Prof. Dr. Jörn Steuding, Miriam Goldschmied, Pascal Stumpf Institut für Mathematik, Universität Würzburg 03.05.2017 Algebraic Graph Theory — 1. Exercise Sheet Exercise 1. A regular dodecahedron is built of twelve pentagons, such that in each of its vertices exactly three pentagons are meeting. We can also see it as a graph with twenty vertices and ? edges, and we refer to this graph as Nicosia for short. (i) Count all ? edges of Nicosia in two different combinatorial ways, only using the information available so far. (ii) List five distinct attributes introduced in the lectures that Nicosia may adorn itself with, and if possible justify each of them briefly. (iii) Can you find a closed walk within Nicosia visiting every vertex exactly once? (iv) What is the minimum number of edges we need to remove from Nicosia, such that the resulting subgraph contains an Euler cycle? Exercise 2. For the classical “double six” set of 28 dominoes Terquem (1849) proved the existence of an Euler cycle for the corresponding graph. (i) What is meant here by corresponding graph? Take a huge sheet of paper and draw the “domino graph” D having 28 vertices, one for each domino tile, and edges reflecting the idea of the game. (ii) Find an Euler cycle in the graph D (or in your double six set). (iii) How many Euler cycles do exist in D? Problems for the first tutorial 1. Starting from the empty graph on n = 5 or 6 vertices without any edges between them yet, try to add as many edges as possible such that our arising graph remains triangle free (no small cycles of length three). 2. Now suppose we are given the complete graph on n > 3 vertices, and we try to remove as few edges as possible such that our subgraph becomes triangle free again. Prove that we need to remove at least 1/3 of all edges in this case. 3. On the other hand, by revisiting the graphs we have met in our course so far, try to find, for every even number n of vertices, one triangle free graph with n2 /4 edges, and also check with help of its adjacency matrix that there are indeed no triangles. Excursion: Determine (without determinants ob.bo) its eigenvalue spectrum. 4. In comparison with the graphs we found, show that a triangle free graph on n > 1 vertices can only have at most bn/2cdn/2e edges, where we get some helpful insight by looking at the neighbourhood of a vertex of maximal degree. 5. Moreover, in spirit of the extremal graphs we found before, also try to construct, for every r > 3 and every number n = a · r (a > 1) of vertices, a r-clique free graph (no r vertices are adjacent to each other) with as many edges as possible. 6. As a generalization of our former bound, let us finally show, for every r > 3, that a r-clique free graph on n > 1 vertices cannot have more than (1 − 1/(r − 1)) · n2 /2 edges, where an induction on the number n of vertices might support us here. 2 Exercise 3. (ob.bo) . . . 17 3 20 r 2 10 be 1 9 7 em 8 16 1 4 ec 2 D 7 5 3 6 14 1 2 2 0 31 2 3 5 3 1 4 12 1 0 2 8 29 2 2 11 19 27 18 26 25 November 2017 6 13 20 27 7 14 21 28 1 8 15 22 29 2 9 16 23 30 3 10 17 24 4 11 18 25 5 12 19 26 5 12 19 26 6 13 20 27 7 14 21 28 1 8 15 22 29 4 11 18 25 7 14 21 28 6 13 20 27 3 10 17 24 31 1 8 15 22 29 5 12 19 26 2 9 16 23 30 3 10 17 24 2 9 16 23 30 7 17 6 14 20 5 3 ay 1 1 4 M 2 20 2 8 1 2 3 1 9 2 10 1 8 1 6 27 1 2 1 9 7 5 8 16 1 4 2 2 15 23 31 22 30 29 7 201 ust 5 6 Aug 3 4 12 13 2 11 1 20 0 1 9 8 19 7 8 16 17 1 26 27 25 5 14 1 23 24 2 21 2 30 31 9 28 2 January 2017 Apr il 2 017 3 4 5 10 1 1 2 6 1 17 1 12 13 7 8 9 8 19 14 1 24 5 1 20 2 5 26 2 21 22 6 7 28 23 29 3 0 4 11 18 25 June 2017 Ju ly 3 20 1 4 0 1 7 5 17 11 6 24 18 12 1 31 25 19 13 1 7 2 8 4 26 20 9 27 21 15 28 22 16 29 23 30 Sep tem b er 201 7 4 5 6 7 1 2 11 1 3 2 13 8 18 1 9 9 20 14 15 1 10 25 2 6 6 27 21 22 2 17 3 28 2 9 30 24 7 201 5 rch Ma 3 4 2 1 2 10 11 1 9 8 9 17 18 1 16 6 7 5 26 24 2 4 15 13 1 22 23 1 3 1 20 2 29 30 8 27 2 Fe br ua 6 ry 1 13 7 20 2 8 20 14 17 3 9 27 21 15 28 22 16 10 1 4 5 1 23 17 24 18 12 25 19 26 17 20 1 8 7 15 6 4 2 O 5 13 1 1 2 9 2 2 4 2 0 3 11 1 9 2 7 28 1 2 2 10 8 6 9 17 1 5 2 2 16 24 23 31 30 r be o ct
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