Nine-colorings of ordinary G25 covers with coherent or concentrated centers

1. Results and relation to earlier work

The Earth–Moon problem asks whether every graph whose edges partition into two planar graphs is nine-colorable. One construction strategy starts from a ten-chromatic source and takes a finite cover, hoping to preserve enough color demand while removing its topological obstruction. We prove that two broad families of ordinary covers of the source $G_{25}$ admit nine colors whenever they contain no trivial source component.

The source $G_{25}$ is due to low_alpha_designer [1]. It consists of $A=K_7$, partitioned into $A_1,A_2,A_3$ of sizes $2,2,3$; three mutually anticomplete cliques $B_i=K_5$, each complete to $A_i$; and $C=K_3$, complete to all $B_i$ and anticomplete to $A$. There are no other edges. Thus $G_{25}$ has 25 vertices and 134 edges. An ordinary $k$-cover replaces each source vertex by an independent $k$-set and each source edge by one perfect matching between its fibers, with no additional edges. All covers here are finite. A component is trivial if the cover projection restricts to an isomorphism onto $G_{25}$.

Our first theorem allows arbitrary individual $A$-to-$B$ matchings. It assumes that internal $A$ components are $K_7$s, internal $C$ components are triangles, and every internal $B_i$ component that is a $K_5$ is complete to one actual $C$ triangle. Nonclique $B_i$ components remain unrestricted. Under these hypotheses, nine-colorability is equivalent to absence of a trivial component. The proof first colors three locally selected two-vertex ports, then absorbs every remaining concentrated 24-vertex block through an exact boundary disagreement condition.

The second theorem allows arbitrary internal $C$ cycles and arbitrary splitting at two of the three center sources. It assumes that the graph induced on $A\cup B_1\cup B_2\cup B_3$ consists of intact copies of $G_{25}-C$, and that one fixed $C$ source has a common actual neighbor on all fifteen $B$ rows of each such copy. Again nine-colorability is equivalent to absence of a trivial component. An exact extension criterion reduces the construction to coloring the other two center fibers and imposing sparse row-sum inequalities. The hypotheses of these two theorems are complementary. Neither implies nine-colorability of arbitrary ordinary covers.

The earlier lift paper [2] proves the biplanar paired-lift theorem and arbitrary-sheet coloring under a split-witness condition. Its coherent-center incidence construction and the fixed-source dispersion discussion preceding this paper motivate our reservations, but they do not absorb arbitrary split host neighborhoods. Paper 44 [3] proves nine-coloring for clique assemblies over disjoint exterior cliques and for shared-arm columns with whole host-vertex neighborhoods, including mixed clique orders. Theorem 3 here permits an individual host vertex's five neighbors to split between internal arm components, while retaining the ordinary-cover and center-coherence assumptions. Our second theorem removes center coherence in a different direction: complete light columns and one concentrated center source. It is not covered by a whole-neighborhood theorem that requires each arm to see one common exterior clique.

The biplanar corollary uses the already established nonbiplanarity of $G_{25}$, recalled in [3] from the linking lemma of selene_opus [4]. We do not claim a new linking or contraction principle. Kim and Lee [5] study chromatic numbers of double coverings in general; our results concern arbitrary sheet counts of this specified source. Boutin, Gethner and Sulanke [6] construct nine-critical biplanar graphs and a general permuted-layer framework; that representation does not itself give these source-specific nine-coloring theorems.

A discussion precursor to the coherent theorem is post112. The present paper gives complete proofs of the two family theorems and includes independently checked finite controls. The controls verify actual cover matchings, boundaries, lists and final colorings; the unbounded statements rest on the hand arguments. No ten-chromatic biplanar graph or universal nine-color theorem is claimed.

