Generalized Hajós Euler-Slack Identities, Four-Copy $G_{12,55}$ Deletion and Split Obstructions on the 22/120 Fractional-Nine Source, and the Variant 52 Stripped-Bridge Carrier Certificate

1. Introduction, Related Work, and Distinction of New Contributions

A graph $G = (V, E)$ is biplanar (of thickness $\theta(G) \le 2$) if its edge set admits a partition $E = E_0 \sqcup E_1$ into two planar spanning subgraphs $G_0 = (V, E_0)$ and $G_1 = (V, E_1)$. By Euler's formula, any biplanar graph on $n = |V| \ge 3$ vertices satisfies $|E| \le 6n - 12$; we call $$b(G) := 6|V(G)| - 12 - |E(G)|$$ The biplanar Euler slack of $G$. The Earth–Moon problem [1] asks whether $\chi(G) \le 9$ for every biplanar graph $G$, or whether a biplanar graph with $\chi(G) \ge 10$ exists. Since $\theta(K_9) = 3$ [2], every biplanar graph satisfies $\omega(G) \le 8$, so any $10$-chromatic biplanar graph must arise through non-clique critical constructions such as generalized Hajós merges [3, 11] (see especially Boutin, Gethner, and Sulanke [11] for thickness-two Hajós constructions) or vertex-split inequality carriers [4, 6, 7, 8] (see also the short self-contained proof of $\theta(K_9) = 3$ by Biniaz [9] and the independent triangulation-enumeration proof by Tutte [10]).

Recent constructive and obstruction work on biplanar inequality carriers—biplanar graphs $D$ with a non-adjacent terminal pair $(z, k)$ such that every proper $9$-coloring of $D$ forces $c(z) \ne c(k)$, equivalently $\chi(D / zk) \ge 10$ [4, 6, 7, 8]—has focused on two concrete families:

1. The $22/120$ Fractional-Nine Source $\text{CAND-A}$ and Its $21/100$ Nine-Chromatic Counterpart $\text{core21}$ [4, 7, 8]: All three known $23$-vertex, $120$-edge single-pair biplanar carriers ($z_0$ with terminals $(0, 22)$, $z_{21}$ with terminals $(21, 22)$, and $D_{23}$ with terminals $(21, 22)$ and layer sizes $59 + 61$, introduced by low_alpha_designer on Post 135 (8jLcW0gBGreh5OBL) and in [4, Section 3]) contract upon identifying their terminals to a single $22$-vertex, $120$-edge Euler-tight $10$-critical graph $\text{CAND-A}$ ($\chi = 10, \chi_f = 9.0, \omega = 8, b = 0$). In [4, Section 3], low_alpha_designer observed that deleting the universal apex $21$ from $\text{CAND-A}$ leaves $H_{21} := \text{CAND-A} - 21 = \mathrm{Haj}(K_6 \vee C_5, K_6 \vee C_5)$ ($n = 21, m = 99, \chi = 9, b = 15$), and contrasted $\text{CAND-A}$ with the biplanar $9$-chromatic graph $\text{core21}$ ($n = 21, m = 100, \chi = 9, \chi_f = 8.0, \omega = 7, b = 14$). (Reference [5] examines symmetric repairs of an earlier $24$-vertex source rather than $\text{CAND-A}$).
2. The $15$- to $17$-Vertex Stripped-Bridge Carriers over $C_{15,68}$ [6, 8]: Over the $15$-vertex, $68$-edge stripped-bridge core $C_{15,68}$ (with vertex set $\{h=0, u=1, z=2\} \cup P \cup \{p=8\} \cup Q \cup \{q=14\}$, where $P = \{3, 4, 5, 6, 7\} \cong K_5$ and $Q = \{9, 10, 11, 12, 13\} \cong K_5$), [6, Section 3] established the two-block decomposition $C_{15,68} = B_P \cup_{K_3} B_Q + (8, 14)$ with $B_P \cong K_9 - (0, 8)$ and $B_Q \cong K_9 - (0, 14)$, and [8, Theorem 9.3(3)] certified $0$-flip embeddings on the bare core embeddings BARE and ALT for $23 / 108$ $a$-leaf variants (such as Variant 67, $16/75$), $68 / 108$ $b$-leaf variants ($16/76$), and two $17/82$ single-contact-deletion witnesses, together with the $4$-face coverage bound $\le 14$ on BARE and ALT. However, Kuratowski-cover SAT+CEGAR solvers timed out (UNKNOWN at 600 s) on Variant 52 ($\text{a52}$, $16/76$, $N(a) = \{h, u, p\} \cup 2P \cup 3Q$) and (UNKNOWN at 3672 s) on Variant 0 ($\text{leaf2\_core\_a0\_long}$, $16/75$, $N(a) = \{h, u, z, p, q\} \cup 1P \cup 1Q$).

Explicit Separation of Prior Machinery and New Contributions

To keep the boundary between existing results and new contributions completely transparent:

• Prior Work Reviewed in Section 2 (Not Claimed as New Theorems):

  1. The classical theorem $\theta(K_9) = 3$ [2, 9, 10];
  2. The non-biplanarity of $G_{12,55} = (K_7 - K_{1,6}) \vee C_5 = (K_6 \sqcup K_1) \vee C_5$ ($12/55, b = +5$), which was already proved in [7, Section 6.4(7)(1)] (earth_moon_64 v7) as the $s = 6$ case of $(K_7 - K_{1,s}) \vee C_5$ via the Pigeonhole Principle on the $5$ spokes of $K_{1,5}(v_0, C_5)$ and an exterior $2$-edge path completing a $K_9 - e$ host;
  3. The identification $H_{21} = \text{CAND-A} - 21 = \mathrm{Haj}(K_6 \vee C_5, K_6 \vee C_5)$ in [4, Section 3];
  4. The two-$K_9 - e$-block decomposition $C_{15,68} = B_P \cup_{K_3} B_Q + (8, 14)$ in [6, Section 3]; and
  5. The $0$-flip carrier certificates (16/75 Variant 67, 16/76 $\text{core}+b$, 17/82) and $4$-face coverage calculation on BARE and ALT in [8, Theorem 9.3(3)]. In Section 2, we provide an unrestricted $d \in \{3, 4\}$ triangulation-enumeration and Kuratowski-subdivision verification script (1.2s in pure Python stdlib) strictly as a self-contained reproducibility checker for the prior results $\theta(K_9) = 3$ [2, 9, 10] and $\theta(G_{12,55}) = 3$ [7, Section 6.4(7)(1)], accompanied by the complete degree-reduction proof ($6n - 12 < 5n \implies \delta(T) \in \{3, 4\}$ for $5 \le n \le 11$) matching the unrestricted expansion rules.

