Can you always split a pile in two, fairly?

Vector balancing in polynomial time: 8272, beaten by 99

A September 2026 paper shows a computer can split any pile into two fair teams. It guarantees a score of 8272. Four days later, another paper reported 99.

By vizmaya desk, 4 November 2026. Paper: Shengtao Guo, Ethan X. Fang and Junwei Lu, arXiv 2609.23540.

Why would anyone care about a score of 8272?

Source: paper p.0 and p.1; arXiv 2609.23540 and 2609.30044

The Komlós problem, in plain words

Split the pile in two so every measure comes out even.

Give each item a team, plus or minus. For each measure, add up the plus items and subtract the minus items.

Your score is the biggest leftover. Smaller is fairer. Every item is small: square its measures, add them, and you get at most 1.

Is there a score you can promise for any pile?

Made-up example: 12 items, 3 measures, all 4,096 splits tried. Typical means the middle one. Not from the paper.

Source: problem statement, paper p.1. The 12 items are invented for this picture.

The old world

Old guarantees crept up as the pile grew.

Early methods promised a score near log n, where n counts the items (p.1). In 1998 Banaszczyk lowered it to the square root of log n. By 2025 the best method was near the fourth root of log n.

These grow slowly. But they keep growing.

Could the line go completely flat?

Shape only: constants set to 1 (computed). Real bounds carry large constants.

Source: paper p.1; Bansal and Jiang, arXiv 2508.03961 (5 Aug 2025); computed from ln n, research file 9d

The crack

A proof said flat is possible. No one could build it.

On 10 September 2026 the same three authors proved a flat guarantee exists: a score below 7.52, for any size of pile (computed from 3 × √(2π)).

But a proof that a fair split exists does not say how to find one. Trying every split takes far too long. Finding it fast was still open (p.1).

So how do you build the split, step by step?

Source: paper p.1; arXiv 2609.11189 (10 Sep 2026); computed 3 × √(2π) = 7.5199

The turn, step 1

Start at zero. Nudge each item toward a team. Freeze it when it lands.

Every item starts undecided, at zero. Little by little it moves toward plus one or minus one. When it lands there, it never moves again (p.4).

When only 16 items are still moving, round them to plus or minus and stop (p.4, p.6).

How do you nudge without wrecking a measure?

Illustration: here the drift is random. The paper chooses each nudge on purpose.

Source: paper p.1, p.4, p.6 (K = 16)

The turn, step 2

Near a safety line, a lock goes on.

Each measure has a safety line. When one gets close, the method forbids any move that would push it closer (p.4).

Every lock uses up one direction of freedom. Too many locks and the method cannot move at all (p.2, p.8).

How many locks can it afford?

Source: paper p.2, p.4, p.8. The squares are an illustration.

The turn, step 3

The trick: prove that few locks are ever needed.

The method adds up how risky every measure is in one table. One number from that table caps the total (p.5).

That cap keeps the locks below 5 out of every 16 free directions, under 31.25% (computed from 5 ÷ 16; p.8).

And if the ground offers no way down?

Illustration: the bucket size is fixed. Each risky measure is a heavy stone, so only a few fit.

Source: paper p.5, p.8 (fewer than 5s/K constraints, K = 16); computed 5 ÷ 16 = 0.3125

The turn, step 4

At the start the ground is flat. So it follows the curve.

At the very start there is no downhill direction: the slope is zero (p.2). So the method steps where the ground curves downward, like the top of a hill (p.2, p.5).

The paper proves this ends after at most about n⁷ steps, where n counts the items (p.11).

What does the finished method promise?

Source: paper p.2, p.5, p.11. The hill is an illustration.

The result

Every measure ends under 8272, for any pile.

It does not matter how many items or measures there are (p.1). The method makes no random choices. Its work grows like a fixed power of the pile: about m × n⁹ + n¹⁰ steps, with m measures and n items (p.6).

