planar graphs prepared by: asa dan, ofer kiselov, hillel mendelson & ofir pupko 049059 - graph...

Planar GraphsPrepared by:

Asa Dan, Ofer Kiselov, Hillel Mendelson & Ofir Pupko049059 - Graph Theory with CE Applications

Technion 2015

Contents

• Plane & Planar graphs• Duality• Euler’s formula• Bridges• Kurtawski Theorem• Surface Embeddings of Graphs

Plane & Planar Graphs

• A Planar Graph is a graph which can be drawn on a plane so that its edges intersect only at vertices.• Also called Embeddable on the plane.

• A planar embedding of has the same vertices and edges.• To distinguish between a planar embedding and a normal graph, vertices are

called points, and edges are called lines.

𝐺~𝐺

Planar embedding

Jordan curve Theorem - glossary

• A curve in the plane is a continuous image of a closed unit line segment.• A closed curve is a continuous image of a circle.• A simple curve is a curve that doesn’t intersect itself.

• An arcwise-connected set of points is a set in which each two points can be connected by a curve lying entirely within the subset.

~𝐺

Jordan curve Theorem

• Any simple closed curve C in the plane partitions the rest of the plane into two disjoint arcwise-connected open sets.• The two sets are the interior and the exterior of a circle.• Denoted int(C) and ext(C).• Int(C) and Ext(C) are their closures – Int(C)∩Ext(C)=C.

• Every path from int(C) to ext(C) goes through C itself.

Jordan curve theorem – example

• How can we use the theorem?• K5 has 5 vertices which are all connected with edges.• We will prove that K5 is non-planar using the Jordan theorem:• Consider V(K5)={v1, v2, v3, v4, v5}.

• We define:• C={v1, v2, v3}• C1={v2, v3, v4}• C2={v1, v3, v4}• C3={v1, v2, v4}

• And suppose, w.l.o.g that v4 int(C4).

1

2 4

3

C1

C3

C2

C

Jordan curve theorem – example cont’• Where will v5 be?• From Jordan Theorem, there must be v5 ext(Ci), where i=1,2,3.• Because edge viv5 exist.• On the other hand, v4 int(C4)• v4v5 must intersect an edge!

• The graph isn’t planar.

1

2 4

3

C1

C3

C2

C

5

Subdivisions

• A graph derived from an edge subdivision of is called a subdivision of .• Proposition: is planar iff every subdivision of is planar.• Proof is straightforward:• If G is planar no two edges collide and edge subdivisions can be made without

harming the planarity of the graph.• If G is not planar, it must have a subdivision which is also not planar – G itself.

~𝐺

Subgraph of a planar graph

• Let be a planar graph.• Can have a non-planar subgraph?• If is planar, it has an embedding of all vertices and edges, and no edge

intersects another.• Removing edges cannot force another intersection.

Spoiler: embedding on different surfaces• We will see that we can embed a planar graph on surfaces other than

the plane.• It’s obvious that we can embed a planar graph on a sphere.• Can we embed a graph embeddable on the sphere onto a plain?

Stereographic projection

• A mapping that projects a sphere onto a plane.• To create it:• Place a plane P below the sphere.• Pick a point Z as the anchor.• Stretch a line L from Z towards P.

• Intersect the desired point Q on the sphere.• Q’ is the intersection between L and P.• Q’ is the embedded coordinates of Q.

P

Z

Q

Q’

L

Back to graphs

• Project all the vertices and edges of the sphere using a stereographic projection.• Pick a Z that’s not a part of the graph.

• The new graph is planar.

