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Fast algorithm to compute the distance matrix for a weighted tree (Stepic/Rosalind/Coursera Problem)

I'm stuck with this Rosalind problem. I'm trying to find time and memory efficient algorithm to compute the distance matrix from a weighted tree.

Input: a weighted tree with n leaves

Output: An n x n matrix (di,j) where di,j is the length of the path between leaves I and j

Sample Input

4

0->4:11
1->4:2
2->5:6
3->5:7
4->0:11
4->1:2
4->5:4
5->4:4
5->3:7
5->2:6

Sample Output

0   13  21  22
13  0   12  13
21  12  0   13
22  13  13  0

Any clues?

rosalind python distance-matrix evolutionary-tree
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updated by Ram Ram • written by m.malkowska • m.malkowska

1 answer

This is the "all paris shortest path" question. The Floyd-Warshall algorithm (https://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm) is a good place to start.

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updated by Ram Ram • written by Ryan Layer • Ryan Layer

Coming back to this years late unfortunately, but I think the Floyd-Warshall algorithm would be a poor choice for this, right? For a graph with V vertices, the algorithm is O(V^3), whereas this problem can be solved in O(V^2) for a binary tree. For example, for a binary tree with n leaves:

M = empty *n* by *n* matrix
for u in tree.postorder_traversal():
    if u is a leaf:
        u.leaf_dists = [(u,0)]
    else:
        u.leaf_dists = empty list
        for c1,d1 in u.left_child.leaf_dists:
            add (c1, d1+c1.edge_length) to u.leaf_dists
            for c2,d2 in u.right_child.leaf_dists:
                add (c2, d2+c2.edge_length) to u.leaf_dists
                M(c1,c2) = M(c2,c1) = d1+c1.edge_length+d2+c2.edge_length

The two extremes are a perfectly balanced binary tree and a perfectly unbalanced (i.e., ladder-like) tree. In the case of a perfectly balanced binary tree, each level of the tree will perform a O(n) operation, and there are O(log n) total levels, making the overall time complexity O(n log n). For a perfectly unbalanced (i.e., ladder-like) tree, the internal nodes will perform n + n-1 + n-2 + ... + 2 operations and each leaf will perform 1 operation (so n total for all the leaves), so the overall time complexity will be n(n+1)/2 - 1 + n = O(n^2)

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written by niemamoshiri • niemamoshiri

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