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    <p>The problem is <a href="http://en.wikipedia.org/wiki/NP-hard" rel="nofollow">NP-Hard</a>, thus the <a href="http://en.wikipedia.org/wiki/Backtracking" rel="nofollow">backtracking</a> solution [or other exponential solution as suggested by @vhallac] will be your best shot, since there is not known [and if P!=NP, there is no existing] polynomial solution for this kind of problem.</p> <p><strong>NP-Hardness proof:</strong> <br>First, we know that a rectangle consists of 4 edges, that are equal in pairs [e1=e2,e3=e4]. <br>We will show that if there is a polynomial algorithm <code>A</code> to this problem, we can also solve the <a href="http://en.wikipedia.org/wiki/Partition_problem" rel="nofollow">Partition Problem</a>, by the following algorithm:</p> <pre><code>input: a group of numbers S=a1,a2,...,an output: true if and only if the numbers can be partitioned algorithm: sum &lt;- a1 + a2 + .. + an lengths &lt;- a1, a2 , ... , an , (sum*5), (sum*5) activate A with lengths. if A answered there is any rectangle [solution is not 0], answer True else answer False </code></pre> <p><strong>Correctness:</strong> <br>(1) if there is a partition to S, let it be S1,S2, there is also a rectangle with edges: <code>(sum*5),(sum*5),S1,S2</code>, and the algorithm will yield True.</p> <p>(2) if the algorithm yields True, there is a rectangle available in lengths, since a1 + a2 + ... + an &lt; sum*5, there are 2 edges with length sum*5, since the 2 other edges must be made using all remaining lengths [as the question specified], each other edge is actually of length <code>(a1 + a2 + ... + an)/2</code>, and thus there is a legal partition to the problem.</p> <p><strong>Conclusion:</strong> There is a reduction <code>PARTITION&lt;=(p) this problem</code>, and thus, this problem is NP-Hard</p> <p><strong>EDIT:</strong> <br>the <strong>backtracking solution</strong> is pretty simple, get all possible rectangles, and check each of them to see which is the best. <br>backtracking solution: pseudo-code:</p> <pre><code>getAllRectangles(S,e1,e2,e3,e4,sol): if S == {}: if legalRectangle(e1,e2,e3,e4): sol.add((e1,e2,e3,e4)) else: //S is not empty elem &lt;- S[0] getAllRectangles(S-elem,e1+elem,e2,e3,e4,sol) getAllRectangles(S-elem,e1,e2+elem,e3,e4,sol) getAllRectangles(S-elem,e1,e2,e3+elem,e4,sol) getAllRectangles(S-elem,e1,e2,e3,e4+elem,sol) getRectangle(S): RECS &lt;- new Set getAllRectangles(S,{},{},{},{},RECS) getBest(RECS) </code></pre> <p><strong>EDIT2:</strong> <br>As discussed in the comments, this answer shows not only this is hard to find the <em>BEST</em> rectangle, it is also hard to find <em>ANY</em> rectangle, making this problem hard for <a href="http://en.wikipedia.org/wiki/Heuristic" rel="nofollow">heuristic</a> solutions as well.</p>
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    1. POConstruct the largest possible rectangle out of line segments of given lengths
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    1. POConstruct the largest possible rectangle out of line segments of given lengths
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    1. COBasically, in geometric terms, we are trying to build something that is as close to *square* as possible, since square is the biggest possible rectangle of given fixed perimeter. And building a square (or getting as close as possible to a square) involves partitioning our segments into 2 equal-total-length sets (or as close as possible) - horizontals and verticals - which is the Partitioning Problem.
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    2. CO@AndreyT: you are right of course, this is the optimization problem for the partition problem, for 4 parts. My answer shows not only finding the BEST rectangle is hard, but also finding if there is ANY rectangle is hard.
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    3. COYou are right. This is actually a pair of "nested" Partition Problems over 4 partitions, that requires obtaining perfect equality within each pair of partitions, and on top of that requires minimizing the difference between the pairs. Each requirement is a Partition Problem by itself: one is in "binary" form, another is in "optimization" form.
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