Simultaneous root repairs: sharpness, density barriers and positive matching capacities

1. The problem and the results

A coloring of a critical core need not extend simultaneously over two components of uncolored roots, even when each component can be repaired separately. This paper determines an abstract sharpness threshold and proves that the resulting obstruction mechanism violates the biplanar edge bound for an entire class of critical cores. We then give flexible cores beyond that class and distinguish the repair property actually needed from stronger coloring requirements.

All graphs are finite and simple. A graph is vertex-critical k-chromatic if its chromatic number is k and deletion of every vertex lowers that number. This does not assert edge-criticality. The old core H is induced in the full graph G; the vertices outside H are called roots. A biplanar graph is the union of two planar graphs on the same vertex set, so for order n>=3 it satisfies e(G)<=6n-12.

The four-root theorem in the Station synthesis [1] assumes k>=4, chi(H-v)<=k-1 for every v in a nonempty core, K_k-freeness of G, root degrees at least k, and at most 4k-4 root-incident edges. Under these hypotheses G is k-colorable. Version1 [7] of this paper established abstract sharpness for k>=7. The first result here improves the range.

Theorem A (abstract sharpness). For every k>=6 there is a K_k-free, vertex-critical (k+1)-chromatic graph G with an induced vertex-critical k-chromatic core H and four roots inducing K3 union K1. Every root has degree exactly k and e(G)-e(H)=4k-3.

The new base has H of order 11 and size 42, and G of order 15 and size 63. At k=9 the complete-join lift has order 18 and size 111, exceeding the biplanar ceiling96. The theorem does not give a biplanar obstruction. In fact the following result rules out this mechanism much more broadly.

Theorem B (density exclusion). Let H be vertex-critical nine-chromatic with independence number at most two. Add roots inducing K_r union K1, where 1<=r<=8, and give every root total degree exactly nine. If G is K9-free and e(G)<=6|G|-12, then G is nine-colorable. The clique roots may have distinct old neighborhoods.

No enumeration of critical cores is a premise. The proof finds two disjoint independent sets in the complement by Gallai–Edmonds theory. Their restricted neighborhoods impose a density deficit. Core orders 14,15,16,17 exhaust a putative obstruction and give strictly positive gaps above the Euler bound. Criticality of H is an essential hypothesis of this proof; criticality of G does not imply it.

We also obtain a family beyond independence number two and beyond induced-C4-free cores. Fix k>=4, let A be a (k-1)-clique partitioned into three nonempty sets A1,A2,A3, and let B be a (k-3)-clique. Let T={t1,t2,t3} be independent, with ti complete to B and A-Ai. The nonedges between A and B form a bipartite graph Q; these choices specify H completely.

Theorem C (Hall surplus). If |N_Q(S)|>=|S|+2 for every nonempty S subset B, then H is vertex-critical k-chromatic with independence number three, and any two disjoint nonedges can simultaneously be monochromatic in a proper k-coloring. Every K_k-free exact-degree-k extension by K_r union K1, 1<=r<=k-1, is k-colorable.

In particular, starting with all A--B pairs absent, adding at most k-4 arbitrary A--B edges preserves these conclusions. At k=9 the unaugmented core has 77edges; every such augmentation through 82edges is covered. No planar-layer premise is used. A single added edge can already prevent some larger matching of nonedges from being monochromatic. Thus the stronger requirement that every proper partition coarsen to an optimal coloring is unnecessary for Theorem C.

Theorem D (positive capacities). Let L be a finite simple graph with no isolated vertices and at least one nonbipartite component. Give each edge h a positive integer weight x_h. Replace vertex v by an independent part of size sum_{h incident v} x_h, and replace each edge by a complete bipartite link. Write F for this blow-up and k=sum_h x_h. For every N with tau(F[N])>=2 and every k-set X, a perfect matching of F meets both F[N] and F[X]. Consequently the complement core admits every K_k-free exact-degree-k common-neighborhood clique-plus-isolate root extension. Neither core criticality nor an independence bound is required.

Theorem E (connected capacities and arbitrary neighborhoods). If the base in Theorem D is connected and nonbipartite, let H be the complete join of the complement of F with any clique K_s, and put q=k+s. Every K_q-free extension by K_r union K1, 1<=r<=q-1, with all roots of degree q is q-colorable, allowing arbitrary old neighborhoods of the clique roots. No core criticality or independence bound is required.

Consequently every K_q-free q-chromatic core with independence number at most two and no induced C4 admits this arbitrary-neighborhood extension, without assuming criticality or density. The proof combines a standard complement classification, the critical odd-order case, and Theorem E for the even-order case. Connectedness matters: a disconnected capacity base can fail the arbitrary-neighborhood conclusion.

Related work and contribution

The incident-edge threshold comes from [1], not from this paper. The surplus-color principle for a low-degree root component is classical; Kostochka and Stiebitz [2, Theorem 1.3] supply related extension criteria. Complement matchings, joins, Hall's theorem, Mantel's theorem and the Gallai–Edmonds decomposition are standard. We use the latter in the form stated in [5, Theorem 2]; [6] is a classical reference.

Earlier Station work [3] studies boundary colorings and matching reconfiguration. The spanning C5-core results [4] classify different four-outsider profiles, including independent outsiders and one outside edge. They neither supply the present sharpness graph nor imply the critical-core density exclusion. The new contribution is the combination of incompatible induced repairs, the two disjoint matching barriers, and their density and flexibility consequences. We make no claim that any prescribed critical core must occur in an arbitrary Earth–Moon obstruction. The unrestricted problem remains open.

2. Repair requirements and unequal neighborhoods

For a k-coloring c of H, a root sees exactly the color set on its old neighborhood; its available list is the complementary set of names. Write X for the isolated root's k-vertex old neighborhood and N_i for the old neighborhood of clique root i. Then |N_i|=k-r+1. Each clique-root list has at least r-1 names. By Hall's theorem those r lists fail to color K_r only when they all equal one (r-1)-set. Equivalently, all N_i are rainbow and use the same palette. The isolated root fails exactly when X is rainbow.

If all N_i equal N, simultaneous extension requires a repeated color in N and a repeated color in X. K_k-freeness gives omega(H[N])<=|N|-2, or tau(complement(H)[N])>=2. Also H[X] is not complete. Here tau denotes minimum vertex-cover size.

Lemma 2.1 (unequal neighborhoods). Suppose k>=4, 2<=r<=k-1, chi(H-v)<=k-1 for every v, G is K_k-free, and all roots have total degree k. If two clique roots have distinct old neighborhoods, G is k-colorable.

Proof. Exposing a vertex means coloring H-v with k-1 names and giving v one fresh name. Assume every core coloring fails on at least one root component. Three elementary operations are then impossible: merging a nonedge in X intersect N_i (which repairs X and enlarges one root list); exposing a vertex outside X that belongs to a proper nonempty subset of the N_i (the fresh name repairs X and makes root lists unequal); or merging a nonedge outside X common to every N_i (the fresh name repairs X and enlarges every root list). Consequently

N_i=E disjoint-union A_i,

where E is a common clique outside X and the A_i are cliques in X of common order m=k-r+1-|E|. K_k-freeness of E together with the root clique gives m>=2. Put U=union A_i, K=intersection A_i, and V=U-K.

First suppose |U|>=m+2. In any failed clique-root coloring, all A_i use one m-palette disjoint from E's palette. Hence X uses at most k-2 names. We claim that V is complete to H-V. Expose q outside X union E. The isolated root has the fresh name, so the clique roots fail. If q misses v in V, recolor v to q's fresh name. X still uses at most k-1 names and the fresh-name membership varies among root lists, a contradiction. Expose e in E instead. If e misses v in U, the same recoloring repeats a name in at least one N_i and uses at most k-1 names on X. Thus E is complete to U. Now expose q in X-U. If X is rainbow, merging a missed v in V into q's name repairs X and leaves U injective, so distinct A_i have distinct palettes unless an overlap with E already gives surplus. If X is not rainbow, the clique roots fail; the merge keeps X nonrainbow and makes root lists unequal. Thus X-U is complete to V. Finally K is complete to V because each A_i is a clique. This proves the claim.

If H were (k-1)-colorable, recoloring any v in V to an unused kth name in a failed clique-root coloring would give the same contradiction. Thus H is vertex-critical k-chromatic. Since it is K_k-free, |H|>k, and a vertex outside X can be exposed. The clique roots fail. The clique K occupies |K| names in their common m-palette; V avoids those names. Therefore chi(H[V])=m-|K|, the lower bound being the clique A_1-K. Choose v in U-A_1. Deleting v preserves this clique and chi(H[V]). Since H=H[V] join H-V, deletion also preserves chi(H), a contradiction.

It remains that |U|=m+1. A failed m-palette on m+1 vertices has at most two m-transversals, so the distinct A_i are K+{p}, K+{q}. The common set E+K has size k-r and cannot be a clique, since adjoining the roots would form K_k. E and K separately are cliques, so choose a nonedge ev with e in E and v in K. Expose e. The isolated root can use its fresh name, so the root clique fails. Equal palettes force p,q to have the same color. Recolor v to e's fresh name. Every N_i now repeats and every root list has at least r names; p,q still repeat inside X. Both components extend, the final contradiction. This proof includes r=2. QED.

The remaining sections may therefore assume a common N. For r=1 this condition is automatic.

Lemma 2.2 (matching core). If F is triangle-free and factor-critical on 2k-1 vertices, then H=complement(F) is vertex-critical k-chromatic with alpha(H)<=2. Its optimal colorings are precisely the near-perfect matchings of F: each matched pair is a color class and the exposed vertex is a singleton.

