Question

Let Q:= {1,2,...,q}. Let G be a graph with the elements of Q^n as vertices and an edge between (a1,a2,...,an) and (b1,b2,...bn) if and only if ai is not equal to bi for exactly one value of i. Show that G is Hamiltonian.

Answer 1

(i) 'Only If': Suppose S, S 2E(G[S]), because the edges in G[S] contributes to the summation twice and the ones in [S : S] only once. Thus it follows easily that |[S : S]| is even. Proof is a cut of G. Then we claim that s deg(v) - ls sl [{v} V(G)\ {v}] is odd. If': If G has an vertex, say v, of odd degree, then the cut (ii) 'Only If': Any cycle alternates between the two vertex classes, so it has even length. If': We may assume that the graph G is connected, since a graph is bipartite if its components are bipartite. Let the distance d(u, v) be the length of shortest path between u and v. Fix u. Define Ui vd(u, v) = i}. Note that an edge of G can join vertices in Ui and Ui only if j = i or j = i+1 or j = i - 1. We claim that there is no edge between vertices in Ui. x-u and y-u have the same length of i. Let z be the last common vertex If ry Ui then the path Then -, -y and ry form a cycle of length 2(i - d(u, v))+1, which contradicts the absence of odd cycles Thus V(G) can be divided into two parts, UoUU2 U U4U... and U1 UU3UU5U... And it is a bipartition of G