Duality

Definitions

• A plane graph G partitions the rest of the plane intoa number of arcwise-connected open sets. These setsare called the faces of G (f1..n).• Each plane graph has exactly one unbounded face, called the outer face (f1)• F(G) = the set of faces • f(G) = the number of faces• Two faces are adjacent if their boundaries have an edge in common• ∂(f) = the boundary of a face (the edge set of the subgraph that is the

boundary)

PropositionLet G be a planar graph, and let f be a face in some planar embedding of G. Then G admits a planar embedding whose outer face has the same boundary as f.ProofThink of a sphere… (homework)

• a planar embedding of a cycle has exactly two faces• a planar embedding of a tree has just one face

Some More Definitions

• A cut edge (e8) in a plane graph has just one incident face, but we may think of the edge as being incident twice with the same face (once from each side). • All other edges are incident with two distinct faces.• The degree, d(f), of a face f is the number of edges in its boundary ∂(f), cut

edges being counted twice.• d(f3) = 6• d(f5) = 5

Subdivision

• To subdivide a face f of G is to add a new edge e joining two vertices on its boundary in such a way that, apart from its endpoints, e lies entirely in the interior of f. • This operation results in a plane graph G + e with exactly one more face

than G; all faces of G except f are also faces of G + e, and the face f is replaced by two new faces, f1 and f2, which meet in the edge e.• In a connected plane graph the boundary of a face can be regarded as a

closed walk in which each cut edge of the graph that lies in the boundary is traversed twice.

TheoremIn a non-separable plane graph other than K1 or K2, each face is bounded by a cycle.

CorollaryIn a loopless 3-connected plane graph, the neighbours of any vertex lie on a common cycle.

Proof Let G be a loopless 3-connected plane graph and let v be a vertex of G. Then G − vis non-separable, so each face of G − v is bounded by a cycle, by the Theorem. If f is the face of G − v in which the vertex v was situated, the neighbours of v lie on its bounding cycle ∂(f).

DualsGiven a plane graph G, one can define a second graph G∗ as follows:• Corresponding to each face f of G there is a vertex f∗ of G∗

• Corresponding to each edge e of G there is an edge e∗ of G∗. • Two vertices f∗ and g∗ are joined by the edge e∗ in G∗ if and only if

their corresponding faces f and g are separated by the edge e in G.• cut edge loop

PropositionThe dual of any plane graph is connected. Proof Let G be a plane graph and G∗ a plane dual of G. Consider any two vertices of G∗. There is a curve in the plane connecting them which avoids all vertices of G. The sequence of faces and edges of G traversed by this curve corresponds in G∗ to a walk connecting the two vertices.

Homework: let G** be the dual of G*Prove: G** ≈ G

Deletion–Contraction Duality: Deletion• Let G be a planar graph and Ĝ a plane embedding of G. • For any edge e of G, a plane embedding of G\e can be obtained by simply

deleting the line e from Ĝ. • Thus, the deletion of an edge from a planar graph results in a planar graph.

~𝐺

Deletion–Contraction Duality: Contraction• The contraction of an edge of a planar graph results in a planar graph. • The line e of Ĝ can be contracted to a single point.• (and the lines incident to its ends redrawn)

• The resulting plane graph is a planar embedding of G/e.

~𝐺

~𝐺

Deletion–Contraction DualityPropositionLet G be a connected plane graph, and let e be an edge of G that is not a cut edge. Then (G\e)* ≈ G*/e* Proof Because e is not a cut edge, the two faces of G incident with e are distinct. denote them by f1 and f2. Deleting e from G results in the amalgamation of f1 and f2 into a single face f. Any face of G that is adjacent to f1 or f2is adjacent in G\e to f. all other faces and adjacencies between them are unaffected by the deletion of e.

Reminder : = ’\‘deletion

= ’/‘contraction

Deletion–Contraction DualityCorrespondingly, in the dual:The two vertices f*1 and f*2 of G* which correspond to the faces f1 and f2 of G are now replaced by a single vertex of (G\e)*: f*.All other vertices of G* are vertices of (G\e)*. Furthermore, any vertex of G* that is adjacent to f*1 or f*2 is adjacent in(G\e)* to f*, and adjacencies between vertices of (G \ e)* other than v are the same as in G*.



