# Independence at most five at the order-twenty-one Earth–Moon frontier

## 1. Result 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 give a restriction at that frontier that requires only
criticality, the clique exclusion, and the Euler edge bound.

**Theorem 1.** Let $G$ be a $K_9$-free $10$-critical graph on $21$ vertices
with $e(G)\leq114$. Then
$$
3\leq\alpha(G)\leq5.
$$

Here a graph is $k$-critical if it has chromatic number $k$ and every proper
subgraph has smaller chromatic number. Thus every $10$-critical biplanar
graph on $21$ vertices has independence number three, four, or five. The
new assertion is the upper bound. The elementary lower bound is recorded
for clarity at the end of the proof.

The principal extension statement is stronger than needed for critical
graphs.

**Theorem 2.** Let $G$ be $K_9$-free on $21$ vertices, with $e(G)\leq114$.
Suppose $X$ induces $K_6\vee C_5$, the remaining ten vertices contain an
independent six-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 is by hand. An independent six-set in a putative critical graph
forces the core $K_6\vee C_5$ and a four-unit budget on the other ten
vertices. A counting lemma colors even a complete $K_4\vee I_6$ from the
resulting lists. Two independently written standard-library programs
check the small numerical tables; neither a graph census nor a SAT
refutation is a premise.

## 2. Related work and distinction

Gethner and Sulanke [3] developed nine-critical thickness-two constructions
and the independence-ratio program. Szeider [4] proved that every biplanar
graph with independence number at most two is nine-colorable. These
results address constructions or small independence number. The present
upper bound on independence number concerns a hypothetical ten-critical
graph at one fixed order.

The order-twenty theorem [2] settles all smaller orders. At order
twenty-one, earth_moon_codex_two [5] excludes nine-critical core orders
twelve through fifteen, bounds sixteen-vertex cores, and treats dense
remainders over $K_6\vee C_5$. Lunar_structure [6] treats prescribed
four-vertex extensions of spanning five-cycle clique blow-ups. Earlier
independence restrictions for sparse seventeen-vertex nine-critical
cores appear in [7]. None of those stated results is the global
independent-six-set exclusion of Theorem 1. Our core-order reduction does
not import the finite classifications or the core exclusions of [5]: a
larger independent set makes Gallai's numerical bound sufficient.

We use the classical critical-core and available-list framework present
in [5], [6], and [7]. The new step is the split-join list count in Section 3 and its
application to the exact four-unit budget in Section 4. In particular,
the theorem permits arbitrary attachments to the eleven-vertex core and
arbitrary missing edges in the remainder within that budget. No planar
embedding or separator assumption is used.

The imported critical-graph facts are Gallai's small-order decomposition
[8] and small-order edge bound, the latter also stated as Theorem 1 of
Kostochka and Yancey [9]. We use them in their following forms:

* A $k$-critical graph of order less than $2k-1$ has disconnected complement.
* For $k\geq4$ and $k+2\leq t\leq2k-1$, a $k$-critical graph of order $t$ has at least
  $$
  f_k(t)=\binom t2-(t-k)^2-1
  $$
  edges.

The second formula is equivalent to
$\tfrac12((k-1)t+(t-k)(2k-t))-1$ in [9]. The degree bound
$\delta(G)\geq k-1$, additivity of chromatic number under joins, and the
classification of three-critical graphs as odd cycles are also used.

## 3. A list-coloring lemma for a split join

Write $P$ for a palette of nine colors. Let $U$ be a clique on four vertices
and $I$ an independent set on six vertices, with every $U$–$I$ edge present.

**Lemma 3.** Give each $u\in U$ a list $L(u)\subseteq P$ and each
$i\in I$ a list $B_i\subseteq P$. Put
$$
a_u=9-|L(u)|,\qquad b_i=4-|B_i|,\qquad
A=\sum_{u\in U}a_u,\qquad B=\sum_{i\in I}b_i.
$$
If
$$
0\leq a_u\leq4,\qquad 0\leq b_i\leq3,\qquad
B\leq4,\qquad A+B\leq8,
\tag{1}
$$
then $K_4\vee I_6$ is colorable from the given lists.

*Proof.* Call a four-element subset $Q\subseteq P$ available if its colors
can be assigned bijectively to the four labeled vertices of $U$, respecting
their lists. Call $Q$ blocked if it contains at least one $B_i$. An
available, unblocked $Q$ colors $U$, after which each $i\in I$ can use a
color of $B_i\setminus Q$. We compare the numbers of available and blocked
four-sets.

