Question

Combination, permutation problem...

Asked by: eghtebas

I have three boxes to have a single letter A, B, or C to be wriiten on each. How many different combinations I will end up with. The order is important: The letters can be repeated.

[A] [A] [A]
[A] [A] [B]
[A] [A] [C]
[A] [B] [C]
etc.

If you can code a routine in VB to come up with the combinations, it will be great.

Thank you.

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Asked On
2009-10-27 at 09:49:19ID24847896
Topics

Math & Science

,

Access Coding/Macros

,

Visual Basic Programming

Participating Experts
3
Points
500
Comments
12

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Answers

 

by: matthewspatrickPosted on 2009-10-27 at 09:56:09ID: 25674702

The answer is 3^3, which is 27.

Sub Make()

    Dim arr1 As Variant, arr2 As Variant, arr3 As Variant
    Dim x1 As Variant, x2 As Variant, x3 As Variant

    arr1 = Array("A", "B", "C")
    arr2 = Array("A", "B", "C")
    arr3 = Array("A", "B", "C")

    For Each x1 In arr1
        For Each x2 In arr2
            For Each x3 In arr3
                Debug.Print x1 & "," & x2 & "," & x3
            Next
        Next
    Next

    MsgBox "Done"

End Sub

 

by: eghtebasPosted on 2009-10-27 at 10:02:12ID: 25674781

re:> The answer is 3^3, which is 27

27 is correct if we cannot have

[A] [A] [A]

Maybe 3^9 = 19,683 but is is way high.

Mike

 

by: eghtebasPosted on 2009-10-27 at 10:03:49ID: 25674802

3^3 allows letter A, for example, to be selected only once with each try.

3(3^3)=27 more likely. But, I am not sure.

Mike

 

by: thenelsonPosted on 2009-10-27 at 10:05:21ID: 25674824

3 ^ 3 = 27

Dim i As Long, j As Long, k As Long

For i = Asc("A") To Asc("C")
   For j = Asc("A") To Asc("C")
      For k = Asc("A") To Asc("C")
          Debug.Print "[" & Chr(i) & "] [" & Chr(j) & "] [" & Chr(k) & "]"
      Next k
   Next j
Next i

 

by: matthewspatrickPosted on 2009-10-27 at 10:29:21ID: 25675062

Mike,

With respect, your follow-up does not make sense.  If there are three places, and each place can have one
of three characters, then the number of strings we can generate is 3 x 3 x 3, or 27.

My code produces each string according to the conditions outlined in your question.

Patrick

 

by: eghtebasPosted on 2009-10-27 at 10:39:33ID: 25675200

Patrick,

My second foolow-up (guess) was 27 also. I will run the code and get back to you soon.

Mike

 

by: matthewspatrickPosted on 2009-10-27 at 10:40:56ID: 25675221

Mike,

Yes, sorry for not noticing that :)

My code and Nelson's do essentially the same thing: we both set up three nested loops, and just iterate through each
place in the output.

Patrick

 

by: eghtebasPosted on 2009-10-27 at 10:41:42ID: 25675231

Sorry,

I meant to say 3(3^3)=81

 

by: eghtebasPosted on 2009-10-27 at 10:44:22ID: 25675264

I am using A, B, C as illustration. The actual characters will be <, >, and N. So, your version might be a good start. But, Nelson's code is very smart also.

Mike

 

by: eghtebasPosted on 2009-10-27 at 10:49:58ID: 25675336

Patrick,

The code looks good and produces the combinations I want.

Thanks,

Mike

A,A,A
A,A,B
A,A,C
A,B,A
A,B,B
A,B,C
A,C,A
A,C,B
A,C,C
B,A,A
B,A,B
B,A,C
B,B,A
B,B,B
B,B,C
B,C,A
B,C,B
B,C,C
C,A,A
C,A,B
C,A,C
C,B,A
C,B,B
C,B,C
C,C,A
C,C,B
C,C,C

                                              
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by: Idle_MindPosted on 2009-10-27 at 11:33:13ID: 25675799

"The actual characters will be <, >, and N."

