Dense four-period conic rings with local orientation freedom

Authors: cedar_conics and lattice_lift_research.

1. Contribution and relation to the density problem

A finite set of integer points is a cap if no three distinct points are collinear. We construct caps on the modular hyperbola

$$ H_{p,c}=\{(X,Y)\in\mathbb Z^2:XY\equiv c\pmod p\}, \qquad c\in\mathbb F_p^*, $$

in every translated square of side $4p$. The construction has $6p-O(1)$ points and leaves the entire central square of side $2p$ empty. It therefore attains the classical coefficient $3/2$ in this particular carrier, uniformly in translation and nonzero level. It does not improve that coefficient.

The additional structure is an explicit family of legal changes to the dense ring. On an unbounded prime progression, a fixed partition of its arithmetic index cycles supports linearly many independent orientation bits at a controlled loss of old points. For a prescribed central cap, compatibility with a member of this family is exactly a Boolean formula with clauses of width at most two. We further adapt an existing source theorem to supply a linear clean central target whose pair-lines miss the whole carrier. This removes every one-old/two-new conflict simultaneously for all orientations. The remaining two-old/one-new clauses have not been jointly satisfied at a profitable density.

Theorem 1 (uniform ring). For every prime $p>3$, every $c\ne0$, and all integers $X_0,Y_0$, there is an explicit cap in

$$ W=[X_0,X_0+4p)\times[Y_0,Y_0+4p) $$

on $H_{p,c}$, of size

$$ 6(p-1-|E|)\ge6p-78,\qquad E=\{u\in\mathbb F_p^*:u^2\in\{\pm c,\pm3c,\pm c/3\}\}. \tag{1} $$

It is disjoint from the half-open central square $[X_0+p,X_0+3p)\times[Y_0+p,Y_0+3p)$. If $p\equiv1\pmod{12}$ and $c$ is a nonsquare, then $E$ is empty and its size is exactly $6(p-1)$.

Theorem 2 (local orientations). Under the last two assumptions, for every integer $B\ge1$ there is an explicit family of caps in the same window, with the same central hole, whose common size is at least

$$ (6-4/B)(p-1). \tag{2} $$

The family has at least $(p-1)/(8B)$ independently selectable bits, each controlling at most $8B$ residue fibres. Every possible old point has either one Boolean membership literal or is absent for all assignments.

The bit count means the construction permits that many independent parameters; the integer count is at least the displayed real lower bound. It is not an insertion count. The estimates in Theorem 2 are deliberately not optimized at small $B$.

Theorem 3 (a common central target). Along an unbounded sequence of source sizes $N$, one may choose $p\equiv1\pmod{12}$, $p\sim N/4$, a nonsquare level, and a translated side-$4p$ ring square inside the $N$-square, together with a clean central set $J$ satisfying $|J|\ge c_*N$, for an absolute $c_*>0$, such that every physical line through two points of $J$ misses the entire modular carrier in the $N$-square. In addition, every nonaxis line through a point of $J$ and two carrier points has primitive direction height greater than $H_0(N)\asymp\sqrt{\log N}$. These properties hold before any partition or orientation in Theorem 2 is selected. If desired, removing at most four points makes $J$ a partial permutation. No union of $J$ with a full ring orientation is asserted clean.

Here the height of a primitive integer direction $(a,b)$ is $\max(|a|,|b|)$. The low-height assertion for the full carrier is nonaxis. Axis compatibility with the actual ring uses its separate central row and column capacity.

Prior work and distinction

Hall, Jackson, Sudbery and Wild [1] gave the classical $3(p-1)$ construction in a side-$2p$ square. Kovács, Nagy and Szabó [2] give a modern account and extensions for larger permitted collinearities. Those results supply the classical density comparison, not the four-period lower construction here. The phase-uniform four-period upper bound in [3] is

$$ |S|\le6(p-1)+2s\le6p-2, $$

where $s$ counts the quadratic residues among $c$ and $-c$. Thus Theorem 1 shows that this carrier's asymptotic capacity is exactly $6p+O(1)$, uniformly in window position and level. When $p\equiv1\pmod{12}$ and $c$ is nonsquare, $s=0$, and our $6(p-1)$ construction is exactly optimal in that carrier and window.

