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    <p>Finding the optimal solution for a specific supermarket is a generalization of the <a href="http://en.wikipedia.org/wiki/Set_cover_problem" rel="nofollow">set cover problem</a>, which is NP-complete. The reduction goes as follows: Given an instance of the set cover problem, just define a cost function assigning 1 to each combination, apply an algorithm that solves your problem, and you obtain an optimal solution of the set cover instance. (Finding the minimal price hence corresponds to finding the minimum number of covering sets.) Thus, your Problem is <a href="http://en.wikipedia.org/wiki/NP-hard" rel="nofollow">NP-hard</a>, and you cannot expect to finde a solution that runs in polynomial time.</p> <p>You really should implement the brute-force method you mentioned. I too recommand you to do this as a first step. If the performance is not sufficient, you can try a using a MIP-formulation and a solver like CPLEX, or you have to devolop a heuristic approach.</p> <p>For a single supermarket, it is rather trivial to find a <a href="http://en.wikipedia.org/wiki/Integer_linear_programming#Integer_unknowns" rel="nofollow">mixed integer program (MIP)</a>. Let <code>x_i</code> be the integer number how often product combination <code>i</code> is contained in a solution, <code>c_i</code> its cost and <code>w_ij</code> the number how often product <code>j</code> is contained in product combination <code>i</code>. Then, you are minimizing </p> <pre><code> sum_i x_i * c_i </code></pre> <p>subject to conditions like</p> <pre><code> sum_i x_i * w_ij >= r_j, </code></pre> <p>where <code>r_j</code> is the number how often product <code>j</code> is required.</p>
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    1. COWhy 1 for each combination when there is an actual price associated to each item?
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