In graph theorie nodes and edges are "dual". Whenever you have a graph with weighted nodes you can construct a corresponding "dual" graph with weighted edges, and vice versa. And all algorithms are equally transformable.
So "why weighting nodes" has the same legitimation as asking "why weighting edges".
It's just a matter of perspective.
PS: Wikipedia restricts this to planar graphs !?! I'm too lazy to verify what normally happens after transposing the incidence matrix... it certainly works for lattices and they are seldom planar graphs.
UPDATE: OK the duality for lattices is very different, from the graph perspective it's just mirroring at the horizontal axis. I'm quote rosted in this 8(
Anyway at least for planar graphs weighting can be "dualized"! | [reply] |
In all the uses I had of graph theory (because in biochemistry, steady state kinetic equations can be mapped into graph theory) they weighted edges. But JavFan is right. What is the graph is of a place where cities charges taxes to enter, regardless which road you come on? How would you model that?
David.
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Could it be that the weight of the edge is a function of the weight of the nodes? (Just thinking out loud)
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Imagine three nodes (e.g. cities), A, B, C.
A and B are connected by multiple paths. B and C have 1 path. A and C have one path. If the question is, given paths of equal length, which two adjacent cities are most important (population?), edge weighting is irrelevant.
Edge weighting isn't bad, but remember edges are defined by two nodes- not 1- which can differentiate a problem definition. | [reply] |
