128 bit hex keygen - vb.net

So my professor gave me a challenge to build a decoder that could break his special formula. It was described to be 32 characters in length, alphanumeric numeric when entered but then "it has a system... the first 106 bits must be 50% 1's and the rest 0's, the remaining 22 bits are basically a hash of the previous bits so that the key can be checked..." were his exact words. Sounds to me like a 128 bit encryption with a twist. I found the below but I need VB2010 or VS2010, this says php.
<?php
function string_random($characters, $length)
{
$string = '';
for ($max = mb_strlen($characters) - 1, $i = 0; $i < $length; ++ $i)
{
$string .= mb_substr($characters, mt_rand(0, $max), 1);
}
return $string;
}
// 128 bits is 16 bytes; 2 hex digits to represent each byte
$random_128_bit_hex = string_random('0123456789abcdef', 32);
// $random_128_bit_hex might be: '4374e7bb02ae5d5bc6d0d85af78aa2ce'
Would that work? Or does it need converting? Please help. Oh and thank you :)
I wasn't promised extra credit but either way I would like to surprise him.

So the first 106 bit are 26 character and the first half of the 27.
You have first of all encode somehow the number of 0 and 1, while building the string you need to keep an eye to the number. An idea would be to build a map like this:
0 = 0000 = -4
1 = 0001 = -2
2 = 0010 = -2
3 = 0011 = 0
4 = -2
5 = 0
6 = 0
7 = +2
8 = -2
9 = 0
a = 0
b = +2
c = 0
d = +2
e = +2
f = +4
then everytime you extract a new random number you check the number associated to it and add it to a variable
balanceOfOneAndZero
your objective is have balanceOfOneAndZero = 0 when you hit your 27th character.
to do that you need a control function, that takes current balanceOfOneAndZero, the proposed character proposedChar, and current string lenght currLenght.
Would be better to split the problem into two part. First is reaching the 26th character of the sequence with balanceOfOneAndZero between -2 and 2. Any other value is not acceptable, because your 27th character can have maximum two 1 or two 0 to completely balance the first 106 characters.
so your function should do something like (I'll write in sort of pseudo code since I don't have an IDE right now)
function checkNextLetter(Dim balanceOfOneAndZero As Integer, Dim proposedChar As Char,
Dim currentLenght as Integer) As Boolean
If( ((26 - currentLenght - 1) * 4 + 2) < MOD(Map.ValueOf(proposedChar) + balanceOfOneAndZero) ) Then
Return true
Else
Return false
ENd If
End function
This function basically check if accepting the new character will still make possible to Balance the number of 0 and 1 before the 26th character.
So your main function should have a loop every time it propose a new character, something like
proposedChar = new RandomChar
While (Not checkNextLetter(balanceOfOneAndZero, proposedChar, len(currentString))
proposedChar = new RandomChar
End While
currentString = currentString & proposedChar
this only until you hit the 26th character.
Than you have to check balanceOfOneAndZero, if its 2 you add a character that begin with 00, if it's 0 you can either have 10 or 01, if it's -2 you have to add a character that begin with 11.
After this I can't help you about the rest 22 character, since there are not enough information. You could brute force the rest
EDIT:
so to brute force the rest (il start from when you reach the 26th character):
Dim stringa1, stringa2, stringa3, stringa4 As String
If balanceOfOneAndZero = 2 Then
stringa1 = currentString & '0'
stringa2 = currentString & '1'
stringa3 = currentString & '2'
stringa4 = currentString & '3'
ELse If balanceOfOneAndZero = 0 Then
stringa1 = currentString & '4'
stringa2 = currentString & '5'
stringa3 = currentString & '6'
stringa4 = currentString & '7'
Else
stringa1 = currentString & 'c'
stringa2 = currentString & 'd'
stringa3 = currentString & 'e'
stringa4 = currentString & 'f'
End if
Function GenerateAllCombination(ByVal iLenght As Integer)
Dim arrayLista As New List(Of String)()
Dim arraySubLista As New List(Of String)()
If (iLenght > 1) Then
arraySubLista = GenerateAllCombination(iLenght -1)
for each objString As String in arraySubLista
for each ele As String in arrayValori
arrayLista.add(objString & ele)
loop
loop
Else
for each ele As String in arrayValori
arrayLista.add(ele)
loop
End If
End Function
Now if you use generateAllCombination you will have a List of string with ALL the combination of 5 character.
Now you just create 4 list by concatenating those combination with your string1 to string4 (string1 & combination) etc..
put all those result on a List of string, and you have 100% that at least ONE of the string will break your teacher code
I forgot, arrayValori must be a List with all values from "0" to "f"

