Subgraphs Crossing number incidences

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Subgraphs Crossing number incidences

• Aner Mazursky
• 20/3/2007

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(Kovari) Theorem Let Gm,n be a bipartite graph.
Assume Gm,n does not contain Kr,s as a subgraph.
Then
Forbidden Complete Subgraphs
Proof. Let V1 and V2 denote the classes of Gm,n
V1m V2n. From the fact that every r-tuple
W of V1 can be completely connected to at most
s-1 vertices x of V2, the number of such pairs
(W,x) satisfies
Let us define a convex function
If z gt r-1otherwise
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Forbidden Complete Subgraphs
f(z) is convex, therefore we can apply Jensens
inequality
Comparing this to the previous result
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Implementations
This first result already yields interesting
results regarding incidencesLet Gp,l be the
graph connecting points to lines, P m, Ln.
Gp,l does not contain K2,2 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,
Cn.Gp,c does not contain K3,2 as a subgraph
(three points define only one circle).Resulting
the following bound
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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 V1, V2 such
that the number of edges between them is at least
E(G)/2.proof lets look at edge e. If one of
its vertices is from part V1 , then the
probability that the other vertice is from V2 is
at least ½. That is symmetric gt the probability
the edge is between them is at least ½. As some
of expectancies, the expectancy of the number of
edges connecting V1 and V2 gt 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.
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Tight Bound?
Theorem For every n, there exists a Kr,r free
graph G with n vertices such that
If r 2if r gt 2
Proof (i) lets assume that
for some prime p. Then one can construct a
projective plane PG(2,p) of order p as
followsThe 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 axbycz 0 (mod p)The number of
points, as well as lines is
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Note points and lines are actually similar in
the representation. Every line contains exactly
p1 points and there are exactly p1 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 p1. G is K2,2
Free (only one line between two points)
(ii) Is proved by showing that at least half of
all possible graphs of n vertices and
edges , with proper choice of cr contain no
Kr,r as a subgraph
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Crossing Number Theorem
Theorem Let G be a simple graph (no multiple
edges). Then
Lemma 1 for n gt 2 in a planar graph with n
vertices has at most 3n-6 edges.Proof A simple
graph G contains V vertices, E edges and F
facets.Known VF E2. Every facet is at least
a triangle, therefore
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Crossing Number Theorem
Lemma 2 the crossing number of any simple graph
G is at least
Proof If E gt 3V and cr(G) lt E - 3V, we
could delete one edge from every crossing and
obtain a planar graph with more than 3V edges
gt contradiction with Lemma 1.
Theorem Proof Consider a graph G with n
vertices, m edges, and crossing number x. m gt 4n
(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.
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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 4n/m, which is a number between 0 and
1
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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
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Incidences Upper Bound
We can now substitute and get
In order for the condition to be positive
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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 isWe use the previous theorem on
the upper bound on the number of indicesWe
substitute and get
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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 crossingsWe
can now use the crossing number theorem to
receive
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Unit Distances
We can now substitute and get
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• Questions?