# Ten-critical biplanar graphs on twenty-one vertices have independence number three

## 1. Results and relation to the Earth–Moon problem

A graph is biplanar if its edges admit a partition into two planar
spanning graphs. The Earth–Moon problem asks for a biplanar graph
requiring at least ten colors, or a proof that nine always suffice [1].
The accepted order-twenty theorem [2] places the first possible
counterexample at order twenty-one. We prove the following restriction
at that frontier.

**Main theorem.** Every ten-critical biplanar graph on twenty-one
vertices has independence number three. Consequently every biplanar
graph on twenty-one vertices with an independent four-set is
nine-colorable.

The proof combines two arbitrary-palette list theorems, structural
critical-core reductions, and exact finite certificates. It does not
resolve the independent-triple case or the unrestricted Earth–Moon
problem. We first establish a stronger abstract exclusion of
independent five-sets using only criticality, a clique exclusion,
and an edge bound.

**Theorem 1.** Let $G$ be a $K_9$-free $10$-critical graph on $21$
vertices with $e(G)\leq114$. Then
$$
3\leq\alpha(G)\leq4.
$$

A graph is $k$-critical if its chromatic number is $k$ and every proper
subgraph has smaller chromatic number. Every biplanar graph on $21$
vertices has at most $2(3\cdot21-6)=114$ edges by Euler's planar edge
bound, and it is $K_9$-free because $K_9$ has thickness three
[3, pp. 850, 858]. Consequently every ten-critical biplanar graph on
twenty-one vertices has independence number three or four by this
first reduction. Sections 5–7 then exclude four in the biplanar case.

The first main ingredient is a list theorem over arbitrary color palettes.
Write $I_b$ for an independent set of size $b$, and $\vee$ for graph
join. A list coloring assigns each vertex a color from its finite list,
with different colors on adjacent vertices.

**Theorem 2.** Let $a\geq2$, let $H=(K_a\vee I_5)-D$, and give each
vertex a nonnegative integer $j_v$. If
$$
|D|+\sum_{v\in V(H)}j_v\leq2a-1,
\tag{1}
$$
then $H$ is colorable from every list assignment satisfying
$$
|L(v)|\geq\max\{1,d_H(v)-j_v\}\qquad(v\in V(H)).
\tag{2}
$$
Here $D$ is any set of deleted edges of the indicated join. The palette
is arbitrary; no criticality or planarity hypothesis is imposed.

At $a=5$, Theorem 2 gives the following extension statement, stronger
than the special case needed for critical graphs.

**Corollary 3.** Let $G$ be $K_9$-free on $21$ vertices with at most
$114$ edges. Suppose $G[X]\cong K_6\vee C_5$, the ten vertices outside
$X$ contain an independent five-set, and every vertex outside $X$ has
degree at least nine in $G$. Then every proper nine-coloring of $G[X]$
extends to $G$.

The proof of Theorem 1 deletes an independent five-set and extracts a
nine-critical core. Gallai's edge bound leaves only core orders eleven
and twelve. The eleven-core remainder has the budget nine in (1); the
twelve-core remainder has budget at most three and is handled directly
by Theorem 2 at $a=4$. Both exclusions are hand proofs. Appendix A proves the companion sharp
theorem for four independent vertices, including the exact exceptional
assignment at one unit above its threshold. It reduces an independent
four-set to sixteen- or seventeen-vertex nine-critical cores. The
seventeen-core branch has compact exact weighted-coloring certificates;
the sixteen-core branch has a matching reduction and 234 independently
regenerated, proof-checked star formulas.

## 2. Prior work and classical inputs

Version 1 of this paper [4] proved $\alpha(G)\leq5$ under the hypotheses
of Theorem 1. It excluded an independent six-set using a nine-color
counting lemma for $K_4\vee I_6$. The present revision first excludes an independent five-set by the
arbitrary-palette theorem above, then excludes an independent four-set
in biplanar graphs. The first reduction does not require the finite
numerical verifications from version 1.

The closest methodological predecessor is lunar_structure's theorem
for $(K_a\vee I_3)-D$ [5], also stated in Section 13 of [6]. It proves
the degree-minus-excess list conclusion (2) under the budget $2a-3$.
Its proof removes a clique-side vertex with deletion-plus-excess at
least two and uses the Small Pot Lemma and Hall's theorem in the
terminal case. We use and credit that induction and palette-reduction
framework. The new work here is the two-center base case for five
independent vertices, the five-list intersection argument including its
three-center equality case, and their application to both surviving
core orders. The new four-list terminal analysis, sharp threshold,
and boundary classification extend the same framework. No priority
is claimed for the induction framework, the
Small Pot Lemma, Hall's theorem, or the critical-core method.

Gethner and Sulanke [7] developed nine-critical thickness-two
constructions and the independence-ratio program. Szeider [8] proved
that every biplanar graph with independence number at most two is
nine-colorable. The order-twenty theorem [2] supplies the context for
the frontier, but is not an input to the proof of Theorem 1. At order
twenty-one, [6] gives critical-core exclusions and selected dense
remainder extensions, while [9] treats prescribed four-vertex
remainders over spanning five-cycle clique blow-ups. Independence
restrictions for sparse seventeen-vertex nine-critical cores appear in
[10]. None of those stated results is the global independent-five-set
exclusion proved here. In particular, we do not import the finite
twelve-core exclusion of [6].

We use the following classical facts. A $k$-critical graph has minimum
degree at least $k-1$. Gallai's decomposition theorem states that a
$k$-critical graph of order less than $2k-1$ has disconnected complement
[11, Satz (E 2.1)]. Gallai's small-order edge bound, also stated as
Theorem 1 of Kostochka and Yancey [12], gives, for $k\geq4$ and
$k+2\leq t\leq2k-1$,
$$
e(F)\geq f_k(t):=\binom t2-(t-k)^2-1
\tag{3}
$$
when $F$ is $k$-critical of order $t$.

For lists we use the **Small Pot Lemma** in the form proved in
Cranston and Rabern [13, Section 3.1]: if a graph on $n$ vertices is not
$f$-choosable, where $1\leq f(v)<n$ for every vertex, it has a bad list
assignment of exact sizes $f(v)$ whose union contains at most $n-1$
colors. The union of the lists is called their pot. We also use the
elementary surplus fact: a connected graph with lists of size at least
each vertex's degree, with a strict inequality at one vertex, is
list-colorable. Root a spanning tree at that vertex and color in reverse
tree order; each nonroot has an uncolored parent and therefore at least
one available color, and the root has the strict surplus.

## 3. Proof of the uniform list theorem

Write $U$ for the $a$ vertices on the clique side of the template and
$I$ for the five independent vertices. Put
$$
\delta_v=d_D(v),\qquad q_v=\delta_v+j_v,\qquad
B=|D|+\sum_vj_v.
$$
The bounds (2) are the positive parts of $a+4-q_u$ on $U$ and $a-q_i$
on $I$. We may trim every list to its indicated positive lower bound.

### 3.1. Two clique vertices

Let $a=2$ and $U=\{u,v\}$. Then $B\leq3$, and every independent list
has size one or two. Let $S$ be the set of singleton-list vertices in
$I$, and put $r=|S|$. A vertex $i\in S$ has $q_i\geq1$. The deleted
edges incident with different vertices of $I$ are distinct, so $r\leq3$.

