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  1. USdave
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    1. CO@Vanwaril The algorithm I provided works for the relaxed version of the problem I have formulated. The algorithm cannot pick up "2,3" first due to the fact that "1,2" is added to the list "pair-areas" before "2,3" in step 1, thus, when sorted in order of ascending pair areas (step 2), "1,2" will still appear before "2,3" even though they share the same value for their area. In order to "break" my algorithm, one would need to recur to an extraneous "black" rectangle labeling scheme, e.g. labeling the "black" rectangles "3, 1, 2, 4", as opposed to "1, 2, 3, 4" as you suggested.
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    2. COMy suggestion for @Louis, still is to review the literature on existing *geometric clustering algorithms* to avoid trying to reinvent the wheel, *perhaps* in light of the algorithm I have provided for a relaxed version of his problem. Alternatively, he could also investigate the possibility of applying a genetic or simulated annealing algorithm to this problem.
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    3. CO@BlueRaja The case you originally stated ("black" rectangles as endpoints of "a cross") can also be solved with two "red" bounding rectangles without overlap resulting in the same area as the solution that involves overlapping. The new case you suggest ("extremely far" fifth square) is optimally solved for total area by use of overlapping as you suggest.
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