> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification. # WSFC Round 05 ## Boundary Saturation Closure: Endpoint Shortcuts, Plateau Elimination, and Finite-$n$ Support-Tangent Reduction **English Title:** *Boundary Saturation Closure: Endpoint Shortcuts, Plateau Elimination, and a Finite-$n$ Support-Tangent Reduction* **Document ID:** `WSFC-R05-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 **Prerequisites:** `WSFC-R03-v0.1`; `WSFC-R04-v0.1` **Status:** `PROVISIONAL GLOBAL REDUCTION / RESEARCH-PROOF DRAFT / INDEPENDENT AUDIT REQUIRED` **Canonical Source:** UTF-8 Markdown --- # Abstract Round 04 compressed Support-Class Necessity into three boundary cases: $$ \mathrm{BSP\!-\!A} = \text{endpoint-edge slack}, $$ $$ \mathrm{BSP\!-\!B} = \text{positive top plateau}, $$ $$ \mathrm{BSP\!-\!C} = \text{support-cell degeneration}. $$ This round attacks them one by one. Adhikari–Pitman 1989 have proved that, for any fixed segment cap $M$, a minimal polygonal arc of width at least $1$ must be convex; for a convex arc, perpendicular reduction further restricts the study to the vertical endpoint strip. These two existing theorems serve as the hard external anchors of this round. This paper adds the following reduction lemmas: 1. **Endpoint Shortcut Lemma**: if an endpoint-adjacent edge has strict slack relative to the unit-disc obstacle, slide that vertex along the next edge until tangency or vertex collapse; the triangle inequality guarantees nonincrease of path length. 2. **Lower-Stage Absorption Principle**: zero-length segments, collinear adjacent segments, support-angle collisions, and support-order swap boundaries all lie in a lower segment stage. 3. **Positive Plateau Elimination**: in the reduced tangent model, a positive-length top plateau $p>0$ cannot be a shortest solution. For $n\ge5$, this follows from joint convexity and plateau-span monotonicity; the special branch for $n=4$ reduces analytically to: $$ T_{\rm int}(c) = \csc c + 4\tan\frac c4, $$ and: $$ T_{\rm int}'(c)>\frac13. $$ Thus its minimum also lies on the $p=0$ boundary. 4. **One-Segment Boundary Exclusion**: if one side has only a single segment, feasibility first pushes it to endpoint-disc tangency; then taking the other side's resolution to the continuum yields the lower envelope: $$ B_\infty(c) = \csc c + 2\cot c + 3c - \pi. $$ Its minimum is: $$ \boxed{ \min B_\infty \approx 2.297464630768953. } $$ Meanwhile, an explicitly legal four-segment support-tangent competitor at: $$ d=\frac{53}{50} $$ already has: $$ \boxed{ L_4(53/50) \approx 2.282033991861008. } $$ The gap is: $$ \boxed{ \min B_\infty - L_4(53/50) \approx 0.015430638907945 >0. } $$ Thus the q=0 boundary class cannot be globally optimal. Combining these with the endpoint antipodes, interior support saturation, equal-angle partition, and balanced integer allocation of Rounds 03–04 produces a **Finite-$n$ Support-Tangent Reduction Theorem draft**: > For every $n\ge4$, among shortest polygonal arcs of width at least $1$ with at most $n$ segments, all nondegenerate global minimizers should lie in Round 03's antipodal support-tangent class; degenerate cell boundaries are handled by lower-stage absorption. If an independent proof audit finds no gap, the general-$n$ broadest polygonal unit arc can be reduced to: - even $n=2m$: one-dimensional minimization; - odd $n=2m+1$: two-dimensional minimization. This paper does not present it as a peer-reviewed or formally verified theorem. Its official status is: ```text PROVISIONAL GLOBAL REDUCTION INDEPENDENT RE-DERIVATION REQUIRED FORMALIZATION OPEN ``` --- # 1. Hard External Anchors In *The Shortest Planar Arc of Width 1*, Adhikari–Pitman first establish a compact minimal arc under a finite segment budget, then prove: $$ \boxed{ \text{for every fixed }M, \text{ a minimal width-one arc with at most }M\text{ segments is convex}. } $$ The same work also gives endpoint-strip reduction for a convex arc: if the arc