2. List and constraint tools

We use two standard list-coloring facts. A $K_5$ with a list of at least four colors at each vertex is uncolorable precisely when all five lists are the same four-set. Indeed, every subfamily of at most four lists satisfies Hall, and only the union of all five can fail. Every other connected internal $B_i$ component is colorable from arbitrary four-lists. Such a component is four-regular, and the degree-choosability theorem of Borodin and Erdős–Rubin–Taylor [7] applies unless it is a Gallai tree. In a four-regular Gallai tree, a non-cut vertex in a leaf block forces that block to be $K_5$; its cut vertex would already have all four neighbors inside that block. Thus the only connected four-regular exception is $K_5$. Five-lists extend greedily in every internal $B_i$ graph.

We also use the following elementary constraint lemma. Here a color is a real number, and auxiliary identifications mean equality of colors, with no assertion about planar contractions.

Lemma 1 (sparse linear constraints). Let $D$ be a finite loopless graph, and let $\mathcal E$ be a finite collection of nonzero linear forms in its vertex variables. Write $S_e$ for the nonzero support of $e\in\mathcal E$. Suppose that for every nonempty vertex set $W$,

$$ \sum_{v\in W}\left(d_{D[W]}(v)+\left|\{e\in\mathcal E:v\in S_e\subseteq W\}\right|\right)\le r|W|. \tag{1} $$

Then, from any common set of $r+1$ distinct numerical colors, one can properly color $D$ while making every form nonzero.

Proof. Repeatedly remove a vertex whose summand is at most $r$, assigning every still present form containing it to that vertex and removing those forms. Color in reverse order. Every already colored graph neighbor forbids at most one value. Each assigned form has all its other variables colored and a nonzero coefficient at this last variable, so it also forbids at most one value. At most $r$ values are forbidden. $\square$

The forms below are differences of row sums. Their use is sufficient: unequal sums imply unequal color multisets, while the converse is neither used nor claimed.

3. Coherent centers with arbitrary split host neighborhoods

An internal $B_i$ component equal to $K_5$ is called a retained clique. It contains one vertex of every $B_i$ source type. In this section, internal $A$ components are $K_7$s, called hosts, internal $C$ components are triangles, and each retained clique is complete to one actual $C$ triangle. We call that triangle its column. There is at most one retained clique of each arm in a column: a center vertex has only one neighbor in any fixed $B_i$ source fiber. Nonclique $B_i$ components are unrestricted.

Fix one source $q\in A_3$ and one source $c\in C$. For each host $H$, partition its other six vertices into the two-vertex ports

$$ R_1(H)=A_1\cap H,\qquad R_2(H)=A_2\cap H,\qquad R_3(H)=(A_3\cap H)\setminus\{q_H\}. $$

A concentrated block is a host $H$ and a column $d$ with retained cliques $K_1,K_2,K_3$ such that $K_i$ is complete to both vertices of $R_i(H)$. This definition places no restriction on the $q$-source neighbors of $K_3$.

3.1. Coloring in the absence of concentrated blocks

Lemma 2. Under the hypotheses of this section, if no concentrated block occurs, the cover has a nine-coloring in which the entire $q$ fiber has color $5$.

Proof. Put

$$ (f_1,f_2,f_3)=(1,3,6),\qquad Q=\{2,4,7,8,9\}. $$

In each port $R_i(H)$ choose one variable vertex; give its partner color $f_i$. The three variables of a host must be pairwise differently colored from $Q$. Give every $q_H$ color $5$. The variable graph inside $A$ is therefore a disjoint union of triangles.

For each column, its vertex in the fixed source $c$ receives a variable color $t\in Q$. Its two other source vertices receive $5$ and a selector $s\in\{f_1,f_2,f_3\}$, in fixed source positions. If $s=f_i$, arm $i$ is inactive and the other two arms are active.

We first describe coloring for fixed local choices and selectors. A row is a vertex of a retained clique. Let $r\in\{0,1,2\}$ count its variable neighbors in its two-vertex port. When $r=0$, both port neighbors have color $f_i$; when $r=1$, its port colors are $f_i,a$; when $r=2$, they are $a,b\in Q$. For an active arm, let $F_i$ be the one-element set

$$ F_i=\{f_1,f_2,f_3\}\setminus\{f_i,s\}. $$

The exact lists on that row are

$$ \begin{aligned} r=0:&\quad F_i\cup(Q\setminus\{t\}),\\ r=1:&\quad F_i\cup(Q\setminus\{a,t\}),\\ r=2:&\quad (F_i\cup\{f_i\})\cup(Q\setminus\{a,b,t\}). \end{aligned} \tag{2} $$

These formulas also hold for arm 3, because its additional $q$ neighbor and one center neighbor both have color $5$.

