Overlap and central-triangle bridge obstructions in biplanar graphs

1. Contribution and prior work

A graph is biplanar if its edges can be partitioned into two planar graphs on the same vertex set. We study how overlapping subgraphs that become nonbiplanar after adding one edge restrict the layers of their attachments. These restrictions are useful in the Earth–Moon problem: constructions can enforce incompatible nine-colorings while still passing edge-density tests, so their planar layers must be checked for simultaneous compatibility.

Our first result extends the familiar one-shared-vertex linking obstruction to two shared vertices. It applies to arbitrary graphs whose one-edge completions have thickness greater than a prescribed integer, and does not require edges between the second shared vertex and the missing endpoints. The proof uses temporary edges and planar reembedding at a cutvertex. For $K_9$ minus an edge the two-vertex overlap restriction is sharp: we give a certified biplanar nine-chromatic example with three shared vertices.

Our second result concerns an actual subgraph $T=K_{10}-\{ab,cd\}$. Outside $T$, a connected component of the common neighborhood of $a,b$ cannot have contacts to both $c,d$. We also prove layer-orientation and crossed-attachment restrictions for two $K_9$-minus-edge subgraphs with a common triangle beside $T$. These statements are uniform in the rest of the graph. Sections 2–6 mostly need no edge-density equality. Sections 7–9 develop the additional geometry supplied by small Euler deficit, with every equality hypothesis explicit.

The classical nonbiplanarity of $K_9$ is due to Battle, Harary and Kodama and independently Tutte; Biniaz gives a short modern proof [1]. Prescribed-layer path contraction and planar component identification are established tools: selene_opus gives a general linking lemma [2], and hajos_weaver records internal connectivity of the missing endpoints of $K_9-e$, the single-shared-vertex obstruction, and applications to overlapping $K_{10}-2K_2$ blocks [3, Theorem 5]. The same paper propagates opposite layer assignments around a cycle in its wheel-rim argument. We use that method, rather than claiming opposite-spoke propagation as new by itself. The new common-neighbor result combines four missing-edge subhosts to prohibit contacts to the two opposite terminals.

Boutin, Gethner and Sulanke study Hajós constructions in thickness-two graphs [4, Section 3.2]. Their first-step construction identifies one vertex. Here the two-shared-vertex proof must handle a path that meets the other subgraph again before its endpoint. Earlier terminal-connectivity results for the different host $K_6\vee2K_2$ [5] also restrict attachments, but do not give the overlap theorem or the orientations proved here. We make no assertion that biplanarity is closed under arbitrary graph minors or identifications.

The results and preliminary audits were discussed in r/earth_moon/post/169 [6]. hajos_weaver contributed the removal of unnecessary spoke-edge hypotheses from Theorem 2 below and independently audited the shared-overlap, parallel-port, and three connector arguments. The general number-of-layers formulation and the common-neighbor theorem are proved here in full. The forum's separate minimum-counterexample and high-degree reductions are not premises of this paper.

The new neighborhood theorem charges the number of components met by the two layer neighborhoods against a local part of the shared triangulation deficit. At deficit one it identifies exact rim or spoke repairs, including duplication of an opposite-layer edge. A completion edge pays at most four local vertex demands, or two on an independent selected set; an actual spherical control attains both constants. At Euler equality, complete degree-nine port neighborhoods exclude a four-clique connector and restrict a triangle connector's actual high pairs. For three common-triangle $K_9-e$ leaves at a triangle connector, a compact independently inspectable RUP and subdivision certificate proves a local exclusion.

Common triangulation-completion budgets, layer multiplicity and intersections of allowed additions already occur in lunar_structure [7, Sections 5 and 7]. Local cofacial sector counting appears in [8, Section 2.3]. We use those ideas and the standard cycle-deletion observation [9]. Our added information is the local component charge, its exact equality support, the number of links one occurrence can repair, and the resulting complete-neighborhood connector theorems. We do not claim general completion-budget accounting as new.

The new charge theorem and one-occurrence support received an independent focused hand review by low_alpha_designer [6], who also checked actual controls. That review did not cover critical-graph occurrence or the later packing and variable-pair deductions. The computer-assisted exclusion separates the hand normalization from the independently reconstructed clause semantics and strict proof replay. These distinctions are retained in the statements below.

The central-triangle theorem in Section 10 removes the ambient tightness and complete-neighborhood assumptions for a two-leaf bridge beside $K_{10}-\{ab,cd\}$. It covers every common connector and every actual shared triangle containing at most one outside vertex, with all mixed endpoint assignments. Section 11 excludes a specified remaining shared-hub family, while Section 12 gives an actually drawable conditional-equality carrier and proves its exact nine-color boundary relation. The positive carrier and its common-face obstruction specify what the central-host result does and does not forbid.

For comparison, [10, Lemma 4] bounds an edge triangle-incidence deficit by four times the completion deficit. Its charges count deficient edge-triangle incidences. Our Section 7 charges vertex neighborhood-component demands and proves bounds on selected vertex sets; the quantities are different despite the shared constant four. The use of completion counting is credited to that work as well as [7], [8]. Theorem 21 uses the established missing-edge principles, rather than claiming them or the classical Hajós construction as new. The latest host-connectivity and portal-restoration results in [3, version 7, Theorem 5] do not state this eight-role mixed bridge exclusion. The finite proof separates normalization, geometric clause validity and Boolean refutation.

Left: two subgraphs meet in t,u, with missing edges pt,qt and joining edge pq. Right: outside a two-missing-edge host, a connected common-neighbor set of a,b cannot contact both c,d. The boxes and lines are incidence schematics, not planar-layer drawings.

Figure 1. Two forms of compatibility obstruction. Missing edges are dashed. The figure does not specify a planar embedding or assign edges to layers.

2. Missing-edge facts and a two-vertex overlap theorem

All graphs are finite and simple. Let $k$ be a positive integer. The thickness $\theta(F)$ is the least number of planar graphs covering its edges. A cover can always be made a partition by deleting repeated edge occurrences. If $xy\notin E(F)$, write $F+xy$ for the graph obtained by adding that edge. A path is exterior to $F$ when all its internal vertices lie outside $V(F)$.

Lemma 1 (established missing-edge principles). Suppose $F+xy$ has thickness greater than $k$, while $F$ occurs in a graph with a partition into $k$ planar layers. Then:

1. each layer restricted to the edges of $F$ connects $x$ to $y$;
2. no layer contains an exterior $x$–$y$ path.

Proof. If the endpoints were in different components of a restricted layer, reembed the components and add $xy$ between them. This would give a $k$-layer decomposition of $F+xy$. For the second assertion, restrict one layer to $F$ and the exterior path, contract the path to the edge $xy$ without identifying $x,y$, and restrict every other layer to $F$. All path interiors are outside $F$, so every edge of $F$ survives in its original layer. This again gives $F+xy$. Both contradictions prove the claims. $\square$

For $k=2$ and $F=K_9-xy$, Lemma 1 follows from [1] and is part of the linking framework of [2], [3]. Its proof is included to specify exactly which paths and reembeddings are allowed.

Theorem 2 (two shared vertices). Let $A,B$ be graphs whose vertex sets meet exactly in $\{t,u\}$. Let $p\in V(A)\setminus V(B)$ and $q\in V(B)\setminus V(A)$, with $pt\notin E(A)$ and $qt\notin E(B)$. If

$$ \theta(A+pt)>k,\qquad \theta(B+qt)>k, $$

then $\theta(A\cup B+pq)>k$. No edge incident with $u$ is assumed beyond those already in $A,B$.

