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    <p>Your algorithm is a sort-based shuffle, as discussed in the Wikipedia article.</p> <p>Generally speaking, the computational complexity of sort-based shuffles is the same as the underlying sort algorithm (e.g. O(<em>n</em> log <em>n</em>) average, O(<em>n</em>²) worst case for a quicksort-based shuffle), and while the distribution is not <em>perfectly</em> uniform, it should approach uniform close enough for most practical purposes.</p> <p>Oleg Kiselyov provides the following article / discussion:</p> <ul> <li><a href="http://okmij.org/ftp/Haskell/#perfect-shuffle" rel="nofollow noreferrer">Provably perfect random shuffling and its pure functional implementations</a></li> </ul> <p>which covers the limitations of sort-based shuffles in more detail, and also offers two adaptations of the Fischer–Yates strategy: a naive O(<em>n</em>²) one, and a binary-tree-based O(<em>n</em> log <em>n</em>) one.</p> <blockquote> <p>Sadly the functional programming world doesn't give you access to mutable state.</p> </blockquote> <p>This is not true: while purely functional programming avoids <em>side effects</em>, it supports access to mutable state with first-class effects, without requiring side effects.</p> <p>In this case, you can use Haskell's mutable arrays to implement the mutating Fischer–Yates algorithm as described in this tutorial:</p> <ul> <li><a href="http://brett-hall.blogspot.com/2008/01/haskell-shuffling.html" rel="nofollow noreferrer">Haskell Shuffling</a> (Brett Hall)</li> </ul> <h2>Addendum</h2> <p>The specific foundation of your shuffle sort is actually an infinite-key <a href="http://en.wikipedia.org/wiki/Radix_sort" rel="nofollow noreferrer">radix sort</a>: as gasche points out, each partition corresponds to a digit grouping.</p> <p>The main disadvantage of this is the same as any other infinite-key sorting shuffle: there is no termination guarantee. Although the likelihood of termination increases as the comparison proceeds, there is never an upper bound: the worst-case complexity is O(∞).</p>
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    1. POWhat, if anything, is wrong with this shuffling algorithm and how can I know?
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    1. COSorry, I was less than precise. *Efficient* access to mutable state. ;)
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    2. COWhat makes you think it's not efficient?
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    3. COIt is possible and easy to fix a simple sort-based shuffled to be *perfectly* uniform (provided the underlying random number generator is also perfectly uniform), without going through the additional complexity of Oleg's pure solution. You lose uniformity when two elements are compared equal during the sorting : an arbitrary choice has to be made to order them. You can pick weights that are guaranteed never to be equal, such as randomly-picked real numbers (floats, or even better lazy boolean streams). Cf [haskell-beginners](http://www.haskell.org/pipermail/beginners/2010-August/005165.html)
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