the uniform selected law, to obtain the explicit inequality (E1) in Theorem 2. Rearranging gives the equivalent bound

$$ \Delta\geq\frac{N}{4Q^2L} \left(\frac{\kappa^4Q^2}{q|R_k|^{L/N}}-1-8\pi^2\right) -\log(2Q). \tag{E9} $$

Proof of Theorem 1. Suppose $|R_k|\leq C_0b^{(2-\delta)k}$. Choose a fixed $L$ such that $\kappa^4b^{\delta L}/q>2(1+8\pi^2)$. As $k\to\infty$, the expression in parentheses in (E9) has a positive lower limit, because $N=k-L+1$ and $C_0^{L/N}\to1$. Thus $\Delta\geq\gamma k-O(1)$ for some $\gamma>0$, proving the claimed power saving. Its finite-depth prefactor absorbs all smaller $k$.

Proof of Theorem 4. Under $M\geq cb^k$, $\Delta\leq\log(1/c)$. Take

$$ L=\left\lfloor\log_b\frac{\sqrt{k}}{\log k}\right\rfloor. $$

For sufficiently large $k$, $Q$ lies within a factor $b$ of $\sqrt{k}/\log k$ and $E=O_{b,c}(1)$. Using $|R_k|\leq C_0b^{(2-\delta)k}$ in (E1) yields

$$ \log q_k\geq\delta L\log b-O(L^2/k)-O(L/k)-O_{b,c}(1). $$

Exponentiation gives the asserted width bound. The useful range is $0<\delta\leq1$; for $\delta>1$ the two size assumptions are eventually incompatible. At $\delta=1$, Theorem 6 gives the stronger bound for deterministic programs.

6. Boundedly many missing-pattern positions

For a nonempty no-three set $S$ and a fixed word length $\ell$, let $B_\ell(S)$ count complete starting positions where some length-$\ell$ paired word is absent from all selected points. We prove the stronger finite form of Theorem 3,

$$ |S|\leq Cb^k e^{-\gamma B_\ell(S)}, \tag{E10} $$

with constants depending only on $b,\ell$. In particular, $|S|\geq c_0b^k$ permits only $O_{b,\ell,c_0}(1)$ missing-pattern positions.

6.1 Explicit test functions

Put $h=1/(4b^\ell)$ and center a triangular tent of half-width $h$ in each coordinate of a length-$\ell$ digit cell. Their product is supported strictly inside the cell and has integral $I=h^2$. The one-dimensional Fourier coefficients have magnitudes $h\operatorname{sinc}^2(\pi ah)$, with $\operatorname{sinc}(u)=\sin(u)/u$ and continuous value at zero. Their absolute sum is one. The product coefficients therefore also have absolute sum one.

Truncate to $|d_1|,|d_2|\leq R$, where

$$ R=\left\lceil\frac{16}{\pi^2h^3}\right\rceil. $$

The uniform error is at most $4/(\pi^2hR)\leq I/4$, by summing the one-dimensional $1/a^2$ tails. Translate this same polynomial to every digit cell. Its nonzero frequency magnitudes $a_d$ do not change under translation, and $A=\sum_{d\ne0}a_d\leq1$. At a position missing one cell, its bump integrates to zero against $\nu_j$, so

$$ \sum_{d\ne0}a_d|\widehat\nu_j(d)|\geq I/2. \tag{E11} $$

Choose an integer $m\geq\ell$ with $b^m\geq8\pi/h^2$. Applying (E4) with block length $m$ gives, at every such position $j\leq k-m$,

$$ \sum_d a_d\sqrt{2D_{d,j,m}}\geq I/4. $$

Weighted Cauchy–Schwarz implies

$$ \sum_d a_dD_{d,j,m}\geq\frac{I^2}{32A}\geq\frac{I^2}{32}. \tag{E12} $$

Here $A>0$: a constant polynomial cannot approximate the nonconstant interior bump within $I/4$.

6.2 Charge entropy only at exceptional positions

Sum (E12) over missing-pattern positions through $k-m$. All terms are nonnegative, so extending each sum to all complete $m$-positions and using (E3) gives

$$ \bigl(B_\ell(S)-(m-\ell)\bigr)\frac{I^2}{32} \leq m\bigl(\Delta+\log(4R+2)\bigr). $$

Indeed every used frequency has $F_d\leq4R+2$, and $A\leq1$. Only the final $m-\ell$ possible positions were discarded. Thus (E10) holds with the explicit choices

$$ \gamma=\frac{h^4}{32m},\qquad C=(4R+2)\exp\bigl(\gamma(m-\ell)\bigr). \tag{E13} $$

If $k<m$, then $B_\ell(S)\leq m-\ell$ (or is zero when $k<\ell$), and $|S|\leq2b^k$ proves the same bound. A positive fraction $\eta k$ of missing positions therefore yields a power saving with exponent $\gamma\eta/\log b$, capped below one if desired. No independence or regularity of the selected digits is used.

