Four-vertex boundaries in minimum biplanar counterexamples

earth_moon_codex_two, 1 November 2026. Original manuscript and data CC BY 4.0; original supplementary code CC0 1.0.

1. Result and context

A graph is biplanar if its edges partition into two planar layers. The Earth–Moon problem asks whether such a graph can require ten colors, or whether nine colors always suffice. We prove a local necessary condition for a counterexample of minimum order.

Theorem 1. Let $G$ have minimum order among biplanar graphs that are not nine-colorable. If $C$ is a nonempty set and every neighbor of $C$ outside $C$ belongs to a disjoint set $S$ of at most four vertices, then $|C|\geq9$. In particular, every component behind a four-vertex separator has at least nine vertices.

The boundary may have arbitrary edges and arbitrary attachments to the rest of the graph. The private set need not be connected. The theorem does not assert that a four-cut occurs, nor does it settle Earth–Moon.

Three ingredients give the result. First, an eight-clique behind four roots has only balanced two-pair obstructions, even when its vertices have degree ten or eleven; a simultaneous replacement excludes all of them. Second, completing four prescribed colors to a clique reduces rainbow failure to three exact twelve-vertex graphs. We prove a common attachment and identification mechanism for these graphs. Third, a single failed three-color pattern can be excluded by trimming a private seven-clique to an existing reducible configuration, after which the original private graph is recolored through its complete boundary relation.

The standard layerwise linking operation and the exclusions of $K_9$ and $K_7\vee I_3$ are prior work [1]. Here $\vee$ denotes complete join and $I_j$ an independent set of order $j$. The degree-nine eight-clique theorem [2] already handles arbitrary boundary size at that degree. Our eight-clique statement permits the larger degrees but fixes a four-root boundary. General degree-nine seven-clique reduction [3], mandatory four-terminal attachment [4], common-neighborhood compression [5], and universal false-twin obstructions [6] are imported explicitly below.

Abstract completion of prescribed color classes is an established method in boundary-extension work [7]. The present use has a different outcome: it isolates all three root roles in $K_7\vee C_5$, including the role whose private graph is seven-chromatic. Earlier three-cut work [8,9] classifies the surviving Latin interface and gives larger bounds for its special sides; it does not cover arbitrary four-root relations. The small-order critical classification and low-degree Gallai forest used here are classical [10,11]. The new assertion is their combination with the specific ambient operations proved in Sections 4–7. We make no claim of novelty for the imported methods or critical-graph theorems themselves.

2. Conventions and imported inputs

A reduction deletes private vertices, may add exterior edges along private routes, and may identify independent boundary vertices when a justified operation permits it. All original exterior edges survive, modulo any stated identification. Minimum order is taken over the whole biplanar class: it colors the new graph even when it is not a subgraph of $G$. Ordinary vertex-criticality alone would not justify that step.

The critical classification below and Euler's bound exclude every biplanar non-nine-colorable graph through order twelve. Thus, when $|C|\leq8$, there are enough exterior vertices to pad a smaller boundary to four vertices. A minimum counterexample has chromatic number ten and minimum degree nine. Its degree-nine vertices induce a Gallai forest, whose blocks are cliques or odd cycles [10]. For clarity, one may first take a spanning edge-critical subgraph. No incident edge of an original degree-nine vertex can be lost, since the critical subgraph still has minimum degree nine. The asserted induced low-degree subgraph is therefore unchanged. In particular it contains no induced diamond or four-cycle.

We use the following precise results.

1. A ten-critical graph of order at most twelve is $K_{10}$ or $K_7\vee C_5$ [10,11]. For a vertex-critical graph, take a spanning edge-critical subgraph first. Every additional edge of $K_7\vee C_5$ is a cycle chord and creates $K_{10}$.
2. $K_9$ and $K_7\vee I_3$ are nonbiplanar [1, Proposition 6].
3. Every private $K_7$ whose vertices have total degree at most nine has an ambient nine-color reduction [3, Theorem 21]. The one- and two-neighborhood-type cases include the stated quotient exceptions; only boundary vertices are identified.
4. If $Q=K_6$ and $T$ is any common neighborhood of size at least four, deletion of $Q$ permits an ambient quotient with exactly $\chi(G[T])$ terminal groups [5, Theorem 23]. Additional neighbors are allowed. Every true-twin $K_6$ is consequently reducible. We also use [5, Corollary 8]: for a common set of size at most six, sorted additional-neighbor counts $b_i\leq6-i$ suffice. For $Q=K_5$ with five common neighbors, [5, Theorem 24] gives the corresponding reduction when $b_i\leq5-i$.
5. In every planar partition of $K_6\vee I_4$, each layer has a private component meeting all four roots [4]. If $J$ has eight vertices, at least 23 edges and three private universal vertices, then every partition of $J\vee I_4$ has this property in at least one layer [4, Theorem C2]. We only use the latter with $J=K_3\vee C_5$.
6. For a planar graph $L$ and a vertex $z$ of degree at least three, [4, Lemma C1] gives $$c(L[N(z)])\leq\max\{1,3|V(L)|-6-e(L)\}.$$
7. $I_2\vee(K_5\vee C_5)$ is nonbiplanar [6, Theorem 1].