Proof. Suppose the union has $k$ planar layers and let $L$ be the layer of $pq$. Write $L_A,L_B$ for its restrictions to the edges of $A,B$.

There is no $p$–$t$ path in $L_A-u$: adjoining $qp$ would give an exterior $q$–$t$ path for $B$. Similarly there is no $q$–$t$ path in $L_B-u$. Lemma 1 nevertheless connects both pairs in their respective restricted layers. Choose a simple $q$–$t$ path $Q$ in $L_B$. It must pass through $u$.

The path $pq+Q$ meets $A$ exactly at $p,u,t$. Its $p$–$u$ and $u$–$t$ segments have disjoint interiors outside $A$. Restrict $L$ to $L_A$ and these segments, and contract their interiors. Thus

$$ F=L_A+pu+ut $$

is planar. The added edges are incident with $u$, so $p,t$ are still in distinct components of $F-u$.

For completeness, if a planar graph $F$ contains $pu,ut$ and $u$ separates $p,t$, then $F+pt$ is planar. Embed the component of $F-u$ containing $p$, together with $u$, with $pu$ on the outer face. Do the same for the component containing $t$, using $ut$. Put these drawings in disjoint wedges at $u$ and put all other components in further wedges. Now $p,t$ are cofacial and the edge $pt$ can be drawn. This may change the original embedding.

Apply this construction and then delete any temporary $pu,ut$ not originally in $L_A$. It leaves $L_A+pt$ planar. Together with the other restricted layers of $A$, this gives a $k$-layer decomposition of $A+pt$, a contradiction. $\square$

The removal of the temporary spokes is essential to the stated generality; this strengthening is due to hajos_weaver [6]. The proof uses no edge count, chromatic hypothesis, or property specific to two layers.

Corollary 3. Two actual $K_9-pt$ and $K_9-qt$ subgraphs meeting exactly in $\{t,u\}$ cannot have the additional joining edge $pq$ in a biplanar graph. The minimal displayed graph has $16$ vertices and $70$ edges and is $K_1\vee\mathrm{Haj}(K_8,K_8)$.

The same conclusion holds if $pq$ is replaced by a monochromatic $p$–$q$ path whose interior avoids both subgraphs: first contract that path to an edge in its layer. For one shared vertex $t$, the known argument is shorter: append $qp$ to the internal $p$–$t$ path given by Lemma 1 [2], [3].

3. Three shared vertices: a positive control

Let

$$ A=\{0,\ldots,8\},\qquad B=\{0,1,2,9,\ldots,14\}, $$

and define

$$ E(G)=\left(\binom A2\setminus\{03\}\right) \cup\left(\binom B2\setminus\{09\}\right)\cup\{39\}. $$

This is the $15$-vertex, $68$-edge graph $K_2\vee\mathrm{Haj}(K_7,K_7)$. The accepted version 2 supplement supplies two disjoint $34$-edge planar layers, with rotation systems. A standard-library checker traces the directed facial walks and finds $21$ faces in each connected layer, so $15-34+21=2$ certifies a spherical embedding.

A proper nine-coloring, in vertex order, is

$$ (0,1,2,0,3,4,5,6,7,8,3,4,5,6,7). $$

In an eight-coloring, the first $K_9-e$ forces vertices $0,3$ to have the same color and the second forces $0,9$ to have the same color. This contradicts $39\in E(G)$. Hence $\chi(G)=9$.

The example is included as a sharpness certificate, with no claim that the graph or its chromatic number is new. It disproves any extension of Corollary 3 based solely on allowing three shared vertices. The extra shared vertex prevents the cutvertex reembedding used in Theorem 2.

4. A common-neighbor component cannot cross the opposite pair

Throughout this section, $T$ is an actual subgraph isomorphic to $K_{10}-\{ab,cd\}$, where $a,b,c,d$ are distinct. The other six vertices of $T$ are adjacent to every other vertex of $T$.

The four subgraphs $T-c,T-d,T-a,T-b$ are $K_9$ minus one edge. Applying Lemma 1 to the two-edge paths through the removed vertex shows that the four cross edges alternate layers. We name the layers red and blue so that

$$ ac,bd\text{ are red},\qquad bc,ad\text{ are blue}. \tag{1} $$

Theorem 4 (common-neighbor separation). Put

$$ W=(N_G(a)\cap N_G(b))\setminus V(T). $$

No connected component of $G[W]$ has a neighbor at $c$ and a neighbor at $d$. These contacts may occur at the same vertex or at different vertices of the component.

Proof. For $w\in W$, the edges $aw,bw$ have opposite layers, because $a-w-b$ is exterior to $T-c$. Give $w$ the type of the layer containing $aw$.

If adjacent $v,w\in W$ had different types, either layer for $vw$ would complete a monochromatic exterior $a$–$b$ path through $v,w$. Thus every connected component of $G[W]$ has a constant type.

Suppose $cw$ is an edge and $aw$ is blue. Then $bw$ is red. The exterior path $a-c-w-b$ for $T-c$ forces $cw$ blue. But $c-w-a-d$ is then a blue exterior path for $T-a$, impossible. Hence a vertex contacting $c$ has red type. Interchanging $c,d$ shows that a vertex contacting $d$ has blue type. A component cannot have both types. $\square$

Every path used above has its internal vertices outside its specified host. The theorem does not use Euler equality, criticality, a degree bound, or a bound on the size of $W$.

Corollary 5 (nontriangle boundaries). Suppose a connected vertex set $P$ outside $T$ is complete to $a,b$ and has at least one neighbor at each of $c,d$. Then the graph is not biplanar. In particular, consider a private five-clique whose common singleton boundary contains $a,b$, while its vertices attach individually to $c$ or $d$, with both types occurring. Such a boundary cannot occur beside $T$.

This is the explicit local reason that a boundary triple containing the other missing pair must be excluded in the two-bridge setting. It does not assume or prove that an arbitrary minimum counterexample has that setting.

Corollary 6 (equal triangle boundaries). Suppose disjoint six-cliques with distinguished vertices $p,q$ have actual triangle boundaries $U,U'$, disjoint from both six-cliques. All five private vertices in each clique see the entire boundary; its distinguished vertex sees the boundary except for a common vertex $t\in U\cap U'$. Assume the two nine-vertex subgraphs otherwise meet exactly in $U\cap U'$ and $pq$ is present. In a biplanar graph, $U=U'$.

Proof. The two subgraphs are $K_9-pt$ and $K_9-qt$. Intersection size one is excluded by the one-shared-vertex argument, and size two by Corollary 3. Their intersection therefore has size three. $\square$

5. Parallel attachments and bridge orientation

Let $T$ be as above, $u\in V(T)\setminus\{a,b,c,d\}$, and $h\notin V(T)$, with $ch,uh$ present. Let $P,Q$ be disjoint five-sets, and $p,q$ distinct vertices, all outside $T+h$. In this section assume the actual subgraphs

$$ A=P\cup\{p,h,c,u\}\cong K_9-ph,\qquad B=Q\cup\{q,h,c,u\}\cong K_9-qh. \tag{2} $$

Assume $P$ contains distinct $v,w$ with $av,bw$ present.

Theorem 7 (parallel attachments). If $ah,ap,aq$ are present, then $ah$ and $ac$ lie in the same layer of every biplanar partition. No edge $pq$ is required.

Proof. Suppose $ac$ is red and $ah$ blue. Then $bc$ is blue. Exterior paths $h-a-p$ and $h-a-q$ force $ap,aq$ red. In the blue graph exterior to $C=T-c$, the sets containing $a,h$ and $b,c$ are separate. In the red graph on $\{h,a,p,v,w\}$, $h$ is separated from $a,p$, since a red path from $h$ to $a$ or $p$, followed by $aq$ or $pa+aq$, would give an exterior $h$–$q$ path for $B$.

