CODEWITHSUNDEEP
vb.netrandommodulo
SSpoke
SSpoke

Reputation: 5836

How do I reverse the Mod operator

I have a simple function that handles overflows with modulus operator

   Private Function RandomizeBlock(seed As Integer, ByVal block() As Byte) As Byte()
        Dim Generator As System.Random = New System.Random(seed)
        Dim newblock(255) As Byte
        Dim i As Integer = 0
        For i = 0 To block.Length - 1
            newblock(i) = (block(i) + Generator.Next(0, 256)) Mod 256
        Next

        Return newblock
    End Function

How do I undo the randomization done to a block?

I know mod works like this:

253,254,255,0,1,2,3,4 overwraps on 0.

Do can I find the inverse of reverse here?

rndValue = Generator.Next(0, 256)
reverse_1 = ((256 - rndValue) + block(i)) Mod 256
reverse_2 = ((256 + rndValue) - block(i)) Mod 256

Upvotes: 0

Views: 1111

Answers (1)

Nico Schertler
Nico Schertler

Reputation: 32627

If you know the random value, then reconstructing the original value is very simple.

You just have to keep in mind that working modulo p, you don't have actual numbers but remainder classes. The first p natural numbers are usually used as the representatives of these classes. Luckily, subtraction and addition are perfectly compatible with remainder classes.

The Mod implementation of VB converts any positive number to the representative of its remainder class. However, it can not do it on negative numbers. You have to do it on your own.

Long story short, this is the code:

Dim reverse As Integer = block(i) - rndValue;
If reverse < 0 Then reverse = reverse + 256 'Convert to representative of remainder class

Upvotes: 1

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