PropositionLet G be a connected plane graph, and let e be a link of G. Then (G/e)* ≈ G*\e* ProofBecause G is connected, G** ≈ G.Also, because e is not a loop of G, the edge e* is not a cut edge of G*, so G*\e* is connected. Then (G*\e*)* ≈ G**/e** ≈ G/e





TheoremThe dual of a non-separable plane graph is non-separable. Proof By induction on the number of edgesLet G be a non-separable plane graph. The theorem is clearly true if G has at most one edge, so we may assume that G has at least two edges, hence no loops or cut edges. Let e be an edge of G. Then either G\e or G/e is non-separable. If G\e is non-separable, so is (G\e)* ≈ G*/e*, by the induction hypothesis and the previous proposition G* is non-separable. The case where G/e is non-separable can be established by an analogous argument.

Directed Dual GraphsThe notion of plane duality can be extended to directed graphs. • Let D be a plane digraph, with underlying plane graph G. Consider a

plane dual G* of G. Each arc a of D separates two faces of G. As a is traversed from its tail to its head. • For each arc a of D, we now orient the edge of G*

that crosses it as an arc a* by going from left to right. • The resulting plane digraph D ∗

is the directed plane dual of D.

Euler’s Formula

v(G) − e(G) + f(G) = 2

Theorem: Euler’s Formula holds for connected plane graphs.Proof: By induction on f(G), the number of faces of G.• If f(G) = 1, each edge of G is a cut edge and so G, being connected, is a

tree. In this case e(G) = v(G) − 1 and the assertion holds.

v(G) − e(G) + f(G) = 2

• Suppose that it is true for all connected plane graphs with fewer than f faces, where f ≥ 2, and let G be a connected plane graph with f faces. • Choose an edge e of G that is not a cut edge. Then G \ e is a

connected plane graph with f−1 faces, because the two faces of G separated by e coalesce to form one face of G \ e. • By the induction hypothesis:

v(G \ e) − e(G \ e) + f(G \ e)=2

v(G) − e(G) + f(G) = 2

• Using the relations:• v(G \ e) = v(G)• e(G \ e) = e(G) − 1• f(G \ e) = f(G) − 1

We obtain the formula.

v(G) − e(G) + f(G) = 2

Corollary: All planar embeddings of a connected planar graph have the same number of faces.Proof: Let G’ be a planar embedding of a planar graph G. By Euler’s we have:• f(G’) = e(G’) − v(G’)+2= e(G) − v(G)+2The number of faces is not depended on the embedding.

v(G) − e(G) + f(G) = 2

Corollary: Let G be a simple planar graph on at least three vertices. |E(G)| = m |V(G)| = n• Then m ≤ 3n − 6

Furthermore, m = 3n − 6 if and only if every planar embedding of G is a triangulation.

Proof: It suffices to prove the corollary for connected graphs. Let G be a simple connected planar graph with n ≥ 3.

v(G) − e(G) + f(G) = 2

• Consider any planar embedding G’ of G. • Because G is simple and connected, on at least three vertices, d(f) ≥ 3

for all f F(G’). Therefore:∈

• Which is equivalent to: m ≤ 3n − 6• Equality is obtained if and only if d(f) = 3 for each f F(G’).∈

m ≤ 3n − 6

Corollary: Every simple planar graph has a vertex of degree at most five.• This is trivial for n < 3. If n ≥ 3, then by the above Corollary:

• It follows that δ ≤ 5

m ≤ 3n − 6

Corollary: K5 is nonplanar.Proof: By the above corollary: 10 = e(K5) ≤ 3v(K5) − 6=9

v(G) − e(G) + f(G) = 2

Corollary: K3,3 is nonplanar.Proof: Suppose that K3,3 is planar and let G be a planar embedding of K3,3. Because K3,3 has no cycle of length less than four, every face of G has degree at least four. Therefore we have:

• Euler’s Formula now implies that:2 = v(G) − e(G) + f(G) ≤ 6 − 9+4=1 (contradiction!)

