Countdown to PDC09: Hard Hat Challenges Are Back! (Sort of)

Sleeping in your car?!?

Congratulations, you just won a 'internal only' PDC poster.

EDIT: link to Ineta in the video is wrong.

Darn, commented then clicked refresh... *tear*

Cool!

There is no link to the blog post about embeding the pdc font into the SL app

New to the hard hat challenges here, but sounded fun. =) I'm pretty sure I'm on the right track... *pretty* sure... lol.

-Stephen

Yeah, nevermind. lol. The quest for the solution continues! I was focusing on groupings of 7 and dot matrix... and somehow fitting in pi along with the fact that you switched positions before showing it. Maybe that'll help someone else... or throw them completely off. lol.

-Stephen

Is the vertical line in the first column, second row meaningful or not? It's on the picture you posted but not on the PDC 2008 video and just wondering.

I see a pretty strong pattern that happens 5 times with certain rows "off by one" but I can't find any use for it. Seems to validate the empty cell at the end, but I can't find any way to relate it to Pie or Pi. Time for dinner, so I guess I'll have to try again in the morning. 

Does the sequence go left to right then next line? or  Left column top to bottom Next Column?

Let's see if this is right.

Is the answer: NOIZE ?

Used letter - number subsitution with the numer of PI

Looks like a bunch of punch cards. http://en.wikipedia.org/wiki/Punch_card

print them out, rotate back and forth and you get I ♥ PDC  

Hmm... every seven rows there is a blank row so there must some pattern, now just have decode it

And what's more, the groups are very symetrical in a vertical way. What I mean is that if a group of seven row begins with a row which has 6 black dots, it also ends with a row with 6 black dots... hmm...

Solved it:

It says SEE YOU AT PDC2008.

I'll post method later.

As promised the method. (Note: The zipped up solution is also available in the SandBox)

Make sure that the textbox is using a Monospaced font, has Multiline and Scrollbars=Vertical

Public Class Form1
 Private Sub ShowSolution(ByRef useThisTextBox As TextBox)
 ' Initialise and array of string lists.
 Dim columns(6) As List(Of String)
 For i As Integer = 0 To 6 : columns(i) = New List(Of String) : Next
 ' Get Digits of PI
 Dim piStr As String = My.Resources.pi.Replace(Environment.NewLine, "") 
' Get PDC Image text representation.
 Dim tfp As New FileIO.TextFieldParser("Resources/PDC2008.txt")
 Dim del() As String = {" "}
 tfp.Delimiters = del
 While tfp.EndOfData = False
 Dim f() = tfp.ReadFields 
' Place field into correct column. 
For i As Integer = 0 To f.Count - 1
 columns(i).Add(f(i))
 Next
 End While tfp.Close()
 ' Collate the columns into a single column.
 For i As Integer = 0 To 5
 columns(6).AddRange(columns(i).ToArray)
 Next 
' Rotate Left by corrisponding PI digit
 For i As Integer = 0 To columns(6).Count - 1
 For j As Integer = 0 To Integer.Parse(piStr(i))
 columns(6)(i) = STR_ROR(columns(6)(i), False)
 Next
 Next
 ' Rotate To Right.
 For i As Integer = 0 To columns(6).Count - 1
 columns(6)(i) = STR_ROR(columns(6)(i), True)
 Next
 ' Build the output string.
 useThisTextBox.Text = ""
 For i As Integer = 0 To columns(6).Count - 1
 useThisTextBox.Text &= columns(6)(i) & ControlChars.NewLine
Next
 ' Hide the ones.
 useThisTextBox.Text = Me.TextBox1.Text.Replace("1", " ")
 ' Replace 0 with # for higher contrast 
useThisTextBox.Text = Me.TextBox1.Text.Replace("0", "#")
End Sub
Private Function STR_ROR(ByVal RotateThisString As String, Optional ByVal RotateToTheRight As Boolean = True) As String 
' A function to do the rotation either 1 place to the right, or to the left.
 If RotateToTheRight Then
 Return RotateThisString(RotateThisString.Length - 1) & RotateThisString.Substring(0, RotateThisString.Length - 1)
 Else
 Return RotateThisString.Substring(1, RotateThisString.Length - 1) & RotateThisString.Substring(0, 1)
 End If
 End Function 
 
Private Sub Form1_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load
 ShowSolution(Me.TextBox1)
 End Sub
End Class 

LINQ-ified Shorter Version.

