Let G=(V,E) be a connected graph and v be a node of G. The eccentricity e(v) of

Question

Let G=(V,E) be a connected graph and v be a node of G. The eccentricity e(v) of v
is the distance to a node farthest from v. Thus:
e(v) = max {d(u,v): u ?V}
where d(u,v) is the shortest distance from v to u.
The radius R(G) is the minimum eccentricity of the nodes, whereas the diameter
D(G) is the maximum eccentricity. Node v is a central node if e(v) = R(G), it is a
peripheral node if e(v) = D(G).
In the following, we assume that G is a weighted graph, n = |V|, and e = |E|.
1- In this part, we assume that the weights are any positive integer:
a) Give an algorithm that finds the set of central nodes and the set of peripheral nodes
in G.
b) Give the time complexity of your algorithm.
2- In this part, we assume that the weights are equal:
a) Give an O(e) single source shortest path algorithm.
b) Using the algorithm in part 1-a, write an algorithm to find the set of central nodes
and the set of peripheral nodes in G. Analyze the time complexity of your algorithm.

Explanation / Answer

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