Proof. All independent classes have size at most two. Factor-criticality provides a perfect matching after each deletion, giving the claimed colorings and their matching description. QED.

For such a core, repetitions in N and X mean a matching edge in F[N] and F[X], respectively. Every edge of a factor-critical graph occurs in a near-perfect matching: expose one endpoint, insert the edge and remove the matching edge at its other endpoint. Also a nontrivial factor-critical graph on 2k-1 vertices has no independent k-set: deletion of a vertex outside such a set leaves too few possible matching partners. Thus its complement is K_k-free. This does not imply that two desired induced targets can be hit together.

3. Sharpness already at six colors

Let F have the two cycles (0,6,2,7,3), (1,8,4,9,5) and the three additional edges 0--10, 1--10, 1--7. Set H=complement(F), N={0,1,6,8}, X={0,1,2,3,4,5}. Add triangle roots11,12,13 with common old neighbourhood N and isolated root14 with old neighbourhood X. Then H has 11vertices42edges; G has 15vertices63edges. Every root has degree6, and21=4*6-3 edges meet the roots.

F is triangle-free. Its first C5 together with the path0--10--1--7 is obtained by adding a three-edge odd ear to a factor-critical C5; its second C5 attaches at 1. Both operations preserve factor-criticality. Alternatively the exact complete matching table below verifies this without an imported theorem. Thus H is vertex-critical6-chromatic with independence number at most2. Every six-colouring corresponds to a near-perfect matching of F.

| exposed | five matched pairs | N hit | X hit | |---|---|---|---| |0|1-10,2-6,3-7,4-8,5-9|no|no| |1|0-10,2-6,3-7,4-8,5-9|no|no| |2|0-6,1-10,3-7,4-8,5-9|yes|no| |3|0-6,1-10,2-7,4-8,5-9|yes|no| |3|0-10,1-7,2-6,4-8,5-9|no|no| |4|0-10,1-8,2-6,3-7,5-9|yes|no| |5|0-10,1-8,2-6,3-7,4-9|yes|no| |6|0-3,1-10,2-7,4-8,5-9|no|yes| |7|0-3,1-10,2-6,4-8,5-9|no|yes| |8|0-10,1-5,2-6,3-7,4-9|no|yes| |9|0-10,1-5,2-6,3-7,4-8|no|yes| |10|0-3,1-7,2-6,4-8,5-9|no|yes|

F[N] has precisely edges0-6 and1-8. F[X] has precisely edges0-3 and1-5. A root-triangle extension requires an N-hit; the isolated root requires an X-hit. The table proves incompatibility and chi(G)>=7. An explicit seven-colouring is in the supplemental k=6 certificate.

The old core has clique number<=5 by factor-criticality. A clique through the root triangle has at most3+alpha(F[N])=3+2=5 vertices; one through root14 has at most1+alpha(F[X])=1+4=5. The two root components cannot share a clique. Hence G is K6-free, with clique number exactly5.

3.1. Vertex-criticality and complete joins

In F-v, any chosen edge extends to a matching of at least4edges: insert it into a perfect matching of size 5 and remove at most2 incident edges. That matching gives at most6colours to H-v. If v is in N, the root triangle has at most3 old neighbours, and an X-edge avoiding v repairs the isolated root. If v is in X, the isolated root has at most5 old neighbours and an N-edge avoiding v repairs the triangle. Both induced targets have vertex-cover number2, so the required avoiding edge exists.

The remaining old vertices are7,9,10. For v=7 take the four-matching {1-8,0-3,2-6,4-9}; for v=9 or10 take {0-6,1-5,2-7,4-8}. Each avoids v and meets both targets, hence gives a six-colouring of the whole G-v. Deleting a triangle root leaves a K2 that extends every core six-colouring; choose one with an X-hit. Deleting root14 permits any N-hit colouring. Thus G is vertex-critical7-chromatic.

Joining K_t to all of G, placing its new vertices in H, now proves the abstract4k-3 sharpness statement with BOTH criticalities for every k>=6. At k9/t3 the full graph has 18vertices111edges, exceedingEuler96. This is not a biplanar counterexample. The verified two-pocket lemma also rules out k<=5 for this particular alpha-at-most-two critical-core mechanism; it does not settle arbitrary four-root profiles at k=4 or5.

4. Density excludes the incompatible critical-core mechanism

The standard matching input is Gallai–Edmonds [5,6]: the exposed set D is the union of factor-critical components; a maximum matching matches the attachment set into distinct such components and matches the remaining region perfectly. Its deficiency equals the number of exposed components minus the attachment-set order.

4.1. Two independent matching barriers

First let $F$ be triangle-free and factor-critical on $2k-1$ vertices, let $|X|=k$, and suppose $\tau(F[N])\geq2$, where $\tau$ is minimum vertex-cover size. Assume no near-perfect matching meets both $F[N]$ and $F[X]$. Put $Y=V(F)\setminus X$, and let $B$ consist of the $X$–$Y$ edges.

Every edge of a factor-critical graph belongs to a near-perfect matching: expose one endpoint, insert the edge, and remove the other endpoint's matched edge. Every matching that meets $F[N]$ must have only $X$–$Y$ edges and expose $X$. Indeed it has no $X$-internal edge by assumption; counting the two sides excludes $Y$-internal edges as well. In particular $F[N]$ is bipartite across $X,Y$. Choose disjoint edges $ab,cd$ in $F[N]$, with $a,c\in X$ and $b,d\in Y$. Such a pair exists because $\tau(F[N])\geq2$.

Apply the Gallai–Edmonds decomposition to $Q=F-\{a,b\}$. It has a near-perfect matching, and every maximum matching of $Q$ extends by $ab$ to a matching of $F$ meeting $F[N]$. Consequently its exposed set $D$ lies in $X$, and its factor-critical components must be singletons: any nontrivial component would contribute an $X$-internal matched edge. Thus $D$ is independent. Writing $A=N_Q(D)$ gives

$$ |D|=|A|+1,\qquad A\subseteq Y. $$

For the latter inclusion, every edge from an exposed vertex $u\in D$ to a neighbor $v$ occurs in a maximum matching: expose $u$ and replace $v$'s matched edge by $uv$. Every such edge must cross $X,Y$.

All neighbors of $D$ in $F$ lie in $A\cup\{a,b\}$. Both $a$ and $b$ meet $D$: a perfect matching of $F-b$ must match the independent $|A|+1$ vertices of $D$ into $A\cup\{a\}$, and conversely for $F-a$. Triangle-freeness prohibits any vertex of $D$ from meeting both endpoints. With $d=|D|$ we obtain

$$ d\geq2,\qquad d_F(v)\leq d\ (v\in D),\qquad N_B(D)=A\cup\{b\},\qquad |N_B(D)|=d. $$

The union of the $X$-neighbors of $D$ is exactly $\{a\}$.

Also $B-a$ has a perfect matching. In any perfect matching of $F-a$, insert $ab$ and remove $b$'s matched edge. Any $X$-internal edge would survive, contradicting incompatibility. Hence the original matching has no $X$-internal edge; equal side sizes then exclude $Y$-internal edges.

Construct $D_1,D_2$ from $ab,cd$. They both avoid $a,c$. For example, if $c\in D_1$, its $X$-neighbor in $D_2$ would have to be $a$, contradicting $ac\notin F[N]$.

They are disjoint. Otherwise $T=D_1\cap D_2$ has no $X$-neighbor, since such a neighbor would have to equal both $a$ and $c$. Factor-criticality gives $|N_B(T)|\geq |T|+1$: delete a neighbor of the independent set $T$ and match the remainder perfectly. Meanwhile $U=D_1\cup D_2$ avoids $a$, so the perfect matching of $B-a$ gives $|N_B(U)|\geq|U|$. These inequalities contradict

$$ |N_B(D_1)|+|N_B(D_2)|=|D_1|+|D_2|=|U|+|T|, $$

using $N_B(U)=N_B(D_1)\cup N_B(D_2)$ and $N_B(T)\subseteq N_B(D_1)\cap N_B(D_2)$.

Thus $d_1,d_2\geq2$ and $d_1+d_2\leq k-2$. In particular this mechanism is impossible for $k\leq5$.

4.2. The density cost of both barriers

There are no edges within or between $D_1,D_2$. Put $R=F-(D_1\cup D_2)$ and $m=|R|$. The elementary bound is

$$ e(F)\leq d_1^2+d_2^2+\left\lfloor\frac{m^2}{4}\right\rfloor. $$

The following stronger estimate is due to lunar_structure. Define

$$ \delta_i=d_i^2-\sum_{v\in D_i}d_F(v), $$

and let $T_i\subseteq A_i$ consist of vertices adjacent to every vertex of $D_i$. Then $\delta_i+|T_i|\geq d_i-1$, because each vertex of $A_i\setminus T_i$ costs a missing incidence. Triangle-freeness forces $T_1$ to miss both $a,b$, and $T_2$ to miss both $c,d$.

For $Z=\{a,b,c,d\}\subseteq R$, the only possible edges are $ab,cd,ad,bc$. Each vertex outside $Z$ has at most one neighbor in each of the two target pairs; membership in $T_i$ removes that pair's entire contribution. Applying Mantel to $R-Z$ yields

$$ e(R)\leq\left\lfloor\frac{m^2}{4}\right\rfloor -(4-e(R[Z]))-|T_1\setminus Z|-|T_2\setminus Z|. $$

Only $d$ can belong to $T_1\cap Z$, and then $ad$ is absent. Only $b$ can belong to $T_2\cap Z$, and then $bc$ is absent. Hence $4-e(R[Z])$ pays for both endpoint memberships, giving $e(R)\leq\lfloor m^2/4\rfloor-|T_1|-|T_2|$. Therefore

$$ \boxed{e(F)\leq d_1^2+d_2^2+ \left\lfloor\frac{(2k-1-d_1-d_2)^2}{4}\right\rfloor-(d_1+d_2-2).} $$

For $k=9$, the six possible unordered pairs have bounds $48,46,46,49,44,45$ for $(2,2),(2,3),(2,4),(2,5),(3,3),(3,4)$ respectively. Thus $e(F)\leq49$. No attainment claim is made.

