> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# WSFC Round 04
## Finite-Segment Convexity Gates, Endpoint-Disc Obstacle Reduction, and Support-Cell Allocation Calculus

**English Title:** *Finite-Segment Convexity Gates, Endpoint-Disc Obstacle Reduction, and Support-Cell Allocation Calculus*  
**Document ID:** `WSFC-R04-v0.1`  
**Version:** v0.1  
**Date:** 2026-09-23  
**Author:** Neo.K  
**AI Research Collaborator:** Aletheia / OpenAI GPT-5.6 Sol  
**Series:** Worm Sequence Freedom Closure  
**Prerequisite:** `WSFC-R03-v0.1`  
**Status:** `SOURCE-GATE CLOSURE / DISCRETE OBSTACLE REDUCTION / EVEN-ALLOCATION THEOREM`  
**Canonical Source:** UTF-8 Markdown

---

# Abstract

Round 03 reduced the main proof debt for the broadest finite polygonal worm to:

$$
\boxed{
\text{global extremizer}
\Rightarrow
\text{antipodal support-tangent class}?
}
$$

and divided its subsidiary responsibilities into:

1. convexity necessity;
2. drape / endpoint-strip necessity;
3. width equalization;
4. antipodal endpoint responsibility;
5. equal-angle tangent partition.

The first result of Round 04 is that the first two gates are not new problems.

In their 1989 study of the shortest width-$1$ planar arc, Adhikari–Pitman first fixed a finite segment cap $M$, considered all polygonal arcs with at most $M$ segments, took a shortest width-$1$ minimizer, and proved that this minimizer must be convex. This is precisely the convexity reduction required by WSFC's finite-stage dual.

The same source also proves in the convex case that, if the convex arc extends left or right relative to the baseline of its endpoints, perpendicular truncation produces a shorter arc of greater width. Thus the shortest convex candidate can necessarily be restricted to the perpendicular strip between the endpoints. This is precisely the drape-type reduction WSFC needs.

More importantly, for endpoints:

$$
A=(0,0),
\qquad
B=(d,0),
$$

and an arc above the endpoint strip, Adhikari–Pitman rewrites the width-$1$ condition as:

- the arc must reach the unit-height line;
- the subpath from $A$ to that contact must not enter the open unit disc centered at $B$;
- symmetrically, the other side must not enter the open unit disc centered at $A$.

Thus Round 03's opposite-endpoint unit-circle tangency is not accidental geometry: it is precisely the finite-segment discretization of this endpoint-disc obstacle formulation.

This round then writes the tangent-partition cost uniformly as:

$$
\boxed{
g(q,\Delta)
=
2q
\tan
\left(
\frac{\Delta}{2q}
\right),
}
$$

where:

- $\Delta$ is the support-angle span that must be traversed around the obstacle;
- $q$ is the number of support-angle gaps on that side.

This round proves:

$$
\boxed{
\frac{\partial g}{\partial q}<0,
}
$$

and:

$$
\boxed{
\frac{\partial^2 g}{\partial q^2}>0.
}
$$

That is:

> Increasing active support cells necessarily reduces the discrete tangent cost, but with strictly diminishing returns.

Furthermore, let:

$$
q_L+q_R=n-2.
$$

This gives the unified reduced obstacle functional:

$$
\boxed{
\mathcal L_n(d,x;q_L,q_R)
=
2\sqrt{d^2-1}
+d
+
g(q_L,\Delta_L)
+
g(q_R,\Delta_R).
}
$$

For even $n$, this round uses:

1. the concavity of $\arctan$ on the positive axis;
2. the fact that $g$ is the perspective of a convex function;
3. the even integer budget $q_L+q_R=n-2$;

to prove, within this reduced obstacle model:

$$
\boxed{
x=\frac d2,
\qquad
q_L=q_R=\frac{n-2}{2}.
}
$$

Round 03's even-stage bilateral symmetry and balanced support partition have therefore advanced from numerical patterns to a reduced-model theorem.

