winter term 2019

Solution sheet 7 Discussion of solutions: December 4, 11:30 am.

Problem 25. 5 points

(a)( 2 points) Let G be a graph with χ(G) ≥ 3. Prove that G contains a K3-minor. What is the maximum number of edges in an n-vertex graph without a K3-minor? (b)( 3 points) The Petersen graph, which we shall denote by H, is the graph with vertex set V (H) = {x0, . . . , x4, y0, . . . , y4} and edge set E(H) = {xixi+1, xiyi, yiyi+2 : i = 0,..., 4}, where subscripts are computed modulo 5. Let G be the graph obtained from H by adding a vertex u disjoint from V (H) and joining u to every vertex in H. Show that G contains an MK6, but does not contain a TK6.

Solution. Part (a)

If χ(G) ≥ 3, then G is not bipartite and hence contains an odd of length at least 3. It is easy to find a subdivision of K3 in any cycle of length at least 3, so G contains a K3-minor, as claimed. Now, if G is an n-vertex graph without a K3-minor, then G cannot contain a cycle. Therefore G is a forest, say, with k components, and so has n − k edges. This is maxi- mized when k = 1, i.e. when G is a with n − 1 edges.

Part (b)

We show that H contains a K5 minor, but does not contain a TK5; this clearly implies the result. To see that H contains a K5 minor, let Xi = {xi, yi} for i = 0,..., 4. Each G[Xi] is isomorphic to K2, so is easily connected. Moreover, it is not hard to check that there is a Xi − Xj edge for each i, j ∈ {0,..., 4} with i 6= j. It follows that H contains a K5 minor. For the second claim, note that subdividing an edge of a graph G with ∆(G) ≥ 2 cannot change the maximum . Thus, if the Petersen graph H contained a subdivision K of K5, we would obtain ∆(H) ≥ ∆(K) = ∆(K5) = 4. However, H is a 3-regular graph, so this is a contradiction. 

Problem 26. 5 points A graph is called outerplanar if it has a drawing in which every vertex lies on the boundary of the outer face. Show that a graph is outerplanar if and only if it contains neither K4 nor K2,3 as a minor. Further, show that any n-vertex outerplanar graph has at most 2n − 3 edges for n > 1, and give an example that shows this bound can be achieved. Solution. Let G = (V,E) be any graph. Consider the graph G0 obtained by adding a new vertex v∗ to G and joining it to all vertices in G. We shall prove that G is outerplanar if and 0 only if G is planar. By Kuratowski’s theorem it then remains to show that TK4 ⊆ G if 0 0 and only if TK5 ⊆ G , and that TK2,3 ⊆ G if and only if TK3,3 ⊆ G .

http://www.math.kit.edu/iag6/lehre/graphtheory2019w/en Graph Theory winter term 2019

Claim. G is outerplanar if and only if G0 is planar. If G is outerplanar, then we can draw v∗ and its adjacent edges into the outer face, obtaining a planar drawing of G0. Conversely, if G0 is planar, then a drawing of it, after removing v∗, is a drawing of G and the face of G where v∗ was has all vertices of G on its boundary, so G is outerplanar. 0 Claim. TK4 ⊆ G if and only if TK5 ⊆ G . ∗ ∗ Let H be a TK4 in G. Because v is adjacent to every node in G, we can add v and the ∗ 0 edges {vv : degH (v) = 3} to H to obtain a TK5 in G . 0 ∗ Conversely, let H be a TK5 in G . If v ∈/ H, then H is already a subgraph of G and ∗ ∗ contains a TK4. If v ∈ H, then H − v is a subgraph of G and contains a subdivision of K4. 0 Claim. TK2,3 ⊆ G if and only if TK3,3 ⊆ G . ∗ ∗ Let H be a TK2,3 in G. Because v is adjacent to every node in G, we can add v and ∗ 0 the edges {vv : degH (v) = 3} to H to obtain a TK3,3 in G . 0 ∗ Conversely, let H be a TK3,3 in G . If v ∈/ H, then H is already a subgraph of G and ∗ ∗ contains a TK2,3. If v ∈ H, then H − v is a subgraph of G and contains a subdivision of K2,3.

