# How asymptotically bad is naive shuffling?

It's well-known that this 'naive' algorithm for shuffling an array by swapping each item with another randomly-chosen one doesn't work correctly:

for (i=0..n-1)
swap(A[i], A[random(n)]);


Specifically, since at each of $n$ iterations, one of $n$ choices is made (with uniform probability), there are $n^n$ possible 'paths' through the computation; because the number of possible permutations $n!$ doesn't divide evenly into the number of paths $n^n$, it's impossible for this algorithm to produce each of the $n!$ permutations with equal probability. (Instead, one should use the so-called Fischer-Yates shuffle, which essentially changes out the call to choose a random number from [0..n) with a call to choose a random number from [i..n); that's moot to my question, though.)

What I'm wondering is, how 'bad' can the naive shuffle be? More specifically, letting $P(n)$ be the set of all permutations and $C(\rho)$ be the number of paths through the naive algorithm that produce the resulting permutation $\rho\in P(n)$, what is the asymptotic behavior of the functions

$\qquad \displaystyle M(n) = \frac{n!}{n^n}\max_{\rho\in P(n)} C(\rho)$

and

$\qquad \displaystyle m(n) = \frac{n!}{n^n}\min_{\rho\in P(n)} C(\rho)$?

The leading factor is to 'normalize' these values: if the naive shuffle is 'asymptotically good' then

$\qquad \displaystyle \lim_{n\to\infty}M(n) = \lim_{n\to\infty}m(n) = 1$.

I suspect (based on some computer simulations I've seen) that the actual values are bounded away from 1, but is it even known if $\lim M(n)$ is finite, or if $\lim m(n)$ is bounded away from 0? What's known about the behavior of these quantities?

• Nice question. I don't know where the best place for this question is. Unless it's clear that another forum is better for it, I think you should leave it here for a week or so, and if you don't get a satisfactory answer, ask it on one of the other forums (and put links in both questions). – Peter Shor Nov 6 '12 at 20:49
• @vzn "Why do hard analysis on a known flawed algorithm?" Because the mathematics is interesting, and you never know where other applications may arise - see Knuth's analysis of Bubble Sort, for instance. Atwood's charts give a rough qualitative analysis of the inhomogenity, but that's a far cry from a mathematically quantitative analysis. (And there are several different equivalent formulations of the Fischer-Yates shuffle - the one I mention works just fine.) – Steven Stadnicki Nov 7 '12 at 19:13
• For the record, OEIS sequence A192053 is max $C(\rho)$ and does not list a closed form. Also, the notes for that entry suggest that min $C(\rho)$ may be $2^{n-1}$, implying that $m(n) \rightarrow 0$. – mhum Nov 8 '12 at 2:00
• @vzn What's wrong with open questions? – Yuval Filmus Nov 9 '12 at 3:52
• @vzn Disagree on your last sentence, there is a lot of analysis of "imperfect" shuffles. For examples, if we make random transpositions, it is known that the threshold for randomness is roughly $(1/2) n\log n$. The present question may be hard, but a priori it is hard to say whether it's "very hard". An answer like mhum's is already very satisfying, showing that the question was appropriate for the forum and did not present an insurmountable barrier (formal proofs set aside). – Yuval Filmus Nov 9 '12 at 12:58

We will show by induction that the permutation $\rho_n = (2,3,4,\ldots, n,1)$ is an example with $C(\rho_n) = 2^{n-1}$. If this is the worst case, as it is for the first few $n$ (see the notes for OEIS sequence A192053), then $m(n) \approx (2/e)^{n}$. So the normalized min, like the normalized max, is 'exponentially bad'.

The base case is easy. For the induction step, we need a lemma:

Lemma: In any path from $(2,3,4, \ldots, n, 1)$ to $(1,2,3, \ldots, n)$, either the first move swaps positions $1$ and $n$, or the last move swaps positions $1$ and $n$.

Proof Sketch: Suppose not. Consider the first move that involves the $n$'th position. Assume that it is the $i$'th move, $i\neq 1$ and $i \neq n$. This move must place the item $1$ in the $i$'th place. Now consider the next move that touches the item $1$. Assume this move is the $j$'th move. This move must swap $i$ and $j$, moving the item $1$ into the $j$'th place, with $i < j$. A similar argument says that the item $1$ can only subsequently be moved to the right. But the item $1$ needs to end up in the first place, a contradiction. $\square$

Now, if the first move swaps the positions $1$ and $n$, the remaining moves must take the permutation $(1, 3,4,5, \ldots, n,2)$ to $(1,2,3,4, \ldots, n)$. If the remaining moves don't touch the first position, then this is the permutation $\rho_{n-1}$ in positions $2 \ldots n$, and we know by induction that there are $C(\rho_{n-1})=2^{n-2}$ paths that do this. An argument similar to the proof of the Lemma says that there is no path that touches the first position, as the item $1$ must then end up in the incorrect position.

