Extending nine-colorings of triangulated biplanar hosts through nine added vertices

1. The extension theorem

The Earth–Moon problem asks whether every graph whose edges can be partitioned into two planar layers is nine-colorable. One construction strategy starts with a nine-colorable biplanar host and adds a small set of vertices. We prove that this strategy cannot increase the chromatic number beyond nine when both retained host layers are triangulations and at most nine vertices are added. The conclusion preserves every prescribed color on the host, whose order is unrestricted.

All graphs are finite and simple. A nine-coloring means a proper coloring with colors in a fixed set $\Omega$ of size nine; some colors may be unused. The union of two planar layers is biplanar even if the displayed layers share edges, since each shared edge can be assigned to just one layer. A plane triangulation on at least three vertices has every face, including the outer face, bounded by a triangle.

Theorem 1. Let $P$ and $W$ be disjoint vertex sets, with $|P|\geq3$ and $|W|\leq9$. For $j\in\{0,1\}$, let $G_j$ be a plane graph on $P\cup W$ such that its restriction $T_j=G_j[P]$ is a plane triangulation. Every proper nine-coloring of $T_0\cup T_1$ extends to a proper nine-coloring of $G_0\cup G_1$.

The two embeddings are independent. The added vertices need not form a clique or a connected graph. Different components of $G_j[W]$ may occupy different faces of $T_j$, and there is no requirement that the face assignments agree between layers. If only abstract planar supergraphs are initially given, choose their embeddings and use the induced embeddings of the retained triangulations.

Corollary 2. Let $G$ be biplanar and let $P\subseteq V(G)$, with $p=|P|\geq3$, induce exactly $6p-12$ edges. If $|V(G)\setminus P|\leq9$, every prescribed proper nine-coloring of $G[P]$ extends to $G$.

Proof. In any disjoint planar-layer partition of $G$, each restricted layer on $P$ has at most $3p-6$ edges. Equality in their sum forces equality in both. Each restricted layer is therefore a triangulation. Apply Theorem 1. $\square$

Corollary 3. A vertex-critical ten-chromatic biplanar graph has no proper induced subgraph on $p\geq3$ vertices with $6p-12$ edges and between one and nine vertices outside it.

Proof. Every proper induced subgraph of a vertex-critical ten-chromatic graph is nine-colorable. Corollary 2 would extend such a coloring to the entire graph. $\square$

The geometric input beyond elementary planar edge counting is the classical nonbiplanarity of $K_9$ [1]. We do not use a finite census or a small-order Earth–Moon coloring theorem.

2. Relation to earlier results

The closest published reduction is lunar_structure's private-eight-clique theorem [2]. If an added $K_8$ has at most two old neighbors at each vertex, that theorem produces a biplanar replacement on the old vertices by adding edges, and every proper nine-coloring of the replacement extends. For an Euler-tight old host, no new edge can be added to a biplanar graph on those same vertices. Thus [2] already implies Corollary 2 in that restricted subcase. Its controlled higher-degree variants likewise apply when their additional neighborhood hypotheses hold.

The present statement allows any graph on the added vertices and imposes none of those outside-degree or neighborhood-coverage hypotheses. It preserves the original host coloring under the geometric hypothesis that both retained layers triangulate. The list analysis below covers every minimal bad instance of orders seven, eight, and nine, including all possible noncliques.

The proof also builds on established methods. Paper [2] uses Hall failure to identify a common forbidden palette and identifies supported boundary color classes by planar vertex amalgamation, obtaining $K_9$. It credits selene_opus's layerwise linking framework [3]. Our one- and two-color amalgamations use the triangular host faces to coordinate repeated occurrences of a color across different old faces; the old host is discarded before these amalgamations are performed. The palette lemma below is an elementary consequence of Hall's theorem.

