在VB.NET中使用停止条件进行递归

N.T.C 发表于 Dev

NTC

我已经在VB.NET中编写了一个函数，该函数使用Bisect Method来查找任意函数的根。

[给定一个连续函数f（x），如果f（x1）和f（x2）具有不同的符号，则它保证在x1和x2之间至少存在一个根。的Bisect Method缩小范围[X1，X2]，直到找到一个根。]

我用两种不同的方式编写了该函数：一种使用while循环，另一种使用递归。

在while循环方法中，我可以轻松地跟踪迭代次数，例如，在经过1000次循环之后，如果找不到根，则函数将停止并退出。

我的问题是：如何在仍能保持尾部调用优化的同时用递归方法插入这种停止条件？先感谢您。

方法1：While循环

Public Function FindOneRoot(lowerBound As Double, upperBound As Double, _
                             tolerance As Double, ByRef Root As Double, _
                             theFunction As Func(Of Double, Double)) As Boolean
  Dim flower As Double = theFunction(lowerBound)
  Dim fupper As Double = theFunction(upperBound)
  Root = (lowerBound + upperBound) / 2
  Dim froot As Double = theFunction(Root)
  Dim count As Integer = 0
  While Math.Abs(froot) > tolerance
     If froot * flower > 0 Then
        lowerBound = Root
     Else
        upperBound = Root
     End If
     Root = (lowerBound + upperBound) / 2
     froot = theFunction(Root)
     count += 1 'keep track of the loops

     If count >= 1000 Then 'stop looping after 1000 iterations
        Exit While
     End If
  End While

  If count < 1000 Then
     Return True
  Else
     Return False
  End If

End Function

方法2：递归

 Public Function FindOneRoot(x1 As Double, x2 As Double, tolerance As Double, _
                                    theFunction As Func(Of Double, Double)) As Double
  Dim x As Double = (x1 + x2) / 2
  If Math.Abs(theFunction(x)) < tolerance Then
     Return x 'found a root
  ElseIf theFunction(x1) * theFunction(x) < 0 Then
     Return FindOneRoot(x1, x, tolerance, Root, theFunction)
  Else
     Return FindOneRoot(x, x2, tolerance, Root, theFunction)
  End If
End Function

沙沙语

递归使用相同条件。但是在函数外部声明计数

添加另一个参数作为Count传递计数并检查函数内部

 Public Function FindOneRoot(x1 As Double, x2 As Double, tolerance As Double, 
 ByRef Root As Double,theFunction As Func(Of Double, Double),
 Count as integer) As Boolean

    If count >= 1000 Then 'stop looping after 1000 iterations
       Return False: Exit Function
    End If

   Dim x As Double = (x1 + x2) / 2
   If Math.Abs(theFunction(x)) < tolerance Then
      Return x 'found a root
   ElseIf theFunction(x1) * theFunction(x) < 0 Then
     Return FindOneRoot(x1, x, tolerance, Root, theFunction,count +1)
  Else
  Return FindOneRoot(x, x2, tolerance, Root, theFunction,count +1)
  End If
  End Function