Bridges

Definition

Let H be a proper subgraph of a connected graph G. The set E(G) \ E(H) may be partitioned into classes (Bridges) as follows:• For each component F of G − V(H), there is a class consisting of the

edges of F together with the edges linking F to H. • Each remaining edge e (that is, one which has both ends in V(H))

defines a singleton class {e} (Trivial Bridge).

~𝐺

Properties

• Bridges of H can intersect only in vertices of H.• For a bridge B of H, the elements of V(B) ∩ V(H) are called its vertices of

attachment to H.• A bridge with k vertices of attachment is called a k-bridge. • Two bridges with the same vertices of attachment are equivalent

bridges.

Bridges of a Cycle

H

• B1 and B2 are equivalent 3-bridges.• B3 and B6 are trivial.

Bridges of Cycles

Henceforth, all bridges are of a cycle C.• The vertices of attachment of a k-bridge B with k ≥ 2 effect a partition

of C into k edge-disjoint paths, called the segments of B.• Two bridges avoid each other if all the vertices of attachment of one

bridge lie in a single segment of the other bridge; otherwise, they overlap.

Bridges of Cycles

• Two bridges B and B' are skew if there are distinct vertices of attachment u,v of B, and u' ,v' of B' , which occur in the cyclic order uu' vv' on C.

Bridges of Cycles

H

• B1 and B4 avoid each other.• B1 and B2 overlap.• B3 and B4 are skew.

Bridges of Cycles

Theorem: Overlapping bridges are either skew or else equivalent 3-bridgesProof: Suppose that bridges B and B’ overlap. • Clearly, each must have at least two vertices of attachment. • If either B or B’ is a 2-bridge, it is easily verified that they must be

skew. • We may therefore assume that both B and B’ have at least three

vertices of attachment.

Bridges of Cycles

• If B and B’ are not equivalent bridges, then B’ has a vertex u’ of attachment between two consecutive vertices of attachment u and v of B.

• Because B and B’ overlap, some vertex of attachment v’ of B’ does not lie in the segment of B connecting u and v. It follows that B and B’ are skew.

• If B and B are equivalent k-bridges, then k ≥ 3. If k ≥ 4, B and B are skew; if k = 3, they are equivalent 3-bridges.

Bridges of Cycles in Plane Graphs

Henceforth, G is a plane graph and C is a cycle in G.• Because C is a simple closed curve in the plane, each bridge of C in G

is contained in one of the two regions Int(C) or Ext(C).• A bridge contained in Int(C) is called an inner bridge, a bridge

contained in Ext(C) an outer bridge.

Bridges of a cycle in a plane graph• B1 and B2 are inner bridges in this embedding.• B3 and B4 are outer bridges in this embedding.

Bridges of Cycles in Plane Graphs

Theorem: In a plane Graph, inner (outer) bridges avoid one another.Proof: Let B and B’ be inner bridges of a cycle C in a plane graph G. Suppose that they overlap. By the last Theorem, they are either skew or equivalent 3-bridges. In both cases, we obtain contradictions.

Bridges of Cycles in Plane Graphs• If skew – Subdivision of K5

Bridges of Cycles in Plane Graphs• If equivalent 3-bridges – Subdivision of K3,3

Unique Plane Embeddings

• We say that two planar embeddings of a planar graph G are equivalent if their face boundaries (regarded as sets of edges) are identical.• A planar graph for which any two planar embeddings are equivalent is

said have an unique embedding in the plane.


A cycle is non-separating if it has no chords (Trivial Bridge) and at most one nontrivial bridge.

Theorem: A cycle in a simple 3-connected plane graph is a facial cycle if and only if it is non-separating.Proof: Let G be a simple 3-connected plane graph and let C be a cycle of G.


• Suppose that C is not a facial cycle of G. Then C has at least one inner bridge and at least one outer bridge.• Because G is simple and connected, these bridges are not loops. Thus

either they are both nontrivial or at least one of them is a chord.• It follows that C is not a non-separating cycle.


