CPSC120B
Fundamentals of Computer Science

Steganography

• Imagine your are a spy in a foriegn country and you need to comunicate with your home country
• Encrypting your messages is a good idea, but people might get suspicious when you have lots of email going to cia.gov
• What you need to do is hide that you are even sending messages
• So you post cute cat pictures to your facebook page and you hide your messages home inside of the images
• Hiding messages inside of other messages is Steganography, and it is your lab activity for today
• There are many ways to do the hiding, the one we are going use slightly modifies colors to encode a message
• If you pull up an online color picker and change the red, green, or blue value of the color, it is really hard to see the difference
• So we will encode the message by making small changes to the color values

• For exmample, this image encodes ‘abc’:

  P3
  1 8
  255
  122
  101
  215
  42
  2
  128
  200
  13
  64
  203
  155
  130
  36
  172
  91
  180
  88
  77
  55
  22
  44
  66
  33
  99

• The message is encoded in the parity of pixel data
• Parity is whether a number is even or odd
• If it is even it encodes 0, if it is odd it encodes 1

• The first eight numbers are

  122, 101, 215, 42, 2, 128, 200, 13

• Which have the parity

  even, odd, odd, even, even, even, even, odd

• And therefore encodes

  0, 1, 1, 0, 0, 0, 0, 1

• But if all we have are zeros and ones how do we get the characters of the message?

• We must convert the number from binary to decimal and then look it up in an ascii table

Binary

• In decimal we have 10 symbols to represent numbers, 0 thorugh 9
• When we want to represent larger numbers larger than 9, we use multiple symbols
• The location of a symbol specifies how much it contributes to the numbers value
• For example in the number 253, the 2 is in the represents hundreds, the 5 represents tens, and the 3 represents 1s

• So it could be rewritten as:

  253 = 2 * 100 + 5 * 10 + 3 * 1

• Or:

  253 = 2 * 10^2 + 5 * 10^1 + 3 * 1^0

• In binary we have only 2 symbols, 0 and 1

• To represent numbers larger than 1, we use multiple symbols

  101 = 1 * 2^2 + 0 * 2^1 + 1 * 2^0 = 1 * 4 + 0 * 2 + 1 * 1 = 4 + 0 + 1 = 5

ASCII

• So now we can convert the encoded sequence 0, 1, 1, 0, 0, 0, 0, 1 to decimal

  01100001 = 2^6 + 2^5 + 2^0 = 63 + 32 + 1 = 97

• Notice it is not in the range 0 to 25 as we have previously used to convert between numbers and characters
• This is because it is using ASCII, a standard for encoding characters that includes lots of other things including punctuation

• So we can look 97 up in an ASCII Table to determine that is ‘a’

Lab