# AMRAL × Lebesgue Universal Covering — Round 03
## Signed Placement Margin and Full Fixed-Cover Finite Certificate

**Document ID:** AMRAL-LUC-FC-R03  
**Version:** v0.1  
**Date:** 2026-09-18  
**Research status:** Round 03 / Formal push / Fixed-cover finite certificate  
**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)  
**Prerequisite documents:** AMRAL-LUC-FC-R00 v0.2; R01 v0.1; R02 v0.1  

---

# 0. Summary and Verdict for This Round

Round 02 already compressed every target shape of diameter at most one into a finite legal dictionary:

$$
\mathscr D_{N,\delta,q},
$$

and gave an explicit Hausdorff error:

$$
\varepsilon_{\mathrm{shape}}.
$$

Round 03 handles the remaining placement continuum:

$$
Q\in O(2),
\qquad
t\in\mathbb R^2,
\qquad
u\in S^1.
$$

This round obtains the following main results.

1. **Translation for a fixed orientation admits a complete dual formulation.**  
   The continuous translation minimax:

   $$
   \min_t
   \max_{u\in S^1}
   [
   a_Q(u)+t\cdot u
   ]
   $$

   equals a zero-barycenter probability-measure dual.

2. **The exact continuous witness for a fixed orientation needs only two or three directions.**  
   By subgradient + Carathéodory:
   - 2-active: must be an antipodal pair;
   - 3-active: the convex hull of the three directions contains the origin.

3. **A branch description from Round 01 is corrected.**  
   Round 01 loosely mentioned one/two/three-active branches; one-active is in fact impossible.  
   This round formally records the correction, without retroactively rewriting history.

4. **The support-direction continuum can be rigorously sandwiched by a finite LP.**  
   For $M$ uniformly spaced directions:

   $$
   m_M(Q)
   \le
   m(Q)
   \le
   m_M(Q)+\eta_M.
   $$

5. **The orientation continuum can be rigorously sandwiched by a finite grid.**  
   Using $P$ orientation samples for each reflection branch gives:

   $$
   G-\rho_P
   \le
   M_U(K)
   \le
   G+\eta_M.
   $$

6. **Reflection is no longer an implicit condition.**  
   $O(2)$ is explicitly split into:

   $$
   \sigma\in\{+1,-1\}
   $$

   and:

   $$
   \phi\in S^1.
   $$

7. **The Round 02 shape dictionary and the Round 03 placement compiler can be merged into a complete fixed-cover certificate.**

This round therefore reaches the verdict:

$$
\boxed{
\text{FIXED-COVER FINITE CERTIFICATE SCHEMA: CLOSED}
}
$$

But:

$$
\boxed{
\text{GLOBAL LEBESGUE OPTIMIZATION: OPEN}
}
$$

because the candidate cover $U$ itself remains an infinite-dimensional optimization variable.

---

# 1. Normalization of the Fixed Candidate Cover

Let:

$$
U\subset\mathbb R^2
$$

be a nonempty compact convex candidate cover.

The placement margin is invariant under translation of $U$.

If:

$$
U_c=U-c,
$$

then:

$$
M_{U_c}(K)=M_U(K).
$$

Hence we may freely take:

$$
c\in U
$$

and assume:

$$
0\in U.
$$

Define a known radius bound:

$$
R_U
\ge
\max_{x\in U}\|x\|.
$$

If only the diameter is known:

$$
D_U=\operatorname{diam}(U),
$$

then after choosing any $c\in U$ it is safe to use:

$$
R_U\le D_U.
$$

For a centered unit constant-width target:

$$
K,
$$

the Steiner point lies in $K$, so:

$$
0\in K.
$$

and:

$$
\operatorname{diam}(K)=1,
$$

so it is safe to take:

$$
R_K\le1.
$$

Later sections may use:

$$
L
=
R_K+R_U.
$$

---

# 2. Signed Placement Margin

For a fixed target $K$ and candidate cover $U$, define:

$$
\boxed{
M_U(K)
=
\min_{Q\in O(2)}
\min_{t\in\mathbb R^2}
\max_{\theta\in S^1}
\left[
h_K(Q^Tu_\theta)
+t\cdot u_\theta
-h_U(\theta)
\right].
}
$$

Then:

$$
M_U(K)\le0
$$

if and only if $K$ can be placed inside $U$ via an isometry.

