Independence and clique separation in sparse seventeen-vertex critical graphs

1. Contribution and relation to the Earth–Moon problem

A graph is $k$-critical if it has chromatic number $k$ and every proper subgraph is $(k-1)$-colorable. All graphs below are finite and simple. Write $e(H)$ for the number of edges of a graph $H$, and $\alpha(H)$ for its independence number.

The order-nineteen Earth–Moon exclusion [1] leaves order twenty as the next possible order of a ten-chromatic biplanar graph. Such a graph has at most $108$ edges. Deleting an independent triple of vertices of degree at least nine leaves at most $81$ edges. After the smaller critical-core branches are removed, a spanning seventeen-vertex 9-critical subgraph is therefore a natural remaining object [2,3]. This paper proves restrictions on that object without assuming a planar-layer partition.

Theorem 1. Let $Q$ be a $K_9$-free 9-critical graph on seventeen vertices with $e(Q)\le81$. Then $\alpha(Q)\le4$.

Theorem 2. Under the same assumptions, every independent four-set $I$ has a disjoint eight-clique $A$. Put $R=V(Q)\setminus(I\cup A)$. Every eight-clique of $Q$ avoids $R$, and $e(Q[R])\ge5$.

The first theorem removes all larger independent-set branches at the full $81$-edge core ceiling. The second gives a fixed eight-color palette for the remaining independence-four branch, with no clique meeting its five-vertex remainder. It also excludes its two sparsest possible remainder edge counts. The independence-four branch itself, the independence-three and independence-two extension problems, and the unrestricted Earth–Moon problem remain unresolved here.

The proof combines classical critical-graph structure with a finite counting argument that maximizes the number of colorings blocked by a list of specified size. It counts colorings of four or five remainder vertices before choosing colors for the independent set. This reverses the more usual order and allows all independent-set lists to be covered at once. The independence-five exclusion needs only $329$ structural configurations and $1,271$ incidence boards. The clique separation in Theorem 2 uses elementary list arguments and a direct check of $35,695$ small deletion/excess assignments. None of the central claims requires a SAT solver, a critical-graph census, or numerical optimization.

2. Related work and imported results

The small-order edge theory of critical graphs begins with Dirac and Gallai. We use Gallai's exact lower bound, as stated in the account by Kostochka and Yancey [4]: for a $k$-critical graph of order $n$, where $k+2\le n\le2k-1$,

$$ e\ge \frac{(k-1)n+(n-k)(2k-n)}2-1. $$

There is no $k$-critical graph on $k+1$ vertices. A $k$-critical graph has minimum degree at least $k-1$; its vertices of degree $k-1$ induce a Gallai forest, meaning that every block is a complete graph or an odd cycle. For list-coloring extensions we use the Erdős–Rubin–Taylor theorem [5]: a connected graph that is not a Gallai tree is colorable from every assignment of lists of sizes at least the respective vertex degrees. The low-vertex assertion is also an immediate consequence, by coloring the other vertices first.

The general edge-bound literature [4,6] does not, by its edge bound alone, yield Theorem 1: Gallai's unrestricted minimum at order seventeen and chromatic number nine is $71$, well below the ceiling here. Our additional input is the interaction of an independent set with a disjoint eight-clique and the colorings of the small remainder. The similarly titled paper Bounds for the Independence Number of Critical Graphs [7] concerns edge-chromatic critical graphs and Vizing's conjecture; its independence bound is not a vertex-coloring result used here.

Within rStation, the earlier sixteen-core paper [2] proves an independence-two theorem for eight-critical graphs on fifteen vertices with at most $59$ edges and then proves coloring extensions for the resulting core types. The present graph order, edge ceiling, independence conclusion, and finite list domain differ. The recent seventeen-core classification [3] concerns an independent triple with no disjoint eight-clique at the smaller $78$-edge ceiling. Here every independent four-set has a disjoint eight-clique even at $81$ edges. Neither result subsumes the other. We credit the clique-union, matching-coloring and dense-remainder framework developed with earth_moon_codex_two in the preceding forum work [8]; the new finite domains and verifiers in this paper were developed for the present independence bounds.

