Fix a configuration and let \(q\ge q_{\mathrm{tail}}\), so \(q{\gt}\sigma _{\max }\ge |\sigma _i|\) for all \(i\). For a moving backward coordinate \(i\) the bound \(b_i\) of Lemma 4 is \(q-|\sigma _i|-\mathrm{jit}_i\ge 0\): for \(e_i=1\) and \(-q{\lt}\sigma _i{\lt}0\), \(\bar\sigma _i=q+\sigma _i\) and the two options give \(|q+t|\in \{ 2q-|\sigma _i|,\, q-|\sigma _i|\} \); for \(\sigma _i=0\) the only option gives \(q\); \(e_i=-1\) is symmetric. If at least two coordinates are backward, \(\sum _ib_i^2\ge 2(q-\sigma _{\max }-1)^2\ge (q-K)^2\), because \(\sqrt2(q-\sigma _{\max }-1)\ge q-K\) is the second condition on \(q_{\mathrm{tail}}\) (and \(K\le \sigma _{\max }+1\le q\)). If exactly one coordinate \(b\) is backward, \(\sum _ib_i^2\ge (q-|\sigma _b|-\mathrm{jit}_b)^2\ge (q-K)^2\). No backward coordinate is excluded by \(K{\lt}\infty \). So every configuration gives \((\sum _ib_i^2)^{1/2}/q\ge 1-K/q\). The configurations of a truncation are among those of \(F\), with the same \(\sigma \) (Lemma 4, capped carry depths), so its \(\mathrm{LB}\) is at least that of \(F\).

Write \(M=F_0\) and consider the carry-free steps \(e=\mathbf{e}_a+\varepsilon \mathbf{e}_b\), \(\varepsilon =\pm 1\), along which only \(a\) and \(b\) move, with \(\sigma _a=M_{aa}+\varepsilon M_{ab}\) and \(\varepsilon \sigma _b=M_{bb}+\varepsilon M_{ba}\). If for some \(\varepsilon \) at most one of \(a\), \(b\) is backward, that configuration costs at least \(\mathrm{jit}=1\), or \(+\infty \). Otherwise \(M_{aa}+\varepsilon M_{ab}\le 0\) for both signs, so \(M_{aa}\le -|M_{ab}|\le -1\), and the one-axis step along \(a\) without carry costs \(-M_{aa}\ge 1\).

Let \(\mu \in \{ 1,\dots ,p-1\} \) be the least nonnegative residue of \((\Phi _{aa})_{0,k-1}\), so \(\mu =p+\tilde M_{aa}\) in (i) and \(\mu =\tilde M_{aa}\) in (ii). Take a cell \(n\) with \(n_a=0\) and its neighbour \(n+e_a\); only the last digit of \(n_a\) changes, so each offset digit of each coordinate changes by a constant modulo \(p\), the first one of coordinate \(u\) by \((\Phi _{ua})_{0,k-1}\). Require (A) the first \(k-1\) offset digits of \(a\) all equal \(p-1\), and (B) for \(u\in \{ b,b'\} \), the first offset digit of \(u\) is not in the set \(W_u\) of the \(|\tilde M_{ua}|\) digit values on which adding \(\tilde M_{ua}\) leaves \(\{ 0,\dots ,p-1\} \). Under (A) the first offset digit of \(a\) becomes \(\mu -1\), the next \(k-2\) (all \(p-1\)) can only decrease, and the remaining digits contribute less than \(p^{1-k}\). So in cell units \(\Delta _a=\mu /p+\theta \) with \(-1/p{\lt}\theta {\lt}p^{1-k}\). Under (B), the first offset digit of \(u\) changes by exactly \(\tilde M_{ua}\), so \(|\Delta _u|{\lt}(|\tilde M_{ua}|+1)/p\). In (i), \(0{\lt}\Delta _a{\lt}1-1/p+p^{1-k}\); in (ii), \(0{\lt}\Delta _a{\lt}\tilde M_{aa}/p+p^{1-k}\le (\tilde M_{aa}+1)/p\) since \(k\ge 2\). It remains to find such a cell. The \(p^{2k}\) indices with \(n_a=0\) are parametrized by \((n_b,n_{b'})\). By the \(t=0\) boxes \((2k-1,k+1,0)\) and \((2k-1,0,k+1)\) (order \(a,b,b'\)), exactly \(p^{k+1}\) of them satisfy (A), and among these the first offset digits of \(b\) and of \(b'\) each take every value exactly \(p^k\) times. Hence at most \((|W_b|+|W_{b'}|)p^k\le (p-1)p^k{\lt}p^{k+1}\) of them violate (B), since \(|\tilde M_{ua}|\le (p-1)/2\). The pair \(n\), \(n+e_a\) lies inside the cube, so it bounds the toroidal and the Euclidean distance.

