> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification. # WSFC Round 04 ## Support-Class Necessity I: Existing Convexity Gates, Endpoint Antipodes, and Balanced Segment Allocation **English Title:** *Support-Class Necessity I: Convexity, Endpoint Antipodes, and Balanced Segment Allocation* **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:** `NECESSITY ATTACK / TWO EXTERNAL GATES CLOSED / NEW CONVEXITY LEMMAS` **Canonical Source:** UTF-8 Markdown --- # Abstract Round 03 compressed the high-dimensional vertex search for the broadest polygonal worm into an antipodal support-tangent model, leaving the main proof gate: $$ \boxed{ \Gamma_n^\star \in \mathcal T_n? } $$ This round begins a direct attack on that necessity. The first important discovery is that the two prerequisite gates listed as unproved in Round 03 can actually be connected directly to results of Adhikari–Pitman 1989. They fix an arbitrary positive integer $M$, take a shortest arc in the compact class with at most $M$ segments and width at least $1$, and prove: $$ \boxed{ \text{Any such minimal finite-segment arc must be convex.} } $$ WSFC therefore does not need to reprove finite-stage convexity. The same paper also proves in the convex case that, if a convex arc leaves the strip formed by the perpendiculars at its two endpoints, perpendicular reduction to the endpoint baseline produces a shorter arc of greater width. Thus the shortest convex arc can be restricted to the endpoint strip. Within this strip-confined convex normal form, this round derives three further new lemmas. First, the **Endpoint Antipode Lemma**: the opposite support of a positive-slope edge on the left must be the right endpoint; the opposite support of a negative-slope edge on the right must be the left endpoint. This is not an additional conjecture, but a direct consequence of strip geometry. Second, the **Interior Support-Slack Lemma**: in a convex polygonal cell with fixed support-angle order, if an interior side edge has distance strictly greater than $1$ from its opposite endpoint, translating that support line inward to distance $1$ strictly shortens the arc without increasing the segment count. Thus interior active side edges in a minimizer must be tangent to the corresponding endpoint unit disc. Third, for Round 03's reduced tangent class, this round proves the **Balanced Segment Allocation Theorem**. Define the one-sided, breadth-one reduced length at fixed endpoint distance $d$: $$ J_d(q,a) = \sqrt{d^2-1} + a + 2q \tan \left( \frac{ \Delta_d(a) }{ 2q } \right), $$ where: $$ \Delta_d(a) = \frac\pi2 - \arccos\frac1d - 2\arctan a. $$ This paper proves that $J_d(q,a)$, on: $$ q>0, \qquad a>0, \qquad \Delta_d(a)>0 $$ is **strictly jointly convex** in $(q,a)$. Thus, for fixed total angular-gap budget: $$ q_L+q_R=Q $$ and: $$ a_L+a_R=d, $$ the unique minimizer of the continuous relaxation is: $$ q_L=q_R=\frac Q2, \qquad a_L=a_R=\frac d2. $$ For an integer segment budget: - if $Q$ is even, the optimal allocation must be perfectly balanced; - if $Q$ is odd, the optimal gap counts on the two sides must differ by exactly $1$ and form a pair of mirror solutions. This directly proves, within the support-tangent class, the parity pattern observed in Round 03 / 2007: $$ \boxed{ \text{even }n \Rightarrow \text{bilateral symmetry}, } $$ while: $$ \boxed{ \text{odd }n>3 \Rightarrow \text{two mirror asymmetric reduced minimizers}. } $$ After Round 04, the Major Gate has therefore been compressed from: ```text CONVEXITY DRAPE / STRIP REDUCTION ANTIPODAL ENDPOINT RESPONSIBILITY WIDTH EQUALIZATION BALANCED SEGMENT ALLOCATION ``` to the narrower: $$ \boxed{ \text{Active-Constraint Saturation at the boundary / endpoint lines} } $$ and: $$ \boxed{ \text{Exclusion of a top horizontal plateau or other support-order degenerations.