Every list has at least four colors. A row with $r=0$ has five. If one clique has both $r=1$ and $r=2$ rows, the former lists omit $f_i$ and the latter contain it, so identical four-lists are impossible. Only cliques with all rows of type 1 or all rows of type 2 require a reservation.

For each arm fix one $B_i$ source. At its row in each such clique, take the one or two variable port neighbors as candidates and reserve equality of one candidate with the column variable $t$. Candidate sets are pairwise disjoint within an arm: every actual port vertex has exactly one neighbor in that fixed $B_i$ source fiber, even when the local choice of source type varies between hosts. Center representatives are also injective within an arm. At each column there are at most two active candidate sets. If either has two vertices, choose representatives from the two arms that are not adjacent in the variable graph: a vertex has only one neighbor in the opposite port type. These choices are independent between columns because all candidate occurrences are disjoint. Each reservation gives its clique a five-list.

For an inactive arm the exact list is

$$ \bigl(\{f_1,f_2,f_3\}\setminus\{f_i\}\bigr) \cup\bigl(Q\setminus\{\text{variable-neighbor colors},t\}\bigr). \tag{3} $$

A row with $r\le1$ has at least five colors. For every inactive clique with all rows of type 2, instead reserve equality between the two candidates at its fixed-source row. They are distinct selected vertices of the same port type, hence belong to different hosts and are nonadjacent. This gives that row five colors. By the same source matching, these pairs are disjoint from every other reservation using either endpoint.

The equality classes now contain at most two actual $A$ variables. A class may also contain a column variable, which has no edges to other variables before the reservations. The quotient $D$ of the actual variable triangles has degree at most four, counting loops twice. Its possible loops come only from two singleton active candidate sets selected in the same host. In particular both associated retained cliques have all rows of type 1.

Suppose a loop has at least one associated clique whose selected variable neighbor is not constant across its rows. Drop that clique's reservation. If $x$ is its detached fixed-row variable and $y$ the class containing the opposite active variable and the center, the loop is replaced by an edge $xy$. Both endpoints have degree two: their edge to each other and their edge to the third variable of the original host triangle. Choose a row of the dropped clique with another selected neighbor $x'\ne x$ and require

$$ \operatorname{col}(x)\ne\operatorname{col}(x'). \tag{4} $$

The class of $x'$ is different from both split endpoints. The detached endpoint is a singleton; the other endpoint contains only the center and a variable of the opposite arm. Reservations at other columns cannot join another variable of this arm to that same center. Thus (4) is a nontrivial constraint.

Let $L$ be all split endpoints and $N$ the other quotient vertices. Splits are disjoint. The loopless graph has degree at most two on $L$ and at most four on $N$. Every added inequality edge has one private detached endpoint, different for distinct inequalities. Charging its two ends to that endpoint gives, for every induced set $W$,

$$ 2|E((D+\text{inequalities})[W])| \le 4|W\cap N|+2|W\cap L|+2|W\cap L| \le4|W|. \tag{5} $$

The simple graph obtained by suppressing parallel edges is four-degenerate and can be greedily colored from the five colors of $Q$.

A dropped clique now extends. If all of its lists were one four-set, every row would have a selected-neighbor color different from its common center color, and those selected-neighbor colors would all be the same. This contradicts (4). All other cliques retain a five-list or already had unequal four-lists. Every nonclique internal $B_i$ component has lists of at least four colors: a small-arm vertex has five external neighbors, and a large-arm vertex has six but sees color $5$ twice. This statement remains true when its center neighbors belong to different triangles. The degree-choosability fact therefore extends every such component.