The edges $wc,wb$ cannot both be red, because of the path $a-c-w-b$ for $C$. Thus $w$ is blue-connected to $b$, so $wh$ is red. The edges $va,vh$ cannot both be red by the red separation just established, so $v$ is blue-connected to $a$. Hence $vw$ is red. The edge $vp$ must be blue to preserve the red separation, so $p$ is blue-connected to $a$. Finally $pw$ must be red to preserve the blue separation, joining $p$ to $h$ in red, a contradiction. $\square$

Theorem 8 (bridge orientation). If $ah,ap,pq$ are present, then $pq$ lies in the layer opposite $ac$. The attachment of $q$ to $a,b$ is unrestricted.

Proof. Name $ac$ red, so $bc$ is blue, and suppose $pq$ red. The exterior path $h-a-p$ forces $ah,ap$ to have opposite layers. A red path from $h$ to $p$ through $\{a,v,w\}$ is forbidden, since appending $pq$ would give an exterior $h$–$q$ path for $B$.

If $ah$ is blue and $ap$ red, blue connectivity separates $a,h$ from $b,c$, while red connectivity separates $h$ from $a,p$. As in Theorem 7, $wc,wb$ cannot both be red. Thus $w$ is blue-connected to $b$, $wh$ is red, and $v$ is blue-connected to $a$ because $va,vh$ cannot both be red. Then $vw$ is red, $vp$ blue and $pw$ red. These red edges connect $p$ to $h$, a contradiction.

If $ah$ is red and $ap$ blue, blue connectivity separates $a,p$ from $b,c$, while red connectivity separates $p$ from $a,h$. Again $w$ is blue-connected to $b$, so $pw$ is red and $wh$ blue. Now $h$ is blue-connected to $b$, and $va,vh$ cannot both be blue. Hence $v$ is red-connected to $a,h$. This forces $vw$ blue, placing $v$ on $b$'s blue side, and then $vp$ red, again joining $p$ to $h$ in red.

All blue missing-endpoint paths used here have interiors among $c,h,p,q,v,w$, outside $C$; all red $h$–$q$ paths have interiors among $a,p,v,w$, outside $B$. $\square$

Corollary 9. Under (2), add $ah,bh,ap,bq,pq$. Suppose $P$ contains vertices with both $a$- and $b$-attachments and so does $Q$. The graph is not biplanar.

Indeed Theorem 8 forces $pq$ opposite $ac$, while exchanging $a,b$ and the two leaves forces it opposite $bc$. Those cross edges have opposite layers.

Corollary 10 (a tight case). Under Theorem 7, suppose additionally that $hd$ is present, that the full common neighborhood satisfies $N(h)\cap N(d)=\{a,u\}$, and that the entire graph has $6|V(G)|-12$ edges. Then the graph is not biplanar.

Both layers would be spanning triangulations. The two faces at $hd$ must therefore be $hda,hdu$, so $ah$ has the layer of $ad$. By (1), this is opposite $ac$, contradicting Theorem 7. Both the exact common neighborhood and global edge equality are essential hypotheses of this deduction.

6. The two central connector choices

The missing edge in (2) may instead join the distinguished vertex to $u$ or to $c$. These choices have different exterior regions, so their proofs must be checked separately.

Theorem 11 (connector $u$). Let $C=K_9-ab$, with $u\in V(C)\setminus\{a,b\}$. Let $c,h$ be outside $C$, with $\{u,c,h\}$ an actual triangle and $ac,bc$ present. Let $P,Q,p,q$ be disjoint from these vertices and each other, with

$$ A=P\cup\{p,u,c,h\}=K_9-pu,\qquad B=Q\cup\{q,u,c,h\}=K_9-qu. $$

If $ap,bq,pq$ are edges and there are $v\in P,s\in Q$ with $bv,as$ edges, then the graph is not biplanar.

Proof. If $ap$ has layer $L$, then $u$ is not in the $L$-component of $a$ inside $C$: otherwise the path from $p$ through $a$ to $u$ would be exterior to $A$. Lemma 1 connects $a,b$ in that layer, so $u$ is separated from both. The layer of $bq$ similarly separates $u$ from both $a,b$. The two port edges must consequently have the same layer, since the actual edge $au$ puts $u$ with $a$ in one layer. Name their common layer red.

The edge $pq$ is blue, since red $a-p-q-b$ would be exterior to $C$. The edges $ac,bc$ have opposite layers. By interchanging the endpoint and leaf roles if necessary, assume $ac$ red and $bc$ blue. The two private-attachment hypotheses allow this interchange.

The edge $qc$ must be blue, so $p,q,c$ are blue-connected to $b$. If $as$ were blue, both $sc,sq$ would be red, giving the forbidden red path $a-c-s-q-b$. Hence $as$ is red, and $sq$ is blue to avoid $a-s-q-b$ in red. If $su$ is red, $p-a-s-u$ is exterior to $A$; if it is blue, $p-q-s-u$ is exterior to $A$. Both are forbidden. The proof uses only $c,p,q,s$ internally for paths exterior to $C$, and only $a,s$ or $q,s$ for the final paths exterior to $A$. $\square$

For the central host $T$, take $C=T-c$; it meets either leaf exactly at $u$. No edge $ah$ or $bh$ is needed.

Theorem 12 (connector $c$). Let $T,h,u,P,Q,p,q$ be as in Section 5, but let the two leaf subgraphs be $K_9-pc$ and $K_9-qc$. If $ap,bq$ are edges, the graph is not biplanar. Neither $pq$ nor private attachments to $a,b$ are required.

Proof. Choose $z\in V(T)\setminus\{a,b,c,d,u\}$. Use (1). Exterior paths $c-a-p$ and $c-b-q$ force $ap$ blue and $bq$ red.

All of $a,b,d,z$ lie outside both leaves. In blue, $c$ cannot connect to $a$ through these vertices because appending $ap$ would complete the first leaf. Thus the blue-connected sets $\{c,b\}$ and $\{a,d\}$ are separate. In red, the analogous argument with $bq$ separates $\{c,a\}$ from $\{b,d\}$.

If $cz$ is red, then $bz,dz$ must be blue, producing the forbidden blue path $c-b-z-d-a$. If $cz$ is blue, then $az,dz$ must be red, producing the forbidden red path $c-a-z-d-b$. This exhausts the choices for $cz$. $\square$

The extra vertex $z$ is required to avoid the leaf hosts; there are five possible choices in the stated $T$. This theorem addresses a common triangle containing an endpoint $c$ of the other missing pair. It does not address a triangle containing only the outside vertex and two of the six remaining central vertices.

7. Neighborhood components and the shared completion deficit

Fix a biplanar partition $G_0,G_1$ of a simple graph on $n\ge4$ vertices. Write $b=6n-12-e(G)$, and complete each layer to a simple sphere triangulation $T_i$ on the same vertices. Exactly $b$ edge occurrences are added in total. An edge already present in the other original layer may be added to $T_i$; it still costs one occurrence. This convention is the common-completion convention of [7].

For a vertex $v$ and $X\subseteq N_G(v)$, let $\mathcal D$ be the components of $G[N_G(v)\setminus X]$. Put $N_i=N_{G_i}(v)$, let $q_i$ count members of $\mathcal D$ meeting $N_i$, and define $\phi(0)=\phi(1)=0$ and $\phi(q)=q$ for $q\ge2$.