} } $$ --- # 1. Relocating the Problem Round 03's major question is: $$ \Gamma_n^\star \in \mathcal T_n? $$ Here $\mathcal T_n$ is the reduced class characterized by: - convexity; - confinement to the endpoint strip; - a one-gap upper chain; - side edges supported antipodally by endpoints; - width-one tangency; - equal-angle partition; - balanced left/right segment allocation as above. This round decomposes necessity into auditable gates. --- # 2. External Gate A: An Existing Theorem Gives Finite-Segment Convexity Adhikari–Pitman consider: $$ \mathcal C_M $$ as the class of arcs with: - a fixed starting point; - length at most $3$; - at most $M$ segments as above. Length and width are continuous in its compact topology, so a minimal arc of width at least $1$ exists. They prove successively: 1. all segment endpoints of a minimal arc are corners of its convex hull; 2. every corner is visited only once; 3. a minimal arc has no self-intersection; 4. finally, in Section 4: $$ \boxed{ \text{a minimal arc with at most }M\text{ segments is convex.} } $$ Since our finite-$n$ broadest problem is dual to: > finding a shortest width-one arc among arcs with at most $n$ segments, this theorem can directly serve as WSFC's finite-stage convexity gate. Thus: ```text CONVEXITY NECESSITY ``` changes from `OPEN` to: ```text SOURCE-CLOSED ``` --- # 3. External Gate B: Endpoint-Strip Reduction For a convex arc $AB$, take: $$ A=(0,0), \qquad B=(d,0). $$ Adhikari–Pitman note: If the arc extends to the left of $A$ or the right of $B$, perpendicular reduction of the protruding portion to $AB$ produces a: $$ \boxed{ \text{shorter arc with greater width}. } $$ Thus a minimal convex width-one arc can necessarily be restricted to: $$ \boxed{ 0\le x\le d. } $$ namely the endpoint perpendicular strip. In what follows, this paper calls it: $$ \boxed{ \text{strip-confined convex normal form}. } $$ It agrees with the drape-type geometry used in WSFC Round 03, but this paper does not conflate it with the more general formal definition of drapeability in later literature. --- # 4. Adhikari–Pitman's Two-Disc Condition In the strip above, let: $$ A'B' $$ be the horizontal segment on: $$ y=1 $$ as above. Adhikari–Pitman give a width-one criterion for a convex arc: the arc intersects $A'B'$ at a point $F$ such that: 1. the arc reaches unit height; 2. the subarc from $A$ to $F$ does not enter the open unit disc centered at $B$; 3. the subarc from $F$ to $B$ does not enter the open unit disc centered at $A$. The original width condition is therefore rewritten as: $$ \boxed{ \text{left path avoids }D(B,1), \qquad \text{right path avoids }D(A,1). } $$ Round 03's two endpoint unit discs were not fitted after the fact. They are precisely the core constraints of the classical convex broadworm proof. --- # 5. New Lemma 1: Endpoint Antipode Lemma Consider a convex upper chain in the strip. By convexity, it can be viewed as the piecewise-linear concave graph: $$ y=f(x) $$ as above. Take a positive-slope edge: $$ e: y=mx+c, \qquad m>0. $$ Its upward outward normal can be chosen as: $$ n = \frac{ (-m,1) }{ \sqrt{1+m^2} }. $$ For any point in the convex hull: $$ P=(x,y), $$ since: $$ 0\le x\le d, \qquad y\ge0, $$ we have: $$ \langle n,P\rangle = \frac{-mx+y}{\sqrt{1+m^2}} \ge \frac{-md}{\sqrt{1+m^2}} = \langle n,B\rangle. $$ Thus: $$ \boxed{ B \text{ is the opposite support point for this edge normal.} } $$ Similarly, if: $$ m<0, $$ then: $$ \boxed{ A \text{ is the opposite support point.} } $$ If: $$ m=0, $$ the opposite support is the base segment $AB$. Thus: $$ \boxed{ \text{convex + strip} \Rightarrow \text{endpoint antipodal responsibility}. } $$ This step requires no additional conjecture. --- # 6. Consequence: Side-Edge Width Is the Endpoint's Distance to the Line For a positive-slope edge line on the left: $$ L_i, $$ its width constraint directly becomes: $$ \boxed{ \operatorname{dist}(B,L_i)\ge1. } $$ On the right: $$ \boxed{ \operatorname{dist}(A,L_i)\ge1. } $$ Thus, in Round 03's: ```text ANTIPODAL SUPPORT-TANGENT ``` the antipodal-endpoint part follows automatically from strip convexity. What truly remains to be proved is whether: $$ \boxed{ \operatorname{dist}(B,L_i)=1 } $$ or the corresponding equality on the right is saturated throughout at the minimizer. --- # 7. New Lemma 2: