# How does Camerini's algorithm for minimum-bottleneck-spanning-tree run in linear time?

I'm having a difficult time understanding Camerini's algorithm because there are very few clear explanations online. The goal is to find a minimum-bottleneck spanning tree in linear time.

Camerini's algorithm does this by splitting the edges by weight into heavy and light halves in O(|E|) time, then building a maximal forest from the light edges. If there is only one tree, then it recurses on the lighter half of edges, but if there are multiple trees, it reduces each tree to a super-vertex, then recurses with the new set of vertices and the heavy edges.

I don't understand how this can run in linear time. If the forest-finding runs in O(|V|+|E|) time, and |V| is not guaranteed to decrease in any predictable fashion, then the algorithm should be bounded by O(|V| log |V| + |E|).

• $|V|$ is not guaranteed to decrease in any predictable fashion: the number of edges decreases in the $i$-th recursion to $|E|/(2^i)$. – hengxin Apr 6 '15 at 5:54
• If you think I am overstepping the use of this clause, I can submit an $O((|V|+|E|)Ack^{-1}(|E|))$ draft I created a few days ago without knowing of this algorithm. – Mark Miller Apr 6 '15 at 20:45
Next, use the fact that (with isolated vertices removed, in each stage after the first) $|V|\le 2|E|$ to simplify the time bound in each stage (after the first) from $O(|V|+|E|)$ to $O(|E|)$.
Finally, observe that at each stage $|E|$ goes down by a factor of two, so all these $O(|E|)$ bounds add in a geometric series to be linear in the size of the original graph.