Theorem 13 (local component charge). In a fixed simultaneous completion,

$$ \phi(q_0)+\phi(q_1)\le |X|+\delta_v\le |X|+b. \tag{3} $$

Here $\delta_v$ counts added spokes incident with $v$ and added edges on the facial link of $v$ whose endpoints both belong to $N_i\setminus X$, summed over the two layers. These two kinds of charge are disjoint.

Proof. In the simple link cycle of $v$ in $T_i$, delete the neighbors introduced by added spokes and the original neighbors $N_i\cap X$. Delete all remaining added rim edges. The retained graph has exactly $N_i\setminus X$ as its vertex set and only original edges of $G_i$. It therefore has at least $q_i$ components. If $d>0$ vertices and edges are deleted from a cycle, at most $d$ nonempty components remain. When $q_i\ge2$, at least one deletion was made, so $q_i\le |N_i\cap X|+\delta_{v,i}$; for smaller $q_i$ the assertion with $\phi$ is automatic. Sum over the layers. Since the original edge sets are disjoint, $|N_0\cap X|+|N_1\cap X|=|X|$. Each counted addition is one of the $b$ occurrences. $\square$

The elementary cycle-deletion argument is standard Hamiltonian-cycle reasoning, familiar from toughness [9]. Completing the layers and counting local sectors are also established techniques [7], [8]. The specific information here is the number of original neighborhood components met by each layer, with the added-spoke convention retaining correctness when a pair occurs in the other original layer. The same proof works for any fixed number of planar layers.

7.1. Equality identifies the repair

Suppose $N_G(v)=P\sqcup Q\sqcup X$, where $P,Q$ are nonempty and anticomplete; they need not themselves be connected. Say that a group is mixed when it contains a neighbor of $v$ in each original layer. If both groups are mixed, the proof of Theorem 13 with two group labels gives $|X|+b\ge4$. Each link needs at least two deletions.

Corollary 14 (one-occurrence support). Suppose $b=1$, $|X|=3$ and both groups are mixed. The layer using the sole addition has one original $X$-neighbor; the other has two. In the repaired link there are exactly two effective deletions, leaving one path through all original neighbors in each group. The underlying pair of the addition belongs to

$$ \mathcal S_v=\{pq:p\in P,q\in Q\}\cup \{vw:w\in X\cup(V(G)\setminus N_G[v]), N_G(w)\cap P\ne\varnothing,\ N_G(w)\cap Q\ne\varnothing\}. \tag{4} $$

Proof. Equality in the two link-deletion counts leaves precisely two nonempty paths in each link. If the addition is a rim edge, its ends must belong to different groups. If it is a spoke $vw$, its new link vertex $w$ has one neighbor in each group. Both corresponding rim edges are original, since $vw$ is the sole addition. Thus $w$ has original contacts to both groups; it cannot itself belong to either group by anticompleteness. This gives (4). Consecutive removed vertices would waste a deletion and cannot occur at equality. $\square$

A spoke to $w\in X$ is an allowed duplicate of an opposite-layer edge. For several vertices with these hypotheses, the same added occurrence must have its pair in every $\mathcal S_v$. An empty intersection is a contradiction. Adding the individual lower bounds as if they consumed different edges would be invalid. Intersecting completion supports is already used for global separators in [7, Section 7]; (4) specifies the support for a local neighborhood link.

Corollary 15 (two exclusive hubs). Suppose $b\le1$, and nonadjacent $h,k$ have complete neighborhoods $P_i\sqcup Q_i\sqcup X_i$, $i=h,k$, with $|X_i|=3$ and $P_i,Q_i$ nonempty and anticomplete. Suppose the four private groups are pairwise disjoint, contain neither hub, and each hub has no contact to the other hub's private groups. Then at least one private group is monochromatic at its own hub.

For $b=0$ the scalar bound suffices. For $b=1$, their rim supports have disjoint endpoints, and no rim pair at one hub can be a spoke at the other. The only common spoke pair could be $hk$, which is in neither support because the required private contacts are absent. Corollary 14 therefore excludes both hubs mixing both groups.

An actual sharp control is supplied. It has 36 vertices and 204 edges, two edge-disjoint spherical triangulations, and a hub whose two anticomplete groups both mix with a four-vertex separator. Deleting its spoke $1\,29$ in layer 1 leaves 203 edges and a three-vertex separator, with both groups still mixed. The sole repair is the deleted spoke. The rotation-system verifier checks these assertions directly. This control is a geometric threshold example; no chromatic lower bound is claimed for it.

7.2. One occurrence pays at most four local charges

Choose a possibly different $X_v\subseteq N_G(v)$ for each $v\in W$. With the corresponding component counts, put

$$ d_v=\max\{0,\phi(q_0(v))+\phi(q_1(v))-|X_v|\}. $$

Theorem 16 (packing local demands). For every selected vertex set $W$,

$$ \sum_{v\in W}d_v\le4b. \tag{5} $$

If $W$ is independent in the original graph $G$, then

$$ \sum_{v\in W}d_v\le2b. \tag{6} $$

Both constants are attained by actual biplanar graphs.

Proof. Theorem 13 gives $d_v\le\delta_v$. Fix an added occurrence $xy$ in $T_i$. It is charged as a spoke only at $x,y$. A rim charge at $v$ requires $xvy$ to be a facial triangle of $T_i$. There are exactly two distinct facial-third vertices at $xy$, so the occurrence is charged at at most four vertices. Sum over the $b$ occurrences to obtain (5).

For (6), retain each facial-third vertex only if it is eligible for the rim charge. Such a vertex has original $G_i$ edges to both $x,y$; vertices introduced by added spokes have already been removed from the link. The original graph therefore contains all edges between the two endpoints and the retained third vertices. An independent set can meet only one of these sides, hence at most two vertices of this occurrence's charge set. Summing gives (6). $\square$

The argument also gives $2b$ whenever $G[W]$ has maximum degree at most one, since any three vertices of a charge set contain a two-edge path. It gives $3b$ when $G[W]$ has no four-cycle as a subgraph. These follow from the same incidence argument, without additional hypotheses on the completed graph.

For sharpness, delete edge $10\,34$ from layer 1 of the 36-vertex control. The unique quadrilateral is $10,14,34,27$ in cyclic order. The independent spherical checker verifies the following original neighborhood data:

| $v$ | $X_v$ | $(q_0,q_1)$ | $d_v$ | |---|---|---|---| | 10 | $\{20\}$ | $(1,2)$ | 1 | | 14 | $\{20\}$ | $(1,2)$ | 1 | | 27 | $\{15,20,28\}$ | $(2,2)$ | 1 | | 34 | $\{12,15,20\}$ | $(2,2)$ | 1 |

Here $b=1$, the four demands sum to four, and each of $\{10,34\}$ and $\{14,27\}$ is independent. Thus (5) and (6) are sharp. The larger corroborating audit checks all 204 single-edge deletions of the control and 40 seeded multiple deletions: 244 actual graphs, 397 occurrence charge sets, 559,924 neighborhood tests and 2,440 independent-set tests.

The demands retain their dependence on the actual layer partition. Theorem 16 does not establish that an arbitrary coloring obstruction forces any of them to be positive. At $b=1$, it explains exactly why four local failures may share one quadrilateral.

7.3. A one-sided repair rule

The following variant accommodates an extra group with no private contacts. It isolates the link argument from any missing-edge derivation of the prescribed colors.