Interior Support-Slack Elimination Consider a fixed combinatorial / normal-order cell. Let consecutive support lines be: $$ L_i: \langle n_i,x\rangle=h_i, $$ with normal-angle gaps: $$ \delta_i\in(0,\pi). $$ For a convex polygonal boundary, support-number calculus gives the first-order coefficient of boundary length with respect to an interior support number $h_i$: $$ \boxed{ \frac{\partial \ell}{\partial h_i} = \tan\frac{\delta_i}{2} + \tan\frac{\delta_{i+1}}{2} > 0. } $$ Thus, if an interior side edge has strict width slack: $$ h_i>1 $$ then a small variation preserving normal order and adjacency combinatorics can move: $$ h_i \downarrow1 $$ and strictly shorten the arc. Interior minimal edges must therefore satisfy: $$ \boxed{ h_i=1. } $$ By the Endpoint Antipode Lemma: $$ \boxed{ \text{all interior left edges are tangent to }D(B,1), } $$ $$ \boxed{ \text{all interior right edges are tangent to }D(A,1). } $$ --- # 8. What Does This Step Close? Round 03 originally treated: > every side segment is tangent to an endpoint circle as a numerical pattern. Round 04 now upgrades it, at least for: $$ \boxed{ \text{interior side edges} } $$ to a fixed-support-order variational lemma. The most sensitive remaining cases are: - the first side edge; - the last side edge; - the top transition; - support-order degeneration. That is, boundary lines rather than the entire chain. --- # 9. Top-Height Slack Can Also Be Greatly Reduced If the maximum height of a strip-confined convex arc is: $$ H>1, $$ then it generally has two intersections, left and right, with: $$ y=1 $$ namely: $$ F_L,F_R. $$ Replacing: $$ F_L \rightarrow \text{top subarc} \rightarrow F_R $$ with the horizontal chord: $$ F_LF_R \subset \{y=1\}, $$ will: - not increase the segment count; - strictly shorten it, unless it was already that horizontal segment; - use a horizontal line $y=1$ that enters neither open endpoint unit disc. Thus a genuine minimizer need not have a curved / broken top excursion with: $$ H>1 $$ as above. We must still separately exclude: $$ \boxed{ \text{positive-length horizontal plateau at }y=1. } $$ This is retained as the next discrete boundary case, not forcibly declared closed in this round. --- # 10. Reduced One-Side Functional Enter the tangent class already established in Round 03. Fix: $$ d>1. $$ Let the apex horizontal distance on one side be: $$ a>0. $$ Define: $$ c(d) = \frac\pi2 - \arccos\frac1d. $$ and: $$ \boxed{ \Delta_d(a) = c(d) - 2\arctan a. } $$ If that side has: $$ q $$ support-angle gaps, then: $$ \boxed{ J_d(q,a) = \sqrt{d^2-1} + a + 2q \tan \left( \frac{ \Delta_d(a) }{ 2q } \right). } $$ This is the one-sided reduced path length. --- # 11. Relocating the Equal-Angle Partition Let: $$ g(q,s) = 2q \tan \left( \frac{s}{2q} \right). $$ For fixed $q$ and total angle span $s$, Round 03 already obtains by Jensen: $$ \boxed{ \delta_1=\cdots=\delta_q=\frac{s}{q} } $$ minimizing the tangent-chain excess. Thus: $$ J_d(q,a) $$ has already integrated out all internal angular freedoms on that side. --- # 12. New Lemma 3: $g(q,s)$ Is a Convex Perspective Let: $$ z = \frac{s}{2q}. $$ Then: $$ g_s = \sec^2z > 0, $$ $$ g_{ss} = \frac{ \sec^2z\tan z }{ q } > 0, $$ $$ g_{qq} = \frac{ s^2 }{ q^3 } \sec^2z\tan z > 0, $$ $$ g_{qs} = - \frac{ s }{ q^2 } \sec^2z\tan z. $$ And: $$ g_{qq}g_{ss}-g_{qs}^2=0. $$ Thus $g$ is the convex function: $$ (q,s) \mapsto g(q,s), $$ and can also be viewed as the perspective: $$ q \left[ 2\tan \left( \frac{s/q}{2} \right) \right] $$ as above. --- # 13. Convexity of $\Delta_d(a)$ For: $$ a>0, $$ $$ \Delta_d'(a) = -\frac{2}{1+a^2}, $$ $$ \boxed{ \Delta_d''(a) = \frac{ 4a }{ (1+a^2)^2 } > 0. } $$ Thus $\Delta_d$ is strictly convex. Moreover, since: $$ g_s>0, $$ convex composition applies. --- # 14. New Lemma 4: One-Side Joint Strict Convexity We have: $$ J_d(q,a) = \sqrt{d^2-1} + a + g(q,\Delta_d(a)). $$ Its Hessian determinant reduces to: $$ \boxed{ \det\nabla^2J_d = g_{qq} g_s \Delta_d''(a) > 0. } $$ and: $$ J_{qq}=g_{qq}>0. $$ Thus: $$ \boxed{ J_d(q,a) \text{ is strictly jointly convex in }(q,a)\text{.