If:

$$
M_U(K)<0,
$$

then there exists uniform support slack.

---

# 3. The Two Branches of $O(2)$

Fix a reflection:

$$
S(x_1,x_2)=(x_1,-x_2).
$$

Every:

$$
Q\in O(2)
$$

can be written as:

$$
Q_{\sigma,\phi}
=
R_\phi S^{(1-\sigma)/2},
$$

where:

$$
\sigma\in\{+1,-1\},
\qquad
\phi\in[0,2\pi).
$$

For a support angle $\theta$:

if:

$$
\sigma=+1,
$$

then:

$$
h_K(Q^Tu_\theta)
=
h_K(\theta-\phi).
$$

if:

$$
\sigma=-1,
$$

then:

$$
h_K(Q^Tu_\theta)
=
h_K(\phi-\theta).
$$

Hence reflection reduces to just two discrete branches.

---

# 4. The Continuous Translation Problem for Fixed Orientation

Fix:

$$
(\sigma,\phi).
$$

Define:

$$
a_{\sigma,\phi}(\theta)
=
h_K(Q_{\sigma,\phi}^Tu_\theta)
-h_U(\theta).
$$

and further define:

$$
F_{\sigma,\phi}(t)
=
\max_{\theta\in S^1}
[
a_{\sigma,\phi}(\theta)
+t\cdot u_\theta
].
$$

The fixed-orientation margin:

$$
\boxed{
m_{\sigma}(\phi)
=
\min_{t\in\mathbb R^2}
F_{\sigma,\phi}(t).
}
$$

So:

$$
M_U(K)
=
\min_{\sigma\in\{\pm1\}}
\min_{\phi\in S^1}
m_\sigma(\phi).
$$

---

# 5. Existence of the Translation Minimizer

Since:

$$
|h_K|\le R_K,
$$

$$
|h_U|\le R_U,
$$

we have:

$$
|a_{\sigma,\phi}(\theta)|
\le
L.
$$

For any:

$$
t\neq0,
$$

take:

$$
u_\theta=\frac{t}{\|t\|}.
$$

Then:

$$
F_{\sigma,\phi}(t)
\ge
\|t\|-L.
$$

and:

$$
F_{\sigma,\phi}(0)\le L.
$$

So:

$$
F_{\sigma,\phi}(t)\to+\infty
$$

as:

$$
\|t\|\to\infty.
$$

Hence a minimizer:

$$
t^\star_{\sigma,\phi}
$$

exists.

and any minimizer may be taken to satisfy:

$$
\boxed{
\|t^\star_{\sigma,\phi}\|
\le2L.
}
$$

---

# 6. Continuous Translation Dual Theorem

Let:

$$
\mathcal P_0(S^1)
=
\left\{
\mu:
\begin{array}{l}
\mu\text{ is a probability measure on }S^1,\\
\int_{S^1}u\,d\mu(u)=0
\end{array}
\right\}.
$$

## Theorem 6.1

$$
\boxed{
m_\sigma(\phi)
=
\max_{\mu\in\mathcal P_0(S^1)}
\int_{S^1}
a_{\sigma,\phi}(u)
\,d\mu(u).
}
$$

### First Direction

For any:

$$
\mu\in\mathcal P_0(S^1)
$$

and any $t$:

$$
\begin{aligned}
\int a\,d\mu
&=
\int
(a+t\cdot u)
\,d\mu\\
&\le
\max_u
(a+t\cdot u).
\end{aligned}
$$

Taking the minimum over $t$:

$$
\int a\,d\mu
\le
m_\sigma(\phi).
$$

Hence the dual value does not exceed the primal value.

---

### Second Direction

Let:

$$
t^\star
$$

be the primal minimizer.

The active direction set:

$$
\mathcal A
=
\left\{
u:
a(u)+t^\star\cdot u
=
m_\sigma(\phi)
\right\}.
$$

By convex subgradient optimality:

$$
0
\in
\partial F(t^\star)
=
\operatorname{conv}(\mathcal A).
$$

So there exist probability weights:

$$
\lambda_i
$$

and active directions:

$$
u_i\in\mathcal A
$$

such that:

$$
\sum_i\lambda_i u_i=0.
$$

Define:

$$
\mu^\star
=
\sum_i\lambda_i\delta_{u_i}.
$$

Then:

$$
\begin{aligned}
\int a\,d\mu^\star
&=
\sum_i\lambda_i
[
m_\sigma(\phi)-t^\star\cdot u_i]\\
&=
m_\sigma(\phi).
\end{aligned}
$$

So the dual value is at least the primal value.

