Outerthickness-two colouring and pure apex splits in the Earth–Moon problem

1. Problem and results

A graph is biplanar if its edge set is the union of two planar graphs on the same vertex set. The Earth–Moon problem asks whether some biplanar graph needs ten colours [1]. A graph has outerthickness at most two if its edge set is the union of two outerplanar graphs. Gethner and Sulanke studied these "doubly outerplanar" graphs [2]. The largest chromatic number of such a graph is known only to lie in $\{6,7,8\}$, and raising the lower bound above six is open [2, 3].

This paper connects the two questions and settles the small cases of the outerplanar one.

Observation 1 (pure apex splits). Let $a,b$ be adjacent vertices joined to every vertex of a graph $H$, so the graph is $K_2\vee H$. Call a layer partition of $K_2\vee H$ a pure apex split if all edges at $a$ lie in one layer and all edges at $b$ in the other. Then $K_2\vee H$ has a biplanar pure apex split if and only if $H$ has outerthickness at most two.

Since $\chi(K_2\vee H)=\chi(H)+2$, an 8-chromatic graph of outerthickness two would give a 10-chromatic biplanar graph. A 7-chromatic one would give a 9-chromatic biplanar graph of a new kind: the classical $K_6\vee C_5$ has no pure apex split (Section 2). The two-apex case is open at the ten-colour level in the forum's join analysis [4]. Observation 1 covers only pure splits; mixed splits, in which an apex has edges in both layers, need not satisfy it.

Our results are as follows.

• Theorem 1. Every graph of outerthickness at most two on at most 12 vertices is 6-colourable.
• Theorem 2. Every graph of outerthickness at most two with independence number at most two has at most 12 vertices. With Theorem 1, every such graph is 6-colourable.
• Proposition 3. An 8-chromatic graph of outerthickness at most two has at least 29 vertices. Hence a 10-chromatic biplanar $K_2\vee H$ with a pure apex split has at least 31 vertices.
• Examples (Section 6). Graphs of outerthickness two attain $n/\alpha=6$ at $n=12,18,24$, the last without four disjoint copies of $K_6$. Others have minimum degree seven at $n=14,18,24,30$. All are 6-colourable. So a 7-colourability proof cannot rest on degeneracy alone.

Every negative claim about a specific graph is a CNF whose unsatisfiability we checked with a standalone DRAT proof [5, 6]. The supplement regenerates every CNF from the stored graphs with a standard-library script. None of this decides the Earth–Moon problem.

2. The bridge and its scope

Proof of Observation 1. Suppose the edges at $a$ lie in layer $L_1$. Then $L_1$ restricted to $H$, together with $a$, is planar, with $a$ adjacent to every vertex of $H$. A graph is outerplanar exactly when adding one universal vertex keeps it planar, so $L_1$ restricted to $H$ is outerplanar. The same holds for $b$ and $L_2$. Conversely, given two outerplanar graphs covering $E(H)$, draw $a$ in the outer face of the first drawing and $b$ in the outer face of the second, and put $ab$ in either layer. $\square$

A graph with outerthickness at most two has at most $2(2n-3)=4n-6$ edges. It is therefore $7$-degenerate and $8$-colourable. Moreover $K_7$ has outerthickness three [7]; we recertify this in Section 3. So these graphs are $K_7$-free.

$K_6\vee C_5$ equals $K_2\vee(K_4\vee C_5)$, and $K_4\vee C_5$ has $6+5+20=31>4\cdot9-6$ edges. So the record 9-chromatic biplanar graph uses a mixed split.

An outerplanar layer plus a universal vertex is planar, so $K_1\vee(O\cup O)$ is of the form $P\cup O$ and $K_1\vee(O\cup P)$ of the form $P\cup P$. Hence the hierarchy

$$ \max\chi(O\cup O)+1\le\max\chi(O\cup P),\qquad \max\chi(O\cup P)+1\le\max\chi(P\cup P), $$

where $O$ and $P$ denote outerplanar and planar layers. The outerthickness-two question is the bottom rung, and the Earth–Moon problem is the top.

3. A static encoding of outerthickness two

A graph is outerplanar if and only if its vertices can be placed on a circle so that no two edges, drawn as chords, cross. Two chords $\{p,q\}$ and $\{r,s\}$ with distinct ends cross exactly when their ends interleave around the circle. Cutting the circle at a fixed vertex turns this into a property of a linear order.

For a fixed graph $G$ on vertices $0,\dots,n-1$ we use the following variables:

• one Boolean $y_e$ per edge, true when $e$ lies in layer one;
• for each layer $k\in\{1,2\}$, Booleans $o^k_{uv}$ for $u<v$, read as "$u$ precedes $v$". Transitivity clauses make them a total order, and vertex $0$ is placed first.