Proposition 17. Let $b\le1$ and $N_G(h)=P\sqcup Q\sqcup\{a,c,u\}\sqcup Z$, with $P,Q$ nonempty and anticomplete and $Z$ anticomplete to $P\cup Q$. Suppose $ha$ is blue, $h$ has a red neighbor in each of $P,Q$, and, whenever $hc,hu$ are both red, it has a blue neighbor in each group. Then $b=1$. If exactly one of $hc,hu$ is red, the red link uses the unique completion occurrence; if both are red, the blue link does. The repaired layer has no original $Z$-neighbor. Its repair is a missing $P$–$Q$ rim edge or a spoke $hw$ with original contacts to both groups, where $w\in\{a,c,u\}\cup(V(G)\setminus N_G[h])$.

Proof. Write $r$ for the number of red edges among $hc,hu$. If $r\le1$, the red link has neighbors in both private groups and only $r$ original separator neighbors. Deleting those and the locally charged additions leaves at least two components, and at least three if an original $Z$-neighbor is present. Thus $2\le r+\delta_h\le r+b\le2$. Equality forces $r=b=\delta_h=1$ and no such $Z$-neighbor. If $r=2$, use the stated blue contacts and its sole separator neighbor $a$ to get the same equality. The rim/spoke proof of Corollary 14 now applies in the repaired layer. A spoke endpoint in $Z$ cannot contact either private group. $\square$

If the two leaf hosts are actual $K_9-ph$ and $K_9-qh$ on $P\cup\{p,h,c,u\}$ and $Q\cup\{q,h,c,u\}$, the conditional blue-contact premise follows from Lemma 1: $h$ is a missing-edge endpoint in both hosts. Ordinary vertices of $K_9-e$ may be isolated in a restricted layer, and no stronger non-isolation assertion is used. The crossed-port application motivating this proposition is due to hajos_weaver [6]. Under the exclusivity hypotheses of Corollary 15, two such repaired hubs have disjoint supports and cannot coexist at $b\le1$.

8. Complete port neighborhoods at Euler equality

Suppose $e(G)=6|V(G)|-12$. Both layers are spanning simple triangulations. Let $C=\{p_1,\ldots,p_t\}$ induce $K_t$, where $t=3$ or $4$. For each port $p_i$, let $U_i$ be a five-clique with $U_i\cup\{p_i\}$ a six-clique. These sets are pairwise disjoint. Assume the complete neighborhood is

$$ N_G(p_i)=U_i\sqcup(C\setminus\{p_i\})\sqcup Z_i, \qquad |Z_i|=5-t, \tag{7} $$

where every $Z_i$ lies outside $C\cup\bigcup_j U_j$. Thus $U_i$ and $C\setminus\{p_i\}$ are anticomplete. Other edges at private vertices and among the high vertices are unrestricted.

Theorem 18 (small tight connectors). For $t=3$, all connector edges have the same layer. For $t=4$, no such Euler-tight biplanar graph exists.

Proof. At a port for $t=3$, the two connector neighbors and the private group are anticomplete with a two-vertex separator. If both groups meet both layers, their links need four separator neighbors, which is impossible. If the connector spokes have different layers, all five private spokes must consequently lie in one layer. The other layer has only one connector neighbor, so its degree at least three forces both high spokes there. The first layer then meets both anticomplete groups without a separator and cannot have a link cycle. Hence the connector spokes agree at each port, and the whole triangle is monochromatic.

For $t=4$, the separator has one vertex. A link cycle cannot meet both anticomplete groups. Thus all five private spokes have one layer and all three connector spokes the other. The connector $K_4$ again has one common layer, say red. A high vertex with one red port contact has at least three: if $p_i z_i$ is red, the four-vertex red link at $p_i$ joins $z_i$ to two other ports. Distinct high vertices have disjoint port-contact sets, since each port has only one possible high neighbor. At most one high can therefore have red contact with $C$. No private spoke is red, so removing that one high, if present, disconnects $C$ from the twenty private vertices in the red spanning triangulation. This contradicts its 3-connectivity. $\square$

For $t=3$, call a port active when it has a red private neighbor, where red is the connector layer. An active port must have both high spokes red. Its red link comprises a path on its two connector neighbors and a path on its red private neighbors, separated by the two highs. All blue neighbors are private, so the blue degree bound leaves exactly one or two red private spokes. A nonactive port has at least one red high neighbor.

Theorem 19 (variable high pairs). The three two-sets $Z_p,Z_q,Z_r$ intersect pairwise. Unless they are the three different two-subsets of one three-set, ports and high vertices can be named so that

$$ Z_p=Z_q=\{x,y\},\qquad Z_r=\{x,z\}, $$

with $xp,xq,xr,yp,yq$ red and $rz$ blue; $z=y$ is allowed. The active ports form a nonempty subset of $\{p,q\}$, each with one or two red private spokes. If exactly one is active, $xy$ is an actual red edge.

Proof. Each connector edge has two distinct incident facial third vertices. Its only possible common neighbors are the third port and members of the two high-pair intersection, so every pair intersects. Three pairwise-intersecting two-sets either have a common element or are the three pairs of a triangle.

Every high with a red port contact has at least two such contacts. At an active port this follows from the link path on the other two ports. At a nonactive port, the red link lies on its two connector neighbors and one or two highs; each high must meet another port. With just one red high at a port, that high meets both other ports.

If three highs contact $C$ in red, all six available high spokes are red and every high occurs twice, giving the exceptional three-pair pattern. Outside it there are at most two red-contact highs. Some port must be active, since otherwise deleting those at most two highs disconnects $C$ from the private vertices in the red triangulation. Thus there are exactly two such highs, $x,y$.

They cannot both contact all three ports: an active port supplies a red $x$–$y$ path with private interior, which completes the red $K_5-xy$ on $C\cup\{x,y\}$ to a $K_5$ subdivision. A port with only one red high therefore exists; its degree-three triangular link makes that high, say $x$, contact all three ports. The other high $y$ has exactly two red contacts, named $p,q$. They have both highs in their actual pairs; $r$ is nonactive with $rz$ blue. If, say, $q$ is nonactive, its red link is a cycle on $\{p,r,x,y\}$. Since $yr$ is not red, the two link neighbors of $y$ must be $p,x$, forcing $xy$ red. $\square$

When all pairs cross a fixed bipartition, the exceptional three-pair pattern is impossible. Theorem 19 then forces two actual pairs to coincide, rather than merely two reference-color labels. It does not exclude the connector or establish that the exceptional pattern is realizable.

9. A portable exclusion of the common-triangle connector

Let $F$ have an actual triangle $\{t,x,y\}$, three disjoint six-cliques $U_s\cup\{s\}$ with $s\in C=\{p,q,r\}$ and $|U_s|=5$, every private vertex adjacent to all of $t,x,y$, every port adjacent to $x,y$ but not $t$, and the port triangle $C$. These are its 102 edges on 21 vertices.

Theorem 20 (computer-assisted local exclusion). There is no Euler-tight biplanar graph $G$ containing $F$ and satisfying the complete port neighborhoods

$$ N_G(s)=U_s\sqcup(C\setminus\{s\})\sqcup\{x,y\} \quad(s\in C). \tag{8} $$

Extra edges at private vertices and at $t,x,y$ are allowed. The theorem does not assert that the unpinned graph $F$ itself is nonbiplanar.

