1. A saturated hydrocarbon is a molecule Cm Hn in which every carbon atom C has four bonds, every hydrogen atom H has one bond, and no sequence of bonds forms a cycle. Show that for every m E N, the molecule Cm Hn can exist only if n = 2m + 2. Solution: View the molecule as a graph with vertices corresponding to atoms and edges corresponding to bonds. Then CmHn = (V, E) is connected, has exactly m + n vertices and is a tree as it contains no cycles. The graph thus has m + n - 1 edges. Cm Hn has m degree 4 vertices and n degree 1 vertices, so ? d(v) = 4m + n. VEV The Handshaking Lemma give that 4m + n = VEV ? d(v) = 2|E| = 2(m + n - 1). Rearranging this gives that n = 2m + 2. 2. Find a minimum weight spanning tree for each of the following graphs. x 2 x a 4 a a 7 d 1 4 6 e 8 6 7 1 2 5 b f 2 5 1 6 4 1 x 6 b 2 8 5 1 6 4 e 5 8 f c 2 b d 8 c c Solution: 2 x x a 4 a 1 1 e 2 f 2 1 b 6 c 2 b d c d a 4 1 6 1 b x 2 4 e f c 3. (a) By adapting the definition of isomorphism for graphs, formulate a definition of what it means for digraphs D = (V, A) and D' = (V', A') to be isomorphic. Solution: There exists a 1-1 correspondence between V and V', and a 1-1 correspondence between A and A' such that if the arc a E A has ends (u, v) and the arc a' E A' corresponding to a has ends (u', v') then u corresponds to u' and v corresponds to v'.
(b) Which pair of the following digraphs are isomorphic? Solution: The first and the last. (c) Explain why no two of the following digraphs are isomorphic. Solution: It is easily seen that isomorphism preserves the indegree and outdegree of each vertex. (More precisely, if q : V -> V' is an isomorphism from D = (V, A) to D' = (V', A '), then for each v E V we have d-(v) = d-(q(v)) and d+ (v) = d+ (q(v)).) It follows that if D and D' are isomorphic then { (d- (v), d+ (v) | v E V} = {(d- (v), d + (v) | v E V'}. (Note that the converse is false.) If D1, ... , D4 are the given digraphs read from left to right, then we have {(d-(v), d'(v)) | v E V(D1)} = {(3,0), (2, 1), (1, 2), (0, 3)} {(d-(v), d'(v)) | v E V(D2)} = {(2, 1), (2, 1), (2, 1), (0, 3)} {(d-(v), d+(v)) | vE V(D3)} ={(3,0), (2, 1), (1, 2),