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    <p>Generally speaking, when you need to do something recursively it helps to start from the "bottom" and think your way up. Consider: You have a set <code>S</code> of <em>n</em> numbers <code>{a,b,c,...}</code>, and a set of four operations <code>{+,-,*,/}</code>. Let's call your recursive function that operates on the set <code>F(S)</code></p> <ul> <li>If <em>n</em> is 1, then <code>F(S)</code> will just be that number. </li> <li>If <em>n</em> is 2, <code>F(S)</code> can be eight things: <ul> <li>pick your left-hand number from <code>S</code> (2 choices)</li> <li>then pick an operation to apply (4 choices)</li> <li>your right-hand number will be whatever is left in the set</li> </ul></li> <li>Now, you can generalize from the <em>n</em>=2 case: <ul> <li>Pick a number <code>x</code> from <code>S</code> to be the left-hand operand (<em>n</em> choices)</li> <li>Pick an operation to apply</li> <li>your right hand number will be <code>F(S-x)</code></li> </ul></li> </ul> <p>I'll let you take it from here. :) </p> <p>edit: Mark poses a valid criticism; the above method won't get absolutely everything. To fix that problem, you need to think about it in a slightly different way: </p> <ul> <li>At each step, you first pick an operation (4 choices), and then</li> <li><em>partition</em> <code>S</code> into two sets, for the left and right hand operands,</li> <li>and recursively apply <code>F</code> to both partitions</li> </ul> <p>Finding all partitions of a set into 2 parts isn't trivial itself, though. </p>
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    1. POComputing target number from numbers in a set
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    1. COThis isn't general enough: if the left-hand side of the operation is always one of the original numbers in S, then this strategy won't ever produce results like `(1+2)*(3+4)`.
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    2. CO@Mark: True, I hadn't thought of that. Updated.
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