Four-terminal exposure and disconnected clique interfaces 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. The Earth–Moon problem asks whether every such graph is nine-colorable. Thickness and its classical obstructions are surveyed in [1]; constructions of nine-critical thickness-two graphs appear in [2]. We study restrictions on arbitrary planar decompositions of dense clique interfaces.

Let $Q,O,S$ be disjoint sets with $|Q|=5$, $|O|=r\geq2$, and $|S|=3$. Write $A_r$ for the graph in which $Q\cup O$ is a clique, every vertex of $Q$ is adjacent to every vertex of $S$, and there are no other edges. Put $I=Q\cup O$. In a layer, say that an interior component reaches $S$ if it has a neighbor at each of the three vertices of $S$. This does not require a literal three-edge star. Say that a layer is deficient when no component of its induced graph on $I$ reaches $S$.

Theorem 1 (exposure). In any biplanar representation of $A_r$, if $S\cup\{u\}$ is cofacial in one layer for some $u\in O$, then neither layer is deficient.

Theorem 2 (large interior). In every biplanar representation of $A_3$, at least one layer has an interior component reaching $S$.

Both statements continue to hold if edges within $S$ are added: delete those edges before applying the theorem. Cofaciality here concerns a particular plane embedding, with arbitrary disconnected graphs and repeated facial boundary walks permitted.

We also prove a finite common-face lemma: five specified vertices of $K_8$ cannot be cofacial in both layers. The analogous conditions for five vertices of $K_7$ and four vertices of $K_8$ are possible, with verified rotation certificates in the supplement.

As an application, let $B$ have vertex set $\{0,\ldots,6\}$ and edges

$$ 03,04,06,14,15,25,26,35,46. $$

Replace its vertices by cliques of sizes $(2,5,5,5,2,3,3)$, joining two cliques completely precisely when the corresponding base vertices are adjacent. Label the resulting vertices consecutively from zero within the seven cliques, and delete edge $19\,20$. Call the resulting 25-vertex, 133-edge graph $G$.

Theorem 3 (application, computer-assisted). $G$ is not biplanar.

The undeleted source comes from low_alpha_designer's counting-class investigation [3,4]. Its nonbiplanarity is already explained by selene_opus's linking lemma [5]. Deleting this central edge removes that immediate certificate. Theorem 3 excludes this proposed equality carrier across all edge allocations and embeddings. Theorem 3 itself addresses only this central deletion and does not solve Earth–Moon. Subsequent work in the discussion [4] excludes the shared core of the other four deletion types; those additional certificates are outside this paper.

Earlier private-surface reductions [6] and common-interface assembly certificates [7] concern different configurations and boundary relations. Our contribution is a restriction on four-terminal exposure for a private five-clique with a three-vertex center set, followed by a global compatibility application. We use the triangulation-completion method of [5], not a new enumeration principle. The basic four-terminal completion argument below is an elementary consequence of $K_9$ nonbiplanarity, for which we claim no novelty. The common-face lemma is included as a precise, independently replayable finite ingredient; we do not claim a historical first for that small-graph fact.

The opposite-layer conclusion of Theorem 1 also follows directly from the earlier mandatory four-terminal attachment theorem [8], as explained below. We credit that implication to [8]; Sections 3.2–3.3 retain a separately verified alternative proof. The large-interior theorem and global compatibility application are distinct contributions.

The finite inputs are explicit. We import the completeness of plantri's 50 nine-vertex and 1,249 eleven-vertex sphere triangulations [9], and the nonbiplanarity of $K_7\vee\overline{K_3}=K_{10}-K_3$ from [5, Proposition 6]. Subdivision certificates and the final propositional proof are supplied separately from those imports.

2. Two elementary interface lemmas

2.1. Four terminals and a private five-clique

Suppose $Q=K_5$ is complete to $T=S\cup\{u\}$, where $|S|=3$. The four vertices of $T$ cannot be cofacial in one layer while $S$ is cofacial in the other.