For odd $n$, full allocation enumeration up to $n=20$ selects the most balanced allocation:

$$
|q_L-q_R|=1,
$$

with $x\neq d/2$ absorbing the unavoidable odd asymmetry; however, this paper has not yet promoted that step to a general theorem.

This round also adds an explicit top-segment red team. If a horizontal top segment at unit height is forcibly retained and $n=5,\ldots,12$ is reoptimized, all optimized top-segment lengths collapse to numerical zero, equivalent to wasting a segment and falling back to the previous-stage candidate. This is strong numerical evidence, but no global theorem is claimed yet.

---

# 1. Scale Duality

For any nondegenerate arc $\gamma$, let:

$$
\ell(\gamma)
$$

be its length,

$$
b(\gamma)
$$

its breadth.

The ratio:

$$
\frac{b(\gamma)}{\ell(\gamma)}
$$

is invariant under uniform scaling.

Thus:

> the maximum breadth of a unit-length $n$-segment arc

is equivalent to:

> the minimum length of an arc of breadth $1$ with at most $n$ segments.

If:

$$
t_n
=
\sup_{\ell(\gamma)=1,\;\#\mathrm{seg}\le n}b(\gamma),
$$

then:

$$
\boxed{
L_n^\star
=
\frac1{t_n}
}
$$

where $L_n^\star$ is the minimum length in the width-$1$ dual.

Adhikari–Pitman's conclusion for a finite-segment width-$1$ minimizer therefore applies directly to WSFC.

---

# 2. Gate A: An Existing Theorem Already Gives Finite-Segment Convexity

Adhikari–Pitman do not simply assume convexity directly in the class of continuous arcs.

They first fix:

$$
M<\infty
$$

and study:

> a shortest polygonal arc with at most $M$ line segments and width at least $1$.

Their Section 3 establishes the standard form of a minimal arc:

- every convex-hull corner is visited only once;
- every segment endpoint is a convex-hull corner;
- a minimal arc has no self-intersection.

Section 4 then proves:

$$
\boxed{
\text{Every finite-}M\text{ minimal width-}1\text{ arc must be convex.}
}
$$

Thus WSFC's:

```text
CONVEXITY NECESSITY
```

can now be marked:

```text
SOURCE-CLOSED
```

rather than OPEN.

---

# 3. Why Does This Theorem Match WSFC Exactly?

Round 03's problem is not an arbitrary shortest arc, but:

$$
L_n^\star
=
\min
\{
\ell(\gamma):
b(\gamma)\ge1,\;
\#\mathrm{seg}(\gamma)\le n
\}.
$$

Let:

$$
M=n.
$$

This is precisely the domain of Adhikari–Pitman's finite-$M$ argument.

A convex minimizer therefore exists:

$$
\boxed{
\gamma_n^\star
\text{ convex}.
}
$$

This step does not need to wait for:

$$
n\rightarrow\infty.
$$

It holds separately for each finite $n$.

---

# 4. Gate B: Endpoint-Strip / Drape Reduction

For a convex arc:

$$
A\leadsto B
$$

make the endpoint line $AB$ horizontal.

Adhikari–Pitman note:

If the arc extends beyond the endpoint strip to the left of $A$ or the right of $B$, truncating by perpendicular projection to $AB$ yields:

- a shorter arc;
- a relevant width that is not smaller and in fact is greater.

Thus the minimizer must be restricted to:

$$
\boxed{
0\le x\le d,
\qquad
d=|AB|.
}
$$

That is:

$$
\boxed{
\text{the finite minimizer can be restricted to a convex drape-type arc}.
}
$$

Thus:

```text
DRAPE / ENDPOINT-STRIP NECESSITY
```

can also be upgraded to:

```text
SOURCE-CLOSED
```

---

# 5. Gate C: The Width Condition Is Compressed to Two Endpoint Discs

Let:

$$
A=(0,0),
\qquad
B=(d,0).
$$

Consider a convex arc above the endpoint strip.

The core width lemma of Adhikari–Pitman uses:

$$
A'B'
$$

to denote the horizontal segment on:

$$
y=1
$$

as above.