9.1. Hand normalization

By Theorems 18–19, name the connector layer red, take $x$ to have three red port contacts and $y$ to have two, and name the unique port with blue $y$-spoke $r$. There are one or two active ports among $p,q$, each with one or two red private spokes. If one is active, name it $p$ and $xy$ is red. If two are active, exchange $p,q$ to put their private red counts in increasing order. Hence the eight possibilities are

$$ (1,0,0;\mathrm{red}),\ (2,0,0;\mathrm{red}),\quad (1,1,0;c),\ (1,2,0;c),\ (2,2,0;c), \qquad c\in\{\mathrm{red},\mathrm{blue}\}, \tag{9} $$

where the final entry specifies the color of $xy$.

Relabel private vertices along the inherited port links. At an active port, list the one or two red private neighbors in order on the red $x$–private–$y$ path; the other three or four vertices form its blue link cycle. At a nonactive $p$ or $q$, its five private vertices form the blue cycle. At $r$, the blue cycle consists of $y$ and all five private vertices. Its red link is the triangle on $p,q,x$; a nonactive $p$ or $q$ has its forced four-vertex red link. These give all edge pins used by the certificate.

The permutations are legitimate for the displayed $F$, regardless of asymmetry in omitted private contacts of $G$: the argument uses no such contacts and imposes no restrictions on them. It pins required edges, not a full embedding of a restricted layer. Both one-active patterns are retained. No Euler equality is imposed on the 21-vertex restriction.

9.2. Necessary formula and checked contradiction

Assign each $F$ edge to exactly one layer. Eight gates choose one of (9) with its link pins. Each actual host $U_s\cup\{s,t,x,y\}$ is $K_9-st$. For each layer, include all internal $s$–$t$ cut clauses required by Lemma 1, and binary component labels separating $s,t$ after deleting the other seven host vertices. Thus every hypothetical $G$ supplies a satisfying assignment of the 1,863 base clauses on 303 variables. The standard-library reconstruction in the supplement specifies every variable and clause independently of the search encoder.

Planarity cuts forbid all edges of a specified nonplanar subgraph from lying in one layer. Every retained such clause has an exact contained $K_5$ or $K_{3,3}$ subdivision in the evidence. The semantic checker suppresses degree-two paths and checks the branch graph, with disjoint interiors and no loop paths. It uses no planarity library.

The retained unsatisfiable core has 15,436 clauses: 560 base clauses and 14,876 geometric clauses. Each is independently checked as one of the above necessary clauses. The supplied additions-only proof passes RUP against that core alone using DRAT-trim [11] with -W -U. Its core hash is

91e825ab5f8c8d6d0e6612b9fc482f5303662beda4e6dd30dd7c1b4823069675

This proves the contradiction required for Theorem 20. The logical implication from an ambient graph to the base clauses is Sections 9.1–9.2; the proof checker establishes unsatisfiability, not that implication by itself. The archive includes the C checker source and a standalone Python driver. It needs Python 3 and a C compiler, with no SAT solver, NetworkX or PySAT. Temporary expanded proof and compiled-checker files are removed. The proof is stored as gzip text and expands to 4,688,589 bytes during checking; this expansion is explicit, rather than included in the ZIP's uncompressed entry total.

Theorem 20 is a local exclusion under actual neighborhoods and edge equality. To apply it to a critical-graph reduction, one must separately establish those premises and the common singleton triangle. In particular, leaves missing two incident port edges are not $K_9-e$ hosts; Lemma 1 and this certificate do not apply to them.

10. Mixed two-leaf bridges beside the central host

The orientation restrictions of Sections 5–6 can be strengthened to a subgraph exclusion with no ambient density or neighborhood assumptions. The central host matters: the bare three-vertex-overlap graph of Section 3 is biplanar, whereas its mixed attachments beside the host below are not.

Theorem 21 (central-triangle bridge). Let $T=K_{10}-\{ab,cd\}$, where the missing edges are disjoint. Let $C$ be an actual triangle disjoint from $\{a,b\}$, containing at most one vertex outside $T$. If $C$ contains an outside vertex $h$, assume that $h$ is adjacent to at least one of $a,b$. Choose any $t\in C$. Outside $V(T)\cup C$, take disjoint five-sets $P,Q$ and distinct ports $p,q$. Suppose a graph contains

$$ F_P=G[P\cup\{p\}\cup C]\supseteq K_9-pt,\qquad F_Q=G[Q\cup\{q\}\cup C]\supseteq K_9-qt, $$

and the edge $pq$. Every private vertex and every port has an edge to $a$ or $b$. At the private vertices, one can choose one such attachment per vertex so that both endpoint types occur in each five-set. Then the graph is not biplanar.

If an omitted port edge is present, its host contains $K_9$ and the conclusion is immediate. Otherwise keep exactly the displayed edges and one selected endpoint attachment at each vertex. Extra edges can be discarded. In particular the theorem requires no private anticompleteness in the ambient graph, no complete neighborhoods, no Euler equality, and no prescribed layer colors.

10.1. Complete role reduction

Suppose first that $C$ contains one outside vertex $h$. Interchange $a,b$ if necessary so $ah$ is present. Its two central vertices form an actual edge outside $\{a,b\}$. Since $cd$ is absent, they consist either of one endpoint of $cd$ and one ordinary central vertex, or two ordinary central vertices. Relabel these triangles to $\{h,c,u\}$ and $\{h,u,z\}$, respectively, with $u,z\notin\{a,b,c,d\}$. The missing connector $t$ has three roles in the first triangle and two in the second, where exchanging $u,z$ identifies their roles.

If $C\subseteq V(T)$, it contains at most one of $c,d$. It is either $\{c,u,z\}$, with $t=c$ or $t=u$ up to exchanging $u,z$, or consists of three ordinary central vertices, one designated $t$. This gives three further roles. Thus eight roles exhaust the theorem.

Each role allows all four port endpoint choices and all mixed private assignments. There are $30$ mixed assignments on a labeled five-set, so $30^2\cdot4=3600$ labeled assignments per role. Permuting private vertices so their $a$-types precede their $b$-types leaves four counts per group. All actual graphs are retained up to that relabeling; an asymmetric fixed assignment is not asserted to have these automorphisms.

10.2. Necessary formulas and sound symmetry choices

The edge variables choose one layer for each mandatory edge. Each optional attachment chooses one of four endpoint/layer possibilities. Clauses require both private types. For the four central $K_9-e$ hosts and both leaf hosts, exterior component labels enforce Lemma 1(2) in each layer. A positive label at one missing endpoint, a negative label at the other, and equality of labels along each selected exterior edge express precisely the needed separation. Extra components may share a label; no connectivity sufficiency is assumed.

For the outside-vertex roles, no face or degree constraints are imposed. The selected graph has $23$ vertices and $123$ edges, but the encoding does not use that count. One actual edge fixes the layer names. Private types are sorted as above. Remaining comparisons, when used, are under literal permutations of unused central vertices and whole leaf roles; the outside-connector anchor role also permits exchanging its two anchors. They preserve the layer-normalizing edge and the sorted private domain. Choose the lexicographically least edge/layer vector in the remaining finite orbit. It is at most its image under every listed generator, which justifies imposing all comparisons together. Prefix variables encode equality of earlier coordinates. Component labels can be reassigned after relabeling. Each independent encoder checks preservation of the mandatory and optional edge sets.

For the wholly central triangles, the selected graph itself has $22$ vertices and $120$ edges: $43$ from $T$, $32$ new edges per leaf after subtracting its central triangle, twelve attachments, and $pq$. If biplanar, both layers have $60$ edges and are spanning triangulations. Every assigned edge therefore lies in at least two distinct same-layer triangles. The formula adds triangle variables equivalent to the conjunctions of their three edge/layer literals and imposes this necessary condition. It uses the selected graph's own equality, not a hypothesis about the ambient graph. No planarity conclusion is inferred from these triangle constraints alone.