• Suppose that C is a facial cycle of G. We may assume that C bounds the outer face of G, so all its bridges are inner bridges.• We’ve seen that Inner bridges avoid one another.• If C had a chord xy, the set {x,y} would be a vertex cut separating the

internal vertices of the two xy-segments of C, contradicting G being 3-connectned.


• If C had two nontrivial bridges, the vertices of attachment of one of these bridges would all lie on a single xy-segment of the other bridge, and {x,y} would be a vertex cut of G separating the internal vertices of the two bridges, contradicting G being 3-connectned.• It follows that C is a non-separating cycle.


Theorem: Every simple 3-connected planar graph has a unique planar embedding.Proof: Let G be a simple 3-connected planar graph.• By last Theorem, the facial cycles in any planar embedding of G are

precisely its non-separating cycles.• Because the latter are defined solely in terms of the abstract structure

of the graph, they are the same for every planar embedding of G.

Kuratowski’s Theorem

Kuratowski’s Theorem

Motivation: Deciding whether a given graph is planar.

We saw that any graph which contains subdivision of K5 or K3,3 is not planar.Kuratowski’s theorem claims it is also a necessary condition (for nonplanarity).

Kuratowski’s theorem: A graph is nonplanar if and only if it contains a subdivision of either K5 or K3,3

A subdivision of K5 or K3,3 is consequently called a Kuratowski subdivisionWe will prove an equivalent theorem- Wagner’s theorem, based on Minors.

Minors

A minor of a graph G is any graph obtainable from G by means of a sequence of vertex and edge deletions and edge contractions

2

1

Ge Contraction

of “e”Deletionof e’e'

v

Deletionof v

Minors of- G

Minors

We will say that “G has a F-Minor”, if G has a minor which is equivalent (isomorphic) to graph F.

In previous example, G has a “3-path”-minor, and a K4-minor

Wagner’s theorem: A graph is nonplanar if and only if it has either K5-minor, or K3,3-minor

Definition: A minor which is equivalent to either K5 or K3,3 is called a Kuratowski minor.

Minors & Subdivisions

Lemma 1.1: If G contains a subdivision of F, then G has a F-Minor.To get this minor, remove all vertices and edges not in the subdivision of F contained in G, and contract each subdivided edge into a single edge.Example:

Graph contains asubdivision of K4

Remove non-relevantvertices and edges

Contract subdividededge


Lemma 1.2: if G has a F-Minor, where F is of maximum degree no more than 3, then G contains a subdivision of F (homework)

Corollary: If G has a K3,3-Minor, then G contains a subdivision of K3,3

Lemma 1.3: If G has a K5-Minor, then G contains a Kuratowski subdivision (homework)


Corollary: G has a Kuratowski minor if and only if it contains a Kuratowski subdivision.Hence, Kuratowski’ theorem (Kuratowski subdivisions are necessary and sufficient for non-planarity) and Wagner’s theorem (Kuratowski minors are necessary and sufficient for non-planarity) are equivalent.

We will prove Wagner’s version.

Proving Wagner’s Theorem

Lemma 2: Minors of planar graphs are planar• Deletions and Contractions keep planarity.

Hence, the sufficient condition for non-planarity applies (G has a Kuratowski’s minor G has a minor which is nonplanar G cannot be planar).

We will prove the necessary part:Claim: Every nonplanar graph has a Kuratowski’s minor.

Proof: Assume we have a nonplanar graph G, with n vertices.


1) Assume G is not-connected.Propositions [Trivial] :

1. Each component of a non-connected graph G, is minor of G.2. Minor of minor of G, is also minor of G.3. If a non-connected graph G is nonplanar, then it has a nonplanar

component (at least one)Corollary: if G is non-connected, and nonplanar, we will take one of its nonplanar components, and prove that it has a Kurtawski’s minor, and therefore G also has such minor.So, it is sufficient to prove the theorem for a connected graph G.