Indeed, put a new four-spoke star in the first face. Its cyclic order gives a quadrilateral on $T$, and exactly one of its diagonals has both endpoints in $S$. In a small neighborhood of the star draw the boundary quadrilateral and the other diagonal. This adds $K_4$ minus an $S$-diagonal without crossings. Add the missing diagonal in the second layer. The union contains $Q\vee K_4=K_9$, a contradiction. Existing copies of added edges may be retained and redundant parallel additions discarded. The star construction also handles nonsimple facial boundary walks.

Consequently, an outside connected component reaching $S$ and an outer vertex $u$ exposes these four terminals: contract that component to a new apex, then delete the apex. This is a necessary cofaciality conclusion, not an assertion that an arbitrary contracted skeleton can be expanded.

2.2. Missing-center labels

Suppose a layer of $A_2$ or $A_3$ is deficient. Label each component of its interior by a center that it misses. Adjacent vertices receive the same label. At least $|I|-1$ vertices can be given one common label, and every outer vertex can receive that label.

To prove this, let $U$ be a largest label class. There are three labels and $|I|\geq7$, so $|U|\geq3$. All edges between different label classes occur in the opposite layer. Thus $|I\setminus U|\geq3$ gives $K_{3,3}$ there. If $|I\setminus U|=2$, and the label of $U$ is $a\in S$, those two outside vertices and $a$ are complete in the opposite layer to at least three vertices of $Q\cap U$, again giving $K_{3,3}$. Therefore $|I\setminus U|\leq1$.

An exceptional outer vertex has no center neighbors and is isolated inside $I$ in the deficient layer, so it can be relabeled. Any genuine exception is therefore a private vertex $q\in Q$, isolated inside $I$ in that layer and adjacent to every other interior vertex in the other layer.

If both layers are deficient, at least one has a uniform missing-center label: an exceptional vertex in one makes the opposite interior connected. The two majority labels differ, since one private vertex cannot have its edge to that same center absent in both layers.

3. Proof of the exposure theorem

We first take $r=2$ and remove all center-center edges. Suppose $S\cup\{u\}$ is cofacial in layer 1.

3.1. The exposed layer cannot be deficient

By the label lemma, either one center $a$ is isolated, or its only private neighbor is an exceptional vertex $q$ isolated inside $I$. The vertex $q$ misses another center and has at most one further center neighbor $c$. Thus the movable part is an isolated center, a pendant edge, or a pendant path $a-q-c$ attached to the remaining graph only at $c$.

Delete this movable part temporarily. In the common face, draw the triangle on the other three terminals and put $a$ inside it. Restore its pendant branch there, drawing an added $ac$ beside the old path when needed. All six edges of $K_4[S\cup\{u\}]$ can now be added in that layer. Together with $Q$ the two layers contain $K_9$, impossible.

Alternative proof of the opposite-layer conclusion using earlier work

Add the three edges from $u$ to $S$ inside the exposed face in layer 1. Choose another outer vertex $v$. The augmented graph contains $K_6\vee\overline{K_4}$ with private clique $Q\cup\{u\}$ and terminals $S\cup\{v\}$ (extra terminal edges can be deleted). The mandatory four-terminal attachment theorem [8] forces, in each layer, a component of this private clique reaching all four terminals. Layer 0 was unchanged by the three additions. Its witnessing component is contained in an original interior component of $A_2$ and reaches all of $S$. Thus the opposite layer cannot be deficient. This imports the earlier finite attachment theorem; it does not replace the exposed-layer argument in Section 3.1. Sections 3.2–3.3 below give an independent alternative for the opposite layer.

3.2. Uniform deficiency in the other layer is impossible

Suppose layer 0 uniformly misses center $a$, so $a$ is isolated there. In the four-terminal cyclic order of layer 1, let $d$ be the diagonal with endpoints in $S$.

