Exercise sheet 5: Solutions - EPFL

Graph Theory Spring 2013

Prof. J´anos Pach Assist. Filip Mori´c

Exercise sheet 5: Solutions Caveat emptor: These are merely extended hints, rather than complete solutions.

1. Prove that in any drawing of a graph with n ≥ 3 vertices and m edges in the plane, there are at least m − 3n + 6 crossings. Solution. For a fixed n, the proof goes by induction on m. If m = 3n − 5, then the graph is not planar, so in any drawing there’s at least one crossing. On the other hand, if we have a drawing of a graph with m ≥ 3n − 4 edges, then, if we remove one edge that contains a crossing, by the inductive hypothesis, we will get a drawing with at least (m − 1) − 3n + 6 crossings. 2. (a) Show that the Petersen graph contains a subdivision of K3,3 . (b) Does it contain a subdivision of K5 ? Does it contain a K5 -minor? (c) Deduce that the Petersen graph is nonplanar. Solution. (a) See the picture. (b) It doesn’t contain a K5 -subdivision, since there’s no vertex of degree at least 4. It does contain a K5 -minor. (c) Follows from Kuratowski’s theorem.



3. (a) Show that every planar graph has a vertex of degree at most 5. (b) Is there a planar graph whose all vertices are of degree at least 5? P Solution. (a) If d(v) ≥ 6 for all v, then 2|E| = d(v) ≥ 6n, contradicting |E| ≤ 3n − 6 . (b) Yes, the graph of an icosahedron.

˜ in which all 4. A graph G is outerplanar if it has a planar embedding G vertices lie on the boundary of its outer face. (a) Show that every outerplanar graph contains a vertex of degree at most 2. (b) What is the maximum number of edges that an outerplanar graph on n vertices can have? Solution. (a) Keep adding edges to G as long as it remains outerplanar. Once we cannot add any more edges, it’s easy to see that (there’s a drawing in which) the outer face is bounded by a cycle v1 v2 . . . vn , that contains all the vertices. Also, all internal faces are triangles. Consider an edge vi vj , not belonging to the outer cycle, for which min{|i − j|, n − |i − j|} is minimum. Clearly, we must have that the minimum is equal to 2, and the unique vertex that lies between vi and vj on the outer cycle is of degree 2. (b) For n = 2 the maximum is 1, and by induction (using (a)), for every n it’s at most 2n − 3. A triangulated n-gon attains exactly that number of edges. 5. The walls of an art gallery form a polygon with n sides. It is desired to place staff at strategic points so that together they are able to survey the entire gallery. (a) Show that bn/3c guards always suffice.



(b) For each n ≥ 3, construct a gallery which requires this number of guards. Solution. (a) Any n-gon can be triangulated. A triangulated n-gon represents a drawing of an outerplanar graph. Any outerplanar graph is 3-colorable. Indeed, we can prove this easily by induction, using the fact that any outerplanar graph has a vertex of degree at most 2 (see the preceding exercise). Now we use this coloring for the vertices of the initial n-gon. At least bn/3c of them will be of the same color. We place a guard at each such vertex. Then the whole gallery is guarded, since the vertices of each triangle in the triangulation are of different colors, and thus one of them is always of the chosen color. (b) For n ≥ 6, polygons like the one in the picture require bn/3c guards.

6. Let S be a set of n points in the plane, where n ≥ 3 and the distance between any two points of S is at least one. Show that no more than 3n − 6 pairs of points of S can be at distance exactly one. Solution. Connect by a straight line segment any two points whose distance is 1. No two such segments can intersect. Indeed, suppose that ab and cd intersect, and let the vertices a, c, b, d appear in that order along the boundary of the convex hull of {a, b, c, d}. Then we would have that 2 = |ab| + |cd| > |ac| + |bd|, with a contradiction. Thus, we have a planar embedding, and the number of edges of the corresponding (planar) graph is at most 3n − 6. 7. Show that every graph can be embedded in R3 with all edges straight. Solution. The shortest way to prove it is probably this: just place the vertices on the curve {(t, t2 , t3 : t ∈ R)}, and draw safely the edges as straight line segments.



8. Show that one can draw K5 on a torus with no crossing. Solution. Even K7 can be embedded on the torus.

9. Dwellers of Flatplanet are hostile creatures. Their countries don’t even share borders, but are well separated from each other. There are six countries in total, three of them are the Allies, and the other three are, not surprisingly, the Axis. For strategic reasons, each country is topologically connected. An embedded correspondent is informing us that the shortest distance from any of the Allies countries to any of the Axis countries is equal to 1 flatkilometer (a unit of length widely used on Flatplanet). Could she be right? Solution. Suppose she’s right. Let the Allies be A1 , A2 , A3 and the Axis B1 , B2 , B3 . For any i, j ∈ {1, 2, 3}, there are points aij ∈ Ai and bij ∈ Bj , such that |aij bij | = 1. Observe that for the same reason as in the solution of Ex.6 of this sheet, no two of the nine segments aij bij intersect each other. For each i ∈ {1, 2, 3}, choose an arbitrary point ai ∈ Ai . Join each ai to ai1 , ai2 , ai3 by using paths that lie completely in Ai . We can do it in such a way that any two of those paths have only an initial segment starting at ai in common (note that we have no information whether the countries have a non-empty interior!). If we do the same for the Axis countries, we will have “almost” a planar embedding of K3,3 , the only problem being that some edges with a common endpoint can have a common initial segment. However, this can clearly be transformed into a true embedding, which is a contradiction.

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