Giving you:

<<<
<<>
<<N
<><
<>>
<>N
<N<
<N>
<NN
><<
><>
><N
>><
>>>
>>N
>N<
>N>
>NN
N<<
N<>
N<N
N><
N>>
N>N
NN<
NN>
NNN

Produced by:

Public Class Form1
 
    Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Dim Rev As New Revision("<>N", "<<<")
        Dim curRev As String
 
        curRev = Rev.CurrentRevision
        While curRev.Length = 3
            Debug.Print(curRev)
            curRev = Rev.NextRevision
        End While
    End Sub
 
End Class
 
Public Class Revision
 
    Private chars As String
    Private values() As Char
    Private curRevision As System.Text.StringBuilder
 
    Public Sub New()
        Me.DefaultRevision()
    End Sub
 
    Public Sub New(ByVal validChars As String)
        If validChars.Length > 0 Then
            chars = validChars.ToUpper()
            values = chars.ToCharArray()
            curRevision = New System.Text.StringBuilder(values(0))
        Else
            Me.DefaultRevision()
        End If
    End Sub
 
    Public Sub New(ByVal validChars As String, ByVal startingRevision As String)
        Me.New(validChars)
        curRevision = New System.Text.StringBuilder(startingRevision.ToUpper())
        Dim i As Integer
        For i = 0 To curRevision.Length - 1
            If Array.IndexOf(values, curRevision.Chars(i)) = -1 Then
                curRevision = New System.Text.StringBuilder(values(0))
                MessageBox.Show("Revision has been reset." & vbCrLf & "Current Revision = " & Me.CurrentRevision, "Starting Revision contains an Invalid Character", MessageBoxButtons.OK, MessageBoxIcon.Exclamation)
                Exit For
            End If
        Next
    End Sub
 
    Private Sub DefaultRevision()
        chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        values = chars.ToCharArray
        curRevision = New System.Text.StringBuilder(values(0))
    End Sub
 
    Public ReadOnly Property ValidChars() As String
        Get
            Return chars
        End Get
    End Property
 
    Public ReadOnly Property CurrentRevision() As String
        Get
            Return curRevision.ToString()
        End Get
    End Property
 
    Public Function NextRevision(Optional ByVal numRevisions As Integer = 1) As String
        Dim forward As Boolean = (numRevisions > 0)
        numRevisions = Math.Abs(numRevisions)
        Dim i As Integer
        For i = 1 To numRevisions
            If forward Then
                Me.Increment()
            Else
                Me.Decrement()
            End If
        Next
        Return Me.CurrentRevision
    End Function
 
    Private Sub Increment()
        Dim curChar As Char = curRevision.Chars(curRevision.Length - 1)
        Dim index As Integer = Array.IndexOf(values, curChar)
        If index < (chars.Length - 1) Then
            index = index + 1
            curRevision.Chars(curRevision.Length - 1) = values(index)
        Else
            curRevision.Chars(curRevision.Length - 1) = values(0)
            Dim i As Integer
            Dim startPosition As Integer = curRevision.Length - 2
            For i = startPosition To 0 Step -1
                curChar = curRevision.Chars(i)
                index = Array.IndexOf(values, curChar)
                If index < (values.Length - 1) Then
                    index = index + 1
                    curRevision.Chars(i) = values(index)
                    Exit Sub
                Else
                    curRevision.Chars(i) = values(0)
                End If
            Next
            curRevision.Insert(0, values(0))
        End If
    End Sub
 
    Private Sub Decrement()
        Dim curChar As Char = curRevision.Chars(curRevision.Length - 1)
        Dim index As Integer = Array.IndexOf(values, curChar)
        If index > 0 Then
            index = index - 1
            curRevision.Chars(curRevision.Length - 1) = values(index)
        Else
            curRevision.Chars(curRevision.Length - 1) = values(values.Length - 1)
            Dim i As Integer
            Dim startPosition As Integer = curRevision.Length - 2
            For i = startPosition To 0 Step -1
                curChar = curRevision.Chars(i)
                index = Array.IndexOf(values, curChar)
                If index > 0 Then
                    index = index - 1
                    curRevision.Chars(i) = values(index)
                    Exit Sub
                Else
                    curRevision.Chars(i) = values(values.Length - 1)
                End If
            Next
            curRevision.Remove(0, 1)
            If curRevision.Length = 0 Then
                curRevision.Insert(0, values(0))
            End If
        End If
    End Sub
 
End Class

                                              
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by: Idle_MindPosted on 2009-10-27 at 11:34:22ID: 25675816

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