Nonseparable Graphs
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چکیده
Proof. We first prove the sufficiency. Since any two vertices of G are connected by two internally disjoint paths, then for each vertex v of G, any two vertices of G − v are connected by at least one path, i.e., c(G − v) = c(G). Hence G has no cut vertex. Next we prove the necessity. Let u, v be two distinct vertices of G. To show that there are internally disjoint paths between u and v, we apply induction on the distance d(u, v) between u and v. When d(u, v) = 1, i.e., u, v are end-vertices of an edge e in G. Since both u and v are not cut vertices, the edge e is not a cut edge. So e is contained in a cycle C. Thus uev and C\e are two internally disjoint paths between u and v. Now assume that any two vertices having distance less than d are connected by two internally disjoint paths, where d ≥ 2. Let d(u, v) = d. Let P := v0e1v1 · · · vd−1edvd be a path from u = v0 to v = vd. Since d(v0, vd−1) = d−1, there are two internally disjoint paths P1 and P2 from v0 to vd−1 in G. Since G has no cut vertex, the subgraph G− vd−1 is connected. Then there is a uv-path P3 in G− vd−1. Let w be the last vertex of P3 that meets P1 ∪ P2. Without loss of generality, we may assume that w lies in P1. Write P1 = P ′ 1Q ′ 1, P3 = P ′ 3Q ′ 3, where P ′ 1 is the sub-path of P from v0 to w, and Q3 is the sub-path of P3 from w to vd; see Figure below.
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تاریخ انتشار 2010