Related

Understanding the steps in making a counter for each letter when a sentence is inputed

I have an example of a program that shows how to set up a counter for how many times each letter of the alphabet was used. I don't understand the syntax of the middle portion of the program.
LET letter$ = MID$(sentence$, LETTERNUMBER, 1)
I have tried searching on youtube and tutorials online
CLS
REM Make Counters for each Letter!!!
DIM Count(ASC("A") TO ASC("Z"))
REM Get the Sentence
INPUT "Enter Sentence:", sentence$
LET sentence$ = UCASE$(sentence$)
FOR I = ASC("A") TO ASC("Z")
LET Count(I) = 0
NEXT I
FOR LETTERNUMBER = 1 TO LEN(sentence$)
LET letter$ = MID$(sentence$, LETTERNUMBER, 1)
IF (letter$ >= "A") AND (letter$ <= "Z") THEN
LET k = ASC(letter$)
LET Count(k) = Count(k) + 1
END IF
NEXT LETTERNUMBER
PRINT
REM Display These Counts Now
LET letterShown = 0
FOR letternum = ASC("A") TO ASC("Z")
LET letter$ = CHR$(letternum)
IF Count(letternum) > 0 THEN
PRINT USING "\\## "; letter$; Count(letternum);
END IF
LET letterShown = letterShown + 1
IF letterShown = 7 THEN
PRINT
LET letterShown = 0
END IF
NEXT letternum
END
A through Z appears with the count of how many times they appeared.
The MID$ function returns a portion of a STRING's value from any position inside a string.
Syntax:
MID$(stringvalue$, startposition%[, bytes%])
Parameters:
stringvalue$
can be any literal or variable STRING value having a length. See LEN.
startposition%
designates the non-zero position of the first character to be returned by the function.
bytes%
(optional) tells the function how many characters to return including the first character when it is used.
Another method to calculate characters in a string:
REM counts and displays characters in a string
DIM count(255) AS INTEGER
PRINT "Enter string";: INPUT s$
' parse string
FOR s = 1 TO LEN(s$)
x = ASC(MID$(s$, s, 1))
count(x) = count(x) + 1
NEXT
' display string values
FOR s = 1 TO 255
PRINT s; "="; count(s); " ";
IF (s MOD 8) = 0 THEN
PRINT
IF (s MOD 20) = 0 THEN
PRINT "Press key:";
WHILE INKEY$ = "": WEND: PRINT
END IF
END IF
NEXT
END

Label a set of objects with (A->Z,AA->ZZ, AAA->ZZZ) in VBA

I have a set which has an unknown number of objects. I want to associate a label to each one of these objects. Instead of labeling each object with a number I want to label them with letters.
For example the first object would be labeled A the second B and so on.
When I get to Z, the next object would be labeled AA
AZ? then BA, BB, BC.
ZZ? then AAA, AAB, AAC and so on.
I'm working using Mapbasic (similar to VBA), but I can't seem to wrap my head around a dynamic solution. My solution assumes that there will be a max number of objects that the set may or may not exceed.
label = pos1 & pos2
Once pos2 reaches ASCII "Z" then pos1 will be "A" and pos2 will be "A". However, if there is another object after "ZZ" this will fail.
How do I overcome this static solution?
Basically what I needed was a Base 26 Counter. The function takes a parameter like "A" or "AAA" and determines the next letter in the sequence.
Function IncrementAlpha(ByVal alpha As String) As String
Dim N As Integer
Dim num As Integer
Dim str As String
Do While Len(alpha)
num = num * 26 + (Asc(alpha) - Asc("A") + 1)
alpha = Mid$(alpha, 2,1)
Loop
N = num + 1
Do While N > 0
str = Chr$(Asc("A") + (N - 1) Mod 26) & str
N = (N - 1) \ 26
Loop
IncrementAlpha = str
End Function
If we need to convert numbers to a "letter format" where:
1 = A
26 = Z
27 = AA
702 = ZZ
703 = AAA etc
...and it needs to be in Excel VBA, then we're in luck. Excel's columns are "numbered" the same way!
Function numToLetters(num As Integer) As String
numToLetters = Split(Cells(1, num).Address(, 0), "$")(0)
End Function
Pass this function a number between 1 and 16384 and it will return a string between A and XFD.
Edit:
I guess I misread; you're not using Excel. If you're using VBA you should still be able to do this will the help of an reference to an Excel Object Library.
This should get you going in terms of the logic. Haven't tested it completely, but you should be able to work from here.
Public Function GenerateLabel(ByVal Number As Long) As String
Const TOKENS As String = "ZABCDEFGHIJKLMNOPQRSTUVWXY"
Dim i As Long
Dim j As Long
Dim Prev As String
j = 1
Prev = ""
Do While Number > 0
i = (Number Mod 26) + 1
GenerateLabel = Prev & Mid(TOKENS, i, 1)
Number = Number - 26
If j > 0 Then Prev = Mid(TOKENS, j + 1, 1)
j = j + Abs(Number Mod 26 = 0)
Loop
End Function

