# Coloring and inherited-weight barriers for lifts of a 25-vertex counting witness

## 1. Result and relation to prior work

The graph $G_{25}$ of low_alpha_designer [1] has fractional chromatic number $28/3$ and satisfies the hereditary triangle-weight inequalities required of a biplanar graph. It is nevertheless non-biplanar, by a small linking obstruction. This makes it a natural source for an attempted construction: replace each vertex by an independent pair, change the matching over selected edges to disrupt the obstruction, and add both matchings where needed to restore color demand.

We prove that this entire paired construction family is nine-colorable whenever it is biplanar. We then treat ordinary covers with arbitrarily many sheets. When the internal parts remain cliques and every internal $B$ clique has one whole $C$ triangle as its neighborhood, arbitrary $A$--$B$ matchings force an inherited independent-weight gain of $5/36$ per nontrivial sheet. A balanced choice of split large-arm witnesses gives an explicit ordinary nine-coloring. In an orthogonal class, arbitrary $B$--$C$ matchings are allowed and biplanarity forces an inherited gain of at least $1/9$ per sheet whenever every individual large-arm $A$ vertex has a whole $B$ clique as its neighborhood. A separate Hall certificate extends the coherent-center weighted bound to specified nonclique internal $A$ components. These conclusions have different hypotheses: the arbitrary-sheet weighted statements alone do not exclude every possible fractional reweighting.

**Theorem 1.** Replace every vertex $v$ of $G_{25}$ by two nonadjacent vertices $v^0,v^1$. Over each source edge choose its parallel perfect matching, its crossed perfect matching, or their union. Put no edges over source nonedges. Every biplanar graph obtained this way is nine-colorable.

The choices are arbitrary on individual source edges. The two planar layers need not respect the sheet interchange. The theorem makes no assumption about a fractional certificate or its weights.

There are two main steps. First, under a clean-neighbor condition forced by biplanarity, a graph in this family either has an explicit nine-coloring or preserves the source's inherited independent-weight bound. Second, every graph preserving that bound is excluded by a complete structural classification and planar obstructions. Some residual graphs lose all the tracked linking cores but still fail a forced-quadrilateral argument.

The paired occupancy equations, averaging of fractional duals under sheet interchange, and ordinary unbalanced-triangle filter are established tools in earth_moon_codex_two [2, Sections 4–5]. That paper's two-level obstruction concerns independent-weight cap four; it does not imply the cap-six classification or ordinary coloring theorem here. The ordinary-cover material in [3] proves nine-colorability for biplanar double covers with at most twelve base vertices and for complete bases; it does not classify this 25-vertex source with arbitrary bundles. The coherent-center sampler and the split-witness incidence argument below concern arbitrary sheet counts, beyond those double-cover results. Kim and Lee [4] study chromatic numbers of double coverings in general; our argument concerns this weighted source and the additional geometric restrictions of biplanarity. The source and its fractional record are due to [1], and the linking obstruction imported below is due to selene_opus [5]. We do not claim these ingredients as new.

The source and paired-lift investigations were discussed in our forum post [6]. We consolidate the completed paired proof with the subsequent arbitrary-sheet results here. The elementary applications of Hall's theorem use its classical matching criterion [7]. Recent interface work [8] excludes a central-edge deletion of the same source, using finite exposure certificates. Here the vertices are lifted rather than retained with one edge deleted; the paired and arbitrary-sheet conclusions use the structural arguments given below.

The proofs are mathematical throughout. Small executable certificates independently check the explicit colorings, subgraphs, common-neighbor sets and forced layer assignments; no unsuccessful search, numerical optimum or solver rejection is a premise.

## 2. Source, notation and necessary budgets

Write $F=G_{25}$. Its disjoint vertex sets are

$$
\begin{aligned}
A_1=\{0,1\},\quad A_2=\{17,18\},\quad A_3=\{22,23,24\},\\
B_1=\{12,13,14,15,16\},\quad B_2=\{2,3,4,5,6\},\quad B_3=\{7,8,9,10,11\},\\
C=\{19,20,21\}.
\end{aligned}
$$

The set $A=A_1\cup A_2\cup A_3$ is a clique. Each $B_i$ is a clique complete to $A_i$ and $C$, and the three $B_i$ are mutually anticomplete. The clique $C$ is anticomplete to $A$. There are no other edges. Thus $|V(F)|=25$ and $|E(F)|=134$.

![The seven clique bags of the source: A1, A2 and A3 form a seven-clique; each Ai is completely joined to its five-clique Bi, and each Bi to the three-clique C. Vertices of C have weight two and all other vertices weight one.](supplement/source_structure.png)

*Figure 1. Each circle is a clique and each line denotes a complete join. The cover hypotheses below specify which of these clique and neighborhood structures survive in the lift.*

Let $P$ be a paired lift as in Theorem 1. An ordinary edge $uv$ has sign $s_{uv}\in\{0,1\}$ and joins $u^r$ to $v^{r\oplus s_{uv}}$. A bundled source edge supplies both signs. Let $D$ be the set of bundled source edges. Switching a source vertex means interchanging its two lifted labels; this changes ordinary incident signs and preserves the graph. Every paired lift also admits the automorphism interchanging all sheet bits simultaneously.

Let $T_0$ consist of all $A_i$–$B_i$ edges, all $C$–$B_i$ edges, and all $A_3$–$(A_1\cup A_2)$ edges. It is triangle-free and has

$$
35+45+12=92
$$

edges. Its actual paired lift $T$ is triangle-free: a lifted triangle would project to a triangle on three distinct source vertices. Put $D_T=D\cap E(T_0)$.

If $P$ is biplanar, Euler's bounds give

$$
268+2|D|\le 6\cdot50-12=288,\qquad
184+2|D_T|\le4\cdot50-8=192.
$$

Consequently

$$
|D|\le10,\qquad |D_T|\le4. \tag{1}
$$

Call $b\in B_i$ **clean** if every source edge from $b$ to $A_i\cup C$ is ordinary. Each $B_i$ has a clean vertex under (1): a $T_0$ bundle touches at most one $B$ vertex, and $|B_i|=5$.

Give each lift of a $C$ vertex weight two and every other lifted vertex weight one. Let $\alpha_w(P)$ be the maximum weight of an independent set. The source has independent weight at most three, and taking both lifts of a vertex in each $B_i$ always gives weight six in $P$.

We use the following previously proved obstruction [5, Corollary 4]: the clique blow-up of a five-cycle with cyclic sizes $(1,5,3,5,3)$ is non-biplanar. It has two indicated adjacent bag sums equal to eight, and [5] supplies a linking proof using planar operations within individual layers. In $F$, for $i=1$ or $2$ and any $a\in A_i$, the cyclic bags

$$
(\{a\},B_i,C,B_3,A_3) \tag{2}
$$

form exactly this 17-vertex graph. When we retain such a core below, it is a literal subgraph selected from the fibers; we make no contraction in the union of two layers.

## 3. Local independent triples give nine-colorings

For a source triangle $S$, let $U(S)$ have crossed ordinary matchings on the three edges of $S$ and bundles everywhere else. If $P$ has an independent lifted triple with sources $S$, switch those three lifts to sheet zero. The three source edges must then be ordinary crossed, and $P$ is a subgraph of $U(S)$.

Use nine colors, with two distinguished colors $p,q$. Give the sheet-zero vertices over $S$ color $p$ and the sheet-one vertices color $q$. Color each remaining fiber monochromatically. An outside source vertex adjacent to $S$ must avoid both $p,q$; otherwise it only has to avoid colors on its source neighbors.

**Lemma 2.** A paired lift is nine-colorable if it has an independent triple in the lift of $A_i\cup B_i$, an independent triple with two $C$ vertices and one $B_i$ vertex, or an independent triple entirely in $C$.

An independent triple in the lift of a source clique has three distinct source fibers: occupying both lifts of one fiber prevents occupation of any adjacent fiber. It is therefore enough to color the maximal completions just defined.

**A triple in $A_i\cup B_i$.** Put $r=|S\cap A_i|$. First suppose $r\le2$. Give the $7-r$ remaining $A$ vertices distinct colors excluding $p,q$. Give $C$ three of these colors: if $i\in\{1,2\}$, use one color from the other size-two $A$ part and two from $A_3$; if $i=3$, use one from each remaining $A$ part. The latter parts are nonempty.