For a convex arc, the responsibility for width at least $1$ can be divided into:

1. the arc reaches the unit-height line;
2. the left half must not enter:

$$
D_B^\circ
=
\{p:|p-B|<1\};
$$

3. the right half must not enter:

$$
D_A^\circ
=
\{p:|p-A|<1\}.
$$

Thus:

$$
\boxed{
\text{global minimum-width constraint}
\rightarrow
\text{two local convex obstacle constraints}.
}
$$

This is the source of the antipodal endpoint circles in Round 03.

---

# 6. WSFC's New Responsibility Decomposition

The global finite-$n$ problem can now be rearranged as:

$$
\boxed{
\text{finite width minimizer}
}
$$

$$
\Downarrow
$$

$$
\boxed{
\text{convex}
}
$$

$$
\Downarrow
$$

$$
\boxed{
\text{endpoint-strip / drape}
}
$$

$$
\Downarrow
$$

$$
\boxed{
\text{left path avoids }D_B^\circ
\quad+\quad
\text{right path avoids }D_A^\circ.
}
$$

Round 03's support structure is therefore no longer a purely numerical conjecture.

The actual remaining issue that is not fully closed is:

> Under a fixed segment budget, must the active-support structure of the obstacle geodesic be a tangent polygon?

---

# 7. Discrete Tangent Cost

Consider the unit-disc obstacle on one side.

If a finite polygonal path uses:

$$
q+1
$$

tangent segments to discretize a total support-angle span:

$$
\Delta,
$$

then the tangent-angle gaps:

$$
\delta_1,\ldots,\delta_q
$$

satisfy:

$$
\sum_{i=1}^{q}\delta_i=\Delta.
$$

The extra length arising from an adjacent tangent intersection relative to the circle-arc envelope is:

$$
2\tan\frac{\delta_i}{2}.
$$

Thus the discrete obstacle cost is:

$$
\sum_{i=1}^{q}
2\tan\frac{\delta_i}{2}.
$$

---

# 8. Equal-Partition Lemma

Since:

$$
f(\delta)
=
2\tan\frac{\delta}{2}
$$

is strictly convex,

for fixed total:

$$
\Delta
$$

as above,

Jensen gives:

$$
\sum_{i=1}^{q}
2\tan\frac{\delta_i}{2}
\ge
2q
\tan
\left(
\frac{\Delta}{2q}
\right).
$$

Equality holds if and only if:

$$
\boxed{
\delta_i=\frac{\Delta}{q}
\quad
\forall i.
}
$$

Thus define:

$$
\boxed{
g(q,\Delta)
=
2q
\tan
\left(
\frac{\Delta}{2q}
\right).
}
$$

This is not an empirical fit.

It is the exact equal-partition cost for a fixed support span and support-cell count.

---

# 9. The Marginal Value of a Support Cell

Let:

$$
z
=
\frac{\Delta}{2q}.
$$

Then:

$$
g(q,\Delta)
=
2q\tan z.
$$

Differentiating with respect to continuous $q$:

$$
\boxed{
\frac{\partial g}{\partial q}
=
2
\left(
\tan z
-
z\sec^2 z
\right).
}
$$

Since:

$$
h(z)
=
z\sec^2 z-\tan z
$$

satisfies:

$$
h(0)=0,
$$

and:

$$
h'(z)
=
2z\sec^2 z\tan z
>
0
$$

for $z>0$,

we have:

$$
\boxed{
\frac{\partial g}{\partial q}<0.
}
$$

Thus adding an active support cell necessarily reduces the tangent discretization cost.

---

# 10. Diminishing Returns

Differentiating again:

$$
\boxed{
\frac{\partial^2 g}{\partial q^2}
=
\frac{
4z^2\sec^2 z\tan z
}{
q
}
>
0.
}
$$

Thus $g$ is strictly convex in $q$.

