Scrambled Words Loop

[OpticTygre]

I need to write a loop that prints all the combination possibilities of a
character array. Basically, taking a scrambled word, or a regular word, and
printing out all the combinations. The letters in the word will only be
used once. It sounds easy, but for some reason, I can't get the logic
straight in my head. The formula is as follows:

If n = the number or letters in the word, the C = the number of
combinations, so:

C = (n)(n - 1)(n - 2)....(n - (n - 1))

Or

C = (n)(n - 1)(n - 2)...(1)

I'm totally screwed up on the logic behind writing a loop (or nested loop)
that would take care of this. Any help is appreciated.

[OpticTygre]

Let's make this even harder for all you hardcore guys. The formula below
only works when each letter is distinct. It will not work if there are
duplicate letters. For example:

There are 24 different combinations for ABCD.

However, if I have AABB, then there are quite a few less:

AABB
ABAB
ABBA
BBAA
BABA
BAAB

So, the problem now becomes, how do I cycle through loops and print out all
DISTINCT possible combinations?
"OpticTygre" <op********@adelphia.net> wrote in message
news:Oo********************@adelphia.com...

I need to write a loop that prints all the combination possibilities of a
character array. Basically, taking a scrambled word, or a regular word,
and printing out all the combinations. The letters in the word will only
be used once. It sounds easy, but for some reason, I can't get the logic
straight in my head. The formula is as follows:

If n = the number or letters in the word, the C = the number of
combinations, so:

C = (n)(n - 1)(n - 2)....(n - (n - 1))

Or

C = (n)(n - 1)(n - 2)...(1)

I'm totally screwed up on the logic behind writing a loop (or nested loop)
that would take care of this. Any help is appreciated.

[Lance Wynn]

Hi, here's a little code that I put together, it seems to work fairly well.
I just used a hashtable to remove duplicates: I'm new to VB.NET, so there
is probably some more efficient way to do this. (I didn't comment any of
the code, but I used descriptive variable names, so it should be easy enough
to read.


Imports System.Collections.Hashtable

Public Class Form1
Inherits System.Windows.Forms.Form

Private mScrambledWords As Hashtable = New Hashtable

Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As
System.EventArgs) Handles MyBase.Load
Dim scrambledword As String
Dim i As Long
Dim currentWord As String
Dim currentEntry As DictionaryEntry
Dim oEnum As IEnumerator
scrambledword = "ABCDEFF"
ScrambleWord("", scrambledword)

oEnum = mScrambledWords.GetEnumerator()

'Write out the words
Debug.WriteLine(mScrambledWords.Count & " Unique Words Found.")
Me.Text = mScrambledWords.Count & " Unique Words Found."
Do Until oEnum.MoveNext = False
currentEntry = oEnum.Current
currentWord = currentEntry.Value
Debug.WriteLine(currentWord)
Loop
End Sub

Private Sub ScrambleWord(ByVal baseLetters As String, ByVal
ScrambledLetters As String)
Dim c As String
Dim subScrambledLetters As String
Dim i As Integer
For i = 1 To Len(ScrambledLetters)
'take the current letter in position 1, and then scramble the
remaining letters
c = Mid(ScrambledLetters, i, 1)
subScrambledLetters = excludeLetter(ScrambledLetters, i)
ScrambleWord(baseLetters & c, subScrambledLetters)
If Len(subScrambledLetters) = 0 Then
addWordToHash(baseLetters & c)
End If
Next
End Sub

Private Sub addWordToHash(ByVal word As String)
Try
mScrambledWords.Add(word, word)
Catch ex As Exception
'Debug.WriteLine("Duplicate Word Found: " & word)
End Try
End Sub

Private Function excludeLetter(ByVal sString As String, ByVal
letterIndex As Integer) As String
If Len(sString) <= 1 Then Exit Function
excludeLetter = sString.Remove(letterIndex - 1, 1)
End Function

End Class

[Chris Dunaway]

To actually calculate C (which I don't think that is what you're
looking for) you could use a recursive function (untested):

Public Function Calc(n as integer) As Integer
If n > 1 Then
Return Calc(n-1)
Else
Return 1
Endif
End Function

Perhaps a different recursive function might help you in getting the
possible combinations of the word. Here is a link that may be useful
to you:

http://mathforum.org/library/drmath/view/60844.html

[OpticTygre]

Hi Chris,

Actually, calculating C is useful, however, the below function only works if
all the letters are different.

'ABCD' has 24 solutions: n! = n(n-1)(n-2)(n-3) = 4(3)(2)(1) = 24

But 'AABB' only has 6 distinct solutions:

AABB
ABAB
ABBA
BBAA
BABA
BAAB

So after lots of math toiling, I came up with the following:

If 'a' represents the count of the letter 'A', and 'b' represents the count
of the letter 'B' and so on, then:

(a + b + c.....)! / (a! * b! * c!....)

Or, more simply

n! / (a! * b! * c!...)

So in the case of 'AABB':

C = (2 + 2)! / (2! * 2!)
C = 4! / 4
C = 24 / 4
C = 6

This formula should work for finding distinct combinations, whether or not
there are duplicate letters.

-Jason
"Chris Dunaway" <du******@gmail.com> wrote in message
news:11**********************@g14g2000cwa.googlegr oups.com...

To actually calculate C (which I don't think that is what you're
looking for) you could use a recursive function (untested):

Public Function Calc(n as integer) As Integer
If n > 1 Then
Return Calc(n-1)
Else
Return 1
Endif
End Function

Perhaps a different recursive function might help you in getting the
possible combinations of the word. Here is a link that may be useful
to you:

http://mathforum.org/library/drmath/view/60844.html