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Suppose we have an undirected, connected graph $G$ where vertices have positive integer weights. Let $\bar{v}$ be a given vertex in $G$. Take a spanning tree $T$ of $G$ rooted at $\bar{v}$ and define its cost as the maximum total weight among all proper subtrees. (Since weights are positive the proper subtree(s) with maximum total weight will be rooted at children of $\bar{v}$.) Is there an efficient algorithm to determine the minimum cost over all spanning trees $T$?

Some motivation: Imagine vertices are cities, weights are population, edges are roads. Everyone wants to travel to $\bar{v}$. A spanning tree is a way of telling people what path to take. The cost of a spanning tree is the max possible traffic on any one road into $\bar{v}$.

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    $\begingroup$ Are you looking for a heuristic or do you want an algorithm to give the exact optimal solution? A plausible heuristic is to delete all edges out of $\bar{v}$, start $k$ copies of Prim's algorithm running in parallel, one per neighbor of $\bar{v}$, and schedule them so that at each step you are running whichever copy has the lowest total cost so far; and never selecting any edge that will connect these $k$ components. I doubt this gives the optimal solution in all cases but it might be a reasonable heuristic? $\endgroup$
    – D.W.
    Oct 21 at 22:14

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This problem is NP-complete. There exists a reduction from the bin packing problem to this problem.

For a set of items $I$ and the number of bins $k$, create a complete bipartite graph of size $k + |I|$. Each left vertex corresponds to a bin and has weight $0$ (or an appropriately small number if strictly-positive weight is required). Each right vertex corresponds to an item and has the item's weight. Finally, a root vertex is added and connected to all $k$ left vertices.

A proper subtree of a spanning tree of the constructed graph corresponds to a set of items packed in a bin. The subtree weight corresponds to the total item weight. A bin packing solution with bin capacity $C$ corresponds to a spanning tree with maximum subtree weight $C$.

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  • $\begingroup$ Thank you. To fill out your argument, for any bin packing with capacity $C$, there's a spanning tree that directly reprsents it and has max subtree weight $C$; for any spanning tree with max subtree weight at most $C$, we can convert it to one where every edge from $\bar{v}$ to the left vertices is taken and that has equal or lesser max subtree weight, i.e. a bin packing with capacity $C$. $\endgroup$
    – Andrew
    Oct 23 at 14:42
  • $\begingroup$ So the problem is in $NP$? $\endgroup$
    – Andrew
    Oct 23 at 14:43
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    $\begingroup$ "To fill out your argument ..." correct. "we can convert it to one where every edge from v¯ to the left vertices is taken" Or if the number of edges from the root to left vertices is $k' < k$ then it is a bin packing with $k'$ bins. "So the problem is in NP?" The problem is in NP because a spanning tree is a certificate of a solution. $\endgroup$
    – pcpthm
    Oct 24 at 4:00
  • $\begingroup$ oh yes of course, whoops $\endgroup$
    – Andrew
    Oct 24 at 12:54

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