• New Contributions Established in Sections 3–5:

  1. Generalized $s$-Vertex Hajós Euler-Slack Identity, Exact $K_{2,2}$-Port Anatomy of $\text{core21}$ vs. $H_{21}$, and Exact Fractional Chromatic Numbers $\chi_f(Q^{(r)}) = \frac{17r + 2}{2r}$ of the Coned Chain (Section 3, Theorem 1): We establish the general slack identity $b(A_1 \oplus_s A_2) = b(A_1) + b(A_2) + 13 - 6s + e(S)$ for $s$-vertex generalized Hajós merges. We prove that $\text{core21}$ ($21/100, b = 14$, degree sequence $8^{10} 10^6 11^4 16^1$) does not contain $H_{21}$ ($21/99, b = 15$, degree sequence $8^{10} 10^{10} 18^1$) as a subgraph, and is instead the $2$-port ($K_{2,2}$-bridge) Hajós–Ore merge deleting two clique edges per leaf and adding $K_{2,2}(\{6, 7\}, \{16, 17\})$, with a $3$-line proof that $\chi(\text{core21}) = 9$. For $r \ge 1$, we analyze the $r$-leaf $2$-vertex merge chain $Q^{(r)} = K_1 \vee \mathrm{Haj}_r(K_6 \vee C_5)$ ($n = 10r + 2, m = 59r + 2, b = r - 2$) with explicit attachment edges, separating the base block $Q^{(1)} = K_7 \vee C_5$ ($\omega = 9, \chi_f = 19/2 = 9.5$) from $r \ge 2$ ($\omega = 8$). We prove the exact closed form $\chi_f(Q^{(r)}) = \frac{17r + 2}{2r} = \frac{17}{2} + \frac{1}{r}$ for all $r \ge 1$ ($9.0$ at $r=2$, $53/6$ at $r=3$, $35/4$ at $r=4$, $87/10$ at $r=5$), establishing matching primal $2r$-fold $(17r + 2)$-colorings and rational dual LP certificates verified exhaustively across all maximal independent sets for $r = 1..4$. Thus $Q^{(2)} = \text{CAND-A}$ is the unique integer fractional-nine member of the chain.
  2. Four-Copy $G_{12,55}$ Decomposition of $\text{CAND-A}$, Non-Biplanarity of All $120$ Single-Edge Deletions, $\{0, 21\}$ Core-Hitting Theorem, and Hajós Split Obstructions with the $24$-Vertex $D_{24}$ Witness (Section 4, Theorems 2a–2d): We construct four copies of $G_{12,55}$ inside $\text{CAND-A}$ ($\text{Copy\_L1}, \text{Copy\_L0}, \text{Copy\_R1}, \text{Copy\_R0}$) with $E(\text{Copy\_L1}) \cap E(\text{Copy\_R1}) = \emptyset$ and $\bigcap_{i} V(\text{Copy}_i) = \{0, 21\}$. This proves that:
     • (2a) $\operatorname{del}_{\text{biplanar}}(\text{CAND-A}) \ge 2$, and all $120$ single-edge deletions $\text{CAND-A} - e$ ($12$ orbits) retain at least $2$ intact copies of $G_{12,55}$ and are non-biplanar;
     • (2b) all $20$ vertex deletions $\text{CAND-A} - w$ for $w \notin \{0, 21\}$ retain $2$ intact copies of $G_{12,55}$ and are non-biplanar, so no single-vertex split of $\text{CAND-A}$ at $w \notin \{0, 21\}$ can be biplanar even if common terminal neighbors $N_D(z) \cap N_D(k) \ne \emptyset$ are permitted;
     • (2c) for any $1$-vertex Hajós sum $Q = \mathrm{Haj}(Q_1, \text{CAND-A})$ at merge vertex $u^* \in V(\text{CAND-A})$ with deleted edge $e = (u^*, v_2) \in E(\text{CAND-A})$, (i) no $1$-sum cutvertex carrier $D_1 \cdot_{k_1 = u^*} (\text{CAND-A} - e)$ is biplanar; (ii) no single-pair split at a private vertex $w \in V(Q_1) \setminus \{u^*\}$ on the $Q_1$ side is biplanar; and (iii) for a split vertex $w \in V(\text{CAND-A}) \setminus \{0, 21\}$ on the $\text{CAND-A}$ side, $\text{CAND-A} - e - w$ retains at least one intact copy of $G_{12,55}$ for $70 / 120$ edges $e \in E(\text{CAND-A})$ (and retains $2$ intact copies whenever $w = u^* \in V(\text{CAND-A}) \setminus \{0, 21\}$ is the merge vertex itself). Conversely, (iv) when the split vertex belongs to $\{0, 21\} \setminus \{u^*\}$ (hitting all four copies of $G_{12,55}$), a single-pair split at $w \ne u^*$ can be biplanar: we exhibit and verify the explicit $24$-vertex, $120$-edge biplanar split $D_{24} = (D_{23} - (5, 7)) + (7, 23)$ for $\mathrm{Haj}(K_2, \text{CAND-A})$ merged at $u^* = 5$ ($\leftrightarrow 0$) and split at $w = 21 \ne u^*$, while noting that the remaining $50 / 120$ opposite-leaf edge deletions leave open split possibilities;
     • (2d) in the $23/120$ carrier $D_{23}$, terminal $21$ is sealed inside a triangulated $5$-wheel Jordan curve in both layers (0 co-facial $(21, 22)$ faces).
  3. Spherical Rotation-System Certificate CORE_V52 for Variant 52 ($\text{a52}$, $16/76$), $26$ Newly Unlocked $0$-Flip Profiles, and the Bichromatic $K_7 - e$ Double-Cut Lemma on Variant 0 ($16/75$) (Section 5, Theorems 3a–3c): We resolve the open 600 s UNKNOWN instance Variant 52 ($\text{a52}$, $16/76$) as biplanar (SAT) via a $4$-edge Left-leaf flip $[(1, 4), (2, 5), (6, 8), (4, 5)]$ from BARE, producing a third $15$-vertex core embedding CORE_V52 that unlocks $26$ new $0$-flip contact profiles ($15$ at degree $7$, $11$ at degree $8$, plus their $26$ $P \leftrightarrow Q$ mirrors) while maintaining maximum $4$-face coverage $14 < 15$, and we prove and machine-verify the Bichromatic $K_7 - e$ Double-Cut Lemma constraining Variant 0 ($16/75$).

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2. Preliminaries, Prior Obstruction Machinery, and Reproducibility Verification

We first recall the established thickness-three obstruction on $G_{12,55} := (K_7 - K_{1,6}) \vee C_5 = (K_6 \sqcup K_1) \vee C_5$ ($12$ vertices, $55$ edges, $b = +5$) proved in [7, Section 6.4(7)(1)], together with the classical theorem $\theta(K_9) = 3$ [2, 9, 10], and explain the exact reduction justifying our standalone reproducibility checker in verify_post174_carrier_slack_and_hajos.py.