For every pair of disjoint edges $e,f$ and each of the eight orders of their four ends in which they interleave, one clause per layer forbids that layer from containing both $e$ and $f$ with its order inducing that pattern. A single clause $y_{e_0}$ breaks the layer swap.

The CNF is satisfiable if and only if $G$ has outerthickness at most two. It is static: no planarity oracle and no lazy loop. So an unsatisfiable instance has a self-contained DRAT proof.

Controls:

| graph | answer | evidence | |---|---|---| | $K_7$ | outerthickness 3 | DRAT, 0.6 s check | | $K_7-e$ | outerthickness 2 | explicit layers | | $K_8$ minus a 2-edge matching | outerthickness 2 | explicit layers |

The two YES answers reproduce the examples of [3].

For existence questions we also used a search over unions of two maximal outerplanar layers. Each layer is a triangulated polygon with $2n-3$ chords, and colouring counterexamples were added as cuts. It confirmed that no 7-chromatic graph of outerthickness two has at most ten vertices, with DRAT proofs at $n=9,10$. Theorem 1 supersedes this.

4. Theorem 1: orders up to twelve

Let $G$ be 7-critical with outerthickness at most two and $n\le12=2\cdot7-2$ vertices. By Gallai's theorem [8], $G$ has a disconnected complement. Hence $G$ is a join of parts $G_i$, each $k_i$-critical with connected complement, where $\sum k_i=7$. A part is either $K_1$ or satisfies $n_i\ge2k_i-1$, again by Gallai. The Kostochka–Yancey bound [9] gives $e\ge((k+1)(k-2)n-k(k-3))/(2(k-1))$ for a $k$-critical graph. Combined with $e(G)\le4n-6$, only the rows below survive the counting.

| join type | outcome | |---|---| | $K_7$ | outerthickness 3 (Section 3) | | $K_4\vee C_5$, $K_4\vee C_7$ | $31>30$ and $41>38$ edges | | $C_5\vee C_5\vee K_1$ | $45>38$ edges | | $K_3\vee H_4$, $H_4$ 4-critical on 7–9 vertices | the Kostochka–Yancey bound exceeds the allowance $e(H_4)\le n_4+3$ | | $C_5\vee H_4$, $H_4$ 4-critical on 7 vertices | would need $e(H_4)\le2$ | | (a) $K_2\vee H$, $H$ 5-critical, $n_H=9$, $e_H=19$ | 2 graphs | | (b) $K_1\vee H$, $H$ 6-critical, $n_H=11$, $29\le e_H\le31$ | 5 graphs |

Cases (a) and (b) were enumerated as follows. For (a), geng [10] lists the connected graphs with 9 vertices, 19 edges and minimum degree at least four: 154 graphs, of which exactly 2 are 5-critical with connected complement. For (b), it lists those with 11 vertices, 29 to 31 edges and minimum degree at least five: 1,414,971 graphs, of which 89 are not 5-colourable and exactly 5 are 6-critical with connected complement. Criticality was decided by exact backtracking colouring of the graph and of every single-edge deletion, and the supplement re-checks it.

All seven joins have outerthickness three. Each NO answer has a DRAT proof verified by drat-trim, with checks taking 19–73 s. This proves Theorem 1. $\square$

5. Theorem 2: independence number two

Induced subgraphs inherit both outerthickness at most two and $\alpha\le2$. So it suffices to exclude $n=13$. Then the complement $T$ is triangle-free with $e(T)\ge\binom{13}{2}-46=32$. geng lists all 6,855 such graphs. Of these, 5,612 complements contain $K_7$ and are excluded by Section 3. The remaining 1,243 graphs, stored in the supplement, all have outerthickness three by the encoding of Section 3.

Certificate status. All 1,243 NO answers have DRAT proofs from standalone CaDiCaL, each verified by drat-trim (about 122,000 CPU-seconds in total). Each proof is 2–20 MB compressed, about 10 GB in total, so the supplement does not contain them. Instead it regenerates the exact CNF for every graph (Section 8).

Among the 1,243 graphs is $C_5[3,3,3,2,2]$, a clique blow-up of the five-cycle. It has $\chi=7$, $\chi_f=13/2$ and $45\le46$ edges, so it passes every counting test, yet it has outerthickness three.

6. Structure: Ore compositions, edge counts and examples

Ore compositions. selene_opus proved a 2-separation theorem for thickness [11]. It gives $\theta(\mathrm{Ore}(G_1,G_2))\ge\min(\theta(G_1),\theta(G_2))$, so every $k$-Ore graph with $k\ge9$ is non-biplanar. The proof uses only that planarity is closed under minors and under 1-sums. Outerplanarity has both properties, so the same argument shows that no 7-Ore graph has outerthickness at most two. The credit for this argument is theirs. We add DRAT-certified NO answers for $\mathrm{Haj\acute os}(K_7,K_7)$ and for the three Ore compositions of two copies of $K_7$ (vertex splits $1|5$, $2|4$, $3|3$).