The coefficient of the corollary is \(\tilde M_{aa}\); let \(|\tilde M_{ba}|,|\tilde M_{b'a}|\le C\), \(C\ge 1\). In (i) of Proposition 79, with \(k\ge 3\), put \(\alpha =1-1/p+p^{1-k}\ge 2/3\) and \(\beta =\bigl((|\tilde M_{ba}|+1)^2+(|\tilde M_{b'a}|+1)^2\bigr)/p^2\le 8C^2/p^2\); then \(c{\lt}\sqrt{\alpha ^2+\beta }\le \alpha +\beta /(2\alpha )\le \alpha +6C^2/p^2\), and \(p^{1-k}\le 1/p^2\le C^2/p^2\), so \(c{\lt}1-1/p+7C^2/p^2\). In (ii), with \(1\le \tilde M_{aa}\le C\), \(c^2{\lt}3(C+1)^2/p^2\le 12C^2/p^2\), so \(c{\lt}4C/p\). (For \(k=2\) the term \(p^{1-k}=1/p\) cancels the \(-1/p\); hence \(k\ge 3\).)

The proof of Proposition 79 applies with all counts multiplied by \(q^{(s-3)k}\) and \(\Delta _a=\mu /q+\theta \), \(\mu =q+\tilde M_{aa}\) in (i), \(\mu =\tilde M_{aa}\) in (ii) (where \(0{\lt}\Delta _a{\lt}(\tilde M_{aa}+1)/q\) as \(k\ge 2\)). The boxes used have depth \(0\) off \(a\) and the transverse coordinate considered, and total depth \(\le 3k\le sk\); no field structure is used; and the union bound over \(u\ne a\) leaves a cell because \(\sum _{u\ne a}|\tilde M_{ua}|{\lt}q\).

(1) Exact interpolation shows that every column of every partial sum \(M^{(j)}\) is admissible (Proposition 18): its polynomial \(g_{d,j}\) has degree \(j\), coefficients in \(\mathbb {Z}[\frac12]\) and leading coefficient \(\pm 2^i\). So the residual problem is solvable over \(R'=\mathbb {Z}[\frac12]\), and Corollary 16 gives the net property with box minors in \(\mathbb {Z}[\frac12]^\times =\{ \pm 2^i\} \), units modulo every odd \(q\); a cell has diameter \(2q^{-k}=2N^{-1/4}\).

(2) (H1) and (H2) of Lemma 4 hold by Proposition 14 and Definition 15, with \(L=\min (k,3)\). Full enumeration of the \(2400\) configurations gives \(K(F)=2\) and \(\sigma _{\max }(F)=7\), so \(q_{\mathrm{tail}}=23\) and Proposition 77 gives \(\mathrm{LB}\ge 1-2/q\) for \(q\ge 23\) and every \(k\ge 1\). For \(q=17,19,21\) the exact relaxation (with \(L=3\), which bounds those with \(L{\lt}3\)) is \(\mathrm{LB}=(q-2)/q\). The bound is attained: the one-axis step along \(\infty \) without carry has \(\sigma =(-2,1,1,1)\), so \(b=q-2\) for the moving coordinate and \(b=0\) for the others.

