By S. Fiorini, Robin J. Wilson
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Additional resources for Edge-colourings of Graphs
8. EULERIAN GRAPHS theorern is true for n = 1 clearly. Suppose n 2:: 2 and the theorern holds for any two vertices of B (d, n - 1). Assurne that x and y are two distinct vertices of B(d, n). First, we consider the case that (x, y) is not an edge of B(d, n). Since B(d,n) = L(B(d,n- 1)), x and y correspond to two edges in B(d,n- 1). Let such two edges be x = (u,u') and y = (v,v'). Then u' -1- v since (x,y) is not an edge of B(d, n). By induction hypothesis, there are d- 1 internally disjoint (u',v)-paths of length at rnost n in B(d,n -1), frorn which we can induce d - l internally disjoint (x, y)-paths of length at rnost n+ 1 in B(d, n).
Example 1. 1 Let G be an undirected graph with J 2:: 2. Then G contains a cycle certainly. Moreover, if G is simple, then G contains a cycle of length at least J + 1. Proof If G contains loops or parallel edges, then the conclusion holds clearly. Suppose that G is simple below and let P = (x 0 , x 1 , · · ·, xk) be a Iongest path in G. Then all neighbors of x 0 must lie on P, that is Note that \Na(xo)\ = dc(xo) 2:: J(G) = J 2:: 2. Thus, there exists Xi E Na(xo), J ~ i ~ k. It follows that (xo,xl, · · · ,Xi-l, 1 Xi, xo) is a cycle of length at least J + 1 in G.
5 Prove that the cartesian product of two eulerian graphs is an eulerian graph, hence 2n-cube Q2n is an eulerian graph. 6 (a) Prove that if Gis a connected digraph, ld(t(x)- da(x)l ~ 1 for any x E V and any edge of G is not contained in odd number of directed cycles, then G is eulerian. (b) Give an example to show that the converse of (a) is not true. 8. 7 Prove that a connected undirected G is eulerian if and only if every edge of G lies on an odd number of cycles. 8 Let G be an eulerian graph and x be a vertex of G.