\chapter{Worked calibration architectures and failure certificates}
\label{ch:examples}

This chapter follows the exact examples distributed with the release. Every
reported certificate value is taken from the executed rational checks in
\path{results/robust_results.json} or from an accompanying planner
certificate. The derivations explain why those values occur, so that the
examples can also be checked without executing the software. The final
conditioning example is an analytic construction, explicitly separate from
the computational test counts.

Vertices and channel indices in the JSON files start at zero. A channel
index is distinct from the coordinate it measures: channel zero may measure
vertex one, for example. We keep this indexing when describing particular
failure sets. The unknown $Q$ is the inverse-gauge parameter obtained from
the aligned discrepancy $DB_0-C=CQ$. All the examples concern that linear
contract. The theorem connecting a sufficiently small $Q$ to a strict
causal alternative is separate from the finite decoding tests.

\section{A complete four-vertex graph: six parameters and three noiseless channels}

The first graph has vertices $0,1,2,3$ and every edge $i\to j$ with $i<j$.
There are six edges. A parent $j$ surrounds a later vertex $i$ because every
child of $i$ is also a child of $j$. The ambiguity supports are consequently
\[
 I_0=\{1,2,3\},\qquad I_1=\{2,3\},\qquad
 I_2=\{3\},\qquad I_3=\varnothing.
\]
There are no installed trusted rows in this example. The residual spaces
are therefore the full coordinate spaces on these supports, and their
dimensions are $(3,2,1,0)$. A general parameter matrix is
\[
 Q=\begin{pmatrix}
 0&0&0&0\\
 a&0&0&0\\
 b&d&0&0\\
 c&e&f&0
 \end{pmatrix}.
\]
The six symbols are independent algebraic parameters. Nevertheless the
noiseless dense-channel minimum is three, because a common three-row
measurement matrix can identify each of its labeled columns. Six unknown
scalar entries do not require six independently designed channels when the
same channel supplies the full labeled operator row.

For one known erasure and one arbitrary corrupted channel, the relevant
fault burden is $1+2=3$. The dense frontier therefore requires
$3+1+2=6$ channels. The test constructs the positive moment rows
\begin{equation}\label{examples:moment}
 C_6=\begin{pmatrix}
 1&1&1&1\\
 1&2&4&8\\
 1&3&9&27\\
 1&4&16&64\\
 1&5&25&125\\
 1&6&36&216
 \end{pmatrix}.
\end{equation}
The first column of $C_6$ does not affect these residual spaces, since the
root coordinate is zero in every residual vector. It remains part of the
known physical readout map and is included in the certificate.

A nonzero vector in $E_0$ has the form $(0,a,b,c)$. Its channel value at
parameter $t$ is $at+bt^2+ct^3=t(a+bt+ct^2)$. At the six distinct positive
parameters, it can vanish at no more than two positions. Hence its measured
support has at least four entries. A vector in $E_1$ gives
$t^2(d+et)$ and has at most one zero, while a nonzero vector in $E_2$ gives
$ft^3$ and has none. The three distances are exactly $4,5,6$.

More generally, every surviving set of $h$ channels has restricted rank
$\min(h,d_j)$ on column $j$. The complete severity profiles are
\[
 (D_{0,1},D_{0,2},D_{0,3})=(4,5,6),\qquad
 (D_{1,1},D_{1,2})=(5,6),\qquad D_{2,1}=6.
\]
The fourth column has an empty profile because it has no ambiguity.
Erasing three channels leaves every column identifiable. Erasing four
leaves one possible ambiguity dimension; erasing five leaves three in
total; erasing all six restores all six. The exact total dimension as a
function of the number $e$ of erased channels is thus
\[
\begin{array}{c|rrrrrrr}
 e&0&1&2&3&4&5&6\\\hline
 \text{unresolved dimension}&0&0&0&0&1&3&6.
\end{array}
\]
Here the value depends only on the number erased, because every restricted
submatrix has the strongest rank compatible with its size. This dependence
on cardinality alone is a property of this design, not a general property
of constrained architectures.

\section{Reading a concrete certificate rather than a genericity claim}

The moment construction examines the residual spaces left by subsets of
previously selected rows and avoids a row that annihilates any active one.
Its executed trace is
\[
\begin{array}{c|rrrrrr}
 \text{step}&1&2&3&4&5&6\\\hline
 \text{active families}&3&5&8&12&17&23\\
 \text{search bound}&10&16&25&37&52&70\\
 \text{chosen parameter}&1&2&3&4&5&6.
\end{array}
\]
The bound is three times the number of active families plus one, reflecting
the maximum polynomial degree in four ambient coordinates. The fact that
small parameters succeed here does not replace the finite search argument
for arbitrary residual spaces.