That is:

$$
\boxed{
\text{support-cell refinement has diminishing returns}.
}
$$

For the first time, this directly turns WSFC's:

$$
\text{Freedom Birth}
$$

into a differentiable cost law.

---

# 11. Unified Left / Right Reduced Functional

Let the apex/contact state be:

$$
F=(x,1).
$$

Define:

$$
C(d)
=
\frac\pi2-\arccos\frac1d.
$$

The support span for the left endpoint-disc obstacle is:

$$
\boxed{
\Delta_L(d,x)
=
C(d)
-
2\arctan(d-x).
}
$$

On the right:

$$
\boxed{
\Delta_R(d,x)
=
C(d)
-
2\arctan x.
}
$$

Let:

$$
q_L+q_R=n-2.
$$

Then the support-tangent reduced functional is:

$$
\boxed{
\mathcal L_n(d,x;q_L,q_R)
=
2\sqrt{d^2-1}
+d
+
g(q_L,\Delta_L)
+
g(q_R,\Delta_R).
}
$$

Round 03's even / odd formulas are both special cases of this expression.

---

# 12. Why Is $q_L+q_R=n-2$?

If one side has:

$$
q_L
$$

tangent-angle gaps,

then it has:

$$
q_L+1
$$

side segments.

The other side is analogous.

The total segment count is:

$$
n
=
(q_L+1)
+
(q_R+1).
$$

Hence:

$$
\boxed{
q_L+q_R=n-2.
}
$$

The actual sequence freedom can therefore be divided into:

1. continuous geometry:

$$
(d,x);
$$

2. an integer support-cell budget:

$$
(q_L,q_R).
$$

---

# 13. $g$ Is a Convex Perspective

Let:

$$
\varphi(s)
=
2\tan\frac s2.
$$

Then:

$$
g(q,\Delta)
=
q
\varphi
\left(
\frac{\Delta}{q}
\right).
$$

This is the perspective of the convex function $\varphi$.

Thus $g$ is jointly convex in:

$$
(q,\Delta)
$$

as above.

Therefore:

$$
\boxed{
g(q_L,\Delta_L)
+
g(q_R,\Delta_R)
\ge
g(q_L+q_R,\Delta_L+\Delta_R).
}
$$

This is the first half of the even allocation proof.

---

# 14. Moving the Apex Away from the Center Increases the Total Support Span

From:

$$
\Delta_L+\Delta_R
=
2C(d)
-
2[
\arctan(d-x)+\arctan x
].
$$

Since $\arctan$ is concave on:

$$
x>0
$$

as above,

when:

$$
x+(d-x)=d
$$

is fixed,

$$
\arctan x
+
\arctan(d-x)
\le
2\arctan\frac d2.
$$

we have:

$$
\boxed{
\Delta_L+\Delta_R
\ge
2\Delta(d),
}
$$

where:

$$
\Delta(d)
=
C(d)
-
2\arctan\frac d2.
$$

Equality holds if and only if:

$$
\boxed{
x=\frac d2.
}
$$

The centered apex therefore minimizes the total support-angle burden.

---

# 15. Even Allocation Theorem

Let:

$$
n=2m.
$$

Then:

$$
Q=n-2=2m-2
$$

is even.

By perspective convexity:

$$
g(q_L,\Delta_L)
+
g(q_R,\Delta_R)
\ge
g(Q,\Delta_L+\Delta_R).
$$

Moreover, since:

$$
\Delta_L+\Delta_R
\ge
2\Delta(d)
$$

and $g$ increases with $\Delta$,

we obtain:

$$
g(q_L,\Delta_L)
+
g(q_R,\Delta_R)
\ge
g(Q,2\Delta(d)).
$$

When:

$$
\boxed{
q_L=q_R=m-1,
}
$$

and:

$$
\boxed{
x=\frac d2
}
$$

hold,

$$
\frac{\Delta_L}{q_L}
=
\frac{\Delta_R}{q_R}
$$

and both inequalities attain equality simultaneously.

Thus, within the support-tangent reduced model:

$$
\boxed{
x^\star=\frac d2,
\qquad
q_L^\star=q_R^\star=m-1.
}
$$

This is not numerical symmetry.

