# How to check for reversed pairings in columns of a matrix?

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Emma Kuttler on 22 Mar 2022
Edited: Jan on 22 Mar 2022
I have a matrix representing arcs in a network, where the first two columns represent the the "to" and "from" nodes and the other eleven columns represent capacities for various products. In my current matrix, there may be both (1,2) and (2,1) in the first two columns, but only (3,5) and not (5,3), for example. What I need to do is to produce a larger matrix that includes both "flipped pairs" in the first two columns. So I would like to have both [1,2] in the first two columns, and [2,1], each with their own values in the remaining columns. But if [3,5] exists and [5,3] doesn't then I want to add a row to the matrix that goes [5,3] followed by the values in the rest of the row for the [3,5] row. So essentially if the flip of the first two columns exists, keep it, and if not, flip the first two columns, keep the values for the rest of the olumns, and add as a new row. For example, if I have matrix A, I want to produce B.
A = [1 2 5 6 0 2 3 5 6 7 0.5 8 0
2 1 7 6 2 0 2 0 0 1 1 1 1 1
3 5 0 1 5 8 2 0 0 2 3 0 9 0.1
4 5 3 6 4 1 7 0 0 1 4 2 1 -2
2 3 0 1 0 1 8 8 2 0 1 5 5 5
3 2 2 5 6 7 1 2 4 2 0 0 9 -3]
B = [1 2 5 6 0 2 3 5 6 7 0.5 8 0
2 1 7 6 2 0 2 0 0 1 1 1 1 1
3 5 0 1 5 8 2 0 0 2 3 0 9 0.1
5 3 0 1 5 8 2 0 0 2 3 0 9 0.1
4 5 3 6 4 1 7 0 0 1 4 2 1 -2
5 4 3 6 4 1 7 0 0 1 4 2 1 -2
2 3 0 1 0 1 8 8 2 0 1 5 5 5
3 2 2 5 6 7 1 2 4 2 0 0 9 -3]
##### 2 CommentsShow 1 older commentHide 1 older comment
Emma Kuttler on 22 Mar 2022
@Jan , edited! I meant larger

Jan on 22 Mar 2022
Edited: Jan on 22 Mar 2022
A = [1 2 5 6 0 2 3 5 6 7 1 8 0 2; ... % Value appended!!!
2 1 7 6 2 0 2 0 0 1 1 1 1 1; ...
3 5 0 1 5 8 2 0 0 2 3 0 9 1; ...
4 5 3 6 4 1 7 0 0 1 4 2 1 -2; ...
2 3 0 1 0 1 8 8 2 0 1 5 5 5; ...
3 2 2 5 6 7 1 2 4 2 0 0 9 -3];
A1 = A(:, 1);
A2 = A(:, 2);
[s1, s2] = size(A);
B = zeros(s1 * 2, s2); % Maximum possible output
iB = 0;
for k = 1:s1
Ak = A(k, :);
iB = iB + 1;
B(iB, :) = Ak;
if ~any(A1 == Ak(2) & A2 == Ak(1)) % If A([2,1],k) is no member of A(1:2, :):
iB = iB + 1;
B(iB, :) = [Ak(2), Ak(1), Ak(3:s2)];
end
end
B = B(1:iB, :) % Crop output
B = 8×14
1 2 5 6 0 2 3 5 6 7 1 8 0 2 2 1 7 6 2 0 2 0 0 1 1 1 1 1 3 5 0 1 5 8 2 0 0 2 3 0 9 1 5 3 0 1 5 8 2 0 0 2 3 0 9 1 4 5 3 6 4 1 7 0 0 1 4 2 1 -2 5 4 3 6 4 1 7 0 0 1 4 2 1 -2 2 3 0 1 0 1 8 8 2 0 1 5 5 5 3 2 2 5 6 7 1 2 4 2 0 0 9 -3

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