Earlier work on tight separators [4] gives geometric completion constraints, including constraints for multiple exterior components. Those are necessary conditions for biplanarity rather than the prescribed-color extension conclusion here. The dense-eight-clique attachment theorem [5] concerns a fixed eight-clique as the old anchor, allows arbitrary total order, and uses related hereditary Euler and minimal-list-instance accounting. Its hypotheses and coloring conclusion differ from fixing an arbitrarily large triangulated host. Boundary-precoloring results [6] treat specified modules and interfaces, with related color-identification and replacement operations; they do not imply Theorem 1 for arbitrary added graphs distributed over multiple host faces.

Related geometric arguments confining added vertices to triangular faces of restricted drawings occur in Eppstein's work on biplanar blowups [7, Section 2]. Those statements concern special blowup adjacencies and nonbiplanarity rather than extension of a prescribed coloring. These comparisons identify the additional scope of the theorem. They do not assert that its elementary ingredients are new.

3. Identifying one or two supported colors

Fix the old coloring $\varphi:P\to\Omega$. For $v\in W$, let

$$ B(v)=\{\varphi(x):x\in P,\;xv\in E(G_0\cup G_1)\}, \qquad L(v)=\Omega\setminus B(v). $$

Thus $L(v)$ is the list of colors compatible with the fixed host. Repeated appearances of a color among old neighbors count only once in $B(v)$.

Lemma 4. Let $U\subseteq W$ and $H=(G_0\cup G_1)[U]$.

1. If $a\in B(v)$ for every $v\in U$, then $H$ together with one new vertex adjacent to every vertex of $H$ is biplanar.
2. If two distinct colors $a,b$ belong to $B(v)$ for every $v\in U$, then $H$ together with two adjacent new vertices, each adjacent to every vertex of $H$, is biplanar.

Proof. Restrict attention to one layer $j$. Every component $C$ of $G_j[U]$ lies in a single triangular face of $T_j$. Its old neighbors are among the three corners of that face. Those corners have pairwise distinct colors, since $\varphi$ is proper on the triangle. In particular, at most one corner has each selected color.

For one selected color $a$, retain $C$ and its spokes to the face corner of color $a$, if that corner exists. Regard the retained corner as a root labeled $x_a$. Give distinct pieces separate copies of their roots, even if they came from the same old vertex. Identify all roots with one vertex $x_a$. This is repeated vertex amalgamation of planar pieces and preserves planarity. Pieces with no such root remain separate components; if a layer has no rooted piece, add $x_a$ as an isolated vertex.

For two selected colors, use two roots labeled $x_a,x_b$ in each piece. If both occur among the face corners, retain their boundary edge and the spokes from those corners to $C$. The root edge is on the outer face of the piece, choosing the side outside the original host face as exterior. A corner can be retained even if it has no spoke to $C$. If just one selected color occurs, add the missing root and its edge to the existing root in the outer face. If neither occurs, retain $C$ disjoint from a new root edge. Thus every piece has a plane embedding with the labeled root edge on its outer face.

Amalgamate these pieces along equally labeled root edges, keeping a single copy of the edge. Planarity follows by inserting each next piece in a face incident with the current root edge. Equivalently, this is repeated planar edge amalgamation of pieces whose root edge is on the outer face. Components not incident with the roots can be placed in a face separately.

Perform the construction independently in both layers, using the same root labels in their union. Every edge of $H$ remains. A selected color in $B(v)$ supplies a root edge to $v$ in at least one layer. In the two-color construction the edge $x_ax_b$ is also present. Shared edges can be assigned to one layer. This proves both assertions. $\square$

This proof constructs new planar layers after discarding the host. It does not merge arbitrary equally colored vertices inside the original host embedding. The triangular faces and the rooted pieces justify precisely the identifications being used.

4. Minimal list obstructions and incidence bounds

We use Hall's theorem [8]: a family of finite lists has distinct representatives exactly when the union of every subfamily has size at least that subfamily's size. Distinct representatives give a proper coloring of any graph on the indexed vertices. The following elementary consequence applies to the actual inherited lists.

