steganography full report

[seminar class]

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Steganography
Steganography is the art and science of hiding information in plain sight.
• For hiding information we need three components:
 Cover Image
 Secret Data
 Stego Image
Embedding Methods in Steganography
• Some of the embedding methods are:
 LSB Method
 PVD Method
Disadvantages in LSB method:
 The No: of message bits embedded in each pixel is same.
 Visual degradation is possible if more amount of data embedded.
PVD Method
• In PVD method the secret data is inserted into every pixel of the image and decides the no: of insertion bits using the difference value between two pixels adjacent to the target pixel.
• Therefore, the no: of insertion bits in each pixel is dependent on whether the target pixel is included in edge area or in a smooth area.
• Embedding Procedure in PVD Method
For Eg:
 Lets consider Two Pixel Method:
• Binary Representation of the message is 1100.
• Consider an image with pixel values as follows
• Target pixel (g) =110
• Upper pixel (g1) = 101
• Left pixel (g2) =100
For d<3:
• Calculate difference value d between the upper pixel (g1) and the left pixel (g2) in a given target pixel (g) by
• d =| g1 - g 2|
• d=|101 -100| = 1
• Calculate n that is the number of the insertion bits in a target pixel P from d
• n= floor (log2 d), if d > 3 and d = odd;
• n= floor (log2 d) -1, if d > 3 and d = even.
• n= 1 , if d < 3
• Calculate a temporary value t from n
• t = b - (g mod 2 ) .
• Where, b is the decimal representation of secret messages as the n bits.
• b = 1, for n=1, therefore, t = 1 – (110 mod 2) -> t = 1
• To make the quality of the image higher, select the nearest value to the target pixel’s value of the cover image by
• t1 = t if [ - <= t <= ]
• t1 = t + 2 ^n if [-(2^n + 1) <= t < ( - )]
• t1 = t – 2^n if [ ((2^ n – 1)/2) <= t < 2^n]
• for n=1 the interval cases will be:
-1/2 to 1/2 , -1 to -1/2, 1/2 to 2
• t =1 lies in 3rd interval
• therefore t1= 1 – 2
t1 = -1,
• Finally, we can get the new pixel value g *
g* = g +t1;
g*= 110 -1
g*=109.
For d>3:
• Target pixel (g) =119
• Upper pixel (g1) = 109
• Left pixel (g2) = 96
• d =| g 1 - g 2 | d=|109 - 96| d= 13
• n = floor(log2 13)
n = 3
• b = 4 ( i.e message bits 100)
• t = 4 – (119 mod 2^3)
• t = -3
• for n = -3 the interval cases will be
• -7/2 to 7/2 , -7 to -7/2 , 7/2 to 8
• t=2 fall in 1st case
• Therefore, t1= - 3
• g*=119 - 3
• Therefore, New pixel value = 116
• Data Extraction:
For d<3:
• Target pixel (g) =109
• Upper pixel (g1) = 101
• Left pixel (g2) = 100
• d =| g 1 - g 2 | d=|101 - 100| =1
• here d=1,
• therefore n=1
• b = mod (g * , 2 ).
• i.e : b = mod (109 , 2)
• b = 1 -> message bit = 1.