# MTH 548 Graph Theory Fall 2003

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1 MTH 548 Graph Theory Fall 2003 Lesson 6 - Planar graphs Planar graph plane drawing - plane graph face length of a face infinite face Thm 12: Euler. Let G be a plane drawing of a connected planar graph. With n, m, f, denote respectively the number of vertices, edges and faces. Then n m + f = 2. Cor 12.1: k components n m + f = k + 1 Cor 12.2: m 3n 6 Cor 12.3: m 2n 4 (triangle free) Thm 13: K 3,3 and K 5 are non-planar. Thm 14: Any graph containing a non-planar subgraph is nonplanar homeomorphic - subdivison, Thm 15: Any graph homeomorphic to a non-planar graph is non-planar. Thm 16: Any graph homeomorphic to either K 3,3 or K 5 is non-planar. Thm 17: Any graph containing a subgraph homeomorphic to either K 3,3 or K 5 is non-planar. Thm 18: (Kuratowski) A graph is planar if and only if it contains no subgraph homeomorphic to K 3,3 or K 5. Thm 19: Every planar graph contains a vertex of degree at most 5 Thm 20: 5 - color theorem Crossing number cr(g) Printed circuits Thickness t(g) 1

2 2 Complexity question: Decide if G is planar.

3 Theorem 12. Euler. Let G be a plane drawing of a connected planar graph. With n, m, f, denote respectively the number of vertices, edges and faces. Then n m + f = 2. Proof: The proof is by induction on the number of edges. The base case is when there are no edges. Then since G is connected, it is the trivial graph with n = 1, m = 0, and f = 1. We see that n m + f = = 2. Let m > 1. We assume that a connected plane graph that has m edges with m > m 1, n vertices and f faces then n m + f = 2. Now take a connected plane graph G on m edges, n vertices, and f faces. If G is a tree then we know that m = n 1 and f = 1. In this case we have: n m + f = n (n 1) + 1 = = 2. If G is not a tree, not every edge is a cut edge. If uv is not a cut edge then uv has is bordered by two different faces, and G uv is connected with 1 less face than G. Then G uv satisfies the induction hypothesis and we have that n (m 1) + f 1 = 2 which implies that n m + f = 2. Corollary Let G be a plane drawing of a planar graph that has k components, n vertices, m edges, and f faces. Then n m + f = k + 1. Proof: Suppose G has k components. For each i [k] suppose component C i has n i vertices, m i edges, and f i faces. Each one is connected, and so by Euler s Formula. For all i [k], n i m i + f i = 2. Adding the k equations we get: n n k (m m k ) + (f f k ) = 2k. Substituting n 1 + +n k = n, m 1 + +m k = m and f 1 + +f k = f+k 1, gives n m + f + k 1 = 2k. Which of course implies, n m + f = 2k k + 1 = k + 1. Definition 2. The length of a face in a plane graph G is the total length of the closed walk(s) in G bounding the face. 3

4 4 Note that a cut edge is bordered on both sides by the same face and contributes 2 times to the length of the face. Corollary If G is a planar graph with n vertices and m edges then m 3n 6. Moreover, if G is maximal planar then, m = 3n 6. Proof: Embed G in the plane. First assume that G is maximal planar. Then G is connected and every face must be bordered by 3 edges. That is the length of each face, F, l(f ) is 3. We have that f i=1 l(f i) = 2m and on the other hand, f i=1 l(f i) = 3f. By Euler s Formula we may substitute f = m n+2. Thus, 2m = 3(m n+2), so that m = 3n 6. If G is not maximal planar, we can add edges to G to obtain G which is maximal planar and has m edges. Then m < m = 3n 6. Corollary If G is a planar graph with n vertices, m edges, and no 3-cycles then m 2n 4. Proof: The argument is similar to that of the last corollary. Except, in this case, we have that the length of each face is at least 4. Thus obtaining 4f 2m and after substituting f = m n+2, we get 4(m n+2) 2m. So that m 2n 4. Theorem 13. K 3,3 and K 5 are non-planar. Proof: Suppose K 3,3 is planar. By Corollary 12.3, for n = 6 and m = 9, 9 2(6) 4 = 8. But this is a contradiction, so K 3,3 cannot be planar. Suppose K 5 is planar. By Corollary 12.2, for n = 5 and m = 10, 10 3(5) 6 = 9. But this is a contradiction, so K 5 cannot be planar. Theorem 14. Any graph containing a non-planar subgraph is non-planar.