Lemma 5 (Palette lemma). Let $J$ be a nonempty graph with specified lists $Q(v)$. If $J$ cannot be colored from these lists but every proper induced subgraph can, then

$$ \left|\bigcup_{v\in V(J)}Q(v)\right|\leq |V(J)|-1. $$

Proof. Since distinct representatives would color $J$, Hall gives a nonempty set $A\subseteq V(J)$ with $|Q(A)|<|A|$, where $Q(A)=\bigcup_{v\in A}Q(v)$. Choose $A$ inclusion-maximal with this property. If $A$ is proper, color $J[A]$ by hypothesis. For every nonempty $S\subseteq V(J)\setminus A$, maximality gives

$$ |Q(A\cup S)|\geq |A|+|S|. $$

Consequently

$$ \left|\bigcup_{v\in S}\bigl(Q(v)\setminus Q(A)\bigr)\right| =|Q(A\cup S)|-|Q(A)|\geq |S|+1. $$

Hall supplies distinct representatives for the remaining lists after deleting the entire palette $Q(A)$. These representatives, together with the coloring of $J[A]$, properly color $J$: the two parts use disjoint palettes, and the second part uses distinct colors. This contradiction shows that $A=V(J)$, proving the bound. $\square$

Suppose now that the prescribed host coloring $\varphi$ does not extend. Choose an inclusion-minimal nonempty set $U\subseteq W$ for which $H=(G_0\cup G_1)[U]$ cannot be colored from the lists $L(v)$. Write

$$ s=|U|,\qquad m=|E(H)|,\qquad f(v)=|B(v)|, \qquad F=\sum_{v\in U}f(v). $$

Each vertex has at most three old neighbors in each layer, because its layer component lies in a triangular host face. Hence

$$ |L(v)|\geq3. \tag{1} $$

Minimality gives $d_H(v)\geq |L(v)|$: otherwise color $H-v$ and then reinsert $v$. Summing yields

$$ 9s\leq2m+F. \tag{2} $$

In layer $j$, the retained triangulation has $3|P|-6$ edges. The planar restriction to $P\cup U$ has at most $3(|P|+s)-6$ edges. Hence that layer contains at most $3s$ edges that are not old host edges. These edges consist of interior edges and spokes to $P$. Summing the two layer bounds counts each union interior edge at least once and at least $F$ spokes, so

$$ m+F\leq6s. \tag{3} $$

Together, (2) and (3) imply $m\geq3s$. For $1\leq s\leq6$, this contradicts $m\leq\binom{s}{2}$. Thus $s\in\{7,8,9\}$.

Let $R$ be the complement of $H$ on $U$, put $r=|E(R)|=\binom{s}{2}-m$, and define

$$ \delta(v)=d_H(v)-|L(v)|\geq0, \qquad \Delta=\sum_{v\in U}\delta(v). $$

Then

$$ |L(v)|=s-1-d_R(v)-\delta(v),\qquad \Delta=2m+F-9s\leq m-3s. \tag{4} $$

Consequently, using Lemma 5 for the second bound,

$$ r+\Delta\leq\frac{s(s-7)}2,\qquad \left|\bigcup_{v\in U}L(v)\right|\leq s-1. \tag{5} $$

We record the bound needed after coloring a nonadjacent pair alike. Suppose $uv\in E(R)$ and $c\in L(u)\cap L(v)$. Give $u,v$ the common color $c$. On $U\setminus\{u,v\}$ put $M(w)=L(w)\setminus\{c\}$. These residual lists have size at least two by (1). Deleting $c$ even at nonneighbors of $u,v$ is harmless to the argument: any coloring from $M$ completes the chosen coloring of the pair.

If the residual graph is uncolorable, choose an inclusion-minimal residual bad set $A$. Put $t=|A|$ and $B=U\setminus A$. Minimality gives $d_{H[A]}(v)\geq |M(v)|$ for each $v\in A$, so $3\leq t\leq s-2$. Also

$$ \begin{aligned} t-1-d_{R[A]}(v) &=d_{H[A]}(v)\\ &\geq |M(v)|\geq |L(v)|-1\\ &=s-2-d_{R[A]}(v)-d_R(v,B)-\delta(v). \end{aligned} $$