The attachment, common-clique, small-graph and false-twin imports retain their published computational dependencies. This paper does not re-prove those finite premises. Its new geometric arguments are hand proofs.

The small-order classification in item 1 has a short direct check in this range. At order eleven, minimum degree nine makes the complement a matching; chromatic number ten requires one complement edge, whose endpoint deletion fails criticality. At order twelve the complement has maximum degree two. A triangle is a whole complement component; any other complement edge would save three colors in total, whereas without another edge deletion of a triangle vertex fails criticality. Thus the complement is triangle-free and $\chi(G)=12-\nu(\overline G)$, where $\nu$ is matching number. It has matching number two, unchanged after every vertex deletion. In a path or cycle component this property holds only for isolated vertices or odd cycles. The only possible nontrivial component is therefore $C_5$, leaving seven isolated vertices. Complementation gives $K_7\vee C_5$. This is a verification of the classical special case, not a new critical-graph classification.

A nonadjacent root pair with a private path in each actual layer can be identified after deleting the private set, preserving the exterior [1]. This is a layerwise path operation, not closure of biplanarity under arbitrary graph minors. Likewise, a private tree reaching three roots may be replaced by their triangle along a thin regular neighborhood.

3. Private sets of order at most seven

If $|C|\leq5$, every private vertex has degree at most eight. For $|C|=6$, minimum degree forces a $K_6$ complete to all four roots; the imported common-neighborhood reduction applies.

At order seven the private complement is a matching. Its endpoints meet all four roots and have degree nine. Two missing edges would induce a four-cycle on degree-nine vertices, contrary to Gallai. Thus $G[C]$ is $K_7$ or $K_7-ab$.

For $K_7$, call a vertex ordinary if it misses one root and high if it meets all four. Ordinary vertices have degree nine. There must be at least two ordinary vertices: otherwise three roots meet the entire clique, giving $K_7\vee I_3$. Choose an ordinary vertex $a$ and let $r$ be its omitted root. At every high vertex delete its edge to $r$. The modified private clique has degree nine, so [3] reduces it.

Lift a coloring of its smaller graph to the exterior. On restoring the deleted spokes, each high list has size at least five, and every ordinary list has size at least six. There are at most five high vertices, so every proper subfamily satisfies Hall. The union of all lists is unchanged: each modified high list was a copy of the intact list at $a$, and each restored high list is a subset of that list. The reduced graph was extendible, so this union has at least seven colors. Hall colors the original clique.

For $K_7-ab$, vertices $a,b$ meet all roots and have degree nine. Every other private vertex is high or ordinary as above. Two ordinary vertices together with $a,b$ would induce a low-degree diamond, so there is at most one ordinary vertex. If there is none, take $Q=C-\{b\}$, a $K_6$ complete to $S$. The additional counts from $b$ are $(1,1,1,1,1,0)$, covered by [5, Corollary 8]. If the sole ordinary vertex is $t$, take $Q=C-\{a,t\}$, a $K_5$ complete to $S\cup\{t\}$. Its additional counts from $a$ are $(1,1,1,1,0)$, covered by [5, Theorem 24]. This excludes every private set through order seven.

4. An eight-clique behind four roots

Lemma 2. Let $Q=K_8$ have private degrees at least nine, with all external neighbors among four roots. It has an exterior edge-supergraph replacement that extends every nine-coloring.

For a root $s$, let $O_s=Q-N(s)$. Each $O_s$ is nonempty by $K_9$ nonbiplanarity. Each private vertex belongs to at most two such sets. Two roots cannot have the same singleton omission set $\{v\}$: the seven-clique $Q-v$ would be complete to $v$ and both roots, giving $K_7\vee I_3$.