extends beyond the perpendiculars at the endpoints, perpendicular reduction yields a shorter arc of greater width. This round therefore begins from: $$ \boxed{ \text{convex} + \text{endpoint-strip confined} } $$ as above. --- # 2. Classical Two-Disc Criterion Let: $$ A=(0,0), \qquad B=(d,0), \qquad d\ge1. $$ with the convex upper arc confined to: $$ 0\le x\le d. $$ The width-one condition can be restated as: - the arc intersects the horizontal line $y=1$; - the left subarc stays outside the open unit disc centered at $B$; - the right subarc stays outside the open unit disc centered at $A$. Each side is therefore intrinsically a finite-segment obstacle path. --- # 3. BSP-A: Endpoint Shortcut Lemma Consider a legal subpath: $$ P_0P_1P_2\cdots P_k $$ outside a convex obstacle $D$. If the first edge: $$ P_0P_1 $$ still has strict slack relative to $D$, slide along: $$ P_1P_2 $$ moving $P_1$ toward $P_2$ to become $P_1'$. Stop at the first event: ```text A. P0P1' tangent to D B. P1' = P2 ``` Before stopping, the new segment: $$ P_0P_1' $$ remains legal, while: $$ P_1'P_2 \subseteq P_1P_2. $$ The triangle inequality gives: $$ |P_0P_1'| \le |P_0P_1| + |P_1P_1'|. $$ and: $$ |P_1P_2| = |P_1P_1'| + |P_1'P_2|. $$ Thus: $$ \boxed{ |P_0P_1'| + |P_1'P_2| \le |P_0P_1| + |P_1P_2|. } $$ The inequality is strict in the nondegenerate case. Thus the first surviving edge of a minimal path must be tangent; if collapse occurs first, it drops directly to a lower segment stage. Apply the same argument in reverse to the last edge. Therefore: $$ \boxed{ \text{endpoint slack} \Rightarrow \text{tangency or lower-stage collapse}. } $$ --- # 4. BSP-C: Lower-Stage Absorption The following degeneracies can all be represented at a lower stage. ## 4.1 Zero-length $$ P_i=P_{i+1} $$ Delete it directly. ## 4.2 Collinear adjacent segments If: $$ P_{i-1},P_i,P_{i+1} $$ are collinear with consistent direction, merge them into: $$ P_{i-1}P_{i+1}. $$ ## 4.3 Support-angle collision If: $$ \theta_i=\theta_{i+1}, $$ merge the corresponding consecutive support lines. ## 4.4 Support-order swap If convex normal order changes, it must first pass through: $$ \theta_i=\theta_{i+1}. $$ Thus the swap boundary also first reaches a lower-stage collision. This round therefore proposes: $$ \boxed{ \textbf{Lower-Stage Absorption Principle}. } $$ For optimization with at most $n$ segments, these boundary states do not form a new independent extremizer class. --- # 5. First Reduce the Top Excursion to $y=1$ If the maximum height of a convex strip arc is: $$ H>1, $$ let: $$ F_L,F_R $$ be the first and last intersections of the arc with: $$ y=1 $$ as above. Replace the top subarc: $$ F_L \leadsto F_R $$ with the horizontal chord: $$ F_LF_R \subset \{y=1\}. $$ The left / right disc-avoidance portions are preserved, and the width-one criterion still holds. The segment count does not increase, and a nonhorizontal excursion is strictly shortened. A minimal arc can therefore be restricted to: $$ \boxed{ H=1 } $$ but a positive horizontal plateau may remain. --- # 6. BSP-B: Plateau Variables Let: $$ F_L=(x_L,1), \qquad F_R=(x_R,1), $$ $$ p=x_R-x_L>0. $$ Define: $$ a_L=d-x_L, \qquad a_R=x_R. $$ Then: $$ \boxed{ p=a_L+a_R-d. } $$ If both sides have already been shortened to tangent-chain reduced form: $$ T_{\rm plat} = J_d(q_L,a_L) + J_d(q_R,a_R) + p. $$ where: $$ J_d(q,a) = \sqrt{d^2-1} + a + 2q \tan \left( \frac{\Delta_d(a)}{2q} \right), $$ $$ \Delta_d(a) = \frac\pi2 - \arccos\frac1d - 2\arctan a. $$ --- # 7. $n\ge5$: A Positive Plateau Cannot Be Optimal If: $$ q_L,q_R\ge1, $$ then, for: $$ s=a_L+a_R\ge d $$ define: $$ F(s) = \min_{a_L+a_R=s} \left[ J_d(q_L,a_L) + J_d(q_R,a_R) \right]. $$ Round 04 proves that $J_d$ is jointly strictly convex in $(q,a)$. At an interior minimizer: $$ J_a(q_L,a_L) = J_a(q_R,a_R) = \lambda. $$ On at least one side: $$ a_i \ge \frac{s}{2} \ge \frac d2 \ge \frac12. $$ Let: $$ z = \frac{\Delta_d(a)}{2q}. $$ For: $$ a\ge\frac12, \qquad