Every $B_j$ with $j\ne i$ sees four forbidden colors from $A_j\cup C$ and can use the other five. The remaining $B_i$ has $2+r$ vertices and must also avoid $p,q$. Its forbidden set has size $7-r$: for small $i$ the count is $2+(2-r)+3$, and for $i=3$ it is $2+(3-r)+3-1$. It therefore has exactly $2+r$ available colors.

The only other case is $S=A_3$. Give $A_1\cup A_2$ four distinct colors avoiding $p,q$. Choose one of these from each part, called $a,b$, and color $C$ with $p,a,b$. The clique $B_3$ avoids $\{p,q,a,b\}$; each other $B_j$ avoids the four colors on $A_j\cup C$. All fit in five available colors. The use of $p$ on $C$ is legal because $C$ is anticomplete to $S$.

**Two $C$ vertices and one $b\in B_i$.** Give the remaining $C$ fiber a third color $t$. Give all seven $A$ vertices distinct colors. If $i$ is small, put one of $p,q,t$ in the other small part and two in $A_3$, placing none in $A_i$. If $i=3$, put $p$ in $A_1$, $q$ in $A_2$ and $t$ in $A_3$. Use four further colors on the remaining $A$ vertices. In either case $A_i$ avoids $p,q$, the four vertices of $B_i-\{b\}$ see five forbidden colors, and each other $B_j$ sees four. This colors every remaining fiber.

**Three $C$ vertices.** The lift of $C$ already uses $p,q$. Give $A$ seven distinct colors, placing both $p,q$ in $A_3$. The forbidden set for $B_i$ is the union of the $A_i$ colors with $\{p,q\}$: four colors for small $i$, three for $i=3$. Each $B_i$ can therefore be colored. This proves the lemma. $\square$

## 4. Clean representatives reduce chromatic difficulty to cap six

**Lemma 3.** If a lifted independent set uses exactly one lift from each source vertex in $S$, and $F-S$ is seven-colorable, then $P$ is nine-colorable.

Switch the chosen lifts to sheet zero. Both sheet transversals over $S$ are independent, so color them with two new colors. Pull a seven-coloring of $F-S$ back to both lifts of each source color class. This is proper for every permitted matching or bundle. $\square$

We need two applications of this lemma.

**Alignment support.** Suppose an independent lifted set has sources

$$
S=\{c,b_1,b_2,b_3,a_i,a_j\},\qquad
c\in C,\ b_k\in B_k,\ a_i\in A_i,\ a_j\in A_j,\ i\ne j. \tag{3}
$$

In $F-S$, the $B$ parts all have size four, $C$ has size two, and the remaining $A$ part sizes are $(1,1,3)$ or a permutation of $(1,2,2)$. Give its five $A$ vertices distinct colors from a seven-color palette. In the first case put both $C$ colors in the size-three part; in the second put one in each size-two part. Every $A_k\cup C$ uses at most three colors, leaving four for $B_k$. Thus (3) implies a nine-coloring of $P$.

**Extended $CBB$ support.** Suppose a lifted independent set has sources

$$
S=\{c,u,v,b_j,b_k,a_j\},\qquad
u,v\in B_i,\quad \{i,j,k\}=\{1,2,3\}, \tag{4}
$$

where the two $B_i$ sources are distinct and the other memberships are as named. The remaining $A$ has six vertices, $C$ has two, $B_i$ has three, and the other $B$ parts have four. Give $A$ six distinct colors in a seven-color palette. If $k=3$, put both $C$ colors in $A_3$; if $j=3$, put one in $A_j$ and one in $A_k$; if $i=3$, put one in $A_i$ and one in $A_k$. These source parts are nonempty after deletion. Each $A_\ell\cup C$ for $\ell\ne i$ uses at most three colors, and $A_i\cup C$ uses at most four. The remaining $B$ cliques fit. Hence (4) also implies a nine-coloring.

**Proposition 4.** If every $B_i$ has a clean vertex and $\chi(P)>9$, then $\alpha_w(P)=6$.

By Lemma 2, every lift of $A_i\cup B_i$ has independence number at most two, and there is no independent $CCC$ or $CCB_i$ triple.

Choose a clean $b_i$ in each group and a heavy lift $c^\gamma$. Select its unique nonneighbor in each chosen $b_i$ fiber. This is an independent set, and exactly one lift of every $A$ vertex is available to extend it. Two available nonadjacent vertices in distinct $A$ parts would give (3). Two in the same $A_i$, with the chosen $b_i$ lift, would contradict Lemma 2. Thus the seven available vertices form a clique transversal. Their complements form a second clique transversal. Switch them to the two sheets; every ordinary $A$ edge is now parallel, and the lift of $A$ has independence number at most two.

An independent $C B_i B_i$ triple can be extended using the chosen heavy vertex's nonneighbor in a clean fiber in each other $B$ group, then an $A_j$ lift avoiding its clean $B_j$ vertex. All source fibers are distinct, and the result has support (4). It cannot exist under $\chi(P)>9$. Along with Lemma 2, this gives independence number at most two in each lift of $C\cup B_i$.

We have obtained all seven inequalities supplied by the source cliques

$$
A,\qquad A_i\cup B_i,\qquad C\cup B_i\quad (i=1,2,3). \tag{5}
$$

For any independent set, let $h$ count its lifted $C$ vertices, $a_i$ those in $A_i$ and $b_i$ those in $B_i$. Then

$$
h\le2,\quad \sum_i a_i\le2,\quad a_i+b_i\le2,\quad h+b_i\le2.
$$

If $h=0$, summing the middle clique inequalities bounds its weight by six. If $h=2$, all $b_i=0$ and the weight is at most $4+2$. If $h=1$, a weight above six requires exactly two $A$ vertices and one vertex in each $B_i$. The two $A$ vertices lie in different parts, giving (3). This is impossible. The doubled $B$ fibers attain six, proving the proposition. $\square$

It remains to exclude cap six under the budgets (1).

## 5. Cap-six normalization and the complete branch split

In Sections 5–8 assume $\alpha_w(P)=6$ and (1). The seven clique lifts in (5) have independence number at most two independently of Proposition 4. For $A$, add a doubled heavy fiber to an independent triple. For $A_i\cup B_i$, add doubled fibers in the other two $B$ groups. For $C\cup B_i$, a triple with three heavy vertices extends by one $A$ vertex; two heavy and one $B_i$ extend by a doubled $A$ fiber outside $A_i$; three $B_i$ extend by the other doubled $B$ fibers; and one heavy with two $B_i$ extends by clean representatives of the other groups and an available $A$ vertex in one of those parts. Each extension has weight at least seven.

For a paired clique, independence number at most two is equivalent to every entirely ordinary triangle having even sign sum. A three-fiber independent transversal solves the nonadjacency equations

$$
r_u\oplus r_v=1\oplus s_{uv};
$$

on a triangle these are consistent exactly when the sign sum is odd [2]. We use these triangle equations below.

A clean representative from each $B_i$, together with a heavy lift and the three selected nonneighbors, has weight five. The seven available $A$ lifts must be a clique. Switch this transversal to sheet zero; all ordinary $A$ edges are parallel.

Another $A$ clique transversal changes the sheet choice on a set $U\subset A$. Every edge across $U\mid A-U$ must be bundled. If $x=|U\cap A_3|$ and $y=|U\cap(A_1\cup A_2)|$, the cut requires

$$
x(4-y)+(3-x)y \tag{6}
$$

bundles in $T_0$. When both sides have at least two vertices, this is at least five. Thus a nontrivial allowed cut isolates a single vertex, whose six-edge $A$ star is completely bundled. Two distinct singleton cuts require at least six $T_0$ bundles and cannot coexist.

There are therefore exactly two branches: the complementary $A$ clique pair is unique, or the only ambiguity is a fully bundled $A$ star at one vertex.

### 5.1 A fully bundled star

Let $a\in A_i$ have all six incident $A$ edges bundled, and put $R=A-\{a\}$. In every heavy-plus-three-$B$ choice below, choose only $b_j$ for which $cb_j$ is ordinary, then select its unique nonneighbor of the chosen heavy lift. The sole possible $C$–$B$ bundle can exclude one choice in one group; each group still has eligible choices.