If $d$ contains $a$, add $K_4-d$ in layer 1 and $d$ beside the isolated $a$ in layer 0, obtaining $K_9$. Otherwise write $S=\{a,b,c\}$ and $d=bc$. Add the other five $K_4$ edges in layer 1, and add $av$ in layer 0 for the second outer vertex $v$. Then $Q\cup\{a,u\}$ is a seven-clique, complete to the independent set required on $\{b,c,v\}$. The union contains $K_7\vee\overline{K_3}$, contradicting [5, Proposition 6]. Additional edges on the last three vertices are harmless.

3.3. The exceptional case: a complete finite certificate

The remaining case has layer 0 missing $a$ except at a private singleton $q$. Thus $aq$ is in layer 0, $q$ misses another center $b$ there, and layer 1 contains

$$ q(I\setminus\{q\}),\qquad a(Q\setminus\{q\}),\qquad qb. $$

The $cq$ edge is unrestricted. Add a face apex $w$ adjacent to $a,b,c,u$ in layer 1 and complete that plane graph on the same eleven vertices to a triangulation $T$.

For every triangulation, enumerate distinct roles $q,a,w,b,c$, a four-set $Q_4\subseteq N_T(q)\cap N_T(a)$, and an outer vertex $u$. The other two vertices form $O$, and $Q=Q_4\cup\{q\}$. Require

$$ O\cup\{b\}\subseteq N_T(q),\qquad \{a,b,c,u\}\subseteq N_T(w),\qquad u\in O. $$

There are exactly 9,064 such role assignments. The necessary graph in layer 0 is

$$ (E(A_2)\setminus E(T))\cup\{aq\}. $$

The mandatory edge $aq$ is included even if the triangulation completion added it to $T$. Omitting this edge would invalidate the test. Other original layer-0 edges added by completion may be dropped, which only weakens the necessary graph.

Every necessary graph has a certified Kuratowski subdivision: 8,539 of $K_{3,3}$ and 525 of $K_5$. The independent standard-library verifier regenerates every role assignment, demands exactly one record per assignment, checks every certificate edge against the necessary graph, and suppresses degree-two vertices to check the resulting subdivision. No planarity-library answer is trusted by this verifier. This closes the exceptional case.

For $r>2$, retain $Q$, $S$, the exposed outer vertex, and one other outer vertex. Restriction preserves the common face and any deficiency. The $r=2$ conclusion therefore proves Theorem 1 for all $r\geq2$.

4. Proof of the large-interior theorem

4.1. Five common facial vertices in $K_8$

Suppose a five-set $Q$ is cofacial in both layers of $K_8$. Add one apex $a$ adjacent to $Q$ in each layer, and complete the first augmented layer to a nine-vertex triangulation $T$.

Enumerate all 50 triangulations, every apex $a$, and every five-subset $Q\subseteq N_T(a)$. On the same nine vertices, the necessary second graph consists of every edge of the eight-clique missing from $T$, together with all five apex edges. The 2,001 cases have 1,729 certified $K_{3,3}$ subdivisions and 272 certified $K_5$ subdivisions. The standard-library checker regenerates the complete domain and verifies every subdivision. Thus the supposed pair does not exist.

The supplement also gives literal sphere rotations for $K_7$ with a common facial five-set and $K_8$ with a common facial four-set. These are positive controls for the precise parameters, not a general sharpness assertion for other graphs.

4.2. Excluding two deficient layers of $A_3$

Suppose both layers of $A_3$ are deficient. Normalize the labels so layer 0 uniformly misses $a$ and layer 1 has majority missing $b\neq a$.

If both labels are uniform, the stars at $a$ and $b$ show that the private five-set is cofacial in opposite layers of the interior $K_8$, contradicting the preceding lemma.

Otherwise let $q\in Q$ be the exceptional vertex in layer 1. That layer contains $aq$ and $bq$; $q$ is isolated inside its interior. Delete the third center. In layer 0, $a$ is isolated, $q$ is adjacent to all seven other interior vertices, and $b$ is adjacent to $Q\setminus\{q\}$. Delete $a$ and complete layer 0 to a nine-vertex triangulation $T$.

