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    <p>I hope I don't misunderstand the problem... Here a solution in SWI-Prolog</p> <pre><code>:- module(subsets, [solve/0]). :- [library(pairs), library(aggregate)]. solve :- problem(R, K, Subsets), once(subset_of_maximal_number(R, K, Subsets, Subset)), writeln(Subset). problem(4, 2, [[1,2,3], [1,2,3], [1,2,4], [1,3,4]]). problem(8, 3, [[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]]). subset_of_maximal_number(R, K, Subsets, Subset) :- flatten(Subsets, Numbers), findall(Num-Count, ( between(1, R, Num), aggregate_all(count, member(Num, Numbers), Count) ), NumToCount), transpose_pairs(NumToCount, CountToNumSortedR), reverse(CountToNumSortedR, CountToNumSorted), length(Subset, K), % list of free vars prefix(SolutionsK, CountToNumSorted), pairs_values(SolutionsK, Subset). </code></pre> <p>test output:</p> <pre><code>?- solve. [1,3] true ; [7,6,2] true. </code></pre> <p><strong>edit:</strong> I think that the above solution is wrong, in the sense that what's returned could not be a subset of any of the input: here (a commented) solution without this problem:</p> <pre><code>:- module(subsets, [solve/0]). :- [library(pairs), library(aggregate), library(ordsets)]. solve :- problem(R, K, Subsets), once(subset_of_maximal_number(R, K, Subsets, Subset)), writeln(Subset). problem(4, 2, [[1,2,3], [1,2,3], [1,2,4], [1,3,4]]). problem(8, 3, [[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]]). subset_of_maximal_number(R, K, Subsets, Subset) :- flatten(Subsets, Numbers), findall(Num-Count, ( between(1, R, Num), aggregate_all(count, member(Num, Numbers), Count) ), NumToCount), % actually sort by ascending # of occurrences transpose_pairs(NumToCount, CountToNumSorted), pairs_values(CountToNumSorted, PreferredRev), % we need higher values first reverse(PreferredRev, Preferred), % empty slots to fill, preferred first length(SubsetP, K), select_k(Preferred, SubsetP), % verify our selection it's an actual subset of any of subsets sort(SubsetP, Subset), once((member(S, Subsets), ord_subtract(Subset, S, []))). select_k(_Subset, []). select_k(Subset, [E|R]) :- select(E, Subset, WithoutE), select_k(WithoutE, R). </code></pre> <p>test:</p> <pre><code>?- solve. [1,3] true ; [2,6,7] true. </code></pre>
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    1. COThanks for your answer. I am not really familiar with Prolog, so I have trouble figuring out your code. I know that Prolog is essentially declarative, so here do you essentially just specify the problem and hope the engine finds a smart way to solve it, or are you actually guiding the search in some way?
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    2. COSets are represented as ordered lists. Most builtins used here are deterministic, i.e. have just a solution. The algorithm it's rather elementary, the search capability of Prolog is used just in **select_k**, that is similar to a permutation generator. **select** take an element from the list. Being sorted, elements are selected in decreasing frequency. Sort is needed to use make use of **ord_subtract**.
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