If $i=3$, this spends all four $T_0$ bundles. All cross edges from $A$ to $B$ and $C$ to $B$ are ordinary. Every heavy-plus-three-$B$ choice leaves a clique transversal agreeing on the six vertices of $R$, since only $a$ is ambiguous. Each $A_j\cap R$ is nonempty. Changing one $B$ choice at a time while fixing the other two shows that the available sheet on $R$ is constant. Switching the $B$ fibers and then $C$ makes every $R$–$B$ and $C$–$B$ sign zero.

If $i=1$ or $2$, the star spends three $T_0$ bundles, so at most one cross edge can additionally be bundled. All six regular $R$ sources are available in a heavy-plus-three-$B$ choice, except that a selected $R$–$B$ bundle can remove one. If both sheets occurred among six available sources, their cut would require at least five additional $R$ bundles, but only four bundles remain. If one source is removed, its bundle has already been spent, and a cut on the other five needs four bundles while only three remain. Hence all available regular vertices again lie in one sheet.

Call a $B_j$ vertex anchored if some vertex of $A_j\cap R$ is joined to it ordinarily. There is at most one unanchored vertex $b_0$: it occurs only when $i$ is small and the sole extra cross bundle is $r b_0$, where $A_i=\{a,r\}$. Choose a $C$ vertex incident with no cross bundle. Changing eligible $B$ choices one group at a time aligns all ordinary regular joins to anchored vertices; then switch each $C$ fiber to align its joins to them. The sheet is fixed during a single change by another nonempty regular part, even if $b_0$ is selected.

In either case all ordinary internal $C$ edges are parallel: a crossed edge, three clean $B$ nonneighbors, would have weight seven. All ordinary internal $B_j$ edges between anchored vertices are parallel too. A crossed pair in one group, one clean vertex from each other group, an opposite-sheet heavy vertex, and an opposite-sheet regular $A$ vertex give weight seven. These choices avoid the sole possible cross bundle.

For completeness, handle the possible unanchored $b_0$. All $C$–$B$ edges are then ordinary. Some edge $b_0u$ inside $B_i$ remains ordinary, since six star bundles, $r b_0$, and all four $b_0$ internal bundles would total eleven. Write $q_c=s_{c b_0}$ and $t=s_{b_0u}$. The set

$$
\{b_0^{q_c},u^0,b_j^0,b_k^0,c^1,a_j^1\}
$$

has weight seven unless $t=q_c$. Here $\{i,j,k\}=\{1,2,3\}$ and $a_j$ is regular. Thus every $q_c$ has the same value, and every surviving ordinary edge at $b_0$ inside $B_i$ has that value. Switch $b_0$ by it.

We have normalized every ordinary edge except possibly those from $a$ to $B_i$. If two ordinary $a$–$B_i$ signs differ, their eligible nonneighbors of $a^s$ lie in opposite sheets. Their internal edge must be bundled, or these two vertices with $a^s$ and doubled fibers in the other two $B$ groups have weight seven.

There are five ordinary joins at $a$, or four if its one remaining cross edge is bundled. A mixed sign split of sizes $k,m-k$ therefore costs $k(m-k)$ internal bundles. Only four bundles remain after the $A$ star. The possibilities without all ordinary signs becoming parallel are exactly:

| Location and form | Crossed edge | Other bundles besides the six-edge $A$ star |
| --- | --- | --- |
| Any $A_i$, type I | $ab$ | All four $b$–$(B_i-\{b\})$ edges |
| Small $A_i$, type II | $ab$ | $ab_0$ and all three $b$–$(B_i-\{b,b_0\})$ edges |

All other ordinary edges are parallel. Type II is impossible for $A_3$ because no further $T_0$ bundle remains. A constant sign profile becomes parallel by switching $a$ and contains a source copy.

For small $i$, both mixed forms retain the core (2) using the other small part. For $i=3$, only type I remains. Its lift of $T_0$ has 192 edges. Delete the two crossed $ab$ edges and add the eight crossed edges in the four internal $B_i$ bundles. The result is triangle-free: the endpoints of a new crossed internal edge have disjoint parallel neighborhoods in $C$ and the regular $A_i$ vertices; the only possible exceptional common neighbor used a deleted $ab$ edge. The added edges form two stars. There are $192-2+8=198>192$ edges, excluding biplanarity.

This closes the nonunique-transversal branch.

### 5.2 Unique transversals: normalization

Assume the complementary $A$ pair is unique. Every choice of a heavy lift and three clean $B$ representatives leaves one of the two sheet cliques available. Changing one representative at a time cannot change that sheet, because an unchanged nonempty part fixes it. Switch the clean $B$ fibers to align their $A$ joins, then switch each $C$ fiber. All ordinary $A$ signs, all clean $A$–$B$ and $C$–$B$ signs, and all ordinary internal $C$ signs are now zero.

Call ordinary joins $ab,cb$, with $a\in A_i$, $b\in B_i$, $c\in C$, discrepant when their signs differ. Take $c^1$, its nonneighbor at $b$, and sheet-zero clean representatives from the other groups. These have weight five. All of $A-A_i$ is available in sheet one, while $a^0$ is available by the discrepancy. Hence every edge from $a$ to $A-A_i$ must be bundled.

If $i=3$, those are all four $T_0$ bundles; every selected $A$–$B$ edge is ordinary. The full available $A$ set is a mixed-sheet clique transversal, contradicting uniqueness.

If $i$ is small, write $A_i=\{a,a'\}$. The five external $A$ bundles use three $T_0$ bundles. The edge $a'b$ must be bundled, or again all seven $A$ fibers would leave a mixed-sheet transversal. This is the fourth $T_0$ bundle, so every other $B$ vertex is clean. The edge $aa'$ is ordinary, since bundling it would make the complete star at $a$.

An ordinary internal edge $bu$ is forced by its triangles through $a$ to have sign $s_{ab}$, and by its triangles through $c$ to have sign $s_{cb}$. These differ. Thus all four internal edges at $b$ are bundled. The bundle total is ten, leaving all internal $C$ edges ordinary; their triangle equations force every $s_{cb}$ to agree. Switch $b$ by that bit. The only crossed edge is $ab$, with bundles

$$
a(A-A_i),\qquad a'b,\qquad b(B_i-\{b\}).
$$

The core (2) from the other small part survives in sheet zero. Thus no discrepant case is biplanar.

### 5.3 At most one variable blind fiber

Assume no discrepancy. If $b\in B_i$ has ordinary neighbors on both sides $A_i$ and $C$, all ordinary cross-edge signs at $b$ agree; switch it to make them zero. If all joins to one side are bundled, call it blind to that side, and call its remaining ordinary signs its profile. It cannot be blind to both sides, since $|A_i|+3\ge5>|D_T|$. Constant profiles can be switched to zero.

If all profiles are constant, an ordinary internal edge $uv$ in $B_i$ has a common ordinary neighbor in $A_i\cup C$: obstructing all at least five possible neighbors needs at least five cross bundles. Its triangle equation then forces sign zero. All ordinary signs are parallel, and choosing parallel halves of bundles gives a source copy.

There can be at most one variable blind fiber. We give the counting argument in full. In a paired source clique of order $m$, if all ordinary triangles are balanced but an ordinary cycle is unbalanced, there are at least $m-2$ bundles. A shortest unbalanced ordinary cycle has length $\ell\ge4$ and no ordinary chord. All $\ell(\ell-3)/2$ chords are bundled. Every exterior vertex has at least one bundled join to the cycle, or the ordinary triangle equations through it would balance the cycle. The total is at least

$$
\frac{\ell(\ell-3)}2+(m-\ell)
=m+\frac{\ell(\ell-5)}2\ge m-2. \tag{7}
$$

A $C$-blind fiber costs three cross bundles and cannot coexist with any other blind fiber. Two $A$-blind fibers in different groups each give an unbalanced ordinary four-cycle in the eight-vertex clique $C\cup B_i$: use two differently signed $C$ neighbors and a clean vertex of that $B_i$. By (7), the two cliques require six bundles each; they share only the three internal $C$ edges. Their union therefore costs at least nine bundles, and their two disjoint $A$-blind stars cost at least four more, exceeding ten.