Definition: Let G be a connected graph, and is a vertex-cut of G. if H is some component of G-S, then we say that is anS-component of G.

SS-component (I) S-component (II)

H


2) Assume G is separable (connected, but not 2-connected).Lemma 3.1: Given a cut-vertex {x} of G, then each of its {x}-components is a minor of G.Proof: Trivial… simply remove vertices and edges of other components.

Lemma 3.2: Given a cut-vertex {x} of G, then G is planar if and only if each of its {x}-components is planar.

x

{x-}component (I)


Proof: • If G is planar, then by lemma 2, all of its minors are planar, and since

from Lemma 3.1 each {x}-component is also a minor of G, it is therefore planar.• If all of its {x}-components are planar, then we can get an embedding

of each such component so that x will touch the outer face, and then we can unite the “x” vertices from all the embeddings while keeping planarity, and so G is planar.


X X

X


Corollary: if G is separable and nonplanar, we will take one of its nonplanar {x}-components, where x is a cut-vertex, and prove that it has a Kuratowski’s minor, and therefore G also has such minor.If the {x}-component we took is itself separable, we will keep doing so iteratively.At some stage we must get a non-separable (2-connected) component (otherwise G is planar).

So, it is sufficient to prove the theorem for a 2-connected graph G.


3) Assume G is 2-connected.Definition: Let G be a graph with 2-vertex cut {x,y}. If H is a {x,y}-component of G, then , where e’ is an edge between x and y, is a marked {x,y}-component of G. The edge e’ is called the marker edge.

x

y

{x,y-}component (I)

x

y

Marked {x,y}-component (I)x

ye'

H


Lemma 4.1: Given a 2-vertex cut {x,y} of G (where G is a 2-connected graph), then each of its marked {x,y}-components is a minor of G.Proof: Let H be an {x, y}-component of G, and define marker edge e’=(x,y). Let xPy be a path in another {x,y}-component of G (and such path exists. Why?).H P ∪ Is a subgraph of G, and it is also a subdivision of H + e’.

x

y

H

P→𝒆 ′

e'


Therefore G contains a subdivision of H+e’, and thus, by Lemma 1.1, G also has a minor of H+e’ as required (H+e’ is a marked {x,y}-component).

Lemma 4.2: Given a 2-vertex cut {x,y} of G, then G is planar if and only if each of its marked {x,y}-components is planar.


Proof: Similar to Lemma 3.2, but now for each marked {x,y}-component, we can build an embedding where the maker edge touches the outer face, and then we can connect the components while keeping planarity.

Is the lemma true for non-marked {x,y} components of G? (homework)

x

ye'


Corollary: if G is 2-connected and nonplanar, we will take one of its nonplanar marked {x,y}-components, where {x,y} are 2-vertex cut, and prove that it contains a Kuratowski’s minor, and therefore G also contains such minor (and doing so iteratively, until getting a 3-connected component).So, it is sufficient to prove the theorem for a 3-connected graph G.

Finally, it is sufficient to prove the claim for 3-connected graphs.Final claim: Every 3-connected nonplanar graph has a Kuratowski minor


Proof: Assume G is a 3-connected and nonplanar. Prove by induction on n.Base: => Always planar, hence trivially true.Induction step: , assume true for

Proposition: every 3-connected graph G with has an edge “e”, s.t. G/e (contraction of “e”) is also 3-connected [See Bondy P.222].

Assume e=(x,y) is such an edge in G, and H=G/e.


If H is nonplanar (relevant only if ; otherwise , and therefore H planar), by induction it has a Kuratowski minor, and since every minor of H is also minor of G, then G has a Kuratowski minor.So we assume H is planar.Denote by z the vertex of H formed by contracting e.Because H is a 3-connected plane graph, the neighbors of z lie on a cycle C, the boundary of some face f of H−z

z

H

C


Denote by Bx and By, respectively, the bridges of C in G\e (deleting e from G) that contain the vertices x and y

If Bx and By avoid each other, we can draw G as planar, in contradiction to the assumption.