(3) As \(|F_0|\le 2{\lt}q/2\), the geometric layer is \(\tilde M=F_0\), and column \(\infty \) of \(F_0\) is \((-2,1,1,1)\). This is the sharpness argument of Theorem A (Section 4.3). Equivalently, it is Corollary 81(i) with \(a=\infty \) without its term \(q^{1-k}\): that term accounts for the deeper offset digits of the moving coordinate, and here they do not change, by (H2). So the own-axis step \(e_\infty \) from a cell whose last digit of \(n_\infty \) is not \(q-1\) and whose first offset digit \(r_\infty \) is at least \(2\) moves coordinate \(\infty \) by exactly \(1-2/q\) cells, and as \(\sum _{u\ne \infty }|F_0[u][\infty ]|=3{\lt}q\), the union bound of that argument leaves such a cell at which every other coordinate \(u\) moves by less than \((|F_0[u][\infty ]|+1)/q\). Hence \(c^2{\lt}(1-2/q)^2+12/q^2\) for every \(k\ge 2\), toroidal and Euclidean (the pair lies inside the unit cube). With \(\alpha =1-2/q\ge 15/17\) we get \(c{\lt}\alpha +6/(\alpha q^2)\le 1-2/q+\frac{102}{15}/q^2{\lt}1-2/q+8/q^2\).

(1) The hypotheses are (H1) and (H2) of Lemma 4 with \(k=L=1\); then Proposition 77. (2) At \(k=1\) Definition 15 realizes the truncated design \((F_0)\) with vanishing deeper diagonal rows, and its configurations are those of \(F\) with all \(j_d=0\), with the same \(\sigma \) and jitter (after Lemma 4). (3) As \(q{\gt}2\max |F_0|\), the centred layer is \(F_0\). Take the step \(e_a\) from a cell with \(r_a\ge K_1\) and \(r_u+F_0[u][a]\in [0,q)\) for every \(u\ne a\). Then \(\Delta r_a=-K_1\) exactly and \(\Delta f_a=0\) (the deeper rows of \(a\) vanish on \(n_a\)), so coordinate \(a\) moves by exactly \(1-K_1/q\) cells, and every \(u\ne a\) by \((F_0[u][a]+\Delta f_u)/q\), of absolute value less than \((|F_0[u][a]|+1)/q\). Such a cell exists: if \(\gcd (\det F_0,q)=1\), \(n\mapsto r=F_0n\) is a bijection of \((\mathbb {Z}/q\mathbb {Z})^s\), so \(r\) can be prescribed. In the second case, for each \(u\ne a\) the box with depths \(2\) on \(a\) and on \(u\), \(0\) on two further coordinates and \(1\) elsewhere (total \(s\)) makes \((n_a,r_a,r_u)\) uniformly distributed (Lemma 1); among the cells with \(n_a\ne q-1\) and \(r_a\ge K_1\) the fraction on which \(r_u\) wraps is \(|F_0[u][a]|/q\), and the union bound leaves a cell as \(\sum _{u\ne a}|F_0[u][a]|{\lt}q\). With \(n_a\ne q-1\) the pair does not cross the boundary of the cube, so the bound holds for the Euclidean distance too.

(1) is a finite computation of \(K_1\). (2) At \(k=1\) the frame is not needed. By Lemma 2, a box minor of a graph form with depths \((1,\dots ,1)\) is the determinant of stacked offset rows restricted to the coordinates of depth \(0\); those that use only the first offset rows are the minors \(F_0[T,Z]\) with disjoint \(T\), \(Z\). If a prime \(p\) divides none of them, every box minor is a nonzero polynomial over \(\mathbb {F}_p\) in the entries of the deeper rows (it is not the zero polynomial: take the deeper rows it uses to be unit vectors that complete the independent rows of \(F_0\)), of degree at most the number of deeper rows it uses, and by the Schwartz–Zippel lemma deeper rows with all box minors nonzero exist modulo \(p\) once \(p\) exceeds the sum \(\mathrm{SZ}(s)\) of these degrees (\(\mathrm{SZ}(5)=105\), \(\mathrm{SZ}(6)=504\)); below it we constructed them explicitly for each admissible prime and certified \(t=0\). Any integer lift of such a completion modulo \(p\) works modulo \(p^a\): every box minor is an integer polynomial in the entries, so it is congruent modulo \(p\) to the minor modulo \(p\), and a determinant prime to \(p\) is a unit modulo \(p^a\). The Chinese remainder theorem combines the lifts for the prime powers of \(q\), and Lemma 1 gives the net property. The separation bounds are Proposition 85 with exact scans below its tail. Computer-assisted are the layers, \(K_1\), the per-prime completions up to \(\mathrm{SZ}(s)\) and the finite scans.