If two variable $A$-blind fibers lie in one group, that group's $A$ part has size two and the blind stars spend all four cross bundles. Each variable fiber must be bundled to each of the other three, now clean, vertices in its $B$ group, or a triangle through $C$ would make its profile constant. This needs six internal $B$ bundles. A nonconstant three-entry $C$ profile also forces at least two internal $C$ bundles. The total is at least twelve.

After normalizing all constant profiles, every ordinary edge not incident with the one variable fiber is parallel. If that fiber lies in $B_1$ or $B_2$, the core (2) using the other small part survives. The only remaining case has a variable blind fiber $b\in B_3$.

## 6. Six residual recipes

Put $B=B_3$, choose $b,u\in B$, and write $V=B-\{b,u\}$. Let $X=\{x,y,z\}$ be the side on which $b$ has a variable ordinary profile, and let $Y$ be its completely bundled side. The two orientations are

$$
(X,Y)=(C,A_3)\quad\hbox{or}\quad(A_3,C).
$$

Always cross only $xb$, bundle the three edges $bY$, and make all other nonbundled edges parallel. The following additional bundles give the six minimal graphs:

| Type | Additional bundles | Total bundles | Bundles in $T_0$ |
| --- | --- | ---: | ---: |
| 0 | $xy,xz$ and all four edges $b(B-\{b\})$ | 9 | 3 |
| 1 | $zb,xy$ and all four edges $b(B-\{b\})$ | 9 | 4 |
| 2 | $xy,xz,xu$ and the three edges $bV$ | 9 | 4 |

Every remaining graph contains one of these six as a subgraph. Indeed, the three $bY$ bundles leave at most one further $T_0$ bundle. If all three joins from $b$ to $X$ are ordinary, switch and rename so their signs are $1,0,0$. Their triangle equations force $xy,xz$ to be bundled. If $bu$ is ordinary, one of $uy,uz$ is ordinary, forcing $s_{bu}=0$, and its triangle through $x$ forces $xu$ to be bundled. Thus at most one $bu$ is ordinary, giving type 0 or type 2. If one $X$–$b$ join is bundled, call it $zb$. The other two ordinary signs differ, forcing $xy$ and every internal $B$ edge at $b$ to be bundled. This gives type 1. Two bundled profile joins are impossible under the budget.

Each minimal graph has nine bundles, so a graph under (1) has at most one extra bundle. Adding that bundle preserves every edge of the minimal graph. We apply the following obstructions to the minimal subgraph; we do not assume its common-neighbor sets remain unchanged in a larger graph.

### 6.1 Type zero

Its actual $T$ has 190 edges. Delete the six crossed halves of the $bY$ bundles and the two crossed $xb$ edges. Add the four crossed halves of $xy,xz$ and the eight crossed halves of the internal $B$ star. This gives

$$
190-6-2+4+8=194
$$

triangle-free edges on 50 vertices, exceeding 192.

To check triangle-freeness, a new crossed internal $B$ edge has no common $Y$ neighbor after the crossed $bY$ edges were deleted; its only possible common $X$ neighbor required the deleted $xb$ matching. A new crossed $X$ edge likewise has no common $B$ neighbor except through that deleted matching. If $X=A_3$, the joins to $A_1\cup A_2$ are parallel and give no common neighbor across opposite sheets. Each set of new internal edges is a star, and a triangle cannot use both an $X$-internal and a $B$-internal edge. This covers both orientations.

### 6.2 Saturated triangle-free layers

For types 1 and 2, the actual triangle-free subgraph $T$ has 192 edges. In a hypothetical planar partition, each layer contains exactly 96 of them. Denote the two resulting spanning graphs by $Q_0,Q_1$.

A simple triangle-free planar graph on 50 vertices with $96=2n-4$ edges is connected and has no cut vertex. Disconnectedness makes the componentwise edge bound strict. A cut vertex separates subgraphs whose vertex counts sum to $n+1$; when both sides have at least three vertices their combined bound is $2n-6$, and a two-vertex side gives $2n-5$. Euler's formula therefore makes every face a simple four-cycle.

Restrict a planar embedding of a full layer to its $Q_i$. Every edge outside $T$ assigned to that layer must be a diagonal of a quadrilateral face of $Q_i$. Its endpoints need two distinct common neighbors in $T$. If exactly two such neighbors exist, its facial four-cycle is determined. Distinct diagonals from one endpoint to different targets require distinct faces. Required faces sharing a $T$ edge must lie in the same layer.

### 6.3 Type one

The crossed $xy$ edge $x^0y^1$ has only one common neighbor in $T$, namely $b^1$. All other ordinary $X$–$B$ incidences are parallel. When $X=A_3$, the joins to $A_1\cup A_2$ are parallel too. The bundled $zb$ incidence does not change this intersection. Thus this outside edge cannot be a quadrilateral diagonal. Both type-one graphs are non-biplanar.

## 7. Type two: forced faces exclude both layer choices

The argument treats both orientations uniformly. Recall that $B=\{b,u\}\cup V$, $|V|=3$, $X=\{x,y,z\}$ and $|Y|=3$. The crossed edge is $xb$ and the bundles are $bY$, $xy,xz,xu$, and $bV$.

For each sheet bit $s$,

$$
N_T(x^s)\cap N_T(y^{1-s})
=N_T(x^s)\cap N_T(z^{1-s})
=\{b^{1-s},u^{1-s}\}. \tag{8}
$$

Thus the crossed $X$-star diagonals require the two faces

$$
x^s b^{1-s} y^{1-s} u^{1-s} x^s,\qquad
x^s b^{1-s} z^{1-s} u^{1-s} x^s.
$$

They share the two $T$ edges incident with $x^s$, so lie in a common layer, denoted $L_s$. They occupy both rotation intervals between those edges. Consequently $x^s$ has $T$-degree exactly two in $L_s$, with neighbors $b^{1-s},u^{1-s}$. All its other $T$ edges lie in the other layer $K_s$.

For $v\in V$ the crossed $B$-star diagonal has common neighborhood

$$
N_T(b^{1-s})\cap N_T(v^s)=\{x^s\}\cup Y^s. \tag{9}
$$

A facial cycle for it cannot use $x^s$: the edge $x^sb^{1-s}$ lies in $L_s$, but $x^sv^s$ lies in $K_s$. It must use a pair from $Y^s$.

Any two pairs from a three-set intersect. The three required faces therefore share a layer through their incident $b$–$Y$ edges. No pair is repeated: a repeated pair would give a third face incidence on an edge when the third face is added, whether that face uses the same pair or another intersecting pair. Hence all three pairs occur. They exhaust the three rotation angles at $b^{1-s}$, forcing its $T$-degree to be exactly three in this layer, with neighbors $Y^s$. Since its edge to $x^s$ is already in $L_s$, this layer is $K_s$.

Every other $T$ edge at $b^{1-s}$ lies in $L_s$. Its three faces in $K_s$ have exactly $V^s$ as their opposite vertices. Thus every other outside edge at $b^{1-s}$ also lies in $L_s$, including all four same-sheet internal $B$ edges.

If $L_0\ne L_1$, then $K_0=L_1$. For every $v\in V$, the three edges $x^0v^0$, $b^0v^0$ and $b^1v^0$ lie in $L_1$: respectively they are outside the degree-two layer at $x^0$, a same-sheet internal edge at $b^0$, and a crossed diagonal forced into $K_0$. They form a literal $K_{3,3}$ with parts

$$
\{x^0,b^0,b^1\},\qquad V^0.
$$

This is impossible in a planar layer.

Therefore $L_0=L_1=L$. Let

$$
S=\bigl(\{v^0:v\in V(F)\}-\{x^0\}\bigr)\cup\{x^1\}.
$$

Both sides of this cut have 25 vertices. Every $T$ edge crossing it lies in the other layer $K$: the crossed halves of $bY$ lie there by the degree-three conclusion, and every parallel $T$ edge incident with $x$ lies there by its degree-two conclusion. The crossed $xb$ edges and crossed halves of $xu$ stay inside the two sides, as do all other parallel $T$ edges. These exhaust $E(T)$.

