1 Plane and Planar Graphs Definition 1 a Graph G(V,E) Is Called Plane If

Home , Cycle (graph theory), Planar graph

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.

1 9 2 4 2 1 9 8 3 3 6 8 7 4 5 6 5 7 G is a planar graph H is a plane graph isomorphic to G

1 The adjacency list of graph F . Is it planar? 1 4 56 8 911 2 9 7 6 103 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 36 10 9 7 4 12 12 10 2 6 8 11 1387 12 9546 What happens if we add edge (1,12)? Or edge (7,4)?

2 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 .

3 Proposition 1

For every face of a given plane graph G, there is a drawing of G for which the face is exterior.

4 2 1 9 3 6 8 5 1 7 8 2 7

4 6 3 5

4 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.

5 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)= X 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.

6 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.

7 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, r 3r ≤ X mi. i=1 On the other hand, since every edge can be in the boundary of at most two faces, r X mi ≤ 2q. i=1 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, r 4r ≤ X mi. i=1 r Together with Pi=1 mi ≤ 2q and p − q + r =2, we get the second part of the Theorem.

8 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.

9 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

10 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. v 3 1 1 1 4 2 v 5 4 3 v2 v 2 3 v 4 v3 1

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.

11 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.

12 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 11 01 10 01 10

11 10 01 00 01 10

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

13 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.

2 1 3 3 2 3 1 2 1 3 1 2 2 1 3 3

1 2

2 1 1 3 3 3 2 1 1 2 1 1 1 3 1 2 1 2 1 3 3

1 2 1

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.

14 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 subgraph.

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 cycle 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.

15 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 component 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.

G 2

G x 1 H 2 x H x 1 H3

G 3

16 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 Hi = Hi ∪xy is planar + (i = 1,...,t). Then, there is an embedding of Hi for which xy is in the outer face. Then we create an embedding of G by + i + + attaching each next Hi+1 to an embedding of ∪j=1Hj , where Hi+1 is embedded into the face that contains xy.

x y

17 Lemma 4 Let G be a graph satisfying the following three conditions: 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.

+ x + H H x x K C KC P

y y y

x K P

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

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

C D y

x z u

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}.

19 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.

CD C D y y v v x x u z u z

20 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.

z x y

v 1 v u u 1 1 1

v 2 v u u 2 2 2

21 Theorem 5 Every graph without Kuratowski subgraph is planar.

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.

planar cases K K 3,3 5