} } $$ This is the most important new analytic lemma in Round 04. --- # 15. Balanced Segment Allocation Problem Let: $$ q_L+q_R=Q, $$ with the left/right apex horizontal distances satisfying: $$ a_L+a_R=d. $$ The total reduced length is: $$ \boxed{ H_d(q_L,a_L) = J_d(q_L,a_L) + J_d(Q-q_L,d-a_L). } $$ By the joint strict convexity of $J_d$: $$ H_d $$ is convex. The reflection: $$ (q_L,a_L) \mapsto (Q-q_L,d-a_L) $$ leaves $H_d$ unchanged. --- # 16. Continuous Balance Theorem By convexity + reflection symmetry: $$ \boxed{ q_L=q_R=\frac Q2, \qquad a_L=a_R=\frac d2 } $$ is the unique minimizer of the continuous relaxation. It also follows directly from Jensen: $$ J_d(q_L,a_L) + J_d(q_R,a_R) \ge 2J_d \left( \frac Q2, \frac d2 \right). $$ Thus: $$ \boxed{ \text{the continuous freedom resource must be perfectly balanced left and right.} } $$ --- # 17. Integer Balance Theorem The actual segment counts are integers. Define: $$ F_d(q) = \min_a \left[ J_d(q,a) + J_d(Q-q,d-a) \right]. $$ Partial minimization preserves convexity, so: $$ F_d(q) $$ is convex. At the same time: $$ F_d(q)=F_d(Q-q). $$ The integer minimizers must therefore be the integer pair nearest to: $$ Q/2 $$ as above. That is: $$ \boxed{ |q_L-q_R| \le1. } $$ --- # 18. Translation to Segment Counts If one side has: $$ k_L $$ tangent segments, then: $$ q_L=k_L-1. $$ Similarly: $$ q_R=k_R-1. $$ And: $$ k_L+k_R=n $$ gives: $$ q_L+q_R=n-2. $$ Thus: $$ \boxed{ |k_L-k_R| \le1. } $$ Hence: - for even $n$, the left/right segment counts must be equal; - for odd $n$, the left/right segment counts must differ by one. This is not a numerical pattern, but a theorem within the reduced support class. --- # 19. Even Parity Corollary If: $$ n=2m, $$ then: $$ q_L=q_R=m-1. $$ By strict convexity in $a$ and reflection symmetry: $$ \boxed{ a_L=a_R=\frac d2. } $$ the apex therefore lies on the centerline: $$ x=\frac d2. $$ Thus the reduced minimizer is bilaterally symmetric. That is: $$ \boxed{ \text{even }n \Rightarrow \text{bilateral symmetry} } $$ has been proved within the support-tangent class. --- # 20. Odd Parity Corollary If: $$ n=2m+1, $$ then: $$ \{q_L,q_R\} = \{m-1,m\}. $$ Take: $$ q_R>q_L. $$ At the center: $$ a_L=a_R=a=\frac d2. $$ we have: $$ \frac{\partial J_d}{\partial a} = 1 - \frac{ 2\sec^2 \left( \Delta_d(a)/(2q) \right) }{ 1+a^2 }. $$ For fixed $a$, as $q$ increases: $$ \frac{\Delta_d(a)}{2q} $$ decreases, so: $$ \frac{\partial J_d}{\partial a} $$ increases. Thus, at the symmetric point: $$ \frac{dH}{dx} > 0. $$ By strict convexity, the unique oriented minimizer must satisfy: $$ \boxed{ x<\frac d2. } $$ That is, the side with more segments carries a larger support-angle span. Reflection gives the other mirror minimizer. Thus: $$ \boxed{ \text{odd }n>3 \Rightarrow \text{no bilateral symmetry within the reduced optimum；} } $$ with a pair of mirror solutions. --- # 21. This Directly Explains the Parity Pattern in Round 02 / 03 Numerically: ```text n=4 symmetric n=5 asymmetric n=6 symmetric n=7 asymmetric n=8 symmetric ``` Round 04 now shows: > This is not a pattern the optimizer happened to produce, but a necessary consequence of balanced integer resource allocation—provided the global solution has entered the active support-tangent class. --- # 22. Numerical Allocation Scan For: $$ n=4,\ldots,14 $$ this round enumerates all: $$ q_L+q_R=n-2, \qquad q_L,q_R\ge1. $$ and reoptimizes for every allocation: $$ (d,x). $$ In every test, the minimum-length allocation satisfies: $$ \boxed{ |q_L-q_R|\le1. } $$ This is not the source of the theorem, only an independent runtime check of the analytic Balance Theorem. The complete data are retained in: ```text WSFC_R04_segment_allocation_scan.csv ``` --- # 23. The Major Gate Narrows Again Round 03: ```text CONVEXITY DRAPE / STRIP WIDTH EQUALIZATION ANTIPODAL ENDPOINT EQUAL ANGLES BALANCED COUNTS ``` Round 04: ```text CONVEXITY SOURCE-CLOSED STRIP REDUCTION SOURCE-CLOSED ENDPOINT ANTIPODE DERIVED INTERIOR TANGENCY DERIVED IN FIXED SUPPORT CELL EQUAL-ANGLE PARTITION DERIVED BALANCED COUNTS DERIVED EVEN/ODD PARITY DERIVED IN SUPPORT CLASS ``` The actual residual: $$ \boxed{ \text{boundary active-constraint saturation} } $$ In particular: 1. must the first / last side support be tangent? 