First I try to use induction.
I think this it like repeat deleting two adjancent edges in graph.
the two delete edge contain three point,supposed they are a,b,c,there are three cases:
- If (a,b) (a,c) exist but (b,c) not exist.If delete (a,b)(a,c),the remained graph is still a graph with even edges
- If (a,b) (b,c) exist but (a,c) not exist. If delete (a,b)(b,c),remained a graph withm even edges
If there is a cycle , such as (a,b)(b,c)(a,c). this has two cases
case 2.1 after deleting, G is still connected , this is ok
case 2.2 after deleting, G become two conponent G1 and G2,for example,if delete (a,c) and (b,c) ,and c has no another edge to the conponent which contains vertice a and vertice b. this again be two case.
case 2.2.1 Both G1 and G2 has even edges, this is ok case 2.2.2 Both G1 and G2 has odd edges, this is bad!!!!
I don't know how to prove in case 2.2 so I come to another way.
First find a edge e1 and then we should find another e2 to match it.After deleting e1 e2 ,there are two case
case 1 G is connected, ok
case 2 w(G)=2, suppose G-e1 = G1 + e2 + G2 , then in G1 must has even edges and G2 has odd edges. the most important one is to find a e2 in G2 .
I don't know how to do when G2-e2 become two unconnect odd conponent!!
can anyone help me ? Thanks~