If a product support $R_k=D_1\times\cdots\times D_k$ contains a selection of size at least $c_0b^k$, every proper paired-digit alphabet $D_i$ gives one missing length-one position. Therefore only $O_{b,c_0}(1)$ alphabets can be proper. Since the product is nonempty,

$$ |R_k|\geq c_1b^{2k} $$

for a constant $c_1>0$. For width-one products, dense selection thus forces positive planar density, stronger than full exponential growth alone.

6.3 Symbolic support interpretation

For comparison, the positive-fraction conclusion also follows directly from Theorem 1. Choose one forbidden length-$\ell$ word at each of at least $\eta k$ positions. A deterministic layered program can check these exclusions by remembering the previous $\ell-1$ paired digits; its width is at most $A^{\ell-1}$, with $A=b^2$. One residue class of starts modulo $\ell$ contains at least $\eta k/\ell$ disjoint excluded windows. Consequently the full containing support satisfies

$$ |R_k|\leq A^k(1-A^{-\ell})^{\eta k/\ell} =b^{(2-\delta)k},\qquad \delta=-\frac\eta{\ell\log b}\log(1-b^{-2\ell})>0. $$

Theorem 1 gives a power saving, although the direct exceptional-position argument proves the sharper finite bound (E10).

Every proper scalar digit subshift omits a finite word, so its no-three capacity has a power saving without a separate support-size hypothesis. A fixed regular support of growth rate strictly below $b^2$ is also covered directly by Theorem 1. Symbolically, trim its automaton to reachable, coaccessible states and fix short initial and final contexts. Distinct labels of length-$t$ trim paths inject into accepted words of length $t+O(1)$. Counts $O(\operatorname{poly}(t)\rho^t)$ for $\rho<b^2$ imply that some finite word is absent from all trim paths.

The geometric input throughout is the bounded modular projection fiber forced by integer no-three sets. Entropy inequalities, Fourier approximation and hidden Markov conditioning are standard tools whose proofs have been included for clarity.

7. Parallel-line products and directional loss

For words $U,V$ one has

$$ [UV]=b^{|V|}[U]+[V]. \tag{6} $$

Lemma 8 (product line cover). At prescribed nonoverlapping block positions of common length $\ell$, allow nonempty word alphabets $W_1,\ldots,W_m$ and fix all other digits. For a direction $d$, let $C_d(W_i)$ be the number of lines parallel to $d$ occupied by the vectors of $W_i$. If $X$ is the resulting Cartesian product point set, then

$$ \alpha(X)\leq2\prod_{i=1}^m C_d(W_i). \tag{7} $$

Proof. Partition each alphabet into its parallel $d$-lines. In one product cell, every block vector is a fixed vector plus a scalar multiple of $d$. Equation (6) shows that every evaluated point in the cell is also a fixed vector plus a scalar multiple of $d$, because both coordinates use the same scalar radix. Thus the whole cell is one physical line. Its points are distinct by fixed-length digit encoding. A no-three subset takes at most two points per cell, and the number of cells is the product in (7). $\square$

A useful special case chooses the same two blocks at many positions: the entire binary product is one line, however far apart the positions are. More generally, for one alphabet $W$ of size $w\geq2$ repeated $t$ times, partition it into a chosen pair and individual outside letters. The exact capacity sum over these binary cells gives

$$ \alpha(W^t)\leq2(w-1)^t-(w-2)^t. $$

The next lemma allows us to use the full alphabets in (7).

Lemma 9 (logarithmic directional loss). Let $W$ consist of $w\geq2$ distinct planar points. Sum over the finitely many directions determined by its point pairs. Then

$$ F(W):=\sum_d\log\frac{w}{C_d(W)}\geq\frac12\log w. \tag{8} $$

Proof. Let $a$ be the maximum line occupancy. In each direction let $l$ range over the occupancies of its parallel lines. Since $l\leq a$ and every point pair belongs to exactly one direction,

$$ \sum_d(w-C_d)=\sum_{d,l}(l-1) \geq\frac2a\sum_{d,l}\binom l2=\frac{w(w-1)}a. $$

Using $\log(w/C_d)\geq(w-C_d)/w$ gives $F\geq(w-1)/a$.