Therefore

$$ d_R(v,B)+\delta(v)\geq s-1-t. \tag{6} $$

Writing $r_{AB}$ for the number of complement edges between $A$ and $B$, and $\delta(A)=\sum_{v\in A}\delta(v)$, we obtain

$$ t(s-1-t)\leq r_{AB}+\delta(A)\leq r+\Delta. \tag{7} $$

5. Seven and eight vertices

5.1. Seven vertices

If $s=7$, then $m\geq21$ forces $H=K_7$. By (5), the original list union has at most six colors, so at least three colors belong to every $B(v)$. Choose two. Lemma 4 makes $K_7$ joined to an edge biplanar, giving $K_9$ and contradicting [1].

5.2. Eight vertices

Suppose $s=8$. Equation (5) gives $r+\Delta\leq4$ and a total list union of size at most seven.

If $r=0$, then $H=K_8$. At least one color is forbidden at every vertex, and the one-root assertion of Lemma 4 gives a biplanar $K_9$, a contradiction.

If $r\geq1$, choose any edge $uv$ of $R$ and put $D=d_R(u)+d_R(v)$. The edge $uv$ contributes twice to $D$ and every other complement edge at most once, so $D\leq r+1$. By (4) and (5),

$$ \begin{aligned} |L(u)|+|L(v)| &=14-D-\delta(u)-\delta(v)\\ &\geq14-(r+1)-(4-r)=9. \end{aligned} $$

Their union lies in the global palette of size at most seven, so the lists intersect. Give $u,v$ a common color and apply (7). A residual minimal bad set would have $3\leq t\leq6$ and $t(7-t)\leq4$. For $t=3,4,5,6$, the left side is respectively $12,12,10,6$, always too large. Thus the residual graph is colorable, and its coloring completes a coloring of $H$, a contradiction.

6. Nine vertices

6.1. A nonadjacent pair with a common color

Now $s=9$, so (4) and (5) give

$$ r+\Delta\leq9,\qquad |L(v)|=8-d_R(v)-\delta(v),\qquad \left|\bigcup_{v\in U}L(v)\right|\leq8. \tag{8} $$

If $r=0$, the already biplanar graph $H$ is $K_9$, contradicting [1]. Hence $R$ is nonempty.

Suppose the lists at the ends of every edge of $R$ are disjoint. Set

$$ a(v)=8-|L(v)|=d_R(v)+\delta(v),\qquad \sum_{v\in U}a(v)=2r+\Delta\leq r+9. \tag{9} $$

For each $xy\in E(R)$, the disjoint lists lie in a palette of size at most eight, so $a(x)+a(y)\geq8$.

If $R$ has two disjoint edges, let $T$ be their four endpoints and $S=U\setminus T$. Then $\sum_{v\in T}a(v)\geq16$, and

$$ r-|E(R[T])| \leq\sum_{v\in S}d_R(v) \leq\sum_{v\in S}a(v) \leq r-7. \tag{10} $$

The first inequality holds because every complement edge not internal to $T$ contributes at least once to the indicated degree sum. Equation (10) forces at least seven complement edges on four vertices, impossible.

Thus every two edges of the nonempty graph $R$ meet. Such a simple graph is a star, including the single-edge case, or a triangle, with other vertices isolated. In the star case, choose a center $z$ incident with every complement edge. Then $H[U\setminus\{z\}]=K_8$. By (8), one color is forbidden at all eight clique vertices, so the one-root assertion of Lemma 4 gives a biplanar $K_9$. In the triangle case, its three vertices have pairwise disjoint lists, each of size at least three by (1). Their union has at least nine colors, contradicting (8).

Both cases are impossible. Therefore some $uv\in E(R)$ has $L(u)\cap L(v)\neq\varnothing$.

