1 Plane and Planar Graphs

Definition 1

A graph G(V,E) is called plane if

• V is a set of points in the plane;

• E is a set of curves in the plane such that

1. every curve contains at most two vertices and these vertices

are the ends of the curve;

2. the intersection of every two curves is either empty, or one,

or two vertices of the graph.

Definition 2

A graph is called planar, if it is isomorphic to a plane graph. The

plane graph which is isomorphic to a given planar graph G is said

to be embedded in the plane. A plane graph isomorphic to G is

called its drawing.

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G is a planar graph H is a plane graph isomorphic to G

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The adjacency list of graph F . Is it planar?

1 4 5 6 8 9 11

2 9 7 6 10 3

3 7 11 8 2

4 1 5 9 12

5 1 12 4

6 1 2 8 10 12

7 2 3 9 11

8 1 11 3 6 10

9 7 4 12 1 2

10 2 6 8

11 1 3 8 7

12 9 5 4 6

What happens if we add edge (1,12)? Or edge (7,4)?

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Definition 3

A set U in a plane is called open, if for every x ∈ U , all points

within some distance r from x belong to U .

A region is an open set U which contains polygonal u, v-curve for

every two points u, v ∈ U .

area without the boundary

Definition 4

Given a plane graph G(V,E), a face of G is a maximal region of

the plane if the vertices and the edges of G are removed.

An unbounded (infinite) face of G is called exterior, or outer face.

The vertices and the edges of G that are incident with a face F form

the boundary of F .

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Proposition 1

For every face of a given plane graph G, there is a drawing of G

for which the face is exterior.

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Dual Plane Graphs

Definition 5

Let G be a plane graph. The dual graph G∗ of G is a new

plane graph having a vertex for each face in G and the edges

that correspond to the edges of G in the following manner:

if e is an edge of G which separates two faces X and Y ,

then the corresponding dual edge e∗ ∈ E(G∗) is an edge

joining the vertices x and y that correspond to X and Y

respectively.

Remark. Different plane drawings (embeddings) of the same

planar graph may have non-isomorphic duals.

Construct duals to these drawings.

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Definition 6

The length of a face F in a plane graph G is the total number

of edges in the closed walks in G that bound the face.

Proposition 2

If l(Fi) denotes the length (the number of edges in its boundary),

of face Fi in a plane graph G, then

2e(G) =∑

l(Fi).

Definition 7

A bond of a graph G is a minimal non-empty edge cut.

Proposition 3

Edges of a plane graph G form a cycle iff the corresponding edges

in G∗ form a bond.

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Theorem 1 (Euler):

If G is a connected plane (p, q)-graph with r faces, then

p− q + r = 2.

Proof. We prove it by induction on q.

Base. If q = 0, then p = 1; obviously, r = 1, and the result

follows.

Inductive Step. Assume that the Euler Theorem holds true for

all connected graphs with fewer than q (q ≥ 1) edges, and let G

be a connected plane graph with q edges.

If G is a tree, then p = q+1 and r = 1 (the only face is exterior)

implying the result.

If G is not a tree, then it has an enclosed face. The edges of the

face form a cycle. Take any edge e on the cycle and consider

graph

H = G− e.

Since q(H) = q − 1, by induction, p(H)− q(H) + r(H) = 2. But

p(H) = p and r(H) = r − 1. The result follows.

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Theorem 2

If G is a planar graph (no parallel edges) with p vertices and q

edges, (q ≥ 3), then q ≤ 3p − 6. If, in addition, G is bipartite,

then q ≤ 2p− 4.

Proof. Let r be the number of faces of G and let mi be the

number of edges in the boundary of the ith face (i = 1, ..., r).

Since every face contains at least three edges,

3r ≤r∑

i=1

mi.

On the other hand, since every edge can be in the boundary of

at most two faces,r∑

i=1

mi ≤ 2q.

Thus, 3r ≤ 2q and by Euler’s Theorem, p−q+2q/3 ≥ 2, implying

q ≤ 3p− 6.