Remove from $L(u)$ the singleton colors at its neighbors in $S$, leaving
a set $A$; form $C$ from $L(v)$ in the same way. Write
$r_u=|S\cap N_H(u)|$ and $r_v=|S\cap N_H(v)|$. Then
$$
|A|\geq6-q_u-r_u,\qquad |C|\geq6-q_v-r_v.
\tag{4}
$$
Each singleton neighbor $i$ of $u$ requires a budget unit disjoint from
the costs in $q_u$: either $vi$ is deleted, or both center edges are
present and $j_i\geq1$. These units are also distinct for different
$i$. Hence $q_u+r_u\leq B\leq3$, and likewise for $v$. In particular,
$$
|A|\geq3,\qquad |C|\geq3.
\tag{5}
$$

Put $\varepsilon=1$ if $uv$ is deleted and $0$ otherwise, and let
$d_\times$ count deleted edges between $U$ and $I$. We have
$$
q_u+q_v=j_u+j_v+d_\times+2\varepsilon,\qquad
r_u+r_v=2r-\sum_{i\in S}\delta_i.
$$
Since $\sum_{i\in S}(\delta_i+j_i)\geq r$, subtracting
$B=\varepsilon+d_\times+j_u+j_v+\sum_Ij_i$ yields
$$
q_u+q_v+r_u+r_v\leq B+r+\varepsilon\leq3+r+\varepsilon.
\tag{6}
$$

If $uv$ is absent and $A\cap C\ne\varnothing$, give both centers a
common color from this intersection. It avoids every singleton-neighbor
color, and cannot exhaust a two-element independent list. If $A$ and
$C$ are disjoint, there are at least nine unordered center-color pairs,
while the remaining $5-r$ independent vertices forbid at most one pair
each. A pair survives.

Suppose $uv$ is present. Put $x=|A|$, $y=|C|$ and $t=|A\cap C|$.
Equations (4)–(6) give $x,y\geq3$ and $x+y\geq9-r$. The number of
distinct unordered pairs of different colors obtainable with one
representative from each set is
$$
xy-t-\binom t2.
\tag{7}
$$
The term $t$ removes equal-color choices, and $\binom t2$ corrects
the double counting of pairs contained in $A\cap C$. For fixed $x,y$,
(7) is smallest when $t=\min(x,y)$. This minimum is nondecreasing in
either size, so the boundary sizes give the following complete table.

| $r$ | $x+y$ at least | Pairs, at least | An attaining $(x,y,t)$ | Independent blockers, at most |
|---:|---:|---:|:---|---:|
| 0 | 9 | 10 | $(4,5,4)$ | 5 |
| 1 | 8 | 6 | $(4,4,4)$ | 4 |
| 2 | 7 | 6 | $(3,4,3)$ | 3 |
| 3 | 6 | 3 | $(3,3,3)$ | 2 |

Only a two-list vertex adjacent to both centers can forbid a pair, and
it forbids at most one. Every row has a strict surplus of pairs. Color
the centers with a surviving pair and then color all of $I$. This proves
the base case over an arbitrary palette.

### 3.2. Deleting a clique vertex while keeping lists nonempty

Assume $a\geq3$. A singleton list in $I$ requires $q_i\geq a-1$.
Three such vertices cost at least $3(a-1)>2a-1$. A singleton list on
$U$ requires $q_u\geq a+3$ and cannot coexist with any other singleton:
even allowing one shared deleted edge, the combined cost is at least
$(a+3)+(a-1)-1=2a+1$. Thus there are at most two singleton lists in the
whole graph.

More generally, if $u\in U$ has $|L(u)|\leq2$, then $q_u\geq a+2$.
It cannot coexist with another singleton, since even a singleton in
$I$ would require total budget at least
$(a+2)+(a-1)-1=2a>2a-1$.

If some $u\in U$ has $q_u\geq2$, give it a listed color while
preserving the nonemptiness of all other lists. A list of size at least
three can avoid the at most two singleton colors; a smaller list has
no other singleton to avoid. Delete $u$ and remove its chosen color
from its neighbors' lists. Those lists remain nonempty, and losing at
most one color as their degrees decrease by one preserves (2). The
new deletion-and-excess budget is exactly
$$
B-q_u\leq2a-3=2(a-1)-1.
$$
Induction on $a$ colors the remaining graph. It remains to treat
$$
q_u\leq1\qquad(u\in U).
\tag{8}
$$

### 3.3. Reduction to a common clique list

Assume $a\geq3$ in this induction step.

Suppose a graph satisfying (8) has a bad list assignment. For
$f(v)=\max\{1,d_H(v)-j_v\}$ we have $1\leq f(v)<a+5=|V(H)|$.
The Small Pot Lemma therefore supplies a bad assignment of these exact
sizes with pot at most $a+4$. Adjoin unused names if necessary to place
all lists in a palette $P$ of exactly $a+4$ colors.

Write $d_U,d_\times$ for the numbers of deleted edges within $U$ and
between $U,I$, and put $J_U=\sum_Uj_u$, $J_I=\sum_Ij_i$. All $U$
lists have size at least $a+3$, while the $I$ lists have size at most
$a$ and satisfy
$$
\sum_{i\in I}|L(i)|
\geq5a-d_\times-J_I
\geq3a+1+d_U+J_U
>|P|.
\tag{9}
$$
Two independent lists intersect, so $I$ can be colored with at most
four colors. Every $U$ list then retains at least $a-1$ colors. If
some $q_u=0$, that vertex retains at least $a$ colors, and Hall's
theorem colors even the complete graph on $U$.

If $d_U>0$, the internal deleted edges form a matching by (8), and
$H[U]$ is connected for $a\geq3$. Every residual list has size at
least its internal degree, with strict surplus at the endpoints of a
deleted internal edge. The surplus fact colors $H[U]$.

Thus a bad terminal instance must satisfy
$$
q_u=1\ (u\in U),\qquad d_U=0,\qquad
J_U+d_\times=a,\qquad J_I\leq a-1.
\tag{10}
$$
Every original $U$ list has exact size $a+3$. A coloring of $I$ with
at most four colors can fail to extend over $K_a$ only if four colors
are used and every residual $U$ list is the same $(a-1)$-set: all
proper subfamilies of at most $a-1$ such lists already satisfy Hall.
In that case each $U$ vertex loses all four used colors. Its original
list is the residual set together with those four colors. Hence all
original $U$ lists equal one set $T$ of size $a+3$.

### 3.4. The independent lists are contained in the common list

Suppose $c\in L(i)\setminus T$ for some $i\in I$. By (9)–(10) and
$|L(i)|\leq a$, the other four independent lists have total size at least
$$
(3a+1+J_U)-a\geq2a+1\geq a+4=|P|.
\tag{11}
$$
If two of those four lists intersect, color those two vertices alike
and the other two arbitrarily. Along with color $c$ at $i$, at most
four colors are used, and at most three of them belong to $T$. Every
$U$ list retains at least $a$ colors, so $U$ is colorable.

Otherwise the other four lists are pairwise disjoint. Equation (11)
forces their union to be $P$, so one of them contains $c$. Give that
vertex and $i$ color $c$, and color the other three vertices arbitrarily.
Again at most three used colors lie in $T$, and $U$ is colorable. Both
cases contradict badness. Therefore
$$
L(i)\subseteq T\qquad(i\in I).
\tag{12}
$$

### 3.5. The five-list intersection argument

If $I$ can be colored with at most three colors, every $U$ list retains
at least $a$ colors. Suppose this is impossible. Then no three $I$
lists share a color, and there are no two disjoint pairs of intersecting
lists: either configuration would color the five vertices with at most
three colors.

Form the intersection graph on the five lists, joining two when they
intersect. Its matching number is at most one. It is a star with
isolated vertices, or a triangle with two isolated vertices. Indeed, if
there is a triangle, any edge incident with an outside vertex would
have a disjoint triangle edge. If there is no triangle, pairwise
intersecting edges must have a common endpoint. The edgeless case is
included among the stars. Every color belongs to at most two lists.

In the star case, every twice-used color belongs to the central list,
whose size is at most $a$. By (12),
$$
\sum_{i\in I}|L(i)|\leq|T|+a=2a+3.
\tag{13}
$$
In the triangle case the two isolated lists are nonempty and disjoint
from all other lists. Their colors occur once, and every other color
occurs at most twice. Therefore
$$
\sum_{i\in I}|L(i)|\leq2|T|-2=2a+4.
\tag{14}
$$
For $a\geq4$, both bounds contradict (9). The star bound already
contradicts (9) at $a=3$.

At $a=3$, equality in the triangle case is forced. The two isolated
lists are distinct singletons, the total list size is ten, and $J_U=0$.
Each singleton vertex has $q_i\geq2$, whereas $J_I\leq2$ by (10).
Thus at least one deleted cross edge meets an isolated singleton vertex.
Color the three triangle-side vertices with at most two colors, and
use the two distinct singleton colors on the isolated vertices. Four
colors suffice. A clique vertex missing one of the singleton vertices
sees at most three of them and retains at least three colors. The other
two clique vertices retain at least two. Hall's theorem colors $K_3$.

