3 个连续项目的堆栈（不允许来自原始序列）
我假设 shuffle(n) 的结果用作 shuffle(n+1) 的起始序列。这不是微不足道的，因为使用相同的起始序列只会导致 7 个有效组合{0, 1, 2, 3}。在应用程序启动时使用固定的启动顺序意味着第一次 shuffle 只能是这 7 个中的一个（可能足够多）。
加扰器类：
Public Class Scrambler
Private rand As Random
Public Sub New()
rand = New Random
End Sub
' FY In-Place integer array shuffle
Public Sub Shuffle(items() As Integer)
Dim tmp As Integer
Dim j As Integer
' hi to low, so the rand result is meaningful
For i As Integer = items.Length - 1 To 0 Step -1
j = rand.Next(0, i + 1) ' NB max param is EXCLUSIVE
tmp = items(j)
' swap j and i
items(j) = items(i)
items(i) = tmp
Next
End Sub
' build a list of bad sequences
' fullfils the "stack of 3 sequential items (from the original sequence..." requirement
' nsize - allows for the "(or any number ..." portion though scanning for
' a series-of-5 may be fruitless
Public Function GetBadList(source As Integer(),
nSize As Integer) As List(Of String)
Dim BList As New List(Of String)
Dim badNums(nSize - 1) As Integer
For n As Integer = 0 To source.Length - nSize
Array.Copy(source, n, badNums, 0, badNums.Length)
BList.Add(String.Join(",", badNums))
Array.Clear(badNums, 0, badNums.Length)
Next
Return BList
End Function
Public Function ScrambleArray(items() As Integer, badSize As Integer) As Integer()
' FY is an inplace shuffler, make a copy
Dim newItems(items.Length - 1) As Integer
Array.Copy(items, newItems, items.Length)
' flags
Dim OrderOk As Boolean = True
Dim AllDiffPositions As Boolean = True
Dim BadList As List(Of String) = GetBadList(items, badSize)
' build the bad list
Do
Shuffle(newItems)
' check if they all moved
AllDiffPositions = True
For n As Integer = 0 To items.Length - 1
If newItems(n) = items(n) Then
AllDiffPositions = False
Exit For
End If
Next
' check for forbidden sequences
If AllDiffPositions Then
Dim thisVersion As String = String.Join(",", newItems)
OrderOk = True
For Each s As String In BadList
If thisVersion.Contains(s) Then
OrderOk = False
Exit For
End If
Next
End If
Loop Until (OrderOk) And (AllDiffPositions)
Return newItems
End Function
End Class
测试代码/使用方法：
' this series is only used once in the test loop
Dim theseItems() As Integer = {0, 1, 2, 3}
Dim SeqMaker As New Scrambler ' allows one RNG used
Dim newItems() As Integer
' reporting
Dim rpt As String = "{0} Before: {1} After: {2} time:{3}"
ListBox1.Items.Clear()
For n As Integer = 0 To 1000
sw.Restart()
newItems = SeqMaker.ScrambleArray(theseItems, 3) ' bad series size==3
sw.Stop()
ListBox1.Items.Add(String.Format(rpt, n.ToString("0000"), String.Join(",", theseItems),
String.Join(",", newItems), sw.ElapsedTicks.ToString))
Console.WriteLine(rpt, n.ToString("0000"), String.Join(",", theseItems),
String.Join(",", newItems), sw.ElapsedTicks.ToString)
' rollover to use this result as next start
Array.Copy(newItems, theseItems, newItems.Length)
Next
一个项目永远不会在它的原始位置上，这在小集合上是有意义的。但对于较大的集合，它排除了大量的合法洗牌（>60%）；在某些情况下，仅仅因为 1 个项目在同一个位置。
Start: {1,2,8,4,5,7,6,3,9,0}
Result: {4,8,2,0,7,1,6,9,5,3}
由于'6'而失败，但它真的是无效的洗牌吗？三个系列规则很少出现在较大的集合中（<1%），这可能是浪费时间。
如果没有列表框和控制台报告（以及一些未显示的分发集合），它会非常快。
Std Shuffle, 10k iterations, 10 elements: 12ms (baseline)
Modified, 10k iterations, 10 elements: 91ms
Modified, 10k iterations, 04 elements: 48ms
修改后的洗牌依赖于重新洗牌，我知道这不会很耗时。因此，当 Rule1 OrElse Rule2 失败时，它只是重新洗牌。10 个元素的 shuffle 必须实际执行 28k shuffle 才能获得 10,000 个“好”的。4 元素 shuffle 实际上具有更高的拒绝率，因为很少的项目（34,000 次拒绝）更容易打破规则。
这不像随机分布那样让我感兴趣，因为如果这些“改进”引入了偏差，那就不好了。10k 4 元素分布：
seq: 3,2,1,0 count: 425
seq: 1,0,2,3 count: 406
seq: 3,2,0,1 count: 449
seq: 2,3,1,0 count: 424
seq: 0,1,3,2 count: 394
seq: 3,0,2,1 count: 371
seq: 1,2,3,0 count: 411
seq: 0,3,1,2 count: 405
seq: 2,1,3,0 count: 388
seq: 0,3,2,1 count: 375
seq: 2,0,1,3 count: 420
seq: 2,1,0,3 count: 362
seq: 3,0,1,2 count: 396
seq: 1,2,0,3 count: 379
seq: 0,1,2,3 count: 463
seq: 1,3,0,2 count: 398
seq: 2,3,0,1 count: 443
seq: 1,0,3,2 count: 451
seq: 3,1,2,0 count: 421
seq: 2,0,3,1 count: 487
seq: 0,2,3,1 count: 394
seq: 3,1,0,2 count: 480
seq: 0,2,1,3 count: 444
seq: 1,3,2,0 count: 414
通过较小的迭代（1K），您可以看到与修改后的形式相比更均匀的分布。但是，如果您拒绝某些合法的洗牌，这是可以预料的。
十个元素的分布是不确定的，因为有太多的可能性（360 万次洗牌）。也就是说，在 10k 次迭代中，往往会有大约 9980 个系列，其中 12-18 的计数为 2。