The planner independently searches the same six-channel architecture. Its
certificate contains a different positive integer matrix:
\[
 C_{\mathrm{plan}}=\begin{pmatrix}
 5&1&5&2\\
 1&1&5&1\\
 3&2&3&3\\
 1&2&1&3\\
 5&2&5&4\\
 3&3&4&2
 \end{pmatrix}.
\]
This matrix passed after sixteen search attempts. The certificate checks
$3\cdot2^6=192$ restricted ranks, one for each subset and each nonzero
column space. Its profiles agree with those of~\eqref{examples:moment}.
The two files therefore demonstrate two valid designs, rather than two
different descriptions of the same numerical matrix.

For example, the planner supplies $v=(0,-5,1,0)\in E_0$ as a first-failure
witness. Direct multiplication gives
\[
 C_{\mathrm{plan}}v=(0,0,-7,-9,-5,-11)^T.
\]
The first two surviving rows do not see it, and deleting rows $2,3,4,5$
makes it invisible. This proves that the distance cannot exceed four.
The all-subset rank certificate proves the matching lower bound. A
witness alone establishes an upper bound; a list of successful subsets
alone needs complete coverage to establish a universal lower bound.

The finite decoding experiment uses three exact vectors,
\[
 x_0=(0,1/3,-2,5/7),\qquad
 x_1=(0,0,2,-1/2),\qquad x_2=(0,0,0,-3/5).
\]
It checks no erasure or one erasure, no error or one error on a retained
channel, and base error amplitudes $-7/3,1,5/2$. The amplitude is multiplied
by $j+1$ for column $j$. There are $19$ scenarios without erasure and
$6\cdot16=96$ with one erasure, giving $115$ scenarios and $345$ successful
column decodes. These are finite implementation checks using rational
arithmetic. They are not empirical accuracy estimates and do not sample
the full continuum of possible errors; the theorem supplies that universal
claim.

\section{Five channels: an explicit pair of indistinguishable observations}

Delete the final row of $C_6$, giving five moment rows. Its root-column
distance is three. That is sufficient for one unknown corrupted channel
alone, but insufficient for one corruption together with one erasure.
The difference is visible in an exact two-model certificate.

Take
\[
 v=(0,2,-3,1),\qquad C_5v=(0,0,6,24,60)^T.
\]
The formula follows from $2t-3t^2+t^3=t(t-1)(t-2)$. Erase channel $2$,
whose parameter is $t=3$, and supply the retained observations
\[
 y=(0,0,\text{erased},24,0)^T.
\]
They have two explanations under exactly the requested budget:
\begin{center}
\begin{tabular}{@{}lll@{}}
\toprule
Parameter & Clean encoded values & Single corrupted channel\\
\midrule
$0$ & $(0,0,0,0,0)$ & channel $3$, error $24$\\
$v$ & $(0,0,6,24,60)$ & channel $4$, error $-60$\\
\bottomrule
\end{tabular}
\end{center}
The erased channel is common to both explanations. A decoder that returns
one of these parameters without reporting the ambiguity makes an unjustified
choice. The supplied exact decoder rejects this input because it does not
have exactly one identifiable candidate.

The example also supplies a local causal impossibility witness. Put $v$ in
column zero of a matrix $\Delta Q$, scale it by a sufficiently small
nonzero $\lambda$, and use $U=(I+\lambda\Delta Q)^{-1}$. The scaled pair
remains a pair of strict admissible models under the finite-context
hypotheses, and the same support pattern of erased and corrupted channels
still makes them indistinguishable. Large error magnitudes in the displayed
integer example are therefore not responsible for the impossibility.

The separate five-channel planner certificate reaches the same architectural
distance three with a different numerical matrix. It reports that the
requested tolerance is impossible even though its chosen design is
certified optimal for that architecture. An optimal design can still be
insufficient for a requested fault budget.

\section{A five-vertex star whose distance is three}

Take a root $0$ with four leaves and no other edges. Its sole nonzero
ambiguity column is the root column. Install the two trusted rows
$[0\ A]$, where
\[
 A=\begin{pmatrix}1&0&1&1\\0&1&1&-1\end{pmatrix}.
\]
The residual leaf vector is in $\ker A$ and has the parametrization
\[
 x=(-u-v,-u+v,u,v).
\]
The basis recorded by the certificate is
$(-1,-1,1,0)$ and $(-1,1,0,1)$, with an initial zero added for the root
coordinate. The residual dimension is two although the leaf support has
four coordinates.