It is a reduced-model theorem.

---

# 16. The Round 03 Even Formula Is Thus Reproved

Substituting:

$$
q_L=q_R=m-1,
$$

gives:

$$
\boxed{
L_{2m}(d)
=
2\sqrt{d^2-1}
+d
+
4(m-1)
\tan
\left(
\frac{\Delta(d)}{2(m-1)}
\right).
}
$$

which is the Round 03 formula.

Round 03's even support compiler therefore no longer needs to assume in advance:

```text
EQUAL LEFT/RIGHT SUPPORT CELL COUNT
```

It follows automatically from this round's allocation theorem.

---

# 17. Odd Stage: The Remaining Integer Allocation Problem

Let:

$$
n=2m+1.
$$

Then:

$$
Q=n-2=2m-1
$$

is odd.

The most balanced integer allocation can only be:

$$
\boxed{
(q_L,q_R)
=
(m-1,m)
}
$$

or its reflection.

Here it is impossible to satisfy both:

$$
q_L=q_R
$$

and:

$$
x=d/2.
$$

Thus the system uses:

$$
x\neq\frac d2
$$

to change slightly:

$$
\Delta_L,
\Delta_R
$$

in compensation for the integer cell imbalance.

This provides a direct mechanism for odd asymmetry.

---

# 18. Odd Allocation Numerical Exhaustion

For:

$$
4\le n\le20
$$

this round enumerates, in the reduced model, all:

$$
q_L,q_R\ge1,
\qquad
q_L+q_R=n-2
$$

and reminimizes, for each allocation:

$$
(d,x).
$$

The result:

$$
\boxed{
|q_L-q_R|
=
\begin{cases}
0,& n\text{ even},\\
1,& n\text{ odd}
\end{cases}
}
$$

holds at all tested stages.

Thus:

```text
ODD BALANCED-AS-POSSIBLE ALLOCATION
```

currently has strong numerical support.

But this paper does not directly promote that finite verification to:

$$
\forall n.
$$

---

# 19. The Next Odd-Stage Proof Target

Since $g(q,\Delta)$ is convex in $q$,

the marginal gain from integer allocation:

$$
g(q,\Delta)-g(q+1,\Delta)
$$

decreases with $q$.

Thus the imbalance:

$$
|q_L-q_R|>1
$$

intuitively means:

> One side has entered a lower-marginal-return region, while the other still has higher-marginal-return support refinement available.

To obtain a theorem, one must also control:

$$
\Delta_L,\Delta_R
$$

and the compensating effect of their dependence on the apex $x$.

This is a natural analytic target for Round 05.

---

# 20. Top-Segment Red Team

Round 03 assumes by default that the top contact is a single point:

$$
F=(x,1).
$$

This round deliberately relaxes it to:

$$
F_L=(x_L,1),
\qquad
F_R=(x_R,1),
$$

and allows a horizontal top segment with:

$$
x_R-x_L>0
$$

as above.

This consumes an additional segment.

---

# 21. Top-Segment Extended Functional

Define:

$$
a=d-x_L,
\qquad
b=x_R.
$$

The total length is:

$$
\begin{aligned}
\mathcal L^{\rm top}
={}&
2\sqrt{d^2-1}
+a+b\\
&+
(x_R-x_L)\\
&+
g(q_L,\Delta(d,a))
+
g(q_R,\Delta(d,b)).
\end{aligned}
$$

If:

$$
x_L=x_R,
$$

the top segment collapses to zero.

---

# 22. Top-Segment Numerical Result

This round performs multi-start optimization for:

$$
n=5,\ldots,12
$$

and all admissible support-cell allocations.

The optimum in every case is:

$$
\boxed{
x_R-x_L
\approx0
}
$$

to machine precision.

The resulting configuration is equivalent to:

> The top segment wastes one segment of the budget and degenerates to the previous-stage candidate with no top segment.

Thus no advantage has been found for:

$$
\boxed{
\text{positive top segment}
}
$$

in the current reduced obstacle model.