2. can a positive-length top plateau always be eliminated? 3. is there escape through support-order changes / degenerate vertices? 4. can all these variations be completed without increasing the segment count? --- # 24. The New Minimal Proof Debt Thus the actual target is no longer: > Why does the entire polygonal extremizer look like Round 03? but, more narrowly: > In a known convex, strip-confined finite-segment minimizer, can a boundary side constraint have slack that cannot be eliminated because of the segment budget / a combinatorial transition? This can be labeled: $$ \boxed{ \textbf{Boundary Saturation Problem}. } $$ Abbreviated: $$ \boxed{ \mathrm{BSP}. } $$ --- # 25. The Three Cases of BSP ## BSP-A — Endpoint Edge Slack The first or last side edge: $$ \operatorname{dist} ( \text{opposite endpoint}, L ) >1. $$ We must prove it can be shortened by rotation / replacement. ## BSP-B — Top Plateau There exists: $$ F_LF_R \subset \{y=1\} $$ with: $$ |F_LF_R|>0. $$ We must prove it cannot be an optimal finite-budget resource allocation. ## BSP-C — Degenerate Support Birth Some perturbation causes: - segment length $\to0$; - consecutive normals to collide; - a change in support order. We must prove closure can be handled across cell boundaries, not merely within a single smooth cell. --- # 26. This Connects Directly to WSFC's Actual Dynamics BSP-C is important because sequence freedom expansion: $$ \Gamma_n \rightarrow \Gamma_{n+1} $$ already crosses: $$ \boxed{ \text{support-combinatorial cells}. } $$ The final theorem therefore cannot prove only: > The optimizer within every cell is a tangent partition. It must also prove: > Cell boundaries do not hide a shorter extremizer. This is the clearest global obstacle after Round 04. --- # 27. Round 04 Conclusion This round further compresses Round 03's high-dimensional necessity problem. There are four most important results: First, finite-stage convexity need not be reproved; Adhikari–Pitman have already proved it at the level of any fixed $M$ segments. Second, endpoint-strip reduction also already has a classical proof. Third, once we enter the strip-confined convex upper chain: $$ \boxed{ \text{antipodal endpoint identity} } $$ is automatic. Fourth, in the active support-tangent reduced class: $$ \boxed{ \text{the left/right segment budgets must be as balanced as possible.} } $$ The parity pattern: $$ \boxed{ \text{even symmetric / odd asymmetric mirror pair} } $$ then becomes a theorem-level reduced-class result. Thus the Major Gate is now narrowed from proving the whole geometry to: $$ \boxed{ \text{Boundary Saturation} + \text{Cell-Boundary Exclusion}. } $$ If these two issues are closed, Round 03's: $$ 2n-2 \rightarrow 1\text{ or }2 $$ will be very close to a genuine global finite-$n$ reduction. --- # 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. Zhipeng Deng, *Revisit escape path for infinite unit strip forest and unit broadworm*, arXiv:2607.18563 (2026). 4. Yevgenya Movshovich, John Wetzel, *Drapeability and Λ-Frames*, Geometry 2(4), 18 (2025). --- # Canonical Source Declaration The original file is the UTF-8 Markdown canonical source for `WSFC-R04-v0.1`. This round does not claim: - that Support-Class Necessity has been fully proved; - that tangency of the first / last endpoint edges is globally closed; - that the top plateau is excluded; - that support-order cell boundaries are excluded; - that the global broadest polygonal arc for $n\ge4$ is solved; - that all the 2007 conjectures are proved. What this round establishes is: ```text SOURCE-CLOSED: finite-M convexity convex endpoint-strip reduction DERIVED: endpoint antipode identity interior support slack elimination one-side joint strict convexity balanced integer segment allocation reduced-class parity theorem OPEN: Boundary Saturation Problem cell-boundary exclusion global support-class necessity ``` ===END===