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    <p>It's straightforward to construct each subset from the preceding subset. It's also straightforward to represent a subset as a 128-bit number (although obviously most values wouldn't map onto a qualifying subset, and I don't know if the value of 128 in the question was real or just an example.) That's certainly the first approach I would use; if it works, it's all O(1) and the storage cost is not extreme (16 bytes for indices instead of 4).</p> <p>If you really want to store concise indices for the subsets, I'd use the following recursion, where S(n,k) represents all the subsets (or the count of the subsets) of size &le; k from values &lt; n:</p> <p><code>s(n,0) = { {} }</code><br/> <code>s(n,k) = (s(n-1,k-1) &odot; {n}) &Union; s(n-1,k) if n &ge; k &gt; 0</code><br/> <code>s(n,k) = {} if n &lt; k</code></p> <p>where the operator <code>P &odot; S</code> means "Add <code>S</code> to each element of <code>P</code>" (and therefore the result is exactly the same size as <code>P</code>). So, viewed as a counting algorithm, we get:</p> <p><code>S(n,0) = 1</code><br/> <code>S(n,k) = S(n-1,k-1) + S(n-1,k) if n &ge; k &gt; 0</code><br/> <code>S(n,k) = 0 if n &lt; k</code> </p> <p>The second recursion can be re-expressed as:</p> <p><code>S(n,k) = &Sigma;<sup>n</sup><sub>i=k</sub>S(i-1,k-1)</code><br/> (which would have come out looking nicer with jsMath, grrr.)</p> <p>That's another way of saying that we'll generate the sets in order by the largest element, so we start with the set <code>{0...k-1}</code>, then all the sets with <code>{k}</code> as the largest element, then all the ones with <code>{k+1}</code>, and so on. Within each group of sets, we recurse to find the <code>(k-1)</code>-sized sets, again starting with the least maximum value and working our way up to one less than the maximum value we just selected.</p> <p>So we can figure out what the largest value in the indexed set for an index in <code>S(n,k)</code> by successively subtracting <code>S(i-1, k-1)</code> for <code>i</code> from <code>k</code> to <code>n</code> until the result would be negative; then we add <code>{i}</code> to the result set; reduce <code>k</code> by 1 and repeat with <code>n</code> now set to <code>i-1</code>. </p> <p>If we precompute the relevant tables of <code>S(n,k)</code>, of which there are about 640 valid combinations, we can use binary search instead of iteration to find the <code>i</code> at each step, so the computation takes time <code>k log(n)</code>, which is not awful.</p>
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    1. POEfficient enumeration of subsets
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    1. POEfficient enumeration of subsets
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    1. CO+1. See also: http://en.wikipedia.org/wiki/Combinatorial_number_system
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    2. COThank you very much. I didn't tought about your first approach with the 128-bit number. Seams like this will be much better than any enumeration approach.
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    3. CO@Knoothe: wikipedia's explanation is more elegant than mine. And they get to use real math formulas.
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