5 Proof: The proof should be clear. Theorem 15. Any graph homeomorphic to a non-planar graph is non-planar. Proof: The proof should be clear. Theorem 16. Any graph homeomorphic to either K 3,3 or K 5 is non-planar. Proof: This is due to Theorem 13 and Theorem 15. Theorem 17. Any graph containing a subgraph homeomorphic to either K 3,3 or K 5 is non-planar. Proof: This is due to Theorem 16 and Theorem 14. Theorem 18. Kuratowski. A graph is planar if and only if it contains no subgraph homeomorphic to K 3,3 or K 5. Theorem 17 gives the sufficiency, ( ). For the necessity, ( ), we must first prove a number of lemmas. Definition 3. A graph is 2-connected if there are no cut vertices. A subgraph of a graph G is called a block if it is a maximal 2-connected subgraph. If a graph is 2-connected then it might be called a block. The meaning of maximal in the above definition is this. If B is a block of G and H is a 2-connected subgraph of G containing B, then B = H. A block of G may contain vertices which are cut-vertices of G but will not be cut vertices in the block. A graph G with at least one cut vertex has more than one block. We notice that the blocks form a sort of tree structure and so there are blocks which behave like leaves in a tree and there will be at least 2 of them. The next lemma says something about cycles in blocks. 5

6 6 ui u ui-1 Figure 1. u i U and u i U Lemma A graph G of order p 3 is a block if and only if every two vertices of G lie on a common cycle of G. Proof: Suppose every 2 vertices of G lie on a common cycle. Then, no vertex could be a cut vertex. For, suppose v is a cut vertex, then u, w such that v is on every u, w-path. But then there is no cycle containing both u and w. Since G has no cut vertex, it is a block. Now suppose G is a block and for the sake of contradiction, suppose for vertex u there is some vertex, not on a cycle with u. Let U be the set containing u and all vertices that are on common cycles with u. Since G u is still connected, every pair of vertices in N(u) are connected by a path in G u. Therefore, N(u) U. We are assuming V U is nonempty. Let v V U. Consider a u, v-path W : u = u 0, u 1,... u n = v. Let i be the smallest integer 2 i n such that u i U. Then there exists a cycle C containing u and u i 1. Because u i 1 is not a cut vertex, there exists a u i, u-path P : u i = v 0, v 1, v 2,... v m = u not containing u i 1. The only vertex common to both P and C cannot be u because if so then following C from u to u i 1 then along edge u i 1 u i to u i followed by P back to u is a cycle containing u.

7 Let j be the smallest integer 1 j m 1 such that v j is on C. Now we have a cycle containing both u and u i contradicting that V U is nonempty. Here is the cycle. Follow P from u i to v j, C toward u, through u back to u i 1, then use the edge u i 1 u i back to u i. Lemma Let G be a connected graph with one or more cut-vertices. Then among the blocks of G, there are at least two, each of which has exactly one cut vertex of G. Proof: Form the following bipartite graph H with bi-partitions A and B. In A, include all the cut vertices of G and in B, include a vertex B i for each block B i of G. We include an edge b i to B j in H whenever b i B j. Claim: H is a tree. Suppose there is a cycle in H. Then it is of the form, a 1, B 1, a 2, B 2,..., B k, a 1 and corresponds to a cycle in G within blocks B 1,..., B k. But then the previous theorem, B 1 B 1 B k is a block of G. So for each i, B i is not maximally 2-connected in G. Thus, H has no cycles. Clearly, H is connected since G is connected. Since H is a tree, there are at least 2 end vertices, which are blocks in G that contain exactly one cut vertex. Definition 4. A block containing one cut vertex is called an end-block. Before we state the next lemma, we give a definition. Definition 5. A block G is called a critical block if v, G v is not a block. Lemma If G is a critical block of order p 4, then G contains a vertex of degree 2. Proof: Clearly no vertex in a block G is of degree 1. For each vertex x of G, there exists another vertex y of G x such 7

