Solutions to Homework 2 : Ma121a - Combinatorics 1) a) Firstly, since every edge bounds two faces, if the number of edges to each face is fi , then 2|E| = X fi . Also, since G is triangle free, each face has at least 4 edges, so that 2|E| = X fi ≥ 4|F |. Putting this bound on |F | into Euler’s formula gives n − |E| + |E| ≥ 2, 2 which is equivalent to the inequality required. b) We can now proceed by induction, noting that the most trivial case, a simple square, is in fact, 2 colorable. If all the vertices have degree ≥ 4, then this implies V ≤ E/2, which is a contradiction, with part (a), hence, there must exist at least one of degree ≤ 3, let this be called v. If we delete v, the result is in fact another planar graph with no triangles, but also, because v is of degree 3, its vertices are one of 3 colors, allowing us to color v the remaining color. 2) We take the graph and label the vertices in any valid way in accordance with the rules of a depth first search tree. It is easy to check that we have a depth first seatrch tree in blue below. The remaining edges have been labelled in red. v3 v5 v9 v4 v0 v2 v6 v8 v1 v7 We now orient the graph so that if i > j and the {vi , vj } ∈ T , then we make that edge (vi , vj ), and (vj , vi ) is added otherwise. It is easy to see this is a depth-first search tree, 3) Following the comments below theorem 3.4, our upper bound is N (4, 4; 2) ≤ N (3, 4; 2) + N (4, 3; 2) = 18 we seek a coloring of a graph with 17 vertices with the right properties. This suggests G = Z17 . If |i−j| ∼ = 1, 2, 4, 8 mod 17, let i, j be blue, otherwise color the edge red. Addition permutes the vertices, so it is sufficient to argue that vertex 0 is not in a monochromatic copy of K4 , then no vertex is. Let us consider the subgraph with just blue. Vertex 0 connected to a blue edge to {1, 2, 4, 8, 9, 13, 15, 16}. The subgraph connected to 0 by a blue edge consists of 8 vertices that are connected to three other vertices, each a possible candidate for another vertex in a copy of K4 , by symmetry arguments, we need only consider {1, 2, 4, 8}, each subgraph containing 2 vertices of degree 4, (0 and the chosen vertex) and 3 others of degree 2, making K4 impossible. The same argument holds on red. 1 2 Just for kicks, the graph is displayed below: 5 4 3 6 2 7 1 8 0 9 16 10 15 11 12 14 13 The upper bound for R(3, 5) is R(3, 5) ≤ R(2, 5; 2) + R(3, 4) = 5 + 9 = 14 we just ned to show there exists a graph on 13 vertices has no blue K3 (triangle) or red K5 . We color edge i, j, blue if |i − j| ∼ = 1, 5 mod 13 and red otherwise. Vertex 0 is connected to 4 vertices via a blue edge, the subgraph is a star shape giving no blue triangle. For a red copy of K5 , we take the subgraph connected to 0 via a red edge namely we take 0 and vertices in {2, 3, 4, 6, 7, 9, 10, 11}, each a possible candidate for a vertexin K5 . Once again, we need only consider {2, 3, 4, 6} by symmetry, but adding any of these results in a subgraph with two vertices of degree 5 and 4 of degree 4, none of which supports a copy of K5 . 4 3 2 5 1 6 0 7 12 8 11 9 10 4) (a) Without loss of generality, we can find a subgraph of size 2k−1 , which we wish to show contains a transitive tournament of order k. We shall prove by induction, with the case k = 1 being trivial. Suppose T is a tournament of order at least 2k and let v be any vertex, either the in-degreee or out-degree of v is greater than 2k−1 . If the out-degree is greater than 2k−1 , by the inductive hypothesis, it contains a transitive tournament of size k − 1, T 0 which v points to every element of T 0 , hence, adding v to T 0 results in a transitive tournament of size k, where v is the vertex whose index is minimal. Similarly, if the in-degree is greater than 2k−1 , then adding v to T 0 as the maximal index vertex results in a transitive tournament of size k. 3 (b) To prove the other part, we use the kind of probabilistic argument we encountered in Theorem 3.5. We wish (k−1) to show there ixists a tournament of order 2 2 that does not contain a transitive subtournament of order k. The probability that any subset of size k forms a transitive tournament is k k!2−(2) , with nk possible choices of k verteces. We simplify k k k−1 n k!2−(2) = n(n − 1) . . . (n − k + 1)2−k(k−1)/2 < n2− 2 , k k−1 If n < 2 2 , then this probability is strictly less than 1, which implies there exists a tournament of order 2(k−1)/2 containing no transitive subtrounament. I.e., k > 1 + 2 log2 n. 5) We wish to count the set of paths of length two, P = {u; v, w : (u, v), (u, w) ∈ E}, which is useful because in every path of length two in a graph without four cycles will have a unique midpoint, u. If du denotes the degree of vertex u, the X |P | = du (du − 1) u∈V where as, picking endpoints defines at most one path, hence X X |P | = du (du − 1) = d2u − du ≤ n(n − 1). u∈V We can now relate the degrees to the number of edges by 2 P du = 2|E|, and using the Cauchy-Swartz inquality 4|E| − 2|E| ≤ n(n − 1)4|E|2 − 2|E|n − n2 (n − 1) ≤ 0 n which is quadratic in |E|, solving gives p √ 2n ± 4n2 + 16n2 (n − 1) n n 4n − 3 = ± |E| ≤ 8 4 4 in which the positive solution is what we are after.