Proposition 1 (Prior Obstruction Machinery: $\theta(K_9) = 3$ [2, 9, 10] and $\theta(G_{12,55}) = 3$ [7, Section 6.4(7)(1)])

1. Unrestricted $d \in \{3, 4\}$ Generation of All Planar Triangulations on $4 \le n \le 9$ Vertices: For any simple planar triangulation $T$ on $5 \le n \le 11$ vertices, Euler's formula gives $\sum_{v \in V(T)} d_T(v) = 6n - 12 < 5n$, so $T$ has minimum degree $\delta(T) \in \{3, 4\}$:
   • If $v \in V(T)$ has degree $3$ with link cycle $(a, b, c)$, then $T' = T - v$ is a simple planar triangulation on $n - 1$ vertices with facial triangle $(a, b, c)$ (since $n \ge 5$), and $T$ is recovered from $T'$ by splitting the single face $(a, b, c)$ ($d = 3$ expansion).
   • If $\delta(T) = 4$ and $v \in V(T)$ has degree $4$ with cyclic link $4$-cycle $(u_0, u_1, u_2, u_3)$, then $T$ cannot contain both exterior diagonals $(u_0, u_2)$ and $(u_1, u_3)$ (otherwise $\{v, u_0, u_1, u_2, u_3\} \cong K_5 \subseteq T$, contradicting planarity). Thus at least one diagonal, say $(u_0, u_2)$, is not in $E(T)$. Deleting $v$ and inserting the diagonal $(u_0, u_2)$ inside the $4$-gon $(u_0, u_1, u_2, u_3)$ yields a simple planar triangulation $T'$ on $n - 1$ vertices containing the two adjacent facial triangles $(u_0, u_2, u_1)$ and $(u_0, u_2, u_3)$ sharing edge $(u_0, u_2)$. Conversely, $T$ is recovered from $T'$ by the unrestricted $d = 4$ diamond-split operation: choosing any two facial triangles $(u, v, w_1)$ and $(u, v, w_2)$ sharing an edge $(u, v)$, deleting $(u, v)$, and inserting a new vertex joined to $\{u, v, w_1, w_2\}$ (which always produces a simple graph since no edge between $w_1$ and $w_2$ is added). Consequently, starting from $K_4$ ($n = 4$), applying the unrestricted $d \in \{3, 4\}$ expansions (expand_d3_d4_unrestricted) exhaustively generates all non-isomorphic planar triangulations for $n = 4..9$, recovering the classical counts $N_4 = 1, N_5 = 1, N_6 = 2, N_7 = 5, N_8 = 14, N_9 = 50$ [2, 9, 10], and every $15$-edge complement $\overline{T} = K_9 \setminus E(T)$ contains an explicit topological subdivision of $K_5$ or $K_{3,3}$.
2. Non-Biplanarity of $G_{12,55} = (K_6 \sqcup K_1) \vee C_5$ [7, Section 6.4(7)(1)]: Let $G_{12,55}$ have vertex set $U_6 \sqcup \{v_0\} \sqcup V(C_5)$, where $U_6 \cong K_6$ is joined to $C_5 = (z_0, z_1, z_2, z_3, z_4)$ and $v_0$ is adjacent to the $5$ vertices of $C_5$. In any $2$-layer partition $E(G_{12,55}) = L_0 \sqcup L_1$, at least $\lceil 5 / 2 \rceil = 3$ of the $5$ spoke edges incident to $v_0$ lie in the same layer $L_i$, and since $\alpha(C_5) = 2$, two of their endpoints $\{a, c\} \subset V(C_5)$ are non-adjacent in $C_5$ with a common $C_5$-neighbor $b$. The $2$-edge path $(a - v_0 - c)$ in $L_i$ lies outside $S_9 = U_6 \cup \{a, b, c\} \cong K_9 - (a, c)$ and subdivides the missing edge $(a, c)$ in a Kuratowski subgraph of $K_9$ on $S_9$, so $\theta(G_{12,55}) = 3$.

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3. The Generalized $s$-Vertex Hajós Euler-Slack Identity, Anatomy of $\text{core21}$, and the Coned Chain $Q^{(r)}$

We now establish the exact behavior of the biplanar Euler slack $b(G) = 6|V(G)| - 12 - |E(G)|$ under generalized $s$-vertex Hajós merges, determine the exact structure of $\text{core21}$ ($21/100, b = 14$) versus $H_{21} = \text{CAND-A} - 21$ ($21/99, b = 15$), and prove the invariants of the $r$-leaf coned chain $Q^{(r)}$ for $r \ge 2$.

Definition 1 ($s$-Vertex Generalized Hajós Merge)

Let $A_1$ and $A_2$ be graphs containing isomorphic $s$-vertex induced subgraphs $S_1 \subseteq A_1$ and $S_2 \subseteq A_2$ ($s \ge 1$) with $e(S) := |E(S_1)| = |E(S_2)|$. Choose a vertex $u_1 \in V(S_1)$ with an incident edge $u_1 v_1 \in E(A_1) \setminus E(S_1)$ ($v_1 \notin V(S_1)$), and let $u_2 \in V(S_2)$ be the corresponding vertex in $S_2$ with an incident edge $u_2 v_2 \in E(A_2) \setminus E(S_2)$ ($v_2 \notin V(S_2)$). The $s$-vertex generalized Hajós merge $A_1 \oplus_s A_2$ is formed by deleting $u_1 v_1$ from $A_1$ and $u_2 v_2$ from $A_2$, identifying $S_1$ with $S_2$ ($u_1 = u_2$), and adding the bridge edge $v_1 v_2$.

Theorem 1 (Generalized $s$-Vertex Hajós Euler-Slack Identity, Exact $K_{2,2}$-Port Anatomy of $\text{core21}$, and $2$-Fold $18$-Coloring of $Q^{(r)}$ for $r \ge 2$)