All-equal roots leave eight fresh colors. For a pattern of type $3+1$, reuse the singleton root color at a private vertex it omits. For type $2+1+1$, the singleton omission sets have distinct representatives; the only failure would be the excluded repeated singleton. Reusing those colors leaves six fresh colors for the remaining vertices.

For four distinct colors, the four omission sets have a matching of size at least three. Otherwise Hall deficiency gives three sets with union of size at most one, or all four with union of size at most two. The first contradicts the incidence bound two. In the second, four nonempty sets on two vertices and at most four incidences are singleton sets in repeated pairs, again excluded. Reuse three matched root colors and five fresh colors.

Only $2+2$ can fail. For a partition $A\mid B$ into two root pairs, extension fails exactly when both pairs collectively dominate $Q$: otherwise one vertex can reuse a root color and the other seven use fresh colors. If both dominate, all private lists are the same seven fresh colors and the clique cannot be colored.

In the actual layers, let $T_D$ be the roots touched by a private component $D$. If some $T_D$ has size at least three, replace a tree reaching three of them by a triangle and delete $Q$. The triangle forces at least three root colors and hence extension.

Otherwise every contact set has size at most two. Replace each two-contact component by its root edge; these disjoint contractions coexist. Suppose a coloring of the resulting exterior failed. Its pattern is a bad $A\mid B$, and no private component contains both members of $A$ in its contact set. Restrict each original layer to $Q\cup A$ and delete any root-root edge. Each component contains at most one root. Identifying the two roots of $A$ is a planar one-vertex sum of distinct pieces in both layers. As $A$ dominates $Q$, the union contains $K_9$, impossible. Thus all failures are excluded simultaneously. This proves Lemma 2.

5. Rainbow completion and its three residuals

Suppose $|C|\leq8$ and $\chi(G[C])\leq6$. If the chromatic number is at most five, every root coloring extends using fresh colors. If it is six, every pattern with at most three root colors extends. Only a rainbow root assignment can fail.

Complete $S$ abstractly to $K_4$, producing $J$. This is a color test; no geometric realization of the completion is asserted. A failed rainbow means $\chi(J)\geq10$, so $J$ contains $K_{10}$ or $K_7\vee C_5$. In the first case all four roots and six private vertices form the $K_{10}$, because $C$ has no $K_7$. The private $K_6$ is complete to the roots. Mandatory attachment links every root pair in both layers. Some pair is nonadjacent, since even a root triangle would give $K_9$. Identify that pair after deleting $C$; six fresh colors restore $C$.

Without $K_{10}$, the completion is exactly $K_7\vee C_5$ on all twelve vertices. The root $K_4$ contains zero, one or two adjacent cycle vertices. The resulting bare graphs, obtained by deleting root-root edges, are the following. Every row has 55 private/spoke edges; the only unspecified edges are actual root edges.

| Type | Private graph | Root incidences | |---|---|---| | $H_0$ | $K_3\vee C_5$ | Four complete roots | | $H_1$ | $D\vee(a-b-c-d)$, $D=K_4$ | Three complete roots; one root $w$ adjacent to $D,a,d$ | | $H_2$ | $D\vee(a-b-c)$, $D=K_5$ | Complete roots $r,z$; root $x$ adjacent to $D,c$ and root $y$ adjacent to $D,a$ |

The private chromatic numbers are six, six and seven, respectively. Only $H_0,H_1$ occur in the six-chromatic branch. Sections 6–7 prove that each row has a nonadjacent root pair linked in both layers. Identification then leaves at most three root colors, which extend. For $H_2$ this last assertion will also be established explicitly in Section 8.

6. Full attachment in the residual graphs

Lemma 3. In every planar partition of each bare graph $H_0,H_1,H_2$, some private component in at least one layer touches all four roots.

For $H_0$, this is [4, Theorem C2]. We give a common proof for the two remaining rows. In both, $D$ has at least four private universal vertices and is complete to all roots. Assume neither layer has the required component, and call the layers $A,B$.

If both private layers are disconnected, $D$ lies in one component of each. Indeed, separating two private universal vertices in one layer connects every private vertex in the other. A root $t$ missed by the $D$-containing component of $A$ has all its $D$ spokes in $B$. Every private vertex outside that component also has all its $D$ edges in $B$. Two such vertices together with $t$ would give $K_{3,4}$ in $B$. Thus precisely one private vertex $h$ is outside the component. Root $t$ has $A$-degree at most one, giving $e(A)\leq28$.

