How to discover the neighbor nodes in an matrix?
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Hello guys,
I'm new here and I need your help. I have this code that I need to translate it into Matlab. I have an idea to create a matrix of distances between nodes, where for example: d11is the distanace between 1 and 1 (which will be 0), d12 is the distance of 1 with 2. Then checking the first row of the matrix compares the distance values with the transmission radius of node 1 and if it is smaller than the radius then these nodes will be included in the vector of neighbors of the first node. this is my idea but I don't know how to write it in Matlab. If someone could help me, I would be very grateful.

2 Kommentare
Jan
am 8 Dez. 2022
John hits the point.
There is a "Neighbor list" as input, a "Neighbor_list" and empty array, a "MyNeighborsSet" and "Neighbors". The pseudocode is not useful and you cannot implement it in Matlab reliably.
Antworten (1)
Jon
am 8 Dez. 2022
Bearbeitet: Jon
am 8 Dez. 2022
Here is a little example that I think does what you are asking
%
% assign threshold for being a neighbor
dMax = 0.6;
% make an example distance matrix where i,j entry gives distance from node
% i to node j
D = rand(5,5);
% make the example matrix symmetric
D = (D + D')/2
% make main diagonal zero, nodes are always neighbors of themselves
D = D - diag(diag(D));
% find the neighbors of each node
% form logical matrix whose (i,j)th entries are true if node j is a
% neighbor of node i
isNeighbor = D <= dMax;
% make neighbor lists
% use cell array because lists may have different lengths
numNodes = size(D,1);
Neighbor = cell(numNodes,1);
for k = 1:numNodes
Neighbor{k} = find(isNeighbor(k,:));
end
display(Neighbor)
You could also do this using MATLAB's graph functions, replacing the last lines of the above code with
% could also do this using a graph
G = graph(isNeighbor);
Neighbor = cell(numNodes,1);
for k = 1:numNodes
Neighbor{k} = neighbors(G,k)';
end
display(Neighbor);
1 Kommentar
Jon
am 8 Dez. 2022
Note also that MATLAB graphs even allow finding nearest neighbors within a radius, so you could probably even further simplify the above.
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