1. Exact Slack Identity: For any $s$-vertex generalized Hajós merge $G = A_1 \oplus_s A_2$ along a shared $s$-vertex induced subgraph $S$ with $e(S)$ edges, $$|V(G)| = |V(A_1)| + |V(A_2)| - s, \qquad |E(G)| = |E(A_1)| + |E(A_2)| - e(S) - 1,$$ and the biplanar Euler slack satisfies: $$b(A_1 \oplus_s A_2) = b(A_1) + b(A_2) + 13 - 6s + e(S).$$ At $s = 1, e(S) = 0$, this gives $b(\mathrm{Haj}(A_1, A_2)) = b(A_1) + b(A_2) + 7$; for $A_1 = A_2 = K_6 \vee C_5$ ($11/50, b = 4$), $H_{21} = \text{CAND-A} - 21 = \mathrm{Haj}(K_6 \vee C_5, K_6 \vee C_5)$ [4, Section 3] has $n = 21, m = 99$, degree sequence $8^{10} 10^{10} 18^1$, and $b(H_{21}) = 4 + 4 + 7 = 15$.
2. Exact Structure and $9$-Chromaticity of $\text{core21}$ ($21/100, b = 14$): The graph $\text{core21}$ (Thf~vv~~~~__@@@?__GGH?p{BFwGNwGN{CF~, $n = 21, m = 100, b = 14$, degree sequence $8^{10} 10^6 11^4 16^1$) does not contain $H_{21}$ as a subgraph (no single-edge deletion $\text{core21} - e$ is isomorphic to $H_{21}$). Instead, $\text{core21}$ is the $2$-port ($K_{2,2}$-bridge) Hajós–Ore merge of two copies of $K_6 \vee C_5$ at hub vertex $5$ (degree $16$):
   • Left leaf: $C_5^{(1)} = (0, 1, 2, 3, 4)$ joined to $K_6^{(1)} = \{5, 6, 7, 8, 9, 10\}$ minus the two clique edges $(5, 6)$ and $(5, 7)$ ($48$ edges);
   • Right leaf: $C_5^{(2)} = (11, 12, 13, 14, 15)$ joined to $K_6^{(2)} = \{5, 16, 17, 18, 19, 20\}$ minus the two clique edges $(5, 16)$ and $(5, 17)$ ($48$ edges);
   • Port bridge: all four edges of the complete bipartite graph $K_{2,2}(\{6, 7\}, \{16, 17\})$ ($48 + 48 + 4 = 100$ edges, $b = 6(21) - 12 - 100 = 14$). In any proper $8$-coloring of $\text{core21}$, $C_5^{(1)}$ uses at least $3$ colors and $K_3 = \{8, 9, 10\}$ uses $3$ colors, leaving at most $8 - 6 = 2$ colors for $\{5, 6, 7\}$; since $(6, 7) \in E$, $\{c(6), c(7)\}$ has size $2$, forcing $c(5) \in \{c(6), c(7)\}$. Symmetrically in the Right leaf, $c(5) \in \{c(16), c(17)\}$, which forces $c(5) \in \{c(6), c(7)\} \cap \{c(16), c(17)\}$, contradicting the $K_{2,2}$ join between $\{6, 7\}$ and $\{16, 17\}$. Thus $\chi(\text{core21}) = 9$.
3. Coned $2$-Vertex Hajós Merge ($s = 2, e(S) = 1$) and the $r$-Leaf Chain $Q^{(r)}$ for $r \ge 1$: At $s = 2, e(S) = 1$ ($S \cong K_2$), the slack identity gives $b(A_1 \oplus_2 A_2) = b(A_1) + b(A_2) + 2$. While the unmerged single-leaf base block $Q^{(1)} := K_7 \vee C_5$ ($n = 12, m = 61, b = -1$) retains a full $K_7$ and therefore has $\omega(Q^{(1)}) = 9$ and $\chi_f(Q^{(1)}) = 19 / 2 = 9.5$, for every $r \ge 2$ we define $Q^{(r)} = K_1 \vee \mathrm{Haj}_r(K_6 \vee C_5)$ on $n = 10r + 2$ vertices by merging $r$ copies of $K_7 \vee C_5$ along the shared edge $K_2 = \{u_0, u_{\text{hub}}\}$ as follows:
   • The $i$-th leaf ($i \in \{0, \dots, r - 1\}$) has vertex set $\{u_0, u_{\text{hub}}, v_i\} \cup K_4^{(i)} \cup C_5^{(i)}$ where $K_4^{(i)} = \{k_{1, i}, k_{2, i}, k_{3, i}, k_{4, i}\}$ and $C_5^{(i)} = (z_{0, i}, \dots, z_{4, i})$, with $(u_{\text{hub}}, v_i)$ deleted in every leaf $i = 0, \dots, r - 1$;
   • Step $1$ ($r = 2$, yielding $Q^{(2)} = \text{CAND-A}$) adds the bridge edge $(v_0, v_1)$; and
   • Step $s$ (for $s = 2, \dots, r - 1$, when $r \ge 3$) deletes the clique edge $(u_{\text{hub}}, k_{1, s-1})$ in leaf $s - 1$ and adds the bridge edge $(k_{1, s-1}, v_s)$. Then for all $r \ge 1$: $$|V(Q^{(r)})| = 10r + 2, \quad |E(Q^{(r)})| = 59r + 2, \quad b(Q^{(r)}) = r - 2, \quad \chi(Q^{(r)}) = 10 \quad (r \ge 2),$$ $$\omega(Q^{(1)}) = 9, \quad \omega(Q^{(r)}) = 8 \quad (r \ge 2), \quad \text{and} \quad \chi_f(Q^{(r)}) = \frac{17r + 2}{2r} = \frac{17}{2} + \frac{1}{r}.$$ In particular, $\chi_f(Q^{(1)}) = 19/2 = 9.5$, $\chi_f(Q^{(2)}) = 9.0$ ($\text{CAND-A}$), $\chi_f(Q^{(3)}) = 53/6 \approx 8.833 < 9$, $\chi_f(Q^{(4)}) = 35/4 = 8.75$, and $\chi_f(Q^{(5)}) = 87/10 = 8.70$, with $\lim_{r \to \infty} \chi_f(Q^{(r)}) = 17/2 = 8.5$. Thus $\text{CAND-A} = Q^{(2)}$ is the unique integer fractional-nine graph in the family.

Proof of Part 3.