Q.E.D.

---

# 7. The Exact Continuous Witness Needs Only Two or Three Directions

From:

$$
0
\in
\operatorname{conv}(\mathcal A)
\subset\mathbb R^2
$$

and the Carathéodory theorem, at most three active directions may be chosen.

A single active direction is impossible, because:

$$
u\neq0.
$$

Hence only the following exist:

$$
\boxed{
2\text{-active}
}
$$

or:

$$
\boxed{
3\text{-active}.
}
$$

---

## 7.1 Two-Active Branch

If:

$$
0
=
\lambda u_1
+
(1-\lambda)u_2,
$$

the two unit vectors must be antipodal:

$$
u_2=-u_1
$$

and:

$$
\lambda=\frac12.
$$

---

## 7.2 Three-Active Branch

The three directions:

$$
u_1,u_2,u_3
$$

must satisfy:

$$
0
\in
\operatorname{conv}\{u_1,u_2,u_3\}.
$$

equivalently, the three directions do not all lie in some open semicircle.

Their barycentric weights:

$$
\lambda_i
$$

give the exact translation dual witness.

---

# 8. Round 01 Correction Ledger

Round 01 Section 22 once loosely listed the translation active-set branches as:

- one-active degeneracy;
- two-active balance;
- three-active balance.

This round corrects this to:

$$
\boxed{
\text{one-active branch impossible}.
}
$$

The correct classification:

$$
\boxed{
\text{2-active antipodal}
\quad\text{or}\quad
\text{3-active origin-enclosing}.
}
$$

Status:

`R01-CORRECTION-001`

This correction does not affect Round 01's main theorem of "at most three active directions"; it only corrects the branch taxonomy.

---

# 9. Finite Support-Direction Grid

Take an even number:

$$
M\ge4.
$$

Define:

$$
\theta_j
=
\frac{2\pi j}{M},
\qquad
j=0,\ldots,M-1.
$$

The maximum nearest angular distance:

$$
\boxed{
\Delta_\theta
=
\frac{\pi}{M}.
}
$$

Define the finite-grid translation value:

$$
\boxed{
m_{\sigma,M}(\phi)
=
\min_{t\in\mathbb R^2}
\max_{0\le j<M}
[
a_{\sigma,\phi}(\theta_j)
+t\cdot u_{\theta_j}
].
}
$$

Clearly:

$$
\boxed{
m_{\sigma,M}(\phi)
\le
m_\sigma(\phi).
}
$$

because the finite grid takes only a subset of the continuous maximum.

---

# 10. Finite-Grid Primal LP

Introduce a scalar:

$$
z.
$$

Then:

$$
m_{\sigma,M}(\phi)
$$

is the following finite LP:

$$
\boxed{
\begin{array}{ll}
\text{minimize}
&
z
\\
\text{subject to}
&
a_{\sigma,\phi}(\theta_j)
+t\cdot u_{\theta_j}
\le
z,
\quad
j=0,\ldots,M-1.
\end{array}
}
$$

The only variables are:

$$
(t_x,t_y,z)\in\mathbb R^3.
$$

---

# 11. Finite-Grid Dual LP

The dual is:

$$
\boxed{
\begin{array}{ll}
\text{maximize}
&
\displaystyle
\sum_{j=0}^{M-1}
\lambda_j
a_{\sigma,\phi}(\theta_j)
\\
\text{subject to}
&
\lambda_j\ge0,
\\
&
\displaystyle
\sum_j\lambda_j=1,
\\
&
\displaystyle
\sum_j\lambda_j u_{\theta_j}=0.
\end{array}
}
$$

Hence:

$$
m_{\sigma,M}(\phi)
$$

can itself already be verified by a finite primal-dual certificate.

An extreme dual solution still needs at most three nonzero:

$$
\lambda_j.
$$

So the translation certificate for every sampled orientation can be compressed into:

- a 2-direction antipodal witness; or
- a 3-direction barycentric witness.

---

# 12. Finite-Grid Translation Bound

Let:

$$
t_M^\star
$$

be the finite LP minimizer.