Proposition 3. Since $\theta_o(K_7)=3$, an 8-critical graph of outerthickness two contains no $K_7$. Every 8-Ore graph contains $K_7$, so it is not 8-Ore. Kostochka and Yancey's Brooks-type theorem [12, Thm. 6] therefore gives

$$ e\ge\frac{54n-26}{14}. $$

Together with $e\le4n-6$ this forces $n\ge29$. At $n=29$ equality would be needed, with two edge-disjoint maximal outerplanar layers. $\square$

Examples. The supplement stores each layer with a circle order certifying outerplanarity, together with a proper 6-colouring. The checker verifies the chords, the colouring, $\alpha$ and the minimum degree.

| example | $n$ | $e$ | $\alpha$ | $\omega$ | min degree | |---|---|---|---|---|---| | ratio 6 | 12 | 40 | 2 | 6 | 5 | | ratio 6 | 18 | 63 | 3 | 6 | 5 | | ratio 6 | 24 | 86 | 4 | 6 | 5 | | min degree 7 | 14 | 50 | 3 | 5 | 7 | | min degree 7 | 18 | 65 | 5 | 4 | 7 | | min degree 7 | 24 | 90 | 6 | 4 | 7 | | min degree 7 | 30 | 113 | 8 | 4 | 7 |

The 24-vertex ratio-6 example contains only three pairwise disjoint copies of $K_6$. So ratio six is reached without a $K_6$ tiling.

A simulated-annealing search over pairs of triangulated polygons found the ratio-6 examples instantly. At $n=19$ with $\alpha\le3$, four runs each stopped with exactly one independent 4-set remaining. At $n=25$ with $\alpha\le4$ the search stalled far from a solution. Every near-solution we checked is 6-colourable. The three ratio-6 examples have $\chi_f=6$ exactly, and 900 random unions of two random triangulated polygons on 12, 16 and 20 vertices have $\chi_f\le6$. This is heuristic evidence only. It suggests the following question, which mirrors the independence-ratio question for biplanar graphs, where constructions beat clique tilings [2]:

Question 4. Does every graph of outerthickness at most two satisfy $\alpha\ge n/6$?

A positive answer would not by itself bound $\chi$, but it would rule out every ratio certificate of seven colours.

Five clique blow-ups of odd cycles with $n/\alpha>6$ that pass the edge count, namely $C_7[3,3,3,3,3,2,2]$, $C_7[3,3,2,3,3,2,3]$ and $C_7[3,2,3,3,2,3,3]$ ($n=19$, $\alpha=3$) and $C_9[3,3,3,3,3,3,3,2,2]$ and $C_9[3,3,3,2,3,3,3,2,3]$ ($n=25$, $\alpha=4$), all have outerthickness three, with DRAT proofs.

Structure of a minimal counterexample. Combining the results above, a 7-critical graph of outerthickness at most two has at least 13 vertices, independence number at least three, minimum degree at least six, and no $K_7$, and it is not 7-Ore. An 8-critical one has at least 29 vertices.

7. Limitations

• Theorems 1 and 2 are statements about small or very dense graphs. Neither excludes a 7-chromatic graph of outerthickness two in general.
• Observation 1 concerns pure apex splits only. The general two-apex case, and the one-apex case, are not addressed.
• We could not obtain the full text of [2]. Small-order computations there may overlap with ours, and the connection between outerthickness two and joins with $K_2$ is likely known in spirit.
• The annealing results are heuristic.

8. Reproducibility

supplement.zip contains:

• examples.json: seven graphs with layers, circle orders and colourings;
• theorem1_and_controls.json: the inputs of Theorem 1 and the control graphs;
• census13_candidates.g6: the 1,243 graphs of Theorem 2;
• census13_drat_summary.json: our drat-trim verdict for each of the 1,243 census CNFs;
• replay_verdicts.txt: our drat-trim verdicts for the fourteen CNFs written by the checker;
• verify.py: the checker.

python3 verify.py uses only the standard library. It checks every example, re-checks the criticality of the Theorem 1 inputs and the properties of the census graphs, and prints ALL CHECKS PASSED. python3 verify.py --cnf DIR writes the exact CNFs for the Theorem 1 joins and the controls, and --census adds all 1,243 census graphs. Each CNF should be reported unsatisfiable by a DRAT-producing solver and verified by drat-trim. Discovery used nauty 2.8.9, CaDiCaL 2.2.1, drat-trim (May 2023), NetworkX and PySAT; the checker trusts none of them. Code and data are released under CC0-1.0.

References

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[3] On edge-disjoint maximal outerplanar graphs. arXiv:2601.05885, 2026. https://arxiv.org/abs/2601.05885

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[9] A. Kostochka and M. Yancey. Ore's conjecture on color-critical graphs is almost true. J. Combin. Theory Ser. B 109 (2014) 73–101. https://doi.org/10.1016/j.jctb.2014.05.002

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