So Bx and By are overlapping, and therefore either skew, or else equivalent 3-bridges

x

G\eC

y e

𝑩𝒙𝑩𝒚


If they are skew:

G contains a K3,3-subdivision, and therefore has a K3,3-minor x

G

C

y

e


If they are equivalent 3-bridges:

G contains a subdivision of K5,and therefore has a K5-minor.

And that’s it…

x

G

C

y

e

Surface embeddings of graphs

Motivation

• Find generalizations of Euler’s Formula and the Four Color conjecture.• Many researches use embeddings on different surfaces for their

properties.• This will be a brief, non-formal presentation of the subject.

Orientable and non-orientable surfaces• A surface is a connected 2D manifold.• An non-orientable surface is a surface on which:• A line can be drawn on the surface.• The line passes through both sides of it.• “Walking” on the line yields different left and right directions.

• A Mobius ring is non-orientable.• A sphere is orientable.

Closed surfaces

• A closed surface is a bounded surface with no boundary.• E.g. a sphere.• Non-closed surface: Mobius ring, a plane.

• Mobius ring has a boundary which is homeomorphic to a circle.• A plane is not bounded.

• Every surface may be constructed from a polygon.• For example: the torus can be constructed by identifying both pairs of

opposite edges.

Adding handles to a sphere

• Adding a handle to a sphere is:• Cutting off two discs of the same radius.• Bending a cylinder with boundaries of that radius.• Attaching the cylinder bases at the discs’ places.

• A sphere with k handles is marked .• is homeomorphic to a torus.• E.g. using discs at the radius of the sphere.

𝑆0 𝑆1 𝑆3

Adding cross-caps to a sphere

• Adding a cross cap to a sphere is:• Cutting off a disc.• Attaching a Mobius band at the position of the disc.• The boundary of the Mobius band is matched to the boundary of the disc.

• A sphere with k cross caps is marked .• Note: this surface intersects itself.

𝑁1 Klein bottle, homeomorphic to

Handles and cross-caps

• Every orientable surface is homeomorphic to for some .• Every closed non-orientable surface is homeomorphic to for some . • Every closed surface belongs to either group.• Theorem: The surface obtained by adding to a sphere k handles and

m cross-caps is homeomorphic to .• So adding both handles and cross caps doesn’t create “new” surfaces.

Cellular embeddings

• An embedding of graph on a surface is cellular if each of the arcwise-connected regions of is homeomorphic to the open disc.• These regions are the faces of .

Cellular embedding example

• Example: embedding K4 on a sphere/torus:

• The face is not homeomorphic to a disc.

Non-cellular embeddingCellular embedding

The Euler Characteristic

• The Euler Characteristic of a surface is defined:

• is a sphere with k handles.• is a sphere with k cross caps.

Generalization of Euler’s formula

• Let be a cellular embedding of a connected graph on a surface . Then:

• Euler’s formula is the case for k = 0. • since the plane is homeomorphic to a sphere.

Corollaries

• All embeddings of a connected graph on a given surface have the same number of faces.• doesn’t change, so can’t either.

• If G is connected and embeddable on , then .

• Using this we can infer which graphs are not embeddable on surfaces.• K8 has 8 vertices and 28 edges.

• It can only be embedded on orientable surfaces homeomorphic to S5 or above.

• K7 has 7 vertices and 21 edges.• It can be embedded on orientable surfaces homeomorphic to S1 or above, e.g. a torus.

The Orientable Embedding Conjecture• An embedding of a graph G on a surface Σ

is a circular embedding if all the faces of are bounded by cycles.• Every loopless 2-connected graph has a circular embedding on some

orientable surface.

The end…

planar graphs prepared by: asa dan, ofer kiselov, hillel mendelson & ofir pupko 049059 - graph...

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