Every pair of columns of $A$ is independent: their $2\times2$ determinants
are nonzero. Some three columns are dependent, since $A$ has rank two.
Thus the smallest support of a nonzero vector in its kernel is three.
The vector $(-1,-1,1,0)$ is an explicit witness. Dedicated leaf sensors
therefore have distance three and profile $(3,4)$.

The planner chooses gains $(1,2,1,2)$ on the four dedicated channels.
Their readings are
\[
 (-u-v,-2u+2v,u,2v).
\]
Nonzero gains cannot change which channel values are zero, so they preserve
the distance. In the certificate, deleting channels $1,2,3$ leaves channel
zero alone and allows the vector $(0,2,-1,1)$ on the leaves. Deleting all
four channels releases both dimensions. Conversely, every two retained
channels identify the two-dimensional kernel, which is why one arbitrary
corrupted channel is correctable.

This is a small instance of the star realization used in the complexity
theorem. Its graph is simple, but the trusted calibration matrix carries
nontrivial linear dependence information. The simplicity of the causal
graph does not make that information disappear.

\section{Doubling support: distance six with two residual parameters}

Replace $A$ by
\[
 A'=\begin{pmatrix}A&0\\I_4&-I_4\end{pmatrix}.
\]
The associated star now has eight leaves and nine vertices. The lower four
equations force its leaf vector to have the form $(x,x)$; the upper two
equations require $Ax=0$. Hence
\[
 \ker A'=\{(x,x):x\in\ker A\}.
\]
Its dimension stays two. Every nonzero support is duplicated, so the first
distance doubles from three to six. The union of the supports of the full
two-dimensional kernel is all eight coordinates, giving profile $(6,8)$.

An explicit minimal-support witness is
\[
 (-1,-1,1,0,-1,-1,1,0).
\]
The numerical certificate also uses the vector
$(0,2,-1,1,0,2,-1,1)$, which vanishes on channels $0$ and $4$ and is
nonzero on the other six. These two surviving channels read the same
underlying coordinate from the two copies. They impose only one
independent condition on the two-dimensional residual space.

The architecture therefore corrects two arbitrary corrupted channels and
tolerates one known erasure: its distance six exceeds $1+2\cdot2=5$.
It cannot achieve distance seven merely by changing its dedicated sensor
gains. The two-copy restriction fixes a six-channel support obstruction.
With seven freely designed dense channels, the same residual dimension
would suffice for this fault budget. This one-channel difference measures
the specified architectural restriction; it does not compare acquisition
costs or physical reliability of the two systems.

The jump from six to eight in the profile is also informative. Six erased
channels can hide one direction, but there is no seven-channel threshold
for releasing the second: at least one surviving dedicated coordinate
always imposes a nonzero functional on this kernel. All eight must be
erased to leave both parameters completely free.

\section{The same zero pattern can have different fault tolerance}

The next pair uses the same four-leaf star and the same dedicated sensors.
Only the trusted coefficients change:
\[
 A_{\mathrm g}=\begin{pmatrix}1&1&1&1\\1&2&3&4\end{pmatrix},
 \qquad
 A_{\mathrm e}=\begin{pmatrix}1&1&1&1\\1&1&3&4\end{pmatrix}.
\]
Both matrices have rank two, both have no zero entries, and both leave a
two-dimensional residual space. Their support diagrams are identical.
For $A_{\mathrm g}$, any two columns have distinct lower entries and thus
a nonzero determinant. Its spark and calibration distance are three, with
profile $(3,4)$. Its witness $(1,-2,1,0)$ is the first basis vector in the
certificate.

For $A_{\mathrm e}$, the first two columns coincide. The vector
$(-1,1,0,0)$ is now in the kernel, so the distance falls to two and the
profile becomes $(2,4)$. The certificate's architectural cut is the pair of
channels $T=\{2,3\}$. Their restricted row spaces have total dimension one;
the cut calculation gives $4-2+1-2+1=2$.

Both planner designs use the same sensor gains $(1,2,1,2)$. The exceptional
witness then has readings $(-1,2,0,0)$. The observation $(-1,0,0,0)$ is
explained either by the zero parameter with an error in channel zero, or
by this witness with an error in channel one. Thus the exceptional case
cannot correct every single-channel corruption. The generic-coefficient
case can. The distinction is a numerical calibration dependency, and a
graph-only support test cannot certify it away.

