\chapter{A depth-one boundary for exact robustness}\label{ch:complexity} \section{A simple graph can contain a hard numerical question} The noiseless channel count depends only on the dimensions of the residual spaces. Robustness depends on how those spaces meet particular coordinate subspaces and sensor kernels. That extra geometry can encode a difficult computational problem even when the causal graph itself is elementary. This chapter constructs that encoding explicitly. Every rational linear kernel can occur as the only nonzero ambiguity column of a star. Dedicated sensors on the leaves then measure that kernel coordinate by coordinate. A sparse kernel vector becomes a small set of channels on which two distinct causal representatives disagree. Exact robustness is therefore exactly a minimum-support question. The reduction is different from the supplied depth-three Set Cover construction. That construction proves hardness of choosing which trusted baseline rows to calibrate. The present construction fixes the trusted rows and the sensor architecture, then asks whether the design can tolerate every fault pattern of a given size. Choosing a design and certifying a fixed design are different computational tasks and have different decision directions. \section{Rational spark and the decision direction} For a rational matrix $A\in\mathbb Q^{p\times n}$ with nonzero kernel, define \[ \spark(A)=\min\{\|x\|_0:0\ne x\in\ker A\}. \] Here $\|x\|_0$ denotes the number of nonzero coordinates. If the kernel is zero, use $+\infty$ in this chapter. For a nontrivial kernel the spark equals the smallest number of linearly dependent columns of $A$. The rational decision problem asking whether $\spark(A)\le k$ is NP-complete; the corresponding optimization is NP-hard \cite{tillmann}. The proof of that foundational complexity result is not reproduced as a new result here. We give a polynomial reduction from it to causal robustness. The word ``whether'' must not obscure the polarity. A small-support null vector is a certificate of failure. It is not a certificate of robustness. Uniform robustness asks whether no such vector exists and therefore takes the complementary decision direction. The resulting positive robustness problem is coNP-complete, not NP-complete, unless the usual complexity classes collapse. For rational input, a failure certificate can be only a subset of columns. Given $S\subseteq[n]$ with $|S|\le k$, exact Gaussian elimination checks whether $A[:,S]$ has rank below $|S|$. A rational nonzero null vector can also be computed and has a representation of polynomial bit length using determinant bounds. Thus no infinite real-number witness is required by the reduction. \section{Realizing every kernel by a causal star} \begin{theorem}[Universal rational-kernel realization]\label{hard:star} Let $A\in\mathbb Q^{p\times n}$ be arbitrary. There is a normalized rich-action latent causal model on a depth-one star with $n+1$ vertices, all baseline edge weights equal to one, and trusted readout matrix \[ M_0=\begin{pmatrix}0&A\end{pmatrix} \] whose only possibly nonzero residual column is \[ E_0=\{(0,x)^T:x\in\ker A\}. \] Every other residual column is zero. With dedicated leaf readouts \[ C=\begin{pmatrix}0&I_n\end{pmatrix}, \] the minimum channel distance is $\spark(A)$ when $\ker A\ne0$. The same distance is the maximum achievable distance for the architecture $H_i=\operatorname{span}\{e_i^T\}$, because nonzero row scaling cannot change support. \end{theorem} \begin{proof} Use vertices $0,1,\ldots,n$ with edges $0\to i$ for every leaf $i$. Every leaf is a sink, so the root surrounds every leaf. No leaf has any child or any outgoing edge. Thus $I_0=\{1,\ldots,n\}$ and $I_j=\varnothing$ for $j>0$. The trusted equations for the root ambiguity column are $Ax=0$. There are no other possible columns. This proves the residual-space assertion. The dedicated readout sends $(0,x)^T$ to $x$, so its smallest nonzero response support is exactly $\spark(A)$. Each admissible nonzero scaling of a dedicated row leaves that support unchanged, whereas setting a row to zero can only decrease distance. To realize the inherited aligned model concretely, take