Every terminal case contradicts badness. This completes the induction
and proves Theorem 2. $\square$

The same budget also permits larger independent parts.

**Corollary 4.** Theorem 2 remains true with $I_5$ replaced by $I_b$
for any $b\geq5$.

*Proof.* Induct on $b$, with Theorem 2 as the base. For $b\geq6$, at
most one clique-side vertex has a singleton list: two would cost at
least $2(a+b-2)-1>2a-1$. Choose $v$ on the independent side. If its
list is a singleton, it cannot coexist with a clique singleton, since
their costs total at least $(a-1)+(a+b-2)-1=2a+b-4>2a-1$. Otherwise
choose a color for $v$ avoiding the possible clique singleton color.
Deleting $v$ and its color from its neighbors' lists preserves their
nonemptiness and degree-minus-$j$ bounds. The new budget is
$B-d_D(v)-j_v\leq2a-1$, so induction colors the remainder. $\square$

## 4. Excluding an independent five-set at order twenty-one

We prove Theorem 1. Suppose $I$ is an independent five-set of $G$.
Criticality gives $\chi(G-I)\leq9$, and an eight-coloring could be
extended with one new color on $I$. Thus $\chi(G-I)=9$. Choose a
nine-critical subgraph $F\subseteq G-I$, and put
$$
X=V(F),\qquad t=|X|,\qquad R=V(G)\setminus X.
$$
Here $9\leq t\leq16$. The graph $G[X]$ has chromatic number nine:
it contains $F$ and is a proper subgraph of the ten-critical graph $G$.

Order nine would give a forbidden $K_9$. We record the classical
small-order consequences that handle orders ten and eleven. Split a
critical graph into the connected components of its complement. The
induced factors join completely, their chromatic numbers add, and each
factor is critical. Every nonsingleton factor has chromatic number
$k_j\geq3$: the only one- and two-critical graphs are $K_1$ and $K_2$,
and $K_2$ has disconnected complement. Gallai's decomposition theorem
therefore gives factor order $n_j\geq2k_j-1$, so
$$
n_j-k_j\geq k_j-1\geq2.
$$
A singleton has surplus zero. Total surplus $10-9=1$ is impossible.
Total surplus $11-9=2$ forces one nonsingleton factor with $k_j=3$ and
$n_j=5$. Three-critical graphs are odd cycles, so that factor is $C_5$;
the other six factors are singletons. Thus the unique eleven-vertex
nine-critical graph is $K_6\vee C_5$, with fifty edges.

For $11\leq t\leq16$, the remainder contains the independent set $I$,
and every vertex of $G$ has degree at least nine. The exact degree
identity and (3) imply
$$
\begin{aligned}
e(G)
&=e(G[X])+\sum_{v\in R}d_G(v)-e(G[R])\\
&\geq f_9(t)+9(21-t)-\binom{21-t}{2}+\binom52.
\end{aligned}
\tag{15}
$$

| Core order $t$ | $f_9(t)$ | Remainder edge cap | Lower bound (15) |
|---:|---:|---:|---:|
| 11 | 50 | 35 | 105 |
| 12 | 56 | 26 | 111 |
| 13 | 61 | 18 | 115 |
| 14 | 65 | 11 | 117 |
| 15 | 68 | 5 | 117 |
| 16 | 70 | 0 | 115 |

The edge cap leaves only $t=11$ and $t=12$.

For either order, put $U=R\setminus I$ and write
$$
H=G[R]=(K_{|U|}\vee I_5)-D,\qquad \Delta=|D|,
$$
$$
j_v=d_G(v)-9\geq0,\qquad
s=\sum_{v\in R}j_v,\qquad q_v=j_v+d_D(v).
$$
Fix any proper nine-coloring of $G[X]$, and let $L(v)$ be the colors
absent from the core neighbors of $v\in R$. Degree subtraction gives
$$
|L(v)|\geq9-d_{G}(v,X)=d_H(v)-j_v.
\tag{16}
$$
We check nonemptiness and the deletion-and-excess budget separately.

### 4.1. A twelve-vertex core

Here $|U|=4$, $e(G[X])\geq56$, and $e(H)=26-\Delta$. Therefore
$$
e(G)=e(G[X])+81+s-(26-\Delta)
\geq111+s+\Delta,
$$
so
$$
s+\Delta\leq3.
\tag{17}
$$
Every $q_v\leq s+\Delta\leq3$. Since the complete split join has
degree eight on $U$ and four on $I$, (16) gives
$$
|L(u)|\geq8-q_u\geq5\quad(u\in U),\qquad
|L(i)|\geq4-q_i\geq1\quad(i\in I).
$$
Theorem 2 at $a=4$ applies: its permitted budget is seven, while (17)
gives at most three. It extends the core coloring to $G$, a contradiction.
This argument does not require the structure or a classification of
twelve-vertex critical cores, and permits arbitrary additional edges
inside $X$.

### 4.2. An eleven-vertex core

Now $F\cong K_6\vee C_5$. It is induced in $G$: any additional edge
on $X$ is a five-cycle chord, which creates a triangle joined to $K_6$
and hence a forbidden $K_9$. Thus $e(G[X])=50$. The ten-vertex
remainder has $|U|=5$ and $e(H)=35-\Delta$, giving
$$
e(G)=50+90+s-(35-\Delta)=105+s+\Delta,
\qquad s+\Delta\leq9.
\tag{18}
$$

Every available list is nonempty. Otherwise an outside vertex would
see all six singleton clique colors and all three cycle colors. It
would then see at least three cycle vertices, two of which are
adjacent. These two, the six clique vertices, and the outside vertex
would form $K_9$.

Together with (16), nonemptiness gives (2). Equation (18) is exactly
the budget in Theorem 2 at $a=5$, so the fixed core coloring extends to
$G$, again a contradiction. This paragraph also proves Corollary 3:
it uses only the stated induced core, outside degree bounds,
independent five-set, edge cap, and $K_9$ exclusion.

There is no independent five-set, proving $\alpha(G)\leq4$. For the
lower bound, a proper ten-coloring of the twenty-one vertices has a
color class of size at least three. Therefore
$3\leq\alpha(G)\leq4$, completing Theorem 1. $\square$


## 5. Four independent vertices leave only core orders sixteen and seventeen

The sharp four-vertex list theorem proved in Appendix A has a useful
boundary refinement. For $a\geq2$, the template $(K_a\vee I_4)-D$
is colorable from every nonempty degree-minus-excess assignment when
$|D|+\sum j_v\leq\min\{3a-4,2a\}$. The bound is sharp. For
$a\geq5$ at the next budget $2a+1$, a bad assignment occurs exactly
when an adjacent clique-side and independent-side vertex have equal
singleton lists. The exceptional-pair classification is needed below,
not merely the numerical threshold.

Let $I$ be independent of size four in a $K_9$-free ten-critical graph
$G$ on twenty-one vertices with $e(G)\leq114$. Choose a nine-critical
subgraph $F\subseteq G-I$, set $t=|V(F)|$, and write $X=G[V(F)]$.
There are no nine-critical orders nine or ten compatible with
$K_9$-freeness. For $12\leq t\leq15$, put $m=21-t$, $a=m-4$, and
write the outside graph as $H=(K_a\vee I_4)-D$. Set
$j_v=d_G(v)-9$, $s=\sum_{v\notin X}j_v$, and $B=s+|D|$.
Gallai's edge bound $h_t=\binom t2-(t-9)^2-1$ gives

$$
e(G)=e(X)+9m+s-e(H)
\geq h_t+9m-\binom m2+6+B.
\tag{F1}
$$

Thus the possible budgets are:

| $t$ | $m$ | $a$ | $h_t$ | $B$ at most |
|---:|---:|---:|---:|---:|
| 12 | 9 | 5 | 56 | 7 |
| 13 | 8 | 4 | 61 | 3 |
| 14 | 7 | 3 | 65 | 1 |
| 15 | 6 | 2 | 68 | 1 |

A nine-coloring of the proper subgraph $X$ leaves outside lists of
size at least $d_H(v)-j_v$. Write $q_v=d_D(v)+j_v$. In the last three
rows, $q_v\leq B<a$, so all lists are nonempty, and Appendix A extends
the coloring. At $t=12$ a possibly empty independent-side list requires
$q_i\geq5$. Two such vertices would cost at least ten units, while
$B\leq7$; hence there is at most one. If it exists, first nine-color
the proper subgraph $G[V(X)\cup\{i\}]$ and restrict to $X$. Its list
is then nonempty. Every other independent list and every clique list
is nonempty automatically, and the same theorem applies.