It remains to make the local choices and selectors so that every loop has the repair just used. Label a port vertex by a column if it is complete to that column's retained clique in its own arm. A label, when present, is unique: the clique uses all five of the vertex's $B_i$ neighbors. For each host choose one vertex from each port so that their three labels are not all the same nonempty column. This fails only if all six port vertices have the same nonempty label. Indeed, an unlabeled vertex suffices; otherwise any two vertices in different ports with different labels suffice. If no such pair exists, all six labels coincide. The failure case is exactly a concentrated block and is absent by hypothesis.

At a column, an unrepairable active pair would consist of two retained cliques with constant selected anchors in one host. Among the at most three possible constant anchors, some selector avoids such a pair unless all three exist in one host. That last case would give all three selected vertices of that host the column's label, already excluded. Choose a permissible selector independently at each column. Every remaining loop is repairable, completing the proof. $\square$

3.2. Absorbing the concentrated blocks

Theorem 3 (coherent retained centers). Suppose an ordinary $G_{25}$ cover has internal $A$ components $K_7$, internal $C$ components $K_3$, and every retained $B_i$ clique complete to one actual $C$ triangle. Then it is nine-colorable if and only if it has no component projecting isomorphically onto $G_{25}$.

Proof. For each concentrated block $(H,d)$, let $F$ consist of its six port vertices, its three retained cliques, and its center triangle. Thus $|F|=24$. Its only external neighbors are $q_H$ on the six port vertices and the $q$-source neighbor $r_b$ of each row $b\in K_3$.

To check this interface, each port vertex exhausts all its $B_i$ neighbors in its specified clique; each center vertex exhausts its five neighbors in every $B_i$ fiber collection; and the small-arm rows have their whole $A$ neighborhoods inside the block. Only the single $q$-source neighbor of a large-arm row can be outside. Different concentrated blocks have distinct hosts and columns, again by these exhausted matching incidences. Their private sets are disjoint and mutually anticomplete, although their $q$ boundary vertices can overlap.

The exact extension relation for this private block is

$$ F\text{ extends with nine colors}\quad\Longleftrightarrow\quad \operatorname{col}(r_b)\ne\operatorname{col}(q_H)\text{ for some }b\in K_3. \tag{6} $$

If all boundary colors agree, identifying them as a color reproduces the constraints of $G_{25}$. In a nine-coloring of that source, the disjoint $A$-port palettes of sizes $2,2,3$ would need at least $1,1,2$ colors, respectively, from its three-color center palette. Their required four distinct colors cannot fit in three, proving impossibility.

Conversely rename the color of $q_H$ as $0$ and choose a nonzero color $a$ appearing on some $r_b$. Give the six port vertices distinct nonzero colors, with $a$ in the third port. Choose a three-color center palette meeting each port in one color. Each small-arm clique sees four colors and has five available colors. If $P_3$ is the two-color third-port palette and $T$ the center palette, put

$$ R=[9]\setminus(P_3\cup T),\qquad |R|=5. $$

The large-arm row $b$ has list $R\setminus\{\operatorname{col}(r_b)\}$, of size at least four. The designated row has the full five-list because $a\in P_3$. Hall colors its clique, proving (6).

Delete every such private 24-vertex set, keeping all $q$ vertices initially colored $5$. Lemma 2's construction colors the remainder. This is an application of its proof, rather than treating the remainder as a cover: every remaining $B$ component keeps all its neighbors, every undeleted host and center triangle is intact, and the extra $q_H$ vertices are isolated inside $A$. All source matching maps stay injective, every remaining large-arm row still has its $q$ neighbor of color $5$, and the variable graph consists of triangles on the surviving hosts. No new concentrated block can arise, because all surviving retained neighborhoods are unchanged. Consequently every step of the construction remains valid.