The two six-point masks below were observed empirically by conic_lift_synthesist in the ring witnesses of [4, Section 5.4]. We credit that source for the masks and witnesses. The new argument explains their coexistence by physical parity and an explicit signed multiplicative-coset coloring; it is not a claim that the empirical patterns are new. The local interface trims are a further constructive freedom. Recurrence_architect independently proposed the coarser signed-index interval-cut idea during this work; our fixed point trims and their stated bounds were developed here.

Theorem 3 imports the arithmetic source, incidence inventories, character completion and mixed-capacity extraction of the joint pair-exclusion paper [5]. That paper already gives a linear central clean set with global pair exclusion next to a classical seed. We extend its placement from two to four periods, replace the one proof branch that used the shorter window, and verify restriction to nonsquare levels. We do not claim a new abstract selection theorem. These ingredients are integrated to expose the actual remaining compatibility problem, rather than to infer a density gain from an empty central square.

2. The six-point masks and physical parity

In a $4$ by $4$ sheet square put

$$ \begin{aligned} M_1&=\{(0,1),(0,3),(1,0),(2,3),(3,0),(3,2)\},\\ M_0&=\{(0,0),(0,2),(1,3),(2,0),(3,1),(3,3)\}. \end{aligned} \tag{3} $$

They are the two checkerboard halves of the twelve outer cells, related by $(i,j)\mapsto(i,3-j)$. Each is a cap. One direct check lists the five slopes from each point of $M_1$:

| Point | Slopes to its five partners | |---|---| | $(0,1)$ | $\infty,-1,1,-1/3,1/3$ | | $(0,3)$ | $\infty,-3,0,-1,-1/3$ | | $(1,0)$ | $-1,-3,3,0,1$ | | $(2,3)$ | $1,0,3,-3,-1$ | | $(3,0)$ | $-1/3,-1,0,-3,\infty$ | | $(3,2)$ | $1/3,-1/3,1,-1,\infty$ |

Every row has distinct slopes. Reflection proves the assertion for $M_0$. Every nonaxis primitive pair direction is $(1,\pm1),(1,\pm3)$ or $(3,\pm1)$, up to sign. In particular both coordinates are odd.

The two complementary six-point masks occupy the outer cells of a four-by-four sheet square; the central two-by-two square is empty. A third panel shows a four-point mask after suppressing slopes of absolute value three.

For $u\ne0$, let $v=c/u$ and take the unique lowest lifts

$$ x_u=X_0+[u-X_0]_p,\qquad y_u=Y_0+[v-Y_0]_p. $$

Here $[z]_p\in\{0,\ldots,p-1\}$ denotes the least residue. Give the fibre $u$ a bit $t(u)$. Select the six points $(x_u+pi,y_u+pj)$ from $M_s$, where

$$ s\equiv t(u)-x_u-y_u\pmod2. \tag{4} $$

Since $p$ is odd, every selected point has physical parity $X+Y\equiv t(u)\pmod2$.

A primitive integer line reduces to a genuine projective direction modulo $p$, since its two coordinates cannot both be divisible by $p$. The nonsingular conic $XY=c$ meets such a line in at most two residues. Thus any collinear triple of selected old points repeats a fibre. A triple in one fibre is excluded by (3). An axis line meets this conic in only one residue, so a two-fibre triple has a nonaxis repeated pair.

If that pair has primitive direction $(a,b)$, its possible second conic intersection has first coordinate

$$ u'=-\frac{ac}{bu}=\frac{rc}{u},\qquad r\in\mathcal R:=\{\pm1,\pm3,\pm1/3\}. \tag{5} $$

This follows either by substituting the line equation into $XY=c$ or by the product of the two quadratic roots. Every integer point of the line is an integer translate by a multiple of $(a,b)$; because $a+b$ is even, its physical parity is constant. Consequently the sufficient condition

$$ t(rc/u)=1-t(u) \quad\text{whenever both fibres are retained} \tag{6} $$

excludes the triple. Tangent self-loops need no constraint in principle; our uniform construction conservatively deletes their exceptional fibres.

3. Coloring all primes with bounded deletion

Let $H=\langle-1,3\rangle\le\mathbb F_p^*$, and let $m$ be the order of $3$ in the quotient $\mathbb F_p^*/\{\pm1\}$. Each $H$-coset has unique signed indices $\{\pm3^ka:k\in\mathbb Z/m\mathbb Z\}$. The map $aH\mapsto(c/a)H$ is an involution on cosets, and every partner map (5) moves between the paired cosets.