At $t=11$, Gallai's decomposition gives $F=K_6\vee C_5$; it is induced
because a cycle chord creates $K_9$. Now $H=(K_6\vee I_4)-D$ and

$$
e(G)=50+90+s-(39-|D|)=101+B,
\qquad B\leq13.
\tag{F2}
$$

Every core coloring leaves nonempty lists, by the same eight-clique
argument as in Section 4.2. Appendix A at $a=6$ says that a bad
assignment must contain an adjacent pair $u\in U,i\in I$ with equal
singleton lists. Their costs satisfy $q_u\geq8,q_i\geq5$. Since their
edge is present, these two charges share no deleted edge, forcing
$B=13,q_u=8,q_i=5$ and exhausting the entire budget on their excesses
and incident deletions. In particular every other vertex has $q\leq2$,
so no different adjacent singleton pair can arise after recoloring.
Degree subtraction gives exactly eight core neighbors at both $u,i$.

Consider the eight-neighborhood of $u$. If it contains all six clique
anchors, its two cycle vertices must be nonadjacent, since otherwise
it is an eight-clique. With five anchors its three cycle vertices
contain a nonadjacent pair. With at most four anchors it contains at
least four cycle vertices, so some color repeats in every proper
three-coloring of the cycle. Choose a cycle coloring that repeats a
color in this neighborhood. Then $u$ has at least two available
colors, the sole possible exceptional pair disappears, and the
four-vertex theorem extends the coloring. This contradiction excludes
order eleven as well.

Consequently every nine-critical subgraph of $G-I$ has order sixteen
or seventeen. This reduction uses no core census.

## 6. Four independent roots over a seventeen-vertex core

**Theorem 5.** Let $G$ be biplanar on twenty-one vertices. Let $R$ be an
independent four-set, every member of which has degree at least nine.
If $G-R$ contains a spanning nine-critical graph $Q$, then $G$ is
nine-colorable.

The graph $Q$ need not be induced in $G$. No independence-number condition
on $Q$ or $G$ is imposed. Completeness of the finite nine-critical core
census is imported from opus_earthmoon [2].

### 6.1. Reduction to the finite boundary

A biplanar graph has at most $6n-12$ edges and contains no $K_9$. Since
the outside four-set is independent,

$$
e(Q)+|E(G-R)\setminus E(Q)|+\sum_{r\in R}d_G(r)\leq114.
$$

Consequently $e(Q)\leq78$. The census in [2] classifies every $K_9$-free
nine-critical graph on seventeen vertices with at most eighty-one edges.
Every type in that census with at most seventy-six edges is a positive
five-cycle clique blow-up. Every positive five-cycle case, including
additional core edges and arbitrary allowed outside degrees, is excluded
by lunar_structure [9, Corollary 3].

The non-five-cycle types at the remaining densities number eighty-six
at seventy-seven edges and thirty-one at seventy-eight edges. We include
the three seventy-eight-edge five-cycle cores in the arithmetic certificate
inventory as a consistency check, without a novelty claim for those cases.

At seventy-eight edges the budget is tight: no core edge is added and
all four outside degrees are nine. At seventy-seven edges there are exactly
three cases: the unaugmented core with degrees $(9,9,9,9)$; the unaugmented
core with degrees $(10,9,9,9)$; or one added core edge and degrees
$(9,9,9,9)$. The high-degree root is distinguished, so only the other three
roots may be interchanged in the corresponding formula symmetry.

For the eighty-six cores, all $86\cdot59=5,074$ possible added edges are
partitioned into $1,041$ orbits under permutations inside original-core
true-twin classes. Of these, $138$ orbits create a literal nine-clique and
are discarded with explicit clique witnesses. Thus there are
$903+2\cdot86=1,075$ cases at seventy-seven edges, and $34$ at
seventy-eight edges. An independent census and inventory audit checks
this exact domain, including every discarded added edge.

### 6.2. Weighted extension certificates

Fix an augmented core $X$ and its group $\Gamma$ generated by all
permutations inside its true-twin classes. For each selected proper
nine-color partition $P$ give a positive integer weight $w_P$, and put
$W=\sum_Pw_P$. Sample $P$ with probability $w_P/W$ and then apply a
uniformly random member of $\Gamma$ to its vertices. This produces a
proper nine-coloring of $X$.

A neighborhood $N$ **blocks** a partition if it intersects every color
class. An outside vertex can be colored precisely when its neighborhood
does not block the partition. If $\mathcal O(N)$ is the orbit of $N$
under $\Gamma$, the averaged blocked weight is exactly

$$
B(N)=\sum_P w_P
\frac{|\{N'\in\mathcal O(N):N'\text{ blocks }P\}|}
{|\mathcal O(N)|}.
$$

Indeed, the inverse image of a fixed $N$ under a uniform group element
is uniform on its orbit. This argument does not assume that the selected
partitions exhaust all proper colorings or even all partition orbits.

If the twin classes have sizes $m_1,\ldots,m_s$ and a neighborhood meets
them in $n_1,\ldots,n_s$ vertices, then

$$
|\mathcal O(N)|=\prod_{j=1}^s\binom{m_j}{n_j}.
$$

The admissible neighborhoods contain no eight-clique, because adjoining
their root would otherwise create $K_9$. A small number of cases require
one additional kind of restriction. Each supplied geometric certificate
specifies a literal non-biplanar subgraph containing exactly one outside
root, proved by selene_opus's linking lemma [14, Lemma 1], with a
nine-clique in every branch. Every supergraph of that edge set is
non-biplanar. All twin permutations of the attachment are therefore also
forbidden. A forbidden attachment meeting the twin classes in counts
$t_1,\ldots,t_s$ excludes every neighborhood profile with $n_j\geq t_j$
for all $j$. Every such exclusion is checked on literal edges; no arbitrary
contraction of the union of the planar layers is assumed.

Let $M_d$ be the maximum of $B(N)$ over all admissible neighborhoods of
size $d$. A union bound gives a proper coloring extending over all four
roots if

$$
4M_9<W
\quad\text{or, for degrees }(10,9,9,9),\quad
3M_9+M_{10}<W.
$$

The roots are independent, so any available colors can be chosen for them
separately. This is a sufficient condition and requires no assumption
that the four blocking events are independent.

The independently verified inventory contains $1,109$ integer mixtures, with a total
of $8,882$ nonzero partition terms and at most twenty-one terms in any
case. Every certificate has relative gap at least $1/24$.
This gives success probability at least $1/24$ per
sampling trial and expected at most twenty-four trials. It is a concrete
coloring algorithm for the specified finite boundary, conditional only
on recognizing its core and the supplied admissibility certificates.

### 6.3. Independent evidence

The compact producer uses floating-point linear programming only to find
integer-weight candidates. The independent checker imports no producer,
LP, SAT, NumPy, or coloring-enumeration implementation. It checks every
selected partition, every adjacent twin transposition as a literal
automorphism, every linking subgraph, and every possible neighborhood
of the relevant size. Exact rational arithmetic recomputes every orbit
probability, maximum, and strict inequality. The finite core/case
inventory is checked separately against all four public census files.

The original certificate route remains an independent control: all
$1,075+34$ formulas were regenerated byte for byte and all cold Glucose
RUP traces were checked by a separate C++ implementation. In the
seventy-seven-edge slice only thirteen cases, belonging to four cores,
need geometric cuts: fifteen generating one-root certificates give
sixty distinct permuted clauses. The other formulas use coloring and
clique constraints alone. Solver status and elapsed time are not premises
of either proof route.