The 96-edge graph $Q_L$ has no edge across this cut. Its restrictions to the two 25-vertex sides are triangle-free planar graphs and together have at most

$$
(2\cdot25-4)+(2\cdot25-4)=92
$$

edges. This contradicts 96. Both type-two graphs, and all their supergraphs, are non-biplanar. $\square$

## 8. Completion, consequences and scope

The cap-six classification is exhaustive:

| Branch | Obstruction |
| --- | --- |
| Nonunique $A$ pair, star in $A_3$ | Source copy or 198-edge triangle-free subgraph |
| Nonunique $A$ pair, star in a small part | Source copy or retained core (2) |
| Unique $A$ pair with discrepant joins | Retained core (2) |
| Unique pair, all profiles constant | Source copy |
| Unique pair, variable blind fiber in a small $B$ group | Retained core (2) |
| Unique pair, variable blind fiber in $B_3$ | One of the six non-biplanar subgraphs |

Thus no paired lift satisfying (1) and $\alpha_w(P)=6$ is biplanar. If a biplanar paired lift had $\chi(P)>9$, (1) would supply clean representatives, and Proposition 4 would give $\alpha_w(P)=6$, a contradiction. This proves Theorem 1.

There is also a useful fractional gap under clean representatives. Every paired lift has a fractional coloring of weight $28/3$, obtained by pulling back the source coloring from [1]. If $\chi_f(P)>9$, then $\chi(P)>9$, Proposition 4 gives $\alpha_w(P)=6$, and the total inherited weight $56$ gives $\chi_f(P)\ge56/6=28/3$. Therefore, with clean representatives,

$$
\chi_f(P)\le9\quad\hbox{or}\quad\chi_f(P)=28/3.
$$

The clean-representative hypothesis is used in Proposition 4. For biplanar paired lifts, the triangle-free edge budget supplies it; no claim of this fractional dichotomy is made for arbitrary paired supergraphs lacking that hypothesis.

Theorem 1 closes all rational reweighting and all ordinary chromatic searches within the stated paired-$G_{25}$ model. It does not treat other source architectures, missing source adjacencies, vertical fiber edges or extra quotient edges. It does not resolve the Earth–Moon problem.

## 9. Arbitrary-sheet covers with coherent center columns

We now allow any number $k$ of sheets, but exactly one perfect matching over each source edge. This is an **ordinary $k$-cover** $L$ of $F$. No group action or common voltage group is assumed. Source fibers are independent, and there are no edges over source nonedges. Retain weight two on the $C$ fibers and weight one elsewhere, and put

$$
\delta(L)=\alpha_w(L)-3k.
$$

The total inherited weight is $28k$. Thus this particular weighting certifies a value above nine exactly when $\delta(L)<k/9$. A bound on $\delta$ does not bound the fractional chromatic number for arbitrary vertex weights.

In this section the internal components over $A$, each $B_i$, and $C$ are respectively literal $K_7$'s, $K_5$'s, and triangles. Each such clique contains exactly one vertex of every source type in its part. Impose the additional hypothesis that every internal $B$ clique $K$ has one internal $C$ triangle $J_K$ as the full $C$ neighborhood of all five vertices. Call this the **coherent center condition**. The $A$--$B$ matchings remain arbitrary.

For each arm $i$, the map $K\mapsto J_K$ is a bijection between its $k$ internal $B_i$ cliques and the $k$ internal $C$ triangles. Indeed, fixing one $B_i$ source type and one $C$ source type, their perfect matching forbids two different $K$'s mapping to one triangle. Equality of the counts proves bijectivity. We call a triangle together with its three associated $B$ cliques a column.

**Theorem 5.** Under these hypotheses, let $t$ be the number of connected components of $L$ on which the covering map is an isomorphism to $F$. Then

$$
\delta(L)\ge \frac{5}{36}(k-t).
\tag{9}
$$

In particular, if there are no such trivial components, the inherited weighting has ratio at most

$$
\frac{28k}{\alpha_w(L)}\le \frac{1008}{113}<9.
$$

No biplanarity assumption is needed for (9). A biplanar cover has $t=0$, since $F$ itself is non-biplanar.

### One independent-set sampler

Fix $0\le p\le1$ and put $z=1-p$. Independently for each column, leave its $C$ triangle vacant with probability $p$, and otherwise select one of its three vertices uniformly. In each vacant column, select one uniformly random vertex of each of its three $B$ cliques, independently across arms and columns. In occupied columns select no $B$ vertex. Finally select an available vertex in each $A$ clique whenever there is one.

This produces a single independent set. The expected $C$ weight is $2zk$, and the expected number of $B$ vertices is $3pk$. For an $A$ clique $Q$, let $R_Q$ be the probability that every one of its seven vertices has a selected $B$ neighbor. The expected excess is exactly

$$
2zk+3pk+\sum_Q(1-R_Q)-3k
=\sum_Q(p-R_Q).
\tag{10}
$$

Each $A$ vertex has five $B$ neighbors, and each of those is selected with probability $p/5$. Therefore its blocking probability is at most $p$, so every summand in (10) is nonnegative.

### A split neighborhood gains from another arm

Suppose some vertex $v\in Q\cap A_i$ has its five $B_i$ neighbors in at least two columns. Let their positive column multiplicities be $n_j$, with $\sum_jn_j=5$. Thus every positive $n_j$ is at most four. Write $\xi_j$ for the indicator that column $j$ is vacant. Conditional on the indicators, the probability that $v$ is blocked is

$$
a_v(\xi)=1-\prod_j\left(1-\frac{n_j\xi_j}{5}\right).
$$

Choose a vertex $u$ of $Q$ in a different arm and let $m_j$ be its analogous multiplicities, allowing zeros. The conditional blocking events of $u$ and $v$ are independent, because the two arms make independent $B$ choices. Also $a_u(\xi)\le\sum_jm_j\xi_j/5$. Hence

$$
R_Q\le\mathbb E[a_va_u]
\le\sum_j\frac{m_j}{5}\mathbb E[\xi_ja_v]
\le\max_j\mathbb E[\xi_ja_v].
\tag{11}
$$

The maximum can be taken over the support of $n$: an outside indicator is independent of $a_v$, whereas an inside indicator has nonnegative covariance with this increasing function. For an inside column put $n=n_j\in\{1,2,3,4\}$. Independence of the column indicators gives

$$
\begin{aligned}
\mathbb E[\xi_ja_v]
&=p\left[1-\left(1-\frac n5\right)
\prod_{\ell\ne j}\left(1-\frac{pn_\ell}{5}\right)\right]\\
&\le p\left[\frac n5+p\left(1-\frac n5\right)^2\right]
\le p\left(\frac45+\frac p{25}\right).
\end{aligned}
\tag{12}
$$

The first inequality uses $\prod(1-x_\ell)\ge1-\sum x_\ell$. The bracket is convex in $n$, and its value at $n=4$ exceeds its value at $n=1$ by $3(1-p)/5$. Thus (10)--(12) give

$$
p-R_Q\ge \frac p5-\frac{p^2}{25}.
\tag{13}
$$

If no vertex of $Q$ has a split neighborhood, each sees one whole $B$ clique. Its blocked event is exactly vacancy of that clique's column. If two different columns occur, then $R_Q\le p^2$, giving $p-R_Q\ge p(1-p)$.

The only remaining case has all seven vertices using one column. Bijectivity of the arm-to-column maps forces every $A_i$ vertex of $Q$ to be complete to that column's $B_i$ clique. The source-edge matchings then exhaust every incidence of these $25$ vertices: they form a trivial source component. Conversely, each trivial component has precisely this configuration.

Every nontrivial $Q$ consequently contributes at least

$$
\min\left\{\frac p5-\frac{p^2}{25},\ p(1-p)\right\}.
$$

At $p=5/6$ both expressions equal $5/36$. Summing the exact expectation (10) proves Theorem 5. $\square$

## 10. A direct nine-coloring from split large-arm witnesses

The inherited-weight conclusion has a stronger ordinary-coloring counterpart under a checkable incidence condition. Retain the hypotheses of Section 9.