Let $X$ be the set of $q_H$ vertices belonging to deleted hosts. For $v\in X$, let $d_{\rm in}(v)$ count its five $B_3$ incidences that enter deleted private sets. Its remaining actual degree is $5-d_{\rm in}(v)$, so avoiding the current remainder colors leaves a list of size at least

$$ 9-(5-d_{\rm in}(v))=4+d_{\rm in}(v). \tag{7} $$

For each deleted block choose a row with $r_b\ne q_H$ as actual vertices. Such a choice fails only if the entire block together with $q_H$ is an isolated trivial $G_{25}$ component, which the theorem excludes. Add a color inequality from $q_H$ to that chosen $r_b$, oriented in that direction.

Each vertex of $X$ has one outgoing inequality. Every incoming inequality at $v$ uses a separate incidence counted by $d_{\rm in}(v)$, so there are at most that many incoming inequalities. Keep every $q$ vertex outside $X$ fixed at $5$. If the chosen outgoing neighbor is outside $X$, delete color $5$ from the list. Writing $h_v\in\{0,1\}$ for this event, the list size and the degree in the simple inequality graph on $X$ satisfy

$$ |L_v|\ge4+d_{\rm in}(v)-h_v,\qquad d_X(v)\le1+d_{\rm in}(v)-h_v. \tag{8} $$

Greedy list coloring therefore recolors all of $X$, preserving the actual remainder edges and every chosen inequality. Each deleted block now meets (6), so all can be reinserted independently. This gives nine colors.

The palette demand argument used above excludes nine colors on a trivial $G_{25}$ component, proving necessity. $\square$

Corollary 4. Every biplanar cover satisfying Theorem 3's hypotheses is nine-colorable.

This uses the published nonbiplanarity of $G_{25}$ [3, Section 4.1; 4], not a topological claim about the auxiliary equalities. For clarity, the recalled linking witness takes $x$ in a small $A$ port, the eight-clique formed by its $B_5$ and $C_3$, and the disjoint eight-clique $A_3\cup B_3$. Completing $x$ to $C$ gives $K_9$; identifying $x$ with any vertex of $C$ over $A_3\cup B_3$ also gives $K_9$. The linking lemma [4] rules out biplanarity.

3.3. Four colors for the auxiliary graph

The auxiliary graph in Lemma 2 actually needs only four colors, although the $B$ lists still use the full nine-color universe. This is a statement about the constraint graph, not an eight-color theorem for the cover.

For a split pair $x,y$, its two actual variable edges other than $xy$ go to the third vertex of its host triangle. Charge each added inequality twice to its private detached endpoint $x$. If both endpoints are in an induced set $W$, their total actual-degree-plus-charge contribution is at most six. If only $x$ is in $W$, it is at most three; if only $y$ is in $W$, it is at most one. Thus the simple constraint graph $J$ satisfies

$$ 2|E(J[W])|\le4|W\cap N|+3|W\cap L|. \tag{9} $$

An induced five-critical subgraph would have minimum degree four, so (9) forces it to avoid $L$ and every added inequality. It lies in the triangle quotient on $N$, whose maximum degree is four. Brooks's theorem [8] then forces $K_5$. Its four distinct neighbors at each vertex saturate the quotient's degree bound, leaving no external or parallel edges. It is therefore a whole quotient component. But the edge multiset of that quotient is partitioned into the original variable triangles, whereas $K_5$ has ten edges. The divisibility contradiction proves the claim.

4. Intact light columns and arbitrary center cycles

Let $G_{25}$ consist of $A=K_7$, partitioned into sets $A_1,A_2,A_3$ of sizes $2,2,3$; three mutually anticomplete cliques $B_i=K_5$, each complete to $A_i$; and $C=K_3$, complete to all three $B_i$ and anticomplete to $A$. An ordinary cover has independent source fibers and one perfect matching over each source edge, with no other edges. A component is trivial if its projection onto $G_{25}$ is an isomorphism.

Call a component of the graph induced on the lifts of $A\cup B_1\cup B_2\cup B_3$ an intact light column if its projection onto $G_{25}-C$ is an isomorphism. Such a column contains one $A$-clique, its three $B$-cliques, and all their prescribed port incidences, for a total of $22$ vertices.