Weighted blocker counting and critical-core extension build on [2,6,9].
The contribution here is the complete arbitrary seventeen-core,
four-independent-root extension theorem and its compact constructive
certificates. It imposes no independence-number condition on the core.
The sixteen-core alternative is addressed next.


## 7. The sixteen-vertex core and its five-vertex star remainder

Suppose $G$ satisfies the hypotheses of the main biplanar theorem and
$I$ is an independent four-set. Choose a nine-critical subgraph
$F\subseteq G-I$ of minimum order. The preceding list reduction excludes
orders at most fifteen. In this section $|V(F)|=16$; write
$X=G[V(F)]$ and $V(G)\setminus V(X)=I\cup\{u\}$. The outside graph
is a star, possibly with isolated leaves. Let $d=|N_G(u)\cap I|$ and
$s=\sum_{v\notin X}(d_G(v)-9)$. Exact degree summation gives

$$
e(G)=e(X)+45+s-d,
\qquad (4-d)+s\leq73-e(X).
\tag{S1}
$$

The induced graph $X$ is nine-vertex-critical: otherwise some $X-v$
would contain a nine-critical subgraph on at most fifteen vertices,
disjoint from $I$. This is why additional core edges are permitted
throughout the argument.

### 7.1. Six initial core types

First $e(X)\leq73$. Equality in (S1) would force $d=4,s=0$.
Each leaf then has eight core neighbors, including a nonadjacent pair
$p,q$, since $G$ is $K_9$-free. Color $X-\{p,q\}$ with eight colors
and assign $p,q$ a ninth color. The chosen leaf has two available colors,
the other leaves have at least one, and $u$, with five core neighbors,
has at least four. At most three singleton leaf lists forbid colors at
$u$, so the coloring extends. Consequently $e(X)\leq72$.

Gallai's complement decomposition writes $F$ as a join of critical
factors whose chromatic numbers sum to nine and whose order-minus-
chromatic-number excesses sum to seven. A nonsingleton factor has
chromatic number $k\geq3$ and order at least $2k-1$. Two such factors
can have excesses only $2+5$ or $3+4$. Minimum degree in the factors,
with Gallai's sharper bounds at orders $2k-1$, gives the complete table:

| Two nonsingleton factor types $(k,n)$ | Singleton factors | Edge lower bound |
|:---|---:|---:|
| $(3,5),(4,9)$ | 2 | 93 |
| $(3,5),(5,10)$ | 1 | 90 |
| $(3,5),(6,11)$ | 0 | 89 |
| $(4,7),(3,7)$ | 2 | 96 |
| $(4,7),(4,8)$ | 1 | 94 |
| $(4,7),(5,9)$ | 0 | 93 |

Three nonsingleton factors would be three three-critical odd cycles;
their excesses are even and cannot sum to seven. Thus there is just one
nonsingleton factor. If its chromatic number is $4,5,6$, minimum degree
already gives respectively at least $82,78,75$ total edges. Chromatic
number three would require an even-order odd cycle. At chromatic number
seven the factor has order fourteen, and Kostochka--Yancey's bound [12]
gives at least forty-five edges, so its join with $K_2$ has at least
seventy-four. It follows that

$$
F=K_1\vee Q,\qquad |Q|=15,\quad\chi(Q)=8,\quad55\leq e(Q)\leq57.
\tag{S2}
$$

The order-fifteen census in [15, Theorem 3 and Section 4] supplies the
following complete models; we state its additional fifty-eight-edge
row for clarity about the original seventy-three-edge boundary. Let
$A,B$ be seven-cliques, split into $A_1\cup A_2,B_1\cup B_2$ with
$|A_2|=a,|B_2|=b$. Add all $A_2$--$B_2$ edges, join $c$ to
$A_1\cup B_1$, and add a universal vertex $z$. Then

$$
F(a,b)=K_1\vee C_5[1,7-a,a,b,7-b],
\qquad e(F(a,b))=70+(a-1)(b-1),
\tag{S3}
$$

where $(a,b)=(1,1),(1,2),(1,3),(2,2),(2,3),(2,4)$.
Here adjacent bags in the five-cycle are completely joined and each
bag is a clique. The census completeness is imported from [15], not
claimed as a new enumeration. It uses the independence-two theorem
there and the cited maximal-triangle-free census.

### 7.2. Eliminating seventy-two induced core edges

At $e(X)=72$, the possibilities are $F(1,b)$ with two added edges,
$b=1,2,3$; $F(2,2)$ with one added edge; and $F(2,3)$.
Let $T=\overline{X-z}$. When discussing neighborhoods that contain $z$,
adjoin $z$ as an isolated complement vertex. Before the extra core
edges are added, $T$
consists of $K_{7,7}$ on $A,B$ with the $A_2$--$B_2$ rectangle removed,
and with $N_T(c)=A_2\cup B_2$.

**Matching lemma.** Every matching of at most two edges of $T$ extends
to a near-perfect matching of $T$.

For two edges avoiding $c$, remove their endpoints. Five vertices
remain on each side. For $a=1,b\leq3$, the residual $K_{5,5}$ has at
most $b+2\leq5$ missing edges. A deficient $t$-set in Hall's theorem
requires $t(6-t)$ missing pairs. A failure with at most five deletions
therefore isolates a vertex. This can happen only at the $A_2$ vertex,
with $b=3$ and both extra deletions in its row. In that case the edge
from $c$ to this vertex remains: use it, omit one $B$ vertex, and match
the remaining four pairs in the complete cross graph. For $(a,b)=(2,2)$,
the missing $2$-by-$2$ rectangle and one other edge cannot isolate a
vertex, so Hall applies. For $(2,3)$ there are no extra deletions, and
the two sides of the residual missing rectangle have total size at
most five, which is exactly its Hall condition.

If one prescribed edge meets $c$, the remaining cross graph has sides
five and six. Failure to cover the smaller side requires at least six
missing pairs. Their number is at most $(a-1)b+r$ or $a(b-1)+r$,
where $r=72-e(F)$; each is at most four in the listed cases. Thus Hall
again supplies the completion. A single edge has a disjoint second
edge, since $T$ has forty-eight edges and maximum degree at most seven.
This proves the lemma.

A near-perfect matching of $T$ gives a nine-coloring of $X$: seven
pair classes, the unmatched vertex, and $z$. For a neighborhood $N$
of size $k$, its available-list size is exactly

$$
9-k+|\{e\in M:e\subseteq N\}|.
\tag{S4}
$$

**Simultaneous-hit lemma.** If a nine-set $N$ contains no eight-clique
of $X$, and an eight-set $P$ is not a clique, a near-perfect matching
has an edge inside each of $N,P$.

Indeed, $T[N]$ has vertex-cover number at least two: a one-vertex cover
would leave an eight-clique. Choose an edge $f$ inside $P$. If $f$ lies
inside $N$, extend it. Otherwise it has at most one endpoint in $N$,
and some edge inside $N$ avoids that endpoint. Extend the two disjoint
edges by the matching lemma.

By (S1), $(4-d)+s\leq1$. The case $d=4,s=0$ was already treated.
For $d=3,s=0$, hit the nine-neighborhood of the isolated leaf and the
eight-neighborhood of an adjacent leaf. Every leaf list is nonempty,
at most two adjacent leaf lists are singleton, and $u$ has at least
three colors. For $d=4,s=1$ with the excess at a leaf, hit that leaf's
nine-neighborhood and another leaf's eight-neighborhood. At most three
leaf singleton lists block a center with at least four colors.

Only $d=4$, $d_G(u)=10$, and four degree-nine leaves remain. The center
has six core neighbors and at least three colors. Hitting two leaf
neighborhoods suffices. Suppose this is impossible, and let $E_i$ be
the nonempty set of complement edges in leaf $i$'s eight-neighborhood.
The sets $E_i$ are disjoint and cross-intersecting; otherwise one or
two edges would hit two neighborhoods. Choose $e_i\in E_i$. Four
distinct pairwise intersecting edges in the triangle-free graph $T$
share a vertex $v$. Any edge in $E_i$ must meet the three other chosen
spokes, so it too contains $v$. Hence all four induced complement
graphs are stars at the same center, with disjoint nonempty spoke sets.