every edge weight to be one and take one old uncalibrated observation row $F_0=(1,\ldots,1)$. Since $\Lambda_0^2=0$, its aligned innovation response is \[ A_0=F_0(I+\Lambda_0)=(n+1,1,\ldots,1). \] Every innovation column is nonzero even though there is only one old observed channel. At leaf $i$, choose a nonzero incoming intervention proportional to $e_0^T$. This spans its one-dimensional parent space and preserves strict edge support for, for example, a positive unit increment. Finally, each residual matrix has only column zero and no root-row entry, so its square is zero. For sufficiently small $Q$, the transformation $(I+Q)^{-1}=I-Q$ gives a genuine strict alternative. The algebraic sparse-support witness is therefore realized by compatible causal models arbitrarily close to the reference. \end{proof} The trusted matrix $M_0$ is allowed to contain arbitrary rational coefficients. It is not claimed to be a free resource obtained from the uncalibrated data. The theorem says that even after those numerical calibrations have been supplied, the resulting robustness question can be hard. The reference edge weights are all one; uncalibrated compatible candidates may have other nonzero edge weights. \section{Sharp certification complexity} \begin{theorem}[Robustness certification on a star]\label{hard:complexity} Consider rational instances of the normalized calibration problem with a specified sensor matrix. The following decision problem is coNP-complete, already on the stars in Theorem~\ref{hard:star}: \emph{Does the calibrated model remain identifiable after every deletion of at most $k$ new sensor channels?} Its complement, the existence of a harmful deletion of at most $k$ channels, is NP-complete. Computing the exact worst-case distance is NP-hard. The same coNP-completeness holds for the question of correcting every pattern of at most $t$ arbitrary corrupted channels, with $t$ part of the input. \end{theorem} \begin{proof} For the star, harmlessness of every deletion of at most $k$ channels is equivalent to $\spark(A)>k$. A harmful deletion exists exactly when a nonzero $x\in\ker A$ is supported on the deleted channels. The mapping from $A$ and $k$ to the graph, trusted matrix, and dedicated sensors is polynomial in the rational input size. The complement is in NP: provide a column index and a deleted sensor set of size at most $k$. Exact rank checking tests whether the retained sensing restriction has nonzero kernel. The rational spark theorem gives NP-hardness of that complement and therefore coNP-hardness of the stated universal problem. An exact distance algorithm would decide the same spark problem by comparing its output with $k$. For error-only correction, distance must exceed $2t$, and an arbitrary threshold $k$ need not be even. Remove this parity issue by replacing $A$ with \[ A^{\langle2\rangle}= \begin{pmatrix}A&0\\I_n&-I_n\end{pmatrix}. \] Its nonzero kernel vectors are exactly $(x,x)$ with $0\ne x\in\ker A$. Consequently \[ \spark(A^{\langle2\rangle})=2\spark(A). \] With $t=k$, failure to correct $t$ errors is equivalent to $\spark(A)\le k$. Membership in NP for the failure question follows by providing a response difference supported on at most $2t$ channels, or simply a retained-row rank-deficiency certificate. This proves the final assertion. \end{proof} If a promised input class requires a nonzero kernel, the reduction can enforce it. Restrict to nontrivial thresholds $1\le k\le n$ and form the block diagonal matrix \[ \widetilde A=\operatorname{diag}\left(A,\begin{pmatrix}I_k&\mathbf1\end{pmatrix}\right). \] The added block has a one-dimensional kernel with minimum support $k+1$. Hence $\widetilde A$ always has a nonzero kernel and \[ \spark(\widetilde A)\le k\quad\Longleftrightarrow\quad\spark(A)\le k. \] The padding size is polynomial because $k\le n$. This avoids an unstated promise in the complexity argument. \section{Architecture feasibility has the same barrier} The cut formula gives a short certificate when no choice of allowed channel weights can meet a specified fault target. For rational bases of $E_j$ and $H_a$, a candidate pair $(j,T)$ permits exact evaluation of \[ m-|T|+r_j(T)-d_j+1. \] If $r_j(T)