vb.net convert number to alphanumeric equivalent

I need to convert a four digit number to a three digit alphanumeric number where
001-999 = 001-999
1000 = 00A
1027 = 01A
etc.. up to whatever = ZZZ
I am very new to programming and cannot figure out how to proceed with this issue. Any help would be GREATLY appreciated.
How does this work for you?
Private Function ConvertMyNumber(toConvert As UShort) As String
Const MaxValueToConvert = 3885S
If (toConvert > MaxValueToConvert) Then Throw New ArgumentException(String.Format("Argument value of {0} exceeds maximum possible value of {1}.", toConvert, MaxValueToConvert))
Const NumeringSchemeThreshold = 1000
Const TotalResultLength = 3
Const PaddingChar = "0"c
If (toConvert < NumeringSchemeThreshold) Then Return toConvert.ToString().PadLeft(TotalResultLength, PaddingChar)
Dim adjustedValue = (toConvert - NumeringSchemeThreshold) 'since we're starting with this numbering scheme at NumeringSchemeThreshold
Const NumbersInAlphabet = 26
Dim moddedValue = (adjustedValue Mod NumbersInAlphabet) '0 to NumbersInAlphabet - 1 based on the cycle
Dim numCycles = Fix(adjustedValue / NumbersInAlphabet) 'What "cycle number" it is, i.e. the number of times around the alphabet loop.
Dim suffix As String = String.Empty
Dim prefix As String = CStr(numCycles)
Const firstStepThreshold = 100
Const secondStepThreshold = 110
Const LastAlphabetChar = "Z"c
If (numCycles >= firstStepThreshold And numCycles < secondStepThreshold) Then
numCycles = numCycles - firstStepThreshold
prefix = CStr(numCycles)
suffix = LastAlphabetChar
ElseIf (numCycles >= secondStepThreshold) Then
numCycles = numCycles - secondStepThreshold
suffix = LastAlphabetChar & LastAlphabetChar
prefix = String.Empty
End If
Const AsciiCharValueOffset = 65
'concat the cycle number to the converted modded letter, and return the zero-padded result.
Return (prefix & CChar(Char.ConvertFromUtf32(moddedValue + AsciiCharValueOffset)) & suffix).PadLeft(TotalResultLength, PaddingChar)
End Function

