Optimización del diseño de un gráfico con distancias de nodo dadas (erróneas)

I have a loosely connected graph. For every edge in this graph, I know the approximate distance d(v,w) between node v y w en posiciones p(v) y p (w) as a vector in R3, not only as an euclidean distance. The error shall be small (lets say < 3%) and the first node is at <0,0,0>.

If there were no errors at all, I can calculate the node-positions this way:

set p(first_node) = <0,0,0>

    for (v,w) in Edges:
        if p(w) is not set:
            set p(w) = p(v) + d(v,w)
    for (u,v) in Edges:
        if p(u) is not set:
            set p(u) = p(v) - d(u,v)

The errors of the distance are not equal. But to keep things simple, assume the relative error (d(v,w)-d'(v,w))/E(v,w) is N(0,1)-normal-distributed. I want to minimize the sum of the squared error

sum( ((p(v)-p(w)) - d(v,w) )^2/E(v,w)^2 ) for all edges

The graph may have a moderate amount of Nodes ( > 100 ) but with just some connections between the nodes and have been "prefiltered" (split into subgraphs, if there is only one connection between these subgraphs).

I have tried a simplistic "physical model" with hooks low but its slow and unstable. Is there a better algorithm or heuristic for this kind of problem?

preguntado el 05 de mayo de 13 a las 21:05

1 Respuestas

Esto parece regresión lineal. Take error terms of the following form, i.e. without squares and split into separate coordinates:

(px(v) - px(w) - dx(v,w))/E(v,w)
(py(v) - py(w) - dy(v,w))/E(v,w)
(pz(v) - pz(w) - dz(v,w))/E(v,w)

If I understood you correctly, you are looking for values px(v), py(v) y pz(v) para todos los nodos v such that the sum of squares of the above terms is minimized.

You can do this by creating a matrix A y un vector b in the following way: every fila corresponds to one of ecuación of the above form, and every visión de conjunto of A corresponde a uno variable, i.e. a single coordinate. For n vértices y m edges, the matrix A tendrá 3m rows (since you separate coordinates) and 3n−3 columns (since you also fix the first node px(0)=py(0)=pz(0)=0).

The row for (px(v) - px(w) - dx(v,w))/E(v,w) would have an entry 1/E(v,w) en la columna de px(v) and an entry -1/E(v,w) en la columna de px(w). All other columns would be zero. The corresponding entry in the vector b sería dx(v,w)/E(v,w).

Now solve the ecuación lineal (AT·A)x = AT·b sin que importe AT denota el transponer of A. The solution vector x will contain the coordinates for your vertices. You can break this into three independent problems, one for each coordinate direction, to keep the size of the linear equation system down.

contestado el 06 de mayo de 13 a las 14:05

I had a similar idea on the next day. Least squares -- yet again;) Good point for solving each dimension on its own. Thank you. - Peter Schneider

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