Contested: "first"

"First polynomial-time algorithm" is the authors' own claim. I did not check every earlier paper.

Where does 8272 come from?

8272

The guaranteed score

Source: paper p.1, p.6, p.12 (Theorem 1.1, Proposition 2.3)

Where 8272 comes from

99% of the 8272 is one safety margin.

The bound adds three pieces: 2 × 16³ + 4 × 16 + 16 = 8272 (p.12). The safety line alone is 8192, or 99.03% (computed).

Entries dropped along the way add 64. Rounding the last 16 items adds 16.

Is 8272 big, or small?

Source: paper p.4, p.12; shares computed: 8192 ÷ 8272 = 99.03%, 64 ÷ 8272 = 0.77%, 16 ÷ 8272 = 0.19%

What it changes for you

A guarantee has two parts: how it grows, and how big it is.

This one does not grow with the pile. But its size is 8272, and no one can promise less than 2.414 (p.1, p.3).

That makes 8272 about 3,426.4 times the floor (computed: 8272 ÷ 2.4142). When you read a breakthrough, read both parts.

Does an older rule still win anywhere?

Source: paper p.1 (Theorem 1.1), p.3 (floor of 1 + √2, Kunisky); computed ratio 3,426.4

The limits, part 1

A 1981 rule still wins, unless every item sits in 68 million sets.

Take sets where each item sits in at most t sets. The 1981 Beck-Fiala rule promises 2t − 1 (p.1). This paper promises 16544 × the square root of t (p.1).

The old rule wins until t passes 68,425,985 (computed).

How fast did the number fall after this?

Source: paper p.1; computed by solving 2t − 1 = 16544 × √t, research file 9c

The limits, part 2

Four days later, another paper reported 99.

On 20 September 2026 this paper posted 8272. On 24 September, Xiaoyu Li posted a method scoring below 99 with far fewer steps. That is a report; I did not check the proof.

Proofs that a split exists went lower: 6.84 on 28 September. The proven floor stays at 2.414.

What did I not check?

Source: arXiv 2609.11189, 2609.20979, 2609.23540, 2609.30044, 2609.38228; Bandeira (blog, paper p.1, p.20); floor p.3. Gaps computed.

The limits, part 3

What I could not check.

A fast result is not a settled result. Here is what stays open.

Who found the proof, and how?

The proofs. I checked the arithmetic and the claims about other papers. I did not check the maths itself.
Peer review. Contested. I found no reviewed version. MathWorld said on 26 October 2026 that review of related proofs was not verified. That is one source.
"First". Contested. The authors claim the first fast method. I did not search every earlier paper.
Real use. The paper has no experiments, no figures and no tables.

Source: research file sections 10, 12 and 14; MathWorld, Komlós conjecture page; paper (no figures or tables)

Credit

Who found the proof?

The paper states that an AI research agent found the first version of the proof. The authors say they checked the final proofs and take full responsibility (p.12).

Where can you check all of this?

The authors

Shengtao Guo, Ethan X. Fang (Duke University) and Junwei Lu (Harvard T.H. Chan School of Public Health). They thank Professor Joel Spencer (p.0, p.12).

The AI agent

The Odin Automatic AI Research Agent found the initial proof and helped revise the paper (p.12). The method builds on Bansal and Jiang, and on Guillen and Kobzar (p.1).

Source: paper p.0, p.1, p.12

Sources and method

Every number, and where it came from.

Photos

Balance scale: Nikodem Nijaki, CC BY-SA 3.0. Needle in a haystack: Dasaptaerwin, CC0. Love Locks Bridge: PicoOrdinalo, CC BY-SA 4.0. Slide rule: Slashme, CC0. All via Wikimedia Commons.

Method

p.N is the printed page of the arXiv v2 PDF (PDF page = N + 1). "Computed" means worked out from the inputs shown. Pictures marked illustration are not data. More: vizmaya stories. By vizmaya desk, 4 November 2026.