1. A lossless data hiding scheme based on three- pixel block differences Ching-Chiuan Lin and Nien-Lin Hsueh Pattern Recognition, Vol. 41(4), April 2008 Pages 1415-1425

2. 2 Outline Introduction Related Work The Proposed Method Experiments Conclusions Comments

3. 3 Lossless data hiding Three-pixel block difference Introduction (1/3)

4. 4 Introduction (2/3) Lossless data hiding Application: medical images, military photos, law enforcement Challenges: Capacity Quality

5. 5 Introduction (3/3) Lossless data hiding in spatial domain: Difference expansion Reversible data embedding using a difference expansion, Jun Tian, IEEE Transactions on Circuits and Systems for Video Technology, vol. 13(8), pp. 890 – 896, Aug. 2003 Reversible watermark using the difference expansion of a generalized integer transform, Alattar, A.M. IEEE Transactions on Image Processing, vol. 13(8), pp. 1147 - 1156, Aug. 2004 Adaptive lossless steganographic scheme with centralized difference expansion, C.C. Lee, H.C. Wu, C.S. Tsai, and Y.P. Chu, Pattern Recognition, Vol. 41, Issue 6, 2008, pp. 2097-2106 Histogram modification Reversible data hiding, Z. Ni, Y.Q. Shi, N. Ansari, and W. Su, IEEE Trans. Circuits Syst. Video Technol., vol.16, no.3, pp.354–362, March 2006 Hiding Data Reversibly in an Image via Increasing Differences between Two Neighboring Pixels, C.C. Lin and N.L. Hsueh, IEICE TRANS. INF. & SYST., VOL.E90–D, NO.12 DE. 2007 A lossless data hiding scheme based on three-pixel block differences, C.C. Lin and N.L. Hsueh, Pattern Recognition 41 (2008) 1415 – 1425

6. 6 Related Work~shift(1/3) 33313 43354 43222 01315 22554 original image 33313 53365 53222 01316 22665 histogram a=3 b=6 peak zero Reversible data hiding, Z. Ni, Y.Q. Shi, N. Ansari, and W. Su, IEEE Trans. Circuits Syst. Video Technol., vol.16, no.3, pp.354–362, March 2006

7. 7 Related Work~embedding (2/3) histogram a=3 b=6 peak zero Secret data: 1 0 1 1 1 0 0 1 43414 54365 53222 01416 22665 stego image 33313 53365 53222 01316 22665

8. 8 Related Work~extraction (3/3) histogram a=3 b=6 Extraction data: 1 0 1 1 1 0 0 1 43414 54365 53222 01416 22665 stego image 33313 43354 43222 01315 22554 original image peakzero

9. 9 The Proposed Methods (1/12) Cover image 1 st Scan Data:0110101… Obtain a optimal histogram of difference values Stego-imageMessage and overhead info. : 01… Remaining message : 01… 2 nd scan Overhead info. + residual message : 10101110…

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11. 11 The Proposed Methods (2/12) Scan Image to get the max. & min. count of difference values 343134542107 889543654367 647766345454 743152123567 245545374368 1x3 Type 0 Difference value Count 02 124 25 36 42 50 60 71 …… 24-1=23 M=1,m=5 g(M)=23,g(m)=0 1x3 Type 1: M=1,m=5,g(M)=16,and g(m)=0 1x3 Type 2: M=1,m=5,g(M)=14,and g(m)=0 3x1 Type 0: M=1,m=7,g(M)=7,and g(m)=0 3x1 Type 1: M=1,m=5,g(M)=8,and g(m)=0 3x1 Type 2: M=1,m=7,g(M)=7,and g(m)=0 M=1,m=5 g(M)=23,g(m)=0 Overhead info. : BO=0,BT=0,M=1,and m=5 will be embeded in the next embedding plane

