# AMRAL × Lebesgue Universal Covering — Round 04 ## Candidate-Cover Boundary Compiler and Global Finite Polygon Bracket **Document ID:** AMRAL-LUC-FC-R04 **Version:** v0.1 **Date:** 2026-09-18 **Research status:** Round 04 / Formal attack phase / Candidate-cover finite compiler **Research mode:** Human-Directed + Semi-Autonomous AI Mathematical Research **Research initiation and methodology source:** Neo.K **AI research collaborator and primary executor:** Aletheia / ChatGPT, GPT-5.6 Sol **Parent methodology:** Relational Constraint–Handoff Methodology (RCHM) **Prior documents:** AMRAL-LUC-FC-R00 v0.2; R01 v0.1; R02 v0.1; R03 v0.1 --- # 0. This round's summary verdict Rounds 01–03 have, in sequence, processed: $$ \text{target domain} \to \text{finite legal target dictionary} \to \text{finite placement certificate}. $$ Round 04 attacks the last major infinite-dimensional domain: $$ \boxed{ \text{candidate universal cover }U. } $$ This round obtains the following main results. 1. **All near-optimal universal covers can be placed into a uniform bounded class.** Since a universal cover must accommodate a diameter-$1$ disk, we may translate it so that: $$ B_{1/2}(0)\subset U. $$ If we additionally restrict: $$ \operatorname{Area}(U)\le\bar A $$ where $\bar A$ is any already-proven universal-cover upper bound, then: $$ \boxed{ U\subset B_{2\bar A}(0). } $$ 2. **Construct a candidate-cover finite outer polygon compiler.** For a finite normal grid: $$ u_j, $$ taking only finitely many support samples of $U$ and quantizing all of them upward, we can construct a finite polygon: $$ P_{M,q}(U) $$ such that: $$ \boxed{ U\subseteq P_{M,q}(U). } $$ Because this is an outer approximation, universality is monotone-safe: $$ U\text{ universal} \Rightarrow P_{M,q}(U)\text{ universal}. $$ 3. **Obtain an explicit Hausdorff outer error.** Let: $$ R=2\bar A, $$ $$ \Delta=\frac{\pi}{M}, $$ $$ d_M=2\sin\frac{\Delta}{2}, $$ $$ R_P=\frac{R+q}{\cos\Delta}. $$ Then: $$ \boxed{ U \subseteq P_{M,q}(U) \subseteq U+\varepsilon_{M,q}B } $$ where: $$ \boxed{ \varepsilon_{M,q} = q+ (R+R_P)d_M. } $$ 4. **Obtain an explicit area inflation bound.** $$ \boxed{ \operatorname{Area}(P_{M,q}(U)) \le \operatorname{Area}(U) + \beta_{M,q} } $$ where: $$ \boxed{ \beta_{M,q} = 2\pi R\varepsilon_{M,q} + \pi\varepsilon_{M,q}^2. } $$ 5. **The infinitude of candidate covers is compressed into a genuinely finite polygon dictionary.** Support heights are only permitted to take finitely many grid values, so the entire candidate dictionary is a finite set. 6. **The Lebesgue constant obtains a finite polygon two-sided bracket.** Let: $$ A^{\mathrm{poly}}_{M,q} $$ be the area of the smallest universal polygon in this finite dictionary. Then: $$ \boxed{ A^{\mathrm{poly}}_{M,q} - \beta_{M,q} \le a_{\mathrm{Leb}} \le A^{\mathrm{poly}}_{M,q}. } $$ Since: $$ \beta_{M,q}\to0 $$ as: $$ M\to\infty, \qquad q\to0, $$ this is a global finite-resolution convergence scheme. 7. **The old removable-region method is upgraded to proof-producing surgery.