If G is bipartite, the shortest cycle is of length at least 4. Thus,

4r ≤r∑

i=1

mi.

Together with∑ri=1mi ≤ 2q and p− q + r = 2, we get the second

part of the Theorem.

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Corollary 1 Every planar graph G contains a vertex of degree

at most 5.

Proof. (HINT: assuming that all degrees are ≥ 6, estimate the

number of edges in the graph, and compare your estimate with

that by Theorem 2 .)

Corollary 2 Graphs K3,3 and K5 are not planar.

Proof. For K5, p = 5 and q = 10. Thus, q > 3p− 6 = 9 and by

Theorem 2, K5 is not planar.

If K3,3 were planar, then the second part of Theorem 2 would

apply, leading to a contradiction, since for this graph p = 6,

q = 9, and q > 2p− 4.

Definition 8 A subdivision of an edge ab in a graph G is an

operation which replaces ab with two edges az and zb where z is

a new vertex different from other vertices of G. The result of

the subdivision is also called a subdivision of G.

A Kuratowski graph is a graph obtained by several subdivisions

from either K5 or K3,3.

Corollary 3 No Kuratowski graph is planar.

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2 Coloring Planar Graphs

Theorem 3 Every planar graph G is 5-colorable.

Proof. By induction on the number n(G) of vertices.

Base. For all planar graphs with n(G) ≤ 5, the statement is

correct.

Inductive step. Let G have more than 5 vertices. Select a

vertex v of degree ≤ 5. It always exists, since else, the number

of edges in the graph would exceed the upper bound of 3p− 6. By

induction, graph G− v is 5-colorable.

Consider a 5-coloring of G − v. If any color, 1 2, 3, 4, 5 is

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not used for vertices adjacent to v, use it for v. Thus, we need

to assume that v has 5 neighbors that are colored using all 5

colors. Let us call those neighbors according to their colors:

v1, v2, v3, v4, v5 (see Figure below).

Consider a bipartite graph H induced by all vertices of G − v

whose colors are 1 or 3, and let C be the connected component

of H which contains vertex v1. If C does not contain vertex v3,

then we re-color C: every vertex of C whose color is 1 (resp.

3) gets color 3 (resp. 1). This recoloring frees color 1 for v,

yielding a 5-coloring of G.

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5 v

v

v

v

v

v

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Finally, assume that C contains vertex v3. Thus, there is a 2-

colored path connecting vertices 1 and 3. This path together with

vertex v forms a cycle which makes it impossible the existence

of a path colored 2 and 4 connecting vertices v2 and v4. Thus,

recoloring the 2-colored connected component containing vertex

v2 makes color 2 available for coloring v.

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From colorings faces of a map to coloring its edges.

The coloring of the faces of an arbitrary map is reduced to that

of a map for which every vertex has three incident edges. See

the Figure below.

An edge coloring of a graph G is an assignment of colors to edges

such that any two edges incident to the same vertex have differ-

ent colors.

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Theorem. (Tait, 1878). A 3-regular plane graph without bridges

is 4-face colorable iff it is 3-edge-colorable.

Proof. Suppose G satisfies the conditions of the theorem and

it is 4-face-colorable. Let the colors be c0 = 00, c1 = 01, c2 =

10, c3 = 11.

Because G is bridge-less, every edge bounds two distinct faces.

Given an edge between faces colored ci and cj, assign this edge

the color obtained by adding ci and cj coordinate-wise modulo 2.

For example: c0 + c1 = c1; c1 + c3 = c2; c2 + c3 = c1.

11

1001

11

0110

1001

00 01 10

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The reverse. Suppose now that G has a proper 3-edge-coloring,

and let the colors be 1, 2, and 3. Since G is 3-regular, every

color appears at every vertex. Let E1, E2, and E3 be the edge-

sets colored 1, 2, and 3, respectively. The union of any two of

them is the union of disjoint cycles. Let H1 = E1 ∪ E2 and

H2 = E1 ∪ E3.

The faces of graph H1 can be 2-colored with two colors α and β.

