periphery of a graph

**xixi** · October 26th 2012, 11:24 AM

Let (eccentricity of v) $e(v)=Max_{u \in V(G)} d(u,v)$ and $diam(G)= Max_{v \in V(G)} e(v)$ , where V(G) is the set of vertices of G and d(u,v) is the distance between u and v. Let (periphery of G) Pe(G) be the graph induced by the vertices of G that have eccentricity equivalent to diam(G). Now suppose that the graph H has n vertices. Prove that H is Pe(G) if and only if $\Delta(H) \le n-2$ . ( $\Delta(H)$ is the maximum degree in H)

**johnsomeone** · October 26th 2012, 11:51 AM

You need an additional constraint/premise of some kind. It's not true as stated.

$\text{If } K_n \text{ is the complete graph on } n \text{ verticies } (n \ge 2), \text{ then}$

$diam(K_n) = 1, \text{ and } e(v) = 1 = diam(K_n) \ \forall \ v \in V(K_n),$

$\text{and so } Pe(K_n) = K_n.$

$\text{But } \Delta(K_n) = n-1 = |V(K_n)| - 1,$

$\text{which contradicts this problem's claim that}$

$H = Pe(K_n) (= K_n) \Rightarrow \Delta(H) \le |V(H)| - 2.$

**xixi** · October 26th 2012, 07:17 PM

You're right. Assume that no vertex in H has eccentricity 1.

**johnsomeone** · October 26th 2012, 11:39 PM

$\text{Let } G \text{ be a connected graph such that } diam(G) > 1 \text{ (i.e connected, not complete, having 3+ verticies.)}$

$\text{ASSUME } \exists v \in H = Pe(G) \ni deg_H(v) = |V(H)|-1.$

$\text{Since } v \in Pe(G), \exists w \in G \ni e(v) = dist(v, w) = diam(G).$

$\text{Since } diam(G) = dist(v, w) \le e(w) \le diam(G),$

$\text{have that } e(w) = diam(G), \text{ and so } w \in H.$

$\text{But then } v, w \in H, \text{ and so since } deg_H(v) = |V(H)|-1,$

$\text{have that } (v,w) \in E(H).$

$\text{Since } H \text{ is an induced subgraph of } G, \text{ have that } (v,w) \in E(G).$

$\text{But } (v,w) \in E(G) \Rightarrow dist(v, w) = 1 \Rightarrow diam(G) = dist(v, w) = 1, \text{ contrary to the premises.}$

$\text{Thus the initial assumption led to a contradiction.}$

$\text{Thus } x \in H \Rightarrow deg_H(x) \le |V(H)|-2.$

$\text{Therefore } \Delta(Pe(G)) \le |V(Pe(G))|-2.$

**xixi** · October 27th 2012, 04:22 AM

what about the converse?

**johnsomeone** · October 27th 2012, 11:13 AM

$\text{Let } H \text{ be a graph such that } \Delta(H) < |V(H)|-1 \text{ and } |H| > 1.$

$\text{Then for all } v \in H \text{ have that } deg_H(v) < |V(H)|-1,$

$\text{and so if } v \in H \text{ there exists } w \in H \text{ such that } (v, w) \notin E(H).$

$\text{Define } G \text{ by } V(G) = V(H) \cup \{ x \}, E(G) = E(H) \cup \{ (x,v) | v \in V(H) \}.$

$\text{Then for all } v \in H \subset G, dist_G(v, x) = 1 \text{ since } (v, x) \in E(G).$

$\text{Thus } e_G(x) = 1.$

$\text{From before, if } v \in H \subset G, \text{ there exists } w \in H \text{ such that } (v,w) \notin E(H),$

$\text{and so by the definition of } E(G), \text{ also have that } (v,w) \notin E(G).$

$\text{So for all } v \in H \subset G, \text{ this proves that } e_G(v) > 1, \text{ since } dist_G(v, w) > 1.$

$\text{Now, for all } v, w \in H \subset G, \{(v, x), (x, w) \} \subset E(G),$

$\text{and so, for all } v, w \in H \subset G, \ dist_G(v, w) \le 2.$

$\text{But from that it follows that, for all } v \in H \subset G, e_G(v) \le 2.$

$\text{(That conclusion also requires also remembering that } dist(v, x) = 1.)$

$\text{So have shown that, for all } v \in H \subset G, 1 < e_G(v) \le 2.$

$\text{Therefore, if } v \in G, v \ne x, \text{ then } e_G(v) = 2. \text{ Also, } e_G(x) = 1.$

$\text{That proves that } diam(G) = 2, \text{ and so also that } Pe(G) = H,$

$\text{since } Pe(G) = \ < \{ v \in V(G) | e_G(v) = diam(G) = 2 \} > \ = \ < H > \ = H.$

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