If the last move swaps the positions $1$ and $n$, the first $n-1$ moves must take the permutation $(2,3,4,\ldots, n,1)$ to the permutation $(n,2, 3,4, \ldots, n-1, 1)$. Again, if these moves don't touch the last position, then this is the permutation $\rho_{n-1}$, and by induction there are $C(\rho_{n-1})=2^{n-2}$ paths that do it. And again, if one of the first $n-1$ moves here touches the last position, the item $1$ can never end up in the correct place.

Thus, $C(\rho_n) = 2C(\rho_{n-1}) = 2^{n-1}$.

• Perfect - the argument behind the lemma looks a lot like the one I had for involutions being the only way of getting the identity permutation, but I'd missed the recursive structure in the explicit swap. Thank you! – Steven Stadnicki Nov 10 '12 at 17:25

After some digging around thanks to mhum's pointer to OEIS, I've finally found an excellent analysis and a nice (relatively) elementary argument (due, as far as I can tell, to Goldstein and Moews ) that $M(n)$ grows superexponentially fast in $n$:

Any involution $\iota$ of $\{1\ldots n\}$ corresponds to a run of the 'naive' shuffling algorithm that produces the identity permutation as its result, since the algorithm will swap $k$ with $\iota(k)$ and subsequently swap $\iota(k)$ with $k$, leaving both unchanged. This means that the number of runs of the algorithm that yield the identity permutation is at least the number of involutions $Q(n)$ (in fact, a little thinking shows that the correspondence is 1-1 and so it's exactly $Q(n)$), and so the maximum in $M(n)$ is bounded from below by $Q(n)$.

$Q(n)$ apparently goes by a number of names, including the telephone numbers : see http://oeis.org/A000085 and http://en.wikipedia.org/wiki/Telephone_number_%28mathematics%29 . The asymptotics are well-known, and it turns out that $Q(n) \approx C\left(\frac{n}{e}\right)^{n/2}e^\sqrt{n}$; from the recurrence relation $Q(n) = Q(n-1)+(n-1)Q(n-2)$ it can be inductively shown that the ratio $R(n) = \frac{Q(n)}{Q(n-1)}$ satisfies $\sqrt{n}\lt R(n)\lt\sqrt{n+1}$ and from there basic analysis gets the leading $n^{n/2}$ term in the asymptotics, though the other terms require a more careful effort. Since the 'scale factor' $\frac{n!}{n^n}$ in the definition of $M(n)$ is only about $C\sqrt{n}e^{-n}$, the leading term of $Q(n)$ dominates and yields (asymptotically) $M(n)\geq Cn^{(n+1)/2}e^{-3n/2+\sqrt{n}}$.

Goldstein and Moews in fact go on to show in  that the identity permutation is the most likely for large $n$, so the $\geq$ is in fact a $\approx$ and the behavior of $M(n)$ is fully settled. This still leaves the question of the behavior of $m(n)$ open; I wouldn't be too surprised if that also yielded to the analysis in their paper, but I haven't had opportunity to read it closely enough yet to really get a grip on their methods, only enough to grok the basic result.

 Goldstein, D. and Moews, D.: "The identity is the most likely exchange shuffle for large n", http://arxiv.org/abs/math/0010066

• It's not too hard to show that the permutation $(2,3,4,\ldots,n,1)$ is an example with $C(\rho) = 2^{n-1}$. If this is the worst case, as it is for the first few $n$, then $m(n) \approx (2/e)^n$. – Peter Shor Nov 9 '12 at 21:21
• @PeterShor Can you give the basic argument? I feel like I'm missing some simple version of the involutions argument that would work, but I'm not quite getting it. I think even if that's not quite minimal that would be good enough; the minimum count seems unlikely to be subexponential in $n$ and just knowing that the normalized max and min are both 'exponentially bad' is a pretty satisfactory answer. – Steven Stadnicki Nov 10 '12 at 0:11
• I added an answer with the argument ... it's too long for a comment. – Peter Shor Nov 10 '12 at 4:07