Choose a line with $a$ points, put $r=w-a$, and denote its direction by $d_0$. Then $C_{d_0}\leq r+1$. Every line in another direction contains at most one of these $a$ points. If it contains that point and $t$ outside points, its defect $l-1=t$ counts exactly its cross pairs. Summing such defects over directions other than $d_0$ counts all $ar$ cross pairs, so

$$ F\geq\log\frac{w}{r+1}+\frac{ar}{w}. \tag{9} $$

If $a\leq w/2$, this gives $F\geq a/2$, alongside $F\geq(w-1)/a$. Hence $F\geq\sqrt{(w-1)/2}\geq\frac12\log w$. The last inequality follows from $(\log w)^2\leq2(w-1)$: the difference vanishes at $w=1$ and has positive derivative for $w\geq1$.

If $a>w/2$, equation (9), $r\geq\log(r+1)$ and $r+1\leq w$ give

$$ F\geq\log\frac{w}{r+1}+\frac r2 \geq\log w-\frac12\log(r+1)\geq\frac12\log w. $$

This includes the collinear case $r=0$. $\square$

For block vectors in $\{0,\ldots,B-1\}^2$, all pair directions belong to a fixed set of size less than $2B^2$: take primitive integer representatives with first coordinate $1,\ldots,B-1$ and second coordinate $-(B-1),\ldots,B-1$, and add the vertical direction $(0,1)$. Counting even the nonprimitive candidates gives $(B-1)(2B-1)+1<2B^2$. Extra directions have $C_d(W)=|W|$ and zero loss.

Let $B=b^\ell$, so $B^2=A^\ell$, and set $H=\sum_i\log|W_i|$. Lemma 9, with zero loss for singleton alphabets, shows that some common direction has total logarithmic loss at least $H/(4A^\ell)$. Lemma 8 therefore gives

$$ \frac{\alpha(X)}{|X|}\leq2\exp\left(-\frac{H}{4A^\ell}\right). $$

8. The block partition and exponential decay

Fix positive integers $k,\ell$, write $k=m\ell+r$ with $0\leq r<\ell$, and assume $m\geq1$. Partition accepted words by their states at every full block boundary and by their final $r$-letter tail. At block position $i$, let $W_{i,u,v}$ be the alphabet of length-$\ell$ words taking boundary state $u$ to boundary state $v$.

Every nonempty class is a Cartesian product of these alphabets followed by its fixed tail. All substitutions preserve the boundary states and acceptance. Determinism makes the classes disjoint, even when transition tables vary with position and depth. There are at most $q^{m+1}A^\ell$ classes; this harmlessly overcounts the initial state and tail length.

For any $h>0$, classes with logarithmic size $H\leq h$ contain at most $q^{m+1}A^\ell e^h$ words altogether. In every other class the preceding product estimate bounds the selected fraction by $2e^{-h/(4A^\ell)}$. Summing proves the finite uniform inequality

$$ \alpha(R_k)\leq q^{m+1}A^\ell e^h +2\exp\left(-\frac{h}{4A^\ell}\right)|R_k|. \tag{10} $$

The constants and inequality do not presume a stationary transition table or fixed width.

To prove Theorem 5(i), choose a positive integer $\ell$ with $q\leq\lambda^{\ell/2}$, and take $h=m\ell\log\lambda/4$. The first term of (10) is at most $qA^\ell\lambda^{3k/4}$. If $|R_k|\geq\lambda^k$ and $k\geq2\ell$, then $m\ell\geq k/2$, giving

$$ \frac{\alpha(R_k)}{|R_k|} \leq qA^\ell e^{-k\log\lambda/4} +2\exp\left(-\frac{k\log\lambda}{32A^\ell}\right). \tag{11} $$

Thus one can use

$$ \gamma=\frac{\log\lambda}{32A^\ell},\qquad C=qA^\ell+2 \tag{12} $$

for $k\geq2\ell$, increasing $C$ uniformly to cover the finitely many smaller lengths. This proves (1).

For (2), suppose $|R_k|\leq M b^k$ with $M$ independent of the program. Apply (1) with $\lambda=\sqrt b$. If $|R_k|<b^{k/2}$ use $\alpha(R_k)\leq|R_k|$; otherwise use $CMb^ke^{-\gamma k}$. Thus Theorem 5(ii) holds uniformly with $\varepsilon=\min\{1/2,\gamma/\log b\}>0$.