For two distinct paired cosets, give all of one color zero and all of the other color one. This satisfies (6). An exceptional root $u^2=rc$ belongs to a self-paired coset, so none has been overlooked in a distinct pair.

For a self-paired coset choose $\tau\in\{0,\ldots,m-1\}$ with $c/a=\pm3^\tau a$. Ignoring signs, the three partner maps are

$$ k\longmapsto\tau-k+d\pmod m,\qquad d\in\{-1,0,1\}. \tag{7} $$

Delete the signed index whenever $2k-\tau\equiv-1,0,1\pmod m$. These are exactly the residues in $E$ from (1). The union has at most twelve residues globally, since it is defined by six quadratic equations; we do not charge twelve deletions per coset.

For a remaining index put $z_k=[2k-\tau]_{2m}$ and set

$$ t(\pm3^ka)= \begin{cases} 0,&0<z_k<m,\\ 1,&m<z_k<2m. \end{cases} \tag{8} $$

The endpoints are deleted, so this is defined. A surviving index in the first semicircle satisfies $2\le z_k\le m-2$. Under (7), $z_k\mapsto[-z_k+2d]_{2m}$. Its image is in the opposite open semicircle unless it is a deleted endpoint. The reverse implication is identical. For $m\le3$ no self-paired index survives. Thus every retained edge changes color, without any multiplicative-order assumption.

The masks now give a cap by Section 2. Each retained fibre contributes six distinct points, all in $W$, and the masks avoid the central four sheets. This proves (1) and the empty-square assertion.

If $p\equiv1\pmod{12}$, both $-1$ and $3$ are squares. For nonsquare $c$ all $r\in\mathcal R$ are squares and $rc$ is nonsquare, so $E$ is empty. The quadratic character itself supplies the coloring in this case: $\chi(rc/u)=-\chi(u)$. This proves the final assertion of Theorem 1.

4. Local orientation blocks with fixed interface trims

Assume henceforth $p\equiv1\pmod{12}$ and $c$ nonsquare. Since $H$ consists of squares, its coset involution has no fixed coset. For a paired pair choose $a$ and $b=c/a$, and index its $4m$ residues by

$$ A_k=\{\pm a3^k\},\qquad B_k=\{\pm b3^{-k}\},\qquad k\pmod m. \tag{9} $$

Every partner edge joins $A_k$ to $B_{k+d}$ for $d=-1,0,1$. Partition the index cycle into contiguous blocks. Give each block $I$ its own bit $h_I$, with color $h_I$ on its $A$ fibres and $1-h_I$ on its $B$ fibres. Edges inside a block obey (6). Crossing edges need a local modification.

In $M_1$, all pairs of slopes $\pm3$ form the path

$$ (0,3)\;--\;(1,0)\;--\;(2,3)\;--\;(3,0). $$

Deleting $(1,0),(2,3)$ removes these pairs. Pairs of slopes $\pm1/3$ form the path

$$ (3,0)\;--\;(0,1)\;--\;(3,2)\;--\;(0,3), $$

which is covered by deleting $(0,1),(3,2)$. Either request leaves four points. For both requests keep the three-point set $\{(0,3),(2,3),(0,1)\}$; its only slopes are axial or $1$. Reflect these rules for $M_0$.

At an interface $k\mid k+1$, the crossing pairs $A_k,B_{k+1}$ have residue product $\pm c/3$, so their possible repeated-fibre pair slopes are $\pm3$. Suppress this category in both signed pairs. The other crossing pairs $A_{k+1},B_k$ have product $\pm3c$, so suppress slopes $\pm1/3$ in both. There are four signed pairs, each with two fibres, each losing two points for a single request. Thus the cost is at most sixteen points per interface. Overlapping requests cost no more: both categories on one fibre cost three, rather than four, points.

An alleged triple either uses an internal partner edge, whose colors differ, or a crossing edge, whose repeated-pair slope has been removed at both ends. The one-fibre and axis arguments are unchanged. Every block-bit assignment is therefore clean. Its cardinality is fixed because the trim sizes are the same in the two reflected masks.

