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Hi all, A Dijkstra solution builds the tree outward from a source node. Is there and easy was to change the sense of the route so the calculated result is based on a tree inward to the a sink node? Is there an example of this anywhere? I could not find one. Thanks, -Steve
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On Wed, 12 May 2010, Stephen Woodbridge wrote:
Hi all,
A Dijkstra solution builds the tree outward from a source node. Is there and easy was to change the sense of the route so the calculated result is based on a tree inward to the a sink node?
Is there an example of this anywhere? I could not find one.
There is a reverse_graph adapter. Would that do what you want? -- Jeremiah Willcock
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Jeremiah Willcock wrote:
On Wed, 12 May 2010, Stephen Woodbridge wrote:
Hi all,
A Dijkstra solution builds the tree outward from a source node. Is there and easy was to change the sense of the route so the calculated result is based on a tree inward to the a sink node?
Is there an example of this anywhere? I could not find one.
There is a reverse_graph adapter. Would that do what you want?
Jeremiah, Thanks I missed that. It may work, I need to think about the use case a little. There are two use cases that come quickly to mind: 1. driving distance/time/cost to a location like a store, to see what coverage area say a 10, 20, 30 min drive time TO the store represents. 2. when doing bi-directional routing, the routing from the destination needs to be reversed sense from the routing from the source. My initial concern is that the reverse_graph adapter creates a new adjacency list and while that may work for small graphs it is not likely a good solution for a say North America or Europe sized graph. My thinking was that the process of routing is searching for nodes that I can go to from my current location and that reverse routing is just searching for nodes that I could come from. I suppose that to be able to do that efficiently you have to have an adjacency list that supports that. I will think about this and try some examples when I get a chance. Let me know if you can think of a better way to tackle this problem. Thanks, -Steve
participants (2)
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Jeremiah Willcock
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Stephen Woodbridge