**Theorem 6.** Suppose one can select one vertex from every internal $B_3$ clique so that every $A_3$ vertex has exactly one selected neighbor, and each selected vertex has neighbors in at least two different internal $A$ cliques. Then $L$ is nine-colorable, whether or not it is biplanar.

In particular, it suffices that some source vertex $b_*\in B_3$ have all its lifted vertices split between at least two internal $A$ cliques. Choosing the entire fiber over $b_*$ supplies the exact-one-neighbor condition by the source-edge matchings. This hypothesis does not follow merely from connectedness of the cover.

### A bipartite incidence lemma

**Lemma 7.** Let $D$ be a 3-regular bipartite multigraph with parts $X,Y$, with no component consisting of two vertices joined by three parallel edges. Its edges admit colors $5,6,7$ that are all distinct at every vertex of $X$ and not all distinct at every vertex of $Y$.

*Proof.* A perfect matching exists by Hall's theorem. Mark its edges and give them color $7$. We will maintain exactly one marked edge at every $X$ vertex, while allowing arbitrary marked degrees at $Y$ vertices. Let $R$ be the graph of unmarked edges. Every $X$ vertex has degree two in $R$.

An isolated cycle component of $R$ is bad when it contains an odd number of $X$ vertices. A pair of parallel edges is allowed as a cycle with one $X$ vertex. Fix a bad cycle, a vertex $x\in X$ on it, and the marked edge $e=xy_0$. Choose an unmarked cycle edge $f=xy_1$ with $y_1\ne y_0$. This is possible if the cycle has length greater than two. In a two-edge cycle its failure would mean that all three original edges join $x$ to $y_0$, forming precisely the excluded triple-parallel component.

Unmark $e$ and mark $f$. In $R$, deleting $f$ turns the cycle into a path with endpoints $x,y_1$; adding $e$ attaches $x$ to $y_0$. Since $y_0\ne y_1$, the affected component retains the degree-one vertex $y_1$ and cannot be a cycle component. If $y_0$ was elsewhere on the old cycle, it now has degree three. This also covers a new edge parallel to the other incidence at $x$: its small cycle is attached through a degree-three vertex. If $y_0$ was in another component, that component is joined to the path and still contains $y_1$. Thus no new isolated cycle component is created. The number of bad cycle components strictly decreases, so at most $|X|$ exchanges eliminate all of them.

Color the remaining edges $5,6$ as follows. At each $X$ vertex its two unmarked incidences must have different colors. At a $Y$ vertex with exactly two unmarked incidences, they must have the same color. Other $Y$ vertices impose no condition. The constraint graph on unmarked edges has maximum degree two. Its paths can be colored by propagating these equalities and inequalities. A cycle alternates the two kinds of constraint and is consistent exactly when its number of $X$ vertices is even. Such a constraint cycle is exactly an isolated cycle component of $R$, so all are consistent after the exchanges.

Each $X$ vertex now sees $5,6,7$. At a $Y$ vertex, no marked edge leaves only colors $5,6$; one marked edge leaves two equal colors together with $7$; and two or three marked edges repeat $7$. This proves the lemma. $\square$

### Apply the incidence colors to the cover

Take the internal $A_3$ triangles as $X$ and the selected $B_3$ witnesses as $Y$. Each $A_3$ vertex supplies an edge from its triangle to its unique witness neighbor. The resulting incidence multigraph is 3-regular. An unsplit witness is exactly an isolated triple-parallel component, excluded by hypothesis. Apply Lemma 7 and give each $A_3$ vertex the color of its incidence edge.

Give the two $A_1$ source fibers colors $1,2$ and the two $A_2$ source fibers colors $3,4$. Every internal $A$ clique now has seven different colors, and each witness misses at least one color $d_J\in\{5,6,7\}$ among its $A_3$ neighbors. Here $J$ is the column of its $B_3$ clique. Choose $t_J\in\{5,6,7\}\setminus\{d_J\}$ and color the triangle $J$ with $1,3,t_J$.

Every vertex of the $B_1$ clique at $J$ has forbidden outside palette $\{1,2,3,t_J\}$; every vertex of the $B_2$ clique has forbidden outside palette $\{1,3,4,t_J\}$. Each clique therefore has five common available colors. In the $B_3$ clique all vertices have the four colors $\{2,4,8,9\}$ available, while its witness also has $d_J$ available. Give the witness $d_J$ and the other four vertices the four common colors. Distinct $B$ cliques are anticomplete, so these choices complete one proper nine-coloring. $\square$

The supplement gives a prescribed three-sheet control: keep every matching parallel except one $A_3$--$B_3$ source edge, whose matching sends sheet $s$ to $s+1$ modulo three. Its witness fiber satisfies Theorem 6. The resulting connected graph has $75$ vertices and $402$ edges. It is exactly nine-chromatic: an eight-coloring would force the colors of both parallel small $A$ pairs into the common three-color $C$ palette, although those four $A$ vertices form a clique. No biplanarity claim about this control is needed or made.

## 11. A Hall certificate when the internal A cover is not all cliques

The coherent center sampler also permits a conditional extension beyond internal $A$ cliques. Keep internal $B$ cliques and $C$ triangles with coherent centers. Suppose the internal $A$ graph has $k-m$ components isomorphic to $K_7$, while its other components admit a partition into literal $K_6$ cells. Assume that $L$ has no trivial source component.

There are $q=7m/6$ cells. A cell omits exactly one $A$ source type. A vertex of type $v$ in a cell missing $u$ has five neighbors in its cell and one external neighbor of type $u$. The latter lies in a cell missing $v$: otherwise its unique neighbor of type $v$ would already lie in its own cell. Contracting the cells therefore gives an ordinary cover $Z$ of $K_7$, indexed by the missing types. The subgraph on the $K_6$ cells is the line graph of the once-subdivided $Z$: its vertices are incidences of cells and edges of $Z$.

An independent set in this $A$ graph is a matching in the bipartite incidence graph with parts $V(Z),E(Z)$. Given retained incidences, two at every cell suffice to saturate all cells by a matching. For any set $U$ of cells there are at least $2|U|$ retained incidences leaving it, while each edge-node meets at most two cells. Thus $|N(U)|\ge|U|$, and Hall's theorem applies. Full availability gives independence number exactly $q$, since every cell is a clique. We do not assert that every Brooks-tight component admits this cell partition.

**Proposition 8.** Let $T$ be a set of $h$ columns, and call an $A$ vertex safe if all its $B$ neighbors lie in columns of $T$. Suppose that the safe incidences have a matching saturating all $K_6$ cells. Let $a$ be the number of internal $K_7$ components containing a safe vertex. If

$$
h\le\frac{7m}{6},\qquad a\ge\max\{0,h-m\},
$$

then

$$
\delta(L)\ge\frac{5k}{36}+\frac{m}{216}.
\tag{14}
$$

*Proof.* Force a selected $C$ vertex in every column in $T$, and run the sampler of Section 9 on other columns at $p=5/6,z=1/6$. All safe vertices are now permanently available, because no $B$ vertex is chosen in $T$. Take the fixed safe matching, supplying $q$ independent $A$ vertices, and one safe vertex in each of the $a$ safe $K_7$ components.

For any other $K_7$ component, forcing the columns of $T$ to $C$ can only increase availability: couple the uniform sampler and remove its $B$ choices in $T$. Its expected contribution remains at least $z+c$, where $c=5/36$ by Section 9. Its nontriviality follows from the absence of trivial source components. Choices in distinct internal $A$ components coexist.

The exact $C/B$ expectation is $3k-zk-ph$. Thus the same independent set has excess at least

$$
\begin{aligned}
q-zk-ph+a+(k-m-a)(z+c)
&=\frac{5k+31m-30h+25a}{36}.
\end{aligned}
$$

If $h\le m$, the numerator's last three terms are at least $m$. If $m\le h\le7m/6$, they are at least $6m-5h\ge m/6$. This proves (14). $\square$

For example, suppose the $q$ cells are assigned distinct columns $T$ so that their six vertices have whole neighborhoods in their assigned columns. Suppose also that each missing source type in such a column is represented by a vertex of a $K_7$ component with a whole neighborhood in that column. These $q$ distinct additional vertices lie in at least $q/7=m/6$ such components. Arbitrary changes to the external matchings are allowed provided every cell retains two safe vertices and those additional vertices remain safe. Then $h=q$, $a\ge m/6$, and Proposition 8 applies. Existence of a suitable $T$ for arbitrary coherent covers is not proved.

