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    <p>I'm not going to give you Objective-C code mainly because:</p> <ul> <li>I only know Objective-C on a very superficial level.</li> <li>I don't have the desire to write all the memory management code required to get this working in a language like C, and it would only detract from the readability anyway.</li> </ul> <p>Instead I will give you some ideas and some code showing how I would implement this in a higher language with generators and garbage collection (Python in this case) and a hint on how to do it without generators. Hopefully someone else may be able to port the code for you if you cannot do it yourself.</p> <p>I would think about your problem in a slightly different way:</p> <ul> <li>How many leading zeros are there in your initial "flushed-right" pattern.</li> <li>How many ways are there to partition that number of zeros into n partitions.</li> </ul> <p>In your last example you have two leading zeros and three partitions with separators '10' and '1':</p> <pre> 2 0 0: 00101 1 1 0: 01001 1 0 1: 01010 0 2 0: 10001 0 1 1: 10010 0 0 2: 10100 </pre> <p>The separators are always of the form <code>111..10</code> except the last which is just <code>111..1</code> without the trailing zero.</p> <p>To enumerate the above partitions use a function like the following in Python:</p> <pre><code>def partitions(n, x): if n == 1: yield [x] else: for i in range(x + 1): for p in partitions(n - 1, x - i): yield [i] + p for p in partitions(3, 2): print p </code></pre> <p>Result:</p> <pre><code>[0, 0, 2] [0, 1, 1] [0, 2, 0] [1, 0, 1] [1, 1, 0] [2, 0, 0] </code></pre> <p>Once you have these partitions it is simple to construct the patterns.</p> <p>One challenge is that Objective-C doesn't have built-in support for the yield construct. The following rewrite of the above function may be easier to convert to Objective-C:</p> <pre><code>def partitions(n, x): if n == 1: return [[x]] else: result = [] for i in range(x + 1): for p in partitions(n - 1, x - i): result.append([i] + p) return result </code></pre> <p>I hope that is of some use to you.</p>
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