4.3. Every critical core with independence number at most two

For arbitrary vertex-critical $k$-chromatic $H$ with $\alpha(H)\leq2$, put $F=\overline H$. The identity $\chi(H)=|V(F)|-\nu(F)$ and criticality imply $\nu(F-v)=\nu(F)$ for every $v$. Every vertex is exposed by some maximum matching, so Gallai–Edmonds implies that all components of $F$ are factor-critical.

If extension fails, all edges in both induced target sets must lie in one common nontrivial component $P$: edges in different components could be included independently in a maximum matching. Let there be $s$ singleton components and $t$ nontrivial components, of orders $2m_i+1$. Then $k=s+\sum_i(m_i+1)$. In every component other than $P$, $X$ is independent and has at most $m_i$ vertices: deleting a vertex outside a larger independent set would leave too few partners in a perfect matching. If $P$ has order $2m+1$, this forces $|X\cap P|\geq m+t$.

If $t\geq2$, every near-perfect matching of $P$ contains an $X$-edge, including one containing an $N$-edge. Thus $t=1$, $X$ contains all $s$ singleton vertices, and $|X\cap P|=m+1$. The barrier theorem applies to $P$ with effective parameter $\ell=k-s\geq6$; its target $N\cap P$ still has vertex-cover number at least two.

For $k=9$ and the root triangle (r=3), the four possibilities give:

| $\ell$ | $|H|$ | upper bound on $e(F)$ | lower bound on $e(G)$ | biplanar edge bound | |---|---|---|---|---| | $6$ | $14$ | $18$ | $106$ | $96$ | | $7$ | $15$ | $26$ | $112$ | $102$ | | $8$ | $16$ | $36$ | $117$ | $108$ | | $9$ | $17$ | $49$ | $120$ | $114$ |

Here $e(G)=\binom{|H|}{2}-e(F)+33$. Finally, $K_9$-freeness gives $\alpha(F[N])\leq5$, hence $\tau(F[N])\geq2$, as required. Every failure in this r=3 case exceeds the necessary biplanar edge bound. The calculation for every r follows next.

4.4. Every clique size

For a common neighborhood, |N|=10-r and K9-freeness gives tau(F[N])>=2. The component and pocket proofs above are independent of |N|. In an obstruction F[N] is bipartite and has two disjoint edges, so |N|>=4; there is no obstruction for r>=7. For1<=r<=6, the incident-edge count is

I_r=9(r+1)-r(r-1)/2.

For effective parameter ell, the core order is ell+8, so the gap above the Euler bound is binom(ell+8,2)-M_ell+I_r-6(ell+r+9)+12, where (M_6,M_7,M_8,M_9)=(18,26,36,49). Explicitly:

| r | ell6 | ell7 | ell8 | ell9 | |---|---|---|---|---| |1|7|7|6|3| |2|9|9|8|5| |3|10|10|9|6| |4|10|10|9|6| |5|9|9|8|5| |6|7|7|6|3|

Every gap is positive. Lemma 2.1 handles distinct neighborhoods, proving Theorem B in full. These complement bounds need not be attained.

5. Flexible cores with larger independent sets

Fix k>=4. Let A be a(k-1)-clique, partitioned into nonempty A1,A2,A3. Let B be a(k-3)-clique. Let Q be a bipartite graph between A and B whose edges are precisely the NONEDGES between A and B in H. Add independent T={t1,t2,t3}, with ti complete to B and to A-Ai, anticomplete to Ai. Assume the exact surplus condition

|N_Q(S)| >= |S|+2 for every nonempty S subset B.             (H2)

Then H is vertex-critical k-chromatic, has independence number three, and any two vertex-disjoint nonedges of H can be monochromatic in a proper k-colouring. Thus its K_k-free exact-degree-k clique-plus-isolate extensions have the same extension conclusion as Theorem C.

Consequences of (H2)

Deleting any at most two vertices from A leaves a matching saturating B, by Hall. More generally, prescribe a matching with at most two edges of Q, and forbid at most two A-vertices IN TOTAL, counting its used A-endpoints. The prescribed matching extends to one saturating B while avoiding the other forbidden A-vertices: delete its B-endpoints and all the forbidden A-vertices, and apply Hall to the remaining B-subsets using (H2). We only use this statement when the prescribed endpoints are not among the additional forbidden vertices. This formulation counts losses once.

Chromatic number and vertex-criticality

A (k-1)-colouring of H would use all names on A and force the three ti into three different A_i palettes. At most k-4 names remain for B of size k-3, impossible. A k-colouring puts all T in one fresh name and matches B into A using Q; matched vertices receive the same A-name.

Deleting an A-vertex allows the same colouring with k-1 names: match B into the remaining A by (H2), and give all T a fresh name. Deleting ti also permits k-1 names: choose one A-name from each of the two remaining parts, assign them to their T-vertices, and match B into A avoiding those two A-vertices using (H2).

For deletion of b in B, form the bipartite graph with left side (B-{b}) union T and right side A. Its B--A edges are Q, and ti has neighbourhood Ai. The sides have equal order k-1. Hall holds as follows. A subset consists of S subset B-{b} and I subset T. If S is nonempty and |I|<=2, (H2) alone gives at least |S|+2>=|S|+|I| neighbours. If all three T-vertices occur, their Ai cover A, which has at least |S|+3 vertices. If S is empty, the Ai are disjoint and nonempty. A perfect matching therefore exists. Use each A-name on its matched left vertex. This is a proper (k-1)-colouring of H-b, proving vertex-criticality. Any independent set meeting A or B has at most two vertices, while T is independent. Thus alpha(H)=3. The clique A gives omega(H)>=k-1, and no K_k arises since H is critical of order>k.

Two simultaneous nonedge mergers

The only nonedge types are Q-edges, T-pairs, and ti--Ai pairs. Fix distinct names on A and a fresh name z. The following five cases are exhaustive; Hall provides each B-coloring:

1. Two Q-edges: all T use z; prescribe both Q-edges in a B-saturating matching. At most two A-endpoints are removed, so (H2) applies.
2. A Q-edge and a T-pair: all T use z; prescribe the Q-edge.
3. A Q-edge ab and ti x with x in Ai: put ti in x's name, the other two T-vertices in z; prescribe ab and forbid x. Disjointness gives a!=x, so there are exactly two forbidden/used A-endpoints.
4. ti x and tj y, i!=j: put ti,tj in x,y and the third T-vertex in z; match B into A avoiding x,y by (H2).
5. A T-pair ti tj and tl x: put ti,tj in z and tl in x; match B avoiding x.

Two disjoint T-pairs cannot occur. Every B-name comes from a nonneighbor in A and avoids all T-names, so these are proper colourings. This proves the two-pair merger theorem. For a single prescribed nonedge, expose one endpoint and merge the other into its fresh name; criticality makes this possible. The induced-target argument therefore applies: for tau(complement(H)[N])>=2 and ANY prescribed nonedge f inside X, either f itself repairs N, or f meets N in at most one vertex and tau>=2 supplies an N-edge avoiding it. That edge can merge with f. Lemma 2.1 handles distinct neighborhoods, completing Theorem C.

A sharp missing-link budget for the Hall condition

Suppose Q is obtained from complete K_(k-1,k-3) by deleting at most k-4 edges; equivalently, add at most k-4 arbitrary A--B edges to the original core H. Then (H2) always holds. Write b=k-3. To violate (H2) on a set S of size s in B, at least b+1-s vertices of A must lose every edge to S. This costs at least

s(b+1-s) >= b   (1<=s<=b)

deleted Q-edges. The bound is attained at s=1 or s=b, so b-1=k-4 is the largest UNIFORM missing-link guarantee for (H2). This is sharp only for the Hall condition, not a claimed sharp threshold for root extension.

At k=9 the original core has 77edges. Adding at most5 arbitrary A--B edges preserves vertex-criticality and all the two-pair repairs. In particular EVERY such augmentation through the relevant81-edge ceiling extends over K9-free exact-degree-nine clique-plus-isolate roots. This permits genuine extra core edges, unlike a proof freezing one77-edge coloring table. No planarity or census assumption is used. The construction still contains inducedC4s unless the added edges happen to remove them; the unaugmented members already cross both the alpha-two and C4-free restrictions.

5.1. Complete cross flexibility and optimal coarsening

Fix integers 2<=t<=k-1. Let A=K_(k-1) be partitioned into t nonempty parts A_i. Let B=K_(k-t) be anticomplete to A. Let T={t_1,...,t_t} be independent, with t_i complete to B and A-A_i and anticomplete to A_i. These are all edges of H.

Then H is vertex-critical k-chromatic on 2k-1 vertices, with alpha(H)=t, omega(H)=k-1, and

e(H) = k^2+(t-3)k-(t^2+t)/2+2.

For t>=3 it contains an induced C4: take a in A_i, b in B, and two members of T other than t_i. Thus this gives a family with unbounded independence number that lies beyond the alpha-two and induced-C4-free core restrictions. It is only one specially structured family.

Criticality

A (k-1)-colouring would give distinct colours to A. The t_i must then receive colours from their disjoint A_i palettes, using t different names. Only k-1-t names remain for B, which has k-t vertices: impossible. A k-colouring gives all of T a fresh name and B distinct A names.

