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    <p>Sort the numbers using the reversed bits as the key.</p> <p>This Python code demonstrates the concept:</p> <pre><code>&gt;&gt;&gt; sorted(range(1,128), key=lambda x: ('{:08b}'.format(x))[::-1]) [64, 32, 96, 16, 80, 48, 112, 8, 72, 40, 104, 24, 88, 56, 120, 4, 68, 36, 100, 20, 84, 52, 116, 12, 76, 44, 108, 28, 92, 60, 124, 2, 66, 34, 98, 18, 82, 50, 114, 10, 74, 42, 106, 26, 90, 58, 122, 6, 70, 38, 102, 22, 86, 54, 118, 14, 78, 46, 110, 30, 94, 62, 126, 1, 65, 33, 97, 17, 81, 49, 113, 9, 73, 41, 105, 25, 89, 57, 121, 5, 69, 37, 101, 21, 85, 53, 117, 13, 77, 45, 109, 29, 93, 61, 125, 3, 67, 35, 99, 19, 83, 51, 115, 11, 75, 43, 107, 27, 91, 59, 123, 7, 71, 39, 103, 23, 87, 55, 119, 15, 79, 47, 111, 31, 95, 63, 127] </code></pre> <p>Looking at the bit patterns for each number shows how/why it works:</p> <pre><code>&gt;&gt;&gt; '{:08b}'.format(64) '01000000' &gt;&gt;&gt; '{:08b}'.format(32) '00100000' &gt;&gt;&gt; '{:08b}'.format(96) '01100000' </code></pre> <p>Note, the process can also be done on-the-fly, not requiring a sort:</p> <pre><code>&gt;&gt;&gt; [int('{:07b}'.format(i)[::-1], 2) for i in range(1, 128)] [64, 32, 96, 16, 80, 48, 112, 8, 72, 40, 104, 24, 88, 56, 120, 4, 68, 36, 100, 20, 84, 52, 116, 12, 76, 44, 108, 28, 92, 60, 124, 2, 66, 34, 98, 18, 82, 50, 114, 10, 74, 42, 106, 26, 90, 58, 122, 6, 70, 38, 102, 22, 86, 54, 118, 14, 78, 46, 110, 30, 94, 62, 126, 1, 65, 33, 97, 17, 81, 49, 113, 9, 73, 41, 105, 25, 89, 57, 121, 5, 69, 37, 101, 21, 85, 53, 117, 13, 77, 45, 109, 29, 93, 61, 125, 3, 67, 35, 99, 19, 83, 51, 115, 11, 75, 43, 107, 27, 91, 59, 123, 7, 71, 39, 103, 23, 87, 55, 119, 15, 79, 47, 111, 31, 95, 63, 127] </code></pre> <p>In C, reversing the bits is a trivial exercise:</p> <pre><code>long reverse(long x) { long result = 0: int i; for (i=0 ; i&lt;32 ; i++) { result &lt;&lt;= 1; result |= x &amp; 1; x &gt;&gt;= 1; } return result; } </code></pre>
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    1. POMost efficient way to count a range of numbers without bias?
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    1. USRaymond Hettinger
    1. USRaymond Hettinger
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    1. COBravo. You are a gentleman and a scholar. (+1)
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      1. USPlatinum Azure
    2. COThanks. It's been a couple years since I did this in college. I know this works but I'm going to need to stare at a few more bit patterns to understand why.
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      1. USJay
    3. COI wonder if there is a faster way—as this isn't particularly fast. You're spending a fair amount of time doing the bit reversal. Then the sort is probably O(n·log n). You could use O(n) sorts, but the constant term is probably fairly high with the small *n*'s we're talking here...
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