** Tension/slack can continue to be responsible for "finding where is worth cutting," but once the cut is made, whether the result is truly universal is decided by the Round 02–03 fixed-cover global certificate. Therefore, Round 04's main verdict is: $$ \boxed{ \text{CANDIDATE-COVER FINITE COMPILER: CLOSED} } $$ and: $$ \boxed{ \text{GLOBAL FINITE-RESOLUTION PROOF GRAPH: ESTABLISHED} } $$ But this still does not amount to exact finite closure. The remaining core problem becomes: $$ \boxed{ \text{saturation / finite branch pruning / exact lower-bound closure}. } $$ --- # 1. External benchmarks and this round's positioning Among the currently published benchmarks: - Gibbs has proven a planar universal cover with upper bound: $$ 0.8440935944; $$ - In 2026, Mishra pushed the lower bound to: $$ 0.8344; $$ - In 2026, Zeng proposed an exact finite-arc hierarchy: $$ \Lambda_M \to a_{\mathrm{Leb}} $$ and gave: $$ 0 \le a_{\mathrm{Leb}}-\Lambda_M \le CM^{-2}. $$ This round's support-polygon outer hierarchy and Zeng's Reuleaux-type hierarchy **are not the same construction**. Round 04 does not claim: - to be stronger than the Zeng hierarchy; - to have a better convergence order; - to be the first to prove that a finite hierarchy exists. What this round genuinely adds to the AMRAL proof graph is: > Using RCHM's handoff logic, the target / placement certificate from Rounds 01–03 is merged with the candidate-cover outer compiler into a single global finite-resolution pipeline. --- # 2. Anchor lemma: a universal cover must accommodate a radius-$1/2$ disk Let: $$ B_{1/2} $$ be the closed disk of diameter $1$. Since: $$ \operatorname{diam}(B_{1/2})=1, $$ for any universal cover $U$ there must exist an isometry $g$ such that: $$ gB_{1/2} \subseteq U. $$ A disk is invariant under rotation/reflection, so $g$ only changes its center. Hence, after translating $U$, we may assume: $$ \boxed{ B_{1/2}(0) \subseteq U. } $$ This anchor is the key to this round's candidate-cover compactification. --- # 3. Certified upper-bound class Let: $$ \bar A $$ be any already-proven universal-cover upper bound. This round may use Gibbs's: $$ \bar A = 0.8440935944. $$ Since there already exists a universal cover of area: $$ \le\bar A, $$ when seeking: $$ a_{\mathrm{Leb}} $$ we may restrict to: $$ \operatorname{Area}(U)\le\bar A. $$ No universal cover of larger area could possibly improve the infimum. --- # 4. Radius bound from area + anchored disk ## Theorem 4.1 If a convex set $U$ satisfies: $$ B_r(0)\subseteq U $$ and: $$ \operatorname{Area}(U)\le A, $$ then: $$ \boxed{ U\subseteq B_{A/r}(0). } $$ ### Proof Take any: $$ x\in U. $$ Let: $$ R_x=\|x\|. $$ If: $$ R_x\le r, $$ the conclusion is trivial. If: $$ R_x>r, $$ rotate coordinates so that: $$ x=(R_x,0). $$ Since: $$ B_r(0)\subseteq U, $$ the two points: $$ (0,r), \qquad (0,-r) $$ both lie in $U$. By convexity, the triangle: $$ T = \operatorname{conv} \{ x,(0,r),(0,-r) \} $$ is contained in $U$. Its area: $$ \operatorname{Area}(T) = rR_x. $$ Hence: $$ A \ge \operatorname{Area}(U) \ge rR_x. $$ Therefore: $$ R_x\le\frac{A}{r}. $$ Q.E.D. --- ## Corollary 4.2 For a universal cover: $$ r=\frac12. $$ If: $$ \operatorname{Area}(U)\le\bar A, $$ then: $$ \boxed{ U\subseteq B_R(0), \qquad R=2\bar A. } $$ With the Gibbs upper bound: $$ R = 1.6881871888. $$ This is not a tight radius bound, only a proof-safe uniform bound. --- # 5. Candidate-cover compact class Define: $$ \boxed{ \mathcal C_{\bar A} = \left\{ U: \begin{array}{l} U\subset\mathbb R^2\text{ compact convex},\\ B_{1/2}(0)\subseteq U,\\ \operatorname{Area}(U)\le\bar A \end{array} \right\}. } $$ From: $$ B_{1/2} \subseteq U \subseteq B_R, $$ this family is precompact under the Hausdorff topology. Area is continuous with respect to Hausdorff convergence of convex bodies, so the area constraint is closed. Hence: $$ \boxed{ \mathcal C_{\bar A} \text{ is compact}. } $$ All near-optimal universal covers can, after translation, be restricted to this family. --- # 6. Uniform support grid Take: $$ M\ge4. $$ Define: $$ \theta_j = \frac{2\pi j}{M}, \qquad j=0,\ldots,M-1, $$ and: $$ u_j = (\cos\theta_j,\sin\theta_j). $$ The maximum angular distance to the nearest grid direction: $$ \boxed{ \Delta = \frac{\pi}{M}. } $$ And define: $$ \boxed{ d_M = 2\sin\frac{\Delta}{2}. } $$ This is the chord distance between two unit vectors at angular distance $\Delta$. --- # 7. Support-height upward quantization Since: $$ B_{1/2}\subseteq U\subseteq B_R, $$ for every direction: $$ \frac12 \le h_U(u) \le R. $$ Choose: $$ q>0. $$ Define the finite level set: $$ \boxed{ \mathcal H_q = \left\{ \frac12+kq: k=0,1,\ldots,K_q \right\}, } $$ where: $$ K_q = \left\lceil \frac{R-\frac12}{q} \right\rceil +1. $$ For each support sample: $$ h_j = h_U(u_j), $$ define the upward quantization: $$ \boxed{ \widehat h_j = \frac12 + q \left\lceil \frac{ h_j-\frac12 }{ q } \right\rceil. } $$ Hence: $$ h_j \le \widehat h_j < h_j+q, $$ and: $$ \widehat h_j \le R+q. $$ --- # 8. Outer polygon compiler Define: $$ \boxed{ P_{M,q}(U) = \bigcap_{j=0}^{M-1} \left\{ x\in\mathbb R^2: x\cdot u_j \le \widehat h_j \right\}. } $$ --- ## Theorem 8.1: Monotone outer handoff $$ \boxed{ U\subseteq P_{M,q}(U). } $$ ### Proof For any: $$ x\in U, $$ we have: $$ x\cdot u_j \le h_U(u_j) = h_j \le \widehat h_j $$ for all $j$. Hence: $$ x\in P_{M,q}(U). $$ Q.E.D. --- ## Corollary 8.2: universality is monotone-safe If: $$ U $$ is a universal cover, then: $$ \boxed{ P_{M,q}(U) \text{ is also a universal cover}. } $$ because any target that fits into $U$ also fits into the larger: $$ P_{M,q}(U). $$ This is the main reason this round chose an outer compiler rather than general smoothing. --- # 9. Outer polygon radius bound Let: $$ P = P_{M,q}(U). $$ Take any: $$ x\in P, \qquad x\neq0. $$ Choose a grid direction: $$ u_j $$ whose angular distance to: $$ \frac{x}{\|x\|} $$ does not exceed: $$ \Delta. $$ Then: $$ x\cdot u_j \ge \|x\| \cos\Delta. $$ On the other hand: $$ x\cdot u_j \le \widehat h_j \le R+q. $$ So: $$ \boxed{ \|x\| \le R_P := \frac{R+q}{\cos\Delta}. } $$ Hence: $$ \boxed{ P\subseteq B_{R_P}(0). } $$ --- # 10. Hausdorff outer error ## Theorem 10.1 $$ \boxed{ U \subseteq P \subseteq U+\varepsilon_{M,q}B } $$ where: $$ \boxed{ \varepsilon_{M,q} = q + (R+R_P)d_M. } $$ ### Proof We already know: $$ U\subseteq P. $$ So it suffices to bound above: $$ h_P(v)-h_U(v). $$ Take any: $$ v\in S^1. $$ Choose the nearest grid direction: $$ u_j $$ such that: $$ \|v-u_j\| \le d_M. $$ Take any: $$ x\in P. $$ We have: $$ \begin{aligned} x\cdot v &= x\cdot u_j + x\cdot(v-u_j)\\ &\le \widehat h_j + \|x\|\,\|v-u_j\|\\ &\le h_U(u_j) + q + R_Pd_M. \end{aligned} $$ And the support function is $R$-Lipschitz in direction: $$ h_U(u_j) \le h_U(v) + R\|u_j-v\|. $$ So: $$ x\cdot v \le h_U(v) + q + (R+R_P)d_M. $$ Taking the maximum over $x\in P$: $$ h_P(v) \le h_U(v) + \varepsilon_{M,q}. $$ Hence: $$ P \subseteq U+\varepsilon_{M,q}B. $$ Q.E.D. --- # 11. Error vanishes As: $$ M\to\infty, $$ we have: $$ \Delta\to0, $$ $$ d_M\to0, $$ $$ R_P\to R $$ if simultaneously: $$ q\to0. $$ Hence: $$ \boxed{ \varepsilon_{M,q}\to0. } $$ This outer compiler's general worst-case error is conservative: $$ O(M^{-1})+O(q). $$ This round does not claim this convergence order is optimal. --- # 12. Area inflation theorem