\section{Exact geometry without nerve-face enumeration} \label{sec:geometry} The duality theorem is useful only if the maximal nerve faces can be obtained reliably. For disks, a finite arrangement construction avoids listing every intersecting subfamily. In this section \[ D_i=\{x\in\R^2:\|x-c_i\|\le r_i\},\qquad r_i\ge0, \] and coincident disks keep their distinct contact labels. \begin{definition} The \emph{witness list} consists of the leftmost point $\ell_i=(c_{i,x}-r_i,c_{i,y})$ of each disk and all intersections of the boundaries of pairs of disks whose supporting circles are distinct. A zero-radius disk contributes its center through $\ell_i$. A coincident pair contributes no additional points. The label of a witness $x$ is $L(x)=\{i:x\in D_i\}$. \end{definition} \begin{theorem}[Quadratic witness bound]\label{thm:witness} Every maximal face of the nerve is the label of a point on the witness list. Hence the inclusion-maximal nonempty witness labels are exactly $\Max N$, and \begin{equation}\label{eq:witness-count} \abs{\Max N}\le n+2\binom n2=n^2. \end{equation} The assertion allows unequal radii, tangency, zero radii and coincident disks. \end{theorem} \begin{proof} Every witness label is a nerve face, because the witness itself lies in all its labeled disks. Let $\sigma$ be a maximal nerve face and set $K=\bigcap_{i\in\sigma}D_i$. This is nonempty and compact. Choose a point $x\in K$ with minimal first coordinate. If $\sigma$ contains a zero-radius disk, then $K$ consists of its center, which is a listed witness. Suppose all its disks have positive radius. At least one constraint is active at $x$: if all were strict, a small open neighborhood of $x$ would lie in $K$, contradicting minimality. If two active constraints have distinct supporting circles, $x$ is one of their boundary intersections and is listed. The remaining case has only one distinct active supporting circle, possibly repeated with several labels. Every other disk of $\sigma$ contains a neighborhood of $x$. In a sufficiently small neighborhood the only effective constraint is therefore the closed disk bounded by that active circle. Unless $x$ is its leftmost point, there are points of that disk arbitrarily close to $x$ with smaller first coordinate. They still satisfy every locally slack constraint, a contradiction. Thus $x$ is the leftmost point of an active disk and is listed. In all cases $\sigma\subseteq L(x)$. The label is a nerve face and $\sigma$ is maximal, so equality follows. Taking maximal labels removes only proper subfaces and duplicates. Each noncoincident pair of circles has at most two intersection points, which proves \cref{eq:witness-count}. \end{proof} \begin{remark} The extremal-point argument is an elementary disk-arrangement method. The bound concerns the number of maximal nerve faces, not the number of all nerve faces or all simplices of the dual complex. A single maximal face on $n$ labels contains $2^n-1$ nonempty nerve faces. Likewise $n^2$ dual vertices alone do not give a quadratic bound on all dual simplices. \end{remark} \subsection{Circle intersections in a quadratic extension} The witness labels can be determined without evaluating square roots approximately. Suppose $c_i\ne c_j$ and put \begin{align} d&=c_j-c_i,\qquad s=d\cdot d>0,\nonumber\\ \alpha&=\frac{s+r_i^2-r_j^2}{2s},\qquad h=\frac{r_i^2}{s}-\alpha^2,\label{eq:circle-parameters}\\ B&=c_i+\alpha d,\qquad P=(-d_y,d_x).\nonumber \end{align} If $h<0$, the boundaries do not meet. If $h=0$, they meet at $B$. If $h>0$, their intersections are \begin{equation}\label{eq:circle-points} x_{\pm}=B\pm\sqrt h\,P. \end{equation} Indeed, subtracting the two circle equations gives the line $(x-c_i)\cdot d=\alpha s$. Substituting $x=B+tP$ in the first equation gives $t^2=h$. This also covers internal and external tangency. For a third disk, write $z=B-c_k$. Membership of $x_\pm$ is the inequality \begin{equation}\label{eq:quadratic-membership} A+C\sqrt h\le0, \quad A=z\cdot z+h(P\cdot P)-r_k^2, \quad C=\pm2z\cdot P. \end{equation} When input numbers are rational, $A,C,h$ are rational. No coordinates of a square-root point need be rounded. \begin{lemma}[Exact radical sign]\label{lem:sign} Let $A,C,h\in\mathbb Q$ with $h\ge0$. The truth of $A+C\sqrt