1 Plane and Planar Graphs. Definition 1 A graph G(V,E) is called plane if

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1 Plane and Planar Graphs Definition 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. every curve contains at most two vertices and these vertices are the ends of the curve;. the intersection of every two curves is either empty, or one, or two vertices of the graph. Definition 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 G is a planar graph H is a plane graph isomorphic to G

2 The adjacency list of graph F. Is it planar? What happens if we add edge (,)? Or edge (7,4)?

3 Definition A set U in a plane is called open, if for every U, all points within some distance r from 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 maimal region of the plane if the vertices and the edges of G are removed. An unbounded (infinite) face of G is called eterior, or outer face. The vertices and the edges of G that are incident with a face F form the boundary of F.

4 Proposition For every face of a given plane graph G, there is a drawing of G for which the face is eterior

5 Dual Plane Graphs Definition 5 Let G be a plane graph. The dual graph G of G is a new plane graph having a verte 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 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

6 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 If l(f i ) denotes the length (the number of edges in its boundary), of face F i in a plane graph G, then Definition 7 e(g) = l(f i ). A bond of a graph G is a minimal non-empty edge cut. Proposition Edges of a plane graph G form a cycle iff the corresponding edges in G form a bond. 6

7 Theorem (Euler): If G is a connected plane (p,q)-graph with r faces, then p q +r =. Proof. We prove it by induction on q. Base. If q = 0, then p = ; obviously, r =, and the result follows. Inductive Step. Assume that the Euler Theorem holds true for all connected graphs with fewer than q (q ) edges, and let G be a connected plane graph with q edges. If G is a tree, then p = q+ and r = (the only face is eterior) 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, by induction, p(h) q(h)+r(h) =. But p(h) = p and r(h) = r. The result follows. 7

8 Theorem If G is a planar graph (no parallel edges) with p vertices and q edges, (q ), then q p 6. If, in addition, G is bipartite, then q p 4. Proof. Let r be the number of faces of G and let m i be the number of edges in the boundary of the i th face (i =,...,r). Since every face contains at least three edges, r r i= m i. On the other hand, since every edge can be in the boundary of at most two faces, r m i q. i= Thus, r q and by Euler s Theorem, p q+q/, implying q p 6. If G is bipartite, the shortest cycle is of length at least 4. Thus, 4r r i= m i. Together with r i= m i q and p q+r =, we get the second part of the Theorem. 8

9 Corollary Every planar graph G contains a verte 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.) Corollary Graphs K, and K 5 are not planar. Proof. For K 5, p = 5 and q = 0. Thus, q > p 6 = 9 and by Theorem, K 5 is not planar. If K, were planar, then the second part of Theorem would apply, leading to a contradiction, since for this graph p = 6, q = 9, and q > p 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 verte 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 K 5 or K,. Corollary No Kuratowski graph is planar. 9

10 Coloring Planar Graphs Theorem 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 verte v of degree 5. It always eists, since else, the number of edges in the graph would eceed the upper bound of p 6. By induction, graph G v is 5-colorable. Consider a 5-coloring of G v. If any color,,, 4, 5 is 0

11 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: v,v,v,v 4,v 5 (see Figure below). Consider a bipartite graph H induced by all vertices of G v whose colors are or, and let C be the connected component of H which contains verte v. If C does not contain verte v, then we re-color C: every verte of C whose color is (resp. ) gets color (resp. ). This recoloring frees color for v, yielding a 5-coloring of G. v 5 v 4 v v v v Finally, assume that C contains verte v. Thus, there is a - colored path connecting vertices and. This path together with verte v forms a cycle which makes it impossible the eistence of a path colored and 4 connecting vertices v and v 4. Thus, recoloring the -colored connected component containing verte v makes color available for coloring v. 4 4

12 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 verte 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 verte have different colors.

13 Theorem. (Tait, 878). A -regular plane graph without bridges is 4-face colorable iff it is -edge-colorable. Proof. Suppose G satisfies the conditions of the theorem and it is 4-face-colorable. Let the colors be c 0 = 00, c = 0, c = 0, c =. Because G is bridge-less, every edge bounds two distinct faces. Given an edge between faces colored c i and c j, assign this edge the color obtained by adding c i and c j coordinate-wise modulo. For eample: c 0 +c = c ; c +c = c ; c +c = c The reverse. Suppose now that G has a proper -edge-coloring, and let the colors be,, and. Since G is -regular, every color appears at every verte. Let E,E, and E be the edgesets colored,, and, respectively. The union of any two of them is the union of disjoint cycles. Let H = E E and H = E E. The faces of graph H can be -colored with two colors α and β. The faces of graph H can be -colored with two colors γ and δ. Then each face of G is a subset of a face in H and a subset of a face in H. Assign to each face in G a pair of colors (,y) where

14 {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. Tutte s conjecture: every bridge-less -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. 4

