Mathematics and Statistics, Part A: Graph Theory Problem Sheet 1, lectures 1-4
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1 1. Draw Mathematics and Statistics, Part A: Graph Theory Problem Sheet 1, lectures 1-4 (i) a simple graph. A simple graph has a non-empty vertex set and no duplicated edges. For example sketch G with V (G) ={1, 2, 3, 4}, E(G) ={{1, 2}, {1, 3}, {2, 3}, {2, 4}}. (ii) a non-simple graph with no loops. Loops are joined vertex pairs {u, u}. For example sketch G with V (G) ={1, 2, 3, 4}, E(G) ={{1, 2}, {1, 3}, {2, 3}, {2, 3}, {2, 4}}. (iii) a non-simple graph with no multiple edges. For example sketch G with V (G) ={1, 2, 3, 4}, E(G) ={{1, 2}, {1, 3}, {2, 3}, {3, 3}, {2, 4}}. 2. (i) By suitably labelling the vertices, show that the first two graphs below are isomorphic. Labelled isomorhpic graphs. (ii) Explain why the third and fourth graphs below are not isomorphic. In the first graph, no vertices of degree 2 are adjacent; in the second graph they are adjacent in pairs. Since isomorphism preserves adjacency of vertices, the graphs are not isomorphic.
2 3*. If G is a simple graph with at least two vertices, prove that G must contain two or more vertices of the same degree. Let G be a graph with n 2 vertices and no two vertices of the same degree. Each vertex has degree n 1. Therefore, there is exactly one vertex of degree k for each k n 1, because there is a total of n vertices. In particular there are vertices u and v of degree 0 and n 1 respectively. It must be that v is adjacent to u, a contradiction. 4. Let G be a connected graph with vertex set {v 1,v 2,...,v n }, m edges, and t triangles. (i) Given that A is is the adjacency matrix of G, prove that the number of walks of length 2 from v i to v j is the i, jth entry of the matrix A 2. In a directed graph A ij = 1 if there is a directed edge [v i,v j ]. The result follows by considering paths v i v k v j and counting with A (2) ij = k A ik A kj (ii) Deduce that 2m = the sum of the diagonal entries of A 2. The sum of the diagonal entries of A 2 is the sum over i and j of the number of paths v i v j v i which is equal to 2m. (iii) Obtain a result for the number of walks of length 3 from v i to v j, and deduce that 6t = the sum of the diagonal entries of A 3. The number of walks of length 3 from v i to v j is the ijth entry of A 3. The sum of the diagonal entries of A 3 is the sum over i, j, k of the number of triangle paths v i v j v k v i. Each triangle is counted 3! = 6 times, the number of ways of permuting i, j, k in the sum. AgraphG is Eulerian with a closed trail containing each edge iff the degree of each vertex is even. 5. (i) For which values of n is K n Eulerian? K n is a complete graph, so each vertex degree is n 1. K n is Eulerian for odd n =1, 3, 5,... (ii) Which complete bipartite graphs are Eulerian? A bipartite graph K r,s has r vertices of degree s and s vertices of degree r so is Eulerian for r and s even.
3 (iv) For which values of n is the wheel W n Eulerian? W n is non-eulerian for all n>1. (v) For which values of k is the k-cube Q k Eulerian? Every vertex in Q k has degree k, soq k is Eulerian for even k. 6. Use Fleury s algorithm to produce an Eulerian trail for the graph below. One solution: Traverse edges in the order indicated. 7. (i) For which values of n is K n Hamiltonian? K n is Hamiltonian for n 3. If V (K n )={1,...,n}, thena Hamiltonian cycle is 1, 2, 3,...,n,1. (ii) Which complete bipartite graphs are Hamiltonian? A spanning cycle in a bipartite graph visits the two partite sets alternately, so there can be no cycle unless the partite sets have the same size. Hence K m,n is Hamiltonian only if m = n. (iii) For which values of n is the wheel W n Hamiltonian? W n is Hamiltonian for all n. If V (W n )={0, 1,...,n 1}, with 0 the center vertex, and consecutive labelling around the circle, then a Hamiltonian cycle is 0, 1, 2,...,n 1, 0. (v) For which values of k is the k cube Q k Hamiltonian? Q k is Hamiltonian for all values of k. 8. (i) Find the two labelled trees corresponding to the Prüfer codes (1, 2, 3, 4) and (3, 3, 3, 3).
4 (ii) Find the Prüfer codes corresponding to the two labelled trees below. (4, 4, 4, 1) and (4, 2, 2, 4). 9. (i) Find the number of trees on n vertices in which a given vertex is an end-vertex. There are n 1 vertices to join the given end-vertex to, and (n 1) n 3 trees without the given vertex and joining edge, so (n 1) n 2 such trees. (ii) Deduce that, if n is large, then the probability that a given vertex of a tree with n vertices is an end-vertex is approximately e 1. The probability is (n 1) n 2 e 1 n n 2 10*. Let T (n) be the number of labelled trees on n vertices. (i) By counting the number of ways of joining a labelled tree on k
5 vertices and one on n k vertices, prove that n 1 ( ) n 2(n 1)T (n) = k(n k)t (k)t (n k) k k=1 Let T (n) be a tree with n vertices. Choose a vertex u in T (n) and remove an incident edge uv. There remains two trees, considered as rooted at vertices u and v. T (n) is obtained by joining roots of rooted trees in this way. There are ( n k) ways of choosing the two label sets, kt(k) and(n k)t (n k) labelledrootedtreeswith respectively k and n k vertices, and each n-tree is overcounted in this construction by the total vertex degree of the tree, which is 2(n 1), since n-trees have n 1edges. (ii) Deduce the identity n 1 ( ) n k k 1 (n k) n k 1 =2(n 1)n n 2 k k=1 This follows because we know that T (n) =n n 2 by other proofs Caley s formula. 11. Show, by drawing, that the following graphs are planar: (i) the wheel W 5 ; (ii) the graph of the octahedron.
6 12. Find three different Gray codes of 4-digit binary words ( 0000) ( 0000) Toggle the first co-ordinate in all the entries to get another code. 13. On a chessboard, a knight always moves two squares in a horizontal or vertical direction and one square in a perpendicular direction. Knight s tour problem: Can a Knight visit each square of a chessboard just once by a sequence of knight s moves, and finish on the same square as it began? (i) Show that there is no knight s tour on a 4 4 ora5 5 chessboard. On a 4 4 chessboard a tour which includes the top left and bottom right corners must contain the cycle shown below. Since it is already a cycle it is impossible to include them as well in a full tour. For the 5 5 chessboard, we use a lemma that a bipartite graph with an odd number of vertices is not Hamiltonian. The graph associated with any chessboard is bipartite, since a knight moves to a square of a different colour from where it started. Thus there
7 RCG 2007 cannot be a knight s tour on chessboards with an odd number of squares. Proof of Lemma: The vertices of any bipartite graph can be split into two sets A and B in such a way that each edge has one end in A and one in B. Any Hamiltonian cycle must alternate between these two sets, ending in the same set as it started. This is impossible if the total number of vertices is odd. (ii) Show that there is a knight s tour on a 8 8 chessboard. Atouris Note that this is a magic square.
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