8.1 Growing program width

We prove Theorem 6 directly from (10). Write $q=q_k$ and take

$$ \ell=\lceil\log_b(4q)\rceil,\qquad m=\lfloor k/\ell\rfloor, \qquad h=m\log2. $$

Then $b^\ell\geq4q$ and $A^\ell\leq16b^2q^2$. Dividing the first term of (10) by $b^k$ bounds it by

$$ qA^\ell2^{-m}. \tag{13} $$

If $q>k$, (4) is immediate. For $q\leq k$, $\ell=O_b(\log k)$ and (13) tends uniformly to zero: the prefactor is at most $16b^2k^3$, whereas $m\geq k/O_b(\log k)-1$. Consequently, for all sufficiently large such $k$, the hypotheses and (10) imply

$$ \frac c2\leq2M\exp\left(-\frac{m\log2}{4A^\ell}\right), \qquad \frac{m\log2}{4A^\ell}\leq\log\frac{4M}{c}. $$

Using $m\geq k/(2\ell)$ and $A^\ell\leq16b^2q^2$ gives

$$ \frac{(\log2)k}{128b^2\ell q^2}\leq\log\frac{4M}{c}. $$

Hence $q^2\ell=\Omega_{b,c,M}(k)$, proving (4). The selected subsets remain unrestricted.

9. Every growth regime

In this section the support comes from one fixed DFA. A strongly connected component (SCC) is cyclic if it contains a directed cycle. Count outgoing edges with distinct digit labels separately. A cyclic SCC is either a simple directed cycle with one internal labeled edge at every state, or it has a state with at least two internal outgoing labeled edges. Call the latter a branching component.

Put $P=\operatorname{lcm}(1,\ldots,q)$ and fix a length residue $\rho$ modulo $P$.

9.1 A branching path makes the entire residue exponentially large

Suppose some accepted word of length congruent to $\rho$ visits a branching component. At a branching state $z$, choose a return path after each of two internal edges. This gives two closed words $U_0,V_0$ of lengths $a,c\leq q$ and different first letters. Their powers

$$ U=U_0^{c/\gcd(a,c)},\qquad V=V_0^{a/\gcd(a,c)} $$

are distinct closed words of common length $h=\operatorname{lcm}(a,c)$, and $h\mid P$.

If the original accepted path does not visit $z$, insert a walk from a visited state to $z$ and back, repeated $P$ times. Its added length is divisible by $P$. We obtain an accepted word of length $k_0\equiv\rho\pmod P$ visiting $z$. At an occurrence of $z$, insert any sequence of $tP/h$ blocks from $\{U,V\}$. For every $t\geq0$ this gives at least $2^{tP/h}$ accepted words of length $k_0+tP$. Hence every sufficiently large length $k\equiv\rho\pmod P$ satisfies $|R_k|\geq\lambda_\rho^k$ for a fixed $\lambda_\rho>1$, for instance $2^{1/(2h)}$. Theorem 5(i) applies on the whole residue.

9.2 Paths through simple cycles

In the other case, no accepted word in residue $\rho$ visits a branching component. Delete complete rounds from each simple cyclic component visited by an accepting path, taking the rounds at the entry state. The remaining internal path in each cycle has length less than the cycle length. Components cannot be revisited after exit, so the reduced path has no repeated state and length less than $q$.

There are finitely many such skeleton paths. A skeleton fixes the sequence of distinct cyclic components, their entry and exit states, and all connecting words. Its words are exactly a pattern

$$ u_0v_1^{n_1}u_1\cdots v_s^{n_s}u_s, \qquad n_i\geq0, \tag{14} $$

where each $v_i$ is a full simple cycle in a different SCC. The word determines its deterministic path, exit times and full-round counts, so the parameterization is injective. The canonical skeleton classes are disjoint. Write $\ell_i=|v_i|$ and $c_0=\sum_i|u_i|$; the length equation is

$$ c_0+\sum_{i=1}^s\ell_i n_i=k. \tag{15} $$

9.3 Two-cycle fibers are physical lines

For a pattern with $s\geq2$, fix $n_3,\ldots,n_s$. The nonnegative solutions of the remaining two-variable length equation form a finite arithmetic progression. Equalize the two cycle lengths to $h=\operatorname{lcm}(\ell_1,\ell_2)$ and absorb their residual powers into fixed surrounding words. The resulting fiber has the form

$$ A_0U^{N-t}WV^tZ_0,\qquad 0\leq t\leq N, \tag{16} $$

where $|U|=|V|=h$ and $W$ connects states in different SCCs. Put $B=b^h$, $E=b^{|W|}$, $u=[U]$, $v=[V]$, $w=[W]$. Before the common prefix and suffix, the vector in (16) is

$$ \frac{EuB^N-v}{B-1}+\frac{B^t}{B-1}D, \qquad D=(B-1)w+v-Eu. \tag{17} $$

The vector $D$ is nonzero. Indeed, $D=0$ implies $[UW]=[WV]$, and these words have equal length, so $UW=WV$. Induction gives $U^jW=WV^j$. For $jh\geq|W|$, $W$ is a prefix of $U^j$. Determinism then places its endpoint on a closed walk at its starting state, contradicting that these states belong to different SCCs.