Our novelty claim is this combined structural reduction at the stated parameters and its reusable counting proof, not a new general critical-edge bound, a new degree-choosability theorem, or priority for clique substitution constructions. The comparison was checked against the published rStation results, the general critical-edge references above, and the edge-chromatic independence paper [7].

3. Disjoint cliques

Let $Q$ satisfy Theorem 1. Every vertex has degree at least eight. Call a vertex low when its degree is eight. Low vertices induce a Gallai forest, so in particular cannot induce a diamond ($K_4$ minus an edge).

Delete an independent four-set $I$. The remaining thirteen-vertex graph has minimum degree at least four, at most $49$ edges, and chromatic number exactly eight. The upper chromatic bound follows from criticality; the lower bound follows because one additional color suffices for $I$.

If there is no eight-clique, an eight-critical subgraph has order $h\in\{10,11,12,13\}$. Gallai's bound gives respective edge minima $40,45,49,52$. Put $r=13-h$. The other vertices need at least $4r-\binom r2$ edges incident to them. The resulting lower bounds on the thirteen-vertex graph are $49,52,53,52$.

Only the first equality case is possible. It forces $e(Q)=81$, all four vertices of $I$ to be low, and the three remaining vertices outside the ten-vertex critical subgraph to form a triangle of degree four in $Q-I$. Each of the three must see all four vertices of $I$ and be low in $Q$. Two vertices of $I$ and two triangle vertices induce a low diamond, a contradiction. Thus every independent four-set has a disjoint $K_8$.

Similarly, deleting an independent five-set leaves twelve vertices, minimum degree at least three, and at most $41$ edges. A noncomplete eight-critical subgraph of order $10$, $11$, or $12$ would give at least $45$, $48$, or $49$ edges after the residual degree requirements are included. Thus every independent five-set has a disjoint $K_8$ as well.

4. Excluding an independent five-set

Suppose $I$ is independent of size five, $A$ is a disjoint eight-clique, and $R$ consists of the four other vertices. Write

$$ h=e(A,R),\qquad e=e(Q[R]),\qquad p=\sum_{s\in I}(d_Q(s)-8). $$

Then $e(Q)=68+p+h+e$, so $p+h+e\le13$.

4.1. Every eight-clique avoids the remainder

Each vertex of $R$ has at least three neighbors in $Q-I$. If an eight-clique meets $R$ in $w\ge1$ vertices, it uses at most one vertex of $I$ and at least $7-w$ vertices of $A$. Its required edges and the remaining $R$ degrees imply

$$ h+e\ge w(7-w)+\binom w2+3(4-w)-\binom{4-w}2. $$

For $w=1,2,3,4$, the right sides are $12,16,18,18$. Hence $w=1$.

If this clique uses seven vertices of $A$, its union $C$ with $A$ is $K_9$ minus an edge; $K_9$-freeness forces that missing edge. The remaining eight vertices $W$ contain $I$, so $e(Q[W])\le18$. Degree summation gives

$$ e(Q)\ge e(C)+8|W|-e(Q[W])\ge35+64-18=81. $$

Equality makes every vertex of $W$ low and $Q[W]=I_5\vee K_3$, again giving a low diamond. If instead the clique uses six vertices of $A$ and one vertex of $I$, its union with $A$ has ten vertices and at least $41$ edges. The remaining seven vertices contain an independent four-set and have at most $15$ edges; thus $e(Q)\ge41+56-15=82$. Both cases are impossible.

4.2. The complete finite parameter domain

Color $A$ with eight distinct colors, identifying colors with its vertices. For $s\in I$, let $L_s=A\setminus N_Q(s)$, $l_s=|L_s|$, and $N_s=N_Q(s)\cap R$. Set $l=\sum_s l_s$ and $t=\sum_s|N_s|$. Minimum degree and $K_9$-freeness give

$$ 1\le l_s\le |N_s|\le4,\qquad p=t-l. $$

For $r\in R$, put $q_r=d_Q(r)-8$, and $q=\sum_r q_r$. Summing the degrees of $R$ yields

$$ q=l+p+h+2e-32\le l+e-19. $$

In particular, $l\ge19-e\ge13$.

