Root d is enough?

0:00Keeping a walk close to home

Take a list of vectors in d-dimensional space, each of length at most one, in a fixed order. Give each one a sign, plus or minus, and follow the walk of partial sums. The goal is one choice of signs that keeps every partial sum short, not just the last one, with a bound that depends only on the dimension, however long the list. A second problem goes with it. If the vectors add up to zero, reorder them, without signs, so that every partial sum stays short. Steinitz studied this to rearrange series of vectors, and Grinberg and Sevast’yanov showed that some order keeps every partial sum within d times the unit ball, for any norm. For Euclidean length, the expected answer is the square root of d. The Steinitz–Bergström conjecture asks for orderings with every partial sum within a constant times √d, and Bansal, Jiang, Meka, Singla and Sinha asked the same for signs in a prescribed order. Banaszczyk proved a bound of order √d + √(log N), which grows slowly with the number of vectors N. A 2026 preprint of Dutta, Jha and Jiang gave an efficient bound of order √d + d^(1/4) times a power of log N. And √d is needed: d + 1 unit vectors pointing to the corners of a regular simplex add up to zero, and in any order, the partial sum halfway through has length at least half of √d. In our test, with 20,000 random unit vectors in 16 dimensions, random signs let the walk drift past 160, while the greedy rule, which picks whichever sign makes the new sum shorter, stays below 4, about √d. But in 2 or more dimensions, a list can be built against any rule that decides one step at a time. Send each new vector perpendicular to the current sum: then both signs give the same length, and after t steps it is √t. For 18 such vectors in the plane, built against the greedy rule, the walk reaches length 4.24. Yet trying all 2¹⁸ sign patterns, with hindsight, keeps every partial sum within √2, the least possible for this list.

2:10The claim

Result 097 is one OpenAI preprint, 29 pages, dated September 24. It claims an absolute constant C such that every list of vectors in the Euclidean unit ball of d-dimensional space, of any length, has one choice of signs keeping every partial sum within C√d. Consequently, every list that adds up to zero can be reordered with every partial sum within the same bound. With the simplex example, this pins down the growth of the Euclidean Steinitz constant, up to a constant factor: it grows like √d. That would settle the Steinitz–Bergström conjecture and the question of Bansal and his coauthors. The constant is not computed. The proof is existential: each sign may depend on the whole list, so it gives no rule that works one step at a time, and no efficient algorithm. The step from signs to orderings is Chobanyan’s transference. Start from the best ordering and sign it with the theorem. Put the plus vectors in that order, then the minus vectors in reverse. Every new partial sum is at most the average of the old bound and C√d, and since the old ordering was already the best, its bound is at most C√d. In our test on 200 zero-sum families with greedy signs, the new order always stayed within that average. OpenAI published a Lean formalization of both statements, 29 files and about 15,000 lines, matched against a separate statement. We have not run the check.

3:41How the proof goes

The proof moves to a space of coefficients, one for each incoming vector plus d for an initial state. A chain of linear maps turns coefficients into states: at each step, an invertible contraction shrinks the old state, and the next coefficient adds a multiple of the incoming vector. The convex body of coefficients whose every state is short, within a constant times √d, will hold the whole walk at once. A shift in each new coefficient exactly undoes the contraction, so that, for the chosen coefficients, every state is exactly a signed partial sum, scaled by a fixed factor. The first ingredient turns a geometric bound into signs. If a symmetric convex body has small Dirichlet energy for every positive diagonal weighting, then signs can be chosen one coordinate at a time, using domains prepared from the whole list, so that the final point lands inside the body, even when each step is also shifted by up to 1/4. This adapts a symmetrization method from a September 2026 preprint by Guo, Fang and Lu, on a related balancing problem, whose authors credit an AI research agent for their proof; the paper proves the version it needs in full. The second ingredient is the energy bound for the body of short states, uniform in the number of steps. The paper runs a stationary Ornstein–Uhlenbeck process in coefficient space and splits the states into dyadic spectral scales. A matrix estimate charges each change to an additive trace budget. At one scale, a group of constraints survives together with probability bounded below, over a matched time interval. The Gaussian correlation inequality combines the groups, scales and intervals, and the survival rate gives the energy bound.

5:32Where it stands

The preprint is not peer reviewed and has not been revised. OpenAI’s Lean formalization covers both the signed and the reordered bound, with an unspecified absolute constant; its files include a proof of the Gaussian correlation inequality. We have not run it. The paper builds on several 2026 preprints on related balancing problems. It is not among the papers OSVerify reviewed, and we have seen no expert review. If it holds, signs for any list, or an ordering for any list that adds up to zero, can keep every partial sum within a constant times √d, however long the list, and √d is the best possible growth. The proof gives no efficient way to find them.