1. 1 Reversible Watermark Using the Difference Expansion of a Generalized Integer Transform Source : IEEE TRANSACTIONS ON IMAGE PROCESSING, VOL. 13, NO. 8, AUGUST 2004, pp.1147-1156 Author : Adnan M. Alattar Speaker : Jen-Bang Feng ( 馮振邦 )

2. 2 Outline Generalized Difference Expansion The Proposed Scheme Embedding Reversible Watermark Reading Watermark and Restoring Image Payload Size Data-Rate Controller Recursive and Cross-Color Embedding Experimental Results Conclusion

3. 3 The Proposed Scheme Hide watermark into protected image Can fully restore the original image Use the difference of neighborhood pixels

4. 4 Generalized Difference Expansion (1/5) vector u=(u 0, u 1, …, u N-1 ) a i is a constant integer, i=0, …,N-1 v 0 is the weighted average of the entities of the vector. v 1, …,v N-1 are the differences between u 1, …,u N-1 and u 0, respectively. The Inverse Difference Expansion Transform f -1 (.) : The Forward Difference Expansion Transform f(.) :

5. 5 (u 0, u 1, u 2, u 3 ) = (8, 10, 14, 15) (a 0, a 1, a 2, a 3 ) = (1, 2, 2, 1) ======================= v 0 = = = 11 v 1 = 10 – 8 = 2 v 2 = 14 – 8 = 6 v 3 = 15 – 8 = 7 ======================= (v 0,v 1,v 2,v 3 ) = (11,2,6,7) (a 0, a 1, a 2, a 3 ) = (1, 2, 2, 1) ======================= u 0 = 11 - = 11- 3 = 8 u 1 = 2 + 8 = 10 u 2 = 6 + 8 = 14 u 3 = 7 + 8 = 15 ======================= (u 0,u 1,u 2,u 3 ) = (8,10,14,15) Generalized Difference Expansion (2/5)

6. 6 Def1. Expandable For all value of b 1, …,b N-1 (u 0, u 1, u 2, u 3 ) = (8, 10, 14, 15) (v 0, v 1, v 2, v 3 ) = (11, 2, 6, 7) (a 0, a 1, a 2, a 3 ) = (1, 2, 2, 1) (b 1, b 2, b 3 ) = (1, 0, 1)  embedded data ======================== v 0 = 11 ṽ 1 = 2 * 2 + 1 = 5 ṽ 2 = 2 * 6 + 0 = 12 ṽ 3 = 2 * 7 + 1 = 15 ũ 0 = 11 – = 11 – 8 = 3 ũ 1 = 5 + 3 = 8 ũ 2 = 12 + 3 = 15 ũ 3 = 15 + 3 = 18 Generalized Difference Expansion (3/5) 000111  001110  00111b

7. 7 Def2. changeable (u 0, u 1, u 2, u 3 ) = (8, 10, 14, 15) (v 0, v 1, v 2, v 3 ) = (11, 2, 6, 7) (a 0, a 1, a 2, a 3 ) = (1, 2, 2, 1) (b 1, b 2, b 3 ) = (1, 0, 1)  embedded data ======================== v 0 = 11 ṽ 1 = 2 * + 1 = 3 ṽ 2 = 2 * + 0 = 6 ṽ 3 = 2 * + 1 = 7 ũ 0 = 11 – = 11 – 4 = 7 ũ 1 = 3 + 7 = 10 ũ 2 = 6 + 7 = 13 ũ 3 = 7 + 7 = 14 Generalized Difference Expansion (4/5) 000111  00011  000110  00011b

8. 8 Generalized Difference Expansion (5/5) S 1 : expandable and |v i | ≤ T i, i=1,…,N-1 S 2 : changeable S 3 : not changeable S 4 = S 1 U S 2 Location map : 1  S 1 0  S 2 or S 3 V 0 is unchanged. An expandable vector is also changeable. A changeable vector remains changeable even after changing the LSBs.