Public Class Form1

 Private Sub ShowSolution(ByRef useThisTextBox As TextBox)
  ' Initialise and array of string lists.
  Dim columns(6) As List(Of String)
  For i As Integer = 0 To 6 : columns(i) = New List(Of String) : Next
  ' Get Digits of PI
  Dim piStr As String = My.Resources.pi.Replace(Environment.NewLine, "")
  ' Get PDC Image text representation.
  Dim tfp As New FileIO.TextFieldParser("Resources/PDC2008.txt")
  Dim del() As String = {" "}
  tfp.Delimiters = del
  While tfp.EndOfData = False
   Dim f() = tfp.ReadFields
   ' Place field into correct column.
   For i As Integer = 0 To f.Count - 1 : columns(i).Add(f(i)) : Next
  End While
  tfp.Close()
  ' Collate the columns into a single column.
  For i As Integer = 0 To 5 : columns(6).AddRange(columns(i).ToArray) : Next
  ' Rotate Left by corrisponding PI digit minus one to rotate right
  For i As Integer = 0 To columns(6).Count - 1 : columns(6)(i) = STR_ROR(columns(6)(i), False, Integer.Parse(piStr(i))- 1) : Next
 ' Build the output string. Hide the ones. Replace 0 with # for higher contrast
  useThisTextBox.Text = ((From T As String In columns(6) Select T).Aggregate( _
                         Function(current As String, nexta As String) current + ControlChars.NewLine + nexta)) _
                         .Replace("1", " ").Replace("0", "#")
 End Sub

Private Function STR_ROR(ByVal RotateThisString As String, _
                          Optional ByVal RotateToTheRight As Boolean = True, _
                          Optional ByVal repeats As Integer = 0) As String
  ' A function to do the rotation either 1 place to the right, or to the left.  While repeats >= 0
   If RotateToTheRight Then
    RotateThisString = RotateThisString(RotateThisString.Length - 1) & RotateThisString.Substring(0, RotateThisString.Length - 1)
   Else
    RotateThisString = RotateThisString.Substring(1, RotateThisString.Length - 1) & RotateThisString.Substring(0, 1)
   End If
   repeats -= 1
  End While
  Return RotateThisString
 End Function

 Private Sub Form1_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load
  ShowSolution(Me.TextBox1)
 End Sub
End Class

Get Byte values Directly From image

Using the image found via link in a previous post.

Public Class Form1

 Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
  Dim bmp As New Bitmap(My.Resources.PDC2008_Hard_Hat_Challenge_8)
  Dim Bytes As New List(Of Byte)
  Dim XPos As Single = 73
  Dim XStep As Single = 21
  Dim YPos As Single = 74
  Dim YStep As Single = 46
  Dim XGap As Single = 63
  Dim PixelBits As Byte = 0
  Dim Base As Single = 0
  For col As Integer = 0 To 5
   YPos = 74
   Dim RowStart = XPos + Base
   For row As Integer = 0 To 19
    PixelBits = 0
    XPos = RowStart
    For octetBit As Integer = 7 To 0 Step -1
     Dim pc = bmp.GetPixel(XPos, YPos)
     If pc.A = 255 AndAlso pc.R = 255 AndAlso pc.B = 255 AndAlso pc.G = 255 Then PixelBits += (2 ^ octetBit)
     bmp.SetPixel(XPos, YPos, Color.YellowGreen)
     XPos += XStep
    Next
    Bytes.Add(PixelBits)
    YPos += YStep
   Next
   Base = XGap - XStep
  Next
  Me.Pic_Image.Image = bmp
  Bytes.RemoveAt(119)
  Dim FinalStr As New System.Text.StringBuilder
  Dim PiDigit As String = My.Resources.pi.Replace(ControlChars.NewLine, "")
  For i As Integer = 0 To Bytes.Count - 1
   Bytes(i) = ROR(Bytes(i), Integer.Parse(PiDigit(i)) - 1)
   FinalStr.AppendLine(ByteToStr(Bytes(i)))
  Next
  Me.Output.Text = FinalStr.ToString.Replace("1", " ").Replace("0", "#")
 End Sub