\section{A direction that several dense-looking sensors cannot see}

The exhaustive architecture test contains the three row spaces
\[
 H_0=H_1=\operatorname{span}\{(1,1)\},\qquad
 H_2=\operatorname{span}\{(1,-1)\}.
\]
Its common certified design is
\[
 C=\begin{pmatrix}2&2\\1&1\\2&-2\end{pmatrix}.
\]
Every row has both coordinates nonzero. Nevertheless the first two channels
measure the same direction. On $E=\mathbb R^2$ the readings of $(a,b)$ are
$(2(a+b),a+b,2(a-b))$. The vector $(1,-1)$ is seen only by channel two,
with readings $(0,0,4)$, so the architecture has distance one and profile
$(1,3)$. Losing that single channel leaves a one-dimensional ambiguity.

On the residual line $\operatorname{span}\{(1,1)\}$, the same matrix has
distance two; on $\operatorname{span}\{(1,-1)\}$ it has distance one. A
sensor can therefore be informative for one residual column and invisible
for another. The data needed for a common certificate are the restricted
row spaces $H_a|_{E_j}$, not merely the ambient support of $H_a$.

For comparison, the test with three unrestricted row spaces supplies
\[
 C_{\mathrm{free}}=\begin{pmatrix}1&4\\3&1\\4&3\end{pmatrix}.
\]
Its three pairwise determinants are $-11,-13,5$, so every pair is
independent on $\mathbb R^2$. Its full-space profile is $(2,3)$. It also
has nonzero values on all three channels for each of the four lines spanned
by $(1,0),(0,1),(1,1),(1,-1)$. It is consequently optimal for all five
spaces simultaneously. A full-space certificate by itself would not have
checked these four additional restrictions.

\section{What the finite architecture enumeration establishes}

The six permitted row-space types in the exhaustive test are zero, the two
coordinate lines, the sum line, the difference line, and the full plane.
Three indexed channels give $6^3=216$ architectures. The prescribed finite
row choices yield $729$ total candidate assignments. Each is checked on all
eight surviving subsets by an independent two-dimensional determinant
oracle, giving $5{,}832$ subset-rank comparisons. Of those assignments,
$449$ attain every predicted subset rank for their architecture.

The additional common-design test uses the full plane and the four lines
just listed. It finds a shared design for all $216$ architectures and checks
$216\cdot5\cdot8=8{,}640$ restricted subset ranks. The largest observed
search-attempt count is eleven. These counts describe the complete bounded
test family, not all possible real row spaces or all matrix sizes. In
particular, the attempt count is not a performance guarantee for the
planner on larger inputs.

The empty cases also have different meanings. If every residual column
space is zero, no calibration parameter remains, and no new channel is
required. Its distance minimum is over an empty family; the certificates
use an empty profile and a null distance entry for such a column. This is
why the zero-dimensional fourth column of the complete graph does not
reduce its overall distance to zero.

In contrast, if $E=\mathbb R^2$ remains unresolved but all three allowed
row spaces are zero, every measurement is zero. The architecture test
correctly returns profile $(0,0)$. Both ambiguity dimensions are already
invisible without any channel failure. An empty parameter space and an
uninformative measurement system must not be merged into the same
software case.

\section{A decoder must reject a feasible family of alternatives}

One regression check takes $E=\mathbb R^2$, $C=I_2$, observation $y=0$,
and allowance for one corrupted channel. The zero parameter is a valid
explanation, but it is not unique: every vector $(a,0)$ can be hidden by
an error in the first channel, and every vector $(0,b)$ by an error in
the second. A decoder that examines only full-rank candidate clean sets
can accidentally retain the zero explanation and skip these
rank-deficient feasible families.

The implementation instead tests consistency first. If a proposed clean
set is consistent but rank deficient, it reports a non-identifiable
feasible family. That behavior is exercised by this regression case. The
remaining input checks reject a negative error budget, a wrong observation
length, and an out-of-range erasure index, while ordinary zero-error
decoding of $(3/2,-7/4)$ is also checked. These are concrete guards against
false recovery claims, rather than additional mathematical assumptions.

\section{Returning to the mixed calibration in the supplied package}

The original package contains a complete five-vertex graph with three
different trusted resources. Its known readout is $(1,2,3,4,5)$, its known
baseline row is the incoming mechanism at target two, and its calibrated
nonunit gain acts at target three. Their combined left-constraint matrix is
\[
 M=\begin{pmatrix}
 1&2&3&4&5\\
 -1&-1&1&0&0\\
 0&0&0&1&0
 \end{pmatrix}.
\]
For the root-column support $I_0=\{1,2,3,4\}$, write the residual vector
as $(0,x_1,x_2,x_3,x_4)$. The baseline equation gives $x_2=x_1$; the
gain equation gives $x_3=0$; and the readout equation then gives
$5x_1+5x_4=0$. Therefore
\[
 E_0=\operatorname{span}\{(0,1,1,0,-1)\}.
\]
For the next column, coordinate one is unavailable. The same baseline
equation forces coordinate two to zero, the gain removes coordinate
three, and the readout removes coordinate four. Thus all later residual
spaces vanish, yielding dimensions $(1,0,0,0,0)$ as in the input
certificate. Its recorded kernel vector is the negative of the displayed
basis vector, which describes the same line.