By Theorem 10.1: $$ P \subseteq U+\varepsilon B, $$ where: $$ \varepsilon = \varepsilon_{M,q}. $$ The planar convex-body Steiner formula: $$ \operatorname{Area}(U+\varepsilon B) = \operatorname{Area}(U) + \varepsilon\operatorname{Per}(U) + \pi\varepsilon^2. $$ And since: $$ U\subseteq B_R, $$ the perimeter of a planar convex body is monotone with respect to inclusion, so: $$ \operatorname{Per}(U) \le 2\pi R. $$ Hence: $$ \boxed{ \operatorname{Area}(P) \le \operatorname{Area}(U) + \beta_{M,q}, } $$ where: $$ \boxed{ \beta_{M,q} = 2\pi R\varepsilon_{M,q} + \pi\varepsilon_{M,q}^2. } $$ and: $$ \boxed{ \beta_{M,q}\to0. } $$ --- # 13. Candidate-cover finite polygon dictionary Define: $$ \boxed{ \mathscr P_{M,q} = \left\{ P(\mathbf b): \mathbf b\in\mathcal H_q^M \right\}, } $$ where: $$ P(\mathbf b) = \bigcap_{j=0}^{M-1} \{x:x\cdot u_j\le b_j\}. $$ Since: $$ b_j\ge\frac12, $$ every polygon contains at least: $$ B_{1/2}(0), $$ so it is nonempty. Because the uniform normals cover the entire circle, and: $$ M\ge4, $$ every polygon is bounded. The dictionary's cardinality: $$ \boxed{ |\mathscr P_{M,q}| \le (K_q+1)^M. } $$ It is typically enormous, but genuinely finite. --- # 14. Every bounded candidate has a dictionary outer representative For any: $$ U\in\mathcal C_{\bar A}, $$ the quantization rule from the previous section produces: $$ P_{M,q}(U) \in \mathscr P_{M,q} $$ such that: $$ \boxed{ U \subseteq P_{M,q}(U) \subseteq U+\varepsilon_{M,q}B } $$ and: $$ \boxed{ \operatorname{Area}(P_{M,q}(U)) \le \operatorname{Area}(U)+\beta_{M,q}. } $$ Hence the infinitude of candidate covers has become a finite outer dictionary at resolution: $$ (M,q). $$ --- # 15. Finite polygon global bracket Define: $$ \boxed{ A^{\mathrm{poly}}_{M,q} = \min \left\{ \operatorname{Area}(P): P\in\mathscr P_{M,q}, \quad P\text{ universal} \right\}. } $$ This minimum exists because the dictionary is finite. The dictionary contains at least one universal polygon: for any already-known universal cover $U$: $$ P_{M,q}(U) $$ is itself universal. --- ## Theorem 15.1: Lebesgue constant finite polygon bracket $$ \boxed{ A^{\mathrm{poly}}_{M,q} - \beta_{M,q} \le a_{\mathrm{Leb}} \le A^{\mathrm{poly}}_{M,q}. } $$ ### Right-hand side $A^{\mathrm{poly}}_{M,q}$ is itself the area of some universal cover, so: $$ a_{\mathrm{Leb}} \le A^{\mathrm{poly}}_{M,q}. $$ ### Left-hand side For any: $$ \eta>0, $$ by the definition of infimum, there exists a universal cover: $$ U_\eta $$ such that: $$ \operatorname{Area}(U_\eta) < a_{\mathrm{Leb}}+\eta. $$ and it may be chosen within: $$ \mathcal C_{\bar A} $$ Its outer polygon: $$ P_{M,q}(U_\eta) $$ is universal, and: $$ \operatorname{Area}(P_{M,q}(U_\eta)) \le \operatorname{Area}(U_\eta) + \beta_{M,q}. $$ Hence: $$ A^{\mathrm{poly}}_{M,q} < a_{\mathrm{Leb}} + \eta + \beta_{M,q}. $$ Letting: $$ \eta\downarrow0, $$ we obtain: $$ A^{\mathrm{poly}}_{M,q} - \beta_{M,q} \le a_{\mathrm{Leb}}. $$ Q.E.D. --- # 16. Convergence Take any sequence: $$ M_n\to\infty, $$ $$ q_n\to0. $$ Then: $$ \beta_{M_n,q_n}\to0. $$ From: $$ A^{\mathrm{poly}}_{M_n,q_n} - \beta_{M_n,q_n} \le a_{\mathrm{Leb}} \le A^{\mathrm{poly}}_{M_n,q_n}, $$ we obtain: $$ \boxed{ A^{\mathrm{poly}}_{M_n,q_n} \to a_{\mathrm{Leb}}. } $$ This is one AMRAL support-polygon global finite-resolution route. This round does not claim that this sequence is: - monotone; - faster than the Zeng hierarchy; - possessed of $O(M^{-2})$ convergence. This