Generate random string in text field

We have that old software (made by one of the first employees many years ago) in company that uses Microsoft Access to run. Boss asked me to add a random string generation in the specific text box on click but i have no idea how to do that. I dont have any Microsoft Access programming experience, thats why i am askin you to help.
I managed to create button and text field so far. Thats where it stops. I also managed to access the code for the button action:
Private Sub command133_Click()
End Sub
This is one way, will work in Access VBA (which is an older basic than vb.net). It will generate a string with letters and numbers.
Sub test()
Dim s As String * 8 'fixed length string with 8 characters
Dim n As Integer
Dim ch As Integer 'the character
For n = 1 To Len(s) 'don't hardcode the length twice
Do
ch = Rnd() * 127 'This could be more efficient.
'48 is '0', 57 is '9', 65 is 'A', 90 is 'Z', 97 is 'a', 122 is 'z'.
Loop While ch < 48 Or ch > 57 And ch < 65 Or ch > 90 And ch < 97 Or ch > 122
Mid(s, n, 1) = Chr(ch) 'bit more efficient than concatenation
Next
Debug.Print s
End Sub
Try this function:
Public Function GetRandomString(ByVal iLength As Integer) As String
Dim sResult As String = ""
Dim rdm As New Random()
For i As Integer = 1 To iLength
sResult &= ChrW(rdm.Next(32, 126))
Next
Return sResult
End Function
Workin on #Bathsheba code, I did this. It will generate a random string with the number of characters you'd like.
Code :
Public Function GenerateUniqueSequence(numberOfCharacters As Integer) As String
Dim random As String ' * 8 'fixed length string with 8 characters
Dim j As Integer
Dim ch As Integer ' each character
random = ""
For j = 1 To numberOfCharacters
random = random & GenerateRandomAlphaNumericCharacter
Next
GenerateUniqueSequence = random
End Function
Public Function GenerateRandomAlphaNumericCharacter() As String
'Numbers : 48 is '0', 57 is '9'
'LETTERS : 65 is 'A', 90 is 'Z'
'letters : 97 is 'a', 122 is 'z'
GenerateRandomAlphaNumericCharacter = ""
Dim i As Integer
Randomize
i = (Rnd() * 2) + 1 'One chance out of 3 to choose one of 3 catégories
Randomize
Select Case i
Case 1 'Numbers
GenerateRandomAlphaNumericCharacter = Chr(Rnd() * 9 + 48)
Case 2 'LETTERS
GenerateRandomAlphaNumericCharacter = Chr(Rnd() * 25 + 65)
Case 3 'letters
GenerateRandomAlphaNumericCharacter = Chr(Rnd() * 25 + 97)
End Select
End Function
I use it with random number of characters, like this :
'Generates random Session ID between 15 and 30 alphanumeric characters
SessionID = GenerateUniqueSequence(Rnd * 15 + 15)
Result :
s8a8qWOmoDvC4jKRjPr5hOY12u 26
TB24qZ4cNfr6EdyY0J 18
6LZRQ9P5WHLNd71LIdqJ 20
KPN0RmlhhJKnVzPTkW 18
R2pNOKWJMKl9KpSoIV2egUNTEb1QC2 30
X8jHuupP6SvEI8Dt2wJi 20
NOTE: This is still not completely random. It will give a higher count of numbers than normal as approx 1/3 of all chars generated will be numbers.
Normally distribution will look like:
10 numbers plus 26 lowercase plus 26 uppercase = 62 possible chars.
Numbers will normally be 10/62 parts of the string or 1/6.2
With the code
i = (Rnd() * 2) + 1 'One chance out of 3 to choose one of 3 catégories
the count of numbers is pushed up to 1/3 (on average)
Probably not too much of a worry - unless you are trying to beat the NSA and then you have decreased your range significantly.

Verify Gamefield VB.NET

So I'm developing a minesweeper game and im assigning the mines, but I've got to check where are the mines now, in order to generate the numbers. The problem is that when I'm verifying the columns and lines I need the program not to get out of the game field.
Here's how my code looks like now:
Public Sub avisinhos(ByVal line, ByVal column)
If mat(line, column) = 0 Then
mat(line, column) = -1
numbandeiras = numbandeiras + 1
End If
For auxlinha = -1 To 1
For auxcolumn = -1 To 1
Next
Next
End Sub
How do I create a IF function to verify that I don't get out of the game field?
Best regards, joao.
pseudo code
int linestart = -1;
int lineend = 1;
int colstart = -1;
int colend = 1;
Assuming a 10 x 10 grid (zero based)
if line < 2 linestart = 0
if line > 8 lineend = 0
if column < 2 colstart = 0
if column > 8 colend = 0
For auxlinha = linestart To lineend
For auxcolumn = colstart To colend
// check
Next
Next
Personally though I wouldn't bother with the loops, they add very little to nothing
HasMineAbove = (line > 1) and (gamefield[line -1,column] = MinePresentValue
would be my approach, do it all in one.
Not to mention the huge potential confusion when auxlinha and auxcolumn are both zero...
I'm not sure exactly what your code is saying. It's a bit cryptic since you're using abbreviations and all lowercase names. You might want to try camelCasing and spelling out the words more completely, intellisense is your friend. =)
But coding style aside, if you are trying to loop through a limited range of values, you can keep your values bounded by using the modulus operator (%). For example, if you need to keep you values between 0-7 and you end up with a value of 12, just take the modulus of 8 to loop back to within range with a value of 4:
12 % 8 = 4
9 % 8 = 1
15 % 8 = 7
24 % 8 = 0
I realize this doesn't answer your specific question, but it's a handy technique might find useful.