Planar Graphs.ppt

7/27/2019 Planar Graphs.ppt

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Graph Theory


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Basic Definitions curve, polygon curve, drawing

crossing, planar, planar embedding, and

plane graph open set

region, face


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Graph


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Proposition: K5 and K3,3 is not planar


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Restricted Jordan Curve Theorem Theorem. A simple closed polygonal curve C consisting of

finite number of segments partitions the plane into exactly2 faces, each have C as boundary


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Dual Graphs Definition:

let function F on a graph F(G) = { faces of G }

if there exits f: V(G*) F(G) so that f is 1-1 & onto and for all

x, y V(G*), there is an edge connects x, y iff there is an edge e inG that f(x) and f(y) are on the different side of e.


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Graph A cut-edge in G becomes a loop in G*

For all x V(G*) and X = f(x), x is in the interior of X

Each edge e in G there is exactly one e* in G* that e and e*

crosses. (G*) *=G iff G is connected

pf:

a) for all G, G* is connected

b) each face in G*

contains exactly one vertex of G Two embeddings of a planar graph may have non-isomorphic

duals.


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Length of a face length of a face is defined as total length of the boundary of

the face.

2e(G) = L (Fi)

Theorem.edges in G form a cycle in G iff the corresponding edges in G*form a bond in G*

Theorem.

the follows are equivalent

A) G is bipartite

B) every face of G has even length

C) G* is Eulerian


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Outerplane graph Def: outerplanar, outerplane graph

a graph is outerplanar if it has an embedding that every vertexis on the boundary of the unbounded face.

The boundary of the outer face of a 2-connected outerplanegraph is a spanning cycle

K4 and K2,3 are not outerplanar.


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Graph Every simple outerplane graph has 2 non-adjacent vertex of

degree at most 2

pf:

1. n(G) < 3, every vertex has degree 22. n(G) = 4 holds. (think about K4 {any edges})

3. n(G) 4


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Eulers Formula

n e + f = 2

All planar embeddings of connected graph G have the samenumber of faces

A graph with k components, n e + f = k+1


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Graph For simple planar graphs, e(G) 3n(G) 6,

if G is triangle free, e(G) 2n(G) 4

pf:

2e = L (Fi) 3f -----(*)

f = e n + 2

=> e 3n 6

for triangle free case, 3f in (*) -> 4f

K5 and K3,3 are not planar


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Maximal Planar Graph Def. Maximal planar graph: a simple planar graph that is not a

spanning subgraph of any other planar graph.

Proposition. The follows are equivalent

A)G has 3n-6 edgesB)G is a triangulation

C)G is a maximal plane graph


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Regular Polyhedra A graph embeds in the plane iff it embeds on a sphere

For a regular polyhedra of degree kand all faces length are l

e( 2/k+ 2/l-1 ) = 2=> (2/k) + (2/l) > 1

=> (k2)(l 2) < 4

hence k and l can only be k l f

3 3 4

3 4 64 3 8

3 5 12

5 3 20


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Graph Theory

Ch6 Planar Graphs(continued)


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Kuratowskis Theorem

Theorem. A graph is planar iff it does not contain a subdivisionof K5 or K3,3.

Kuratowski subgraph: a subgraph contains a subdivision of K5 or

K3,3. minimal nonplanar graph: a nonplanar graph that every proper

subgraph is planar


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Lemma 1

if F is the edge set of a face in a planar embedding of G, then Ghas an embedding with F being the edge set of the unbounded

face.


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Lemma 2

every minimal planar graph is 2-connected.

Lemma 3

let S = {x, y} be a separating set of G, if G is nonplanar, thereExist some S-lobe adding (x, y) is nonplanar.


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Lemma 4

if G is a graph with Fewest Edges among all nonplanar graphswithout Kuratowski subgraphs G is 3-connected


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Convex embedding: planar embedding that each face boundaryis a convex polygon

Theorem. Every 3-connected planar graph has a convex

embedding Theorem. Every 3-connected graph G with at least 5 vertices

has an edge e such that Ge is 3-connected.


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Lemma 5.

if G has no Kuratowski subgraph, Ge has no Kuratowskisubgraph


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Theorem. (Tutte1960) if G is a 3-connected graph withoutsubdivision of K5 or K3,3, then G has a convex embedding in theplane with no three vertices on a line

Pf: induction on n(G)K4:


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n(G) > 4:

exist e that Ge is 3-connected. Ge has no Kuratowskisubgraph.

e z, H = GeH-z is 2-connected.


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Definition: H is a minor of G if a copy of H can beobtained by deleting or contracting edges of G.

G is planar iff neither K5 nor K3,3 is a minor of G.

Nonseparating

Let G be a subdivisions of a 3-connected graph.G is planar iffevery edge e exactly lies in 2 nonseparating cycles.


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H-fragment

Conflict

Planarity testing

If a planar embedding of H can be extended to a planarembedding of G, then in that extension every H-fragment of Gappears inside a single face of H.


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Planarity testing

1. find a cycle G0

2. for each Gi-fragment B, determine all faces of Gi that contain

all vertices of attachment of B. call it F(B)3. if F(B) is empty for some B, stop (FAIL). Else, choose one.

4. choose a path P between 2 vertices of attachment of B.embed P across F(B). Result in Gi+1.

Planar Graphs.ppt

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