Finally, suppose |G| = n > 1. The graph G0 is planar so we know that |E(G0)| ≤ 3(n+1)−6 = 3n−3. Therefore, the number of edges of G is at most 3n−3−n = 2n−3. So we see that any n-vertex outerplanar graph has at most 2n − 3 edges. There are several possible examples that show this bound is tight. For instance, consider a cycle C of length n, pick a vertex v ∈ C, and join v to every vertex of C. We see that this graph has n+(n−3) = 2n−3 edges, and it clearly has a drawing in the plane with every vertex on the boundary of the outer face. Another example can be constructed inductively, as follows: start with a triangle in the plane; for each edge e on the boundary, add a vertex ve in the outer face, and join this vertex to the endpoints of e (‘erect a triangle on each side of the first triangle’); repeat this procedure for each new erected triangle. This graph is clearly outerplanar and it has the requisite number of edges. Indeed, the triangle has 2 · 3 − 3 = 3 edges. Suppose we obtain a graph on n0 vertices and 2n0 − 3 edges at some later stage in the construction. We obtain our graph in the next step by adding n0 new vertices and 2n0 edges. Accordingly, with n := 2n0, we have a graph on n vertices and 0 0 0 (2n − 3) + 2n = 2(2n ) − 3 = 2n − 3 edges. 

Problem 27. 5 points Show that adding a new edge to a maximal of order at least 5 always produces a TK5 subgraph (you may use the fact that every face is bounded by a cycle in a 2-connected plane graph). Solution. Since G is maximal planar, it follows by Kuratowski’s theorem that G is edge maximal without a K5 or K3,3 topological minor. Since |G| > 4, we may apply a lemma from lecture which shows that G is necessarily 3-connected. Pick two vertices u, v ∈ V (G) with uv∈ / E(G). By the global version of Menger’s theorem there are three independent u − v paths P1,P2,P3 in G. Let xi ∈ V (Pi) ∩ N(u) for i = 1, 2, 3. We may assume that the paths P1,P2,P3 do not contain any other vertices of N(u) by choosing each path to have minimum length.

http://www.math.kit.edu/iag6/lehre/graphtheory2019w/en Graph Theory winter term 2019

Claim 1. G[N(u)] contains a spanning cycle. Proof. As G − u is still planar, consider a drawing of G − u in the plane and let f denote the face containing u. Let C denote the boundary of f. As u ∈ f, we have that the neighborhood of u is contained in C. Since G is 3-connected, G − u is 2-connected, and therefore C is actually a cycle. As G is maximally planar (and hence maximally plane as well), u must be joined to every vertex of C. With the above claim, we are now done. Indeed, let C be the cycle spanning N(u). Then each of the xi are pairwise connected by the subpaths Q1 := x1Cx2 (the arc not containing x3), Q2 := x2Cx3 (not containing x1), Q3 := x3Cx1 (not containing x2). These subpaths are pairwise internally vertex disjoint. Moreover, since N(u) ∩ V (Pi) = {xi} for i = 1, 2, 3, each path Qi intersects Pj only in {x1, x2, x3} for i, j ∈ [3]. Therefore, after adding the edge uv we obtain a copy of TK5 where the vertices of our K5 are u, v, x1, x2, x3. 

Problem 28. 5 points Prove that every outerplanar graph is 3-colorable (without appealing to the 4-color the- orem). Solution. We prove this by induction on the number of vertices in G. The overall strategy is to reduce this problem to the 2-connected case, and to then use the fact that any face in a 2-connected planar graph is bounded by a cycle. Note that any subgraph of an outerplanar graph is again outerplanar. As induction base consider the 1-vertex graph, which is clearly 3-colorable. So let G be outerplanar with a fixed outerplanar embedding and with |V (G)| ≥ 2. If G is disconnected then we may simply combine 3-colorings of its components, which exists by induction. We now have two cases to consider: G is connected, but has a cut-vertex, or G is 2-connected. Case 1. G is connected but not 2-connected. Similar to the previous case consider a 3-coloring of every block of G (which exist by induction). By permuting the colors we can assume without loss of generality that any two intersecting blocks agree on the same color of the cut vertex they have in common. We remark that we use here that the block-cut-vertex graph is a tree and that every tree can be built up from a single vertex by adding leaves one at a time. Thus, all 3-colorings combined give a 3-coloring of G. Case 2. G is 2-connected. Then every face of G is bounded by a cycle. In particular, the outer face is bounded by a cycle, which contains all vertices of G, and we denote it by C. A cycle is 3-colorable, so assume there is an edge e = uv∈ / C. Let P1, P2 be the two paths linking u and v on C. Both these paths have lengths at least two. Consider the induced subgraphs Gi = G[Pi], i = 1, 2, both of which have fewer vertices than G. If we can show that E(G) = E(G1) ∪ E(G2), then we can take 3-colorings of both of them, possibly permute the colors to agree on V (G1) ∩ V (G2) = {u, v} and combine them to a 3-coloring of G. 0 0 So let e = xy ∈ E(G) and assume e ∈/ E(G1) ∪ E(G2), i.e., without loss of generality 0 x ∈ P1 \ P2 and y ∈ P2 \ P1. In particular, x, y∈ / {u, v}. Because e ∈/ E(C), the interior 2 of w lies in the same region of R \ (P1 ∪ P2) that contains the interior of e = uv. But this implies that e and e0 cross, a contradiction to the planarity of G. 

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