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copied!<p>This solution is going to involve a bit of <a href="http://docs.python.org/library/itertools.html#module-itertools" rel="nofollow"><code>itertools.permutations</code></a> usage, <a href="http://docs.python.org/library/stdtypes.html#set" rel="nofollow"><code>set()</code></a>, and some good ol' fashioned set difference. Bear in mind, the runtime for finding a permutation will still be O(n!). My solution won't do it in-line, either, but there <em>may</em> be a much more elegant solution that allows you to do so (and doesn't involve <code>itertools.permutations</code>). For this purpose, this is the straightforward way to accomplish the task.</p> <p>Step 1: Algorithm for generating cycles, using the first element given. For a list <code>[1, 1, 2, 2]</code>, this will give us <code>[1, 1, 2, 2], [1, 2, 2, 1], [2, 1, 1, 2], [2, 2, 1, 1]</code>.</p> <pre><code>def rotations(li): count = 0 while count < len(li): yield tuple(li) li = li[1:] + [li[0]] count += 1 </code></pre> <p>Step 2: Importing <code>itertools.permutations</code> to give us the permutations in the first place, then setting up its results into a <code>set</code>.</p> <pre><code>from itertools import permutations perm = set(permutations([1, 1, 2, 2])) </code></pre> <p>Step 3: Using the generator to give us our own set, with the cycles (something we want to rid ourselves of).</p> <pre><code>cycles = set(((i for i in rotations([1, 1, 2, 2])))) </code></pre> <p>Step 4: Apply set difference to each and the cycles are removed.</p> <pre><code>perm = perm.difference(cycles) </code></pre> <p>Hopefully this will help you out. I'm open to suggestions and/or corrections.</p>

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