How to convert the X & Y value in a matrix format?
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Asyran Abdullah
el 26 de Sept. de 2018
Comentada: Asyran Abdullah
el 26 de Sept. de 2018
Hi, Given X and Y as follow:
area_length = 100;
area_width = area_length ;
x = round((area_length - 1) * (rand - 0) / (1 - 0) + 1);
y = round((area_width - 1) * (rand - 0) / (1 - 0) + 1);
How to convert the X & Y value into matrix format? Because i want to add a weights for make i possible to perform a shortest path.
x= ;
y= ;
weights = rand ;%-ok
G = graph(x,y,weights);
plot(G,'EdgeLabel',G.Edges.Weight);
Thanks for your help
6 comentarios
KSSV
el 26 de Sept. de 2018
YOu need to have those points to get the shortest path........do you have those points?
Respuesta aceptada
Walter Roberson
el 26 de Sept. de 2018
Editada: Walter Roberson
el 26 de Sept. de 2018
Your x and y calculated like the above define the coordinates of a single point. To do shortest path, you need multiple points.
I suggest,
N = 50; %nodes
cutoff = 25;
area_length = 100;
area_width = area_length ;
x = round((area_length - 1) * (rand(N,1) - 0) / (1 - 0) + 1);
y = round((area_width - 1) * (rand(N,1) - 0) / (1 - 0) + 1);
A = squareform( pdist([x, y]) );
A(A > cutoff) = 0;
%and do any other tweaking for the A matrix
G = digraph(A);
plot(G, 'XData', x, 'YData', y);
and you can do shortest path calculations on G
With the above code, you could also use G = graph(A) to create undirected graphs, but if you do so then you would not be able to have asymmetric distance matrix. In the real world, there are cases where the transmit and receive path might not be the same.
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KSSV
el 26 de Sept. de 2018
Simple rand..gives you a single number....try rand(N,1) with N as desired number of points.
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