 Private Function ByteToStr(ByVal x As Byte) As String
  Dim r As String = ""
  For i As Integer = 7 To 0 Step -1
   r &= If(((2 ^ i) And x) > 0, "1", "0")
  Next
  Return r
 End Function

 Private Function ROR(ByVal x As Byte, ByVal r As Integer) As Byte
  While r >= 0
   x = ((x And &H80) >> 7) + ((x And &H7F) << 1)
   r -= 1
  End While
  Return x
 End Function
End Class

Thank you, I've sent the email and if you want to "ruin" the poster it upto you.

Sadly I won't be attending PDC2009 in body but I'll be there in spirit and vicariously through the sessions webcasts.

 Maybe I'll go sometime in the future.

Have a great PDC2009.

Edit: Topic Posting as </Dream.In.Code> containing

A description of the effort.

Yes, congratulations!

I'm just wondering, how this could be solved on paper as you mentioned in last years PDC video comment

Whoops!  Typo in the video for INETA url for Code Challenge.  The correct url is: ineta.org/codechallenge

I'm dissapointed.  I think Tommy needs more camera time!

To do it manually. You need sheets of Paper, Scissors, Pen and Glue.

Follow this method.

1. Print out the image
2. Label the columns Left to Right Rotate (Their positions rotated anticlockwise) A B C D E F
3. Cut out the columns. (Ask a responsiable adult to do this for you.)
4. Write on the righthand side of each, column row the corrisponding digit of PI (What they ate.)
5. So the first few are 3, 1, 4, 1. 5, 9 etc
6. Cut out row 1 of column A.
7. Cut out the first 3 digits (Take One if doing on computer (Off by one))
8.  Note: 1 -> 1 digits etc, 0 & 8 -> 0 digts, 9 ->1digit. (Eg Row A1 = [001] [10000] )
9. Glue the two peices to another peice of paper. First part last. (Eg Row A1 = [10000] [001] )
10. Repeat step 6 - 9 for the other rows and columns.
11. Stand back about from it & squint.
12. On the answer is spelt out vertically.

Simples.

The bold are the clues from the video.

Here a the first 8 row with cut corrispoding cut points.

[001][10000] [10000][001] [.####][##.]
[0][0011110] [0011110][0] [##....#][#]
[1111][0011] [0011][1111] [##..][....]
[1][1000000] [1000000][1] [.######][ ]
[11100][111] [111][11100] [...][...##]
[0][0011110] [0011110][0] [##.....][#]
[01][100000] [100000][01] [.#####][#.]
[000000][00] [00][000000] [..][......]

If you interested in the full text with cut points, change the code so the output.text is this.

 Dim o As String = ""
  For q As Integer = 0 To Bytes.Count - 1
   Dim rr = Integer.Parse(PiDigit(q)) Mod 8
   Dim bs = Convert.ToString(Bytes(q), 2).PadLeft(8, "0")
   o &= String.Format("[{1}][{0}]{2}", bs.Substring(0, rr), bs.Substring(rr), ControlChars.NewLine)
  Next
  Me.Output.Text = o

I think this section of code maybe possible to compact this down single line of code. (using LINQ)

It is. (So if you exclude the image extraction code, or you use the manual text file)

Solve in a single line of vb.net (LINQ) code.

Me.Output.Text = ((Bytes.Select(Of String) _
                     (New System.Func(Of Byte, Integer, String) _
                      (Function(lb As Byte, li As Integer) _
                         String.Format("[{1}][{0}]", _
                                       Convert.ToString(Bytes(li), 2).PadLeft(8, "0"). _
                                       Substring(0, Integer.Parse(PiDigit(li)) Mod 8), _
                                       Convert.ToString(Bytes(li), 2).PadLeft(8, "0"). _
                                       Substring(Integer.Parse(PiDigit(li)) Mod 8) _
                                       ) _
                      ) _
                     ) _
                    ).Aggregate( _
                     Function(current, nexta) current & ControlChars.NewLine & nexta) _
                   ).ToString.Replace("1", " ").Replace("0", "#")

Now I've refactored the code to its limit. I'm stepping away.