Prove that if every edge of a graph G lies on an odd number of cycles, then G is Eulerian.
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- Prove that connecting two nodes u and v in a graph G by a new edge creates a new cycle if and only if u and v are in the same connected component of G.Let G be a graph. Prove that if G has a closed walk of odd length, that the vertices and edges of the walk contain a cycle.Let G be a graph of order n ≥ 3. Prove that G is 2-connected if and only if for any vertex v and edge e of G, v and e lie on a common cycle of G.
- Give an upper bound on the number e of edges of G in terms of n and g if G is a connected plane graph with n vertices and girth g.Let G be a graph. Prove that G is Eulerian if and only if G has an orientation D where D is an Eulerian digraph.Prove that if u is a vertex of odd degree in a graph, then there exists a path from u to another vertex v of the graph where v also has odd degree.
- The symmetric difference graph of two graphs G1 = (V, E1) and G2 = (V, E2) on the samevertex set is defined as G1△G2 := (V, E1△E2). Remember that E1△E2 := (E1 \ E2) ∪(E2 \ E1). If G1 and G2 are eulerian, show that every vertex in G1△G2 has even degreeLet G be a simple graph with 11 vertices, each of degree 5 or 6. Prove that G has at least 7 vertices of degree 8 or at least 6 vertices of degree 7. Do not use the planar equation e <= 3v - 6.Prove that if v0 and v1 are distinct vertices of a graph G = (V,E) and a path exists in G from v0 to v1 , then there is a simple path in G from v0 to v1 .