8 8 H w 1 w 2 u G2 G1 v Figure 2. w 2 V (H) that G x y is disconnected. Note that each component of G x y must contain some vertices adjacent to x and some adjacent to y. Otherwise we would have a cut vertex. Among all such pairs x, y of vertices of G, let u, v be a pair such that G u v contains a component G 1 of minimum order n. If n = 1 then that vertex in G 1 is of degree 2, since it must be adjacent to both u and v. Assume n 2. Let H =< V (G 1 ) {u, v} >, and G 2 = G V (H). Let w 1 V (G 1 ). Let w 2 G w 1 be such that G w 1 w 2 is disconnected. Consider the following cases. Case 1. Suppose w 2 V (H). See Figure 2. Since we know that no vertices of G 2 are adjacent to w 1 and < V (G 2 ) {u} > and < V (G 2 ) {v} > are connected, some component of G w 1 w 2 is smaller than G 1. So this case cannot happen. Case 2. Suppose w 2 V (G 2 ). See Figure 3. Each component of G w 1 w 2 contains some vertices of H and some vertices of G 2 due to the fact that each component must have vertices adjacent to each of w 1 and w 2. Since G u v has the component H u v and G 2 is the union of the remaining components, at least one of the vertices

9 9 G2 H u w 2 w 1 G1 v Figure 3. w 2 V (G 2 ) u and v are on every path between vertices in H u v and G 2. Thus if u and v were in the same component of G w 1 w 2, there would be some component of G w 1 w 2 without vertices adjacent to w 1 or without vertices adjacent to w 2. So, H w 1 must contain at least 2 components, one containing u, call it H u and one containing v, call it H v. If H w 1 contains another component then again, that component would have no vertices adjacent to w 2. If either H u or H v is trivial, then its only vertex would be of degree 2, adjacent to both w 1 and w 2. So assume H u and H v are nontrivial. Now consider G w 1 u. This has a component contained in H w 1 u, namely H u u. We have H = n + 2, the sets H u, H v, {w 1 } form a partition of H, and H v 2. So H u u n 1. We again reach a contradiction to the way u and v were chosen. Lemma A graph is planar if and only if each of its blocks is planar. Proof: The graph G is planar if and only if each of its components is planar, so assume G is connected. Trivially, if G is planar then each block is planar. It remains to show that if each

10 10 block is planar then G is planar. This is proved by induction on k, the number of blocks. Suppose k = 1. Then G is planar. Now suppose G is a graph with k 2 blocks and suppose for all graphs H with k 1 planar blocks, H is planar. Let B be an end-block of G with cut vertex v. Then H = G (B v) has k 1 planar blocks and so is planar by the induction assumption. Take any region R of the planar embedding of H which has v on its border. Consider a planar embedding of B with v on its exterior region. Now, B can be embedded in R while identifying v in B with v in H. Lemma If a block contains no subdivision of K 5 or K 3,3, then it is planar. Proof: Suppose it s not true. Of all non-planar blocks that contain no subdivision of K 5 or K 3,3, pick H of minimum size, number of edges. Claim: δ(h) 3. If δ(h) = 1 then H is not a block. Suppose v V (H) such that deg v = 2 with vu, vw E(H). Then we would have one of the following two cases. Case 1. uw E(H). In this case, H v is still a block and still has no subdivision of K 5 or K 3,3. But H v is smaller in size than H and so must be planar because of the way we chose H. Clearly if H v can be embedded in the plane then so can H. Case 2. uw E(H). Again, in this case, H v + uw is still a block and still has no subdivision of K 5 or K 3,3. For, suppose H 1 is a subdivision of

