正如@FelixCQ提到,你正在寻找的洗牌被称为紊乱。构建均匀随机分布的紊乱并不是一个小问题,但是一些结果在文献中是已知的。排除错误的最明显的方法是拒绝方法:使用像Fisher-Yates这样的算法生成均匀随机分布的排列,然后排除固定点的排列。该程序的平均运行时间是e * n + o(n),其中e是欧拉常数2.71828 ...这可能适用于您的情况。
产生紊乱的另一个主要方法是使用递归算法。但是,与Fisher-Yates不同,我们有两个分支:算法列表中的最后一项可以与另一个项目交换(即双周期的一部分),也可以是更大周期的一部分。所以在每一步中,递归算法都必须进行分支以产生所有可能的紊乱。此外,是否采取一个分支或另一个分支的决定必须以正确的概率进行。
设D(n)为n个项目的排列次数。在每个阶段,最后一个项目到两个周期的分支数目是(n-1)D(n-2),并且最后一个项目到较大周期的分支数目是(n-1)D(n -1)。这为我们提供了计算排列数的递归方法,即D(n)=(n-1)(D(n-2)+ D(n-1)),并给出了分支到二(n-1)D(n-2)/ D(n-1)。
现在我们可以通过确定最后一个元素属于哪种类型的循环来构造紊乱,将最后一个元素交换到n-1个其他位置之一并重复。然而,跟踪所有分支可能很复杂,因此在2008年,一些研究人员利用这些想法开发了一种简化的算法。您可以在http://www.cs.upc.edu/~conrado/research/talks/analco08.pdf上看到演练。该算法的运行时间与2n + O(log^2 n)成正比,比拒绝方法提高了36%的速度。
我已经在Java中实现了他们的算法。使用长期股票可以支撑22点左右。使用BigIntegers将算法扩展到n = 170左右。使用BigInteger和BigDecimals将算法扩展到n = 40000左右(限制取决于程序其余部分的内存使用情况)。
package io.github.edoolittle.combinatorics;
import java.math.BigInteger;
import java.math.BigDecimal;
import java.math.MathContext;
import java.util.Random;
import java.util.HashMap;
import java.util.TreeMap;
public final class Derangements {
// cache calculated values to speed up recursive algorithm
private static HashMap<Integer,BigInteger> numberOfDerangementsMap
= new HashMap<Integer,BigInteger>();
private static int greatestNCached = -1;
// load numberOfDerangementsMap with initial values D(0)=1 and D(1)=0
static {
numberOfDerangementsMap.put(0,BigInteger.valueOf(1));
numberOfDerangementsMap.put(1,BigInteger.valueOf(0));
greatestNCached = 1;
}
private static Random rand = new Random();
// private default constructor so class isn't accidentally instantiated
private Derangements() { }
public static BigInteger numberOfDerangements(int n)
throws IllegalArgumentException {
if (numberOfDerangementsMap.containsKey(n)) {
return numberOfDerangementsMap.get(n);
} else if (n>=2) {
// pre-load the cache to avoid stack overflow (occurs near n=5000)
for (int i=greatestNCached+1; i<n; i++) numberOfDerangements(i);
greatestNCached = n-1;
// recursion for derangements: D(n) = (n-1)*(D(n-1) + D(n-2))
BigInteger Dn_1 = numberOfDerangements(n-1);
BigInteger Dn_2 = numberOfDerangements(n-2);
BigInteger Dn = (Dn_1.add(Dn_2)).multiply(BigInteger.valueOf(n-1));
numberOfDerangementsMap.put(n,Dn);
greatestNCached = n;
return Dn;
} else {
throw new IllegalArgumentException("argument must be >= 0 but was " + n);
}
}
public static int[] randomDerangement(int n)
throws IllegalArgumentException {
if (n<2)
throw new IllegalArgumentException("argument must be >= 2 but was " + n);
int[] result = new int[n];
boolean[] mark = new boolean[n];
for (int i=0; i<n; i++) {
result[i] = i;
mark[i] = false;
}
int unmarked = n;
for (int i=n-1; i>=0; i--) {
if (unmarked<2) break; // can't move anything else
if (mark[i]) continue; // can't move item at i if marked
// use the rejection method to generate random unmarked index j < i;
// this could be replaced by more straightforward technique
int j;
while (mark[j=rand.nextInt(i)]);
// swap two elements of the array
int temp = result[i];
result[i] = result[j];
result[j] = temp;
// mark position j as end of cycle with probability (u-1)D(u-2)/D(u)
double probability
= (new BigDecimal(numberOfDerangements(unmarked-2))).
multiply(new BigDecimal(unmarked-1)).
divide(new BigDecimal(numberOfDerangements(unmarked)),
MathContext.DECIMAL64).doubleValue();
if (rand.nextDouble() < probability) {
mark[j] = true;
unmarked--;
}
// position i now becomes out of play so we could mark it
//mark[i] = true;
// but we don't need to because loop won't touch it from now on
// however we do have to decrement unmarked
unmarked--;
}
return result;
}
// unit tests
public static void main(String[] args) {
// test derangement numbers D(i)
for (int i=0; i<100; i++) {
System.out.println("D(" + i + ") = " + numberOfDerangements(i));
}
System.out.println();
// test quantity (u-1)D_(u-2)/D_u for overflow, inaccuracy
for (int u=2; u<100; u++) {
double d = numberOfDerangements(u-2).doubleValue() * (u-1)/
numberOfDerangements(u).doubleValue();
System.out.println((u-1) + " * D(" + (u-2) + ")/D(" + u + ") = " + d);
}
System.out.println();
// test derangements for correctness, uniform distribution
int size = 5;
long reps = 10000000;
TreeMap<String,Integer> countMap = new TreeMap<String,Integer>();
System.out.println("Derangement\tCount");
System.out.println("-----------\t-----");
for (long rep = 0; rep < reps; rep++) {
int[] d = randomDerangement(size);
String s = "";
String sep = "";
if (size > 10) sep = " ";
for (int i=0; i<d.length; i++) {
s += d[i] + sep;
}
if (countMap.containsKey(s)) {
countMap.put(s,countMap.get(s)+1);
} else {
countMap.put(s,1);
}
}
for (String key : countMap.keySet()) {
System.out.println(key + "\t\t" + countMap.get(key));
}
System.out.println();
// large random derangement
int size1 = 1000;
System.out.println("Random derangement of " + size1 + " elements:");
int[] d1 = randomDerangement(size1);
for (int i=0; i<d1.length; i++) {
System.out.print(d1[i] + " ");
}
System.out.println();
System.out.println();
System.out.println("We start to run into memory issues around u=40000:");
{
// increase this number from 40000 to around 50000 to trigger
// out of memory-type exceptions
int u = 40003;
BigDecimal d = (new BigDecimal(numberOfDerangements(u-2))).
multiply(new BigDecimal(u-1)).
divide(new BigDecimal(numberOfDerangements(u)),MathContext.DECIMAL64);
System.out.println((u-1) + " * D(" + (u-2) + ")/D(" + u + ") = " + d);
}
}
}
将每个项目随机放入另一个位置。有一个很小的机会,你不能找到最后一个位置,但刚刚开始。 – adrianm
https://en.wikipedia.org/wiki/Sattolo's_algorithm – Bergi
有限递推将从数学上证明算法的工作原理:在迭代i结束时,位置i处的元素不再是原始元素。当迭代n-2时,数据[n-2]自动与数据[n-1]混洗。因此,如果data [n-1]仍然保持其原始值,则在最后一次迭代时它将被交换。数据[n-1]也是如此。 – Rerito