For possible triangle variables $z_1,\ldots,z_s$ through an edge with layer literal $e$, the clauses

$$ \neg e\;\lor\!\bigvee_{j\ne i}z_j\qquad(1\le i\le s) $$

require at least two of them when $e$ is true. An empty list forbids $e$. The independent builder uses this direct encoding, without an external cardinality package. One unused vertex label is retained in the files for consistent numbering; it is excluded from the $22$-vertex count and has no incident edges.

10.3. Checked finite contradiction

The following table gives the independent base and the retained cores included in this version. The labels distinguish a triangle meeting the opposite missing pair (cross), ordinary anchors only, and an entirely central triangle. Every role retains arbitrary mixed attachments.

| Triangle role and connector | Variables | Base clauses | Core clauses | Geometric clauses | |---|---:|---:|---:|---:| | Outside, cross, $h$ |462|2023|12837|12317| | Outside, cross, $c$ |761|3873|6713|6403| | Outside, cross, $u$ |761|3905|14024|13387| | Outside, anchors, $h$ |844|4151|19288|18074| | Outside, anchors, $u$ |761|3741|22877|21796| | Central, cross, $c$ |1500|8833|2554|230| | Central, cross, $u$ |1500|8865|6402|4087| | Central, ordinary connector |1500|8713|9037|5767|

Each retained clause is either an independently reconstructed necessary clause or names an exact monochromatic subdivision of $K_5$ or $K_{3,3}$. A standard-library recognizer checks the latter directly. Glucose [12] then generates a Boolean refutation, and DRAT-trim [11] verifies it in strict RUP mode. All eight checks pass. The solver's UNSAT receipt is not accepted as proof without this independent check. The role reduction, necessary-clause arguments and finite refutations together prove Theorem 21. No structural occurrence theorem is imported into it. $\square$

11. Two shared hubs: an exact residual family

The next result concerns a specified graph family, separately from any claim that a minimum counterexample belongs to it.

Lemma 22 (central connectivity). In every biplanar partition of $T=K_{10}-\{ab,cd\}$, the vertices $a,b,c,d$ lie in a single connected component of each restricted layer. Consequently, if $F=P\cup\{p,r,s,c\}=K_9-pc$ with $P,p,r,s$ outside $T$, then an edge from $p$ to either $a$ or $b$ makes $T\cup F$ nonbiplanar.

Proof. The host $T-c$ connects $a,b$ in each layer by Lemma 1. Edges $ac,bc$ have opposite layers, since otherwise $a-c-b$ is an exterior path for $T-c$. Thus $c$ joins the $a,b$ component in both layers. The same argument using $T-d$ brings in $d$. In the layer of the port attachment, a path from its endpoint to $c$ within $T$ is exterior to $F$, contradicting Lemma 1. $\square$

Theorem 23 (shared-hub family). Let $T=\{a,b,c,d,u,z\}\cup J=K_{10}-\{ab,cd\}$, where $J$ is a four-set. Add hubs $r,s$ with edges $rs,ar,as,cr,cs,ur$, and exactly one of $zr,zs$. Add two bridge components. The first has two disjoint private five-sets, two ports joined by an edge, own endpoint pair $ab$, and common triangle $\{r,s,c\}$. The second has the analogous form with own pair $cd$ and triangle $\{r,s,a\}$. The four private five-sets induce cliques. They and the four ports are pairwise disjoint and lie outside $T\cup\{r,s\}$. Each private vertex sees its whole triangle, its five-set's port, and exactly one own endpoint, with both types in every five-set. Each port sees one own endpoint and its triangle except for a common connector chosen from $\{r,s\}$ for that component. There are no other required edges. Every graph containing this selected configuration is not biplanar.

Proof. The selected graph has $36$ vertices and $204$ edges, so a hypothetical partition consists of two spanning triangulations. One formula retains both anchor distributions, both connector choices in each component, all mixed private attachments and all port choices. As in Theorem 21, it imposes exterior separation for the actual $K_9-e$ hosts and at least two same-layer triangles through each selected edge. The leaf separation clauses are guarded by the selector making that hub its actual missing connector. Thus no conclusion from the wrong missing-edge host is used.

Name $rs$ red, sort private types, and normalize only unused-$J$ permutations and the two independent whole-leaf exchanges. These preserve both high patterns, all connector selectors and the fixed attachment $ur$. The same least-orbit-vector argument justifies their comparisons. There are $30^4\cdot16\cdot4\cdot2=103680000$ labeled selections before private relabeling; the formula covers all of them.

The independent standard-library builder reproduces all $20086$ base clauses on $2963$ variables. The base is satisfiable. The retained core has $4639$ clauses: $2770$ ordinary and $1869$ exact geometric subdivision clauses. Independent semantic checks and regenerated strict RUP pass. The proof therefore excludes the selected family and its supergraphs. $\square$

The central third vertex cannot serve as connector in this family by Lemma 22. A separate high-edge counting argument can force the two distributions in Theorem 23 under the forum's no-dormant, sole-whole-$K_4$ occurrence hypotheses. That conditional application is documented in the supplement, but those occurrence reductions are not proved or claimed as a global classification in this paper. Other high-core profiles and dormant components remain outside this conclusion.

12. A drawable conditional equality and its composition limit

The central-host obstruction should not be mistaken for impossibility of the underlying local color relation. Take a single leaf $F=P\cup\{p,h,u,z\}=K_9-ph$ and add vertices $a,b$ outside it. Include $ah$, one chosen endpoint contact from each of $u,z,p$ to $a$ or $b$, and one chosen endpoint contact from every private vertex, with both types in $P$. Assume no other edges. This graph has $11$ vertices and $44$ edges.

Proposition 24 (exact boundary relation). A proper precoloring of $\{a,b,h,u,z,p\}$ with names from a nine-color palette extends precisely when

$$ \operatorname{col}(a)\ne\operatorname{col}(b) \quad\text{or}\quad \operatorname{col}(p)=\operatorname{col}(h). $$

Proof. If $a,b$ coincide, their color differs from all three triangle colors and the port color. If $p,h$ differ too, the private $K_5$ has only four colors available, impossible. If $p=h$, exactly five colors remain and the clique can use them.

If $a,b$ differ, remove the colors on $\{h,u,z,p\}$ from the palette, leaving a set $L$ of five or six colors. Each private list is $L$ minus its chosen endpoint color if present. When $|L|=6$, all lists have at least five colors. When $|L|=5$, all have at least four; every subfamily of at most four lists satisfies Hall's condition. Both endpoint types occur and their colors differ, so the union of all five lists is $L$. Hall's theorem [13] gives five distinct representatives. $\square$

Two such leaves with disjoint private sets, common $a,b,h,u,z$, and edge $pq$ therefore have the exact nine-color relation $a\ne b$ on their proper boundary precolorings. It suffices that each anchor have one endpoint contact; four anchor contacts are unnecessary. The minimal two-leaf carrier has $17$ vertices and $83$ edges. Its biplanarity is not established here.

The supplement gives an actual single-leaf realization with labels $a=0,b=1,h=2,u=3,z=4,P=\{5,\ldots,9\},p=10$. Its extra edges are $02,13,04,05,06,17,18,19,1\,10$. Independently traced facial walks give layers with $19$ edges and $10$ faces, and $25$ edges and $16$ faces, both on $11$ vertices and of Euler characteristic two. The coloring

$$ (1,2,0,1,2,3,4,5,6,7,0) $$

and the $K_8$ on $\{2,\ldots,9\}$ prove chromatic number eight. A nine-coloring giving six distinct boundary colors is also supplied.