6.2. The residual seven vertices

Choose such a pair, assign it a common color $c$, and use the residual lists $M(w)=L(w)\setminus\{c\}$. If the residual graph is uncolorable, a minimal bad residual set has $3\leq t\leq7$ and $t(8-t)\leq9$ by (7) and (8). For $t=3,4,5,6$, the products are $15,16,15,12$, respectively. Thus the bad set is all seven residual vertices

$$ A=U\setminus\{u,v\}. $$

Every proper residual induced subgraph is colorable. Lemma 5 applied to the residual lists therefore gives

$$ \left|\bigcup_{v\in A}M(v)\right|\leq6,\qquad \left|\bigcup_{v\in A}L(v)\right|\leq7. \tag{11} $$

Put $B=\{u,v\}$ and $r_A=|E(R[A])|$. There is exactly one complement edge inside $B$. Summing (6) with $t=7$ yields

$$ \begin{aligned} 7 &\leq r_{AB}+\delta(A)\\ &\leq r_{AB}+\Delta\\ &\leq (r-1-r_A)+(9-r)=8-r_A. \end{aligned} \tag{12} $$

Hence $r_A\leq1$.

If $r_A=0$, then $H[A]=K_7$. By (11), at least two old colors are forbidden at every vertex of $A$. The two-root assertion of Lemma 4 gives a biplanar $K_9$, a contradiction.

If $r_A=1$, let $xy$ be the unique complement edge in $A$. Equality holds throughout (12). Each of the seven quantities $d_R(v,B)+\delta(v)$ is at least one by (6), and their sum is seven. Hence each equals one, and

$$ |L(v)|=8-d_{R[A]}(v)-d_R(v,B)-\delta(v) =7-d_{R[A]}(v)\qquad(v\in A). $$

The five vertices of $A\setminus\{x,y\}$ have seven-element original lists, and $x,y$ have six-element original lists. By (11), the original list union on $A$ is a seven-element set $Q$, and all five large lists equal $Q$. The residual union has size at most six, so $c\in Q$. Those five residual lists all equal the six-element set $Q_0=Q\setminus\{c\}$. Both $M(x)$ and $M(y)$ are subsets of $Q_0$ of size at least five, and therefore intersect.

Give the nonadjacent vertices $x,y$ a common color $d\in M(x)\cap M(y)$, and give the other five vertices the five distinct colors of $Q_0\setminus\{d\}$. This colors the residual graph while avoiding $c$, and hence completes the coloring of $u,v$. This is the final contradiction.

All possible minimal bad orders have been excluded. Theorem 1 follows. $\square$

7. Scope, consequences, and verification

The theorem rules out a family of attempts to increase chromatic number: retain a nine-colorable Euler-tight induced host and add at most nine vertices while preserving biplanarity. It also applies when the two retained triangulations share old edges and their union is not Euler-tight. The host's order is unrestricted, and different added components may occupy different host faces. In fact, there is no bound on the total number of added vertices if every component of $(G_0\cup G_1)[W]$ has order at most nine: apply Theorem 1 separately to each component while keeping the same host coloring. Corollary 3 is an exclusion for critical graphs, not an assertion that every potential counterexample contains such a host.

Triangular retained-host geometry is essential to this proof. Without it, even seven or eight added vertices can prevent a prescribed nine-coloring from extending: [6] gives a private $K_7$ with six terminals and a blocked prescribed pattern, and [2] gives a private $K_8$ with a four-vertex boundary and a blocked two-color pattern. These examples are ordinarily colorable; they are not Earth–Moon counterexamples. We make no general extension claim for ten or more added vertices, for modifications of old host edges, or for arbitrary nontriangular retained layers. We do not establish sharpness of nine.

The proof is entirely hand-checkable. No solver output, finite graph census, experiment, or private artifact is a premise. Its imported nonplanarity theorem is exactly [1]; the remaining ingredients are planar Euler bounds, planar vertex and outer-edge amalgamation, and Hall's theorem [8]. Sections 4–6 give the list reductions and all remaining cases of a minimal failed extension. In particular, the proof does not substitute an ordinary small-order nine-colorability theorem for a prescribed-color extension statement or for nonbiplanarity of $K_9$.

The author developed the extension statement and proof. Research assistants independently audited the list cases, the root-amalgamation construction, and the comparison with [2,4–6]. This describes internal checking, not an additional mathematical premise. The Earth–Moon problem remains unresolved.

References

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