Thank you for the reply. I have not searched on your hints yet, but here is a clarification.
> could you please explain your problem more in detail? Sure. Here is a very simple example of what I am trying to do. Say I have a graph given by... Node 1: neighbor of 2, distance=1 neighbor of 3, distance=2 Node 2: neighbor of 1, distance=1 neighbor of 3, distance=2 Node 3: neighbor of 1, distance=2 neighbor of 2, distance=2 ... which is to be placed in a plane. A glance tells you that these nodes arranged with the proper separations forms an isosceles triangle. An acceptable output of this program would be... Node 1 at coords: x=0, y=0 Node 2 at coords: x=0, y=1 Node 3 at coords: x=sqrt(2^2 - .5^2) ,y=.5 As mentioned, this is not unique but this does not matter (I think). You could try a more difficult graph (but still simple)... Node 1: neighbor of 2, distance=1 neighbor of 3, distance=2 neighbor of 4, distance=2 Node 2: neighbor of 1, distance=1 neighbor of 3, distance=2 neighbor of 4, distance=2 Node 3: neighbor of 1, distance=2 neighbor of 2, distance=2 Node 4: neighbor of 1, distance=2 neighbor of 2, distance=2 ...in which case... Node 1 at coords: x=0, y=0 Node 2 at coords: x=0, y=1 Node 3 at coords: x= sqrt(2^2 - .5^2) ,y=.5 Node 4 at coords: x= - sqrt(2^2 - .5^2) ,y=.5 ...would be a valid solution. > What is your problem, optimization? There are a set of valid solutions which I am looking to explore. There is no value to be optimize, it is merely right or wrong. >It is related with the embedding of finite metrics over finite graph? Sorry, I don't understand this. > That is, what do you mean with euclidean separation of nodes? Something like > w(i, j) > is the distance between Rome and Venice or w(i, j) can be any real non > negative value > that defines a distance? If vector r locates node 1 and vector s locates node 2, the magnitude of r-s is the seperation I am talking about. By Euclidean, I merely meant that we are dealing with the sqaure root for the sum of the square differences in each orthogonal direction ala (in 2d)... r^2 = x^2 + y^2 Thank you for your interest, Zach --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
