Saf karışma ne kadar asimptotik olarak kötü?

Her öğeyi rastgele seçilen bir tane ile değiştirerek bir diziyi karıştırmak için kullanılan bu 'naif' algoritmanın doğru çalışmadığı bilinmektedir:

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

Her başlangıcı Spesifik olarak, $n$ yineleme, bir $n$ seçenek (düzgün bir olasılık ile) yapılır, orada $n^n$ bir hesaplama ile mümkün 'yolları'; çünkü olası permütasyon sayısı $n!$yolların sayısı eşit bir şekilde bölündüğü vermez $n^n$ bu algoritma her üretmek için, bu imkansız $n!$eşit olasılıklı permütasyonlar. (Bunun yerine, kişi Fischer-Yates karıştırmasını kullanmalıdır; bu , esasen [0..n) arasından bir rastgele sayı seçmek için yapılan çağrıyı [i.n) arasından bir rastgele sayı seçmek için yapılan çağrıyla değiştirir; benim soruma göre bu olsa bile.)

Merak ettiğim şey, saf karıştırmanın ne kadar 'kötü' olduğu? Daha spesifik olarak, izin $P(n)$ tüm permütasyon grubu olabilir ve $C(\rho)$ elde edilen üretmek naif algoritma üzerinden yolların sayısı olduğu bir permütasyon $\rho\in P(n)$ , fonksiyonlarının asimptotik davranış nedir

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

ve

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

Başlıca faktör, bu değerleri “normalleştirmek” tir: eğer saf karıştırma, "asimptotik olarak iyi" ise, o zaman

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

(Gördüğüm bazı bilgisayar simülasyonlarına dayanarak) gerçek değerlerin 1'den uzakta olduğundan şüpheleniyorum, ancak lim M ( n ) sonlu ise veya lim m ( n ) 0'dan uzakta ise bile biliniyor mu? Bu miktarların davranışları hakkında bilinenler nelerdir?$\lim M(n)$$\lim m(n)$

İndüksiyonla ρ n = ( 2 , 3 , 4 , … , n , 1 ) permütasyonunun C ( ρ n ) = 2 n - 1 olan bir örnek olduğunu göstereceğiz . Bu en kötü durumsa , ilk n için olduğu gibi ( OEIS dizisi A192053 için notlara bakın ), sonra m ( n ) ≈ ( 2 / e ) $\rho_n = (2,3,4,\ldots, n,1)$$C(\rho_n) = 2^{n-1}$$n$$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, $C(\rho_n) = 2C(\rho_{n-1}) = 2^{n-1}$.

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 [1]) 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 [1] 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.

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