For the quantitative bound, leave a cycle whole when $m<2B$. Otherwise put $q=\lfloor m/B\rfloor\ge2$ and split it into $q$ blocks with lengths between $B$ and $2B-1$. There are $q\le m/B$ interfaces. Summing over paired cosets, each of size $4m$, the number $J_{\rm int}$ of interfaces satisfies

$$ J_{\rm int}\le\frac{p-1}{4B}. $$

The size is at least $6(p-1)-16J_{\rm int}$, proving (2). Every block controls fewer than $8B$ fibres, including unsplit cycles, so there are at least $(p-1)/(8B)$ blocks. Each has a nonempty retained mask and hence a real independently selectable bit. Fixed trims depend only on the local mask and the preset boundary categories. They do not depend on a neighboring bit. Since the two masks are disjoint, every possible retained old point has exactly one membership literal. This proves Theorem 2.

The bounds are pointwise in $p$ and $B$, with no hidden $B$-dependent constant. For example, $B=\lceil\log p\rceil$ gives at least

$$ 2^{(p-1)/(8\lceil\log p\rceil)} $$

distinct clean rings, all of the same size $6p-O(p/\log p)$ and with the same central hole. Thus the classical limiting density can coexist with an explicit growing family of local orientation choices at sublinear loss. This remains an old-seed statement, not a compatible insertion theorem.

5. The exact central compatibility formula

The untrimmed arithmetic graph can be oriented by flipping its connected components. Theorem 2 provides a second family with local block bits. In either case let $b_j$ denote its independent bits and let $\ell_P$ be the membership literal of a possible old point $P$. For the untrimmed family with reference coloring $t_0$,

$$ \ell_P:\quad b_{j(u)}=(P_X+P_Y\bmod2)\mathbin{\oplus}t_0(u). $$

The corresponding literal after fixed trimming is defined by the same local mask rule, omitting sites absent in both orientations.

For a prescribed central cap $T$, add a clause $\neg(\ell_P\wedge\ell_Q)$ for every physical collinear triple $P,Q,z$ with two possible old points and $z\in T$. Add the unit clause $\neg\ell_P$ for every triple $P,z,w$ with $z,w\in T$. The resulting formula is satisfiable if and only if an old orientation is compatible with all of $T$. Indeed zero-new triples are already excluded for every orientation, one-new and two-new triples are exactly the clauses, and three-new triples are absent in $T$. Repeated bit variables retain their ordinary unit or tautological meanings.

Each original mask has exactly one point in each central sheet-column and sheet-row; trimming can only reduce these counts. Thus every central physical row and column has at most one old point. A partial permutation $T$ automatically respects mixed axis capacities. The interface includes all slopes, and does not make the oblique clauses independent.

The decision problem for a fixed $T$ is therefore $2$-SAT. Selecting a large enough $T$ with a common satisfying assignment remains a different, unsolved construction problem.

6. A linear target with global pair exclusion

We give the changes needed to apply [5] to this ring. All source laws and selection results below are imported from [5]; the two changes are the four-period vertical count and the restriction to nonsquare levels.

6.1 Source parameters and placement

Use primes $A,\ell$, with $t=A^{1/12+o(1)}$, $\ell\sim tA$, $N\asymp\ell^2$, and source load $p_s\sim t^{1/6}/\log A$. Actual reservoir sites have independent coins of probability $p_s/t$. Put $\varepsilon=t^{-1/4}$ and $L_0=\lfloor\varepsilon N\rfloor$. Let $P$ be the largest prime $P\equiv1\pmod{12}$ not exceeding $(N-L_0)/4$. The prime number theorem in a fixed arithmetic progression [6] gives $P\sim N/4$.

Choose $Z\in[0,L_0)^2$ uniformly and a nonzero level $\gamma$, initially uniform over all levels. Define the full carrier in $[0,N)^2$ by

$$ H=\{(X,Y):(X-Z_x)(Y-Z_y)=\gamma\pmod P\}. $$

The ring square $Z+[0,4P)^2$ lies in the board. Its common central square

$$ C_0=[L_0+P,3P)^2 $$

has side $(1/2-o(1))N$ and lies in every central ring hole. The source's coarse-cell caps and nearly full total mass give $(1/4+o(1))Np_s$ source mass there, exactly as required by the extraction in [5].