15 Kuratowski theorem Definition 9 A subgraph H of a graph G which is a subdivision of K 5 or K, 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 Kuratowski subgraphs. If e = y is an edge in G, then we form a new graph H by contracting e into a verte z, which is adjacent to all vertices that were adjacent to at least of one of 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 = y all of whose neighbors lie on cycle C =... t. Obviously, the way and y are adjacent to the vertices of C creates the non-planarity of G. Analyzing the pattern of these adjacencies, we may find Kuratowski subgraphs in G. 5

16 4 Proof of Kuratowski s theorem Definition 0 A minimal non-planar graph is a non-planar graph F such that every proper subgraph of F is planar. Lemma For every face of a given plane graph G, there is a drawing of G for which the face is eterior. Lemma Every minimal non-planar graph is -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 net component inside a face of the embedding of the previous component (under some ordering of the components). Let G have a cut-verte and let G,G,...,G t be the connected components of G. Let H i be the subgraph induced on V(G i ) {}. By minimality of G, each H i is planar. Let us embed each H i so that verte is in the outer face and then squeeze H i into an angle < π/t at verte. Combining those embeddings together yields an embedding of G. G G H H H G 6

17 Definition Let S be a separating setof agraph G and G,...,G t be the connected components of G S. The lobes of S in G are subgraphs H,...,H t that are induced on V(G ) S,V(G ) S,...,V(G t ) S, respectively. Lemma Let S = {,y} be a separating set of G. If G is nonplanar, then at least for one lobe H i [S], adding edge y creates a non-planar graph. Proof. Suppose for every lobe H i, graph H + i = H i y is planar (i =,...,t). Then, there is an embedding of H + i for which y is in the outer face. Then we create an embedding of G by attaching each net H + i+ to an embedding of i j=h + j, where H + i+ is embedded into the face that contains y. y 7

18 Lemma 4 Let G be a graph satisfying the following three conditions: ) G doesn t contain a Kuratowski subgraph; ) G is non-planar; and ) among the graphs satisfying () and (), G has the smallest number of edges. Then G is -connected. Proof. The condition () implies that the removal of any edge violates either () or (). 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, G is at least -connected. Suppose G is not - connected, and let S = {,y} be a separating set. Then by Lemma, adding edge y to some S-lobe yields a non-planar graph. Let H be that lobe and H + = H {y}. 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 different from H. H + K C H + K P C y y y K P y Find a path P connecting and y in H. Now, we replace edge y in K with P. Clearly, K y P is a Kuratowski subgraph of G. The contradiction at least for one lobe H i [S], adding edge 8

19 y creates a non-planar graph. implies that G has no -cut, i.e. G is -connected. Definition The operation of contraction of an edge e = (,y) of a graph G removes e, and y from the graph, and adds a new verte z, which is adjacent to any old verte u iff u was adjacent to at least one of or y. The resulting graph is denoted G/(,y). Lemma 5 Every -connected graph G with at least five vertices has an edge e such that G/e is -connected. Proof. Take any edgee = y and assume G/e isnot -connected. Then G/e has aseparating set S of size. SinceGis-connected, S contains the verte replacing and y. Let z be the other verte. Then, {,y,z} is a -separating set of G. C y D z u If G has no edge whose contraction yields a -connected graph, every edge y has a verte z, call it a partner, such that the removal of,y and z disconnects the graph. Among all edges of G, we take edge y and a partner z whose removal from G creates the largest connected component C. Let D be the other component of G {,y,z}. 9

20 Since {,y,z} is a minimal separating set of G, each of the vertices,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 {,y,z} is disconnected. If v is outside of set V(C) {,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) {,y}, then the removal of z, v disconnects G, contradicting the fact that G is -connected. This completes the proof. C y z u v D C v y z u D 0

21 Lemma 6 If G/e has a Kuratowski subgraph, then G has Kuratowski subgraph. Proof. Let H be a Kuratowski subgraph of G/e, where e = y and let z be the verte of obtained from contracting y. If z V(H), H is a Kuratowski subgraph of G. If z V(H) is one of the vertices used to subdivide an edge in K 5, or K, to obtain H, then replacing z with, or y, or y yields a Kuratowski subgraph of G. If z V(H) and among the edges incident to z at most one is incident to (resp. y) (this must happen if H is a subdivision of K, ), then replacing z with y (resp. with ) reveals a Kuratowski subgraph in G. Finally, let H be a subdivision of K 5, and z be adjacent to four edges, two of which are adjacent to and the remaining two are adjacent to y. Let u and u (resp. v and v ) be the two vertices of degree 4 in H that are connected with z through the two edges incident to (resp. y). In this case, we find a subgraph of G which is a subdivided K, : vertices u,u, and y form one partition of the subgraph, and v,v, and form the other; the paths connecting them are those from H. z y u v u v u v u v

22 Theorem 5 Every graph without Kuratowski subgraph is planar. Proof. Because of lemma, we can consider -connected graphs only. We prove the Theorem by induction on n the number of vertices of G. Base. If n 4, the the only -connected graph is K 4, which is planar. Inductive Step. Let n 5 and e = y be an edge such that H = G/e is -connected. planar cases K, K 5

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