h\le0$ is determined by rational comparisons as follows: \begin{enumerate} \item if $h=0$ or $C=0$, test $A\le0$; \item if $A=0$ and $h>0$, test $C\le0$; \item if $A<0$ and $C<0$, the inequality holds; if $A>0$ and $C>0$, it fails; \item if $A>0$ and $C<0$, test $A^2\le C^2h$; \item if $A<0$ and $C>0$, test $C^2h\le A^2$. \end{enumerate} \end{lemma} \begin{proof} The first three cases follow from signs. In the fourth case the desired inequality is $A\le(-C)\sqrt h$, with both sides nonnegative, so squaring preserves its direction and equivalence. In the fifth case it is $C\sqrt h\le -A$, again with nonnegative sides. These cases exhaust all possibilities. Equality is retained because the disks are closed. \end{proof} \subsection{Enumeration and chain construction} \begin{algorithm}[Exact facet computation]\label{alg:facets} \leavevmode \begin{enumerate} \item Interpret the finite coordinates and radii as rationals; reject negative radii or nonfinite input. \item Label the $n$ leftmost points by squared-distance comparisons. \item For each pair of distinct centers, compute \cref{eq:circle-parameters}. If $h\ge0$, label its one or two boundary intersections using \cref{lem:sign}. Pairs with equal centers require no boundary computation. \item Delete empty labels and duplicates. Sort the remaining masks by decreasing cardinality, and retain a mask exactly when it is not contained in a previously retained mask. \end{enumerate} The output is $\Max N$ by \cref{thm:witness}. \end{algorithm} With $m=\abs\M$ and a winning predicate $w$, the full dual two-skeleton is constructed as follows. Retain $M\in\M$ as a vertex if $w(M)$ is true. An edge $\{M_i,M_j\}$ is retained if $w(M_i\cap M_j)$ is true. A candidate triangle is retained if all three edges exist and $w(M_i\cap M_j\cap M_k)$ is true. The edge checks are redundant mathematically for an upward-closed rule, but they avoid unnecessary work and ensure a consistent boundary indexing. Each triangle gives a boundary column with three nonzero edge entries over $\FF$. Gaussian elimination determines its rank, while a union--find structure computes the graph components. \Cref{eq:betti-formula} then gives both Betti numbers. \begin{theorem}[Finite exact low-degree computation]\label{thm:algorithm} For $n$ closed planar disks with rational input data, and an exactly evaluable monotone winning predicate, \cref{alg:facets} and the complete dual two-skeleton compute $H_0$ and $H_1$ of the surviving region exactly over $\FF$. Witness labeling uses $O(n^3)$ rational sign tests. A naive maximal-label filter uses $O(n^4)$ mask-containment tests. The dual two-skeleton uses at most $n^2$ vertices, $\binom{n^2}{2}$ edge predicates and $\binom{n^2}{3}$ triangle predicates. \end{theorem} \begin{proof} There are at most $n^2$ witnesses and $n$ membership tests per witness. A quadratic comparison of at most $n^2$ labels bounds the naive filtering step. Correctness of those steps follows from \cref{thm:witness,lem:sign}. The simplex count follows from $m\le n^2$, and every simplex through dimension two is included by its defining predicate. \Cref{thm:duality,prop:two-skeleton} identify the resulting homology with that of the surviving region. \end{proof} This is an arithmetic-operation and oracle-query bound. It is not a unit-time bit-complexity bound: rational numerator sizes, arbitrary-precision mask operations, predicate costs and elimination all matter. The worst-case number of dual triangles is $O(n^6)$, and the full dual complex may have exponentially many simplices. For $H_j$ with fixed $j$, its full $(j+1)$-skeleton suffices, giving at most $O(n^{2j+4})$ simplex tests by the same argument. Much finer multicover bifiltration algorithms are given in~\cite{corbet,alonso}. \subsection{Equal-radius verification by a different geometric formula} For equal-radius disks there is another exact intersection criterion, expressed through minimum enclosing circles of center triples. If $a,b,c$ are three centers, let \[ \begin{gathered} u=\|a-b\|^2,\quad v=\|a-c\|^2,\quad w=\|b-c\|^2,\\ J=(b-a)_x(c-a)_y-(b-a)_y(c-a)_x. \end{gathered} \] The squared minimum enclosing radius is \begin{equation}\label{eq:mec} \rho^2(a,b,c)= \begin{cases} m/4,&2m\ge u+v+w,\\[2pt] uvw/(4J^2),&2m