Subgraphs Crossing number incidences Aner Mazursky 2032007 Forbidden

Subgraphs Crossing number incidences Aner Mazursky 20/3/2007

Forbidden Complete Subgraphs (Kovari) Theorem: Let Gm, n be a bipartite graph. Assume Gm, n does not contain Kr, s as a subgraph. Then: Proof. Let V 1 and V 2 denote the classes of Gm, n |V 1|=m |V 2|=n. From the fact that every r-tuple W of V 1 can be completely connected to at most s-1 vertices x of V 2, the number of such pairs (W, x) satisfies: Let us define a convex function: If z > r-1 otherwise

Forbidden Complete Subgraphs f(z) is convex, therefore we can apply Jensen’s inequality: Comparing this to the previous result:

Implementations This first result already yields interesting results regarding incidences: Let Gp, l be the graph connecting points to lines, |P| = m, |L|=n. Gp, l does not contain K 2, 2 as a subgraph (between two points passes only one line…). Resulting the following bound: We can get a bound on the number of incidences between points and circles. Let Gp, c be the graph connecting points to circles, |P| = m, |C|=n. Gp, c does not contain K 3, 2 as a subgraph (three points define only one circle…). Resulting the following bound:

Generalization Let us take a look at general graphs. Lemma: The vertex set of any graph G can be decomposed into 2 disjoint equal sized (+-1) parts V 1, V 2 such that the number of edges between them is at least |E(G)|/2. proof: let’s look at edge e. If one of it’s vertices is from part V 1 , then the probability that the other vertice is from V 2 is at least ½. That is symmetric => the probability the edge is between them is at least ½. As some of expectancies, the expectancy of the number of edges connecting V 1 and V 2 >= ½. Therefore, there exists such a decomposition. Corollary: Let G be a Kr, s free graph with n vertices. Then: Proof: by the Lemma, there is a bipartite graph satisfying: Applying Kovari theorem gives the result.

Tight Bound? Theorem: For every n, there exists a Kr, r free graph G with n vertices such that: If r = 2 if r > 2 Proof: (i) lets assume that for some prime p. Then one can construct a projective plane PG(2, p) of order p as follows: The points are represented by triplets (a, b, c) where a, b, c are of Zp and not all of them are 0. Two points represent the same point they are a constant multiplication of each other. A line in PG(2, p) consists of all triplets (x, y, z) satisfying: ax+by+cz = 0 (mod p) The number of points, as well as lines is:

Note: points and lines are actually similar in the representation. Every line contains exactly p+1 points and there are exactly p+1 lines passing through every point. Let G be graph whose vertices are the points of PG(2, p). two distinct points (a, b, c) and (x, y, z) will be joined be an edge ax + by + cz = 0 (mod p). The neighbors of (a, b, c) in G obviously form a line in PG(2, p), which may contain (a, b, c) itself. Therefore every vertex in G has a degree p or p+1. G is K 2, 2 Free (only one line between two points): (ii) Is proved by showing that at least half of all possible graphs of n vertices and a subgraph edges , with proper choice of cr contain no Kr, r as

Crossing Number Theorem: Let G be a simple graph (no multiple edges). Then: Lemma 1: for n > 2 in a planar graph with n vertices has at most 3 n-6 edges. Proof: A simple graph G contains V vertices, E edges and F facets. Known: V+F = E+2. Every facet is at least a triangle, therefore:

Crossing Number Theorem Lemma 2: the crossing number of any simple graph G is at least: Proof: If |E| > 3|V| and cr(G) < |E| - 3|V|, we could delete one edge from every crossing and obtain a planar graph with more than 3|V| edges => contradiction with Lemma 1. Theorem Proof: Consider a graph G with n vertices, m edges, and crossing number x. m > 4 n (so claim is positive). We choose a random subset V’ by including each vertex v belonging to V in V’ at probability p. Let G’ (n’, m’, x’) be the graph induced by V’.

Crossing Number Theorem we recall Lemma 2: This relation of course holds for the expectations as well: We can now substitute with earlier results, and choose p = 4 n/m, which is a number between 0 and 1:

Incidences Upper Bound (Szekely) Theorem: For n points and l lines in the Euclidean plane, the number of incidences is at most: Proof: Define a graph G in the plane such that the vertex set of G is our n given points, and join two points with an edge drawn as a straight line if the points are consecutive points on one of the lines. This drawing shows that: (each line crosses another line at most once) The number of points on any of the lines is one greater than the number of edges drawn along that line. The number of incidences among the points and the lines is at most l greater than the number of edges. We can now use the crossing number theorem to receive:

Incidences Upper Bound We can now substitute and get: In order for the condition to be positive:

Incidences Upper Bound Theorem: Let . For n points in the Euclidean plane, the number l of lines containing at least k of them is at most : Proof: Since there are l lines each of which contain at least k points, the number of incidences x is: We use the previous theorem on the upper bound on the number of indices: We substitute and get:

Unit Distances Theorem: The number of unit distances among n points in the plane is at most: Proof: Similar proof to the incidences upper bound proof. We define G, whose vertex set is the set of n points. We draw a unit circle around each point, and draw an edge between to consecutive points on the unit circle. Define the number of unit distances: x. The number of edges in G is at most n less than the unit distances: Every two unit circles can cross at most twice. Therefore the maximum number of crossings: We can now use the crossing number theorem to receive:

Unit Distances We can now substitute and get:

Questions?