VB.NET - permutations - Asked By George n t on 02-Oct-08 02:10 AM

how to generate the permutation  of any set of numbers

11,22,23,45,34

without repetition in vb.net

Hi,

I used following code to generate the permuation of given number.
Suppose number given : 123

Output will be:
123
132
213
231
312
321

Code
----
' Generate permutations of the values in the
' values collection.
' Return the result through a collection of
' collections that each hold a permutation.
Private Function GeneratePermutations(ByVal values As _
    Collection) As Collection
    Dim results As New Collection

    ' See if there is only one value.
    If values.Count = 1 Then
        ' Return a collection containing one
        ' permutation equal to the single value.
        results.Add(New Collection)
        results.Item(1).Add(values.Item(1))
        Return results
    End If

    ' Build permutations starting with
    ' each possible first item.
    results = New Collection
    Dim num_values As Integer = values.Count
    For i As Integer = 1 To num_values
        ' Save this value.
        Dim first_value As Object = values.Item(i)

        ' Remove the item.
        values.Remove(i)

        ' Generate the permutations of the
        ' remaining values.
        Dim new_permutations As Collection = _
            GeneratePermutations(values)

        ' Make permutations by adding first_value
        ' to the beginning of each of the new
        ' permutations.
        For j As Integer = 1 To new_permutations.Count
            ' Add the first item.
            Dim new_result As New Collection
            new_result.Add(first_value)

            ' Add the rest of the items in the jth
            ' new permutation.
            For k As Integer = 1 To _
                new_permutations(j).Count
                new_result.Add(new_permutations(j).Item(k))
            Next k

            ' Add this new permutation to the results.
            results.Add(new_result)
        Next j

        ' Restore the removed value.
        If i > values.Count Then
            values.Add(first_value)
        Else
            values.Add(first_value, , i)
        End If
    Next i

    ' Return the results.
    Return results
End Function

Regards,
Megha

Hi,

You can generate permutation by using following program which uses recursive method.

Private m_NumValues As Integer
Private m_Used() As Integer
Private m_CurrentSolution() As Integer

Private Sub EnumerateValues(ByVal index As Integer)
Dim result As String
Dim i As Integer

    ' See if there are any values left to try.
    If index > m_NumValues Then
        ' All values are m_Used.
        ' Get a string for the current solution.
        For i = 1 To m_NumValues
            result = result & Format$(m_CurrentSolution(i)) _
                & " "
        Next i

        ' Add the current solution to the list.
        txtResults.Text = txtResults.Text & _
            result & vbCrLf
        Exit Sub
    End If

    ' Examine each value.
    For i = 1 To m_NumValues
        ' See if this value has been m_Used yet.
        If Not m_Used(i) Then
            ' It is unm_Used. Try using it.
            m_Used(i) = True
            m_CurrentSolution(index) = i

            EnumerateValues index + 1

            m_Used(i) = False
        End If
    Next i
End Sub

Regards,
Megha

list = originalString.split('')
index = (0,0)
list = [""]
for iteration n in 1 to y:
  index = (index[1], len(list))
  for string s in list.subset(index[0] to end):
    for character c in originalString:
      list.add(s + c)

Private Function GeneratePermutations(ByVal values As _
    Collection) As Collection
    Dim results As New Collection

    ' See if there is only one value.
    If values.Count = 1 Then
        ' Return a collection containing one
        ' permutation equal to the single value.
        results.Add(New Collection)
        results.Item(1).Add(values.Item(1))
        Return results
    End If

    ' Build permutations starting with
    ' each possible first item.
    results = New Collection
    Dim num_values As Integer = values.Count
    For i As Integer = 1 To num_values
        ' Save this value.
        Dim first_value As Object = values.Item(i)

        ' Remove the item.
        values.Remove(i)

        ' Generate the permutations of the
        ' remaining values.
        Dim new_permutations As Collection = _
            GeneratePermutations(values)

        ' Make permutations by adding first_value
        ' to the beginning of each of the new
        ' permutations.
        For j As Integer = 1 To new_permutations.Count
            ' Add the first item.
            Dim new_result As New Collection
            new_result.Add(first_value)

            ' Add the rest of the items in the jth
            ' new permutation.
            For k As Integer = 1 To _
                new_permutations(j).Count
                new_result.Add(new_permutations(j).Item(k))
            Next k

            ' Add this new permutation to the results.
            results.Add(new_result)
        Next j

        ' Restore the removed value.
        If i > values.Count Then
            values.Add(first_value)
        Else
            values.Add(first_value, , i)
        End If
    Next i

    ' Return the results.
    Return results
End Function

how to call the above  function & dispaly the permutaion of numbers   32,21,14

You main code should looks like below:

Sub MyMain()

    Dim Item As String;

    Dim Resulsts As Collection

    Results = GeneratePermutations(123);
    For Each Item In Results
        ListBox1.Items.Add(Item)
    Next

End Sub

Regards,

Megha

how to call the above  function & dispaly the permutaion of numbers   32,21,14

After finding the call method suggested did not work. The following might help some of you struggling with this.  The function looks for a collection of characters.  Convert your string into a collection of characters. Pass that collection in and then after the call put the collection back together.  I used a list instead of a collection for what I wanted to do.