round proves only: $$ \boxed{ \text{finite dictionary + explicit vanishing bracket width}. } $$ --- # 17. Fixed-cover worst-margin stability in the cover variable Round 03 defines: $$ W(U) = \max_{K\in\mathcal W_1^0} M_U(K). $$ ## Theorem 17.1 For any compact convex $U,V$: $$ \boxed{ |W(U)-W(V)| \le d_H(U,V). } $$ ### Proof For any fixed target $K$ and fixed placement: in the support residual, $U$ appears only through: $$ -h_U $$ Hence: $$ |M_U(K)-M_V(K)| \le \|h_U-h_V\|_\infty = d_H(U,V). $$ Taking the maximum over $K$ gives the conclusion. Q.E.D. --- # 18. Directionality of the outer compiler's effect on the universality margin If: $$ U\subseteq P, $$ then: $$ h_U\le h_P. $$ So: $$ M_P(K) \le M_U(K) $$ for all $K$. Hence: $$ \boxed{ W(P)\le W(U). } $$ And by the Hausdorff Lipschitz property: $$ W(P) \ge W(U)-\varepsilon_{M,q}. $$ So: $$ \boxed{ W(U)-\varepsilon_{M,q} \le W(P) \le W(U). } $$ This shows the outer compiler cannot manufacture false non-universality. --- # 19. Candidate cell pruning theorem For a fixed support quantization vector: $$ \mathbf b, $$ consider all candidate covers that the same upward quantization maps to: $$ P(\mathbf b) $$ Every such $U$ satisfies: $$ U\subseteq P(\mathbf b). $$ Hence: ## Theorem 19.1 If: $$ P(\mathbf b) $$ has itself already been proven non-universal, then every $U$ in that cell is non-universal. That is: $$ \boxed{ P(\mathbf b)\text{ non-universal} \Rightarrow \text{entire candidate cell pruned}. } $$ This provides a genuine branch-and-bound pruning rule for the candidate space. --- # 20. Why a universal polygon's zero margin creates a verification problem If a polygon $P$ is exactly universal, then typically: $$ W(P)=0 $$ which may be saturated by active targets. If Round 03's finite error upper certificate has no strict negative slack, proving directly that: $$ W(P)\le0. $$ might require infinite refinement. This round therefore introduces a certification buffer. --- # 21. Outer safety inflation Take: $$ \gamma>0. $$ Define: $$ \boxed{ P^\gamma = P+\gamma B. } $$ Its support function: $$ h_{P^\gamma} = h_P+\gamma. $$ Hence for any target: $$ M_{P^\gamma}(K) = M_P(K)-\gamma. $$ So: $$ \boxed{ W(P^\gamma) = W(P)-\gamma. } $$ If $P$ is universal: $$ W(P)\le0, $$ then: $$ \boxed{ W(P^\gamma)\le-\gamma. } $$ Hence there exists strict certificate slack. --- # 22. Safety inflation area overhead Since: $$ P\subseteq B_{R_P}, $$ we have: $$ \operatorname{Per}(P)\le2\pi R_P. $$ Steiner formula: $$ \operatorname{Area}(P^\gamma) = \operatorname{Area}(P) + \gamma\operatorname{Per}(P) + \pi\gamma^2. $$ Hence: $$ \boxed{ \operatorname{Area}(P^\gamma) \le \operatorname{Area}(P) + 2\pi R_P\gamma + \pi\gamma^2. } $$ Combining with the outer polygon bound: $$ \operatorname{Area}(P^\gamma) \le \operatorname{Area}(U) + \beta_{M,q} + 2\pi R_P\gamma + \pi\gamma^2. $$ So proof-producing universal certificates can use an arbitrarily small: $$ \gamma. $$ --- # 23. Practical finite upper hierarchy Round 03's fixed-cover certificate, applied to the strict margin: $$ W(P^\gamma)\le-\gamma $$ yields a negative upper bound at sufficiently fine target / placement resolution. Hence: > For every true universal dictionary polygon $P$, its outer inflation by any $\gamma>0$ can ultimately be verified as universal by the Round 02–03 certificate using finite data. So a practical finite upper search need not wait for an exact zero-margin classification. --- # 24. Practical