11 K 5 or K 3,3 in H v + uw. Then H would contain a subdivision of H 1. Thus again we see that there would be a planar embedding of H in this case as well. This proves the claim. Now we know by Theorem 18.3 that H is not a critical block. Thus, there exists an edge e = uv such that H e is still a block. Set H = H e. H is planar since H is of minimum size. By Theorem 18.1, we know that H contains a cycle containing both u and v. Among all planar embeddings of H, choose one that has a cycle C with the maximum number of interior regions. Let C : u = v 0, v 1,..., v i = v, v i+1,..., v n = u where 1 < i < n 1. Observation 1: H is non-planar so both interior and exterior subgraphs are nonempty. Otherwise we can add the edge uv. Observation 2: No 2 vertices are connected by a path embedded in the exterior of C among {v 0, v 1,..., v i } or {v i, v i+1,..., v n }. For, such a path cannot be embedded in the exterior region due to our choice of C having the most number of interior regions. By these two observations, we see that there must exist a v j v k path P in the exterior with 0 < j < i < k < n that doesn t intersect C anywhere else or have any other paths connecting P to C at vertices other than v j or v k. See Figure 4 Consider the subgraph H (C {v j, v k }). Let H 1 be the component of this subgraph that contains the path P. By the choice of C, H 1 cannot be switched to the interior of C in a 11

12 12 P v j v 0 v i C v k Figure 4. Path P plane manner. This fact and the fact the edge e = uv cannot be inserted in a plane manner imply that the interior of the subgraph H must contain one of the following: (1) A v r, v s -path, Q, 0 < r < j, i < s < k, (or, equivalently, j < r < i and k < s < n) none of whose vertices different from v r and v s belongs to C. (2) A vertex w not on C that is connected to C by three internally disjoint paths such that the end-vertex of one such path P is one of v 0, v j, v i, and v k. If P ends at v 0, the end-vertices of the other paths are v r and v s with j r < i and i < s k but not both r = j and s = k hold. If P ends at any of the other vertices, v j, v i or v k, there are three analogous cases. (3) A vertex w not on C that is connected to C by three internally disjoint paths P 1, P 2, P 3 such that the end-vertices of the paths are three of the four vertices v 0, v j, v i, v k. Say v k is the missing one, and P 1 connects v 0 to w and P 2 connects v i to w. Then there is also a vertex x v 0, w, v i on one of the paths P 1 or P 2 and a path P 4 from x to

13 13 P v j P v j v 0 v i v 0 v i C C v k v k Case 1 Case 2 P v j P v j v 0 v i v 0 v i C C v k v k Case 3 Case 4 Figure 5 v k. The remaining choices of the missing vertex produce analogous results. (4) A vertex w not on C that is connected to the vertices v 0, v j, v i, v k by four internally disjoint paths. Figure 5 depicts the four situations given above. First, convince yourself that these are the only possibilities. Then observe that when we include the edge e = uv, in the first three cases, a subdivision of K 3,3 can be found and in the fourth case a subdivision of K 5 exists. This is of course a contradiction and completes the proof. Proof of Theorem 18: Given that G contains no subgraph homeomorphic to either K 5 or K 3,3, we know that none of its

14 14 blocks contains one of these subgraphs. So all the blocks of G are planar by Lemma Then also G is planar by Lemma Theorem 19. Every planar graph has a vertex of degree at most 5. Proof: If we suppose not, then 6n < n i=1 deg(v i) = 2m. But this contradicts Corollary Theorem 20. Every planar graph is 5-colorable. Proof: The proof is by induction on the order, n, of the graph. The base case is when n = 1 and the theorem is trivially true for any graph on 1 vertex. Now suppose n > 1 and any planar graph on fewer than n vertices can by 5-colored. Suppose G is a planar. If G has a vertex v of degree at most 4, we can color G v by induction and then use a color not used on N(v) for v. If not, then G has a vertex, v, of degree 5. We color G v by induction and look at the colors used on N(v). If fewer than 5 colors are used, we can color v with one of our 5 colors. So assume all 5 colors c 1, c 2, c 3, c 4, c 5 are used on the neighbors of G, v 1, v 2, v 3, v 4, v 5 and that c(v i ) = c i. Moving clockwise through the edges incident to v, we choose one neighbor to call v 1, the next v 2, and so on until we reach v 5. For {i, j} [5], let H(i, j) be the subgraph of G induced on the vertices that are colored with a color in {c i, c j }. If for some i j, the vertices v i and v j are in separate components of H(i, j), then we can pick the component containing v i, exchange the colors c i and c j in that component only. Now both v i and v j are colored with color c j so that c i can be used for v.

15 We know that this has to happen for 2 of the colors since, if there is a path joining v 1 to v 3 in H(1, 3) then there is no path joining v 2 to v 4 in H(2, 4). 15

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