Dima - I've got a 2D grid-structure, of some size "n x n". Each point in the grid is connected to it's four neighboring and adjacent cells in the grid by an edge. The edges (and their weights) represent the delay to travel between those two points in the grid. The weight of the edges is non-uniform, and determined at run-time. I have a set of {X1,Y1} and {X2,Y2} pairs that I want to find the shortest path between. The set is 'large'. Rather than running the standard dijkstra_shortest_path, and therefore having a run-time of O (num_pairs * V^2), I use johnson to find the all_pairs paths, at a cost of O(V*E*log(V)). Since E=2*V, that's effectively (V^2 log(V)). For any num_pairs > log(V), the johnson algorithm /should/ be faster. I do this a number of times (10 million), constructing a new graph each time, putting the edges in, and then solving the all_pairs paths. Profling the code shows that the graph construction is fairly trivial compared to the actual johnson algorithm (and the BFS_Visitor beneath that). Anything else that might help? Thanks for your interest, - Greg On May 11, 2006, at 10:53 AM, Dmitry Bufistov wrote:

Greg Link wrote:

...

I'm using BGL for the johnson_all_pairs_shortest_path algorithm, [snip] and I'm wondering if that's the best choice. My graph has (E=2*V), and all nodes are of constant degree, with 4 edges incident on them.

Hi Grek, Probably there is better algorithm for your original problem). Could you describe in brief it? Regards, --dima

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Dima - I've got a 2D grid-structure, of some size "n x n". Each point in the grid is connected to it's four neighboring and adjacent cells in the grid by an edge. The edges (and their weights) represent the delay to travel between those two points in the grid. The weight of the edges is non-uniform, and determined at run-time.

I have a set of {X1,Y1} and {X2,Y2} pairs that I want to find the shortest path between. The set is 'large'. Rather than running the standard dijkstra_shortest_path, and therefore having a run-time of O (num_pairs * V^2), I use johnson to find the all_pairs paths, at a cost of O(V*E*log(V)). Since E=2*V, that's effectively (V^2 log(V)). For any num_pairs > log(V), the johnson algorithm /should/ be faster.

I do this a number of times (10 million) If every time your delays being changed at the same value, then you

Greg Link wrote: probably don´t need to rerun shortest pathes algorithm every time, you just can recalculate all you shortest-pathes trees in linear time as described here http://arxiv.org/PS_cache/cs/pdf/0205/0205041.pdf (Note sure that even in this case it will increase perfomance, but if you are really concerned about it then you can try) --dima