There are $9\cdot8\cdot7\cdot6=3024$ injective assignments from $U$ to
$P$. Each forbidden vertex-color pair occurs in exactly
$8\cdot7\cdot6=336$ assignments. By the union bound at least
$3024-336A$ assignments respect the lists. A four-set is the image of at
most $4!=24$ assignments. Hence there are at least
$$
126-14A
\tag{2}
$$
available four-sets.

Only $A=5$ requires a stronger estimate. Order the deficits as
$a_1\geq a_2\geq a_3\geq a_4$. Greedy distinct-color choice in this
order gives at least $\prod_{j=1}^4(10-j-a_j)$ valid assignments. The
complete list of partitions of five into four parts, each at most four, is

| Deficits $(a_1,a_2,a_3,a_4)$ | Greedy lower bound |
|:---|---:|
| $(4,1,0,0)$ | $5\cdot7\cdot7\cdot6=1470$ |
| $(3,2,0,0)$ | $6\cdot6\cdot7\cdot6=1512$ |
| $(3,1,1,0)$ | $6\cdot7\cdot6\cdot6=1512$ |
| $(2,2,1,0)$ | $7\cdot6\cdot6\cdot6=1512$ |
| $(2,1,1,1)$ | $7\cdot7\cdot6\cdot5=1470$ |

Thus at least $\lceil1470/24\rceil=62$ four-sets are available when
$A=5$.

A list of size $\ell\leq4$ is contained in
$\binom{9-\ell}{4-\ell}$ four-sets. For deficits $b=0,1,2,3$, the
counts are respectively $1,6,21,56$. Six lists with total deficit at most
$B=0,1,2,3,4$ therefore block at most
$$
M(B)=6,\ 11,\ 26,\ 61,\ 66,
\tag{3}
$$
respectively. To check these maxima by hand, start with six zero deficits
and sum six. The successive gains from the first, second, and third deficit
units on one list are $5,15,35$. Through total three, concentrating the
units on one list maximizes the sum. For total four the positive deficit
partitions $(3,1)$, $(2,2)$, $(2,1,1)$, and $(1,1,1,1)$ give respectively
$66,46,36,26$.

The strict comparison is now

| $A$ | Largest allowed $B$ | Available four-sets, at least | Blocked four-sets, at most |
|---:|---:|---:|---:|
| $0,\ldots,4$ | 4 | 70 | 66 |
| 5 | 3 | 62 | 61 |
| 6 | 2 | 42 | 26 |
| 7 | 1 | 28 | 11 |
| 8 | 0 | 14 | 6 |

An available, unblocked set exists in every row, proving the lemma. $\square$

## 4. Extension over the eleven-vertex core

We prove Theorem 2. Write $R=V(G)\setminus X=U\mathbin{\dot\cup}I$,
where $|U|=4$ and $I$ is the given independent six-set. Let $D$ be the
edges missing from the split join $K_4\vee I_6$ on $R$, and set
$$
\Delta=|D|,\qquad x_v=d_G(v)-9,\qquad
s=\sum_{v\in R}x_v,\qquad q_v=x_v+d_D(v).
$$
The $x_v$ are nonnegative. Since $e(G[X])=50$ and $e(G[R])=30-\Delta$,
degree summation gives the exact identity
$$
e(G)=e(G[X])+\sum_{v\in R}d_G(v)-e(G[R])
    =110+s+\Delta.
\tag{4}
$$
Consequently
$$
s+\Delta\leq4,\qquad q_v\leq4,\qquad
\sum_{v\in R}q_v=s+2\Delta\leq8,\qquad
\sum_{i\in I}q_i\leq s+\Delta\leq4.
\tag{5}
$$
The last inequality holds because each edge of $D$ has at most one endpoint
in $I$.

Fix any proper nine-coloring of $G[X]=K_6\vee C_5$. For each outside
vertex $v$, let $L(v)$ be the colors absent from its neighbors in $X$.
Every $L(v)$ is nonempty. Otherwise $v$ would see all six singleton clique
colors and all three cycle colors, hence at least three cycle vertices.
Two of those vertices are adjacent, and with $v$ and the $K_6$ they form
a forbidden $K_9$.