finite lower hierarchy If a polygon $P$ is non-universal, then: $$ W(P)>0. $$ Since Round 01 has already proven the target family compact, the worst-case maximum is attained. The Round 02–03 lower certificate converges, as resolution is refined, to: $$ W(P). $$ So: > Every true non-universal dictionary polygon can ultimately be excluded by a finite positive witness certificate. Because each layer of the dictionary is finite, for any area threshold below: $$ A^{\mathrm{poly}}_{M,q} $$ all candidate polygons can ultimately be excluded, one by one, in finitely many steps. This is the theoretical interface for lower-bound exhaustive certification. --- # 25. Global lower-certificate rule If, for some: $$ L $$ it has been proven that: > all: $$ P\in\mathscr P_{M,q} $$ with: $$ \operatorname{Area}(P) < L+\beta_{M,q} $$ are non-universal, then: $$ \boxed{ a_{\mathrm{Leb}} \ge L. } $$ ### Proof Suppose, for contradiction: $$ a_{\mathrm{Leb}}0, $$ an inward cut can be made under that finite placed family. At the time, this could only establish: > For the current finite test family, this cut is safe. It could not derive: > that after the cut it remains a Lebesgue universal cover. --- # 28. Certificate-safe boundary surgery Round 04 now upgrades Q6 into: ## Search stage Using: - active tension; - finite placed witnesses; - support slack; - low-activity boundary directions; propose an inward cut: $$ U' = U \cap \left\{ x: x\cdot v \le h_U(v)-\tau \right\}. $$ ## Proof stage Do not trust the search heuristic. Apply the Round 02 + Round 03 fixed-cover global certificate directly to: $$ U' $$ If we obtain: $$ \boxed{ W^+(U')\le0, } $$ then: $$ U' $$ is rigorously proven universal. If, at the same time: $$ \operatorname{Area}(U') < \operatorname{Area}(U), $$ we obtain a genuine new upper-bound candidate. Hence: $$ \boxed{ \text{tension search} \to \text{boundary surgery} \to \text{global independent certificate} } $$ becomes a replayable research cycle. --- # 29. Area–coverage dual tension This round formally organizes the old intuition into two quantities. ## Coverage margin $$ W(U). $$ Universal condition: $$ W(U)\le0. $$ ## Area objective $$ A(U) = \operatorname{Area}(U). $$ An inward perturbation of the candidate body: $$ U_\tau \subset U $$ produces: $$ \Delta A = A(U)-A(U_\tau)>0 $$ and: $$ \Delta W = W(U_\tau)-W(U)\ge0 $$ which typically holds. Define the finite-difference tension ratio: $$ \boxed{ \rho = \frac{ \Delta W }{ \Delta A }. } $$ The search algorithm can preferentially select inward surgeries with low: $$ \rho $$ But: $$ \rho $$ is responsible only for candidate generation. Final correctness is always decided by: $$ W^+(U_\tau)\le0 $$ --- # 30. Current status of the three-layer finite compiler After combining Rounds 01–04: ## Layer T: target $$ \text{all diameter-}\le1\text{ targets} $$ is compressed into: $$ \text{finite legal constant-width dictionary} + \varepsilon_{\mathrm{target}}. $$ ## Layer G: placement $$ O(2)\times\mathbb R^2\times S^1 $$ is compressed into: $$ \text{finite reflection/orientation/support LPs} + \varepsilon_{\mathrm{placement}}. $$ ## Layer U: candidate cover $$ \mathcal C_{\bar A} $$ is compressed into: $$ \mathscr P_{M,q} + \varepsilon_{\mathrm{cover}} + \beta_{\mathrm{area}}. $$ Hence: $$ \boxed{ \text{target infinity} + \text{placement infinity} + \text{candidate-cover infinity} } $$ have all now entered the same finite-resolution proof graph. --- # 