Reversible watermarking method based on difference expansion

1. Divide the image into pixel pairs (x, y) x,y∈Z,0≤x,y≤255

Personally think that , This splitting process needs special records , It records which two pixels in the image form this pixel pair , This is to restore the watermark later .

2. Define its integer mean l And the difference h 

                                   

              Formula 1                                                                           Formula 2

 3. Send secret bit information m It is embedded into the difference by the method of difference expansion h in ：

 4. What will be obtained h′ Substitute into formula 2 , Get a new image pixel pair , Form an image after embedding secret information .

5. In order to ensure that the data after reversible hiding does not overflow , Set the carrier image as 8bit Grayscale image of , The conditions for reversible digital watermarking of a pixel pair can be obtained .

Take any pixel pair as an example

The encryption process ：

（1） Suppose an image pair is （206,201） namely x=206,y=201

（2）l = floor((x+y)/2) = 203     h=206-201=5

（3） take h Convert to binary 101

（4） Suppose we have a secret message to be embedded in this pixel pair m=1, Then put 1 Add to 101 Of LSB Get on h'

（5） obtain h' = 1011 namely h' = h*2+m = 2*5+1 = 11

（6） hold h' Bring in formula 2 to get ：x' = 203+ floor((11+1)/2) = 203+6 = 209  y'=203 - floor((11)/2) = 198

（7） Therefore, the encrypted information is （x',y'）=（209,198）

The decryption process ：

（1） It is known that （x',y'）=（209,198）

（2） Calculation l' = floor((x'+y')/2) = 203  h' = x' - y' = 11

（3）m = LSB(h') = LSB(1011) = 1     // Is o h' The last bit of

（4）h = floor(h'/2) = 5

（5）x =  l'+ floor((h+1)/2) = 203 + 3 = 206    y = l'-floor(h/2) = 203-2 = 201

（6） So restore the original pixel pair (x,y) = (206,201) Secret news m = 1

reference ：

Jun Tian   Reversible Data Embedding Using a Difference Expansion