I am trying to find a fast algorithm with modest space requirements to solve the following problem.
For each vertex of a DAG find the sum of its in-degree and out-degree in the DAG's transitive closure.
Given this DAG:
I expect the following result:
Vertex # Reacability Count Reachable Vertices in closure
7 5 (11, 8, 2, 9, 10)
5 4 (11, 2, 9, 10)
3 3 (8, 9, 10)
11 5 (7, 5, 2, 9, 10)
8 3 (7, 3, 9)
2 3 (7, 5, 11)
9 5 (7, 5, 11, 8, 3)
10 4 (7, 5, 11, 3)
It seems to me that this should be possible without actually constructing the transitive closure. I haven't been able to find anything on the net that exactly describes this problem. I've got some ideas about how to do this, but I wanted to see what the SO crowd could come up with.
This can be done by first finding Strongly Connected Components (SCC), which can be done in O(|V|+|E|) . Then, build a new graph, G' , for the SCCs (each SCC is a node in the graph), where each node has value which is the sum of the nodes in that SCC.
The reachability relation of a DAG can be formalized as a partial order ≤ on the vertices of the DAG. In this partial order, two vertices u and v are ordered as u ≤ v exactly when there exists a directed path from u to v in the DAG; that is, when u can reach v (or v is reachable from u).
Reachable Nodes In Subdivided Graph. You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1 . You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.
For an exact answer, I think it's going to be hard to beat KennyTM's algorithm. If you're willing to settle for an approximation, then the tank counting method ( http://www.guardian.co.uk/world/2006/jul/20/secondworldwar.tvandradio ) may help.
Assign each vertex a random number in the range [0, 1). Use a linear-time dynamic program like polygenelubricants's to compute for each vertex v the minimum number minreach(v) reachable from v. Then estimate the number of vertices reachable from v as 1/minreach(v) - 1. For better accuracy, repeat several times and take a median of means at each vertex.
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