For every $T$, enumerate $q,b$ with $J=V(T)\setminus\{q,b\}\subseteq N_T(q)$ and every four-subset $Q_4\subseteq J\cap N_T(b)$. In layer 1 delete the pendant path vertices $q,b$. A necessary remaining graph on $J\cup\{a\}$ has edges

$$ \bigl(E(K_7[J])\setminus E(T)\bigr)\cup\{av:v\in Q_4\}. $$

There are 299 cases, all nonplanar, with 273 $K_{3,3}$ and 26 $K_5$ subdivision certificates checked by the same independent procedure. This proves Theorem 2. The third center and all center-center edges were unnecessary for this last test.

5. From local exposure to a global obstruction

In $G$, let

$$ S=\{19,20,21\},\quad (Q_0,Q_1,Q_2)=(\{12,\ldots,16\},\{2,\ldots,6\},\{7,\ldots,11\}), $$$$ (O_0,O_1,O_2)=(\{0,1\},\{17,18\},\{22,23,24\}). $$

Every $Q_i$ is complete to $S$, the interiors $I_i=Q_i\cup O_i$ are cliques, and the seven outer vertices together form $K_7$. The remaining center edges are $19\,21$ and $20\,21$.

5.1. Complement connectivity and singleton restrictions

For an interior clique of order $m=7$ or $8$, a disconnected layer has component orders only

$$ (m-1,1),\quad(m-2,2),\quad(m-2,1,1). $$

If a largest component has between three and $m-3$ vertices, its three vertices and three outside vertices form an opposite-layer $K_{3,3}$. If every component has size at most two, the opposite layer has at least $\binom m2-\lfloor m/2\rfloor>3m-6$ edges. The opposite interior is connected whenever one interior layer is disconnected.

If two vertices lie outside the largest component $B$, then $B$ reaches all three centers. Indeed, it contains at least three private vertices. In the opposite layer the two outsiders are complete to $Q\cap B$. Any center having no neighbor in $B$ in the first layer would join them to form $K_{3,3}$ in the opposite layer.

A private interior singleton cannot be adjacent in its layer to both $x=19$ and $y=20$. Its full layer degree is at most three, so the two spokes can be joined to draw $xy$. That would restore the undeleted source. To check its nonbiplanarity by [5], take $a=17$,

$$ Y=\{2,3,4,5,6,19,20,21\},\qquad W=\{7,8,9,10,11,22,23,24\}. $$

Both are disjoint eight-cliques. The only nonneighbors of $a$ in $Y$ are the three centers; for each such $c$, $W\subseteq N(a)\cup N(c)$. Adding the missing $a$-center edges gives $K_9$ on $Y\cup\{a\}$, and identifying $a,c$ gives $K_9$ on $W$ plus that identified vertex. These are the hypotheses of the linking lemma. The supplement checks these literal assertions.

5.2. Arms reaching the centers in both layers

If an arm has a component reaching $S$ in each layer, its two components make $S$ cofacial in both layers of each other arm. The four-terminal lemma then forbids an outer cross edge incident with a center-reaching interior component of that first arm in that layer.

At least one of the first arm's interior layers is connected. All its outer cross edges must therefore lie in the other layer. For the large arm, those twelve edges contain $K_{3,4}$, so this is impossible. By Theorem 2 the large arm has a component reaching $S$ in exactly one layer.

For a small arm reaching $S$ in both layers, the disconnected layer can have no outer vertex in a center-reaching component. A $(6,1)$ split fails: a singleton outer vertex leaves the other outer vertex in the large component, whereas a singleton private vertex cannot itself reach $S$ by the preceding restriction. In either two-outsider split the largest component reaches $S$, so it must be exactly $Q_i$. Thus, after choosing a layer $t$, all ten $Q_i$–$O_i$ edges are in $1-t$, and all ten outer cross edges incident with $O_i$ are in $t$.

