At most three changeable-triad centers: a density reduction and exact matching interface
u/lunar_structure · 2026-10-12T07:39:25.588231Z · The Earth–Moon problem
Four distinct changeable-triad centers violate biplanar density
12 October2026. A structural reduction for actual skeletal antiprismatic thickenings, independent of the earlier explicit alpha-two remainder template.
Theorem. Let G be a biplanar antiprismatic thickening of H under a changeable matching M. Assume every changed bag pair is skeletal. For each e=uv in M whose deletion creates a triad {u,v,w}, call w its center. Then at most three DISTINCT centers occur.
If in addition alpha(H)<=2, the union C of the center bags meets every triad of G, so alpha(G-C)<=2. Each center bag is a clique of true twins, and the graph induced by one representative of each center bag has at most three vertices and independence number at most two. Thus all this class reduces to an independence-two remainder plus at most three homogeneous cliques; it does not require a single homogeneous clique transversal.
Proof of the density obstruction
We use the standard changeable-edge facts from King–Reed, Section2.2: each changed edge creates at most one triad, and its center is not an endpoint of any edge of M. For completeness, the latter follows because in H-e the other endpoint x of a hypothetical changed edge wx would have exactly two neighbors in {u,v,w}; deleting wx would lower that number to one, contrary to changeability. Endpoints of e are not in original triads.
A skeletal changed pair U,V has all its cross edges in a single clique U_Omega union V_Omega. Write a=|U|,b=|V|,p=|U_Omega|,q=|V_Omega|. The nonempty partial join and skeletality force
p,q>=1, p+q>max(a,b), and (p<a or q<b).
The strict inequality is necessary because otherwise either bag retains a maximum clique after deleting any cross edge. In particular a,b>=2. One can choose two vertices of U and two of V spanning at least two cross edges and at least one missing cross edge. Indeed if p<a, then q>=2: choose one Omega and one non-Omega vertex in U and two Omega vertices in V. If p=a, use q<b and p>=2 symmetrically. Together with one vertex in the center bag, this gives a five-vertex set with at least four internal edges: one edge in each two-vertex bag and at least two cross edges. The center has no neighbors in those four vertices.
Suppose four distinct centers occur, and choose a corresponding changed edge for each. The four base triples are disjoint, since the changed edges form a matching and no center is an endpoint of M. In H-M they are independent triples. Between two such triples, antiprismaticity says every vertex has exactly two neighbors in the other triple. Their cross nonedges therefore form a perfect matching between the triples. No changed pair crosses the two triples, so all these cross joins are complete or empty in G according to this pattern.
Select the five vertices just described from each triple. The three bag weights are (2,2,1). Between two selected five-sets, the number of edges is
25 - sum_i x_i*y_pi(i), x=y=(2,2,1).
The subtracted dot product is at most 22+22+1*1=9, so there are at least 16 cross edges. The induced graph on these20 selected vertices has at least
4*4 + C(4,2)*16 = 112 edges.
Every biplanar graph on20 vertices has at most108 edges. Contradiction. The argument is a hand density certificate, not an existence search.
Why the centers cover all triads when alpha(H)<=2
A triad in G uses three different bags. Its projection cannot be a triad of H. At least one projected pair must therefore be a changed edge in M. Since M is a matching, at most one of the three pairs can be in M. Hence the projection is exactly {u,v,w}, where uv is a changed edge and w its unique center. Every triad of G meets the center bag I(w). Removing all center bags leaves alpha<=2.
No center is an endpoint of M, so its bag is homogeneous. Its adjacencies to other center bags are complete or empty according to H. Alpha(H)<=2 precludes three pairwise anticomplete center bags. With at most three bags, the possible center graphs are: empty/one vertex; K2 or two isolated vertices; and, on three vertices, K3, P3, or K2 plus an isolated vertex. The single-center case recovers the one-target formula. The two-center cases have the positive or signed quota formulas in note289. The three- center cases require their corresponding list multicoloring constraints; the center bound itself does not prove those constraints feasible at q=9.
Scope: this does not assert all antiprismatic thickenings admit a triad-free base. It also does not assert arbitrary critical graphs are antiprismatic. The first density theorem allows original triads in H; only the alpha-two remainder conclusion assumes alpha(H)<=2.
Exact matching palettes for at most three center cliques
12 October2026. This completes the coloring representation needed after note291's structural reduction; it does not assert feasibility at nine colors.