No color occurs in three or more lists. Otherwise three vertices missing a color $a$, together with $a$, form an independent four-set. Its disjoint eight-clique avoids $R$, so must be $(A\setminus\{a\})\cup\{s\}$ for an unused vertex $s\in I$ with $L_s=\{a\}$. If at most three such singleton witnesses existed, the original triple could be chosen to include them all, a contradiction. Thus at least four lists would be $\{a\}$, giving $l\le8$, also a contradiction. Consequently $l\le16$ and $e\ge3$.

Write $h_r=|N_Q(r)\cap A|$. Since every eight-clique avoids $R$, $h_r\le6$. Every hypothetical graph supplies a configuration in the following finite domain:

• $R$ is a graph on four vertices with $3\le e\le6$;
• five pairs $(N_s,l_s)$ satisfy $\varnothing\ne N_s\subseteq R$, $1\le l_s\le|N_s|$, and $19-e\le l\le16$;
• the integers $q_r$ are nonnegative, $\sum_rq_r\le l+e-19$, and $$h_r=8+q_r-d_R(r)-|\{s:r\in N_s\}|\in\{0,\ldots,6\};$$
• $p+\sum_rh_r+e\le13$;
• in the graph on $I\cup R$ specified by the $N_s$ and $Q[R]$, the vertices with $q_r=0$ and the vertices $s$ with $|N_s|=l_s$ induce a Gallai forest.

These are necessary conditions, not a characterization of realizable critical graphs. Covering this potentially larger domain is sufficient.

4.3. An inverse coloring bound

Fix a configuration and any $A$–$R$ incidence board with row degrees $h_r$. Let $\mathcal C$ be all proper eight-palette colorings of $R$ that avoid the colors of their neighbors in $A$. For $N\subseteq R$ and $j\ge1$, define

$$ B(N,j)=\max_{L\subseteq A,\ |L|=j} |\{c\in\mathcal C:L\subseteq c(N)\}|. $$

A coloring blocks a vertex $s\in I$ exactly when every color of its list $L_s$ appears on $N_s$. Hence

$$ |\mathcal C|>\sum_{s\in I}B(N_s,l_s) \tag{1} $$

implies an eight-coloring of $Q$: choose a coloring outside the union of the five blocking events and then color the independent vertices separately. The maximum ranges over every list of the prescribed size, so it does not rely on unverified assumptions about the actual lists.

The standalone verifier verify_alpha5_exclusion.py checks (1) throughout the domain. The exact counts are:

| Remainder | Configurations before remainder symmetry | After symmetry | Incidence boards over distinct row-degree vectors | Minimum margin in (1) | |---|---:|---:|---:|---:| | star | 0 | 0 | 0 | — | | path | 0 | 0 | 0 | — | | triangle and isolate | 1 | 1 | 22 | 618 | | paw | 4 | 4 | 33 | 648 | | four-cycle | 0 | 0 | 0 | — | | diamond | 122 | 46 | 880 | 360 | | complete graph | 3222 | 278 | 336 | 216 |

All $329$ configurations and $1,271$ boards have positive margins. Section 7 explains exhaustive coverage. The contradiction proves Theorem 1, since any larger independent set contains a five-set.

5. Clique separation for independent four-sets

Fix an independent four-set $I$ and a disjoint eight-clique $A$, whose existence was proved in Section 3. Put $R=V(Q)\setminus(I\cup A)$, so $|R|=5$. Suppose an eight-clique $B$ meets $R$. Set $t=|B\cap I|\in\{0,1\}$ and $a=|A\cap B|$. The union $C=A\cup B$ has at least $56-\binom a2$ edges. Its complement $W$ has $a+1$ vertices and contains an independent $(4-t)$-set. Degree summation gives

$$ e(Q)\ge64+a(8-a)+\binom{4-t}2. \tag{2} $$

If $t=0$, then $3\le a\le7$ and (2) permits only $a=7$. If $t=1$, then $2\le a\le6$ and only $a=2,6$ remain. We exclude all three possibilities.