For any $t\neq0$, there exists a grid direction $u_{\theta_j}$ whose angular distance to $t/\|t\|$ is at most:

$$
\Delta_\theta.
$$

Hence:

$$
t\cdot u_{\theta_j}
\ge
\|t\|
\cos\Delta_\theta.
$$

and:

$$
a_{\sigma,\phi}(\theta_j)\ge-L.
$$

Therefore:

$$
F_M(t)
\ge
\|t\|
\cos\Delta_\theta
-L.
$$

and:

$$
F_M(0)\le L.
$$

So:

$$
\boxed{
\|t_M^\star\|
\le
\frac{
2L
}{
\cos\Delta_\theta
}.
}
$$

---

# 13. Angular Lipschitz Bound

The support function of a body contained in a ball of radius $R$ satisfies:

$$
|h(\theta)-h(\psi)|
\le
R|\theta-\psi|.
$$

Therefore:

$$
a_{\sigma,\phi}
$$

is $L$-Lipschitz:

$$
|a(\theta)-a(\psi)|
\le
L|\theta-\psi|.
$$

For the finite-grid minimizer:

$$
t_M^\star,
$$

the function:

$$
a(\theta)+t_M^\star\cdot u_\theta
$$

has angular Lipschitz constant at most:

$$
L+\|t_M^\star\|.
$$

So:

$$
m_\sigma(\phi)
\le
m_{\sigma,M}(\phi)
+
\eta_M,
$$

where:

$$
\boxed{
\eta_M
=
L
\left(
1+
\frac{2}{\cos(\pi/M)}
\right)
\frac{\pi}{M}.
}
$$

Combining with the lower bound:

$$
\boxed{
m_{\sigma,M}(\phi)
\le
m_\sigma(\phi)
\le
m_{\sigma,M}(\phi)+\eta_M.
}
$$

This is Round 03's support-direction finite certificate.

---

# 14. Using the Actual $t_M$ to Improve the Bound

If the certificate itself already provides a verified:

$$
\|t_M^\star\|
\le
T_M,
$$

then one may instead use:

$$
\boxed{
\eta_M^{\mathrm{local}}
=
(L+T_M)\frac{\pi}{M}.
}
$$

Hence the global:

$$
\eta_M
$$

is only a universal safe bound that does not require solving for the translation norm.

---

# 15. Orientation Lipschitz Theorem

Fix the reflection branch $\sigma$.

For:

$$
\phi,\psi\in S^1,
$$

only the target support term changes.

Since:

$$
K\subseteq B(0,R_K),
$$

we have:

$$
\left|
h_K(Q_{\sigma,\phi}^Tu)
-
h_K(Q_{\sigma,\psi}^Tu)
\right|
\le
R_K|\phi-\psi|.
$$

So:

$$
\boxed{
|m_\sigma(\phi)-m_\sigma(\psi)|
\le
R_K|\phi-\psi|.
}
$$

For a centered unit target:

$$
R_K\le1.
$$

---

# 16. Finite Orientation Grid

Take:

$$
P\ge2.
$$

Define:

$$
\phi_\ell
=
\frac{2\pi\ell}{P},
\qquad
\ell=0,\ldots,P-1.
$$

The nearest orientation gap:

$$
\boxed{
\Delta_\phi
=
\frac{\pi}{P}.
}
$$

Define the exact finite-grid minimum:

$$
G_{M,P}(K,U)
=
\min_{\sigma\in\{\pm1\}}
\min_{0\le\ell<P}
m_{\sigma,M}(\phi_\ell).
$$

---

# 17. Full Placement Bracket

## Theorem 17.1

$$
\boxed{
G_{M,P}
-
\rho_P
\le
M_U(K)
\le
G_{M,P}
+
\eta_M,
}
$$

where:

$$
\boxed{
\rho_P
=
R_K\frac{\pi}{P}
\le
\frac{\pi}{P}.
}
$$

### Right-Hand Side

Take the achieving sampled orientation:

$$
(\widehat\sigma,\widehat\phi).
$$

Then:

$$
M_U(K)
\le
m_{\widehat\sigma}(\widehat\phi)
\le
m_{\widehat\sigma,M}(\widehat\phi)
+\eta_M
=
G_{M,P}+\eta_M.
$$

---

### Left-Hand Side

Let:

$$
(\sigma^\star,\phi^\star)
$$

be the continuous orientation minimizer.