9. 9 Embedding Reversible Watermark (1/7) Image I(i, j, k) R G B U R ={u 00,u 01,u 02 } U G ={u 10,u 11,u 12 } U B ={u 20,u 21,u 22 } Secret key U={u 00,u 10,u 20 u 01,u 11,u 21 u 02,u 12,u 22 } Assume threshold=30

10. 10 Embedding Reversible Watermark (2/7) a1221 U150161151155 V1541115 Back150161151155 Expanda ble 15423311 CU1144167147155 Changea ble 1541115 CU2150161151155 a1221 U197198199200 V198123 Back197198199200 Expanda ble 198357 CU1195198200202 Changea ble 198133 CU2197198200 a1221 U100210110101 V140110101 Back100210110101 Expanda ble 140221213 CU1592808062 Changea ble 140111111 CU2100211111101 u 00 (expandable)u 01 (expandable) u 02 (changeable)

11. 11 a1221 U200100110101 V120-100-90-99 Back200100110101 Expanda ble 120-199-179-197 CU12798010082 Changea ble 120-99-89-99 CU2200101111101 a1221 U200100190101 V146-100-10-99 Back200100190101 Expanda ble 146-199-19-197 CU12525323355 Changea ble 146-99-9-99 CU2199100190100 a1221 U150145155153 V150-553 Back150145155153 Expanda ble 150-9117 CU1149140160156 Changea ble 150-553 CU2150145155153 u 10 (changeable) u 11 (changeable) u 12 (expandable) Embedding Reversible Watermark (3/7)

12. 12 a1221 U255 V 000 Back255 Expanda ble 255111 CU1255256 Changea ble 255111 CU2255256 a1221 U0000 V0000 Back0000 Expanda ble 0111 CU10111 Changea ble 0111 CU20111 a1221 U510811 V8536 Back510811 Expanda ble 811713 CU1011713 Changea ble 8537 CU2510812 u 20 (non-changeable) u 21 (expandable) u 22 (expandable) Embedding Reversible Watermark (4/7)

13. 13 Embedding Reversible Watermark (5/7) U={u 00,u 10,u 20 u 01,u 11,u 21 u 02,u 12,u 22 } V = f(U) V={v 00,v 10,v 20 v 01,v 11,v 21 v 02,v 12,v 22 } S 1 ={v 00,v 01,v 21,v 12,v 22 } S 2 ={v 10,v 11,v 02 } S 3 ={v 20 } S 4 ={v 00,v 10,v 01,v 11,v 21, v 02,v 12,v 22 } Location Map = 100101011 B 2 = 001001001 Compress location map to form B 1,and append EOS. B 3 :watermark B=B 1 B 2 B 3

14. 14 Embedding Reversible Watermark (6/7) S 4 ={v 00,v 10,v 01,v 11,v 21, v 02,v 12,v 22 } Location Map = 100101011 v 00 :expandable b=(1,0,0) u 00 =(197,198,199,200) v 00 =(198,1,2,3) ṽ 00 =(198,3,4,6) ũ 00 =(195,198,199,201) v 10 :changeable b=(1,0,1) u 10 =(200,100,110,101) v 10 =(120,-100,-90,-99) ṽ 10 =(120,-99,-90,-99) ũ 10 =(200,101,110,101) v 01 :expandable b=(0,1,1) u 01 =(150,161,151,155) v 01 =(154,11,1,5) ṽ 01 =(154,22,3,11) ũ 01 =(144,166,147,155)

15. 15 Ŝ 1 ={ṽ 00, ṽ 01, ṽ 21, ṽ 12, ṽ 22 } Ŝ 2 ={ṽ 10, ṽ 11, ṽ 02 } ῦ = f -1 (Ṽ) ũ 00, ũ 01, ũ 21, ũ 12, ũ 22 ũ 10, ũ 11, ũ 02 R G B watermarked image I w (i, j, k) Embedding Reversible Watermark (7/7)

16. 16 Reading Watermark and Restoring Image (1/3) R G B watermarked image I w (i, j, k) Secret key Ũ R ={ũ 00, ũ 01, ũ 02 } Ũ G ={ũ 10, ũ 11, ũ 12 } Ũ B ={u 20, ũ 21, ũ 22 } Ũ={ũ 00, ũ 10, u 20 ũ 01, ũ 11, ũ 21 ũ 02, ũ 12, ũ 22 }