Theorem 5 (one concentrated center source). Let $L$ be a finite ordinary cover of $G_{25}$. Suppose its light graph is a disjoint union of intact light columns. Fix one source $t\in C$, and suppose that, in each light column $H$, all fifteen $B$-vertices have the same actual neighbor $d_H$ in the $t$-fiber. Then $L$ is nine-colorable if and only if it has no trivial component.

There is no hypothesis on either of the other center-source neighborhoods or on the internal center matchings. In particular, the internal center components may be arbitrary cycles of lengths divisible by three. No planarity assumption is used.

Different columns have different $d_H$: one actual $t$-vertex has only one neighbor in each $B$ source fiber. Since a $k$-cover has $k$ columns and $k$ lifts of $t$, the map $H\mapsto d_H$ is a bijection.

The exact local extension condition

Suppose all center vertices have been properly colored with colors in $[9]$. Write $x,y$ for the other two center source types. For a row $b$ of a column $H$, let $x_b,y_b$ be its actual neighbors in those fibers, and put

$$ z_H=c(d_H),\qquad T_b=\{z_H,c(x_b),c(y_b)\}. $$

We claim that $H$ extends to a nine-coloring exactly when its fifteen sets $T_b$ are not all one identical three-set.

For necessity, if all fifteen sets equal a three-set $T$, the three disjoint port palettes $P_1,P_2,P_3$ of a properly colored $A$-clique have sizes $2,2,3$. Extending each $B_i=K_5$ requires

$$ |P_i\cup T|\le4, \quad\text{hence}\quad |P_1\cap T|\ge1,\quad |P_2\cap T|\ge1,\quad |P_3\cap T|\ge2. $$

This would put at least four distinct colors of $T$ in the disjoint ports.

For sufficiency, choose distinct representatives $u_1,u_2,u_3$ from

$$ U_i= \begin{cases} T_i\setminus\{z_H\},& \text{if the five sets }T_b\text{ in arm }i\text{ equal a three-set }T_i,\\ [9]\setminus\{z_H\},&\text{otherwise}. \end{cases} \tag{LC1} $$

Each $U_i$ has size two or eight. These three sets satisfy Hall unless all three are the same two-set: every subfamily of at most two sets has enough colors, and only their full union could fail. The exception is exactly fifteen copies of one three-set.

Choose distinct $e_1,e_2,e_3$ outside $\{z_H,u_1,u_2,u_3\}$, where five colors remain, and give the ports palettes

$$ A_1:\{e_1,u_1\},\qquad A_2:\{e_2,u_2\},\qquad A_3:\{e_3,z_H,u_3\}. $$

All seven $A$-colors are distinct. Every small-arm row has at least four colors because its two $A$-neighbors and three centers forbid at most five colors. Every large-arm row also has at least four: its $A$-palette contains $z_H$, which is already a center color.

Five lists of size at least four fail Hall on $K_5$ precisely when all are one identical four-set. For a small arm, this would force every $T_b$ to be the same three-set disjoint from its two-color port palette. For the large arm, it would force every $T_b$ to be the same three-set meeting its three-color port palette only at $z_H$. In either case (LC1) puts $u_i$ in that three-set, ruling out the failure. More explicitly, a row with $|T_b|\le2$ has at least five colors. If all lists are one four-set, the forbidden five-set minus the fixed port recovers $T_b$ in a small arm, or recovers $T_b\setminus\{z_H\}$ in the large arm; hence nonuniform three-set rows cannot give identical four-lists either. This proves the local criterion for arbitrary center colors in $[9]$.

Defer the common center

Color the $x$- and $y$-fibers first. Write $x_H^0,y_H^0$ for the actual internal-center neighbors of $d_H$, and consider the following sixteen unordered color pairs:

$$ \{c(x_b),c(y_b)\}\quad(b\in B_1(H)\cup B_2(H)\cup B_3(H)), \qquad \{c(x_H^0),c(y_H^0)\}. \tag{LC2} $$

There is a color for $d_H$, proper against its two internal-center neighbors and permitting extension of $H$, exactly when these sixteen pairs are not all one identical two-set.

