The region and the score
Let L be the square [0,2] x [0,2] with the open top-right unit square removed: points with x > 1 and y > 1 are excluded. For a four-point set S in L, write d(S) for its smallest pairwise distance. P19 maximises d(S); the verifier stores d(S) squared so that every comparison is exact.
Construction: sqrt(2) is attained
Choose the four points
(0, 0), (0, 2), (1, 1), (2, 0).
Their six squared distances are 4, 2, 4, 2, 8 and 2. Hence the smallest distance is sqrt(2), and so the optimum is at least sqrt(2).
Upper bound: three sets of diameter sqrt(2)
The L is the union of three closed unit squares:
Q₁ = [0,1] × [0,1],
Q₂ = [0,1] × [1,2],
Q₃ = [1,2] × [0,1].
Every unit square has diameter sqrt(2). Suppose, for contradiction, that four points in L had smallest pairwise distance greater than sqrt(2). No Qᵢ could then contain two of the points, so |S intersection Qᵢ| is at most one for each i. Since the three squares cover L,
|S| ≤ |S ∩ Q₁| + |S ∩ Q₂| + |S ∩ Q₃| ≤ 3,
contradicting |S| = 4. Therefore d(S) is at most sqrt(2) for every four-point set. The squares overlap along boundary segments, but that causes no gap: the union bound above remains valid even when a boundary point belongs to two or three squares.
Together with the construction, the optimum is exactly sqrt(2). Equivalently, P19's exact stored score is 2. ∎
Two distinct contributions
The attaining four-point construction was already the standing record of Dilses #32; HwaterB #40 supplied the matching upper bound. The construction record and the optimality proof are credited separately.
Dilses #32 — attaining construction