The sum row $c=(1,1,1,1,1)$ evaluates to one on the displayed basis. It
identifies the scalar residual parameter $\lambda$ without faults.
Three separately indexed copies return $(\lambda,\lambda,\lambda)$ and
correct one arbitrary corrupted channel; the median is an exact decoder
in that noiseless one-error case. Four copies have distance four and
also tolerate the combined budget of one erasure and one corruption.
This is the dense frontier with $d_*=1$, whereas the complete four-vertex
example without the installed calibrations has $d_*=3$ and requires six
channels for the same combined budget.

The comparison accounts only for the new channels. The trusted mechanism,
gain, and readout used to reach the one-dimensional space are additional
resources and have not been assigned zero physical cost. It illustrates
how the residual spaces summarize their mathematical effect, allowing the
robust channel calculation to begin after those resources have been
specified.

This example also makes the inverse-coordinate step especially transparent.
If $Q=\lambda w e_0^T$ with $w=(0,1,1,0,-1)^T$, then $Q^2=0$ because
$e_0^Tw=0$. Therefore $(I+Q)^{-1}=I-Q$ exactly. Recovering $\lambda$ from
the calibrated readings determines the forward transformation without an
iterative matrix inversion. The finite strict-support conditions still
restrict how large $\lambda$ may be for a particular reference model;
the nilpotent identity itself is valid for every real $\lambda$.

\section{Positive coefficients do not supply a uniform noise margin}

Consider the analytic family
\[
 C_\epsilon=\begin{pmatrix}
 1&1\\1&1+\epsilon\\1&1+2\epsilon
 \end{pmatrix},\qquad\epsilon>0,
 \qquad E=\mathbb R^2.
\]
All coefficients are positive. Every pair of rows is independent, so each
matrix has distance two and tolerates one arbitrary erasure. These exact
integer-valued properties do not deteriorate as $\epsilon$ approaches zero.
The rows remain bounded and their norms approach $\sqrt2$, so no vanishing
global gain is being used to manufacture the example.

However, for the unit vector $w=(1,-1)/\sqrt2$,
\[
 C_\epsilon w=(0,-\epsilon/\sqrt2,-2\epsilon/\sqrt2)^T.
\]
The full-map minimum gain is at most $\epsilon\sqrt{5/2}$. After erasing
the last channel, the gain on this direction is only
$\epsilon/\sqrt2$. Thus the one-erasure robust singular-value margin is
at most that quantity and tends to zero. Small observation error can
produce arbitrarily large relative parameter uncertainty even though
every matrix passes the exact distance certificate.

This final example separates three reportable statements. A positive row
construction establishes a sign constraint. An all-subset rational rank
certificate establishes exact fault tolerance. A numerical lower bound on
the relevant restricted singular values establishes a quantitative noise
margin. Each answers a different question, and none of the first two
should be presented as evidence for the third.

\section{Figures derived from the executed examples}
\begin{figure}[htbp]
\centering
\includegraphics[width=0.95\textwidth]{../figures/complete_four_failure_profiles.pdf}
\caption{Exact deletion thresholds for the three nonzero residual columns of the complete four-node example. The plotted data are the rationally verified integer profiles. The curves describe individual columns; a common deletion can affect several columns simultaneously.}
\end{figure}
\begin{figure}[htbp]
\centering
\includegraphics[width=0.95\textwidth]{../figures/same_support_coefficient_dependence.pdf}
\caption{The trusted calibration matrices have identical nonzero-entry patterns and equal rank, but different dedicated-sensor distances. The difference comes from an exact numerical column dependency.}
\end{figure}
\begin{figure}[htbp]
\centering
\includegraphics[width=0.95\textwidth]{../figures/normalized_conditioning_margin.pdf}
\caption{Floating-point illustration of a proved conditioning obstruction: five strictly positive convex-average rows have exact distance four for every positive separation parameter, while the worst retained two-row singular-value margin tends to zero. Numerical values in this plot are illustrative rather than exact rank certificates.}
\end{figure}