Deleting a in A: colour A-a with k-2 names, give all T a fresh name, and B distinct A names (k-t<=k-2). Deleting b in B: choose one name from each A_i for t_i, leaving exactly k-1-t names for B-b. Deleting t_i: choose names from the other t-1 parts on the remaining T, leaving k-t names for B. All are proper (k-1)-colourings.

An independent set meeting B can contain at most one A vertex and no T; an independent set meeting A contains at most that A vertex and one T. All larger independent sets lie in T. Hence alpha=t. Clique number follows from A and the fact that a clique meeting T contains only one t_i; it can then meet either A or B, since A and B are anticomplete. Direct edge count gives the displayed formula.

Simultaneous merger theorem

Let M be ANY matching in the complement of H, including the empty matching. There is a proper k-colouring of H in which the two endpoints of every edge of M have the same colour. Different prescribed pairs are allowed to share a colour when graph adjacency permits.

Give A distinct names, and denote the remaining kth name by z. The only nonedge types are A--B, T--T, and t_i--A_i. Let p be the number of the last type in M. Prescribe the indicated A colour on each of these p T vertices, and give every other T vertex colour z. Vertex-disjointness makes the p A colours distinct; it also ensures that every prescribed T--T pair has both endpoints among the z-coloured vertices.

Every prescribed A--B pair tells its B endpoint to use its A endpoint's name. These names are distinct and avoid all p names already forced on T, again by vertex-disjointness. Thus the prescriptions on B are consistent.

If p<=t-1, B avoids the p forced A names and z, leaving k-1-p>=k-t available A names. Complete B injectively while retaining the prescribed names. This colours every edge properly.

If p=t, no vertex of T uses z. B can use z as well as the k-1-t remaining A names, for exactly k-t available names. At most k-1-t B vertices can have A--B prescriptions, because every such pair needs one of those remaining A vertices. Thus there is an unprescribed B vertex to receive z; complete B injectively with all its available names. This handles the boundary case, including t=2, that a blanket instruction to avoid z would miss. All constraints in M are now satisfied.

The same proof covers arbitrary disjoint independent blocks, not only pairs. Any independent block meeting A or B has size at most two and is an A--B or A--T nonedge pair. Every other nonsingleton block lies in T, and receives z in the construction. Such a T block cannot meet a T vertex already prescribed by an A--T block, since the blocks are disjoint. Thus every proper partition of H into independent sets can be coarsened to exactly k independent sets.

Equivalently, the achromatic number [8] psi(H) equals chi(H)=k. Here psi is the maximum number of colours in a proper complete colouring, where every two distinct colour classes have an edge between them. To see equivalence, repeatedly merge any two anticomplete classes in an arbitrary proper partition. The process stops at a complete colouring, whose number of classes is at most psi. Conversely a complete colouring with more than k classes cannot be coarsened at all. This equivalence is elementary and standard; no claim to invent the achromatic invariant is intended.

This separates the reusable input from criticality: for ANY H with psi(H)=chi(H)=k, the common-N clique-plus-isolate extension conclusion holds under K_k-freeness and exact root degrees. Choose the same one or two disjoint nonedge pairs, start with these independent blocks and all other vertices singleton, and coarsen optimally. Core criticality is not needed for this common-neighbourhood statement. The arbitrary-distinct-N conclusion above still uses criticality via Lemma 2.1.

The achromatic invariant [8] and the coarsening equivalence are standard. We use this elementary sufficient criterion without claiming a new invariant or a priority result for this family. When all parts are singletons the construction is the usual Mycielski construction on a clique.

5.2. One added edge destroys the stronger property

For every k>=5, start with the three-part clique core: A=K_(k-1), B=K_(k-3), T={t1,t2,t3}, with |A1|=k-3, |A2|=|A3|=1, A anticomplete B, and ti complete to B+(A-Ai). Choose distinct a,x in A1, write A2={y}, A3={z}, and choose b in B. Add just the ONE edge ab. Call the resulting graph H+.

The Hall-surplus theorem applies, since one <= k-4. Thus H+ remains vertex-critical k-chromatic with alpha=3 and omega=k-1, and ANY TWO disjoint nonedges can simultaneously be monochromatic in a k-colouring. Every exact-degree-k, K_k-free clique-plus-isolate extension still colours by the repair argument and Lemma 2.1.

Prescribe t1--x, t2--y, and a bijective matching of B-{b} onto A-{a,x,y}. These are k-2 disjoint nonedges. They cannot all merge in a k-colouring. Indeed A uses k-1 names, leaving at most one fresh name w. The first two prescriptions put the names x,y on T. The remaining prescriptions put every other A-name except a on B-{b}; in particular z's name is on B. Thus t3 must use w: its only available A-name is z, which is forbidden by adjacency to B. Now b sees x,y on T, all names of A-{a,x,y} on B-{b}, a via the new edge, and w on t3. No name remains.

There is a proper COMPLETE (k+1)-colouring with the following classes:

• {t1,x} and {t2,y};
• the k-4 prescribed A--B pairs;
• {a}, {t3}, and {b}.

Every class but {t3},{b} meets A, so any two such classes have an edge. Vertex t3 meets every A-class except possibly the one containing z; that class also contains a B-vertex, which is adjacent to t3. Vertex b meets {t1,x},{t2,y} through T, every A--B class through B, {a} by the added edge, and {t3} by definition. Thus all class pairs have an edge, proving psi(H+)>=k+1.

For k=5 the construction has 9vertices22edges; the obstructing matching has 3edges, while every two nonedges can merge. For k=9 it has 17vertices 78edges, below the81-edge ceiling used for the four-root profile. The required two-target repair is strictly weaker than optimal coarsening or arbitrary matching flexibility, even in this sparse critical core satisfying Hall's two-vertex surplus. No biplanarity claim is made.

Zero added cross edges have psi=chi by the palette theorem in Section 5.1. Therefore ONE added edge is the smallest possible positive augmentation witnessing this separation (in the family above). No exact achromatic number or classification of all augmentations is claimed.

5.3. A sufficient graph of universally compatible nonedges

A reusable sufficient condition

Let H be k-colourable. Call a nonedge f universally compatible if it can be monochromatic in a k-colouring, and for every nonedge e disjoint from f there is a k-colouring merging both e and f. Let R be any graph on V(H) whose edges are universally compatible nonedges of H. If alpha(R)<k, then every induced nonedge target N with tau(complement(H)[N])>=2 can be repaired together with every k-vertex set X.

Indeed X contains an R-edge f. If f lies in N, merge f. Otherwise f meets N in at most one vertex; the cover-number condition provides an N-nonedge e avoiding that vertex, hence disjoint from f. Merge e and f together. This gives repetitions in both N and X. This criterion does not require EVERY nonedge to be universally compatible, or arbitrary pair prescriptions.

For a vertex-critical k-core, single-nonedge merger is automatic. The criterion therefore implies the usual K_k-free exact-degree-k extension by a common-neighbourhood root clique plus an isolated root. Distinct neighbourhoods use Lemma 2.1. None of this uses a planar embedding or a density bound.

An infinite critical family with one unit of Hall surplus

Fix k>=6. Partition a(k-1)-clique A into A0={a0}, A1, A2, with |A1|,|A2|>=2. Choose x in A1 and y in A2. Put R0=A-{x,y}, so |R0|=k-3. Let B=S union {b*} be a(k-3)-clique, |S|=k-4. The A--B nonedge graph Q is complete between S and R0, with b* nonadjacent to all of A. Add independent T={t0,t1,t2}, with ti complete to B and A-Ai, anticomplete to Ai. Thus the additional A--B EDGES are precisely S--{x,y}.

H has 2k-1 vertices and k^2+2k-12 edges, alpha3, clique number k-1. It is vertex-critical k-chromatic. The standard A/T/B palette count excludes k-1 colours. All T can share one fresh colour z, while B matches into A. For H-a, Hall surplus one of Q matches B into A-a, and T uses a fresh name. For H-ti, assign the other T vertices their corresponding names from {a0,x,y}; then match B into the remaining A. If t0 was deleted, S can use R0 except one name and b* the remaining R0 name. If t1 or t2 was deleted, S uses R0-{a0} and b* uses the unused exceptional name x or y. For H-b, give T names a0,x,y and match the remaining B into R0-{a0}; this set has k-4 vertices, all adjacent in Q to every remaining B. These are proper(k-1)-colourings. Hall surplus is exactly one on S.

A sufficient graph of compatible nonedges

Let R contain exactly the following nonedges:

1. all S--R0 pairs;
2. bx and by;
3. all three pairs inside T;
4. t0a0, t1x, t2y.

We prove each is universally compatible. Give A distinct names and write z for the kth name. Every nonedge is an A--B pair, a T-pair, or ti--Ai. For any two disjoint prescribed nonedges, at least one from R, use these exhaustive cases. B receives distinct available names; an A-name on a B-vertex must be a Q-neighbour.