The faces of graph H2 can be 2-colored with two colors γ and δ.

Then each face of G is a subset of a face in H1 and a subset of a

face in H2. Assign to each face in G a pair of colors (x, y) where

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x ∈ {α, β} and y ∈ {γ, δ}.

It is easy to prove that any two adjacent faces of G get pairs

off colors that are differ in at least one coordinate. Thus the

assignment is a 4-coloring of the faces of G.

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Tutte’s conjecture: every bridge-less 3-regular graph G which

is not three-edge colorable has the Petersen graph as its minor,

that is the Petersen graph can be obtained from G by contracting

and removing edges.

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3 Kuratowski theorem

Definition 9 A subgraph H of a graph G which is a subdivision

of K5 or K3,3 is called a Kuratowski graph.

Theorem 4 Every non-planar graph contains a Kuratowski sub-

graph.

The Idea of a Proof.

If the theorem is incorrect, let us take a smallest graph for which

it fails. Thus, G is the smallest non-planar graph without Kura-

towski subgraphs.

If e = xy is an edge in G, then we form a new graph H by

“contracting” e into a vertex z, which is adjacent to all vertices

that were adjacent to at least of one of x and y. It turns out that

such an operation does not create Kuratowski subgraphs (must

be proved), thus by minimality of G, graph H is planar.

Let us assume (must be checked if this can always be done)

that an embedding of H − z contains a face bounded by a cy-

cle C, which contains all vertices adjacent to z. Then, the graph

G has an “almost” plane embedding, where the only non-plane

section is the edge e = xy all of whose neighbors lie on cycle

C = x1x2 . . . xt. Obviously, the way x and y are adjacent to the

vertices of C creates the non-planarity of G. Analyzing the pat-

tern of these adjacencies, we may find Kuratowski subgraphs in

G.

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4 Proof of Kuratowski’s theorem

Definition 10 A minimal non-planar graph is a non-planar

graph F such that every proper subgraph of F is planar.

Lemma 1 For every face of a given plane graph G, there is a

drawing of G for which the face is exterior.

Lemma 2 Every minimal non-planar graph is 2-connected.

Proof. If G is disconnected, then by minimality, each compo-

nent of G is planar, and the whole graph is planar, since we can

embed every next component inside a face of the embedding of the

previous component (under some ordering of the components).

Let G have a cut-vertex x and let G1, G2, . . . , Gt be the connected

components of G−x. Let Hi be the subgraph induced on V (Gi)∪

{x}. By minimality of G, each Hi is planar. Let us embed each

Hi so that vertex x is in the outer face and then squeeze Hi

into an angle < 2π/t at vertex x. Combining those embeddings

together yields an embedding of G.

H

G1

x

G2

G3

x1H

x

2Hx

3

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Definition 11 Let S be a separating set of a graph G and G1, . . . , Gt

be the connected components of G−S. The lobes of S in G are

subgraphs H1, . . . , Ht that are induced on V (G1) ∪ S, V (G2) ∪

S, . . . , V (Gt) ∪ S, respectively.

Lemma 3 Let S = {x, y} be a separating set of G. If G is non-

planar, then at least for one lobe Hi[S], adding edge xy creates

a non-planar graph.

Proof. Suppose for every lobe Hi, graph H+i = Hi∪xy is planar

(i = 1, . . . , t). Then, there is an embedding of H+i for which

xy is in the outer face. Then we create an embedding of G by

attaching each next H+i+1 to an embedding of ∪i

j=1H+j , where H

+i+1

is embedded into the face that contains xy.

yx

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Lemma 4 Let G be a graph satisfying the following three con-

ditions: 1) G doesn’t contain a Kuratowski subgraph; 2) G is

non-planar; and 3) among the graphs satisfying (1) and (2), G

has the smallest number of edges.

Then G is 3-connected.