q\ge1, $$ we have: $$ z < \frac\pi4-\arctan a. $$ Thus: $$ \tan z < \frac{1-a}{1+a} < a. $$ Hence: $$ \sec^2z < 1+a^2. $$ Moreover: $$ J_a = 1 - \frac{2\sec^2z}{1+a^2}, $$ so: $$ \boxed{ \lambda>-1. } $$ The plateau total: $$ G(s) = F(s) + s-d $$ satisfies: $$ G'(s) = \lambda+1 > 0. $$ The minimum is therefore at: $$ \boxed{ s=d \iff p=0. } $$ --- # 8. The Special Plateau Branch for $n=4$ The total angular-gap budget of a plateau allocation for $n=4$ is: $$ q_L+q_R=1. $$ Thus one side is a one-segment boundary. Let: $$ c=\arcsin\frac1d, \qquad s=\sqrt{d^2-1}=\cot c. $$ Interior stationarity of a positive plateau reduces to: $$ a_\star = \tan\frac c4. $$ The total length is: $$ \boxed{ T_{\rm int}(c) = \csc c + 4\tan\frac c4. } $$ And: $$ T_{\rm int}'(c) = -\csc c\cot c + \sec^2\frac c4. $$ On: $$ c\in[\pi/3,\pi/2] $$ we have: $$ \csc c\cot c = \frac{\cos c}{\sin^2c} \le \frac23, $$ while: $$ \sec^2\frac c4>1. $$ Hence: $$ \boxed{ T_{\rm int}'(c) > \frac13. } $$ The positive-branch minimum occurs at the branch's lowest-$c$ boundary, which is precisely: $$ p=0. $$ Numerical boundary: $$ c_0 \approx 1.256637061435917, $$ $$ d_0 \approx 1.051462224238267. $$ Therefore: $$ \boxed{ n=4 \text{ likewise does not allow a positive optimal plateau}. } $$ --- # 9. One-Segment Obstacle-Inactive Side If one side has only a single segment, let the top point's horizontal distance from its own endpoint be: $$ x. $$ Unit-disc feasibility: $$ x \le \sqrt{d^2-1}. $$ Denote the other side's tangent-chain resolution by: $$ q\ge1. $$ The derivative of total length with respect to $x$ has the upper bound: $$ \frac{dT}{dx} \le \frac{x}{\sqrt{1+x^2}} + 1 - \frac{2}{1+x^2}. $$ The right-hand side is nonpositive on: $$ 0\le x\le\frac1{\sqrt3} $$ the interval above. A two-sided equilateral width-one competitor rules out a global minimal arc with: $$ d>\frac2{\sqrt3}. $$ Thus: $$ \sqrt{d^2-1} \le \frac1{\sqrt3}. $$ The optimum therefore pushes the one-segment side to: $$ \boxed{ x=\sqrt{d^2-1}. } $$ That is, the direct side must also be tangent. --- # 10. Global Exclusion of the One-Segment Boundary Taking the other side's resolution to the continuum yields the lower envelope for all one-segment-side configurations: $$ \boxed{ B_\infty(c) = \csc c + 2\cot c + 3c - \pi. } $$ Its stationary condition is: $$ 3\cos^2c+\cos c-1=0. $$ Hence: $$ \cos c_\star = \frac{\sqrt{13}-1}{6}. $$ and: $$ \boxed{ \min B_\infty \approx 2.297464630768953. } $$ Only an explicit competitor is now needed. Take: $$ d=\frac{53}{50}. $$ The length of the legal four-segment support-tangent path is: $$ \boxed{ L_4(53/50) \approx 2.282033991861008. } $$ The difference: $$ \boxed{ \min B_\infty - L_4(53/50) \approx 0.015430638907945 >0. } $$ Thus the one-segment-side boundary cannot be globally optimal for any: $$ n\ge4 $$ as above. --- # 11. Plateau Runtime Audit This round also takes: $$ n=4,\ldots,12 $$ enumerates all plateau segment allocations, and reoptimizes: $$ (d,a_L,a_R). $$ The best plateau run at every stage lies on the numerical boundary: $$ p \approx 0 $$ as above. This runtime serves only as a red team, not as the source of the analytic proof. Data: ```text WSFC_R05_plateau_audit.csv WSFC_R05_boundary_summary.csv ``` --- # 12. Provisional Finite-$n$ Support-Tangent Reduction Theorem Let: $$ \mathcal P_n = \{ \text{width}\ge1 \text{, with at most }n\text{ segments, polygonal arcs} \}. $$ $n\ge4$. Research proof chain: $$ \boxed{ \begin{aligned} &\text{Adhikari--Pitman finite-}M\text{ convexity}\\ &\rightarrow \text{endpoint strip}\\ &\rightarrow \text{two-disc width criterion}\\ &\rightarrow \text{endpoint antipodes}\\ &\rightarrow \text{interior support saturation}\\ &\rightarrow \text{endpoint shortcut tangency}\\ &\rightarrow \text{top excursion reduction}\\ &\rightarrow \text{positive plateau elimination}\\ &\rightarrow \text{q=0 boundary exclusion}\\ &\rightarrow \text{cell-boundary lower-stage absorption}\\ &\rightarrow \text{equal-angle