Take the nearest orientation sample:

$$
\phi_\ell
$$

such that:

$$
|\phi_\ell-\phi^\star|
\le
\Delta_\phi.
$$

By orientation Lipschitz:

$$
m_{\sigma^\star}(\phi_\ell)
\le
M_U(K)
+
\rho_P.
$$

and:

$$
m_{\sigma^\star,M}(\phi_\ell)
\le
m_{\sigma^\star}(\phi_\ell).
$$

Therefore:

$$
G_{M,P}
\le
M_U(K)+\rho_P.
$$

that is:

$$
G_{M,P}-\rho_P
\le
M_U(K).
$$

Q.E.D.

---

# 18. Numerical / Interval LP Enclosure Version

In practice, the certificate does not require the LP value to be solved exactly in floating point.

For each:

$$
(\sigma,\ell)
$$

it suffices to provide a verified interval:

$$
m^-_{\sigma,\ell}
\le
m_{\sigma,M}(\phi_\ell)
\le
m^+_{\sigma,\ell}.
$$

This can be produced by:

- outward-rounded interval arithmetic;
- exact rationalized trigonometric enclosure;
- a verified primal feasible point;
- verified dual feasible weights;
- an independent LP checker.

Define:

$$
G^-
=
\min_{\sigma,\ell}
m^-_{\sigma,\ell},
$$

$$
G^+
=
\min_{\sigma,\ell}
m^+_{\sigma,\ell}.
$$

Then:

$$
\boxed{
G^- -\rho_P
\le
M_U(K)
\le
G^+ +\eta_M.
}
$$

This is the practical certificate form.

---

# 19. Primal-Dual Finite Witness Schema

For each sampled orientation, one may store:

## Primal Witness

$$
(t_x,t_y,z)
$$

satisfying all of:

$$
a_j+t\cdot u_j\le z.
$$

This provides:

$$
m_{\sigma,M}(\phi_\ell)\le z.
$$

## Dual Witness

At most three:

$$
(j_1,j_2,j_3)
$$

and:

$$
\lambda_i\ge0,
$$

such that:

$$
\sum_i\lambda_i=1,
$$

$$
\sum_i\lambda_i u_{j_i}=0.
$$

Then:

$$
\sum_i\lambda_i a_{j_i}
\le
m_{\sigma,M}(\phi_\ell).
$$

If the gap between the primal / dual objectives is less than:

$$
\gamma,
$$

then a certified LP interval width is obtained:

$$
\le\gamma.
$$

---

# 20. Merging with the Round 02 Finite Shape Dictionary

Round 02 provides the finite legal dictionary:

$$
\mathscr D
$$

and the shape error:

$$
\varepsilon_{\mathrm{shape}}.
$$

That is, for every genuine centered unit constant-width body $K$ there exists:

$$
\widehat K\in\mathscr D
$$

such that:

$$
d_H(K,\widehat K)
\le
\varepsilon_{\mathrm{shape}}.
$$

and:

$$
|M_U(K)-M_U(\widehat K)|
\le
d_H(K,\widehat K).
$$

So:

$$
\boxed{
|M_U(K)-M_U(\widehat K)|
\le
\varepsilon_{\mathrm{shape}}.
}
$$

---

# 21. Full Fixed-Cover Worst-Case Margin

Define:

$$
\boxed{
W(U)
=
\max_{K\in\mathcal W_1^0}
M_U(K).
}
$$

By Round 01:

$$
U\text{ universal}
\iff
W(U)\le0.
$$

For each:

$$
\widehat K\in\mathscr D
$$

let:

$$
L_{\widehat K}
=
G^-_{\widehat K}
-
\rho_P,
$$

$$
U_{\widehat K}
=
G^+_{\widehat K}
+
\eta_M.
$$

Then:

$$
L_{\widehat K}
\le
M_U(\widehat K)
\le
U_{\widehat K}.
$$

Therefore:

$$
\boxed{
\max_{\widehat K\in\mathscr D}
L_{\widehat K}
\le
W(U)
\le
\max_{\widehat K\in\mathscr D}
U_{\widehat K}
+
\varepsilon_{\mathrm{shape}}.
}
$$

This is Round 03's full fixed-cover certificate theorem.

---

# 22. Universal Certificate Rule

If:

$$
\boxed{
\max_{\widehat K\in\mathscr D}
U_{\widehat K}
+
\varepsilon_{\mathrm{shape}}
\le0,
}
$$

then:

$$
W(U)\le0.
$$

Therefore:

$$
\boxed{
U\text{ is rigorously universal}.
}
$$

This conclusion simultaneously covers:

- all planar targets with diameter $\le1$;
- all constant-width targets;
- all reflections;
- all rotations;
- all translations;
- all support directions;

with all error sources fully itemized.