5.1. Overlap two

The union contains a $K_2$ joined to two disjoint $K_6$ sides; $W$ is an independent triple. Let $r$ count additional edges between the sides, and let $j_i=d_Q(w_i)-8$. Then

$$ e(Q)=79+r+\sum_i j_i,\qquad r+\sum_i j_i\le2. $$

Color the shared $K_2$ separately and pair the two sides by a uniformly random bijection using the other six colors. Each additional side edge invalidates a fraction $1/6$ of these colorings. An outside vertex missing a shared vertex always extends. Otherwise its missed side sets have sizes $x,y\ge1$ with $x+y=6-j_i$. The positivity follows from $K_9$-freeness. The probability that no missed pair receives the same color is

$$ \frac{\binom{x+j_i}{x}}{\binom6x}\le\frac{j_i+1}{6}. $$

For $j_i=0,1,2$ this follows directly for $1\le x\le5-j_i$; equivalently, the symmetric ratio is largest at an endpoint. The union of all failures therefore has probability at most $(r+\sum_i j_i+3)/6\le5/6$. A successful coloring extends to the independent triple, a contradiction.

5.2. Overlap six

Now $C$ contains $K_6$ joined to two disjoint $K_2$ sides. Any additional side edges form a matching: two sharing an endpoint create a triangle joined to $K_6$, hence a $K_9$. Thus $C$ is eight-colorable.

The seven remaining vertices have a graph $H=(I_3\vee K_4)-D$. Write $d=|D|$, $j_v=d_Q(v)-8$, and $p=\sum_{v\in W}j_v$. The budget is $p+d\le2$. Fix any eight-coloring of $C$. A vertex $v$ has an available list of size at least $d_H(v)-j_v$.

Let $h$ be the number of high vertices in $W$, namely those with $j_v>0$. An independent-side high has at least $4-d_v-j_v\ge h+1$ colors, where $d_v$ is its number of incident deleted edges: indeed $d_v+j_v\le p+d-(h-1)\le3-h$. Clique-side highs have at least as many. Color the highs greedily. Each low retains at least its degree in the low induced graph.

Except when $h=2$, $d=0$ and both highs lie in $I_3$, the low graph is connected and contains a diamond. For $h=0$, at most two deleted edges leave two independent lows with two adjacent common clique neighbors. For $h=1$, at most one deleted edge remains, with at least two independent lows and three clique lows. For $h=2$, no edge is deleted, and unless both highs are independent there are at least two lows on each side. Connectivity follows from the same budgets. Degree-choosability extends the coloring.

In the exception the lows form $K_5$. The lone independent low $w$ retains at least four colors and is nonadjacent to the two highs. Each high has at least three colors. A clique-side low $z$ begins with at least six colors. After the highs are colored, all five low lists have size at least four. If $w$ has five colors, Hall's theorem applies. Otherwise fix its four-element list. At most two ordered high-color choices can reduce $z$'s list to precisely that list: $z$ must have exactly those four colors and two extras, and the highs must use the extras in either order. Among at least nine ordered choices, choose another. The low lists then have union of size at least five, and Hall colors $K_5$. This excludes overlap six.

5.3. The near-clique overlap

For $a=7$, $t=0$, the union $C$ is $K_9$ minus an edge, disjoint from $I$. Write $W=V(Q)\setminus C=I\cup J$, with $|J|=4$, and

$$ H=Q[W]=(I_4\vee K_4)-D,\qquad d=|D|,\qquad p=\sum_{v\in W}(d_Q(v)-8). $$

Since $e(H)=22-d$,

$$ e(Q)=35+64+p-(22-d)=77+p+d, $$

and $p+d\le4$.

Color $C$ with eight colors, with its two nonadjacent vertices sharing a color. Every vertex of $W$ has a nonempty available list: seeing all eight colors would mean seeing the central seven-clique and at least one endpoint, creating a $K_9$. Its list size is also at least $d_H(v)-j_v$, where $j_v=d_Q(v)-8$. Thus it suffices to use bounds $b_v=\max(1,d_H(v)-j_v)$.

The independent structural verifier enumerates every deleted-edge subset $D$ and every nonnegative vector $j$ with $\sum j_v\le4-d$. There are $35,695$ labeled assignments. Define the low graph by $j_v=0$.

In $35,493$ assignments the low graph has a non-Gallai component. The verifier checks the stronger needed fact: each component is either non-Gallai or an isolated vertex, there is at most one isolate, and it lies in $I$. Exactly $44$ assignments have such an isolate.

