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<div class=Section1><pre>The discussion of randomly permuting a list comes up every few years. Here’s <o:p></o:p></pre><pre>what I wrote last time (2005):<o:p></o:p></pre><pre><o:p> </o:p></pre><pre>“Clearly, you can do a perfect shuffle in O(N) time using mutable arrays,<o:p></o:p></pre><pre>using the standard imperative algorithm. You can also do it in O(N)<o:p></o:p></pre><pre>expected time using *immutable* arrays, using Haskell's bulk array<o:p></o:p></pre><pre>operations.<o:p></o:p></pre><pre><o:p> </o:p></pre><pre>1. Start with a list of length N. If N < 2, stop.<o:p></o:p></pre><pre>2. Pair each element with a random index in the range [1,2N] (or<o:p></o:p></pre><pre>[0,2N-1]).<o:p></o:p></pre><pre>3. Use "accum" to create an array of size 2N, where slot I contains a<o:p></o:p></pre><pre>list of all the elements paired with I.<o:p></o:p></pre><pre>4. Map the shuffle function recursively across the array to re-shuffle<o:p></o:p></pre><pre>any slots that got more than one element.<o:p></o:p></pre><pre>5. Append all the slots together into the result list.<o:p></o:p></pre><pre><o:p> </o:p></pre><pre>If you leave out step 4, you get an algorithm that runs in O(N)<o:p></o:p></pre><pre>worst-case time, rather than expected time, but you no longer have a<o:p></o:p></pre><pre>perfect shuffle (although it might be good enough). Upping the size of<o:p></o:p></pre><pre>the array from 2N to 3N or 4N will help reduce collisions, but will not<o:p></o:p></pre><pre>entirely eliminate them.<o:p></o:p></pre><pre><o:p> </o:p></pre><pre>With step 4, you can imagine pathological cases where everything gets<o:p></o:p></pre><pre>put in the same slot, but with a reasonable random number generator, it<o:p></o:p></pre><pre>is vanishingly unlikely that you will have enough collisions to drive<o:p></o:p></pre><pre>the cost higher than linear. Again, upping the size of the array from<o:p></o:p></pre><pre>2N to 3N or 4N makes this even more unlikely.”<o:p></o:p></pre><pre><o:p> </o:p></pre><pre>Now, you can consider it cheating to say this is O(N) expected time<o:p></o:p></pre><pre>because it uses O(N log N) random bits. Fair enough. It is O(N) to the<o:p></o:p></pre><pre>same extent that the ordinary imperative algorithm is O(N), albeit expected<o:p></o:p></pre><pre>rather than worst case.<o:p></o:p></pre><pre><o:p> </o:p></pre><pre>-- Chris<o:p></o:p></pre><pre><o:p> </o:p></pre></div>
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