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A Quadratic-Residue-based Fragile Watermarking Scheme

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Presentation on theme: "A Quadratic-Residue-based Fragile Watermarking Scheme"— Presentation transcript:

1 A Quadratic-Residue-based Fragile Watermarking Scheme
Chair Professor Chin-Chen Chang Feng Chia University National Chung Cheng University National Tsing Hua University

2 Outline Introduction The proposed scheme Experimental results
Conclusions

3 The Fields of Image Authentication
Media communications Embedding Verifying Authentication code

4 (Tampered areas located Image)
Introduction Authentication bits: (Cover Image) (Watermarked Image) (Tampered Image) (Tampered areas located Image) (Restored Image)

5 Requirements Effectiveness:
Differentiation: Security: Recoverability: Provide a high probability of tamper detection Can distinguish between an innocent adjustment and replacing or adding features on purpose Only a selected group of people having a secret key can perform the detection Can recover back to the correct image from the modification

6 The Proposed Method Three techniques are employed Watermark Generating
Watermark Embedding Tamper Detecting and Recovering

7 The Proposed Scheme . Embedding Processes … . . .
1. Divide the image into non-overlapping 1x3 blocks 50 60 61 175 30 150 100 95 203 90 93 . . . .

8 The Proposed Scheme 2. Generate authentication bits:
Y = (PM)2e-1 mod N N = r × s = 13x19 = 247 < 255 (grey-level) PM = the decimal value of the first pixel’s high nibble e-1(a randomly chosen modifier) = the inverse of e mod (r × s), where Case 1 a: the first element satisfied and Case 2 b: the first element satisfied and Case 3 c: the first element satisfied and Case 4 d: the first element satisfied and

9 The Proposed Scheme Ex: 28 90 93 PM = 1
PM = 1 R = {2 mod r | 1 ≤  ≤ r – 1} = {1, 3, 4, 9, 10, 12} S = {2 mod s | 1 ≤  ≤ s – 1} = {1, 4, 5, 6, 7, 9, 11, 16, 17} a = 4, b = 3, c = 5, d = 2 If e = c = 5, then e-1 = 99 Y = (PM)2e-1 mod N = (1)2 × 99 mod 247=99

10 The Proposed Scheme 3. Embed authentication bits into cover pixels
28 90 93 28 86 83 (binary representation) Y = 99

11 The Proposed Scheme Verifying Processes PM = 1 Y = 99 0110 0011
1. Extract PM and the embedded Y 28 86 83 (binary representation) PM = 1 Y = 99

12 The Proposed Scheme 2. Compute , where t = 0
e = 5 (e = c, c is selected by a pseudo-random number generator with private key K ) N = 247

13 The Proposed Scheme = 1 PM = 1 3. Verify 12 mod 247 = 12 mod 247
 pass (Otherwise, mark this block as tampered block)

14 The Proposed Scheme Recovering Processes n1 n2 n3 P(n1,n2,n3)

15 Recovering Processes . … . . . x1 = P(61, 175, 203) = 61
50 60 61 175 30 150 100 95 203 x1 x2 x3 . . . . x1 = P(61, 175, 203) = 61 x2 = P(175, 30, 61) = 144 x3 = P(30, 150, 144) = 36

16 Experimental Results Original Images (512 X 512)

17 Experimental Results Modified Images (512 X 512)

18 Experimental Results

19 Experimental Results

20 Wet Paper Coding Key 1 1 1 Fridrich, J. Goljan, M., Lisonek, P. and Soukal, D.,  “Writing on Wet Paper,” IEEE Transactions on Signal Processing, vol. 53, no. 10, pp   

21 The important area is marked as wet pixel
Wet Paper Coding (2/2) The important area is marked as wet pixel 21 30 30 Cover Image × = ? Random Matrix LSB of Cover Image Secret Data 20 30 31 Stego-image

22 Wet Paper Coding with XOR Operation
Key 30 35 31 33 34 32 Eight groups {31}, {35, 31, 32}, {34, 35, 33}, {32}, {33}, {35, 35}, {33, 33, 34}, {32, 32} Secrets: LSB(31) {30} Stego-pixels At least one dry pixel LSB(35) ⊕LSB(31) ⊕ LSB(32) 1 {35, 31, 33}

23 30 35 31 33 34 32 Secret Extracting LSB(30) = 0 LSB(35) ⊕LSB(31) ⊕ LSB(33) =1 LSB(34) ⊕LSB(35) ⊕LSB(33) = 0 LSB(33) = 1 LSB(32) = 0 LSB(35) ⊕LSB(34) = 1 LSB(33) ⊕LSB(33) ⊕LSB(35)= 1 LSB(32) ⊕LSB(33) = 1

24 Conclusions Simple Efficient Good stego image quality

25 Thanks for your listening~


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