Creating a dilution rule matrix - help needed

Hey there,

I have some difficulty with what seems to be a pretty simple code. I am trying to create a small code that makes a dilution rule for pipe salinity leaving a node. for example: if I have a [ 1; 1; 1; 2; 2; 2; 3] node vector for pipes leaving nodes (the pipe vector is: [1;2;3;4;5;6;7])

I would like to get a matrix:

[1 -1 0 0 0 0 0; 1 0 -1 0 0 0 0; 0 0 0 1 -1 0 0; 0 0 0 1 0 -1 0]

this matrix will then be multiplied by the salinity vector in pipes and I will make sure it equals zero. The question is how to get the matrix above.

this is what I have:

node_1 = [1; 1; 1; 2; 2; 2 ; 3];
pipe=1:7;
count=1;
B0=[];
for i=1:size((node_1),1)
    for j=i+1:size(node_1,1)
        if node_1(i)==node_1(j)
            B0(count,pipe(i))=1;
            B0(count,pipe(j))=-1;
            count=count+1;
        end
    end
end

of course this doesn't work because I get two extra rows for the 2 and 3 pipes and the 5 and 6 pipes. In addition I keep on repeating the B0(count,pipe(i))=1; for a node, that has more than 2 pipes leaving it, which is wasteful.

I tried different kinds of solutions but they all seem faulty.

I hope the description of the problem is clear.

Thanks in advance, Asher

Answer 1

[Guillaume]

The bit you should have explained better is the construction rule of your matrix. The way I understand it, for numbers that are repeated, create a row in your matrix for each repeat where:

• first occurence occuring on row fi is marked 1 in column fi
• row ri of subsequent occurences are marked -1 in column ri

This may work for this:

nodes = [1; 1; 1; 2; 2; 2; 3];
pipes = 1:numel(nodes);
B0 = [];
[uniq, pos] = unique(nodes);
for node = uniq'
   nodepipes = pipes(nodes == node);
   if numel(nodepipes) > 1
      pipeout = nodepipes(1);
      for pipein = nodepipes(2:end)
         Brow = zeros(1, numel(nodes));
         Brow(pipeout) = 1;
         Brow(pipein) = -1;
         B0(end + 1, :) = Brow;
      end
   end
end

[Asher Metzger]

Works great. Thanks.