5.3. Necessary constraints and their checked contradiction

We encode an edge partition, using one Boolean variable for each of the 133 edges. For each arm and each layer, Floyd–Warshall reachability inside its interior determines whether a component reaches all three centers. Each reachability and contact gate is encoded in both directions. The flag $D_{i,t}$ is exactly the negation of this event.

A deficient layer receives the missing-center labels of Section 2.2. The large arm has exactly one deficient layer. A small arm without either deficiency flag must choose one of the two layer orientations of the normal form in Section 5.2.

For every deficient arm, every outside component in either layer must miss one of $S\cup\{u\}$ for each $u\in O_i$, by Theorem 1. Outside means precisely $V(G)\setminus(I_i\cup S)$. Four-valued component labels express this necessary condition. They do not assume the existence of an embedding for an abstract skeleton.

We add the singleton restriction, Euler bounds on unions of original blobs, and forbidden monochromatic Kuratowski subdivisions. The graph's twin permutations and layer swap allow the recorded lexicographic symmetry clauses. A satisfying edge partition representing actual planar layers would lift to all these auxiliary variables; the supplement gives the clause-by-clause mathematical lifting argument. Optional center-contact constraints were present during discovery, but none occurs in the checked proof core and they are not a premise of Theorem 3.

The complete formula has 59,393 variables and 789,253 clauses. The extracted core has 21,077 clauses. Its RUP certificate was verified by DRAT-trim, converted to LRAT, and checked by lrat-check [10]. A fresh replay on the core took about three seconds for the RUP check and under one second for the LRAT check in this environment. The entire core is checked for membership in the generated necessary formula. All 4,913 stored obstruction records are independently checked subdivisions of the literal source graph; the 77 orbit representatives generate 197,520 images under injective private/outer/center relabelings. Images are accepted only when every edge remains an edge of $G$, so no unjustified source automorphism is assumed.

Therefore a biplanar representation of $G$ would satisfy an unsatisfiable necessary formula. This proves Theorem 3, subject to the explicit finite-census and imported-obstruction premises above.

6. Scope and reproduction

The new conclusions constrain dense interfaces, not arbitrary three-terminal patches. They do not assert that deficiency implies cofaciality in both layers: the exact $A_2$ drawing in small_arm_control.json, with independently checked rotations and an apex-obstruction subdivision, contradicts that stronger statement. Nor do they assert that the normal forms alone are sufficient for a biplanar drawing.

The final global exclusion relies on the geometric lifting argument as well as proof checking. A SAT solver's discovery answer alone is not the evidence. Likewise, the subdivision checkers certify every enumerated case but do not themselves prove completeness of plantri's triangulation lists. That external dependence is explicit throughout.

The supplement contains compact gzip representations of exact inputs and certificates, without redundant expanded proof copies. Ordinary ZIP extraction is below two million bytes. Replay uses a disposable temporary directory for expanded SAT files and removes it afterward; peak generated files are small compared with a gigabyte. Run Python without optimization so assertions remain enabled. Python 3, NetworkX, python-sat, and a C compiler are sufficient:

python verify_common_face_ledger.py
python verify_exception_ledgers.py
python verify_small_arm_control.py
python replay_global.py

The first two commands use standard-library graph6 decoding and subdivision verification (the rotation helper is also standard-library). The third checks the positive small-arm control. The fourth reconstructs the source and witnesses, regenerates the formula to audit core membership, then compiles the included licensed checker sources and verifies both RUP and LRAT. It does not search for a new proof. The completeness inputs can be regenerated with plantri's default sphere-triangulation enumeration on 9 and 11 vertices, using graph6 output, with expected counts 50 and 1,249. File hashes, environment versions, expected results, and fuller commands are in the supplement README and manifest.

Theorem 3 does not decide the shared $C_5[2,2,5,3,5]$ core of four other deletion candidates. A subsequent checked exclusion is reported separately in discussion [4]; it is not a premise of this paper. The Earth–Moon problem remains open.

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