31. This is still not exact finite closure RCHM must distinguish between: $$ \text{finite-resolution convergence} $$ and: $$ \text{finite exact closure}. $$ This round only proves that: for any precision, a finite dictionary / finite certificates exist. It still has not proven that there exists some fixed: $$ (M,q,N,\delta,\ldots) $$ such that: $$ a_{\mathrm{Leb}} $$ becomes, at some finite layer, exactly equal to that layer's answer. So: $$ \boxed{ \text{finite-resolution global closure: YES} } $$ but: $$ \boxed{ \text{finite exact saturation: OPEN}. } $$ --- # 32. Relationship to the Zeng hierarchy Zeng's 2026 hierarchy has publicly claimed: $$ a_{\mathrm{Leb}} = \lim_{M\to\infty}\Lambda_M $$ and: $$ 0 \le a_{\mathrm{Leb}}-\Lambda_M \le CM^{-2}. $$ Round 04's: $$ A^{\mathrm{poly}}_{M,q} $$ route runs in a different direction: - Zeng: a finite Reuleaux / finite-arc variational hierarchy; - AMRAL Round 04: a finite support-normal outer polygon dictionary + full placement verifier. Hence, for now, this round claims only to be: > an independent finite representation / cross-check route. Whether there is genuine mathematical novelty must wait until Zeng's full text and external review have been fully compared. --- # 33. Sanity example Using a legal candidate support: $$ h_U(\theta) = 0.515 + 0.01\cos2\theta. $$ its curvature density: $$ h_U+h_U'' = 0.515 - 0.03\cos2\theta > 0. $$ and: $$ h_U(\theta) \ge0.505>\frac12, $$ so: $$ B_{1/2}(0)\subset U. $$ its exact Fourier-area formula: $$ \operatorname{Area}(U) = \frac12 \int_0^{2\pi} (h_U^2-h_U'^2)\,d\theta. $$ This round uses: $$ M=128, \qquad q=5\times10^{-4} $$ to build the outward-quantized polygon. Numerical check: 1. all boundary samples satisfy the polygon halfspaces; 2. the polygon's numerical Hausdorff excess is smaller than the analytic: $$ \varepsilon_{M,q}; $$ 3. the polygon's area inflation is smaller than: $$ \beta_{M,q}. $$ `SANITY: PASS` The analytic bound uses the global safety radius: $$ R=2\bar A $$ so it is very conservative; the actual error is far smaller than the theorem bound. --- # 34. COMPUTE-DEFERRED The following tasks can now genuinely be handed off to local compute. ## C04-1: Known upper-cover polygonization Given the Gibbs 2018 candidate cover as input: - build an exact / interval support oracle; - outer-approximate it into: $$ \mathscr P_{M,q}; $$ - re-verify universality using the Round 03 certificate. ## C04-2: Candidate-cell pruning Enumerate low support-resolution cells: - outer polygon; - fixed-cover nonuniversality certificate; - prune the entire cell. ## C04-3: Certificate-safe inward surgery For the current best universal candidate: 1. use support slack to find a cut; 2. make the cut; 3. recompute the area; 4. apply Round 03 global verification; 5. if it passes, record the new upper-bound candidate. ## C04-4: Area lower-cell bound For each support quantization cell, establish a certified lower bound on: $$ \inf \operatorname{Area}(U) $$ for use in branch-and-bound ordering. All marked: `COMPUTE-DEFERRED` --- # 35. RCHM freedom ledger After Round 03: $$ \mathcal F_3 = \{ \text{candidate-cover boundary}, \text{resolution limit}, \text{saturation} \}. $$ Round 04: ## Candidate-cover boundary From: $$ \text{infinite-dimensional convex-body domain} $$ compressed into: $$ \boxed{ \text{finite outward support-height dictionary} } $$ with: $$ \varepsilon_{M,q}, \qquad \beta_{M,q}. $$ ## Resolution limit Still requires: $$ M\to\infty, \qquad q\to0. $$ So this is not an exact finite layer. ## Saturation The largest current obstruction has now clearly become: $$ \boxed{ \text{Can all active branches / the exact extremizer be identified at a finite layer?