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    <p>So, the simple way is to permute through all possible combinations. This is slightly tricky, the order of the numbers can be important, and certainly the order of operations is.</p> <p>One observation is that you are trying to generate all possible expression trees with certain properties. One property is that the tree will always have exactly 4 leaves. This means the tree will also always have exactly 3 internal nodes. There are only 3 possible shapes for such a tree:</p> <pre><code> A / \ N A / \ (and the mirror image) N A / \ N N A / \ N A / \ A N (and the mirror image) / \ N N A /` `\ A A / \ / \ N N N N </code></pre> <p>In each spot for A you can have any one of the 4 operations. In each spot for N you can have any one of the numbers. But each number can only appear for one N.</p> <p>Coding this as a brute force search shouldn't be too hard, and I think that after you have things done this way it will become easier to think about optimizations.</p> <p>For example, <code>+</code> and <code>*</code> are commutative. This means that mirrors that flip the left and right children of those operations will have no effect. It might be possible to cut down searching through all such flips.</p> <p>Someone else mentioned RPN notation. The trees directly map to this. Here is a list of all possible trees in RPN:</p> <pre><code>N N N N A A A N N N A N A A N N N A A N A N N A N N A A N N A N A N A </code></pre> <p>That's 4*3*2 = 24 possibilities for numbers, 4*4*4 = 64 possibilities for operations, 24 * 64 * 5 = 7680 total possibilities for a given set of 4 numbers. Easily countable and can be evaluated in a tiny fraction of a second on a modern system. Heck, even in basic on my old Atari 8 bit I bet this problem would only take minutes for a given group of 4 numbers.</p>
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    1. CO+1: for enumerating RPN possibilites and mapping to tree.
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    2. COAlright, I think I'll start working with this suggestion since it was well explained and a few other people seem to like the RPN approach. This is only my second homework assignment in my Intro to Software engineering class, so its taking me some time to catch up with what everyone's been suggesting. :)
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    3. CO@hahuang65: I have code I just wrote based on this approach, and it works for several cases I tried. :-)
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