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    <p>Your mention of rows and columns suggests that you envisaged arranging the rectangles in a grid, possibly with a few spaces (e.g. some of the bottom row) unfilled. Assuming this is the case:</p> <p>Suppose you scale the objects such that (an as-yet unknown number) <code>n</code> of them fit across the screen. Then</p> <pre><code>objectScale=screenWidth/(n*objectWidth) </code></pre> <p>Now suppose there are N objects, so there will be </p> <pre><code>nRows = ceil(N/n) </code></pre> <p>rows of objects (where ceil is the <a href="http://en.wikipedia.org/wiki/Floor_and_ceiling_functions" rel="nofollow noreferrer">Ceiling function</a>), which will take up </p> <pre><code>nRows*objectScale*objectHeight </code></pre> <p>of vertical height. We need to find <code>n</code>, and want to choose the smallest <code>n</code> such that this distance is smaller than <code>screenHeight</code>.</p> <p>A simple mathematical expression for <code>n</code> is made trickier by the presence of the ceiling function. If the number of columns is going to be fairly small, probably the easiest way to find <code>n</code> is just to loop through increasing <code>n</code> until the inequality is satisfied. </p> <p><strong>Edit:</strong> We can start the loop with the upper bound of</p> <pre><code>floor(sqrt(N*objectHeight*screenWidth/(screenHeight*objectWidth))) </code></pre> <p>for n, and work down: the solution is then found in O(sqrt(N)). An O(1) solution is to assume that</p> <pre><code>nRows = N/n + 1 </code></pre> <p>or to take</p> <pre><code>n=ceil(sqrt(N*objectHeight*screenWidth/(screenHeight*objectWidth))) </code></pre> <p>(the solution of Matthieu M.) but these have the disadvantage that the value of <code>n</code> may not be optimal.</p> <p>Border cases occur when <code>N=0</code>, and when <code>N=1</code> and the aspect ratio of the objects is such that <code>objectHeight/objectWidth &gt; screenHeight/screenWidth</code> - both of these are easy to deal with.</p>
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    1. POHow to optimize the layout of rectangles
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    1. USChris Johnson
    1. USChris Johnson
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    1. COThe problem though is how to optimally calculate 'n' also. Of course, if I know how many objects I'm going to fit across the screen then it's simple to work out everything else - but this is an unknown. The problem is how to calculate the optimal 'n' and therefore also the number of rows required - and then the resulting object width and height... Cheers, Jamie.
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      1. USbadmanj
    2. COAbsolutely - see the edit: I don't think an analytic expression for the optimal n is possible, but one can minimise the searching.
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      1. USChris Johnson
 

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