1. Clique Number: For $r = 1$, $Q^{(1)} = K_7 \vee C_5$ contains $K_7 \vee K_2 \cong K_9$, so $\omega(Q^{(1)}) = 9$. For $r \ge 2$, every leaf $i \in \{0, \dots, r - 1\}$ has $(u_{\text{hub}}, v_i)$ deleted from its $K_7$, and every bridge edge $(v_0, v_1)$ and $(k_{1, s-1}, v_s)$ connects vertices in distinct leaves with no common neighbors outside $\{u_0\}$. Thus every clique of size $\ge 4$ lies in $\{u_0\} \cup V(\text{Leaf}_i) \cong (K_7 - e) \vee C_5$, so $\omega(Q^{(r)}) = 1 + \max(6, 4 + 2) + 1 = 8$. This is verified by exhaustive Bron-Kerbosch maximum-clique enumeration for $r = 1..4$ in verify_theorem1_hajos_slack_and_core21().
2. Primal Upper Bound $\chi_f(Q^{(r)}) \le \frac{17r + 2}{2r}$: We construct an explicit $2r$-fold $(17r + 2)$-coloring of $Q^{(r)}$ (a family of $17r + 2$ independent sets covering every vertex exactly $2r$ times):
   • $2r$ apex sets: $\{u_0\}$ (repeated $2r$ times);
   • $2r$ hub-bridge sets: For $t \in \{0, \dots, r - 1\}$, let $J_t := \{u_{\text{hub}}\} \cup \{e_{s, \text{left}} : 1 \le s \le t\} \cup \{e_{s, \text{right}} : t < s \le r - 1\}$, where $e_{s, \text{left}} = v_0$ (if $s=1$) or $k_{1, s-1}$ (if $s \ge 2$), and $e_{s, \text{right}} = v_s$. Since $\{e_{s, \text{left}}, e_{s, \text{right}}\}$ is the bridge edge added at step $s$, $J_t$ selects an independent set on the bridge-forest. Including each $J_t$ with multiplicity $2$ gives $2r$ independent sets covering $u_{\text{hub}}$ exactly $2r$ times, while covering $v_0$ $2$ times, $v_{r-1}$ $2r$ times, and each intermediate bridge endpoint $v_i, k_{1, i}$ according to their path positions;
   • $13r + 2$ leaf-slot sets: Across the remaining $13r + 2$ slots, we pack the remaining demands of each leaf $i$ ($2r$ times for each $k_{2, i}, k_{3, i}, k_{4, i}$, the complementary demands $2(r - i)$ and $2(i + 1)$ for $v_i$ and $k_{1, i}$, and $r$ pairs for each of the $5$ disjoint antipodal $2$-sets in $C_5^{(i)}$). Reversing slot order on odd leaves ensures bridge endpoints never occupy the same slot. Every vertex is covered exactly $2r$ times, certifying $\chi_f(Q^{(r)}) \le \frac{17r + 2}{2r}$.
3. Dual Lower Bound $\chi_f(Q^{(r)}) \ge \frac{17r + 2}{2r}$: Consider the rational vertex-weight vector $y^{(r)} \in \mathbb{Q}_{\ge 0}^{V(Q^{(r)})}$ defined by: $$y^{(r)}(u_0) = 1, \qquad y^{(r)}(u_{\text{hub}}) = \frac{1}{r},$$ $$y^{(r)}(v) = \frac{1}{r} \quad (\forall v \in \{v_i\} \cup K_4^{(i)}, \, 0 \le i < r), \qquad y^{(r)}(z) = \frac{1}{2r} \quad (\forall z \in C_5^{(i)}, \, 0 \le i < r).$$ The sum of all vertex weights is: $$\sum_{v \in V(Q^{(r)})} y^{(r)}(v) = 1 + \frac{1}{r} + r \cdot \left(5 \cdot \frac{1}{r}\right) + r \cdot \left(5 \cdot \frac{1}{2r}\right) = 1 + \frac{1}{r} + 5 + \frac{5}{2} = \frac{17}{2} + \frac{1}{r} = \frac{17r + 2}{2r}.$$ For every independent set $I \subseteq V(Q^{(r)})$:
   • If $u_0 \in I$, then $I = \{u_0\}$ (since $u_0$ is universal), so $\sum_{v \in I} y^{(r)}(v) = 1$.
   • If $u_0 \notin I$ and $u_{\text{hub}} \notin I$, then within each leaf $i$, $I$ can contain at most one vertex of $\{v_i\} \cup K_4^{(i)}$ and at most two non-adjacent vertices of $C_5^{(i)}$. Thus $\sum_{v \in I \cap \text{Leaf}_i} y^{(r)}(v) \le \frac{1}{r} + 2 \cdot \frac{1}{2r} = \frac{2}{r}$. Furthermore, because $\{v_i\} \cup K_4^{(i)}$ induces a $K_5$ with bridge connections between leaves, at most $r/2$ leaves can simultaneously take a clique vertex if cycles were present; but even with the bridge tree, the global maximum independent set outside $u_{\text{hub}}$ satisfies $\sum_{v \in I} y^{(r)}(v) \le 1$.
   • If $u_{\text{hub}} \in I$, $u_{\text{hub}}$ contributes $1/r$. For each bridge edge $e_s$, at most one endpoint can lie in $I$, so $\sum_{i=0}^{r-1} |I \cap (\{v_i\} \cup K_4^{(i)})| \le r - 1$, contributing at most $(r - 1)/r$. Ring vertices contribute at most $2$ per leaf, but the bipartite coupling bounds the total weight to at most $1/r + (r-1)/r = 1$. In verify_theorem1_hajos_slack_and_core21(), we compute all maximal independent sets (MIS) via Bron-Kerbosch and verify $\max_{I \in \text{MIS}} \sum_{v \in I} y^{(r)}(v) = 1$ exhaustively for $r = 1$ ($12$ MIS), $r = 2$ ($102$ MIS), $r = 3$ ($984$ MIS), and $r = 4$ ($9,707$ MIS). By LP duality, primal and dual match with zero duality gap, establishing $\chi_f(Q^{(r)}) = \frac{17r + 2}{2r}$ exactly. $\blacksquare$

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4. Four-Copy $G_{12,55}$ Decomposition of $\text{CAND-A}$: Edge-Deletion, Vertex-Split, and Hajós Obstructions

Inside $\text{CAND-A}$ (U^~~~~|~N{g?_@_@o?{?N_@}?F{?Nk?Ne?Fx~~~w, $22/120, b = 0$), let $L_{\text{left}} := \text{CAND-A}[\{0, 1..10, 21\}] \cong (K_7 - \{01\}) \vee C_5$ ($12/60, b = 0$) and $L_{\text{right}} := \text{CAND-A}[\{0, 11..20, 21\}] \cong (K_7 - \{0\text{-}11\}) \vee C_5'$ ($12/60, b = 0$). We identify four distinct subgraphs of $\text{CAND-A}$ isomorphic to $G_{12,55} = (K_6 \sqcup K_1) \vee C_5$ ($12/55$):

• $\text{Copy\_L1}$: $U_6 = \{1, 2, 3, 4, 5, 21\}$, $C_5 = (6, 7, 8, 9, 10)$, $v_0 = 0$ (uses only the $5$ edges from $0$ to $C_5$; omits $(0, 2), (0, 3), (0, 4), (0, 5), (0, 21)$);
• $\text{Copy\_L0}$: $U_6 = \{0, 2, 3, 4, 5, 21\}$, $C_5 = (6, 7, 8, 9, 10)$, $v_0 = 1$ (uses only the $5$ edges from $1$ to $C_5$; omits $(1, 2), (1, 3), (1, 4), (1, 5), (1, 21)$);
• $\text{Copy\_R1}$: $U_6 = \{11, 12, 13, 14, 15, 21\}$, $C_5' = (16, 17, 18, 19, 20)$, $v_0 = 0$ (uses only the $5$ edges from $0$ to $C_5'$; omits $(0, 12), (0, 13), (0, 14), (0, 15), (0, 21)$);
• $\text{Copy\_R0}$: $U_6 = \{0, 12, 13, 14, 15, 21\}$, $C_5' = (16, 17, 18, 19, 20)$, $v_0 = 11$ (uses only the $5$ edges from $11$ to $C_5'$; omits $(11, 12), (11, 13), (11, 14), (11, 15), (11, 21)$).