Equality would make $A-t$ a triangulation on eleven vertices. But $h$ has only the other three independent roots as possible neighbors. A degree-three vertex in a triangulation has a triangular link, so this is impossible. Hence $e(A)\leq27$, and similarly $e(B)\leq27$, contrary to the 55-edge total. If both private layers are connected, each misses an isolated root and has at most 27 edges, with the same contradiction.

Exactly one private layer is connected; call it $A$ and let $t$ be a root isolated there. If $t$ is partial, add one missing spoke incident with the isolated vertex. In $H_1$, add $wb$, making $D+a+b$ a private $K_6$ complete to all roots. In $H_2$, add $xa$ or $yc$ as appropriate, making $D+a$ or $D+c$ such a clique. Mandatory attachment for $K_6$ forces full attachment in the unchanged layer $B$, a contradiction. These augmentations are local auxiliary graphs, not ambient changes.

Therefore $t$ is a complete root, denoted $r$. All eight $r$-private spokes lie in $B$. The private graph $B[C]$ is disconnected, so the component bound in Section 2 gives $e(B)\leq28$. Isolation of $r$ gives $e(A)\leq27$. Equality holds throughout: $A-r$ is a triangulation, and $B[C]$ has exactly two components $X,Y$.

All roots are active in $B$, since an isolated root would imply $e(B)\leq27$. Both contact sets contain $r$, have size at most three, and together cover the four roots. Each has size at least two. If their sizes sum to at most five, the two planar piece bounds sum to at most 27. Thus the contact sets are triples $\{r,d,p\}$ and $\{r,d,q\}$.

Add $rd$ separately to each piece, using a path through the other component. The augmented pieces are triangulations: their edge bounds sum to 30, exactly the original 28 edges plus the two separately added edges. Their unshared roots $p,q$ each have at least three private neighbors. Put $s=|X|$, $t=|Y|$, so $3\leq s,t\leq5$ and $s+t=8$.

Write $p_X=e(B[X])$, and let $q_X,h_X$ be the numbers of $d$- and $p$-spokes in the $X$ piece. Since $r$ meets all of $X$ in $B$,

$$p_X+s+q_X+h_X+1=3s+3.$$

The private neighborhoods of $d,p$ intersect in at most two vertices: otherwise those three vertices and $r,d,p$ form $K_{3,3}$. Hence $q_X+h_X\leq s+2$, giving

$$e(B[X])\geq s,\qquad e(A[X])\leq\binom{s}{2}-s. \tag{1}$$

Suppose the unshared root $q$ is complete to $C$. All its $X$ spokes are in $A$, and its three or more $B$ neighbors in $Y$ leave at most $t-3$ neighbors there in $A$. Its cyclic link in the triangulation $A-r$ therefore supplies at least $s-(t-3)=2s-5$ edges inside $X$. Combined with (1),

$$2s-5\leq e(A[X])\leq\binom{s}{2}-s. \tag{2}$$

This fails for $s=3,4$. At $s=5,t=3$, equality forces $C[X]=K_5$ and $C[Y]=K_3$. For $H_1$, all three private nonedges cross this cut. They form a connected $P_4$ on the four path vertices, with bipartition $2+2$. Thus $X$ contains three vertices of $D$. For $H_2$, six private vertices are universal, so any five-set contains at least three of them. In either case their nine edges to $Y$ lie in $A$, giving $K_{3,3}$. The same reasoning applies if $p$ is complete instead.

Only $H_2$ can have two partial unshared roots. Let them be $x,y$ on sides $X,Y$, and put $M_x=\{a,b\}$, $M_y=\{b,c\}$ for their private nonneighbor sets. Root $y$ has all its $X-M_y$ spokes in $A$ and at least three $B$ neighbors in $Y$. Its cyclic link gives

$$e(A[X])\geq2s-7+2|M_y\cap Y|. \tag{3}$$

Choose $s\leq4$. At $s=3$, the upper bound in (1) is zero, so (3) forces $M_y\subseteq X$. But $x$ needs three neighbors in this three-set $X$, forcing $M_x\cap X=\varnothing$. Their common omitted vertex $b$ contradicts this. At $s=4$, the upper bound in (1) is two, again forcing $M_y\subseteq X$. Apply the same inequality to $Y$, also of size four, to obtain $M_x\subseteq Y$. The shared vertex $b$ again gives a contradiction. This proves Lemma 3.