Proof. The condition (3) implies that the removal of any edge

violates either (1) or (2). Since it cannot create a Kuratowski

subgraph, the removal of any edge creates a planar subgraph of

G. Thus, G is a minimal non-planar graph.

By Lemma 2, G is at least 2-connected. Suppose G is not 3-

connected, and let S = {x, y} be a separating set. Then by

Lemma 3, adding edge xy to some S-lobe yields a non-planar

graph. Let H be that lobe and H+ = H ∪ {xy}. Since H+ is

non-planar and has fewer edges than G does, H+ must contain

a Kuratowski subgraph, say K ⊆ H+. Let H ′ be an S-lobe dif-

ferent from H.

PK

x

y

H +

PKx

y

CxH +

y

K Cx

y

Find a path P connecting x and y in H ′. Now, we replace edge

xy in K with P . Clearly, K − xy ∪ P is a Kuratowski subgraph

of G. The contradiction at least for one lobe Hi[S], adding edge

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xy creates a non-planar graph. implies that G has no 2-cut, i.e.

G is 3-connected.

Definition 12 The operation of contraction of an edge e =

(x, y) of a graph G removes e, x and y from the graph, and adds

a new vertex z, which is adjacent to any old vertex u iff u was

adjacent to at least one of x or y. The resulting graph is denoted

G/(x, y).

Lemma 5 Every 3-connected graph G with at least five vertices

has an edge e such that G/e is 3-connected.

Proof. Take any edge e = xy and assume G/e is not 3-connected.

Then G/e has a separating set S of size 2. Since G is 3-connected,

S contains the vertex replacing x and y. Let z be the other ver-

tex. Then, {x, y, z} is a 3-separating set of G.

u

x

z

yDC

If G has no edge whose contraction yields a 3-connected graph,

every edge xy has a vertex z, call it a partner, such that the

removal of x, y and z disconnects the graph.

Among all edges of G, we take edge xy and a partner z whose

removal from G creates the largest connected component C. Let

D be the other component of G− {x, y, z}.

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Since {x, y, z} is a minimal separating set of G, each of the

vertices x, y, z has neighbors in C and in D. Let u ∈ D be a

neighbor of z. and let v be the partner of edge zu.

By definition, G−{x, y, z} is disconnected. If v is outside of set

V (C) ∪ {x, y}, then the removal of {z, u, v} creates a connected

component which is larger than C, contradicting the choice of

C. On the other hand, if v ∈ V (C)∪{x, y}, then the removal of

z, v disconnects G, contradicting the fact that G is 3-connected.

This completes the proof.

u

v x

z

y

x

z

y

u

v

DCC D

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Lemma 6 If G/e has a Kuratowski subgraph, then G has Ku-

ratowski subgraph.

Proof. Let H be a Kuratowski subgraph of G/e, where e = xy

and let z be the vertex of obtained from contracting xy.

If z 6∈ V (H), H is a Kuratowski subgraph of G.

If z ∈ V (H) is one of the vertices used to subdivide an edge in

K5, or K3,3 to obtain H, then replacing z with x, or y, or xy

yields a Kuratowski subgraph of G.

If z ∈ V (H) and among the edges incident to z at most one is

incident to x (resp. y) (this must happen if H is a subdivision of

K3,3), then replacing z with y (resp. with x) reveals a Kuratowski

subgraph in G.

Finally, let H be a subdivision of K5, and z be adjacent to four

edges, two of which are adjacent to x and the remaining two are

adjacent to y. Let u1 and u2 (resp. v1 and v2) be the two vertices

of degree 4 in H that are connected with z through the two edges

incident to x (resp. y). In this case, we find a subgraph of

G which is a subdivided K3,3: vertices u1, u2, and y form one

partition of the subgraph, and v1, v2, and x form the other; the

paths connecting them are those from H.

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Theorem 5 Every graph without Kuratowski subgraph is pla-

nar.

Proof. Because of lemma 3, we can consider 3-connected graphs

only. We prove the Theorem by induction on n the number of

vertices of G.

Base. If n ≤ 4, the the only 3-connected graph is K4, which is

planar.

Inductive Step. Let n ≥ 5 and e = xy be an edge such that

H = G/e is 3-connected.

5Kplanar cases K

3,3

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