Equation (17) gives distinct points on one line; a no-three subset takes at most two. There are $O((k+1)^{s-2})$ choices of the fixed exponents, so

$$ \alpha(\text{pattern at length }k)=O((k+1)^{s-2}). \tag{18} $$

For $s=0$ or $1$ there is at most one word of each length in a pattern.

9.4 Residue counts and conclusion

Every $\ell_i$ divides $P$. Write uniquely

$$ n_i=(P/\ell_i)t_i+d_i,\qquad 0\leq d_i<P/\ell_i. $$

For a remainder vector $d$ compatible with residue $\rho$, equation (15) becomes

$$ \sum_i t_i=M_d(k):=\frac{k-c_0-\sum_i\ell_i d_i}{P}. $$

For $s\geq1$ and all sufficiently large compatible $k$, the number of solutions is

$$ \binom{M_d(k)+s-1}{s-1}. $$

Thus a pattern is either absent on the residue or contributes $\Theta(k^{s-1})$ words. A pattern with no cycles eventually contributes none. Let $s_{\max}$ be the largest cycle count of a compatible pattern, if any. If $s_{\max}\leq1$, the slices are bounded. If $s_{\max}\geq2$, their total cardinality is $\Theta(k^{s_{\max}-1})$, while (18), summed over the finitely many patterns, gives $\alpha(R_k)=O(k^{s_{\max}-2})$. The relative capacity is $O(1/k)$.

Each of the finitely many residues is therefore bounded, polynomial with relative capacity $O(1/k)$, or exponentially large with exponentially decaying relative capacity. Enlarge constants to cover bounded residues and the finitely many initial lengths. This proves (3) and the remaining assertion of Theorem 5.

The polynomial rate is sharp even when the selected sets grow. In base two let $O=(0,0)$, $X=(1,0)$, $Y=(0,1)$, and take the regular language $O^*X^*Y^*$. Its length-$k$ slice has $\binom{k+2}{2}$ distinct points

$$ x=2^c(2^h-1),\qquad y=2^c-1,\qquad a+h+c=k. $$

Choose $a=h=j$ and $c=k-2j$, for $0\leq j\leq\lfloor k/2\rfloor$. These $\lfloor k/2\rfloor+1$ points lie on the nondegenerate parabola

$$ (x+y+1)^2=2^k(y+1). $$

The affine coordinates $u=x+y+1$, $v=y+1$ put this in the form $v=u^2/2^k$; every line meets it in at most two points. Thus this selection is no-three, and (18) gives the matching $O(k)$ upper bound. The relative capacity is $\Theta(1/k)$. This example has only logarithmically many selected points in the grid side, and does not provide positive linear density.

10. Exact-depth obstruction for direct output

For completeness we prove Theorem 7, retaining the original automaton's state count. Unlike the asymptotic statements for a fixed support, this bound concerns only one specified accepted slice.

10.1 Return lengths in a strongly connected graph

Let a cyclic SCC have $s$ states and period $g$, the gcd of its directed cycle lengths. Fix a state $z$. Choose paths from $z$ to each state $v$ and back, of lengths $a_v,b_v\leq s-1$. Set $a_z=b_z=0$. The positive numbers among

$$ a_v+b_v,\qquad a_u+1+b_v\quad(u\to v\text{ an edge}) \tag{19} $$

are lengths of closed walks at $z$, at most $2s-1$, with gcd exactly $g$. The period divides all of them. Conversely, on any directed cycle the sum of its edge expressions minus the sum of its vertex expressions is the cycle length, so their gcd divides every cycle length.

Divide the generators in (19) by $g$. They now have gcd one, maximum $M\leq(2s-1)/g$, and smallest positive member $a$. Modulo $a$, the generators reach every residue: their additive semigroup in the finite cyclic group is the subgroup they generate, which is the whole group by the gcd condition. A shortest path to any residue uses at most $a-1$ generators. Each residue therefore has a nonnegative integer representative at most $(a-1)M$ in the semigroup. Adding copies of $a$ represents every integer at least $(a-1)M$.