---

# 23. Non-Universal Certificate Rule

If there exists a legal dictionary shape:

$$
\widehat K\in\mathscr D
$$

such that:

$$
\boxed{
L_{\widehat K}>0,
}
$$

then:

$$
M_U(\widehat K)>0.
$$

so this target cannot be placed inside $U$.

Therefore:

$$
\boxed{
U\text{ is rigorously non-universal}.
}
$$

No shape-net error is needed, because $\widehat K$ is itself a legal target.

---

# 24. Ambiguous Zone

If:

$$
\max L_{\widehat K}\le0
$$

but:

$$
\max U_{\widehat K}
+
\varepsilon_{\mathrm{shape}}
>0,
$$

then the finite certificate cannot yet decide.

At this point one may refine by:

1. increasing the shape compiler $N$;
2. decreasing $\delta$;
3. decreasing the coefficient grid $q$;
4. increasing the support grid $M$;
5. increasing the orientation grid $P$;
6. lowering the LP primal-dual gap;
7. performing local refinement only on near-active dictionary cells.

The system therefore naturally supports adaptive branch-and-bound.

---

# 25. Error Ledger

The main error sources of the fixed-cover universality upper certificate:

## Shape Approximation

$$
\varepsilon_{\mathrm{shape}}.
$$

## Support-Direction Discretization

$$
\eta_M
=
L
\left(
1+\frac{2}{\cos(\pi/M)}
\right)
\frac{\pi}{M}.
$$

## Finite LP Certification

already included in:

$$
G^+.
$$

Note:

orientation discretization:

$$
\rho_P
$$

appears only in the non-universality lower certificate.

The reason is that universality only needs to find one legal sampled orientation; the orientation sample is itself a genuine orientation.

This asymmetry is an important structural feature of this round.

---

# 26. Why the Upper / Lower Errors Are Asymmetric

## To Prove Universal

For each target it suffices to exhibit:

$$
\exists g
$$

such that containment holds.

Hence a sampled orientation suffices as a legal witness.

The only question is:

> Do the sampled support directions miss a protrusion in an un-sampled direction?

So the upper certificate mainly pays:

$$
\eta_M.
$$

---

## To Prove Non-Universal

One must prove:

$$
\forall g
$$

all fail.

Hence the entire orientation continuum must be ruled out.

A finite orientation grid needs to pay:

$$
\rho_P.
$$

This corresponds exactly to the original problem's quantifier:

$$
\forall K\exists g.
$$

Round 03 directly reflects this quantifier asymmetry in the certificate error budget.

---

# 27. Branch Compression Within the Support-Direction LP Itself

The dual LP extremum for every sampled orientation can be witnessed by at most three directions.

So even if the support grid has:

$$
M
$$

directions, the actual certificate can be compressed to:

$$
\boxed{
\le3\text{ active support indices}.
}
$$

This means future certificate archives need not store the entire dual vector.

It suffices to store:

- active indices;
- weights;
- primal translation;
- objective interval;
- support-value hashes.

---

# 28. Certificate Record Schema

For fixed $U$, the complete certificate package should contain at least:

## Global Metadata

- candidate cover definition;
- cover normalization / anchor;
- certified $R_U$;
- shape dictionary parameters:

$$
(N,\delta,q);
$$

- shape error:

$$
\varepsilon_{\mathrm{shape}};
$$

- support-grid size:

$$
M;
$$

- orientation-grid size:

$$
P.
$$

## Per Dictionary Shape

- coefficient vector;
- legality certificate;
- dictionary shape hash.

## Per Reflection/Orientation Sample

- $\sigma$;
- $\phi_\ell$;
- support sample intervals;
- primal LP witness;
- dual 2/3-active witness;
- primal-dual verified interval.

## Final Aggregation

- each target's lower/upper margin;
- global worst-case lower/upper;
- verdict:
  - UNIVERSAL;
  - NON-UNIVERSAL;
  - UNRESOLVED / REFINE.