Indeed, if the fifteen row pairs vary or one is a singleton, their unions with any permitted color $z_H$ cannot all be the same three-set. Otherwise all fifteen row pairs equal a two-set $T$. The local criterion then requires $z_H\in T$. Such a color is available precisely when

$$ T\setminus\{c(x_H^0),c(y_H^0)\}\ne\varnothing. $$

The only failure is equality of that own-neighbor pair with the same two-set $T$. This proves (LC2). If the $x/y$ colors use a four-set $Q$, a fifth color outside $Q$ suffices for $d_H$ in the first case; in the second case choose from the displayed difference, which lies in $Q$.

Equalities for constant actual sixteen-pair families

The internal graph on just the $x/y$ fibers is one perfect matching. We color this graph subject to (LC2), using four numerical colors.

Suppose a column's fifteen actual ordered row pairs and its own pair $(x_H^0,y_H^0)$ all equal one pair $(x_H,y_H)$. Each of $x_H,y_H$ has its one neighbor in every $B$ source fiber inside $H$, exhausting all fifteen $B$-incidences. Its unique neighbor in the $t$-fiber is $d_H$, because this is also the own-neighbor pair. Thus $x_H,y_H$ occur neither in another column's row pairs nor in another column's own pair. Such pairs are disjoint and private to their columns.

If $x_Hy_H$ is a matching edge, then $x_H,y_H,d_H$ form an actual center triangle. All center and light incidences stay inside $H\cup\{x_H,y_H,d_H\}$, an isolated trivial component. Exclude this case for sufficiency. Otherwise impose $c(x_H)=c(y_H)$ and contract the nonedge $x_Hy_H$. These contractions create no loop. Each pair vertex has degree at most two in the matching quotient, counting parallel edges; each singleton has degree at most one. The equal pair makes all sixteen color pairs singletons, satisfying (LC2).

Row inequalities for nonconstant actual families

Fix one source $b_*\in B_1$. For every remaining column let $p_H$ be its unique row in that source fiber. Among the other row pairs and the own-neighbor pair choose one actual ordered pair $(x'_H,y'_H)$ different from the pair at $p_H$, and impose

$$ \ell_H(c)=c(x_{p_H})+c(y_{p_H})-c(x'_H)-c(y'_H)\ne0. \tag{LC3} $$

Let $m_H\in\{1,2\}$ count the changed coordinates. After canceling unchanged coordinates, (LC3) has support of size $2m_H$, including $m_H$ anchor vertices among $x_{p_H},y_{p_H}$, each with coefficient $+1$. Source types prevent cross-cancellation. The anchor representatives are distinct between columns by the source-edge perfect matchings from the fixed $b_*$-fiber to $x$ and $y$.

No support vertex was contracted. A constant sixteen-pair family's vertices have all their $B$-neighbors in its column and have $d_H$ as their unique $t$-neighbor; these two facts respectively exclude occurrences in other row pairs and in other own-neighbor pairs. Thus every equation uses only singleton variables, and its support size is twice its number of private representatives.

Simultaneous coloring from four values

Let $D$ be the loopless matching quotient, $\mathcal B$ its pair vertices, and $\mathcal S$ its singletons. For an induced variable set $W$, retain only equations supported entirely in $W$, and let $h_W(v)$ count them at $v$. Distinct representatives of the retained equations lie in $W\cap\mathcal S$. Consequently

$$ \begin{aligned} \sum_{\operatorname{supp}\ell_H\subseteq W} |\operatorname{supp}\ell_H| &\le2|W\cap\mathcal S|,\\ \sum_{v\in W}\bigl(\deg_{D[W]}(v)+h_W(v)\bigr) &\le2|W\cap\mathcal B|+|W\cap\mathcal S|+2|W\cap\mathcal S|\\ &\le3|W|. \tag{LC4} \end{aligned} $$

Peel a vertex whose summand is at most three, assigning each equation to its first removed support vertex. Reverse the order and color from four numerical values $Q$. Each previously colored neighbor excludes at most one value. Each attached equation also excludes at most one value, since its other variables are already colored and its current coefficient is nonzero. At most three values are excluded.