17. 17 Ṽ = f(Ũ) Ṽ ={ṽ 00, ṽ 10,v 20 ṽ 01, ṽ 11, ṽ 21 ṽ 02, ṽ 12, ṽ 22 } S 3 ={v 20 } Ŝ 4 ={ṽ 00, ṽ 10, ṽ 01, ṽ 11, ṽ 21, ṽ 02, ṽ 12, ṽ 22 } Extract LSBs of each vector in Ŝ 4 B=B 1 B 2 B 3 Identify EOS Extract B 1 Location map =100101011 Ŝ 1 ={ṽ 00, ṽ 01, ṽ 21, ṽ 12, ṽ 22 } Ŝ 2 ={ṽ 10, ṽ 11, ṽ 02 } Reading Watermark and Restoring Image (2/3)

18. 18 Ŝ 1 ={ṽ 00, ṽ 01, ṽ 21, ṽ 12, ṽ 22 } Ŝ 2 ={ṽ 10, ṽ 11, ṽ 02 } b 1, …,b N-1 are taken from B to restore V. S 1 ={v 00, v 01, v 21, v 12, v 22 } S 2 ={v 10, v 11, v 02 } U = f -1 (V) u 00, u 01, u 21, u 12, u 22 u 10, u 11, u 02 R G B Image I(i, j, k) The remainder B B 3 :watermark Reading Watermark and Restoring Image (3/3)

19. 19 Payload Size (1/2) ||x|| indicates number of elements in x ||B 2 || = ( N – 1 ) ||S 2 || Factors of the size 1. Number of expandable vectors. 2. How well the location map can be compressed.

20. 20 Payload Size (2/2) w x h host image, (w x h) / N vectors α є [0,1] : the percent of the expandable vector in image. β є [0,1] : the compression rate of the location map ||S 1 || = α((w x h) / N) ||B 1 || = β ((w x h) / N) N increases  The number of expandable vectors may decrease.  Thresholds must be increased to maintain the same number of selected expandable vectors.  The quality of the embedded image decreases.

21. 21 Data-Rate Controller Assume N is fixed. C is the payload length. λє (0,1) : constant T(n)={T 1 (n),…,T n-1 (n)} T(0) is a preset value.

22. 22 Recursive Embedding

23. 23 Cross-Color Embedding cross-color triple u = (R,G,B) cross-color quad u = (R,G,G,B) ||B 3 || = 2||S 1 || - ||B 1 || ||B 3 || = (2α – β) x w x h Example. R=2 G=1 B=1 V 0 = = 1 V 1 = R – G = 1 V 2 = B – G = 0 ====================== G = V 0 – = 1 – 0 = 1 R = V 1 + G = 1 + 1 = 2 B = V 2 + G = 0 + 1 = 1

24. 24 Experimental Results (1/6) Comparison between the performance of the spatial, triplet-based algorithm and the spatial, quad-based algorithm applied to the image once.

25. 25 Experimental Results (2/6) Comparison between the performance of the spatial, triplet-based algorithm and the spatial, quad-based algorithm applied to the image twice.

26. 26 Experimental Results (3/6) Embedded using the cross-spectral, triplet-based algorithm. Embedded using the cross-spectral, quad-based algorithm.

27. 27 Experimental Results (4/6) Payload size versus PSNR for Lena Size of compressed map versus PSNR for Lena

28. 28 Experimental Results (5/6) Comparison results between spatial, triplet-based algorithm and Tian’s using grayscale images. Comparison results between spatial, quad- based algorithm and Tian’s using grayscale images.

29. 29 Experimental Results (6/6) Comparison results between spatial, quad- based algorithm and Celik’s using grayscale images. Comparison results between spatial, triplet- based algorithm and Celik’s using grayscale images.

30. 30 Conclusion A reversible watermarking scheme The amount of data embedded into a image depends highly on the nature of the image

31. 31 Comments Integrated issues But unknown purpose Low robustness No distortion allowed Low capacity