---

# 29. Fixed-Cover Compiler Pseudo-Algorithm

```text
INPUT:
    convex candidate cover U
    certified support oracle for U
    shape compiler parameters N, delta, q
    support grid M
    orientation grid P

PREPROCESS:
    translate U so 0 in U
    certify R_U
    build finite legal target dictionary D
    compute epsilon_shape

FOR each K_hat in D:
    FOR sigma in {+1, -1}:
        FOR l = 0,...,P-1:
            phi_l = 2*pi*l/P
            evaluate/enclose support samples a_j
            solve finite primal LP
            solve finite dual LP
            verify 2/3-active dual witness
            store interval [m_lower, m_upper]

    G_minus = min all m_lower
    G_plus  = min all m_upper

    target_lower = G_minus - rho_P
    target_upper = G_plus + eta_M

GLOBAL:
    W_lower = max target_lower
    W_upper = max target_upper + epsilon_shape

IF W_upper <= 0:
    verdict = UNIVERSAL

ELSE IF W_lower > 0:
    verdict = NON-UNIVERSAL

ELSE:
    verdict = REFINE
```

This is a completely finite fixed-cover certificate workflow.

---

# 30. Sanity Validation Problem

This round uses an anisotropic ellipse as the candidate cover:

$$
U
=
\left\{
(x,y):
\frac{x^2}{0.72^2}
+
\frac{y^2}{0.52^2}
\le1
\right\}.
$$

Its support function:

$$
h_U(\theta)
=
\sqrt{
0.72^2\cos^2\theta
+
0.52^2\sin^2\theta
}.
$$

The target uses a legal centered unit constant-width trigonometric body:

$$
h_K(\theta)
=
\frac12
-
\frac{0.22}{8}\cos3\theta
-
\frac{0.10}{24}\sin5\theta.
$$

that is, the curvature density:

$$
r_K(\theta)
=
\frac12
+
0.22\cos3\theta
+
0.10\sin5\theta.
$$

Take:

$$
M=64,
\qquad
P=48.
$$

The coarse finite certificate produces:

$$
G_{M,P}.
$$

Separately, using a denser:

$$
M_{\mathrm{ref}}=512,
\qquad
P_{\mathrm{ref}}=192
$$

as a non-proof numerical reference.

The experiment confirms that the dense reference falls within Round 03's analytic bracket.

Hence:

`SANITY: PASS`

---

# 31. COMPUTE-DEFERRED

Round 03's core theorems do not depend on large-scale exhaustive computation.

But the following can already be handed off to local compute.

## C03-1: Pal/Gibbs Candidate Cover Certificate Dry Run

For a publicly known candidate cover $U$:

- build a support oracle;
- run a small dictionary;
- verify the certificate pipeline;
- do not initially pursue the ultimate final resolution.

## C03-2: Active Pair/Triple Census

Tally:

- 2-active antipodal;
- 3-active;

and their proportion and bifurcation across hard targets / orientations.

## C03-3: Adaptive Orientation Refinement

Refine only in the region where:

$$
m_\sigma(\phi)
$$

is close to the current minimum.

## C03-4: Independent Verifier

Build a certificate checker that shares no search code:

- support interval;
- primal feasibility;
- dual feasibility;
- aggregation inequality.

All marked:

`COMPUTE-DEFERRED`

---

# 32. RCHM Freedom Ledger

At the end of Round 02:

$$
\mathcal F_2
=
\{
\text{rotation},
\text{translation},
\text{support direction},
\text{active switching},
\text{container boundary},
\text{resolution limit}
\}.
$$

Round 03, for fixed $U$:

## Translation

From continuous $\mathbb R^2$:

$$
\longrightarrow
$$

finite LP / 2-or-3 active dual witness.

## Support Direction

From continuous $S^1$:

$$
\longrightarrow
$$

finite $M$ grid + explicit $\eta_M$.

## Rotation

From continuous $S^1$:

$$
\longrightarrow
$$

finite $P$ grid + explicit lower-side $\rho_P$.

## Reflection

From a hidden branch:

$$
\longrightarrow
$$

two explicit discrete branches.

So for fixed $U$:

$$
\boxed{
\mathcal F_{\mathrm{fixed}\ U}
}
$$

has become:

$$
\boxed{
\text{finite certificate data}
+
\text{explicit vanishing resolution errors}.
}
$$

---

# 33. The Largest Remaining Degree of Freedom

After Round 03, the main obstruction is no longer the target or the placement.