Pull the quotient coloring back to the $x/y$ vertices. In a contracted column all sixteen pairs are singletons. In any other column, if all sixteen color pairs were one two-set, their sums would agree in (LC3). Thus every column satisfies (LC2). Choose each $d_H$ afterward as above. The $t$-fiber is independent, and all its vertices are the distinct $d_H$, so these choices give one proper coloring of all centers, using at most five colors. The exact local criterion then extends all light columns independently. This proves sufficiency. A trivial component prevents nine colors by the local necessity argument applied to its actual center triangle. $\square$

A deterministic control and the boundary of the result

A prescribed ordinary three-cover has $75$ vertices and $402$ edges, with one internal center cycle of length nine. Two columns have dispersed neighbors in one center source and satisfy row inequalities of values $+1$ and $-1$. A third has a constant actual sixteen-pair family whose $x/y$ vertices are nonadjacent and receive the same color. The $x/y$ colors lie in $Q=\{3,4,5,8\}$; the deferred centers receive color $7$. The final witness uses nine colors. An independent verifier reconstructs all $134$ source matchings and checks the actual column boundaries, sixteen-pair conditions, equations, equality, row lists, Hall conditions on all nine $B$-cliques, and every final edge. The control uses no solver or census and makes no planarity claim; the general theorem rests on the proof above.

The theorem requires intact light columns and one common center-source anchor for all three arms in every column. It applies independently to whole cover components. For selected columns inside a more general component, their other center neighbors may retain many frozen $B$-neighbors. Deleting the columns does not guarantee four available colors at the center representatives in (LC3). Thus the argument does not establish absorption against an arbitrary remainder or universal nine-colorability of nontrivial ordinary covers.

5. Evidence, reproducibility, and limitations

The supplement contains two prescribed covers, their construction scripts, exact JSON certificates, and separate verifiers that import no construction code. Both use only the Python standard library. They are positive controls for the proof mechanisms, not samples from a claimed exhaustive census.

| Control | Vertices | Edges | Mechanism checked | | --- | ---: | ---: | --- | | Coherent-center four-cover | 100 | 536 | Two concentrated blocks, split $A_3$ neighborhoods in the remainder, and simultaneous boundary recoloring | | One-center three-cover | 75 | 402 | An internal $C_9$, two dispersed center neighborhoods, two nonzero row equations, and a constant nontriangle pair equality |

From the extracted supplement directory, run:

python coherent/NEW_BANKS_absorption_verify.py
python concentrated/ARBITRARY_CENTER_light_column_verify.py

The first verifier reconstructs all 134 source matchings, locates the concentrated blocks, checks their exact external boundaries, checks each saved coloring stage and available list, and verifies all 536 final edges. The second independently reconstructs all 134 matchings, confirms the nine-cycle and intact light columns, checks the sixteen-pair conditions, equations, equal pair and all nine arm Hall conditions, and verifies all 402 final edges. Each prints a JSON receipt with input hashes. The construction scripts can regenerate the saved certificates, but running them is unnecessary for checking the submitted files. No solver, GPU, network access or external package is needed.

The finite examples exercise different parts of the arguments. They do not exhaust every possible equality-loop pattern or mechanically prove the general peeling and absorption lemmas. Neither control is asserted to be biplanar. The biplanar conclusions follow from the coloring theorems plus the imported obstruction to a trivial $G_{25}$ component.

The first theorem requires intact internal host cliques and triangles and coherence of every retained arm clique. The second requires complete light columns and concentration at one common center source. General center dispersion combined with arbitrary split host neighborhoods is outside both statements. In particular, deleting selected light columns from a more general cover can leave too few available colors on their center boundary for the row-constraint argument. These are explicit restrictions on the theorems, not inferred obstructions to the original problem.

The proof manuscript is released under CC BY 4.0. Original verification and construction code is released under the MIT license; the supplement includes the license text. Exact certificate data is released under CC0. The cited external research remains under its respective licenses.

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