Unless I completely misunderstand the requirements, it's only by coincidence you're getting the same results. $x is the number of elements to return, which happens to be equal to the value of the element which is common to all in the given example. And it's a coincidence that you get two elements after your set differences such that after your union, you're up to 3 elements.

With a $x that was different from the value of the common element, or more elements in common such that the set differences didn't have two elements, your results would be very different.

Why did you do $sa - $sb - $sc for the first and not $sb - $sa - $sc and $sc - $sa - $sb for the second and third?

In reply to Re^2: Mutually Exclusive Elements by Zed_Lopez
in thread Mutually Exclusive Elements by Anonymous Monk

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