What genuinely remains:

$$
\boxed{
\text{the infinite-dimensional optimization of the candidate cover }U\text{ itself.}
}
$$

That is:

$$
a_{\mathrm{Leb}}
=
\inf_U
\operatorname{Area}(U)
\quad
\text{s.t.}
\quad
W(U)\le0.
$$

We now already have a finite-resolution oracle:

$$
U
\mapsto
[W^-(U),W^+(U)].
$$

The natural next step becomes:

> How can the candidate-cover boundary itself also be compressed into a legality-preserving finite branch family?

---

# 34. Assigned Topic for Round 04

## AMRAL-LUC-FC-R04
### Candidate-Cover Boundary Compiler and Area-Coverage Dual Tension

Main objectives:

1. Build a finite compiler for the support function of $U$;
2. Preserve:
   - convexity;
   - area computability;
   - support-oracle certification;
3. Establish the error propagation from candidate-cover shape error to:

$$
W(U)
$$

and:

$$
\operatorname{Area}(U)
$$

;

4. Compress:

$$
\inf_U
$$

into a finite branch / adaptive certificate structure;

5. Reconnect with the removable-region / active-tension idea from earlier research.

If Round 04 succeeds, then:

$$
\boxed{
\text{target infinity}
+
\text{placement infinity}
+
\text{container infinity}
}
$$

all three layers will, for the first time, enter the same finite-resolution proof graph.

---

# 35. Reproducibility Checklist for This Round

## Continuous Translation Dual

`PROVED`

## Active Support Size

`2 OR 3 ONLY`

## Round 01 Active-Branch Correction

`RECORDED`

## Finite Support LP

`PROVED`

## Support-Grid Error

`PROVED`

## Orientation Lipschitz

`PROVED`

## Reflection Branch

`EXPLICIT`

## Full Placement Bracket

`PROVED`

## Shape + Placement Fixed-Cover Certificate

`PROVED`

## Numerical Sanity

`PASS`

## New Lebesgue Bound

`NONE`

## Global Problem

`OPEN`

---

# 36. Shortest Handoff Conclusion

Round 03 completes the following compression:

$$
\boxed{
\min_{Q\in O(2)}
\min_{t\in\mathbb R^2}
\max_{\theta\in S^1}
}
$$

is converted into:

$$
\boxed{
\text{2 reflection branches}
\times
P\text{ orientations}
\times
M\text{-direction LPs}
}
$$

along with an explicit error:

$$
\boxed{
G^- -\rho_P
\le
M_U(K)
\le
G^+ +\eta_M.
}
$$

Merging further with the Round 02 shape error:

$$
\boxed{
\max_{\widehat K\in\mathscr D}
L_{\widehat K}
\le
W(U)
\le
\max_{\widehat K\in\mathscr D}
U_{\widehat K}
+
\varepsilon_{\mathrm{shape}}.
}
$$

Therefore, for any **fixed convex candidate cover $U$ with a certified support oracle**, Lebesgue universality can already be converted into a completely finite, replayable, independently verifiable certificate workflow.

The next genuine obstruction:

$$
\boxed{
\text{candidate-cover infinity}.
}
$$

---

# References

1. J. C. Baez, K. Bagdasaryan, P. Gibbs, *The Lebesgue Universal Covering Problem*, arXiv:1502.01251.
2. U. Mishra, *Curves of constant width and Lebesgue's covering problem*, arXiv:2608.30538, 2026.
3. S. Zeng, *An exact hierarchy for Lebesgue's universal covering constant and a certified 0.834 lower bound*, arXiv:2609.01284, 2026.
4. R. T. Rockafellar, *Convex Analysis*.
5. R. Schneider, *Convex Bodies: The Brunn–Minkowski Theory*.
6. Neo.K + Aletheia, *AMRAL × Lebesgue Universal Covering — Round 01*, 2026-09-18.
7. Neo.K + Aletheia, *AMRAL × Lebesgue Universal Covering — Round 02*, 2026-09-18.

---

# 37. Declaration

This round does not claim to solve the Lebesgue universal covering problem.

This round does not improve the rigorous upper and lower bounds of:

$$
a_{\mathrm{Leb}}
$$

.

What this round accomplishes is:

$$
\boxed{
\text{a finite certificate compiler for all targets + all placements, over a fixed candidate cover.}
}
$$

Its purpose is so that subsequent candidate-cover optimization no longer needs to reprocess the full infinite quantifiers over the target and $E(2)$.