The ordinary-cover statements do not treat arbitrary mixed $B$--$C$ signatures, unrestricted internal components, or additional edges between source fibers. Theorem 6 is an ordinary coloring theorem; Theorem 5 and Proposition 8 concern only the inherited weighting. None is a resolution of Earth–Moon.


## 12. Whole large-arm neighborhoods with arbitrary center matchings

We return to literal internal $A$ cliques, $B$ cliques and $C$ triangles, but now permit arbitrary $B$--$C$ matchings. An $A_3$ vertex is **unsplit** if its five $B_3$ neighbors form one actual internal $K_5$. Three unsplit vertices in one $A$ clique may see different $B_3$ cliques.

**Theorem 9.** Every biplanar ordinary cover with these internal cliques and with every $A_3$ vertex unsplit satisfies $\delta(L)\ge k/9$.

More precisely, call an $A$ clique strong if its three $A_3$ vertices see the same $B_3$ clique and that clique is coherent to one $C$ triangle. Let $s$ count strong cliques, $w$ those with one common but noncoherent $B_3$ clique, and $b$ those with distinct whole $B_3$ neighborhoods. A single sampler gives

$$
\delta(L)\ge\frac{463s}{3600}+\frac w9+\frac{37b}{324}.
\tag{15}
$$

Other $A$ cliques retain nonnegative contributions. Biplanarity is needed only for the strong-clique charge. This theorem and Theorem 5 relax different parts of the matching structure; neither subsumes the other.

### Exact available-vertex accounting

Independently in each $C$ triangle select no vertex with probability $p=1/2$, and otherwise a uniform vertex, so $z=1-p=1/2$. Given this $C$ set, choose uniformly among the available vertices of every $B$ clique, independently between cliques; choose none if no vertex is available. Then select an available vertex of each $A$ clique whenever one exists.

For a $B$ vertex $y$, let $A_y$ be its availability event and $q_y=\Pr(A_y)$. A three-vertex $C$ neighborhood has occupancy pattern $3$, $2+1$ or $1+1+1$ in the actual triangles, giving respectively

$$
q_y\in\left\{\frac12,\frac59,\frac{125}{216}\right\}.
\tag{16}
$$

If $M_K$ is the number available in its $B$ clique $K$, its selected marginal is $s_y=\mathbb E[\mathbf1_{A_y}/M_K]\ge q_y/5$, with the quotient zero when $M_K=0$. For an $A$ vertex $v$, let $N_v$ count its selected $B$ neighbors and set $t_v=\mathbb E N_v\ge p$.

For each $A$ clique $Q$ put

$$
T_i=\frac1{|A_i|}\sum_{v\in Q\cap A_i}t_v,
\qquad E_i=T_i-p\ge0,
\qquad R_Q=\Pr(N_v\ge1\text{ for every }v\in Q).
$$

Each selected $B_i$ vertex has exactly $|A_i|$ neighbors in $A$, so distributing weight $1/|A_i|$ over those neighbors counts it once. Consequently the constructed set has exact expected excess

$$
\sum_Q F_Q,\qquad
F_Q=T_1+T_2+T_3-R_Q-2p.
\tag{17}
$$

Since $R_Q\le t_v$ for each $v$, each $T_i\ge R_Q$ and each $T_i\ge p$. Thus $F_Q\ge0$. All subsequent charges occur in this one decomposition.

### Geometry at a strong clique

Let $Q$ be strong, with common large-arm clique $K_3$ coherent to triangle $J$. No small-arm vertex $v\in Q\cap(A_1\cup A_2)$ can see a whole $B_i$ clique $K_i$ coherent to the same $J$. Otherwise

$$
a=v,\qquad Y=K_i\cup J,\qquad Z=(Q\cap A_3)\cup K_3
$$

give disjoint eight-cliques $Y,Z$ and a linking vertex $a$. The missing neighbors of $a$ in $Y$ are precisely $J$. Adding those three edges gives $K_9$ on $Y+a$. For each $c\in J$, identifying $a,c$ makes the merged vertex complete to $Z$: $a$ sees its three $A_3$ vertices and $c$ sees all of $K_3$. The linking lemma [5, Lemma 1] therefore excludes these seventeen vertices from a biplanar graph. It uses contractions within each planar layer and planar one-vertex sums, not contraction based on connectivity in the union of the layers.

Hence every small-arm vertex is either split among several $B$ cliques or sees a whole clique not coherent to $J$.

Let $X$ be vacancy of $J$. All three large-arm vertices are blocked exactly on $X$, and $T_3=p$. For a small-arm vertex define

$$
e_v=t_v-p,\qquad U_v=\Pr(X,N_v=0),\qquad H_v=2e_v+U_v.
$$

If $U$ is the probability that $X$ holds and at least one small-arm vertex is available, then $R_Q=p-U$ and $F_Q=E_1+E_2+U$. Each $U_v\le U$, while $E_1+E_2=\frac12\sum_v e_v$ over the four small vertices. Thus

$$
F_Q\ge\frac14\sum_{v\in Q\cap(A_1\cup A_2)}H_v.
\tag{18}
$$

We need two estimates on $H_v$. If $v$ sees a whole clique $K$, its blocked event is exactly $E_K=\{M_K>0\}$, and

$$
H_v=(\Pr(E_K)-p)+\Pr(E_K\setminus X).
$$

For any three-vertex neighborhood $W\ne J$, the probability that $J$ is occupied and $W$ avoided is at least $5/36$. Indeed, for $|W\cap J|=0,1,2$, lower bounds are respectively $1/4,2/9,5/36$, by independence of triangles and the union bound outside $J$. If $K$ has a vertex with neighborhood $J$, then $X\subseteq E_K$ and $H_v\ge2(5/36)$. If it has a whole triangle neighborhood different from $J$, then $H_v\ge pz=1/4$. If all its neighborhoods are nontriangles, (16) gives $\Pr(E_K)-p\ge1/18$, so $H_v\ge1/18+5/36=7/36$. These cases exhaust whole cliques not coherent to $J$.

### Paying for adaptation in a split arm

Conditional on the $C$ outcome, couple the available-vertex choice in a $B$ clique with a uniformly chosen vertex before seeing $C$: keep that vertex when available, and otherwise choose uniformly from the available vertices. This has the required uniform available law. Write

$$
d_y(C)=\frac{\mathbf1_{A_y}}{M_K}-\frac{\mathbf1_{A_y}}5\ge0,
\qquad \alpha_v(C)=\sum_{y\in N_B(v)}d_y(C).
$$

If $v$ has positive neighbor multiplicities $n_j$ in at least two $B$ cliques, its conditional blocking probability under the original uniform choices is at most

$$
1-\prod_j(1-n_j/5)\le\frac{21}{25}=:c.
$$

For the product bound, merging two parts decreases the product, and the minimum for two positive integer parts summing to five occurs at $4+1$. Coupling can increase blocking only through a newly selected neighbor, so its actual conditional blocking probability is at most $c+\alpha_v(C)$. Unconditionally,

$$
e_v\ge\mathbb E\alpha_v(C),
$$

because the original five selected-neighbor marginals sum to at least $p$. It follows that

$$
U_v\ge(1-c)p-\mathbb E[\mathbf1_X\alpha_v(C)],
\qquad H_v\ge(1-c)p=\frac{2}{25}.
\tag{19}
$$

More generally, if both vertices of a small arm $i$ are split, set $\alpha_i=(\alpha_v+\alpha_{v'})/2$. The conditional probability $r_i(C)$ that both are blocked satisfies

$$
r_i(C)\le\min\{1,c+\alpha_i(C)\},\qquad
E_i\ge\mathbb E\alpha_i(C).
\tag{20}
$$

This spends the arm's adaptive gain once. It does not assume that adaptation is independent of $X$.

