最小直径生成树的算法

Question

给定无向图和连通图G，找到直径最小的生成树。

Answer 1

singhsumit linked the relevant paper by Hassin and Tamir , entitled "On the minimum diameter spanning tree problem", but his answer is currently deleted. The main idea from the paper is that finding a minimum diameter spanning tree in an undirected graph can be accomplished by finding the "absolute 1-center" of the graph and returning a shortest path tree rooted there.

The absolute 1-center is the point, either on a vertex or an edge, from which the distance to the furthest vertex is minimum. This can be found via an algorithm of Kariv and Hakimi (An Algorithmic Approach to Network Location Problems. I: the p-Centers) or an earlier algorithm of Hakimi, Schmeichel, and Pierce (On p-Centers in Networks), which I will attempt to reconstruct from just the running time and decades of hindsight. (Stupid pay walls.)

Use Floyd--Warshall or Johnson's algorithm to compute all-pairs distances. For each edge u--v, find the best candidate for a 1-center on that edge as follows.

Let the points on the edge u--v be indexed by µ ranging from 0 (u itself) to len(u--v) (v itself). The distance from the point at index µ to a vertex w is

min(µ + d(u, w), len(u--v) - µ + d(v, w)).

As a function of µ, this quantity is increasing and then decreasing, with the maximum at

µ = (len(u--v) + d(v, w) - d(u, w))/2.

Sort the vertices by this argmax. For each partition of the array into a left subarray and a right subarray, compute the interval [a, b] of µ that induce the same argmax partition. Intersect this interval to [0, len(u--v)]; if the intersection is empty, then move on. Otherwise, find the maximum distance L in the left subarray from the point on u--v indexed by a, and the maximum distance R in the right subarray from the point on u--v indexed by b. (The cost of computing these maximums can be amortized to O(1) for each partition, by scanning left-to-right and right-to-left at the beginning.) The best choice is the µ in [a, b] that minimizes max(L - (µ - a), R + (b - µ)).