ordinary line: one that goes through exactly two points.
sylvester gallai: existence of an ordinary line given all don't lie on a line. (first stated in 1893, proved in 1940, stated again by erdos in 1943, proved by gallai)
proof1: take the pair p,L such that p does not lie on L and distance between p and L is the smallest. L is ordinary
proof2: take the dual of the given point set. Now every point is a line and every line a point. Consider the graph formed by these points and lines in the dual. It is planar, so there exists a vertex with degree 4. The dual of this was the ordinary line.
Ex1: show that in R^3, there exist n point sets (for all n >=6) such that the plane defined by every 3 points not on a line contains at least one more point of the point set. (give a construction of such a set)
In any two colouring of the edges of a planar graph, there exists a vertex with at most two colour changes. (we count colour changes by going around a vertex and counting how many times the edge colour changes).
proof: assume false. then 4n <= number of colour changes <= 2f_3 + 4f_4 + 4f_5 + ... <= 4e -4f
given any finite collection of black and white points not all on a line, there always exists a monochromatic line.
proof: look at the dual and apply lemma above.
there needn't exist an ordinary monochromatic line.
silva and fukuda conj: for every set of red and blue points not all on a line such that the red and blue point are separable by a line and the cardinality of red and blue sets differs by at most 1, there exists an ordinary bichromatic line.
9 point counter example can be found online. but for larger point sets, it is not known.
HW: show both conditions of the silva fukuda conjecture are necessary, i.e, 1) there exist (arbitrarily large) point sets consisting of red and blue points such that the red and blue are separable by a line and there does not exist a bichromatic line 2) there exist (arbitrarily large) point sets consisting of red and blue points such that the cardinality of red and blue differ by at most 1 and there does not exist a bichromatic line. (for all statements above, both the red and blue point sets are non empty)
how many ordinary lines?
motzkin(51) >= sqrt(n)
dirac(51) conjecture >= lower integer(n/2) : this is tight. even the lower integer of n/2 is important as there are counter examples otherwise.
kelly moser (58) >= 3n/7
csima sawyer (93) >= 6n/13
erdos -de bruijn: Every n point set in the plane, not all on the line spans at least n lines.
proof by induction: delete a vertex from the ordinary line.
linear algebra proof: incidence matrix, A, has rank >= n because A^T A is positive definite.
this also gives that partitioning K_n into cliques requires at least n cliques.
(and n is enough because we can take K_(n-1) and all remaining edges)
EX2: show that any partition of K_n into complete bipartite graphs requires at least n-1 such graphs.