Status:

```text
STRONG NUMERICAL EVIDENCE
NOT YET GLOBAL THEOREM
```

---

# 23. The Responsibility of Endpoint Tangency

In Adhikari–Pitman's continuous convex solution, if the direct segment would cross the opposite-endpoint disc, the shortest path takes the form:

$$
\boxed{
\text{tangent}
\rightarrow
\text{circle boundary}
\rightarrow
\text{tangent}.
}
$$

A finite segment budget does not permit a continuous circular arc.

The core interpretation of Round 03 / 04 is:

$$
\boxed{
\text{circle boundary arc}
\rightarrow
\text{finite circumscribed tangent partition}.
}
$$

Thus:

$$
\boxed{
\text{antipodal endpoint responsibility}
}
$$

is not an arbitrary guess from a vertex pattern,

but the natural discretization of a finite-stage obstacle geodesic.

---

# 24. The Bridge That Is Still Not Fully Proved

We must distinguish very precisely:

Adhikari–Pitman have proved:

```text
GLOBAL FINITE MINIMIZER
    -> CONVEX
    -> ENDPOINT STRIP
    -> ENDPOINT-DISC OBSTACLE FORMULATION
```

This round has proved:

```text
IF ONE USES ACTIVE TANGENT SUPPORT CELLS
    -> EQUAL ANGLE PARTITION
    -> g(q,Delta)
    -> EVEN BALANCED ALLOCATION
```

Still missing:

```text
FINITE-SEGMENT OBSTACLE MINIMIZER
    -> EVERY FREE SIDE SEGMENT IS AN ACTIVE TANGENT
```

This is the main bridge currently required for Support-Class Necessity.

---

# 25. Obstacle Support-Activation Lemma Candidate

Candidate lemma:

> Fix the endpoints, chosen side, unit-disc obstacle, and polygonal segment budget. If a polygonal path is the shortest convex feasible path in that class, then, apart from structures forced by the endpoints, the supporting line of every free side segment must contact the obstacle.

The geometric intuition:

If some side support line has positive slack from the disc,

then the support line can be tightened toward the disc.

The obstacle is still not entered, while the convex envelope is reduced, so the boundary path should become shorter.

If this lemma holds rigorously:

$$
\boxed{
\text{Disc Avoidance}
\Rightarrow
\text{Tangent Activation}.
}
$$

then Round 03's antipodal support-tangent class necessity will move much closer to closure.

Current label:

```text
PROOF-SKETCH / FORMAL DEBT
```

---

# 26. Correspondence with Deng's 2026 Antipodal Constraints

Deng's continuum convex formulation uses:

$$
(
\mathbf r(t)-\mathbf r(t+\pi)
)
\cdot
n(t)
\ge1.
$$

This writes the width responsibility directly as an antipodal support inequality.

Round 04's finite support cells:

$$
(\theta_i,\text{opposite endpoint})
$$

can be seen as:

$$
\boxed{
\text{finite active antipodal constraints}.
}
$$

The finite polygonal sequence and the continuum convex program therefore also begin to align in their constraint language.

---

# 27. Round 04 Gate Ledger

```text
SCALE DUALITY                         CLOSED
FINITE-STAGE EXISTENCE                SOURCE-SUPPORTED
CONVEXITY NECESSITY                   SOURCE-CLOSED
ENDPOINT-STRIP / DRAPE REDUCTION      SOURCE-CLOSED
DISC-OBSTACLE CHARACTERIZATION        SOURCE-CLOSED / ADAPTED
EQUAL-ANGLE PARTITION                 CLOSED
g(q,Delta) DECREASING IN q            CLOSED
g(q,Delta) CONVEX IN q                CLOSED
EVEN BALANCED CELL ALLOCATION         CLOSED IN REDUCED MODEL
ODD BALANCED-AS-POSSIBLE ALLOCATION   NUMERICALLY STRONG / OPEN THEOREM
POSITIVE TOP SEGMENT NECESSITY        EVIDENCE SAYS NO / OPEN THEOREM
TANGENT-ACTIVATION NECESSITY          PROOF-SKETCH / OPEN FORMAL GATE
GLOBAL SUPPORT-CLASS NECESSITY        PARTIALLY CLOSED
```

---

# 28. What Has Support-Class Necessity Now Been Reduced To?

In Round 03:

$$
\boxed{
\text{Convexity}
+
\text{Drapeability}
+
\text{Width Equalization}
+
\text{Antipodal Endpoint Responsibility}
}
$$

all appeared open.