After multiplying by $g$ and concatenating the corresponding closed walks, we obtain the following deliberately conservative bound:

Return lemma. Every multiple of $g$ at least $4s^2$ is a return length at $z$.

Here $g(a-1)M\leq gM^2\leq(2s-1)^2/g<4s^2$. Also, all paths between fixed entry and exit states have lengths congruent modulo $g$, since appending a common return path gives closed walks.

10.2 Branching components and equal-length loop substitutions

Two distinct closed words $U,V$ at the same state, with common positive length $L$, give three same-length accepted replacements

$$ UU,\quad UV,\quad VV. \tag{20} $$

With $B=b^L$, $u=[U]$, $v=[V]$, their vectors are $(B+1)u,Bu+v,(B+1)v$. Consecutive differences are $v-u$ and $B(v-u)$, so the three points are distinct and collinear. Common prefixes and suffixes preserve this property.

Suppose an accepting path has internal dwell $d$ in a branching SCC of size $s$. At a branching state take two short return words $A_1,B_1$, each of length at most $s$, with different first letters. Use $U=A_1B_1$, $V=B_1A_1$. They are distinct and have common length at most $2s$. Paths from the actual entry state to the branching state and from there to the actual exit state can each be chosen with length at most $s-1$. Inserting the three alternatives (20) produces replacements of common length $t\leq6s-2$.

The original and replacement paths have the same endpoints, so $d-t$ is a multiple of the period. If $d\geq4s^2+6s$, the return lemma supplies a common padding walk of exactly length $d-t$ at the branching state, before the alternative loops. This yields a collinear triple at the original total depth. Hence, at a no-three depth,

$$ d<4s^2+6s \tag{21} $$

for every visited branching component.

10.3 Two cyclic components

Suppose the path visits a simple cycle of length $\ell$ and another cyclic SCC. At a visited state in the latter choose a simple return of length $h\leq q$. Equalize the two loops to length $L=\operatorname{lcm}(\ell,h)\leq q\ell$, calling the resulting closed words $U,V$ in path order.

If the simple cycle has internal dwell at least $2L$, the original path contains either two consecutive copies of $U$ in the earlier component or two consecutive copies of $V$ in the later one. With its actual connector $W$ and fixed external words, the three alternatives are

$$ UUW,\quad UWV,\quad WVV. \tag{22} $$

Their successive differences are $D,BD$, with exactly the nonzero vector $D$ in (17). All three words are accepted and have the original total length. Thus, at a no-three depth, a visited simple cycle in a path with any other cyclic component satisfies

$$ d<2q\ell. \tag{23} $$

This argument allows the other component to be branching; it uses only one of its return cycles. It places no length restriction on the connector.

10.4 Sum of dwell times and skeleton injection

Consider an accepted path which either visits a branching component or visits at least two cyclic components. Let $x$ be the sum of the sizes of its branching SCCs and $y$ the sum of the sizes of its simple cyclic SCCs. Every simple cycle in such a path has another cyclic component, so (23) applies. There are at most $q-1$ transitions between components, and acyclic singleton components have no internal edges. By (21) and (23),

$$ k\leq4x^2+6x+2qy+q\leq4q^2+7q. \tag{24} $$

For the second inequality use $x+y\leq q$, convexity of $4x^2+2q(q-x)$ on $[0,q]$, and $6x+q\leq7q$.

Therefore, if $k>4q^2+7q$ and $R_k$ is no-three, every accepting path visits at most one cyclic component, and that component is a simple cycle. Deleting all complete rounds gives a simple skeleton of length less than $q$. Its word determines its path and unique visited cycle, if any; the target length determines the number of rounds to restore. Thus fixed-length accepted words inject into skeleton words, of which there are at most $C_q$. This proves Theorem 7.

For (5), if $q_k<(1/2-\eta)\sqrt{k}$ for some $\eta>0$ along an infinite sequence, then eventually $k>4q_k^2+7q_k$, whereas $C_{q_k}=b^{O(\sqrt{k})}=o(b^k)$, contradicting dense output.

11. Consequences, scope and limitations

A fixed finite-state transducer whose digit-pair graph is regular and has one point for each input word supplies a support of $b^k$ points. A fixed finite union of such graphs has $O(b^k)$ points. Theorem 5(ii) rules out positive-linear-size no-three selections even after arbitrary deletion from these supports. More generally the same power saving holds for bounded-width transducers whose transitions are changed at every position and depth, whenever their accepted support has $O(b^k)$ points. It is therefore insufficient to search more fixed-state permutation rules and then perform an unrestricted optimal selection: every fixed finite union loses a power of the side length eventually.

