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POCalculate which products together would deliver the requested power
 

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  1. POCalculate which products together would deliver the requested power
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    <p>Let's say I've got three products:</p> <p><strong>Product A</strong> Will deliver 5 power. Costs 50.</p> <p><strong>Product B</strong> Will deliver 9 power. Costs 80.</p> <p><strong>Product C</strong> Will deliver 15 power. Costs 140.</p> <p>I want to know what combination of products I could buy when I need 7 power. I could buy two of <strong>A</strong> but one of <strong>B</strong> is cheaper.</p> <p>When I'd need 65 power. I would need 4 times <strong>C</strong> and 1 time <strong>A</strong> (costs 680). But I could also go for seven <strong>B</strong> products and one <strong>A</strong> (costs 610).</p> <p>I am looking for a way to calculate the possible combinations of products for the given amount of power I need.</p> <p>The way I tried doing this doesn't give me what I want:</p> <pre><code>// $products are sorted DESC on their $power $power = 65 while( $power &gt; 0 ) { foreach( $products as $productPower ) { if( ( $productPower &gt; $power &amp;&amp; $power - $productPower &gt; 0 ) || $productPower == end( $products ) ) { // Add product to list $power -= $productPower; break; } } } </code></pre> <p>This sample code will only give me 4 times <strong>C</strong> and one time <strong>A</strong>. How should I go about it?</p> <p><strong>EDIT</strong> The number of products is variable. Also, the specific cost and power is variable. So there may be 10 products with cheeper and more expensive price tags.</p> <p><strong>EDIT 2</strong> As I said above, I want to calculate the possible <strong>combinations</strong> (plural). Some people seem to have missed that in my description.</p>
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    1. COVery interesting question. I'll put my thought into it.
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    2. COJust to be clear, do you want all possible combinations? Or only the cheaper one?
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    3. COIt seems you've run into a variation of the [Knapsack Problem.](http://en.wikipedia.org/wiki/Knapsack_problem) The good news is, there's a lot of various algorithms to help you with. The bad news is that you won't be able to do it quickly. This is a category of problems where you'll need to compare every solution to every other solution (known as [combinatorial optimization](http://en.wikipedia.org/wiki/Combinatorial_optimization)).
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