Hi Joel
Your current strategy effectively identifies combinations where VID1 and VID2 contain different vehicle IDs by utilizing intersect to check for common elements.
You can use several strategies to handle unique elements and intersections. First, using sets is effective; set operations like intersect, union, and setdiff are well-suited for tasks involving unique elements, and your current use of intersect is appropriate. Second, logical indexing can sometimes be faster for numeric arrays, especially for simple inclusion checks, though it may be less straightforward for cell arrays of IDs.
To extend your function to handle three vehicles (VID1, VID2, VID3) refer the below code snippet:
% Initialize vehicle ID cell arrays with example data
VID1 = {173, [150 155], 160};
VID2 = {[173 178], 190, [165 170]};
VID3 = {[180 185], [195 200], 175};
% Example matrix A for combinations (simplified for this example)
% Let's just manually specify some combinations for simplicity
A = [1 2 3; 2 3 1; 3 1 2]; % Simplified combination matrix
% Call the function
IDCELL = c3(VID1, VID2, VID3, A);
% Display the result
disp(IDCELL);
function [IDCELL] = c3(VID1, VID2, VID3, A)
IDCELL = {}; % Initialize output cell array
for oz = 1:size(A, 1)
index = A(oz,:);
% Adjusted logic for three vehicles
if numel(intersect(VID1{index(1)}, VID2{index(2)})) == 0 && ...
numel(intersect(union(VID1{index(1)}, VID2{index(2)}), VID3{index(3)})) == 0
IDCELL{end+1} = [VID1{index(1)} 0 VID2{index(2)} 0 VID3{index(3)}];
else
continue
end
end
end
Hope it helps!