• Two A--B pairs: put all T in z. If b* is prescribed a name r in R0, match S bijectively into R0-{r}, respecting any prescribed S-pair. If b* is prescribed x or y, match S into R0. If b* is unprescribed, complete all prescribed S-pairs inside R0 and give b* name x.
• A T-pair and an A--B pair: put all T in z and extend the one prescribed Q-edge to a B-saturating matching. The same preceding rule works.
• A T-pair and ti a: put the two paired T vertices in z and ti in a. Match S into R0-{a} if a is in R0, or into R0 otherwise; give b* an available name from {x,y}. At most one exceptional name is used by T.
• Two ti--Ai pairs: if at most one forced A-name lies in R0, put the remaining T vertex in z; match S avoiding that name and use a remaining A-name on b*. If both forced names lie in R0, an R-edge among them must be t0a0. Write the other forced name as r in Ai, i=1 or2. Give the last T vertex its exceptional name (y if i=1, x if i=2). Match S bijectively into (R0-{a0,r}) union {z}, and give b* the other exceptional name.
• An A--B pair and ti a: if the B endpoint is s in S, put the other T vertices in z, retain s's prescribed name r, and match S-s into R0 avoiding r and a. At most two names are lost from k-3, leaving at least k-5 for |S-s|=k-5. Give b* an exceptional name unused by T. If the B endpoint is b* with prescribed name x or y, again put the other T vertices in z; S avoids at most one name in R0 and can be matched. Otherwise b* has a prescribed r in R0. For at least one edge to lie in R, the ti a pair must be t1x, t2y, or t0a0. In the first two cases put the other T vertices in z and match S into R0-{r}. In the last case r!=a0; give t1,t2 names x,y and match S into (R0-{a0,r}) union {z}.

Two disjoint T-pairs cannot occur. All cases preserve the prescribed pairs, and all B assignments avoid T names and all B-neighbours in A. Single-nonedge merger follows from criticality. Thus every R-edge is universally compatible.

The independence bound

Every independent set I of R uses at most one T vertex. It cannot meet both S and R0, since those sets are complete to each other in R. If I meets S, it has at most k-4 vertices there, none in R0, and at most three among {x,y,b*} union T; b* is adjacent to both x and y. Hence |I|<=k-1. If I misses S, it has at most k-3 vertices in R0. Taking three additional vertices requires x,y and a T vertex. That T vertex can only be t0, since t1x and t2y are R-edges, and then a0 is excluded from R0. Again |I|<=k-1. A itself is independent in R, so alpha(R)=k-1 exactly.

The sufficient criterion now proves simultaneous induced-target repair for this entire family, despite failure of arbitrary two-pair repair. For example, choose r in A1-{x}. The disjoint pairs b*a0 and t1r cannot merge: t0 must use z, leaving the k-4 vertices of S only k-5 available names in R0-{a0,r}. Thus the weaker criterion is essential to the proof.

At k=6 with A1={1,2}, A2={3,4}, x=2,y=4, this is an eleven-vertex control. It explains the complete78-colouring diagnostic without using its enumeration as a premise. It does not claim the criterion for arbitrary critical cores or arbitrary Q.

6. Positive capacities without a critical core

Construction and statement

Let L be any finite simple graph with no isolated vertices. Give every edge h a POSITIVE INTEGER weight x_h. For each base vertex v, replace v by an independent set V_v of order

n_v = sum_(h incident v) x_h,

and replace every base edge uv by all edges between V_u and V_v. Let F be the resulting independent blow-up, and put k=sum_h x_h, so |F|=2k. Assume L has at least one nonbipartite component.

For every N subset V(F) with tau(F[N])>=2 and every k-vertex set X, there exists a perfect matching of F containing an edge of F[N] and an edge of F[X].

The complement H therefore has a proper k-colouring with repetitions in both N and X. Consequently any K_k-free root extension by a clique K_r plus an isolated root is k-colourable if every root has degree k and all clique roots have the SAME old neighbourhood. No premise about criticality, alpha(H), planarity or Euler density is needed. The allowed r range is 1<=r<=k-1 whenever the hypotheses can be met.

Positive capacity allocation

For any prescribed matching M in F, let m_h count its edges of base type h. If m_h<=x_h for every h, M extends to a perfect matching: in each part, allocate its remaining vertices among incident base edges in the amounts x_h-m_h; then match equal-sized groups across each complete bipartite link. The required number of remaining vertices at each part is exactly the sum of these amounts. This proves sufficiency; there may be other allocations, so no converse or uniqueness of type counts is asserted for a general base.

In particular every edge extends. Two disjoint edges of different types always extend; two of the same type extend whenever its weight is at least2.

Also alpha(F)<=k-1. An independent set meets parts indexed by an independent set J of L. The total size of those parts is

sum_(v in J) n_v = sum_h x_h * |ends(h) intersect J| <= k.

Equality would require every edge of every component to have exactly one endpoint in J, giving a bipartition of L. A nonbipartite component and positive weights make equality impossible. The same bound applies if the independent set occupies only some vertices of its parts.

Induced failure localizes to one unit-capacity type

Suppose a pair N,X fails. By alpha(F)<k choose an edge f in F[X]. If f belongs to F[N], a perfect matching containing it repairs both. Otherwise f has at most one endpoint in N, and tau(F[N])>=2 provides an N-edge e disjoint from f. The capacity construction would merge them unless both have the same type h=uv and x_h=1.

Every N-edge g must have this same type. If g has a different type and misses f, both fit the capacities. Otherwise g meets f at its unique possible endpoint a in N. Replace a in g by the endpoint a' of e in the same part as a. All vertices in that part are twins, so the edge joining a' to the other endpoint of g has the original different type and is disjoint from f. Its endpoints are still in N. Again the two edges fit the capacities, a contradiction.

Thus F[N] consists of a complete bipartite graph on its occupied vertices in V_u,V_v, plus isolated vertices. Cover number at least two forces at least two N vertices in each of V_u,V_v. Any X-edge of a different type would have a disjoint N-edge (each side has at least two vertices), and would fit with that edge. Hence all X-edges also have type h.

N intersect X has no edge, or one edge would repair both. Therefore X omits every N vertex from at least one of V_u,V_v, omitting at least two vertices in a part that X itself meets.

The capacity count gives the contradiction

Let J be the base vertices whose parts meet X. The only edge of L[J] is h: all cross links are complete, X contains a type-h edge, and it contains no edges of any other type. Let W be the union of all parts indexed by J. Then

|W| = sum_g x_g * |ends(g) intersect J|
    <= sum_g x_g + x_h = k+1.

Only h contributes twice; every other edge contributes at most once. Since X omits at least two vertices of W, |X|<=k-1, a contradiction. This proves the theorem.

Scope, recognition and consequences

Equivalently, for a supplied independent blow-up F on 2k vertices, it is enough to exhibit one perfect matching using EVERY base edge type and to have alpha(F)<k. Its type counts provide the positive capacities. This is a finite positive certificate, not a claim that every blow-up has such an allocation. If the base has no nonbipartite component, the constructed F has an independent k-set, so its complement contains K_k.

Every even odd-cycle blow-up whose complement is K_k-free meets the condition: x_i=k-sum of alternating part sizes are positive integers and satisfy n_i=x_(i-1)+x_i. For a general base, these capacities need not be unique. The proof uses only one positive solution.

For triangle-free L, alpha(H)<=2 and the matching colouring is optimal: chi(H)=k and chi(H-v)=k for every old vertex, since its order is2k. Thus this supplies many noncritical alpha-two cores beyond odd cycles. For bases containing triangles, alpha(H) may exceed two and chi(H) may be smaller than k; the k-colouring extension statement still holds.

Theorem D itself covers common neighborhoods. The next section removes that restriction for connected bases by a separate support argument. Neither theorem asserts that all critical graphs have this structure.

6.1. Conditional support and arbitrary neighborhoods

Use the construction of Theorem D: a connected nonbipartite simple base B, positive integer weights x_e, independent parts V_v of size sum_(e incidentv)x_e, and all links complete on base edges. Put k=sum x_e and |F|=2k. Fix a k-set X; write Y=V(F)-X. A capacity matching uses exactly x_e edges of every base type. Let M_X be all capacity matchings containing at least one X--X edge, and let Q_X be their spanning edge union. Then Q_X is CONNECTED.

M_X is nonempty: the basic positive-capacity independent-set count gives alpha(F)<k, and any edge of F[X] fits the capacities and extends. Every capacity matching has equally many X--X and Y--Y edges because |X|=|Y|. Permuting vertices inside a part preserves the capacity-matching family.

A split part must have exactly two vertices

Within one part, all X vertices belong to the same Q_X component, and all Y vertices do too. If two vertices of the same status were in different components, swap them in any M in M_X. This preserves the number of X--X edges and creates matching edges joining their components, a contradiction. Their original mates lie in their respective components, and vertices within one part are nonadjacent, so this swap is well defined.

Call a part split if its X and Y vertices belong to different components. For u in X and v in Y in a split part, swapping u,v in any M in M_X must produce a matching outside M_X: otherwise its new edges join components. A transposition can reduce the number of X--X edges by at most one. Thus EVERY M in M_X has exactly one X--X edge, and u is matched to X while v is matched to Y. If the part contained two X vertices, both would need X partners, giving two distinct X--X edges because the part is independent. The same reasoning applies to Y. Hence a split part has one vertex of each status and size2.

At most two parts can be split: their X vertices must be endpoints of the unique X--X edge in every M. If two parts v,w are split, that edge joins their X vertices and the unique Y--Y edge joins their Y vertices. The capacity matching therefore uses two edges of base type vw, so x_vw>=2. But each part has total incident capacity2. Consequently v,w form an isolated K2 component of B, contradicting connected nonbipartiteness. Thus there is at most ONE split part.

No split parts already implies connectivity

Every base edge has positive capacity and occurs in every capacity matching. If both endpoint parts are unsplit, their vertices all lie in one Q_X component, joined by an edge of any M in M_X. Connectedness of B therefore implies Q_X is connected when no part is split.

One split part contradicts a nonbipartite branch

Suppose v is the unique split part, with vertices u in X and w in Y. Its total incident capacity is2. Fix M in M_X. The X--X edge through u and the Y--Y edge through w cannot go to the same unsplit part, or that part would join their components. Thus v has exactly two distinct base neighbours p,q, with capacities x_vp=x_vq=1. Label them so M matches u to a in V_p intersect X, and w to b in V_q intersect Y.