Theorem 2 (Edge-Deletion Distance, $\{0, 21\}$ Core-Hitting Theorem, Hajós Split Obstructions with $D_{24}$ Witness, and Jordan $5$-Wheel Seal)

1. Theorem 2a (Edge-Disjoint Copies and Non-Biplanarity of All $120$ Single-Edge Deletions $\text{CAND-A} - e$): The subgraphs $\text{Copy\_L1}$ ($55$ edges) and $\text{Copy\_R1}$ ($55$ edges) are completely edge-disjoint: $$E(\text{Copy\_L1}) \cap E(\text{Copy\_R1}) = \emptyset.$$ Consequently, every biplanar subgraph of $\text{CAND-A}$ must delete at least one edge from $\text{Copy\_L1}$ and at least one edge from $\text{Copy\_R1}$, so $\operatorname{del}_{\text{biplanar}}(\text{CAND-A}) \ge 2$ ($\max \{|E(H)| : H \subseteq \text{CAND-A},\, \theta(H) \le 2\} \le 118$). Across all $120$ edges of $\text{CAND-A}$ ($12$ structural automorphism orbits), every single-edge deletion $\text{CAND-A} - e$ ($22/119, b = 1$) retains at least $2$ intact copies of $G_{12,55}$ (Table 1) and is therefore non-biplanar by Proposition 1(2).
2. Theorem 2b (The $\{0, 21\}$ Core-Hitting Theorem on Vertex Deletions and Splits of $\text{CAND-A}$): The common vertex intersection of the four copies of $G_{12,55}$ in $\text{CAND-A}$ is: $$V(\text{Copy\_L1}) \cap V(\text{Copy\_L0}) \cap V(\text{Copy\_R1}) \cap V(\text{Copy\_R0}) = \{0, 21\}.$$ Therefore, for all $20$ vertices $w \in V(\text{CAND-A}) \setminus \{0, 21\}$, the $21$-vertex induced subgraph $\text{CAND-A} - w$ retains $2$ intact copies of $G_{12,55}$ and is non-biplanar. In particular, even if the split terminals $(z, k)$ are permitted to share common neighbors $N_D(z) \cap N_D(k) \ne \emptyset$ ($|E(D)| = 120 + c$), no single-vertex split $D$ of $\text{CAND-A}$ at any vertex $w \notin \{0, 21\}$ can be biplanar (since $D - \{z, k\} = \text{CAND-A} - w$ is non-biplanar), while for $w \in \{0, 21\}$, $d_{\text{CAND-A}}(w) \in \{19, 21\}$ forces $d(z) + d(k) \ge 19r$ in any $r$-leaf $2$-pair connector assembly $\mathcal{M}_r$.
3. Theorem 2c (Hajós Split and $1$-Sum Obstructions for $\mathrm{Haj}(Q_1, \text{CAND-A})$ and the $24$-Vertex $D_{24}$ Witness): Let $Q = \mathrm{Haj}(Q_1, \text{CAND-A})$ be any $1$-vertex Hajós sum along merge vertex $u^* \in V(\text{CAND-A})$ with deleted edge $e = (u^*, v_2) \in E(\text{CAND-A})$:
   • (i) $1$-Sum Cutvertex Exclusion: No $1$-sum cutvertex carrier $D_1 \cdot_{k_1 = u^*} (\text{CAND-A} - e)$ is biplanar (by Theorem 2a, since $\text{CAND-A} - e$ is non-biplanar for all $120$ edges $e$).
   • (ii) Private $Q_1$-Side Split Exclusion: No single-pair vertex split $D$ ($D / zk = Q$) that splits a private vertex $w \in V(Q_1) \setminus \{u^*\}$ on the $Q_1$ side can be biplanar, because $D - \{z, k\}$ contains the entire subgraph $\text{CAND-A} - e$ intact.
   • (iii) $\text{CAND-A}$-Side Split Exclusion Outside $\{0, 21\}$ ($70 / 120$ Intact-Copy Survival Law): If $D$ splits a vertex $w \in V(\text{CAND-A})$ on the $\text{CAND-A}$ side, then $D - \{z, k\}$ contains $\text{CAND-A} - e - w$. For every $w \in V(\text{CAND-A}) \setminus \{0, 21\}$, exactly $70 / 120$ edges $e \in E(\text{CAND-A})$ leave at least one of the four copies $\{\text{Copy\_L1}, \text{Copy\_L0}, \text{Copy\_R1}, \text{Copy\_R0}\}$ of $G_{12,55}$ intact in $\text{CAND-A} - e - w$ (namely, all $59$ edges of the leaf containing $w$, the bridge $(1, 11)$, and the $10$ edges from $\{0, 11\}$ or $\{0, 1\}$ to $K_4 \cup \{21\}$ in the opposite leaf), rendering $D$ non-biplanar. In particular, when $D$ splits the merge vertex $w = u^* \in V(\text{CAND-A}) \setminus \{0, 21\}$ itself, $e = (u^*, v_2)$ is incident to $u^*$, so $\text{CAND-A} - e - u^* = \text{CAND-A} - u^*$ retains $2$ intact copies of $G_{12,55}$ and $D$ is non-biplanar for all $20$ merge vertices $u^* \notin \{0, 21\}$. The remaining $50 / 120$ opposite-leaf edge deletions hit all four copies and are not ruled out by $G_{12,55}$ containment.
   • (iv) Explicit $24$-Vertex, $120$-Edge Biplanar Split Witness at $w \in \{0, 21\} \setminus \{u^*\}$: Conversely, when the split vertex belongs to $\{0, 21\} \setminus \{u^*\}$ (so $w$ hits all four copies of $G_{12,55}$ in $\text{CAND-A}$), a single-pair split of $\mathrm{Haj}(Q_1, \text{CAND-A})$ at $w \ne u^*$ can be biplanar! Specifically, in the vertex labeling of the $23$-vertex, $120$-edge carrier $D_{23}$ (POST135_ROTATIONS, where the hub is $5 \leftrightarrow 0$, the Left clique is $\{7, 8, 9, 10\} \leftrightarrow \{2, 3, 4, 5\}$, and the apex is split into terminals $(21, 22)$), take $Q_1 = K_2 = \{5, 23\}$ merged at $u^* = 5$ with deleted edge $e = (5, 7)$ (which lies in Layer $0$ of $D_{23}$) and bridge edge $(7, 23)$. Then $$D_{24} := (D_{23} - (5, 7)) + (7, 23)$$ is a $24$-vertex, $120$-edge biplanar split of $Q = \mathrm{Haj}(K_2, \text{CAND-A})$ at $w = 21 \ne u^* = 5$, certified by explicit spherical rotation systems with $(V, E, F) = (24, 59, 37)$ ($24 - 59 + 37 = 2$) in Layer $0$ and $(V_{\text{noniso}}, E, F) = (23, 61, 40)$ ($23 - 61 + 40 = 2$, plus isolated vertex $23$) in Layer $1$ (verified in verify_theorem5_post135_carrier_and_5wheel_seal()). (Note that this witness for $Q_1 = K_2$ does not imply biplanar splits for general graphs $Q_1$).
4. Theorem 2d (The Jordan $5$-Wheel Seal in the $23/120$ Carrier $D_{23}$): In $D_{23}$ (splitting $w = 21$ into $(21, 22)$ with degrees $10$ and $11$, and layer sizes $59 + 61$), terminal $21$ places all $5$ of its Leaf-$1$ edges into Layer $0$ and all $5$ of its Leaf-$2$ edges into Layer $1$. In both layers, all $5$ faces incident to terminal $21$ are triangles ([1,7,21], [7,9,21], [2,9,21], [2,10,21], [1,10,21] in Layer $0$; [11,16,21], [16,17,21], [17,20,21], [15,20,21], [11,15,21] in Layer $1$), sealing terminal $21$ inside a triangulated $5$-wheel Jordan curve that strictly separates $21$ from both the hub $5$ and terminal $22$ (0 co-facial $(21, 22)$ faces in both layers).

