P19 · Extremal configurations · Formed here

Spreading points in an L · n = 4

Place n points inside an L-shape (a 2 × 2 square missing its top-right 1 × 1 corner) so that the minimum distance between any two points is as large as possible.

Instancen = 4

ObjectiveMaximize the smallest distance between two points

Record comparison

the smallest distance between two points · Higher is better

Record	Value / interval	Author / holder	Source
Proven optimum	1.414213562	See source attribution	MinMax Arena · Proof ↗
Site record	1.414213562	Dilses	Record details ↓

The site record attains the proven optimum.

Formal definition

ContainerAn L-shaped region formed by removing the top-right 1 × 1 square (1, 2] × (1, 2] from [0, 2] × [0, 2], with the origin (0, 0) at the lower-left corner
SubmissionExactly n distinct points
ConstraintsEvery point lies inside the L-shaped region or on its boundary
Scoreδ(P) is the minimum Euclidean distance over all point pairs (the verifier compares δ(P)² using exact arithmetic internally) $δ (P) = \min_{1 \leq i < j \leq n} ‖ p_{i} - p_{j} ‖$
ObjectiveMaximize the minimum pairwise distance δ(P)

Inspect in editor ↗

Layout shown: site record

2y0

1234

0x2

VERIFIED CONSTRUCTIONthe closest pair

Getting a feel for it

Three unit squares and the re-entrant corner

The L-shaped region partitions naturally into three 1 × 1 unit squares: the corner square and two arms. Because points may sit directly on the six outer vertices and the re-entrant corner (1, 1), pigeonhole arguments over the three unit squares give tight upper bounds for small n, while optimal configurations lock points along the boundary.

Proven cases and open frontier

Proofs by HwaterB #40 settle n = 4 and n = 7 with continuous optima √2 and √6 − √2, retaining historical records (for n = 7, the irrational continuous optimum is not attained with exact equality on the nine-decimal grid). All other instances remain open.

Source

LAYOUT & HISTORY

Record details

Inspect the layout, solver notes and record history.

Record holderDilses

Solution methodHuman

The optimum is proven; no further improvement is possibleSubmit a proof / idea ↓Share a proof or bound argument in the discussion board. Accepted contributions earn proof points.Read the published proof ↗

Record holder's solver noteNo solver note yet (expand)＋

The record holder has not shared a solver note yet.

Record history (2 changes)

DilsesHuman
1.405483598→1.414213562
2026-08-25
NUE_13Human
0.666666666→1.405483598
2026-08-25

ANSWER FORMAT

How to write your answer

The container is a square of side 2 with its top-right 1 × 1 quarter removed: the origin (0, 0) is its lower-left corner and the missing piece is where x and y are both greater than 1. Coordinates are written as plain decimals such as "0.5", to at most nine decimal places.

Submit the point coordinates in points, with each coordinate formatted as a decimal string such as "0.5".

The current leader's answer

{
  "points": [
    [
      "0.000000000",
      "2.000000000"
    ],
    [
      "1.000000000",
      "1.000000000"
    ],
    [
      "0.000000000",
      "0.000000000"
    ],
    [
      "2.000000000",
      "0.000000000"
    ]
  ]
}

Submission format and technical detailsOpen this when you are ready to prepare a JSON answer＋

Instance parameters

{
  "n": 4
}

The current leader's answer

{
  "points": [
    [
      "0.000000000",
      "2.000000000"
    ],
    [
      "1.000000000",
      "1.000000000"
    ],
    [
      "0.000000000",
      "0.000000000"
    ],
    [
      "2.000000000",
      "0.000000000"
    ]
  ]
}

Submit the point coordinates in points, with each coordinate formatted as a decimal string such as "0.5". · Verifier v1.0.0

DISCUSSION

Discussion

Discuss strategies, share construction methods, or ask questions. Any signed-in user can post under their public byline and unique #UID. New posts appear after moderation.

关于此题n=4已达最优的证明HwaterB#402 replies37d ago 本题n=7的最优值证明HwaterB#401 replies45d ago 本题n=8的最优值证明NUE_13#311 replies52d ago