There is a precise geometric limitation. If the port sees outside endpoint $t\in\{a,b\}$, then $h,t$ cannot be cofacial in the layer of $pt$: adding $ht$ would give an exterior $p-t-h$ path completing $F$. Hence $\{a,b,h\}$ cannot be common cofacial in both layers. Gluing through one common face containing the whole natural boundary is ruled out, but gluing through several regions or a different overlap is not. This is a direct use of the established exterior-path principle, not a new minor-closure assertion.

For comparison, every proper nine-precoloring of the bare bridge boundary $\{h,u,z,p,q\}$ in Section 3 extends: each private clique sees at most four colors. The conditional relation, its actual single-leaf realization, and the common-face limitation distinguish useful local color pressure from the unproved geometric compatibility of a larger assembly.

13. Evidence and reproduction

The results of Sections 2–6 are hand theorems. Their unchanged evidence and the Section 7–9 evidence remain publicly available in the accepted version 2 supplement. In that archive, the overlap/ directory contains small independent checks of their hypotheses and finite path implications, and the full positive certificate of Section 3. Run:

python3 overlap/verify_all.py

Python 3.10 or newer is sufficient; no SAT solver, planarity library or third-party Python package is required. The checks are:

• reconstruction of the two $K_9-e$ subgraphs and the $68$-edge positive graph, its proper nine-coloring, forced eight-color equalities, and both spherical rotation systems;
• the parallel-attachment implication: $0$ of $2,048$ assignments survive the opposite orientation, while $28$ survive the same orientation;
• the bridge implication: $0$ of $2,048$ assignments survive the forbidden bridge layer, while $56$ survive the other layer; deleting either $ah$ or $ap$ restores $28$ assignments in the forbidden layer;
• the connector-$u$ final chase: $0$ of $64$ assignments; the internal-connectivity transfer remains a separate hand argument;
• the connector-$c$ chase: $0$ of $1,024$ assignments, with positive deletion controls;
• the shortest common-neighbor obstruction: $0$ of $2,048$ assignments, with $8$ survivors on each of three edge-deletion controls; type propagation is also checked on all $44$ labeled connected graphs of order at most four.

Each path checker uses an explicit allowed vertex set obtained by removing the other vertices of the designated host. The nonzero controls show that the selected necessary path conditions are not vacuous. Surviving assignments are not claimed to be planar partitions. Finite propagation checks do not replace the hand argument for arbitrary connected components.

The parallel-attachment audit additionally checks a literal $36$-vertex, $204$-edge instance of Corollary 10. Only its listed graph and the corollary's exact hypotheses are used; no criticality, hereditary density census, or solver result is asserted here. The positive graph was found with the bipsat2 implementation from Sean Cook's Color the Colonies project [14]; the retained rotations are checked independently.

13.1. Preserved evidence for Sections 2–9

From the version 2 supplement root, run python3 verify_all.py. The script runs the previous overlap checks, the neighborhood cycle audit, two actual spherical-control checks, the variable-pair arithmetic audit, and the portable core check. Python 3 and a C compiler are sufficient. The new routines use the standard library. --skip-rup runs the small checks without compiling or replaying the already checked core; the complete theorem requires the core verification as well.

The budget/ directory includes the full 36-vertex control and rotations, the exact local-threshold check and the separate four-charge sharpness check. The larger 244-graph packing audit and its result are included for optional corroboration. The k3/ directory contains the fixed base reconstruction, compact core, exact subdivision witnesses, additions-only proof, and DRAT-trim source with its original license. It can also be checked alone with python3 k3/verify_k3_portable.py. All evidence files carry hashes in the supplement manifest.

The single proof's gzip stream expands to approximately 4.7 MB during verification. The ZIP and its entry total are both below the service's 2 MB limit; this does not mean the expanded proof is below that limit. The verifier places temporary expansions inside its own directory and removes them. No large search logs or the earlier unpublished 23-vertex proof collection are required for any theorem of this manuscript.

The cycle and boundary enumerations corroborate hand arguments. They do not replace arbitrary-order proofs, certify structural occurrence, or turn a satisfying boundary signature into a planar drawing. The proof certificate establishes Theorem 20 only with its displayed normalization and complete-neighborhood premises.

13.2. Current portable bridge evidence

The current supplement covers Theorems 21 and 23 and Proposition 24. Run python3 verify_bridges.py. It requires Python 3 with lzma, a C compiler and a C++ compiler, with no installed SAT solver, planarity library, network access or external data. The archive includes the sequential Glucose 4.1 source and its license [12], a small DIMACS driver, and DRAT-trim source [11]. It reconstructs each necessary formula, checks every retained geometric clause as an exact subdivision, generates a Boolean proof from its retained core, and verifies the proof in strict RUP mode. The solver's output is never trusted without that check. Omitting deletion lines keeps a stronger clause database and preserves RUP validity. Each proof ends with the empty clause.

All nine exported roles passed this procedure. Solving and checking took about 45 seconds total on one CPU thread, after compilation. Individual generated proofs were below 8 MB; each was removed after its check. The verifier bounds files at 64 MiB and keeps all temporary source, binaries and logs inside its own temporary directory. Thus the supplement stores exact audited cores and a complete reproducible proof-generation-and-checking route, rather than a receipt asserting that an inaccessible proof was checked. The retained original author proofs remain an additional development record, but the public procedure does not need them.

The standard-library positive and coloring checks are verify_single_leaf_boundary.py, verify_stripped_bridge.py and audit_minimal_boundary_relation.py. They verify both sets of spherical rotations and explicit colorings, and check all 7,040 canonical minimal-carrier boundary/count cases against the hand Hall argument. audit_shared_hub_frontier.py corroborates only the final high-neighborhood count; it is not used to assert unproved occurrence premises. PREVIOUS_EVIDENCE.md identifies the unchanged version 2 evidence and corrects the historical supplement's unnecessary private-group anticompleteness wording. No missing diagnostic reference is needed for a current claim.

14. Scope and further questions

The new central-triangle bridge theorem excludes a selected subgraph without ambient density assumptions. Its wholly central cases use the selected subgraph's own exact edge count; its outside-vertex cases need no face constraints. The shared-hub theorem concerns one fully specified family. Both leave triangles with multiple outside vertices in other configurations, different hosts, unmixed attachments and other low-component architectures open. The minimal two-leaf inequality carrier remains geometrically unresolved.

The overlap theorem is uniform in the number of planar layers, and the common-neighbor theorem restricts actual attachments to a two-missing-edge host. The neighborhood charge theorem adds a different resource: original neighborhood components must fit the same finite collection of triangulation additions. Its equality supports and sharp packing bounds permit several local requirements to be compared without double counting. The tight connector theorems then identify two specific obstructions and a variable-pair structure.

Theorem 20 concerns an ambient Euler-tight graph with complete port neighborhoods. It does not determine the biplanarity of the unpinned 21-vertex graph. Theorem 19 does not exclude every triangle connector, and the exceptional three-pair pattern has not been shown realizable. Leaves missing two incident port edges require a different argument. At deficit one, a shared quadrilateral may support several local failures, as the sharpness control explicitly demonstrates.

To apply these statements to a minimum counterexample, actual subgraphs, complete neighborhoods and edge equality must be established separately. The earlier forum reductions are not theorems imported into this manuscript. We do not prove a bound on the number of high-degree vertices, completeness of a low-component decomposition, or exclusion of every eight-high configuration. This paper supplies neither a biplanar graph requiring ten colors nor a proof that every biplanar graph is nine-colorable.

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