After Round 04:

$$
\boxed{
\text{Convexity}
}
$$

and:

$$
\boxed{
\text{Endpoint-Strip / Drape Reduction}
}
$$

are closed directly by existing theorems.

The source also supplies the endpoint discs.

Thus what actually remains is:

$$
\boxed{
\text{Obstacle Support Activation}
}
$$

and a general proof for odd integer allocation.

This is a substantial contraction of proof debt.

---

# 29. A New Understanding of WSFC's Concept of Freedom

We can now distinguish three kinds of freedom:

### Representation Freedom

$$
2n-2.
$$

### Obstacle Support Freedom

$$
(q_L,q_R,\theta_i).
$$

### Reduced Responsibility Freedom

even:

$$
(n,d),
$$

odd:

$$
(n,d,x).
$$

Thus:

$$
\boxed{
\text{Freedoms are not deleted,
but quotiented out by theorem responsibility.}
}
$$

---

# 30. Round 04 Conclusion

The most important advance of this round is not a new $t_n$.

Instead, it relocates Round 03's main question:

$$
\boxed{
\text{global finite polygonal problem}
}
$$

has already been reduced, through existing theory, to:

$$
\boxed{
\text{convex endpoint-disc obstacle problem}.
}
$$

The finite freedom within the obstacle problem:

$$
q
$$

in turn has an exact cost:

$$
\boxed{
g(q,\Delta)
=
2q\tan\frac{\Delta}{2q}.
}
$$

It satisfies:

$$
g_q<0,
\qquad
g_{qq}>0.
$$

Thus every new active support cell:

- is necessarily useful;
- but has diminishing marginal returns.

For even stages, one can further prove:

$$
\boxed{
\text{centered apex}
+
\text{perfectly balanced support-cell allocation}.
}
$$

The 2007 even-symmetry conjecture is therefore not yet fully proved for the global problem, but has a very clear analytic explanation in the support obstacle model currently reduced through the source.

The next round should genuinely concentrate on:

$$
\boxed{
\textbf{Obstacle Support-Activation Lemma}.
}
$$

Once:

$$
\text{finite shortest disc-avoiding polygon}
\Rightarrow
\text{all free sides tangent}
$$

is made rigorous, Round 03's support compiler will advance another level beyond a reduced candidate model, toward a genuine finite-$n$ theorem.

---

# References

1. Ani Adhikari, Jim Pitman, *The Shortest Planar Arc of Width 1*, American Mathematical Monthly 96(4), 309–327 (1989).
2. John M. Maki, John E. Wetzel, Wacharin Wichiramala, *Drapeability*, Discrete & Computational Geometry 34(4), 637–657 (2005).
3. Chatchawan Panraksa, John E. Wetzel, Wacharin Wichiramala, *The broadest three-segment unit arc*, Periodica Mathematica Hungarica 55(2), 157–168 (2007).
4. Zhipeng Deng, *Revisit escape path for infinite unit strip forest and unit broadworm*, arXiv:2607.18563 (2026).

---

# Canonical Source Declaration

The original file is the UTF-8 Markdown canonical source for `WSFC-R04-v0.1`.

This round particularly distinguishes:

```text
SOURCE-CLOSED
DERIVED-CLOSED
REDUCED-MODEL THEOREM
NUMERICAL EVIDENCE
PROOF-SKETCH
GLOBAL OPEN
```

A reduced-model theorem must not be directly equated with a global finite-$n$ theorem.

===END===