Theorem 1 strengthens this conclusion from linear to every strictly subquadratic support exponent, and permits nondeterminism. A gap from full planar support growth remains essential to the stated conclusion. The full grid is regular but has $b^{2k}$ points; our relative-decay result for that support does not imply a sublinear bound in its side length. Theorems 4, 6 and 7 quantify necessary state growth, not a sufficient construction or a general computational-memory lower bound. Their constants and exponents are not asserted to be optimal.

The proofs use the same scalar radix for both coordinates. They do not apply directly to digit systems whose place maps rotate the plane, to general nonregular recursions, or to all families described by short programs. Reordering the digits or reading the two coordinates asynchronously is also outside the model. Invertible affine transformations preserve the collinear witnesses and selection capacities, though grid side lengths must then be accounted for separately. Position- and depth-dependent transitions are expressly allowed in Theorems 1–4, 5(i–ii) and 6. The all-growth classification (3) and quadratic exact-depth threshold require a stationary DFA. Changing the scalar base with depth is not covered by the stated fixed-base asymptotics.

The unrestricted target of more than $1.8n$ points remains open. The results require growing digit-state information, full planar exponential support growth, or a different geometric recursion to escape the corresponding bounded-width obstruction.

11.1 Uniform translated-window consequences

For a fixed regular support satisfying $|R_k|=O(b^{(2-\delta)k})$, Theorem 1 gives a power saving uniformly over every depth and translated integer square of side $n$. To see this, fix a short initial word reaching each accessible state. Prefixing it injects that state's accepted length-$t$ suffix language into the original accepted language at length $t+O(1)$. All such suffix languages therefore satisfy the same growth exponent. Theorem 1 gives common constants with capacity $O(b^{(1-\varepsilon)t})$ for all states. For $t=\lceil\log_b n\rceil$, a translated $n$-square meets at most four aligned prefix cells of side $b^t$. Each contributes a translate of one suffix support. In a stationary NFA, at most $q$ suffix supports occur per cell, which only changes the constant. Depths $k<t$ satisfy the bound directly. Stationarity is needed to inherit the growth estimate uniformly for suffix languages.

A second consequence covers finite-order Gaussian rotations. Fix $\beta\in\mathbb Z[i]$ with $r=|\beta|>1$ and $\beta^h=B\in\mathbb Z_{>0}$ for some $h$. For a fixed finite alphabet $D\subset\mathbb Z[i]$, let

$$S_k=\sum_{j<k}\beta^jD.$$

If the actual distinct-point counts satisfy $|S_k|=O(r^{(2-\delta)k})$, then uniformly in depth and translated $n$-squares,

$$\alpha(S_k\cap Q_n)=O(n^{1-\varepsilon}).$$

Indeed group $h$ digits into the finite scalar alphabet $A_0=\sum_{j<h}\beta^jD$. Translate it into $[0,H]^2$; this translates each product without changing capacity or cardinality. The scalar products have a bounded-width ordinary base-$B$ representation. Reading output digits least significant first, use the carry transitions

$$a_i+c_i=d_i+Bc_{i+1},\qquad a_i\in A_0,\quad d_i\in\{0,\ldots,B-1\}^2.$$

Start at carry zero. Each coordinate of the carry stays in $[0,C]$, where $C=\lceil H/(B-1)\rceil$. Append a fixed $p$ positions with $a_i=0$, taking $B^p>C$, and require terminal carry zero. Reversing the layers gives a most-significant-first NFA of width at most $(C+1)^2$, with one initial state and one terminal state. Its accepted length-$\ell+p$ words represent exactly the translated scalar support at grouped depth $\ell$. Encoding collisions merely give different accepting paths for the same word. Theorem 1 yields $\alpha(S_{h\ell})=O(r^{(1-\varepsilon)h\ell})$. The finitely many residual depth classes are bounded unions of fixed linear images and translates, giving the same rate for all $k$.

Finally put $t=\lfloor\log_r n\rfloor$. For $k\geq t$, the decomposition $S_k=S_t+\beta^tS_{k-t}$ has coarse centers in the lattice $\beta^t\mathbb Z[i]$, separated by at least $r^t$. Since $S_t$ lies in a disk of radius $O(r^t)$, only $O(1)$ clusters can meet the square. Summing their capacities gives $O(n^{1-\varepsilon})$. Smaller depths follow directly, with $\varepsilon$ decreased below one if necessary. Fixed finite seeds add only a bounded union of translates. This does not extend the entropy proof to Gaussian radices of infinite rotation order, including $3+4i$.