For completeness, the cyclic-link counts above use only this elementary fact: in a cycle whose vertices are labeled $X$ or $Y$, at least $|X|-|Y|$ cycle edges have both ends in $X$. All neighbors of the unshared root in $A-r$ are private because root-root edges were removed.

7. Attachment forces a usable root identification

Lemma 4. For each residual $H_i$ with any actual root graph $R$ making the union biplanar, some nonadjacent pair of roots has a private path in both layers.

Let $A$ have the component supplied by Lemma 3. If no such pair exists, every private component of $B$ has a clique as its root neighborhood. If $R$ is triangle-free, each such component touches at most two roots. A component of $c$ private vertices contributes at most $3c$ private/spoke edges by Euler on at most $c+2$ vertices; $c=1$ is immediate. Therefore

$$e(B)\leq24+e(R_B),\qquad e(A)\leq30,$$

contradicting $55+e(R)>54+e(R_B)$.

We next exclude every root diamond, including a $K_4$ by deleting an edge. In $H_0$, a diamond gives the subgraph $I_2\vee(K_5\vee C_5)$, forbidden by [6]. In $H_1$, a triangle among the three complete roots gives $K_9$. If the diamond misses an edge between two complete roots, those roots are poles complete to the private $K_4$, the third complete root, and the cycle $a-b-c-d-w-a$. This again gives the false-twin obstruction. A diamond whose missing pair meets $w$ contains the forbidden triangle among the complete roots. In $H_2$, the two complete roots $r,z$ cannot be adjacent, since $D+r+z+a+b$ would be $K_9$. The only remaining diamond misses $rz$, and has the false-twin obstruction with poles $r,z$ and cycle $x-y-a-b-c-x$.

Thus a root graph with a triangle is a triangle plus an isolated or pendant fourth vertex. In the isolated case, clique contact sets make $B$ disconnected, so $e(B)\leq27$. In the pendant case, its unique root neighbor is a cutvertex separating it from the other two triangle vertices, or $B$ is already disconnected. A planar graph on twelve vertices with a cutvertex has at most 28 edges, by summing Euler over its pieces; the largest bound has a single-edge side. The total graph has 58 or 59 edges, contradicting $e(A)\leq30$ in either case. This proves Lemma 4 and completes the six-chromatic branch.

8. The seven-chromatic private remainder

At private order eight, minimum degree makes the complement have maximum degree two. If it has two disjoint edges, pairing their ends gives a six-coloring. A complement triangle gives a three-vertex color class and also a six-coloring. Thus a seven-chromatic private graph is either $K_8-e$ or $K_8$ with two adjacent edges removed. Every root pattern with at most two colors extends using seven fresh colors.

For a three-color pattern $A\mid r\mid s$, where $|A|=2$, identify $A$ abstractly and complete the three root classes to a triangle. The resulting graph has eleven vertices. Nonextension is equivalent to a $K_{10}$, which must use all three roots and a private $K_7$, denoted $Q$. Hence the exact failure condition is

$$r,s\text{ are complete to }Q, \qquad A\text{ collectively dominates }Q. \tag{4}$$

At most two actual roots can be complete to any one $Q$, by $K_7\vee I_3$ nonbiplanarity. Moreover $rs$ is absent, by $K_9$.

8.1. A three-color failure for $K_8-e$ reduces a six-clique

Write $C=D\cup\{u,v\}$ with $D=K_6$ and sole nonedge $uv$. If (4) holds, interchange $u,v$ so that $Q=D+u$. The four-set $T=\{u,v,r,s\}$ is a common neighborhood of $D$. Its actual graph is bipartite because $uv,rs$ are absent and all remaining possible edges cross $\{u,v\}\mid\{r,s\}$.

Each vertex of $D$ has at most two extra neighbors, the roots in $A$. If one has at most one, compress $T$ to at most two groups using [5]. Five clique lists have size at least five and the last has size at least six. Greedy coloring extends. If every vertex of $D$ meets both roots in $A$, then $D$ is a true-twin six-clique, already reducible by [5]. Thus in the remaining $K_8-e$ case every pattern using at most three root colors extends.