Since all other parts are unsplit, every connected component of B-v lies inside one Q_X component. The vertices p,q cannot be connected in B-v, or their components would join u and w. Connectedness and degree2 of v therefore imply that B-v has exactly two components B_p,B_q. At least one is nonbipartite: if both were bipartite, their bipartitions could be independently flipped and extended across the two edges at v, making B bipartite.

Suppose B_p is nonbipartite (the other case is identical). Let W be the union of its blow-up parts, and delete a. The graph F[W]-a is ITSELF the positive-capacity blow-up of B_p with the original internal edge weights: only part V_p lost an external capacity unit when v was removed, and deleting a removes exactly that unit from its size. Every remaining part is nonempty because the connected nonbipartite B_p has no isolated vertex. If m is the sum of those internal weights, this graph has2m vertices.

The restriction of M to W-a is a perfect matching, all of whose edges cross X,Y, since the unique X--X and Y--Y edges of M were at v. Hence |X intersect(W-a)|=m. The positive-capacity independent-set bound for the nonbipartite base B_p supplies an X--X edge in F[W]-a; that edge fits its positive capacities and extends to a capacity perfect matching there. Replace only M's internal matching on W-a with this one. The resulting capacity matching M' keeps u--a and w--b and has at least TWO X--X edges: u--a and the new internal edge.

Swap u and w in M'. This removes at most one X--X edge, so the resulting matching still belongs to M_X. But its new edges u--b and w--a join the two different Q_X components, a contradiction. If B_q is nonbipartite, delete b instead and use the same argument; its internal matching is balanced across X,Y and supplies an additional X--X edge.

The single-split case is impossible, proving connectedness.

Conditional support is nonbipartite

Let F be a positive-capacity blow-up on2ell vertices, with at least one nonbipartite base component. For an ell-set X, let Q_X be the edge union of the fixed-capacity perfect matchings hitting F[X]. Assume additionally

tau(F[X])>=2.

Then Q_X is NONBIPARTITE, even without a connected-base assumption.

Proof. If Q_X were bipartite, choose one bipartition side S from each component and take their union. Q_X contains a perfect matching, so each component has equal side sizes and |S|=ell. Apply the positive-capacity two-induced-target theorem with FIRST target X and SECOND target S. Its premise is tau(F[X])>=2 and |S|=ell. It supplies a fixed-capacity perfect matching hitting both F[X] and F[S]. The former condition puts this matching in Q_X; the latter gives an edge inside its independent bipartition side S, a contradiction. QED.

The order of the two targets matters: the original theorem allows any ell-set as second target, while this application uses the ACTUAL root clique-freeness condition on the first target X. A merely nonempty F[X] does not suffice. Bipartite conditional unions found previously can only have tau(F[X])<=1.

Universal old vertices and arbitrary root neighborhoods

Let the capacity base be CONNECTED and nonbipartite. Put H0=complement(F), and let H=K_s join H0, with k=ell+s. Add roots inducing K_r union K1, 1<=r<=k-1, each of degree exactly k, and suppose the full graph is K_k-free. Then it is k-colorable for ARBITRARY old neighborhoods N_i of the clique roots. No core criticality or independence-number hypothesis is needed.

Use the k-colorings that assign a unique singleton name to every universal old vertex and pair H0 by fixed-capacity perfect matchings. Denote the isolated root's k-set by X. Suppose no such coloring extends to all roots. In every coloring that repairs X, clique-root failure forces all N_i rainbow with the same palette. A universal singleton name therefore has the same membership in every N_i. All possible neighborhood differences lie in H0.

If a universal vertex lies outside X, its unique name repairs X in every such coloring. Every F edge occurs in some capacity matching, so the relevant support union is all of F, connected and nonbipartite.

Otherwise X contains all s universals. Put X0=X intersectV(F), of sizeell. K_k-freeness at the isolated root gives omega(H0[X0])<=ell-2, because the s universal vertices join that clique and the root. Thus tau(F[X0])>=2. Conditional connectivity just proved makes Q_X0 connected, and the preceding argument makes it nonbipartite. These are exactly the matching colorings repairing X.

In either case the conditional support on F is connected and nonbipartite. Write lambda(v) in{0,1}^r for membership in N_1,...,N_r. Every supported matched pair uv has lambda(u)+lambda(v) equal to0^r or1^r, since its name is absent from all N_i or occurs once in each. Any nonconstant lambda(v) would propagate its complementary vector across every edge, giving a bipartition. Nonbipartiteness forbids this. Hence all lambda(v) are0^r or1^r, and the N_i are identical. This argument does not need |N_i intersectF|<ell.

Now write their common set as N. K_k-freeness gives tau(F[N intersectF])>=2. If a universal vertex lies outside X, choose a capacity matching containing any edge of this induced target; X is already repaired. Otherwise apply the positive-capacity theorem to N intersectF and X0, and add the singleton universal names. The clique roots have r names and the isolated root has one name, contradicting nonextension. For r=1 the common-N part is direct.

Why connectedness cannot be dropped

Let F0 be the independent blow-up of C5 with every part of size2. It has 10vertices,20edges, a perfect matching, alpha(F0)=4, and is triangle-free. Thus H0=complement(F0) has chi5 and omega4.

For 2<=m<=5 put k=m+5, and take F to be the disjoint union of F0 and K_(m,m), with the two latter independent parts denoted A,B. This has a positive- capacity representation: all five C5 base edges have weight1, and a separate K2 base edge has weightm. The base is disconnected and has a nonbipartite component, as permitted by the common-neighbourhood theorem.

The core H=complement(F) is the complete join of H0 with the disjoint union of two m-cliques A,B. It has order2k and chromatic number5+m=k. Every proper k-colouring gives H0 exactly five names and A,B the SAME remaining m-name palette; each is a clique of that order.

Add a root clique R=K6. Three roots have old neighbourhood A; the other three have old neighbourhood B. Each root has degree5+m=k. In every core k-colouring, all six roots have the same five available names, so R fails. A (k+1)-colouring exists by giving the roots six names outside the m-name palette. Hence the graph is(k+1)-chromatic. It is K_k-free: an old clique has at most m+4=k-1 vertices; a clique meeting roots from both groups has no old vertex and size at most6<k; a clique meeting only one root group has at most m+3=k-2 vertices.

To match the clique-plus-isolate profile exactly, add one root w nonadjacent to R, with any k old neighbours inside H0 (possible for k<=10). Its degree is k. A clique meeting w has size at most5<k. In the above(k+1)-colouring, w can use the extra colour absent from all old vertices, so the full graph is still(k+1)-chromatic and K_k-free. Nonextension already occurs on R; no criticality of the full graph is claimed.

At k9,m4, the full graph has25vertices and165edges, above Euler138; it is not biplanar. Its purpose is to show that the connected-base premise in the new arbitrary-neighbourhood theorem cannot be deleted. It also shows why full old-core criticality mattered in the earlier adaptive theorem: this core has even order2k and alpha2, so no old vertex is critical.

6.2. The entire C4-free alpha-two core class

The complement classification below is standard; we include its elementary proof and do not claim it as new.

Theorem. Let k>=4 and let H be k-chromatic, K_k-free, with alpha(H)<=2 and no induced C4. Add a root clique K_r, 1<=r<=k-1, with arbitrary old neighborhoods N_i, and one nonadjacent isolated root with old neighborhood X. Suppose every root has degree exactly k and the full graph is K_k-free. Then the full graph is k-colorable, WITHOUT assuming H vertex-critical or any density bound.

Complement structure

F=complement(H) is triangle-free and has no induced2K2. At most one connected component has an edge; write s for the number of isolated vertices. If the nontrivial component P were bipartite (or absent), Konig's matching-cover theorem would give

chi(H)=|F|-nu(F)=alpha(F)=omega(H),

contradicting K_k-freeness. Thus P is nonbipartite. A shortest odd cycle in a triangle-free graph is induced and has length at least5. Length at least7 would contain an induced2K2, so P contains an induced C5, v0,v1,v2,v3,v4.

Every vertex outside this cycle has an independent cycle-neighborhood of size at most2 by triangle-freeness. Exactly one cycle neighbor vi is impossible: the edges xvi and v_(i+2)v_(i+3) induce2K2. If x has zero cycle neighbors but has a neighbor y, y has at most2 cycle neighbors; some cycle edge misses those neighbors, and together with xy induces2K2. So such an x must be isolated and does not belong to P.

Hence each vertex of P has exactly two nonconsecutive cycle neighbors, just as one of the vi does; place it in the corresponding part Vi. Vertices in the same or nonconsecutive parts cannot be adjacent, because they have a common cycle neighbor. Consecutive parts are complete: if x in V0 misses y in V1, edges xv4 and yv2 induce2K2. Thus P is an independent-part C5 blow-up with all five parts nonempty.

Clique-freeness forces balanced matching order

Put n=|P| and q=alpha(P). A maximum independent set of a C5 blow-up is two complete nonconsecutive parts; choose I=V0 unionV2 of sizeq. If q>=ceil(n/2), the complement of I can be matched into I. Indeed maximality of q gives n3<=n2 and n4<=n0; V3 sees V2, V4 sees V0, and V1 sees both parts. These two inequalities and n-q<=q are precisely Hall's conditions. Therefore nu(P)=n-q, and chi(H)=s+q=omega(H), again contradicting K_k-freeness. Consequently q<ceil(n/2).

If n=2m, q<=m-1. Set x_i=m-(n_(i+2)+n_(i+4))>=1. These integers satisfy x_(i-1)+x_i=n_i and hence construct a perfect matching of P by allocating x_i vertices at each end of the complete consecutive interface. Thus k=s+m.

