Polylogarithmic Full-Chord Buffon Discrepancy
Steinerberger introduced the Buffon discrepancy problem, asking how accurately a one-dimensional set of length in a convex body can match the Crofton-predicted line-intersection counts, and proved an upper bound via a Steinhaus longimeter construction. His third open question asks whether restricting to sets built from full chords - intersections of lines with , the class containing every Steinhaus set - fundamentally changes the problem.
It does. Using the Aistleitner-Bilyk-Nikolov star-discrepancy theorem for arbitrary measures, full-chord constructions with discrepancy are shown to exist for every compact convex body with finite piecewise boundary. In the disk, every full-chord construction is shown to have discrepancy at least , via Schmidt's two-dimensional rectangle lower bound - where Steinerberger's concentric-circle construction, which is not full-chord, achieves discrepancy at most .
- Result
- Proved(see note)
- Status
- Partial result
- AI contribution
- AI-assisted
- Method
- Argument
- Field
- Discrepancy theory / integral geometry
- Posed by
- Stefan Steinerberger
- Year posed
- 2026
- Years open
- 0y
- Solved
- 2026-05-18
- Model
- GPT-5.5
- Vendor
- OpenAI
- Collaborators
- Samuel Korsky
- Verification
- Unreviewed
- Publication
- Preprint
- Significance
- 5 / 100
- Disclosed cost
- —
- Wikipedia
- No dedicated article
What was actually shown
This settles Steinerberger's third open question and separates the two models: in the disk, full chords cost you a factor growing like over what is achievable without the restriction. It also improves the Steinhaus-type to polylogarithmic within the full-chord class.
It does not settle the Buffon discrepancy problem itself. Steinerberger's first question - whether every convex body admits a set of discrepancy , and if not what the truth is - is untouched, and the paper's closing line names it as the natural next question. Inside the full-chord model the order is pinned only between and , and the lower bound is proved for the disk alone. The paper says the exponents are unlikely to be sharp.
The upper bound is an existence statement: it inherits the Aistleitner-Bilyk-Nikolov theorem, which is proved by transference and supplies no explicit construction.
What the AI did
While the proof strategy (apply weighted versions of known discrepancy results in the necessary ways) was due to the author, GPT performed the bulk of the technical details and deserves a substantial amount of credit here.
Verification
A preprint by a single author, not refereed and not endorsed by anyone independent, so this stays Unreviewed.
This site checked the reduction the note rests on. Lemma 3.1 says the chords crossing a test line form a union of two rectangles in endpoint-pair space, of measure - the identity that turns a Buffon problem into a two-dimensional rectangle discrepancy problem. It was confirmed exactly for the disk by quadrature at five arc widths (agreement to ), and for an ellipse by sampling the kinematic measure in coordinates, which know nothing about endpoint pairs, giving agreement within 0.2% and an implied of 4.6012 against a perimeter of 4.6026. The Aistleitner-Bilyk-Nikolov bound is quoted faithfully: their at is . An independent exact-supremum harness reproduces both known growth rates: for Steinhaus-type constructions against the proved , and for i.i.d. chords against the square root.
Neither theorem itself was checked. The upper bound rests on an existence result with no explicit construction, and the closest thing this site could build - a Halton set pushed through the Rosenblatt transform of - fits , no better than Steinhaus. That is a limitation of the proxy, not evidence against the theorem. The lower bound is below the resolution of any feasible experiment.
Sources
- PaperarXivKorsky - Randomly shifted Steinhaus longimeters and Buffon discrepancy (the earlier attempt)Aistleitner, Bilyk, Nikolov - the arbitrary-measure star-discrepancy theorem behind the upper bound
- Problem recordSteinerberger - Buffon discrepancy and the Steinhaus longimeter (the problem, open question 3)
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