generating a series for all 2 power n combinations in

Posted on 2011-09-28
Last Modified: 2012-05-12
I have n items and each item can be assigned a 1 or a 2. So I would like to get the matrix result that would generate all possible combinations.
For eg. if n= 3 , then the possible outcomes are : I  need an algorithm that can generate this series for n . Please help thanks. ideally i would like to store the result in a datatable

1 1 1
1 1 2
1 2 1
2 1 1
2 1 2
1 2 2
2 2 1
2 2 2
Question by:TrialUser
  • 3
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LVL 12

Assisted Solution

jagssidurala earned 100 total points
ID: 36719097
Logic is

2^0 2^0 2^0
2^0 2^0 2^1
2^0 2^1 2^0
2^0 2^1 2^1
2^1 2^0 2^0
2^1 2^0 2^1
2^1 2^1 2^0
2^1 2^1 2^1

The power to the num 2 is binary seris


Accepted Solution

Ambusy earned 400 total points
ID: 36719344
        Dim HighestValue As Integer = 2 ' max value
        Dim NrOfValues As Integer = 3 ' nr of values in one result
        Dim Values(NrOfValues) As Integer
        Dim i As Integer
        For i = 0 To NrOfValues - 1
            Values(i) = 1
        Values(NrOfValues - 1) = 0 ' to generate first as ALL 1
        For i = 1 To HighestValue ^ NrOfValues
            Values(NrOfValues - 1) += 1
            For j As Integer = NrOfValues - 1 To 0 Step -1
                If Values(j) > HighestValue Then
                    Values(j) = 1
                    Values(j - 1) += 1
                End If
            Dim Result As String = ""
            For j As Integer = 0 To NrOfValues - 1
                Result = Result & CStr(Values(j))

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Author Comment

ID: 36814309
Thank you both for the suggestoins. Both the suggestions are great and made me understand the concept behind.

Ambusy, the code you provided just works great and does exactly what I wanted.

Just one question, if I change     HighestValue , and NrOfValues  then this logic should just work for any number of items and any number of possible values for each item right?

thanks again.
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Expert Comment

ID: 36817258
Exactly, that's why I wrote them as variables. I tested them with values in a textbox to be certain it workes.

Expert Comment

ID: 36896300
If my solution is correct, please accept it.

Author Closing Comment

ID: 36948593
the solution works great. does exactly what I want

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