The degrees in the complete split join are nine on $U$ and four on $I$.
Thus $v$ has $q_v$ core neighbors when $v\in U$ and $5+q_v$ core
neighbors when $v\in I$. It follows that
$$
|L(u)|\geq9-q_u\geq5\quad(u\in U),\qquad
|L(i)|\geq\max\{1,4-q_i\}\quad(i\in I).
\tag{6}
$$

For each $i\in I$ retain any sublist $B_i\subseteq L(i)$ of size
$\min\{4,|L(i)|\}$. Set $a_u=9-|L(u)|$ and $b_i=4-|B_i|$.
Equations (5)–(6) give all hypotheses (1), since $a_u\leq q_u$ and
$b_i\leq q_i$. Lemma 3 colors the supergraph $K_4\vee I_6$, so it also
colors $G[R]$. Its colors avoid the already colored core by definition of
the lists. This proves Theorem 2. $\square$

## 5. Excluding an independent six-set

We prove Theorem 1. Suppose $I$ is an independent six-set of $G$.
Criticality gives $\chi(G-I)\leq9$, and an eight-coloring would extend
with one new color on $I$. Hence $\chi(G-I)=9$. Choose a nine-critical
subgraph $F\subseteq G-I$ and write $t=|V(F)|$.

The case $t=9$ would give $K_9$. There is no nine-critical graph on ten
vertices. One way to see this, also useful at order eleven, is to split a
critical graph into the connected components of its complement. The
induced factors join completely and are themselves critical. Every
nonsingleton factor has chromatic number $k_j\geq3$; the only one- and
two-critical graphs are $K_1$ and $K_2$, and the latter has disconnected
complement. Gallai's decomposition theorem applied to the factor yields
order $n_j\geq2k_j-1$. Thus each nonsingleton factor has surplus
$n_j-k_j\geq k_j-1\geq2$, while a singleton has surplus zero. Total
surplus one is impossible.

We have $11\leq t\leq15$. The remainder $R=V(G)\setminus V(F)$ contains
$I$, and every vertex of $G$ has degree at least nine. Gallai's edge bound
therefore gives
$$
\begin{aligned}
e(G)
&=e(G[V(F)])+\sum_{v\in R}d_G(v)-e(G[R])\\
&\geq f_9(t)+9(21-t)-\binom{21-t}{2}+\binom62.
\end{aligned}
\tag{7}
$$

| Core order $t$ | $f_9(t)$ | Lower bound (7) on $e(G)$ |
|---:|---:|---:|
| 11 | 50 | 110 |
| 12 | 56 | 116 |
| 13 | 61 | 120 |
| 14 | 65 | 122 |
| 15 | 68 | 122 |

The cap $114$ forces $t=11$. In the complement-component decomposition
of $F$, total surplus $11-9=2$ forces exactly one nonsingleton factor,
with $k_j=3$ and $n_j=5$. It is $C_5$; the other six factors are
singletons. Hence $F\cong K_6\vee C_5$.

Moreover, $F$ is induced in $G$. Any added edge on its vertex set is a
chord of the five-cycle. The chord creates a triangle which, together
with $K_6$, gives $K_9$. Theorem 2 now extends a nine-coloring of $F$ to
$G$, contradicting ten-criticality. Thus $\alpha(G)\leq5$.

For the lower bound, if $\alpha(G)\leq2$, then $\overline G$ is
triangle-free. Every neighborhood in $\overline G$ is independent there
and hence a clique in $G$. The $K_9$ exclusion gives
$\Delta(\overline G)\leq8$, so
$$
e(G)\geq\binom{21}{2}-4\cdot21=126>114.
$$
This proves $3\leq\alpha(G)\leq5$. $\square$

## 6. Verification, scope, and next boundary

The proof above is complete without computation. The supplement supplies
two independent arithmetic checks:

```sh
python check_alpha_six_list_lemma.py
python check_alpha_five_audit.py
```

The first checks all 355 labeled clique-deficit vectors, all 204 labeled
independent-side vectors, and the 21,197 compatible pairs, as well as the
injection multiplicities and the core-edge table. The smallest strict
counting margin is one. The second checks the deficit tables by a separate
product calculation and verifies core uniqueness by enumerating the 156
unlabeled graphs of maximum degree at most two on eleven vertices as
unions of paths and cycles. It evaluates their clique partition numbers
and independence numbers directly. The sole retained complement is a
five-cycle and six isolated vertices. This independently checks the
small core conclusion using $\delta(F)\geq8$ rather than trusting the
factor enumeration in the first program.

Both programs use only the Python standard library and finish in under a
second in the recorded CPU run. They verify the stated small calculations,
not Gallai's imported theorems, all list assignments, or all graphs on
twenty-one vertices. The manuscript and original supplemental code are
released under CC BY 4.0 and the MIT License, respectively.

The result leaves independence numbers three, four, and five at order
twenty-one. It does not exclude every eleven-vertex core attachment, prove
that a surviving graph is biplanar, improve the known order lower bound,
or resolve the unrestricted Earth–Moon problem. For an independent
five-set, the same degree count permits an eleven-vertex core with
$G[R]\subseteq K_5\vee I_5$ and $s+\Delta\leq9$; coloring under this
larger budget is a distinct remaining extension problem.

## References

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