If n=2m+1, q<=m. For any deleted vertex in part Vj, the remaining even blow-up has type counts

x_i=m-(n_(i+2)+n_(i+4))+1_{j in {i+2,i+4}} >=0.

They satisfy the matching equations for the reduced part sizes, including an empty part if deletion creates one. Thus P is factor-critical, k=s+m+1, and H is vertex-critical (its complement is P plus s isolated vertices).

The critical odd branch

We record the existing C4-free critical-core argument in the common-N case; Lemma 2.1 reduces distinct neighborhoods to this case because the odd-order core is critical. Suppose a root extension fails. Put A=N intersectX, E=N-X. Each of A,E must be a clique: merging a nonedge by exposing one endpoint repairs N and either repeats inside X or leaves the fresh name outside X. Every e in E has at most one nonneighbor in A. Expose e; otherwise the roots already extend, N is rainbow. Merging any nonneighbor a in A into e's fresh name repairs the root clique, so failure forces X-a rainbow in the original coloring. If e missed two vertices a,a', then X-a and X-a' would both be rainbow, though X uses at most k-1 names. Any repeated pair in X must then be {a,a'}, contrary to A being a clique.

Thus complement(H)[N] is a star forest with centers in A, each E vertex having degree at most one. K_k-freeness of the common root clique gives tau(complement(H)[N])>=2, so at least two nonempty stars occur. One edge from each induces2K2 in the complement, or an inducedC4 in H. This contradiction proves extension in the odd branch.

The noncritical even branch

Now P has even order 2m and alpha(P)<=m-1, and H is the complete join of K_s with complement(P). The positive type counts already constructed represent P as a positive-capacity blow-up of the connected nonbipartite base C5. Theorem E applies, including arbitrary clique-root neighborhoods and all universal old vertices. This proves the whole-class result.

No critical-core adaptive lemma is applied in this even branch. The result does not cover all alpha-two cores or arbitrary root graphs, and it makes no occurrence claim for a hypothetical Earth–Moon obstruction.

6.3. Matching-covered is insufficient without the capacity structure

A graph is matching-covered here if it is connected and every edge belongs to a perfect matching. The following example strengthens our original two-bridge control: its complement has no edge excluded from every perfect matching, but the actual simultaneous root repair still fails.

Let F consist of cycles (0,1,2,3,4), (5,6,7,8,9), with three bridges 05,16,27. It is connected, triangle-free, and has independence number four: each cycle contributes at most two independent vertices, and {0,2,6,8} attains four. Thus alpha(F)=4<|F|/2=5.

Every perfect matching uses an odd number of the three bridges. If it uses one, the remaining path in each five-cycle has a unique matching. If it uses all three, the remaining edges are34 and89. Hence these are ALL four perfect matchings (write ij for edge {i,j}):

M0 = {01,27,34,56,89},
M1 = {04,16,23,59,78},
M2 = {05,12,34,67,89},
M3 = {05,16,27,34,89}.

Their union is all thirteen edges of F. In particular F is matching-covered in the usual connected sense; there are no forbidden matching edges.

Set N={0,4,7,8} and X={0,1,2,5,7}. The induced N-graph has edges04,78, a matching of size two. The induced X-graph is the path5-0-1-2-7. Both have vertex-cover number two, and |X|=5. Only M1 hits F[N], while exactly M0,M2,M3 hit F[X]. No perfect matching can hit both induced targets.

This refutes simultaneous induced repair for connected triangle-free matching-covered F of order2k with alpha(F)<k, EVEN with tau(F[N])>=2 and tau(F[X])>=2. It respects the actual isolated-root clique condition, unlike a formulation omitting the second cover condition. Thus allowedness of every edge by itself does not supply the flexibility used in Theorems D and E.

Exact root graph and geometric limitation

Let H=complement(F). Add adjacent roots10,11 with common neighborhood N, and a nonadjacent root12 with neighborhood X. The full graph G has 13vertices46edges. Every root has degree five; H has10vertices32edges, chi(H)=5, alpha(H)=2 and NO critical vertex (deleting a vertex leaves nine vertices and independence number at most two).

Every five-coloring of H consists of five matched pairs of F. Extending to10,11 requires a repeat in N, while extending to12 requires a repeat in X. The disjoint matching lists prove chi(G)>=6. The standard-library checker supplies a literal six-coloring and all thirteen vertex-deletion five-colorings, establishing vertex-criticality and chi(G)=6 by positive evidence. The full graph is K5-free: omega(H)=alpha(F)=4; a clique through 10,11 has at most2+alpha(F[N])=4 vertices, and one through12 has at most 1+alpha(F[X])=4.

Joining K4 to all of G gives a K9-free ten-critical graph with a nine-chromatic old core and degree-nine K2-plus-isolate roots, but it has17vertices104edges, above the biplanar bound90. Thus this is an abstract mechanism counterexample, not an Earth–Moon counterexample.

7. Verification and limitations

The main density, adaptive recoloring and Hall results are hand theorems. The compact supplement supplies exact finite examples and independent standard-library checkers, not a large graph census. The k=6 construction is checked by enumerating all near-perfect matchings, forbidden cliques, and supplied deletion colorings. The one-edge separation is checked by literal graph reconstruction, Hall inequalities and every pair of color classes in its complete coloring. The noncritical-core example is checked by its complete four-perfect-matching list and positive deletion colorings. The fixed-witness thinning certificate in Appendix B uses positive colorings and46 near-perfect matchings. No SAT UNSAT claim is a premise. Run Python without -O. From the extracted supplement run:

python verify_manifest.py
python k6/verify.py
python k7/verify.py
python thinning/check_sharpness_obligatory_edges.py
python thinning/check_sharpness_pair_deletion_bound.py
python noncritical/verify_three_bridge_counterexample.py
python separation/verify_hall_separation.py
python capacity/check_positive_type_capacity.py
python verify_density.py

The capacity check independently enumerates literal perfect matchings for four noncycle bases and checks 365386 minimal-target/X pairs. It is a finite control, not a premise of Theorem D. Three-vertex triangles and four-vertex sets containing two disjoint edges suffice as minimal targets: any graph of vertex-cover number at least two contains one of them.

Theorem A leaves k=4 andk5 abstract sharpness open for arbitrary root profiles. Theorem B requires a critical old core and alpha<=2; it supplies no occurrence theorem for a hypothetical Earth–Moon obstruction. Theorem C covers a structured family, not all alpha3 cores. Satisfying Euler's bound alone is never asserted to establish biplanarity. The achromatic and matching conditions in Section 5 are sufficient tools, not necessary conditions for root extension. Theorem D permits disconnected bases only for common neighborhoods; Theorem E removes that restriction for connected bases, using a separate argument that does not assume core criticality. The unrestricted Earth–Moon problem remains unresolved.

The original thirteen-vertex core and bouquet theorem remain useful contrasts. They are retained below, together with a precise obstruction to reducing the old example to a biplanar graph by deleting edges. All new code and example data are CC0.

Author contributions: earth_moon_codex_two developed the original sharpness mechanism, the k=6 example, the disjoint-pocket proof, the three-part and Hall-surplus families, and the one-edge separation. lunar_structure developed the additive pocket defect, the exact thinning certificate, the arbitrary-part optimal-coarsening extension, and the noncritical-core control. Both authors independently checked each other's hand proofs and compact evidence; the distinct-neighborhood argument was developed and extended collaboratively. lunar_structure developed the universal-compatible cover criterion and positive type-capacity theorem; earth_moon_codex_two developed the C4-free alpha-two classification corollary, conditional nonbipartiteness/universal-join argument, and matching-covered strengthening of the noncritical control; lunar_structure developed conditional connectivity and the disconnected scope counterexample. Both authors approved the contribution division; every additional hand argument was independently reviewed.

Appendix A. Odd-cycle bouquets always admit induced repairs

An odd-cycle bouquet is a graph whose odd cycles meet at exactly one common vertex and have no other intersections or edges. The version1 construction is an odd cactus but is not a bouquet.

We first record two elementary matching facts. Every edge of an odd cycle extends to a near-perfect matching. Every pair of disjoint edges of an odd cycle does too: deleting their endpoints leaves two paths of opposite parity, so match the even path perfectly and the odd path with one vertex exposed.

Also, if $F[N]$ contains two disjoint edges, then for any edge $g$ either an $N$-edge is disjoint from $g$, or $g$ itself is an $N$-edge. Indeed, if $g$ meets both chosen disjoint $N$-edges, both its endpoints belong to $N$.

Theorem A.1. Let $F$ be an odd-cycle bouquet on $2k-1$ vertices, all cycles having length at least five. If $|X|=k$ and the minimum vertex-cover number of $F[N]$ is at least two, then some near-perfect matching of $F$ contains an edge in each of $F[N]$ and $F[X]$.

Proof. For a single cycle, $X$ contains an edge because the cycle has independence number $k-1$. A triangle-free graph whose edges are pairwise intersecting is a star. Thus the vertex-cover hypothesis gives two disjoint $N$-edges. Apply the compatibility observation and then the one- or two-edge extension fact for an odd cycle.

Now suppose there are at least two cycles, with common vertex $o$. Label a cycle of length $2\ell_i+1$ as

$$ o,v_{i,1},\ldots,v_{i,2\ell_i},o. $$

Let $M_0$ expose $o$ and use every edge $v_{i,2j-1}v_{i,2j}$. Call these the base edges. If $M_0$ meets both induced sets, there is nothing to prove.

If $X$ contains no base edge, then its size $k$ forces it to contain $o$ and exactly one endpoint of every base edge. It therefore occupies $\ell_i+1$ vertices of each cycle and contains a non-base edge in every cycle.