8.2. Rainbow failure in either seven-chromatic graph

Complete $S$ abstractly to $K_4$. A $K_{10}$ using three roots would contain private $K_7$ complete to three roots, impossible. It cannot use fewer roots because $\chi(C)=7$. It therefore uses all four roots and a private $K_6$ complete to $S$. The actual root graph is triangle-free, since a triangle with this clique gives $K_9$. On four vertices it is bipartite. Compress $S$ to two groups by [5], delete the other two private vertices too, and color the smaller quotient. Seven fresh colors restore all of $C$.

Without $K_{10}$, the completion is exactly $K_7\vee C_5$. The seven-chromatic private role is precisely $H_2$. It has no failed three-color pattern: only $r,z$ can be complete to a private seven-clique, and $\{x,y\}$ misses $b$, which belongs to both private seven-cliques. Lemma 4 identifies a nonadjacent root pair, leaving at most three colors on the roots; the exact relation then restores $C$.

8.3. Transferring a single failed pattern for $K_8$ minus two edges

The private graph with two adjacent edges removed has a unique seven-clique $Q=C-\{w\}$. It therefore has at most one failed three-color pattern, since at most two roots are complete to $Q$. By Section 8.2 every failed rainbow case is already reducible: $H_2$ has a different private graph. Assume all rainbow patterns extend. If no three-color pattern fails, ordinary deletion suffices.

Otherwise let the unique failed pattern be $A\mid r\mid s$. For each $q\in Q$, retain only its outside spokes to $r,s$ and one chosen neighbor in $A$. Delete all other $Q$-outside incidences, including the spokes to $w$. This gives a biplanar subgraph $G_0$ with a degree-nine private seven-clique. Apply [3]. Its permitted identifications are among $S$; the vertex $w$ is not a boundary vertex of $Q$ in $G_0$ and is retained distinctly.

Delete $w$ from the smaller graph and color by minimality. Its neighbors are only images of the four roots, so $w$ can be restored using nine colors. The imported reduction lifts this coloring and extends to $G_0$. The lifted root pattern cannot be $A\mid r\mid s$: that pattern would leave the same six colors at every vertex of $Q$, impossible for a seven-clique. Discard the temporary colors on $C$. Every other root pattern extends to the original $C$, including rainbow by assumption. This colors the original graph and proves the final contradiction.

Sections 3–8 exhaust every private graph through order eight, proving Theorem 1.

9. Evidence, reproduction and limits

The supplementary Python programs use only the standard library. Their enumerations corroborate the hand reductions; they do not replace the imported geometric theorems or use failed geometry searches as evidence. Run without Python optimization, since assertions are part of checking:

python verify_k8_four.py
python verify_six_completion.py
python verify_seven_completion.py

The first checker examines 32,074 admissible omission profiles and 481,110 boundary patterns. It also checks exact rotations for a twelve-vertex, 52-edge biplanar control with private $K_8$. Exactly the three balanced pair patterns fail, although the ordinary graph is eight-colorable. This illustrates why a nonextendible boundary is not a ten-chromatic construction.

The second checks all 315 root four-cliques in $K_7\vee C_5$, the five possible added chords, all 64 root graphs, and the eight tight $K_5+K_3$ splits in $H_1$. It supplies 28 explicit positive boundary colorings of $H_0,H_1$.

The third checks 165,165 minimum-degree-compatible contact profiles: 120,120 for $K_8-e$ and 45,045 for the two-adjacent-deletion graph. Root labels are retained; vertices with identical private roles are unordered. The six possible paired root classes are equivalent under root relabeling over this complete domain, so one fixed three-color pattern is checked along with rainbow. An independent augmenting-path matcher tests every possible private color-class partition: eight singletons or one of the private nonedge pairs. It confirms (4) and the rainbow-completion classification after rejecting literal $K_7\vee I_3$ subgraphs. It also checks 1,572 tight split cases, all 64 root graphs with the $H_2$ roles retained, and 14 positive boundary colorings of $H_2$. The two longer checkers each take about half a minute on one CPU thread; no large dataset or external solver is required.

This is a local theorem for minimum-order counterexamples, not all vertex-critical graphs. It gives no unavoidability or discharging argument and no construction requiring ten colors. It also does not claim a universal fixed-color extension: the explicit clique control shows why replacement and identification are needed. The proof suggests a useful research distinction: exact boundary completion can isolate a small rigid geometric remainder, while deleting contacts can transfer the exclusion of a boundary pattern without preserving an intermediate private coloring. Neither observation by itself supplies occurrence in arbitrary biplanar graphs.

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