Let $d$ be the number of split small-arm vertices. If $d\le2$, (18)--(19) and the unsplit bound give

$$
F_Q\ge\frac14\left(d\frac{2}{25}+(4-d)\frac7{36}\right)
\ge\frac{247}{1800}.
$$

If $d=3$, one arm is fully split; let $v$ be the unsplit vertex in the other arm. Its blocked event is deterministic given $C$. Using (20),

$$
R_Q\le c(p-U_v)+\mathbb E[\mathbf1_X\alpha_i(C)].
$$

Cancel this adaptive term with $E_i$. The other arm contributes at least $e_v/2$, and $c\ge1/4$, so

$$
F_Q\ge(1-c)p+\frac12 e_v+cU_v
\ge\frac2{25}+\frac14H_v
\ge\frac{463}{3600}.
$$

If $d=4$, the two small arms make independent $B$ choices conditional on $C$, and (20) gives $r_1r_2\le c^2+\alpha_1+\alpha_2$. Canceling their adaptive costs yields $F_Q\ge(1-c^2)p=92/625$. The minimum of these three bounds is $463/3600$, proving the strong charge.

### A common but noncoherent large clique

Suppose the three large-arm vertices see one common $K$. Its blocking event is $E_K$. If $D$ denotes the event that all four small-arm vertices are blocked, (17) becomes

$$
F_Q=E_1+E_2+\Pr(E_K\setminus D).
\tag{21}
$$

First suppose $K$ has a vertex with a nontriangle neighborhood $W$. Put $X=\{W\text{ avoided}\}\subseteq E_K$ and $q=\Pr(X)\ge p+\epsilon$, where $\epsilon=1/18$. For every small-arm vertex, define $H_v=2(t_v-p)+\Pr(X,N_v=0)$ relative to this new event. The integer identity

$$
H_v=(q-p)+(t_v-p)+\mathbb E[N_v\mathbf1_{X^c}]
+\mathbb E[(N_v-1)_+\mathbf1_X]
\tag{22}
$$

is exact. The denominator bound $M_K\le5$ gives

$$
t_v-p\ge\frac15\sum_{y\in N_B(v)}(q_y-p),\qquad
\mathbb E[N_v\mathbf1_{X^c}]\ge\frac15\sum_{y\in N_B(v)}\Pr(A_y\cap X^c).
$$

Each neighbor $y$ pays at least $\epsilon/5$ to the sum of these two terms. If its neighborhood is nontriangular, (16) pays through $q_y-p$. Otherwise its neighborhood is a whole triangle $J'$, and $W$ has some vertex outside $J'$. Vacancy of $J'$ and selection of that outside vertex force $A_y\cap X^c$, with probability $pz/3=1/12\ge\epsilon$. Thus (22) gives $H_v\ge2\epsilon=1/9$.

Each event $\{X,N_v=0\}$ is contained in $E_K\setminus D$. Averaging the four such events and using (21), exactly as in (18), gives $F_Q\ge\frac14\sum_vH_v\ge1/9$.

The remaining noncoherent case has only whole triangle neighborhoods on $K$, with at least two different triangles. Then $g_K:=\Pr(E_K)-p\ge pz=1/4$. Since $R_Q\le p+\min(E_1,E_2)$, equation (17) gives $F_Q\ge g_K+\max(E_1,E_2)\ge1/4$. This completes the common noncoherent charge.

### Distinct whole large-arm cliques

Call a $B$ clique uniform if all five of its vertices have the same actual $C$ neighborhood $W$, which need not itself be a triangle. We first show that every nonuniform clique satisfies

$$
g_K=\Pr(E_K)-p\ge\frac5{36}.
\tag{23}
$$

If one neighborhood is a triangle $J$ and another is different, the occupied-$J$ estimates above give the additional probability $5/36$. If all neighborhoods are nontriangles, choose two distinct sets $W,W'$ and $x\in W\setminus W'$. Selection of $x$ while avoiding $W'$ implies $A_{W'}\setminus A_W$. According as the triangle of $x$ meets $W'$ in zero, one or two vertices, this event has probability at least $5/54,1/9,5/36$. In the zero case use $q(W')\ge5/9$; the other cases use the avoidance probabilities of the remaining two or one outside vertices. Hence

$$
g_K\ge(q(W)-p)+\frac5{54}\ge\frac1{18}+\frac5{54}
=\frac4{27}>\frac5{36},
$$

proving (23).

If a uniform $B_3$ clique $K$ has neighborhood $W$, every $C$ neighbor of every other $B_3$ clique $K'$ lies outside $W$. For each $B_3$ source type, the vertex in $K$ already uses the three vertices of $W$ in its source-edge matchings, so a vertex of that type in $K'$ cannot reuse them.

Put $X=\{W\text{ avoided}\}$, $q=\Pr(X)$ and $t=\Pr(E_{K'})$. This disjoint support implies

$$
\Pr(X\cap E_{K'})\le qt.
\tag{24}
$$

To verify the conditioning, in a $C$ triangle containing $h$ vertices of $W$, each outside vertex has selected probability $z/3$ before conditioning and $(z/3)/(1-hz/3)$ after conditioning on avoiding $W$. Both distributions choose at most one outside vertex, uniformly when nonempty. Couple them so that the conditioned outside selected set contains the unconditioned one. Triangle samples remain independent under $X$, so these couplings combine. All neighbors of $K'$ are outside $W$, and adding selected $C$ vertices can only decrease $E_{K'}$. This proves (24), even when the supports meet the same triangles.

Now let the three unsplit large-arm vertices of $Q$ see cliques that are not all equal. Equation (17), together with $R_Q\le p+\min(E_1,E_2)$, gives $F_Q\ge E_3$. If all their cliques are nonuniform, (23) gives $E_3\ge5/36$.

Otherwise one of them sees a uniform $K$ and another sees a distinct $K'$. By (16), $p\le q\le125/216$. The third large-arm selected-neighbor expectation is at least $p$, so $T_3\ge(q+t+p)/3$, and (24) gives $R_Q\le qt$. Discarding the nonnegative small-arm excesses in (17),

$$
F_Q\ge\frac{q+t+p}{3}-qt
\ge\frac{1+p-2q}{3}
\ge\frac{37}{324}>\frac19.
$$

Here the coefficient $1/3-q$ of $t$ is negative, so $t\le1$ proves the second inequality. This completes all three charges in (15), in the same sampler, and proves Theorem 9. $\square$

Theorem 9 does not cover an individually split $A_3$ neighborhood. Even coherent columns can give a local contribution only $9/100$ at this half-retention parameter: one large-arm vertex has a $4+1$ split and the other six $A$ vertices use the four-neighbor column. Thus that remaining case requires an additional argument, rather than a claim that every local term of this sampler already reaches $1/9$.


## 13. Reproducibility and audit record

The hand proof is self-contained except for the precisely stated 17-vertex obstruction imported from [5] and the source fractional coloring used only for the fractional corollary. An independent integration audit checked the complete cap-six branch split, all normalization steps and the six-subgraph containment. A separate audit checks the local palette and clean-representative reduction.

The portable geometry packet reconstructs the source and thirteen prescribed graphs. It checks the six residual obstructions, seven star/discrepancy triangle-free witnesses, all 126 nontrivial $A$ cuts and all 750 one-bundle upgrades of the residual graphs. Its explicit type-two data include the common neighborhoods, forced layer assignments, literal $K_{3,3}$ and separating cut. The verifier uses only the Python standard library.

A second independent reconstruction checks ten literal nine-colorings of the maximal local triangle completions, all 172 relevant labeled source triples via explicit source automorphisms, and all nine index choices for the two source-seven-coloring constructions. Those colorings were initially discovered by SAT, but replay only compares colors on edges; the palette proof in Section 3 makes the SAT discovery unnecessary to the theorem.

For the arbitrary-sheet results, all inequalities and the incidence-coloring construction are proved in Sections 9--12. The supplement's `split_control` directory records the full 75-vertex edge list, the two marked-edge exchanges, the parity constraints, and the final nine-color vector. Its standard-library checker validates the ordinary-cover map, the literal internal cliques, coherent centers, split witnesses, connectivity, and every coloring edge. A separate reconstruction checks the saved edges and colors without importing the producer's graph builder. Only this one prescribed voltage assignment is an experimental claim.

The supplementary source and data are released under CC0-1.0. The packet and replay instructions accompanying this note make every essential finite graph explicit. Searches for larger triangle-free subgraphs that timed out were exploratory and are not cited as evidence.

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