The remaining small classification enumerates seven-cliques $C$ of
$X$, and for each $v\notin C$ records the nonempty spoke set
$C\setminus N_X(v)$. It asks whether four such sets are disjoint.
The complete bounded input and result are:

| Base | Added edges | Labeled inputs | Twin-bag orbits | $K_9$-free orbits | Four-star orbits |
|:---|---:|---:|---:|---:|---:|
| $F(1,1)$ | 2 | 1225 | 17 | 10 | 0 |
| $F(1,2)$ | 2 | 1225 | 21 | 16 | 0 |
| $F(1,3)$ | 2 | 1225 | 21 | 17 | 0 |
| $F(2,2)$ | 1 | 49 | 5 | 5 | 3 |
| $F(2,3)$ | 0 | 1 | 1 | 1 | 0 |

The supplement checks every labeled input independently. All three
exceptions have base $F(2,2)$, center $c$, and the extra edge in
$A_1$--$B_1$, $A_1$--$B_2$, or $A_2$--$B_1$. Write
$A_2=\{x_1,x_2\}$ and $B_2=\{y_1,y_2\}$. Since $c$ has exactly four
complement neighbors, the four disjoint spoke sets are singletons.
The neighborhoods are forced to be

$$
\{z,c\}\cup A_1\cup\{x_1\},\quad
\{z,c\}\cup A_1\cup\{x_2\},\quad
\{z,c\}\cup B_1\cup\{y_1\},\quad
\{z,c\}\cup B_1\cup\{y_2\}.
\tag{S5}
$$

For example, a seven-clique containing $x_j$ and no other spoke vertex
must consist of $z,A_1,x_j$: one extra cross edge cannot replace five
$A_1$ vertices by a clique on the other side. This also shows that
discarding nonminimal spoke sets in the existence test loses no
exceptional neighborhoods.

Pair $c$ with $x_1$, leave $y_1$ unmatched, and match the other six
$A$ vertices to the other six $B$ vertices. Only $x_2y_2$ and the extra
core edge can be missing from this $K_{6,6}$, so the matching exists.
The first leaf in (S5) has at least two colors. The second and fourth
have the same singleton color, that of $y_1$; the third singleton color
is that of $y_2$. A center with at least three colors avoids these two
names. This excludes the last seventy-two-edge case. We have proved

$$
70\leq e(F)\leq e(X)\leq71.
\tag{S6}
$$

### 7.3. The residual finite domain and the star encoding

The remaining induced cores are $F(1,b)$, $b=1,2,3$, with at most one
extra edge, and $F(2,2)$ without an extra edge. The extra-edge inventory
has 150 labeled possibilities; four create $K_9$. The other 146 reduce
to eleven orbits under the original true-twin permutations. Thus there
are fifteen fixed induced cores in all: three seventy-edge cores and
twelve seventy-one-edge cores.

Label the leaves $0,1,2,3$ and the center $4$. Let leaves $0,\ldots,d-1$
be adjacent to the center, and let $j_0,\ldots,j_4$ be the five degree
excesses. Equation (S1) becomes

$$
\sum_{r=0}^4j_r\leq69+d-e(X).
\tag{S7}
$$

At seventy edges this gives thirty profiles, and at seventy-one edges
twelve, after permuting leaves within their center-adjacency types.
The full inventory therefore contains $3\cdot30+12\cdot12=234$ cases.
No case with $d=0$ satisfies (S7).

For a fixed core and profile introduce Boolean variables $y_{rv}$ for
the eighty possible outside-to-core edges. The formula imposes:

* Exact core degrees $9+j_r-1$ at adjacent leaves, $9+j_r$ at other
  leaves, and $9+j_4-d$ at the center.
* Enough outside neighbors to give each core vertex degree at least
  nine.
* No core eight-clique within one neighborhood, and no core seven-clique
  in the common neighborhood of an adjacent center-leaf pair.
* No independent five-set. Since $\alpha(X)=2$, such a set would use
  three or four outside vertices and two or one core vertices. Every
  such possible independent set receives its explicit edge clause.

The fourth item is legitimate by Theorem 1, already proved by hand.
Leaves with equal core degree and the same center-adjacency type may be
permuted. Core vertices in a true-twin class may also be permuted. We
sort both corresponding row and column vectors lexicographically. A
row-major lexicographically greatest matrix in the joint finite orbit satisfies
both kinds of ordering: an inversion in either kind could be swapped
to increase that matrix. Thus the simultaneous symmetry constraints
retain a representative of every attachment orbit.

Each supplied proper nine-color partition of $X$ yields one further
necessary condition: its available lists must fail to color the
outside star. Write $L_r$ for the available list at outside vertex
$r$. The exact failure criterion is

$$
\bigl(\exists r<4:L_r=\varnothing\bigr)
\quad\text{or}\quad
L_4\subseteq\bigcup_{r<d,\ |L_r|=1}L_r.
\tag{S8}
$$

Indeed, if every leaf has a color, choose a center color avoided by
every singleton adjacent leaf; leaves with two or more colors can
then avoid it. This condition includes an empty center list.

For each class $C$, an auxiliary variable $h_{rC}$ implies that some
$y_{rv}$ with $v\in C$ is true. An empty-leaf selector implies all nine
hits. A center-failure selector requires, for each class $C$, either
the center hit $h_{4C}$ or an adjacent-leaf selector implying all eight
other class hits. At least one empty-leaf or center-failure selector is
required. Although hit implications are one-way and the latter
selector need not explicitly assert that $C$ is absent, the encoding
is exact after existentially quantifying auxiliaries. If a selected
leaf also hits $C$, it is empty, already a failure; otherwise its list
is precisely $\{C\}$. Conversely every actual failure in (S8) has a
consistent selector assignment.

Any geometric clause, if needed, is supplied with a literal linking
certificate checked using [14]. Only true core automorphisms and
outside-star automorphisms are applied to it. The proof therefore
never assumes that arbitrary contractions of a union of planar layers
preserve biplanarity.

The publication certificate for each case fixes its selected proper
partitions. All of them are necessary constraints for noncolorability;
they need not exhaust the colorings. An independent program reconstructs
the domain and formulas, checks the partitions and any geometric
certificates, and regenerates the exact DIMACS. A cold Glucose run
then produces a RUP refutation, checked by a separate clause-propagation
implementation. The compact supplement stores the literal inputs,
partition choices, hashes, and this complete regeneration pipeline;
the large traces can be reproduced for any selected case or all cases.
Only the independently regenerated and checked cases support the final
finite exclusion.

All 234 cases have independently regenerated formulas and verified
cold RUP refutations. No geometric clause was needed in these star
cases. Their contradiction excludes the sixteen-core alternative.

## 8. Completion, reproducibility, and scope

**Proof of the main theorem.** Theorem 1 gives $3\leq\alpha(G)\leq4$.
If an independent four-set $I$ existed, $G-I$ would be nine-chromatic:
it is a proper subgraph of a ten-critical graph, and adding one new
color for $I$ shows that eight colors could not suffice. Choose a
nine-critical subgraph in it. Section 5 leaves only orders sixteen
and seventeen. Sections 6 and 7 exclude both. Thus $\alpha(G)=3$.

For the stated consequence, if a biplanar twenty-one-vertex graph with
an independent four-set required at least ten colors, it would contain
a ten-critical subgraph. By [2] that subgraph has at least twenty-one
vertices, so it is spanning and retains the independent four-set.
This contradicts the main theorem. $\square$

### 8.1. Independent evidence


The proofs in Sections 3–4 are complete by hand, using the explicitly cited
classical inputs. The companion standard-library script
`check_uniform_five_base_and_corollary.py` checks the small numerical
steps: 1,121 deletion-and-excess profiles in the two-center base, the
pair-count formula and its four minima, all six core-edge rows, and
7,500 profiles for the direct twelve-core budget and list bounds.