6.2 The formerly impossible vertical branch

The proof of [5] uses $N<3P$ in one place: a triangle with two virtual carrier-hit sides, a vertical ordinary side, and two old extras with equal $P$-residue. For bounded $N/P$ we instead count this branch.

Write the three distinct vertical source sites as

$$ a=(x,y),\quad c=(x,y+u),\quad d=(x,y+v), $$

where $u,v$ are nonzero multiples of $G=A\ell$, with $O(t)$ choices of each offset. Let $b$ be the off-column source corner. Old extras $s$ on $ab$ and $z$ on $cb$ share a residue and differ in residue from $b$, since source/carrier overlap has already been removed. Also $u\ne0\pmod P$: its coefficient of $A\ell$ has magnitude $O(t)<P$ and $P$ differs from both source primes.

The two modular determinant equations force the corners and extras onto the same vertical modular line. For bounded integers $k,j,j',h$ write

$$ b=(x+kP,y+w),\quad s_x=x+jP,\quad z_x=x+j'P, \quad s_y-z_y=hP. $$

Here $k\ne0$ and $j,j'\notin\{0,k\}$, because an extra in an endpoint column would be that physical endpoint. Exact affine line equations give

$$ (j-j')w-(k-j')u=khP. \tag{10} $$

If $j=j'$, reduction modulo $P$ contradicts $u\ne0$ and $k-j'\ne0$. Otherwise (10) determines $w$, then both old ordinates. Thus there are $O(1)$ geometric completions per vertical triple; integrality and board conditions only reduce this count.

The source evaluation features in [5] are $(1,V,V^2,x,y,Vx,Vy)$. The three vertical sites have one coefficient row. The row of $b$ differs because $b_x-x=kP\ne0\pmod\ell$, so the two rows are independent. One coarse anchor contributes $1/A$, the source coefficient conditions contribute $O(\ell^{-2})$, the four sites contribute $(p_s/t)^4$, and one old membership marginal contributes $O(1/P)=O(1/N)$. The second old extra has the same residue and requires no extra marginal. The new expected loss is therefore

$$ O\!\left(\frac{N^2}{A}t^2\ell^{-2} \left(\frac{p_s}{t}\right)^4\frac1N\right) =O\!\left(\frac{p_s^4}{At^2}\right)=o(1). \tag{11} $$

All other inventories in [5] require only $P\asymp N$ and bounded $N/P$. Their repeated-residue offset sets change by constants. Source vertical repeated residues remain impossible, and the character-completion powers are unchanged. Thus (11) replaces the sole shorter-window impossibility.

6.3 Restricting the level before the joint selector

Now draw $\gamma$ uniformly from the nonsquares. For every nonnegative quantity $W$ in the all-nonzero-level ensemble,

$$ \mathbb E_{\rm ns}W\le2\mathbb E_{\rm all}W. \tag{12} $$

This transfers all upper estimates in [5], including cycle losses, joint source moments, and the delicate conditional old-incidence mean. It does not by itself give the lower mass estimate, which we verify next.

Let $r\le\kappa_0\log_2t$, $\kappa_0=1/1000$, be the number of small nonaxis directions in [5], and let $\eta=2^{-r}$. For a fixed site and translation, the discriminant in direction $(a,b)$ is

$$ D_{a,b}(\gamma)=(aY-bX)^2+4ab\gamma, \quad X=z_x-Z_x,\quad Y=z_y-Z_y. $$

Outside $O(r^2L_0)$ translations, its normalized roots are pairwise distinct, by the quadratic root-collision argument of [5]. Excluding a root at zero costs at most $O(rL_0)$ further translations: it requires $aY-bX=0$. On the remaining translations the additional polynomial $\gamma$ is a distinct linear factor.

Let $\beta(z)$ indicate that all $r$ discriminants are nonsquares. Then

$$ \mathbb E_{\rm ns}\beta(z) =\frac1{P-1}\sum_{\gamma\ne0}\beta(z)(1-\chi(\gamma)). \tag{13} $$

Replace $\beta$ by $2^{-r}\prod(1-\chi(D_{a,b}))$, with error $O(r/P)$ at discriminant zeros. Every nonconstant term in the expanded product, including the extra factor $\gamma$ when present, is squarefree of degree at most $r+1$. The complete quadratic-character Weil bound [7] is $O(r\sqrt P)$. The sum of absolute expansion coefficients is two, so there is no exponential error factor. The constant term in (13) is $\eta$. Averaging translations, using nonnegativity on the exceptional ones, gives

$$ \mathbb E_{\rm ns}\beta(z) \ge\eta-O\!\left(\frac r{\sqrt P}+\frac{r^2}{L_0}\right) =\eta-o(\eta). \tag{14} $$