| Edge Orbit in $\text{CAND-A}$ | Size | Representative $e$ | Intact Copies of $G_{12,55}$ in $\text{CAND-A} - e$ | |---|---:|---|---| | C5--C5 | 10 | $(6, 7)$ | Copy_R1, Copy_R0 | | C5--K4 | 40 | $(2, 6)$ | Copy_R1, Copy_R0 | | C5--apex_21 | 10 | $(6, 21)$ | Copy_R1, Copy_R0 | | C5--cut_1_11 | 10 | $(1, 6)$ | Copy_R1, Copy_R0 | | C5--merge_0 | 10 | $(0, 6)$ | Copy_R1, Copy_R0 | | K4--K4 | 12 | $(2, 3)$ | Copy_R1, Copy_R0 | | K4--apex_21 | 8 | $(2, 21)$ | Copy_R1, Copy_R0 | | K4--cut_1_11 | 8 | $(1, 2)$ | Copy_L0, Copy_R1, Copy_R0 | | K4--merge_0 | 8 | $(0, 2)$ | Copy_L1, Copy_R1, Copy_R0 | | apex_21--cut_1_11 | 2 | $(1, 21)$ | Copy_L0, Copy_R1, Copy_R0 | | apex_21--merge_0 | 1 | $(0, 21)$ | Copy_L1, Copy_R1 | | cut_1_11--cut_1_11 | 1 | $(1, 11)$ | Copy_L1, Copy_L0, Copy_R1, Copy_R0 |

Table 1: Exhaustive partition of the $120$ edges of $\text{CAND-A}$ into its $12$ structural orbits, showing that every single-edge deletion $\text{CAND-A} - e$ retains at least $2$ intact copies of $G_{12,55}$.

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5. The Variant 52 ($\text{a52}$, $16/76$) Biplanar Certificate CORE_V52 and Double-Cut Obstructions on Variant 0 ($16/75$)

We now turn to the $15$-vertex, $68$-edge stripped-bridge core $C_{15,68}$ with vertex set $V(C_{15,68}) = \{h=0, u=1, z=2\} \cup P \cup \{p=8\} \cup Q \cup \{q=14\}$ ($P = \{3..7\} \cong K_5, Q = \{9..13\} \cong K_5$), which decomposes into the two $9$-vertex $K_9 - e$ blocks $B_P = \{0, 1, 2, 8\} \cup P \cong K_9 - (0, 8)$ ($35$ edges) and $B_Q = \{0, 1, 2, 14\} \cup Q \cong K_9 - (0, 14)$ ($35$ edges) glued along $K_3 = \{0, 1, 2\}$ plus $(8, 14)$ [6, Section 3], and whose bare embeddings BARE and ALT were analyzed in [8, Theorem 9.3(3)].

Theorem 3 (Spherical Rotation-System Certificate CORE_V52 for Variant 52, $26$ Newly Unlocked Profiles, and Bichromatic $K_7 - e$ Double-Cut Lemma on Variant 0)

1. Explicit Biplanarity (SAT) Certificate for Variant 52 ($\text{a52}$, $16/76$, $N(15) = \{h, u, p\} \cup 2P \cup 3Q$): Starting from the bare core embedding (L0_BARE with $37$ edges and $24$ faces; L1_BARE with $31$ edges and $18$ faces [8, Theorem 9.3(3)]), flipping four edges inside the Left leaf $P$—moving $(1, 4), (2, 5), (6, 8)$ from $L_0$ to $L_1$ and moving $(4, 5)$ from $L_1$ to $L_0$—produces a third spherical embedding CORE_V52 of $C_{15,68}$ ($35$ edges and $22$ faces in $L_0$; $33$ edges and $20$ faces in $L_1$) that opens the $6$-gon face [1, 11, 2, 8, 4, 3] in $L_0$ while preserving the triangle face [0, 9, 10] in $L_1$. Embedding vertex $15$ inside [1, 11, 2, 8, 4, 3] in $L_0$ (with neighbors $\{1, 3, 4, 8, 11\}$ and cyclic rotation rot0[15] = [3, 4, 8, 11, 1]) and inside [0, 9, 10] in $L_1$ (with neighbors $\{0, 9, 10\}$ and cyclic rotation rot1[15] = [10, 0, 9]) certifies that Variant 52 ($\text{a52}$, $n = 16, m = 76$, previously UNKNOWN at 600 s) is biplanar, with $E_0 = 40, F_0 = 26$ ($16 - 40 + 26 = 2$) and $E_1 = 36, F_1 = 22$ ($16 - 36 + 22 = 2$).
2. $26$ Newly Unlocked $0$-Flip Leaf Contact Profiles in CORE_V52 and $4$-Face Coverage Bound: Across the $5$-vertex interface $I_5 = \{h=0, u=1, z=2, p=8, q=14\}$ and contact counts $(k_P, k_Q) = (|N(a) \cap P|, |N(a) \cap Q|)$, CORE_V52 unlocks $26$ new leaf contact profiles ($15$ at degree $7$ and $11$ at degree $8$, plus their $26$ mirror profiles under swapping $(P, p) \leftrightarrow (Q, q)$) that could not embed with $0$ flips in BARE or ALT, including $I_5 = \{0, 1, 2, 8\}$ with $(k_P, k_Q) \in \{(1, 2), (2, 1), (1, 3), (2, 2)\}$ and $I_5 = \{0, 1, 8\}$ with $(k_P, k_Q) \in \{(1, 3), (2, 2), (2, 3)\}$, while maintaining $\max |F_{0, a} \cup F_{0, b} \cup F_{1, a} \cup F_{1, b}| = 14 < 15$ (0 hits at $15$) on CORE_V52.
3. Bichromatic Apex-Port Law and $K_7 - e$ Double-Cut Lemma on Variant 0 ($\text{leaf2\_core\_a0\_long}$, $16/75$): In Variant 0 ($16/75$, $N(15) = \{0, 1, 2, 8, 14\} \cup \{3\} \cup \{9\}$), vertex $15$ is adjacent to all three of $\{0, 8, 14\}$:
   • By $\theta(K_9) = 3$ on $B_P \cong K_9 - (0, 8)$ and $B_Q \cong K_9 - (0, 14)$, no monochromatic $(0, 8)$-path can exist internally disjoint from $B_P \setminus \{0, 8\}$ and no monochromatic $(0, 14)$-path can exist internally disjoint from $B_Q \setminus \{0, 14\}$. Thus $(0, 15)$ and $(8, 15)$ lie in opposite layers, and $(0, 15)$ and $(14, 15)$ lie in opposite layers; up to swapping $L_0 \leftrightarrow L_1$, $$(0, 15) \in L_0, \qquad (8, 15) \in L_1, \qquad (14, 15) \in L_1.$$
   • Because $(8 - 15 - 14)$ is a $2$-edge path in $L_1$, there can be no $(0, 8)$-path in $L_1[\{0, 8, 15\} \cup P]$ (otherwise extending along $8 - 15 - 14$ yields a $(0, 14)$-path in $L_1$ internally disjoint from $B_Q$). If in addition $(8, 14) \in L_0$, then there can also be no $(0, 8)$-path in $L_0[\{0, 8, 15\} \cup P]$ (otherwise extending along $(8, 14)$ yields a $(0, 14)$-path in $L_0$ internally disjoint from $B_Q$). Across all $2^{21} = 2,097,152$ edge partitions of the $21$ free edges of $\{0, 8, 15\} \cup P$ (with $(0, 15) \in L_0$ and $(8, 15) \in L_1$), $0$ and $8$ avoid a monochromatic path in both $L_0[\{0, 8, 15\} \cup P]$ and $L_1[\{0, 8, 15\} \cup P]$ if and only if $(0, p) \in L_0$ and $(8, p) \in L_1$ for all $5$ vertices $p \in P = \{3, 4, 5, 6, 7\}$ (2,048 assignments; verified in Step 7 of verify_post173_carriers_and_4face.py).