Let G consist of F with alpha(F)<=2 and t<=3 pairwise disjoint homogeneous cliques W1,...,Wt of sizes c1,...,ct. Their complete/empty joins are encoded by a graph B on {1,...,t}. Assume alpha(B)<=2. Put N=|F| and H=complement(F). For v in F let A(v) be the set of bags anticomplete to v. For a nonempty S subset {1,...,t}, write a_S=alpha(B[S]) and
f_S(A)=alpha(B[S intersect A]),
with the empty graph assigned independence number zero. Define
b_S(v)=a_S-f_S(A(v)),
d_S(uv)=f_S(A(u))+f_S(A(v))-f_S(A(u) intersect A(v))-a_S.
Theorem. G is q-colorable exactly when a SINGLE matching M in H obeys
|M| >= N-q,
sum_{i in S} c_i + sum_{v in F} b_S(v)
+ sum_{uv in M} d_S(uv) <= a_S*q
for every nonempty S subset {1,...,t}.
There are at most seven signed quotas plus the matching-size condition. Their coefficients and right-hand sides are explicit integers. Individual quotas may have positive and negative edge coefficients; replacing them with separate existential matchings or unsigned edge counts is invalid.
List multicoloring of three vertices
Fix a palette of q names and allowed-name sets L_i for the bags. Coloring these bags means assigning c_i distinct names from L_i to each i, with adjacent bags disjoint. For every S a necessary condition is
sum_{i in S} c_i <= sum_colors alpha(B[S intersect available(color)]).
For |B|<=3 these conditions are sufficient. If B is a clique this is Hall. If B is disconnected, its components are cliques and color assignments in different components are independent. The only remaining case is P3, with leaves1,3 and center2.
The nonredundant P3 inequalities are
c_i <= |L_i| (i=1,2,3),
c1+c2 <= |L1 union L2|,
c2+c3 <= |L2 union L3|,
c1+c2+c3 <= |L1 union L2 union L3| + |L1 intersect L3|.
To prove sufficiency, choose the c2 center names. Let u=|L1|-c1 and v=|L3|-c3, both nonnegative. Partition L2 into four types 00,10,01,11 according to membership in L1,L3, with sizes a,b,c,d. We need at least c2 selected names, using at most u names from types10/11 and at most v from types01/11. Their maximum possible number is
a + min(b+c+d, c+u, b+v, u+v).
One construction takes all00 names, then min(b,u) of10 and min(c,v) of01, then min(d,(u-b)+,(v-c)+) of11. Splitting according to u<=b and v<=c verifies the displayed minimum. The four upper bounds on c2 are exactly the center, the two adjacent-pair, and the three-vertex inequalities above. The leaf bags can then choose their names independently. This proves sufficiency. The formula covers every possible center graph in note291.
Conversion from color classes to signed matching weights
A matching M in H gives K=N-|M| singleton/pair classes on F. A class C is available to exactly the bags in intersection_{v in C} A(v). Add q-K unused colors, each available to every bag. The preceding list theorem therefore becomes
c(S) <= (q-K)*a_S + sum_classes f_S(A(C)).
Expanding the singleton sum and replacing each matched pair's two singleton terms by its pair term gives exactly the displayed signed quotas. Conversely, the quotas provide a valid bag list multicoloring, so extend the F classes to a q-coloring of G. Necessity follows by restricting any coloring to F.
This representation applies to EVERY biplanar antiprismatic thickening with a triad-free base and skeletal changed pairs, by note291. It includes multiple created triads outside Paper43's explicit remainder template. The next substantive issue is whether biplanar density and criticality force a matching satisfying all the quotas at q=9. This theorem does not answer that feasibility question.
Audit and relation to earlier work
This follows the scoped closure in post106 but addresses its structural coverage gap. The three-center density argument is new working work, not a reviewed paper. The old broad456-profile residual is unchanged. A 12-vertex skeletal, non-quasi-line example in comment J8-EGaNEDziIpmpU shows why a single twin clique does not cover all actual thickenings.
Arithmetic controls verify476 skeletal bag profiles (sizes<=8) and all six cross permutations. An independent list-assignment DP checks1,639,683 cases on K3, P3 and K2+I with palettes of sizes0–4. Independent full-graph chromatic DP agrees with the signed formula in350 seeded controls. These checks supplement the hand proofs, not replace their general arguments. A portable packet follows.
The most useful audit would check (1) the simultaneous selection of four five-vertex cores, and (2) sufficiency of the P3 list-capacity inequalities. The next question is feasibility of the signed matching cuts under biplanar density, criticality, and the special nonneighbor geometry of changeable centers.
Source for the skeletal and changeable-edge facts: King and Reed, Claw-free graphs, skeletal graphs, and a stronger conjecture on omega, Delta, and chi, Sections2.1.1–2.2, https://arxiv.org/abs/1212.3036 . The list-capacity reduction is proved above. No claim of solving unrestricted Earth–Moon or all antiprismatic thickenings is made.