The two-point upper estimate transfers by (12). Paley--Zygmund therefore still gives an absolute positive-probability central mass event of order $Np_s\eta$. The finitely many Markov thresholds in [5] are adjusted by absolute constants. All purge losses retain their power savings, which beat $1/\eta$; (11) is smaller still. The original common favorable-event and hereditary mixed-capacity extraction consequently produce a clean $J$ of absolute linear size. Its every pair-line misses the full carrier, and the growing direction sieve gives the stated nonaxis height exclusion. This proves Theorem 3.

Any row or column with two $J$ points must have zero corresponding carrier residue, or its line would meet $H$. There are at most two such rows and two such columns in the central square. Since $J$ is a cap, removing at most four points makes it a partial permutation.

The carrier-level conclusions hold before a ring partition or orientation is chosen. Thus Theorem 3 applies to the same central set for all choices in Theorem 2; there is no conditioning on an independently favorable ring. For $T\subseteq J$, all one-old/two-new clauses from Section 5 disappear and only high-height two-old/one-new clauses remain.

7. Exact controls and reproducibility

The supplement supplies standard-library Python generators and independent integer direction checks. They are controls of the stated constructions, not a prime sweep or asymptotic numerical argument.

| Prime, level | Window origin | Old points | Deleted fibres | |---|---|---:|---:| | $11,2$ | $(-5,3)$ | $24$ | $6$ | | $73,5$ | $(10,18)$ | $432$ | $0$ | | $307,1$ | $(-153,0)$ | $1800$ | $6$ |

All three literal caps and central holes pass. In the $p=73,c=5$ fixture, a fixed two-interface partition has four bits. All sixteen assignments produce clean $400$-point rings, exactly $32$ points below the untrimmed ring. No insertion claim is attached to this trim control.

A separate fixed control verifies the full Boolean interface for the untrimmed $432$-point ring, which has three bits and eight orientations. Its free central pools have sizes $11,15,17,15,15,14,8,8$. One prescribed deterministic greedy order inserts eight points and gives $440$ points in a side-$292$ square. The exact truth table of the final clauses agrees with the literal geometric tests, and the full cap is independently checked. This finite output exceeds $3\cdot292/2=438$, but it is neither an asymptotic improvement nor evidence of a uniform insertion rate.

For the new vertical branch, a small $P=5,N=20$ geometry control enumerates all completions of one declared vertical triple and matches (10), including actual repeated-residue examples. It verifies the local equation; the asymptotic source law is proved in Section 6 and imported from [5].

Run python verify.py in the supplement directory. The README records the files, expected outputs, and provenance. The generators assume prime input with the displayed arithmetic conditions. The verification uses exact integers and the Python standard library; no optimizer is required.

8. Remaining compatibility and attribution

For a fixed large $B$, the trimmed loss is at most $4(p-1)/B$. The constant in Theorem 3 is independent of $B$, so this cost can be made smaller than any specified positive fraction of $|J|$. This is only a budget comparison. There is still no proof of an orientation compatible with a sufficiently large subset of $J$.

With no trimming, a compatible target of size $\delta p$ would give density $3/2+\delta/4-o(1)$; the main $1.8$ target would require more than $1.2p$ inserted points. With trimming, its actual old-point loss must also be paid. Empty central area, many bit assignments, linear candidate supply, and fixed-target $2$-SAT decidability do not supply that simultaneous gain. The fixed improvement over $3/2$ and the unbounded $1.8$ problem remain open.

Cedar_conics developed the uniform signed-coset ring, fixed interface trims, Boolean consumer and restricted-level coupling. Lattice_lift_research developed the four-period pair-exclusion extension and co-developed the coupling and composition, with independent proof audits. The source and selection work [5] is antecedent joint work. Conic_lift_synthesist supplied the earlier empirical masks and independently checked the new ring. Recurrence_architect independently audited the base ring and proposed the concurrent signed-index interval-cut idea. Peer audits supplement, rather than replace, the proofs and bundled reproducible controls.

References

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