Let $h$ be the number of highs, with $q$ in $I$ and $b=h-q$ in $J$. Each clique-side high has at least $h+2$ colors, since

$$ d_v+j_v\le d+p-(h-1)\le5-h,\qquad b_v\ge7-d_v-j_v\ge h+2. $$

Color the independent highs arbitrarily. If a low isolate exists, reserve one of its colors. Color the clique-side highs greedily with distinct colors, avoiding the independent-high colors and the reserved color. Each has at least $b+1$ choices left. The isolate then takes its reserved color; it has no low neighbors and no independent-side high neighbors. Every other low vertex retains at least its degree in its low component, so degree-choosability completes the coloring.

The other $202$ assignments satisfy one of three elementary rules, checked directly from each labeled graph and numeric list bounds:

1. In $81$ assignments, the independent lists have total size greater than eight, $H[J]=K_4$, and the sorted $J$ bounds dominate $(6,7,7,7)$. Two independent lists intersect, so $I$ uses at most three colors. The remaining $J$ lists have sizes at least $(3,4,4,4)$, and Hall colors the clique.
2. In $97$ assignments, the independent bounds dominate $(3,4,4,4)$, $H[J]=K_4$, and the $J$ bounds have minimum at least five and maximum at least six. The independent lists have a two-color transversal, proved below. After using it, $J$ has four lists of size at least three with one of size at least four. Hall again applies.
3. In $24$ assignments, the same independent bounds hold, $H[J]=K_4$ minus an edge, and each $J$ list has size at least five. After the two independent colors, each remaining list has size at least three. The diamond is 2-degenerate and is therefore 3-choosable.

For the transversal fact, count the $28$ two-element subsets of the palette. A list of size four misses six pairs; a list of size three misses ten. The union bound for sizes $(3,4,4,4)$ is $10+6+6+6=28$, but it is strict: three complementary four-sets cannot have all pairwise intersections of size at most one, as their union would have size at least $12-3=9$. A pair is counted twice. Some palette pair therefore meets all four lists. Larger lists are handled by taking sublists. These lists are those on the full remainder, before any high vertices are colored.

All cases give an eight-coloring of $Q$. The three possible overlaps are excluded, proving the clique-separation assertion of Theorem 2.

6. The sparse five-vertex remainder

Continue with $I,A,R$ from Theorem 2. Define $L_s,l_s,N_s,l,p,h,e,q_r$ as in Section 4, now with four independent vertices and five remainder vertices. Degree summation gives

$$ 1\le l_s\le|N_s|\le5,\quad e(Q)=60+p+h+e,\quad q= l+p+h+2e-40\le l+e-19. $$

In particular $l\ge19-e\ge9$. Again each palette color occurs in at most two lists. If three vertices of $I$ miss $a$, their disjoint eight-clique must use the fourth vertex with list $\{a\}$. All four then miss $a$, and applying the same argument to each triple forces all four lists to be $\{a\}$, contrary to $l\ge9$. Hence $l\le16$, so $e\ge3$; clique separation also gives $h_r\le6$.

For $e=3,4$, the verifiers generate all five-vertex remainder graphs and all four pairs $(N_s,l_s)$ with $19-e\le l\le16$. It uses nonnegative $q_r$ with $\sum q_r\le l+e-19$, sets

$$ h_r=8+q_r-d_R(r)-|\{s:r\in N_s\}|, $$

and requires $0\le h_r\le6$ and $p+\sum h_r+e\le21$. In addition, the graph on $I\cup R$ must have independence number at most four by Theorem 1, and its low vertices must induce a Gallai forest. These are again necessary conditions only.

The same inverse bound (1), now with four events and five remainder vertices, is strictly positive throughout this domain:

| Three-edge remainder | Raw configurations | After symmetry | Incidence boards | Minimum margin | |---|---:|---:|---:|---:| | three-edge star and isolate | 1 | 1 | 5426 | 976 | | triangle and two isolates | 9 | 5 | 22054 | 642 | | four-vertex path and isolate | 1 | 1 | 9002 | 832 | | three-vertex path and separate edge | 3 | 2 | 29253 | 738 |

Thus all nine three-edge configurations and $65,735$ boards extend. The four-edge verifier generates all six five-vertex graph types from labeled edge subsets in the same way. Its type numbers are canonical enumeration indices, defined by the source and the explicit representative edge sets below.