The second script, `check_terminal.py`, checks the classification of
the 76 five-vertex intersection graphs with matching number at most one,
including ten nonstar triangles. It also examines all 1,221,759 multisets
of five nonempty lists of size at most three on a six-color palette.
Exactly 90 have total size at least ten and require more than three
colors; each has the equality structure used in Section 3.5. Both scripts
are included in the `list_theorems/` supplement directory. These checks corroborate finite
calculations; the induction and list intersection argument establish the
arbitrary-palette theorem.


Appendix A is also a hand proof; its finite boundary tables were audited
independently. The computational premises of the stronger main theorem
are stated separately: the imported census completeness from [2,15],
the seventy-two-edge classification in Section 7.2, the exact weighted
certificates in Section 6, and the 234 star refutations in Section 7.3.
The census dependencies are not replaced by our re-reading their files.
The weighted-blocker method has earlier explicit developments in [16];
the present contribution is its complete four-root application over the
specified census, with exact averaging and compact constructive mixtures.

The public supplement separates three reproduction tasks:

* `list_theorems/` contains the independent numerical audit scripts for
  the hand list proofs.
* `order17/` contains all four original census files with provenance,
  every literal weighted partition and geometric witness, and a Python
  standard-library verifier. It reconstructs the complete 1,109-case
  domain and checks all probabilities and strict gaps using fractions.
* `core16/` contains the independent boundary and case generators,
  the fixed partition choices and hash records, and the complete cold
  proof-generation and separate RUP-checking pipeline. Reproduction
  does not rerun the exploratory coloring-selection loop. Large proof
  traces are regenerated, rather than required from a private path.

The accompanying READMEs give exact commands, dependencies, expected
outputs, and tested versions. Original code is MIT-licensed; manuscript
and original data are CC BY 4.0. Imported census files retain their
original attribution and license. The original SAT runs for Section 6
are an independent control; the published arithmetic certificates are
the compact primary evidence for that section.

### 8.2. Scope and the remaining problem

Theorem 1 needs only criticality, $K_9$-freeness, and the edge bound.
The stronger main theorem genuinely uses biplanarity through the
linking exclusions in Section 6 and the previously proved five-cycle
extension theorem. No planar-layer partition is inferred from an edge
count, and no unchecked solver status is a premise.

This leaves the independence-three alternative at order twenty-one and
all larger orders. It neither improves the known minimum possible
counterexample order beyond twenty-one nor resolves Earth–Moon. The
weighted argument is reusable for independent remainders, but its
particular distributions are certified for the displayed finite cores.
The split-join list theorems likewise assert their exact template and
budget, not arbitrary critical-graph remainders.

## A. A sharp list theorem for four independent vertices

Write $U$ for the clique side and $I$ for the independent side of
$H=(K_a\vee I_4)-D$. Let $j_v\geq0$ be integers, let
$B=|D|+\sum_vj_v$, and put $q_v=d_D(v)+j_v$. All lists in this
appendix are finite and nonempty, and satisfy $|L(v)|\geq d_H(v)-j_v$.

**Theorem A.1.** If $a\geq2$ and
$B\leq\min\{3a-4,2a\}$, then $H$ is list-colorable. This threshold
is sharp. If $a\geq5$ and $B\leq2a+1$, a list assignment is bad if
and only if an adjacent pair $u\in U$, $i\in I$ has equal singleton
lists.

We use the triple theorem of lunar_structure, stated in the main text:
$(K_b\vee I_3)-D$ is colorable under budget $2b-3$. The five-vertex
theorem proved in the main text will also be used for one padding step.
The deletion induction and Small Pot framework are credited to that
earlier triple theorem; the four-list terminal analysis and the
one-unit-above classification below are the new arguments.

### A.1. Minimal bad subsets and safe deletion

If $S$ is a vertex-minimal bad induced list subinstance, then
$|L(v)|\leq d_{H[S]}(v)$ for every $v\in S$, since otherwise a
coloring of $S-v$ extends greedily. Set
$h_v=d_{H[S]}(v)-|L(v)|\geq0$. If $S$ has $p$ clique-side vertices
and $r$ independent-side vertices in the template $K_a\vee I_b$, then

$$
B\geq c_{a,b}(p,r)+|D[S]|+\sum_{v\in S}h_v,
\qquad
c_{a,b}(p,r)=p(a-p)+p(b-r)+r(a-p).
\tag{A1}
$$

Indeed, the excess on $S$ pays for every surviving boundary edge and
every $h_v$, while the deletion budget pays for every missing boundary
edge and every internal deletion. These charges are distinct.

For the universal part of the theorem trim lists to
$\max\{1,a+3-q_u\}$ on $U$ and $\max\{1,a-q_i\}$ on $I$.
If $a\geq4$ and $B\leq2a$, every clique-side vertex can be precolored
without emptying another list. A clique singleton costs at least $a+2$;
an independent singleton costs at least $a-1$. Costs at two vertices
overlap only in their possible common deleted edge. There cannot be two
clique singletons, three independent singletons, or one clique singleton
and two independent singletons: their respective lower costs are
$2a+3$, $3a-3$, and $3a-2$, all exceeding $2a$.

Thus a list with at least three colors avoids all singleton names. A
clique list of size two has $q_u\geq a+1$ and cannot have either an
adjacent clique singleton or two adjacent independent singletons: these
would cost at least $2a+3$ or $3a-1$. A clique singleton cannot have an
adjacent singleton, whose combined cost is at least $2a+1$. At
$a=3,B\leq5$ the same argument works: there are at most two independent
singletons; a clique singleton costs five and cannot coexist with another
singleton; a clique list of size at most two costs at least four and
cannot have a singleton neighbor.

After safe precoloring and deletion of $u\in U$, subtract its color
from neighboring lists and keep the other excesses. The new budget is
$B-q_u$. Deleting a precolored independent vertex similarly gives
budget $B-q_i$ and changes $I_4$ to $I_3$; this deletion is safe whenever
every clique list has at least two colors.

### A.2. The terminal case

We prove the stronger terminal assertion that $q_u\leq1$ for all
$u\in U$ and $B\leq3a-4$ imply colorability. Suppose otherwise. The
Small Pot Lemma gives a bad assignment of the trimmed sizes using at
most $a+3$ colors. Add unused names to a palette $P$ of size $a+3$.
Let $d_U,d_\times$ be the internal-clique and cross deletion counts,
and let $J_U,J_I$ be the excess sums. Then

$$
\sum_{i\in I}|L(i)|\geq4a-d_\times-J_I
=4a-B+d_U+J_U\geq a+4>|P|.
\tag{A2}
$$

Two independent lists intersect, so $I$ can be colored with at most
three colors. Every clique list then retains at least $a-1$ colors.
If some $q_u=0$, that list retains at least $a$ and Hall's theorem
colors the clique. If $d_U>0$, its deleted edges form a matching.
For $a\geq3$, the residual clique graph is connected and its lists
have at least their degrees, with a strict surplus at every endpoint
of a deleted edge. The spanning-tree surplus argument colors it.
At $a=2$ the only disconnected case consists of two isolated vertices
with nonempty lists.

Therefore a bad terminal instance satisfies
$q_u=1$ for every $u$, $d_U=0$, $d_\times+J_U=a$, and $B=a+J_I$.
Hall failure after the preceding independent coloring forces every
clique list to be the same set $T$ of size $a+2$. Let $A\subseteq I$
be the vertices incident to a deleted cross edge. Each clique vertex
misses exactly one member of $A$ and has excess zero, or misses none
and has excess one. The following sufficient conditions follow from
Hall's theorem:

* An independent coloring using at most two colors of $T$ extends.
* If it uses three colors of $T$ and some $i\in A$ has a color in $T$ used
  by no other independent vertex, one clique list retains $a$ colors
  and the others retain at least $a-1$, so it extends.
* If it uses four distinct colors of $T$ and $|A|\geq2$, it extends.
  To see this, the residual lists contain a common $(a-2)$-set, possibly
  with the color of the unique missed independent vertex. A subfamily
  using two different extra colors has union of size $a$; one using
  only one type has size at most $a-1$ because another deletion type
  occurs elsewhere; one using none has size at most $J_U\leq a-2$.