} } $$ --- # 36. Round 05 assigned topic ## AMRAL-LUC-FC-R05 ### Global Candidate-Cell Branch-and-Bound and Saturation Gate The next round no longer establishes new representations. It begins research directly on: 1. area lower bounds for candidate support cells; 2. outer nonuniversality pruning; 3. universal upper candidates; 4. active-cell refinement; 5. whether finite branch stabilization exists; 6. the cross-map between the Reuleaux hierarchy and the support-polygon hierarchy; 7. exact-saturation failure modes. Round 05's core question: $$ \boxed{ \text{When can a finite-resolution hierarchy stop, rather than only ever refining indefinitely?} } $$ This is precisely: $$ \boxed{ \text{Saturation Gate}. } $$ --- # 37. Reproducibility checklist ## Universal disk anchor `PROVED` ## Uniform radius bound `PROVED` ## Finite outward support compiler `PROVED` ## Universality monotonicity `PROVED` ## Explicit Hausdorff error `PROVED` ## Explicit area inflation `PROVED` ## Finite candidate dictionary `PROVED` ## Global finite polygon bracket `PROVED` ## Cover-margin Hausdorff stability `PROVED` ## Candidate-cell prune rule `PROVED` ## Certificate-safe surgery `PROOF SCHEMA ESTABLISHED` ## Numerical sanity `PASS` ## New numerical bound `NONE` ## Exact finite saturation `OPEN` --- # 38. Shortest handoff conclusion Round 04 establishes: $$ \boxed{ U \to P_{M,q}(U) \in \mathscr P_{M,q} } $$ where: $$ U\subseteq P_{M,q}(U), $$ $$ d_H(U,P_{M,q}(U)) \le \varepsilon_{M,q}, $$ and: $$ \operatorname{Area}(P_{M,q}(U)) - \operatorname{Area}(U) \le \beta_{M,q}. $$ Hence: $$ \boxed{ A^{\mathrm{poly}}_{M,q} - \beta_{M,q} \le a_{\mathrm{Leb}} \le A^{\mathrm{poly}}_{M,q}. } $$ Rounds 01–04 have, at this point, completed the finite-resolution compiler for all three layers: $$ \boxed{ \text{target} \to \text{placement} \to \text{candidate cover} } $$ The next phase no longer asks: > How to finitize? but instead asks: $$ \boxed{ \text{How can we prove that the finite branches have been saturated?} } $$ --- # References 1. J. C. Baez, K. Bagdasaryan, P. Gibbs, *The Lebesgue Universal Covering Problem*, arXiv:1502.01251. 2. P. Gibbs, *An Upper Bound for Lebesgue's Covering Problem*, arXiv:1810.10089, 2018. 3. U. Mishra, *Curves of constant width and Lebesgue's covering problem*, arXiv:2608.30538, 2026. 4. S. Zeng, *An exact hierarchy for Lebesgue's universal covering constant and a certified 0.834 lower bound*, arXiv:2609.01284, 2026. 5. R. Schneider, *Convex Bodies: The Brunn–Minkowski Theory*. 6. Neo.K + Aletheia, *AMRAL × Lebesgue Universal Covering — Round 00–03*, 2026-09-18. --- # 39. Declaration This round does not claim to solve the Lebesgue universal covering problem. This round does not improve on the currently published upper and lower bounds. What this round accomplishes is: $$ \boxed{ \text{candidate-cover infinite domain} \to \text{finite outward polygon dictionary} + \text{explicit geometric error} + \text{explicit area error}. } $$ Combined with Rounds 01–03, AMRAL now has a global finite-resolution proof architecture. Whether the convergence hierarchy can be further compressed into finite exact closure is left to the saturation program after Round 05.