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Math-2889 · mathematics · Probability

The length exponent of critical honeycomb chords with fixed endpoints

Status OpenLLM status Open unverified

Statement

Definitions and mathematical statement

Statement recorded in the supplied source.
For critical honeycomb chords weighted per visited vertex in convex pure lattice hexagons of diameter R converging to a regular hexagon, with prescribed side-interior boundary endpoints separated macroscopically, does length lie in $[R^{4/3-\epsilon },R^{4/3+\epsilon }]$ with probability tending to one for every $\epsilon >0$? Ask the same in the specified nondegenerate half-plane, comparable-width strip and pi/3 or 2pi/3 wedge geometries, uniformly on compact families, with partition mass at least $cR^{-5/4}$.

Short English statement

For critical honeycomb chords weighted per visited vertex in convex pure lattice hexagons of diameter R converging to a regular hexagon, with prescribed side-interior boundary endpoints separated macroscopically, does length lie in $[R^{4/3-\epsilon },R^{4/3+\epsilon }]$ with probability tending to one for every $\epsilon >0$? Ask the same in the specified nondegenerate half-plane, comparable-width strip and pi/3 or 2pi/3 wedge geometries, uniformly on compact families, with partition mass at least $cR^{-5/4}$.

Sources

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-2889-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-2889

Chords with prescribed separated ports are claimed to have length $R^{4/3+o(1)}$ in regular hexagons, with analogous half-plane and cut laws. The question retains these same parameters and conventions. Within these assumptions, this is a claimed affirmative answer to the stated question.

The length exponent of critical honeycomb chords with fixed endpoints

OpenAI's claim

OpenAI presents a proof in the selected manuscript. Chords with prescribed separated ports are claimed to have length $R^{4/3+o(1)}$ in regular hexagons, with analogous half-plane and cut laws.

Mathematical statement of the claimed result

Result 1.
A honeycomb chord joins two boundary-edge midpoints, enters through triangle centres, visits each centre at most once, and makes no additional boundary exit. Its length $L$ counts visited centres. Give each chord probability proportional to $x_*^L$, where $x_*=(2+\sqrt2)^{-1/2}$. Let $D_R$ be convex hexagons built from complete lattice rows with $R^{-1}D_R$ tending to a regular hexagon. Suppose the prescribed endpoints $a_R,b_R$, after division by $R$, tend to different interior points of sides. For every $\varepsilon>0$, \[ \mathbb P_{D_R;a_R,b_R}\{R^{4/3-\varepsilon}\le L\le R^{4/3+\varepsilon}\}\longrightarrow1. \] The conclusion also applies in a lattice half-plane with endpoint distance $R$; in a strip whose height and $R$ have comparable ratio, with endpoints on opposite cuts and separation $O(R)$; and in a wedge of angle $\pi/3$ or $2\pi/3$ with one endpoint on each ray, distance $O(R)$ from the apex, and distance comparable to $R$ from the opposite cut. Bounds are uniform over compact families of these nondegenerate rescaled configurations. Each partition sum is at least $cR^{-5/4}$.

Outline of the argument

Finite estimates under the irreducible bridge law first control a fixed source and terminal intervals. Strip partition regularity and exact height/displacement atoms sharpen the result to specified endpoints, while parallel and nonparallel cut comparisons give the different geometries and averaged-height mean statements.

Formal verification

For family 237, the selected Lean statements give critical-strip mass and displacement exponents, bridge summability, and existence of pulled free energy. The $4/3$ bridge-length and $3/4$ spatial laws and the other polygon/chord exponents are outside these statements. No Comparator is assigned to the selected manuscript here; the family page selects other papers or supporting statements. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

Sources

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