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    <p>Assuming I understand your question I believe this is straightforward for fairly small sets.</p> <p>I will use <em>Mathematica</em> code for illustration, but the concept is universal.</p> <p>I generate <code>10</code> random subsets of length <code>4</code>, from the set {1 .. 8}:</p> <pre><code>ss = Subsets[Range@8, {4}] ~RandomSample~ 10 </code></pre> <blockquote> <pre><code>{{1, 3, 4, 6}, {2, 6, 7, 8}, {3, 5, 6, 7}, {2, 4, 6, 7}, {1, 4, 5, 8}, {2, 4, 6, 8}, {1, 2, 3, 8}, {1, 6, 7, 8}, {1, 2, 4, 7}, {1, 2, 5, 7}} </code></pre> </blockquote> <p>I convert these to a binary array of the presence of each number in each subset:</p> <pre><code>a = Normal@SparseArray[Join @@ MapIndexed[Tuples[{##}] &amp;, ss] -&gt; 1]; Grid[a] </code></pre> <p><img src="https://i.stack.imgur.com/aXmiz.png" alt="Mathematica graphics"></p> <p>That is ten columns for ten subsets, and eight rows for elements {1 .. 8}.</p> <p>Now generate all possible target subsets (size <code>3</code>):</p> <pre><code>keys = Subsets[Union @@ ss, {3}]; </code></pre> <p>Take a "key" and extract those rows from the array and do a BitAnd operation (return <code>1</code> iff all columns equal <code>1</code>), then count the number of ones. For example, for key <code>{1, 6, 8}</code> we have:</p> <pre><code>a[[{1, 6, 8}]] </code></pre> <p><img src="https://i.stack.imgur.com/b03J5.png" alt="Mathematica graphics"></p> <p>After BitAnd:</p> <p><img src="https://i.stack.imgur.com/2PJYZ.png" alt="Mathematica graphics"></p> <p>Do this for each key:</p> <pre><code>counts = Tr[BitAnd @@ a[[#]]] &amp; /@ keys; </code></pre> <p>Then find the position(s) of the maximum element of that list, and extract the corresponding parts of <code>keys</code>:</p> <pre><code>keys ~Extract~ Position[counts, Max@counts] </code></pre> <blockquote> <pre><code>{{1, 2, 7}, {2, 4, 6}, {2, 4, 7}, {2, 6, 7}, {2, 6, 8}, {6, 7, 8}} </code></pre> </blockquote> <p>With adequate memory this process works quickly for a larger set. Starting with 50,000 randomly selected subsets of length <code>7</code> from {1 .. 30}:</p> <pre><code>ss = Subsets[Range@30, {7}] ~RandomSample~ 50000; </code></pre> <p>The maximum sub-subsets of length <code>4</code> are calculated in about nine seconds:</p> <pre><code>AbsoluteTiming[ a = Normal@SparseArray[Join @@ MapIndexed[Tuples[{##}] &amp;, ss] -&gt; 1]; keys = Subsets[Union @@ ss, {4}]; counts = Tr[BitAnd @@ a[[#]]] &amp; /@ keys; keys~Extract~Position[counts, Max@counts] ] </code></pre> <blockquote> <pre><code> {8.8205045, {{2, 3, 4, 20}, {7, 10, 15, 18}, {7, 13, 16, 26}, {11, 21, 26, 28}}} </code></pre> </blockquote> <p>I should add that <em>Mathematica</em> is a high level language and these operations are on generic objects, therefore if this is done truly at the binary level this should be much faster, and more memory efficient.</p>
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    1. COThanks for your reply. In my case, I really want to avoid generating all possible subsets of size `k` since for example I have an instance with `N=1000` and `k=100`, which means I would have to generate about 10^140 subsets.
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    2. CO@mitchus are your subsets S1..Sn *all possible* subsets of that size, or an arbitrary group of subsets?
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    3. COthey are an arbitrary group, possibly with several repeating instances (in my example S1 and S2 are the same).
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