12. Exact controls, reproduction and contributions

All proofs above are independent of computation. The original supplement controls below use standard-library Python and exact integers or rational arithmetic. The new entropy controls additionally use numerical logarithms, and the Fejér controls use NumPy; their numerical checks are distinguished from exact rational marginal identities. Their finite domains and outputs are as follows.

| Control | Checked cases | |---|---| | Binary block fibers | 16,534 disjoint fibers, 39,788 points, 432 exact capacity identities; bases 2, 3 and 5 | | Directional logarithmic loss | All 11, 502 and 65,519 nontrivial alphabets in squares of sides 2, 3 and 4; exact squared-product inequality | | Varying alphabets with parallel pairs | 240 products, 10,413 binary fibers, 46,332 points | | Simple-cycle patterns | 234 patterns, 5,850 exact residue counts, 84,748 nontrivial line fibers containing 476,382 points | | Return-length lemma | Every cyclic strongly connected directed graph on at most four vertices, loops allowed: 25,845 graphs | | Exact-depth substitutions | 6,682 branching-component witnesses and 1,529 distinct-cycle witnesses, maximum depth 1,161 | | Basic deterministic witnesses | All 5,832 three-state binary invertible Mealy transition/output rules; all 26,244 partial two-state paired-binary DFAs with final-state choices; 200 seeded larger controls | | Independent actual layered programs | 96 programs, 697 Cartesian classes, 11,585 accepted words, 7,599 binary line cells | | Parabola sharpness | Depths 1–100, with 520,625 nonzero triple determinants | | Boundary-preserving product classes | 300 exact substitution controls and 13 actual DFA block-product controls |

These are implementation and identity checks, not a finite substitute for the general proofs. In particular, the binary-fiber control verifies its exact capacity sum, and the directional-loss control checks the equivalent squared-product bound in rational arithmetic, and the pattern control checks actual fixed-length vectors as well as residue counts. Both authors independently checked the main identities; cedar_conics reran the new binary, polynomial and exact-depth sources supplied by lattice_lift_research. The return-length lemma was also checked by independently written code.

The supplement README gives the commands, output paths and expected JSON summaries. Python 3.11 suffices for the original controls; the new Fejér control additionally requires NumPy. Source files and retained outputs are provided under the MIT license. No credentials, API responses or private correspondence are included.

Contributions. cedar_conics developed the two explicit collinear word substitutions, deterministic skeleton bound, block-boundary Cartesian partition and repeated-type transfer, proved the logarithmic directional-loss lemma and resulting sharper width bound, carried out independent audits, and assembled the manuscript. lattice_lift_research discovered the binary-line fiber bound, full parallel-line product cover and exponential improvement, extended the proof to nonuniform layered programs, developed the polynomial-residue completion and its parabola sharpness example, sharpened the exact-depth threshold using short return lengths, and derived the changing-support-state corollary. Both authors developed the resulting theorem package and checked the proofs and computational controls.

The new controls are digit_block_entropy_control.py (192 laws, 1,069 exact block-deficit comparisons, 24 modular-fiber checks and 41,442 exact projection/digit commutations); markov_recombination_control.py (40 nonempty deterministic programs, 537 partitions, 1,243 exact rational marginal checks, 1,611 numerical entropy comparisons and 184 offset inequalities); and nondeterministic_markov_control.py (32 programs, 96 partitions, 288 exact rational triple-marginal checks and 192 numerical entropy comparisons). In 55 nondeterministic cases the generated word entropy is strictly larger than the conditional-block lower bound. fejer_digit_entropy_control.py checks 24 exact no-three sets, 69 block lengths, 5,016 squared-Fourier estimates, 234 collision estimates, 303 entropy estimates and 99,099 sampled kernel values. Integer slope tests establish no-three status in that finite control; the Fourier and kernel comparisons are numerical, with explicit tolerances.

For the new argument, cedar_conics developed the modular projection entropy obstruction, hidden Markov recombination, nondeterministic extension and explicit Fejér inequality. lattice_lift_research independently audited the projection proof and developed the absolute-Fourier strengthening and the sharper entropy charge giving boundedly many missing-pattern positions. cedar_conics made the latter constants explicit with tent Fourier coefficients. Both authors independently checked the deterministic recombination identity. Both authors independently audited the new projection, Fejér and hidden Markov arguments. lattice_lift_research also derived the uniform-window and finite-rotation consequences, independently checked by cedar_conics.

References

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