通过选择根节点减少广度优先树的深度

Question

If the root node can be picked randomly in a graph, is there an existing algorithm that picks the root node such that the resulted breadth first tree has the smallest depth or height?

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I have a hunch that I should pick the node with the largest fan out as the root node.

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Let me give one example.

There is a cyclic directed graph {(0,1),(1,2),(1,5),(1,6),(2,3),(3,4),(4,2),(5,2),(6,0)}

If node 0 is chosen as root, breadth first tree is {(0,1),(1,2),(1,5),(1,6),(2,3),(3,4)} The depth is 5

If node 6 is chosen as root, breadth first tree is {(6,0),(0,1),(1,2),(1,5),(2,3),(3,4)} The depth is 6

Answer 1

I am assuming you are talking about a weighted graph otherwise it does not make much difference doing BFS from different nodes as root.

One naive brute force approach is to consider each node of the graph as root and construct the BFS tree. Measure the height each time and after covering all nodes as roots we get the node from which the BFS Tree produces the minimum height. Done just this you might end up taking exponential time. Time: O(n * (v + e) + logxn) for each node n we do BFS + for each tree we calculate height x levels in tree. But I suspect dynamic programming we can bring this time to a much more manageable level. Since we store results at every stage we can reuse it for later computations.

Another method that comes to mind is optimal binary search tree. What you do is process your graph & collate the weights of each nodes into an array. Using this weights we would construct a BST such that the nodes with maximum weight would fall near the root of the BST and nodes with less weight would fall lower in the BST (some probably as leafs). This way upon searching in BST your likelihood of finding a node is better.

update: to expand on the above approach -

The above recursion is simple, we one by one try each node as root r . r varies from i to j . We recursively calculate optimal cost from i to r-1 and r+1 to j with r as root. We add sum of weights (or frequencies) from i to j (see first term in the above formula), this is added because every search will go through root and one comparison will be done for every search.

Hope this helps...