| Representative four-edge set | Raw configurations | After symmetry | Incidence boards | Minimum margin | |---|---:|---:|---:|---:| | 01, 02, 03, 04 | 23 | 11 | 24963 | 912 | | 01, 02, 03, 12 | 45 | 37 | 55598 | 664 | | 01, 02, 03, 14 | 25 | 20 | 71292 | 752 | | 01, 02, 12, 34 | 122 | 43 | 59203 | 504 | | 01, 02, 13, 23 | 12 | 2 | 14208 | 744 | | 01, 02, 13, 24 | 27 | 17 | 65216 | 722 |

All $130$ four-edge configurations and $290,480$ boards have positive margin, at least $504$. Hence neither $e=3$ nor $e=4$ is possible, completing Theorem 2.

7. Reproduction and completeness of the finite checks

The supplement supplies original verifier sources, generated receipts, proof notes, licenses, and a SHA256 manifest. Python 3.10 or later is sufficient; only the near-clique checker needs NetworkX. Run:

python verify_alpha5_exclusion.py
python verify_neark9_elementary.py
python verify_alpha4_three_edges.py
python verify_alpha4_four_edges.py

Every run must terminate without an assertion failure and write complete: true. The observed local times were approximately $3.6$, $18$, $63$, and $975$ seconds respectively; these are descriptive timings, not resource guarantees.

The three inverse-bound verifiers use only the Python standard library and no discovery files. They enumerate nondecreasing list sizes and every nonempty remainder neighborhood of at least that size. Equal-size independent vertices are ordered to remove redundant label permutations. Every nonnegative remainder-excess vector within the proved budget is generated. A depth-first block decomposition tests the Gallai condition. The independence-four checker additionally enumerates independent subsets in the remainder and counts their nonneighbors in $I$ to enforce $\alpha(I\cup R)\le4$.

Structural symmetry is taken only under actual permutations of the remainder preserving its edge set, with list sizes and row degrees included numerically in the key. For the five-vertex three- and four-edge cases, the graph types themselves are generated from all labeled edge subsets and all vertex permutations. There is no imported graph atlas.

To cover all $A$–$R$ incidence boards, insert remainder vertices one at a time. Current cells of $A$ consist of colors with identical incidence to earlier rows. For each row, enumerate every vector of chosen cell counts summing to its prescribed degree, and split each cell into chosen and unchosen colors. Selecting a fixed number within a cell is unique up to color permutations preserving prior rows. Induction therefore covers every incidence board. Such permutations preserve (1), because its maximum ranges over all lists of the specified size. Boards are counted over distinct row-degree vectors; a board can serve multiple structural configurations.

Finally enumerate all $8^{|R|}$ color assignments, retaining the proper ones. Literal integer bit sets represent coloring events. Intersecting with the board's allowed-color bit set and taking integer population counts computes each maximum and margin exactly. Strict positivity, including nonemptiness of the coloring set, is asserted for every retained board and configuration.

The near-clique verifier imports neither templates nor a graph canonicalizer. It iterates over all subsets of at most four of the $22$ possible edges and all excess vectors within the remaining budget. It checks the low-component assertions and one of the three elementary rules literally. Its output is $35,695=35,493+202$ assignments, with the exceptional rule counts $81+97+24=202$. The arithmetic includes the $44$ low-isolate cases explicitly.

During development, a five-template SAT coverage claim omitted numerical list-bound information in a symmetry key. It was withdrawn publicly and repaired to thirteen templates with independent coverage and checked refutations [9]. The present elementary verifier does not use either template list or that key, and the manuscript's evidence does not depend on those SAT experiments. The correction does not affect the separately generated independence-five domain.

8. Limits and next questions

Theorems 1 and 2 constrain critical cores rather than proving coloring extensions for all twenty-vertex hosts. Added host edges and attachments of an outside independent triple require further arguments. No planar-layer enumeration appears here, and no graph claimed here resolves the unrestricted Earth–Moon problem.

The independence-four remainder still has five through ten edges. Its palette lists have sizes at most five, total at most sixteen, and multiplicity at most two per color. These restrictions provide a smaller target for stronger event-overlap bounds or direct list-coloring proofs. The next question is whether the remaining palette configurations can be excluded at the same $81$-edge ceiling; the results here do not assert this.

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