'
' Zig-zag Cipher
' Problem 2, AIC 2004 (Senior)
' Pascal Sample Solution
'

'
' This solution deciphers the message by:
'  - first determining the length of the original message, textLength;
'  - encrypting the integers 0 .. textLength-1 according to the scheme in the
'    question.
'  - reading the numbers left-to-right, top-to-bottom from the step above to
'    generate a table that links index in the encrypted text to an index in
'    the plain text.
'  - use this table to decrypt the message.
'
' For example, to decrypt COPATGAHRRY in 3 rows, we consider the 12 integers
' 0, 1, 2, ... 11. Encrypting this into a 2-D array, dummyText, makes the
' array look like:
'
'   0 5 6 11
'   1 4 7 10
'   2 3 8 9
'
' We can then generate a table, decryptMap, that matches the position in the
' encrypted text to their index in the original text:
'
'            i  0    1    2    3    4    5    6    7    8    9   10   11
' decryptMap[i] 0    5    6   11    1    4    7   10    2    3    8    9
'
' We can then read the original message by printing out the characters in the
' order given by the table.
'
'

const MAXMESSAGE = 10000
const MAXROWS = 100

' decryptMap(i) = ciphertext index for plaintext(i).
Dim decryptMap(MAXMESSAGE) As Long
Dim dummyText(MAXROWS, MAXMESSAGE) As Long
Dim rowLen(MAXROWS) As Long
Dim rows As Long
Dim ciphertext As String
Dim textLength As Long

' currRow keeps track of the current row we are on (starting from 0)
Dim currRow As Long

' currDir keeps track of the direction the text is currently heading
' (-1 for upwards, -1 for downwards). }
Dim currDir As Long

Dim i As Long
Dim dummyIndex As Long
Dim inputLine As String

Sub Main()
    ' Open the I/O files.
    Open "zigin.txt" For Input As #1
    Open "zigout.txt" For Output As #2

    ' Read the number of rows the deciphered text will fill.
    Input #1, rows

    ' Start with an empty rowLen array - assume all rows are empty.
    For i = 1 To rows
        rowLen(i) = 0
    Next i

    ' Read in the first line of text.
    Line Input #1, inputLine
    ciphertext = ""
    ' If the current line of text is a "#", then we have the entire input.
    While inputLine <> "#"
        ' We have a string which is not just a #
        ' Append it to the existing string.
        cipherText = cipherText & inputLine

        ' Read in the next line.
        Line Input #1, inputLine
    Wend
    Close #1

    textLength = Len(ciphertext)

    ' We know the length of the encrypted text and the number of rows that it
    ' fills.

    ' Fill in the dummyText array with the encrypted version of the integers
    ' 0 .. textLength-1.

    ' Starting at row 0.
    currRow = 1

    ' Heading down to begin with.
    currDir = 1

    For i = 1 To textLength
        ' Mark this position in the dummy array.
        rowLen(currRow) = rowLen(currRow) + 1
        dummyText(currRow, rowLen(currRow)) = i

        ' Move to the next row for the next character.
        currRow = currRow + currDir

        ' If we are at the top or the bottom, change direction.
        If currRow = rows then
            currDir = -1
        ElseIf currRow = 1 then
            currDir = 1
        End If
    Next i

    ' Construct the decryptMap array from the dummyText.
    dummyIndex = 1
    For currRow = 1 To Rows
        For i = 1 to rowLen(currRow)
            decryptMap(dummyText(currRow, i)) = dummyIndex
            dummyIndex = dummyIndex + 1
        Next i
    Next currRow

    ' We now know the order to print out the letters from the plain text.
    For dummyIndex = 1 to textLength
        Print #2, Mid$(ciphertext, decryptMap(dummyIndex), 1);
    Next dummyIndex
    Print #2, ""
    Close #2
End Sub