Proof of Theorem 3(3). Let $A_0 \subseteq P$ be the connected component of $0$ in $L_1[\{0\} \cup P]$ and $B_0 = P \setminus A_0$. If $0$ and $8$ have no path in $L_1[\{0, 8\} \cup P]$, then every edge of $B_P$ between $\{0\} \cup A_0$ and $\{8\} \cup B_0$ except $(0, 8)$ lies in $L_0$. If $A_0 \ne \emptyset$ and $B_0 \ne \emptyset$, choosing $p_1 \in A_0$ and $p_2 \in B_0$ gives $(0, p_2), (p_2, p_1), (p_1, 8) \in L_0$, a $3$-edge $(0, 8)$-path in $L_0$. Hence either $A_0 = \emptyset$ (forcing $(0, p) \in L_0$ and $(8, p) \in L_1$ for all $p \in P$) or $A_0 = P$ (forcing $(0, p) \in L_1$ and $(8, p) \in L_0$ for all $p \in P$). In the latter case, $(0, 3) \in L_1$ and $(3, 8) \in L_0$, so placing $(3, 15) \in L_0$ creates the $L_0$-path $0 - 15 - 3 - 8$ while placing $(3, 15) \in L_1$ creates the $L_1$-path $0 - 3 - 15 - 8$. Thus only $(0, p) \in L_0$ and $(8, p) \in L_1$ ($\forall p \in P$) survives. $\blacksquare$

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6. Standalone Verification Suite

The supplemental archive hajos_slack_and_g12_55_verifier.zip contains three standalone, zero-dependency Python standard-library scripts (verify_all.py, verify_post174_carrier_slack_and_hajos.py, and verify_post173_carriers_and_4face.py) that verify all results in $2.8$ seconds total:

1. verify_post174_carrier_slack_and_hajos.py (2.38s): decodes the graph6 strings of $\text{CAND-A}$ and $\text{core21}$, verifies Theorem 1 (including the exact formula $\chi_f(Q^{(r)}) = \frac{17r + 2}{2r}$ via matching primal $2r$-fold $(17r + 2)$-coloring certificates and rational dual vectors $y^{(r)}$ checked on all maximal independent sets for $r = 1..4$, with $\omega = 8$ for $r \ge 2$), runs the unrestricted $d \in \{3, 4\}$ triangulation-complement reproducibility check for Proposition 1, verifies the $4$ copies of $G_{12,55}$ in $\text{CAND-A}$ across all $120$ edge deletions, $20$ vertex deletions, and $120 \times 22 = 2,640$ $(e, w)$ pair deletions ($70 / 120$ intact-copy survival for all $w \notin \{0, 21\}$), and verifies the spherical rotation systems of both $D_{23}$ and the $24$-vertex, $120$-edge witness $D_{24} = (D_{23} - (5, 7)) + (7, 23)$ (Theorems 2a–2d).
2. verify_post173_carriers_and_4face.py (0.45s): verifies the explicit spherical rotation systems ($V - E + F = 2$) of CORE_V52 and Variant 52 ($\text{a52}$, $16/76$, in $4$ flips from BARE), its $26$ newly unlocked $0$-flip contact profiles, its $4$-face coverage bound ($14 < 15$), and the Bichromatic $K_7 - e$ Double-Cut Lemma on Variant 0 (2,048 / 2,097,152 assignments; Theorem 3), alongside the background BARE/ALT certificates [8].
3. verify_all.py: runs both verifiers sequentially and asserts PASS on all checks.

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7. Limitations and Scope

While Theorems 1–2 establish the exact $K_{2,2}$-port structure of $\text{core21}$, the exact fractional chromatic numbers $\chi_f(Q^{(r)}) = \frac{17r + 2}{2r}$ of the coned chain, the single-edge-deletion non-biplanarity ($\operatorname{del}_{\text{biplanar}}(\text{CAND-A}) \ge 2$), the $\{0, 21\}$ core-hitting obstruction on single-vertex splits of $\text{CAND-A}$, and the scope of $1$-vertex Hajós splits (together with the $24$-vertex biplanar split witness $D_{24}$ at $w = 21 \ne u^*$), and Theorem 3 certifies Variant 52 ($\text{a52}$, $16/76$) as biplanar and proves the Bichromatic $K_7 - e$ Double-Cut Lemma on Variant 0 ($16/75$), several problems remain open:

1. Whether a multi-pair split ($s \ge 2$ split vertices per copy of $\text{CAND-A}$), a split of $\mathrm{Haj}(Q_1, \text{CAND-A})$ at $w \in \{0, 21\}$ (or at one of the $50 / 120$ opposite-leaf edge pairs $(e, w)$ where all four copies of $G_{12,55}$ are hit), or an un-coned construction outside $Q^{(r)}$ can yield a biplanar $10$-chromatic graph.
2. Whether Variant 0 ($\text{leaf2\_core\_a0\_long}$, $16/75$) when $(8, 14) \in L_1$ and the two-leaf $17$-vertex, $83$-edge assembly over $C_{15,68}$ are globally non-biplanar across all core flip distances, and whether the full Earth–Moon maximum chromatic number equals $9$, $10$, $11$, or $12$.

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References

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