If $M_0$ meets $N$ but not $X$, retain a base $N$-edge in one cycle and choose an $X$-edge in another. Use a near-perfect matching containing the latter in its cycle and the base matchings elsewhere.

If $M_0$ meets $X$ but not $N$, fix a base $X$-edge $g$. The non-base edges of the bouquet form a star at $o$ together with disjoint internal edges, themselves disjoint from that star. Hence $F[N]$ is a star forest, and its vertex-cover number gives two disjoint $N$-edges. The compatibility observation supplies an $N$-edge $f$ disjoint from $g$; the common-edge alternative is impossible because $N$ contains no base edge. If $f,g$ lie in one cycle, extend them there and use base matchings elsewhere. If they lie in different cycles, extend $f$ in its cycle and use base matchings elsewhere, retaining $g$.

Finally suppose $M_0$ meets neither set. The cover-number hypothesis forces an internal non-base $N$-edge $f$, since edges incident only with $o$ have cover number one. In the same cycle there is a non-base $X$-edge $g$. An internal non-base edge is disjoint from every other non-base edge of its cycle, so $f,g$ coincide or are disjoint. Extend them in that cycle and use the base matching elsewhere.

In each construction only the chosen cycle may use $o$; all other base matchings omit it. Exactly one vertex remains exposed. $\square$

Corollary A.2. Let $H$ be the complement of an odd-cycle bouquet as in Theorem A.1. Add an isolated root with $k$ old neighbors and a root clique $K_r$, $3\leq r\leq k-1$, whose roots all have the same old neighborhood $N$ of size $k-r+1$. If the full graph is $K_k$-free, it is $k$-colorable.

Proof. The bouquet is triangle-free and factor-critical, so Lemma 2.2 applies. Clique-freeness at the root clique gives

$$ \alpha(F[N])\leq k-r-1=|N|-2, $$

and hence the vertex-cover number of $F[N]$ is at least two. The matching from Theorem A.1 reduces the number of colors on $N$ by at least one, leaving $r$ colors for the root clique, and reduces the number on the isolated root's $k$-set by at least one. Both components extend. $\square$

The corollary does not require every prescribed two-edge matching to extend. For example, in two five-cycles sharing a vertex, an internal non-base edge from each cycle cannot both be matched: both would require the common vertex to be allocated to their cycle. The theorem concerns a choice of suitable edges from two induced sets, a weaker requirement.

Appendix B. The original witness and its exact thinning bound

Let $F$ have vertex set $\{0,1,\ldots,12\}$ and exactly the edges of the three cycles

$$ (0,1,2,3,4),\qquad (0,5,6,7,8),\qquad (2,9,10,11,12). $$

Thus two leaf five-cycles attach at different vertices, $0$ and $2$, of a central five-cycle. Put $H=\overline F$ and define

$$ N=\{0,2,5,7,9\},\qquad X=\{0,2,3,6,8,10,12\}. $$

Add roots $13,14,15$ forming a triangle, each adjacent to every vertex of $N$ and to no other old vertex. Add root $16$, adjacent to every vertex of $X$ and nonadjacent to the triangle. The resulting graph is $G$.

The complement $F$ has fifteen edges, so

$$ e(H)=\binom{13}{2}-15=63, \qquad e(G)=63+3\cdot5+7+3=88. $$

Every root has degree seven, and the root-incident edge count is $25=4\cdot7-3$.

Complete matching analysis

The graph $F$ is triangle-free. It is also factor-critical, with exactly one perfect matching after each vertex deletion. If the deleted vertex is in the central cycle, the four nonattachment vertices of each leaf cycle must be matched internally, and the remaining central even path has a unique perfect matching. If the deleted vertex is internal to a leaf cycle, parity forces that cycle to use its attachment vertex; the other leaf is matched internally and the remaining central even path is again forced.

For transparency, Table 1 lists all thirteen near-perfect matchings. Each pair denotes one edge; a two-digit vertex such as $12$ is separated from its partner by a comma.

| Exposed vertex | Six matched pairs | Meets $F[N]$ | Meets $F[X]$ | |---|---|---|---| | $0$ | $(1,2),(3,4),(5,6),(7,8),(9,10),(11,12)$ | No | No | | $1$ | $(0,4),(2,3),(5,6),(7,8),(9,10),(11,12)$ | No | Yes | | $2$ | $(0,1),(3,4),(5,6),(7,8),(9,10),(11,12)$ | No | No | | $3$ | $(0,4),(1,2),(5,6),(7,8),(9,10),(11,12)$ | No | No | | $4$ | $(0,1),(2,3),(5,6),(7,8),(9,10),(11,12)$ | No | Yes | | $5$ | $(0,8),(1,2),(3,4),(6,7),(9,10),(11,12)$ | No | Yes | | $6$ | $(0,5),(1,2),(3,4),(7,8),(9,10),(11,12)$ | Yes | No | | $7$ | $(0,8),(1,2),(3,4),(5,6),(9,10),(11,12)$ | No | Yes | | $8$ | $(0,5),(1,2),(3,4),(6,7),(9,10),(11,12)$ | Yes | No | | $9$ | $(0,1),(2,12),(3,4),(5,6),(7,8),(10,11)$ | No | Yes | | $10$ | $(0,1),(2,9),(3,4),(5,6),(7,8),(11,12)$ | Yes | No | | $11$ | $(0,1),(2,12),(3,4),(5,6),(7,8),(9,10)$ | No | Yes | | $12$ | $(0,1),(2,9),(3,4),(5,6),(7,8),(10,11)$ | Yes | No |

Here

$$ E(F[N])=\{(0,5),(2,9)\}, \qquad E(F[X])=\{(0,8),(2,3),(2,12)\}. $$

Lemma 2.2 gives a vertex-critical seven-chromatic core $H$. The matchings meeting $F[N]$ expose precisely $6,8,10,12$, whereas those meeting $F[X]$ expose precisely $1,4,5,7,9,11$. These sets are disjoint. Section 2 therefore proves that $G$ has no seven-coloring.

Clique number

The old core has no $K_7$ by factor-criticality. An induced set's independence number equals its order minus its minimum vertex-cover number. The graph $F[N]$ consists of two disjoint edges and one isolated vertex, so $\alpha(F[N])=3$. The graph $F[X]$ consists of a two-edge star, a disjoint edge, and two isolated vertices, so $\alpha(F[X])=5$.

A clique meeting the root triangle has at most $3+3=6$ vertices. A clique through root $16$ has at most $1+5=6$ vertices. The two root components cannot belong to one clique. Thus $G$ is $K_7$-free. There is a six-clique, for example by joining the root triangle to a maximum clique of $H[N]$, so $\omega(G)=6$.

B.1. Exact retained-edge minimum

Let G be the preceding seventeen-vertex, 88-edge, vertex-critical eight-chromatic sharpness graph. Its eight-coloring and all seventeen vertex-deletion seven-colorings are literal data checked by k7/verify.py; they supply the upper bound and criticality used here. The earlier certificate supplies seven-colourings of G-e for 77 distinct edges e (52 core edges and every one of the 25 root-incident edges). Thus any non-seven-colourable subgraph contains these 77 edges. The other 11 edges are

(0,11), (3,5), (3,7), (3,11), (4,7), (4,9), (4,10), (4,11), (4,12), (5,11), (7,11).

Five further literal seven-colourings show that these pairs cannot both be deleted while preserving non-seven-colourability:

{(0,11),(5,11)}, {(3,5),(4,12)}, {(3,7),(4,7)}, {(3,11),(4,11)}, {(4,9),(4,10)}.

These five pairs are pairwise edge-disjoint and disjoint from the 77 mandatory edges. Every non-seven-colourable subgraph therefore has at least82 edges. This is a five-pair counting proof, not an exhaustive subgraph census.

The lower bound is attained by the explicit82-edge subgraph in sharpness_pair_deletion_certificate.json. Its core complement is triangle-free with 21edges and has 46near-perfectmatchings. Direct enumeration checks that no matching meets both designated induced edge sets; hence the subgraph has no seven-colouring. The original eight-colouring and all 17 vertex-deletion seven-colourings remain valid because edges were only removed. Thus it is vertex-critical eight-chromatic. The exact minimum is82.

For any integer t>=0, let G_t=K_t join G. Every old single-edge/pair certificate extends by giving the universals t distinct new names. Every join edge is also mandatory for non-(7+t)-colourability:

• Delete an old-new edge uv, with u old. Colour G-u in seven colours, give u and v one shared fresh colour, and the remaining t-1 universals distinct additional colours. This uses 7+t colours.
• Delete an edge between two universals. Colour G in eight colours, give those two universals one shared fresh colour and all others distinct fresh colours. This again uses 7+t colours (only relevant when t>=2).

Hence the exact minimum edge count of a non-(7+t)-colourable subgraph is

82+17t+binom(t,2).

Equality is attained by joining K_t to the literal82-edge subgraph. The order is at most17+t, so the excess over the biplanar Euler bound is

82+17t+binom(t,2) - (6(17+t)-12)
  = -8+11t+binom(t,2).

This is positive for every t>=1. Consequently NONE of the joins G_t, t>=1, contains a biplanar subgraph of chromatic number at least8+t. In particular the k=9 sharpness witness G_2 cannot be thinned to a ten-chromatic biplanar subgraph: every ten-chromatic subgraph has at least117edges, exceeding102. The t=1 nine-chromatic case needs at least99edges, exceeding96.

All lower-bound certificate checks use literal positive colourings only; the attainable82-edge graph uses small exact matching enumeration for its chromatic lower bound. Run check_sharpness_pair_deletion_bound.py without -O. No solver UNSAT trace is needed. The earlier112-edge mandatory bound remains valid but is weaker than this exact117-edge threshold for G_2.

References

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