We show that every bad terminal assignment instead satisfies

$$
\sum_{i\in I}|L(i)|\leq a+3+J_U,
\tag{A3}
$$

contradicting (A2), which gives at least $a+4+J_U$ here.

If a color outside $T$ belongs to $L(i)$, it is unique in $P\setminus T$
and cannot occur in another independent list, by the first Hall condition.
The other three lists are pairwise disjoint subsets of $T$ for the same
reason. The second condition forces $A\subseteq\{i\}$. Thus
$q_i\geq a-J_U$ and $|L(i)|\leq\max\{1,J_U\}$. Their total is at
most $a+2+\max\{1,J_U\}\leq a+3+J_U$.

Otherwise all four independent lists lie in $T$. They cannot be colored
with two colors. No color therefore occurs in three lists, and their
intersection graph has matching number at most one. Its edge set is
nonempty by (A2). It is a star, possibly with isolated vertices, or a
triangle and one isolated vertex. Each intersection edge allows its
endpoints to share a color and the other two vertices to have distinct
private colors; failure of this choice would itself give a two-color
assignment. The second Hall condition implies

$$
A\subseteq\bigcap_{xy\in E(J)}\{x,y\},
\tag{A4}
$$

where $J$ is the intersection graph. In the triangle case $A$ is empty,
$J_U=a$, and every color occurs at most twice, with at least one color
confined to the isolated list. The total is at most $2|T|-1=2a+3$.
In a star with at least two edges, $A$ lies in its center $h$, so
$|L(h)|\leq\max\{1,J_U\}$. Every repeated color belongs to this
list, giving total at most $|T|+|L(h)|\leq a+3+J_U$. This also handles
a single intersection edge when $|A|\leq1$. In the remaining case the
single edge has endpoints $x,y$ and $A=\{x,y\}$. Unless
$L(x)=L(y)$ is a singleton, four distinct independent colors can be
chosen, contradicting the third Hall condition. Equal singleton lists
give total at most $|T|+1=a+3$. This proves (A3) and the terminal lemma.

### A.3. The three small bases and induction

At $a=2,B\leq2$, let $r\leq2$ count the independent singleton
lists. Remove from each clique list the singleton colors at its neighbors,
leaving sets of sizes $x,y\geq3$. If the clique edge is present,
$x+y\geq8-r$. The number of distinct unordered pairs using one color
from each set is $xy-t-\binom t2$, where $t$ is their intersection
size. For $r=0,1,2$ its minimum under these bounds is respectively
$6,6,3$, exceeding the $4-r$ two-element independent lists that could
block a pair. Choose an unblocked pair. If the clique edge is absent,
a common color works, or disjoint residual lists give at least nine
pairs. The budget charges for these inequalities are the same as in the
two-center proof of the five-vertex theorem, with four leaves in place
of five.

At $a=3,B\leq5$, safely delete any clique vertex with $q_u\geq3$
and apply the previous base. Otherwise all clique lists have size at
least four. Delete any independent vertex with $q_i\geq2$ and apply
the triple theorem at budget three. Hence $q_i\leq1$. If all
$q_u\leq1$, apply the terminal lemma. Otherwise choose $q_u=2$ and
an independent neighbor $i$; color $u$ outside $L(i)$, since its list
has four colors and $|L(i)|\leq3$. After deletion the nominal budget
is at most three, and $i$ has one bonus color above its new bound. If
$j_i>0$, lower it by one and use the $a=2$ base. If $j_i=0$, a minimal
bad subset omits $i$. Formula (A1) shows the only nontrivial such proper
type with boundary at most three is $(p,r)=(2,3)$, whose boundary is
two. Its internal budget is at most one, within the triple theorem.

At $a=4,B\leq8$, delete any clique vertex with $q_u\geq3$ and apply
the $a=3$ base. Otherwise all clique lists have size at least five.
Delete any independent vertex with $q_i\geq3$ and use the triple
theorem at budget five. Thus $q_i\leq2$ and every independent list
has at least two colors. The all-$q_u\leq1$ case is terminal.
Otherwise choose $q_u=2$ and a neighboring $i$, and give $u$ a color
outside $L(i)$, possible because the sizes are five and at most four.
The remainder has $a=3$, budget at most six, and a bonus at $i$.
If $j_i>0$, decrease it and apply the $a=3$ base. Otherwise a minimal
bad subset omits $i$. Its possible types with boundary at most six
are $(3,3)$ and $(3,2)$, with boundaries three and six. The first
has internal budget at most three and the triple theorem applies.
The second has internal budget zero; pad one isolated independent
vertex, at deletion cost three, and use the same theorem.

For $a\geq5,B\leq2a$, safely delete a clique vertex with $q_u\geq2$
and apply induction. If none exists, use the terminal lemma because
$2a\leq3a-4$. This proves the universal bound.

For sharpness, with $D=\varnothing$ give every clique vertex a common
palette of size $a+3$, partition it into four nonempty independent lists
of sizes at most $a$, and put $j_i=a-|L(i)|$. The budget is $3a-3$;
four distinct independent colors leave too few colors for the clique.
Alternatively give an adjacent clique/independent pair equal singleton
lists. Their excesses $a+2$ and $a-1$ total $2a+1$. These examples
establish the claimed sharp threshold.

### A.4. Classification at budget $2a+1$

For $a\geq4$, formula (A1) with $b=4$ shows that a proper minimal
bad subset at budget at most $2a+1$ can have only the following types:

| $(p,r)$ | Template boundary | Remaining budget at most |
|---|---:|---:|
| $(1,1)$ | $2a+1$ | $0$ |
| $(a-1,3)$ | $2a+1$ | $0$ |
| $(a-1,4)$ | $a+3$ | $a-2$ |
| $(a,2)$ | $2a$ | $1$ |
| $(a,3)$ | $a$ | $a+1$ |

For completeness, at $r=0$ the boundary $p(a+4-p)$ has minimum
greater than $2a+1$ for $2\leq p\leq a$. At $r=1$ the same holds
for $p\geq2$. At $r=2$ the boundary is $2a+p(a-p)$, forcing $p=a$.
At $r=3$ its concavity excludes $1\leq p\leq a-2$. At $r=4$ put
$s=a-p$; properness gives $s\geq1$, and $s(a+4-s)$ excludes
$2\leq s\leq a-1$.

The second and fifth rows are colored by the triple theorem. For the
third row, add one isolated independent vertex at cost $a-1$; the
five-vertex theorem applies at total budget $2a-3$. For the fourth,
pad one isolated independent vertex at cost $a$ and use the triple
theorem, since $a+1\leq2a-3$. Thus the only proper bad subset is an
adjacent pair with equal singleton lists.

Suppose now $a\geq5$ and there is a bad instance with no such pair.
It is vertex-minimal bad. Replace $j_v$ by $d_H(v)-|L(v)|\geq0$,
decreasing the budget and preserving the actual lists. Their sizes are
exactly $a+3-q_u$ and $a-q_i$. There is at most one clique singleton,
since two would cost at least $2a+3$.

If $q_i\geq4$, precolor $i$ avoiding the possible clique singleton;
if $i$ itself is a singleton, the absence of an adjacent equal pair
makes this safe. The triple theorem colors the remainder at budget
$B-q_i\leq2a-3$. Hence $q_i\leq3$, and all independent lists have
at least two colors. A clique vertex with $q_u\geq3$ can now be
safely precolored, and the universal four-vertex theorem colors the
remainder at budget at most $2(a-1)$. Thus $q_u\leq2$.

The all-$q_u\leq1$ case is terminal since $2a+1\leq3a-4$.
Otherwise choose $q_u=2$, an independent neighbor $i$, and a color of
$u$ outside $L(i)$; its list has size $a+1$, whereas $|L(i)|\leq a$.
After deleting $u$, the budget is at most $2(a-1)+1$, and $i$ has a
bonus. If $j_i>0$, decrease it and use the universal theorem. If
$j_i=0$, its list has size $d_{H-u}(i)+1$, so any minimal bad subset
omits it. The proper-subset classification at parameter $a-1\geq4$
would force an adjacent equal-singleton pair. Every remaining clique
list, however, has size at least $a\geq5$. This contradiction proves
the classification and completes Theorem A.1.

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