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POoptimal solution for a modified version of the bin-packing task
 

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  1. POoptimal solution for a modified version of the bin-packing task
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    <p><strong>Problem</strong></p> <p>Suppose we have a set of <em>N</em> real numbers <em>A = {x_1, x_2, ..., x_N}</em>.</p> <p>The goal is to partition this set into <em>A_1, A_2, ..., A_L</em> subsets with the limitation of <em>sum( A_i ) &lt;= T</em> and <strong>minimizing</strong> this term:</p> <p><em>Cost := sum( abs( sum(A_i) - T ) )</em></p> <p>where <em>sum(A_i)</em> denotes the summation of numbers in <em>A_i</em> and <em>T</em> is a given threshold.</p> <p><em>I am looking for a non-evolutionary optimal algorithm.</em></p> <p><strong>Update:</strong> <em>x_i</em> are real positive numbers and not greater than <em>T</em> ( <em>0 &lt; x_i &lt;= T</em> ).</p> <p><strong>Update 2:</strong> The cost function fixed.</p> <hr> <p><strong>Nice try, Greedy algorithm!</strong></p> <p>A simple idea is to use a <a href="http://en.wikipedia.org/wiki/Greedy_algorithm" rel="nofollow">Greedy</a> approach to solve the problem. Here is a pseudocode:</p> <pre><code> 1. create subset A_1 and set i=1. 2. remove the largest number x from A. 3. If sum(A_i) + x &lt;= T * put x into A_i 4. Else * create a new subset A_i+1, * put x into A_i+1 * set i=i+1 5. If A is non-empty * goto step 2. 6. Else * return all created A_i s </code></pre> <p>The problem is this solution is not optimal. For example, there are cases that it is better not to put two largest numbers, <em>x1</em> and <em>x2</em>, in the first subset <em>A_1</em> even they don't exceed <em>T</em>, because there is no other *x_i* available to add to that set and make its sum closer to <em>T</em>. In the other hand, if we had put <em>x1</em> and <em>x2</em> in the separate sets, a better solution could be found (a solution with smaller <em>Cost</em> value).</p> <p><strong>Possible solutions</strong></p> <p>I have thought of using a <a href="http://en.wikipedia.org/wiki/Backtracking" rel="nofollow">Backtracking</a> algorithm which can find the optimal solution too, but I guess its complexity in this problem would be high.</p> <p>I have read some article on Wikipedia like <a href="http://en.wikipedia.org/wiki/Bin_packing_problem" rel="nofollow">Bin packing problem</a> (NP-hard <em>[sighs...]</em>) and <a href="http://en.wikipedia.org/wiki/Cutting_stock_problem" rel="nofollow">Cutting stock problem</a> and apparently my problem is very similar to this standard problems, but I am not sure which one matches my case.</p>
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    1. COFor what i understood, it looks like you have to solve several knapsack problems, i mean, solve a knapsack for T, so you get a set A_1 that has min cost, but close enought to T, and then you run a knapsack again (removing the A_0 elements from A) so you get a new A_1 optimal and close to T, then you repeat again until you find all the A_i
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      1. UShiguaro
    2. CO@h3nr1x: I guess this is what greedy solution does. While this approach finds optimal solutions for each knapsack, it fails to find a global optimum solution (An example is given in the question).
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    3. COThere is something wrong with the term to minimize because it is constant regardless of how the sets are partitioned. It is always equivalent to the sum of all N numbers minus T times L.
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      1. POoptimal solution for a modified version of the bin-packing task
      1. USAntti Huima
 

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