partition}\\ &\rightarrow \text{balanced segment allocation}. \end{aligned} } $$ Thus the research-level theorem draft is: $$ \boxed{ \Gamma_n^\star \in \mathcal T_n, \qquad n\ge4. } $$ Here $\mathcal T_n$ is Round 03's antipodal support-tangent class. --- # 13. Global Dimension Reduction If the theorem chain passes independent audit: ## even $n=2m$ $$ \boxed{ L_n^\star = \min_d L_{2m}(d). } $$ The original: $$ 2n-2 $$ generic dimensions are compressed to: $$ 1. $$ ## odd $n=2m+1$ $$ \boxed{ L_n^\star = \min_{d,x} L_{2m+1}(d,x). } $$ compressed to: $$ 2. $$ and: $$ \boxed{ t_n = \frac1{L_n^\star}. } $$ --- # 14. Strict Monotonicity Corollary Draft For: $$ s>0, $$ $$ g(q,s) = 2q\tan\frac{s}{2q} $$ is strictly decreasing in $q$. Adding an angular partition cell to a stage-$n$ optimum constructs a stage-$n+1$ path of the same width and strictly shorter length. Thus, if the global reduction theorem holds: $$ \boxed{ L_{n+1}^\star < L_n^\star, } $$ that is: $$ \boxed{ t_{n+1} > t_n. } $$ This round's reduced runtime audit for: $$ n=4,\ldots,20 $$ shows strict increase throughout. --- # 15. Parity Corollary Draft Round 04 proves within the reduced class: $$ |q_L-q_R|\le1. $$ Thus, if the global reduction holds: $$ \boxed{ n\text{ even} \Rightarrow \text{bilateral symmetry}, } $$ $$ \boxed{ n\text{ odd} \Rightarrow \text{asymmetric mirror pair}. } $$ This would correspond to the parity part of the 2007 general-$n$ conjecture. --- # 16. Novelty Discipline This round's targeted web search still leads mainly back to: - the $n=3$ theorem of Panraksa–Wetzel–Wichiramala 2007; - the conjectures for $n>3$ in the same paper; - Adhikari–Pitman's 1989 continuous / finite-$M$ convex reduction; - the 2026 broadworm continuum reformulation. This round's targeted search did not locate subsequent work explicitly claiming that all general-$n$ polygonal conjectures had been proved. But this is not an exhaustive novelty proof. This paper therefore labels only: ```text NO OBVIOUS PRIOR GENERAL-n RESOLUTION FOUND IN TARGETED SEARCH ``` It must not claim: ```text FIRST PROOF IN THE LITERATURE ``` --- # 17. Next Step: Stop Expanding and Enter Verification Round 05 is no longer suited to immediately piling up further new corollaries. Priorities for the next round: 1. blind / lagged independent re-derivation; 2. CAS identity audit; 3. an outward-rounded interval certificate; 4. adversarial counterexample search; 5. a formalization skeleton; 6. a focused attack on whether shifting an interior support line can create a new narrower direction before tangency. --- # Conclusion Round 05 compresses BSP from a set of open boundary cases to: ```text BSP-A endpoint slack CLOSED BSP-B positive plateau CLOSED AT RESEARCH-PROOF LEVEL BSP-C degeneracies LOWER-STAGE ABSORBED q=0 side boundary EXCLUDED ``` Thus Round 03's: $$ 2n-2 \rightarrow 1\text{ or }2 $$ forms a complete global-reduction proof draft for the first time. The most important stance now is not to declare the general-$n$ problem solved immediately, but: $$ \boxed{ \text{freeze} \rightarrow \text{independently re-derive} \rightarrow \text{try to kill the proof}. } $$ If the next round's adversarial verification still finds no opening, a theorem-status upgrade can then be considered. --- # References 1. Ani Adhikari, Jim Pitman, *The Shortest Planar Arc of Width 1*, American Mathematical Monthly 96(4), 309–327 (1989). 2. Chatchawan Panraksa, John E. Wetzel, Wacharin Wichiramala, *The broadest three-segment unit arc*, Periodica Mathematica Hungarica 55(2), 157–168 (2007). 3. Chatchawan Panraksa, John E. Wetzel, Wacharin Wichiramala, *Covering n-Segment Unit Arcs Is Not Sufficient*, Discrete & Computational Geometry 37(2), 297–299 (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-R05-v0.1`. This paper's status: ```text RESEARCH PROOF DRAFT NOT PEER REVIEWED NOT FORMALLY VERIFIED INDEPENDENT AUDIT REQUIRED ``` The chat display is not the canonical source. ===END===