12. 12 1 st scan( 處理有 1 或 254 的 block) For each block i satisfying the following two conditions:  (1) 1 ≦ b i0, b i1, b i2 ≦ 254  (2)max(b i0, b i1, b i2 )=254 or min(b i0, b i1, b i2 )=1 Call the embedding_procedure to embed the message After invoking the embedding procedure, if max(b i0, b i1, b i2 ) = 255 or min(b i0, b i1, b i2 ) = 0, record block i as an FB(fresh boundary block) in the overhead information For example, if M = 1 and m = 20, (2, 1, 2)  (1,1)  (2,2)  (2, 0, 2) if it embeds“11”  FB (253, 254, 240)  (1,4)  (1,5)  (253, 254, 239) if it embeds “0” (246, 243, 254)  (3,11)  (4,12)  (246, 242,254) embedding_procedure: 1.increase d i0 by 1 if M+1 ≦ d i0 ≦ m−1, and increase d i1 by 1 if M + 1 ≦ d i1 ≦ m − 1; 2.embed data bits into block i if d i0 = M or d i1 = M.

13. 13 2 nd scan ( 處理其它 block~ 沒有 1 或 254) For each block i with 2 ≦ b i0, b i1, b i2 ≦ 253 call the procedure embedding_procedure to first embed the overhead information, and then the residual message. For example, if M = 1 and m = 20, (3, 4, 6)  (1,2)  (1,3)  (3, 4, 7) if it embeds“0”

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16. 16 The Proposed Methods (3/12) 343134542107 889543654367 647766345454 743152123567 245545374368 histogram M=1 m=5 peak zero 343034641107 889543653267 637766245454 842162124567 146535384269 Difference value Secret data: 001 001 00001 0 01 0001 0011 1 stego image 1 st embed #FB IFB BO BT M m 2 nd embed

17. 17 The Proposed Methods (4/12) Embed (0,0)Extraction 789 789

18. 18 The Proposed Methods (5/12) Embed (0,1)Extraction 7 89 78 10 2 9 1

19. 19 The Proposed Methods (6/12) Embed (1,0)Extraction 789 6 2 6 89 1 7

20. 20 The Proposed Methods (7/12) Embed (1,1)Extraction 7 8 9 6 2 6 8 10 1 7 2 9 1

21. 21 The Proposed Methods (8/12) Embed (0,1)Extraction 7 8 6 6 7 8

22. 22 The Proposed Methods (9/12) Embed (1,0)Extraction 7 8 6 6 7 8

23. 23 The Proposed Methods (10/12) Embed (1,1)Extraction 7 8 9 9 7 8

24. 24 The Proposed Methods (11/12) Overhead Information  Secret data: (0011) 2  BO´ =0, BT´ =1, M´ =1, and m´ =3 343134542107 889543654367 647766345454 743152123567 245545374368 1 st scan: max(b i0,b i1,b i2 )=254 or min(b i0,b i1,b i2 )=1 1,3,41,5,21,2,3 0,3,4 1,6,21,2,4 Embed ‘0’Embed xEmbed ’01’ Recorded as a FB Overhead information : #FBs ∥ IFB ∥ BO´ ∥ BT´ ∥ M ´ ∥ m ´

25. 25 The Proposed Methods (12/12) Overhead Information Overhead info.1Plane 1: Embed Message Plane 2: Embed Message + Overhead info.1 Overhead info.2 Plane n: Embed Message + Overhead info.n-1Overhead info.n ●●●●●● #FBs ∥ IFBs ∥ BO ∥ BT ∥ M ∥ m #FBs 1 ∥ IFB 1 #FBs 2 ∥ IFB 2 ∥ BO 1 ∥ BT 1 ∥ M 1 ∥ m 1 #FBs n ∥ IFB n ∥ Bo n-1 ∥ BT n-1 ∥ M n-1 ∥ m n-1 || #EPs Bo n ∥ BT n ∥ M n ∥ m n

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27. 27 Extraction Follow the BO and the BT to divide the stego-image into blocks. For each block i with 1 ≦ b i0, b i1, b i2 ≦ 254, perform extraction and recovery Save the extracted message bits in the list List-1 if min(bi0, bi1, bi2) = 1 or max(bi0, bi1, bi2) = 254 Save the extracted message bits in the list List-2 if 2 ≦ b i0, b i1, b i2 ≦ 253. List-2 contains the overhead information (List-2’) and the message(List-2’’) Decode the overhead information in List-2’ For each FB, which is recorded in List-2, perform extraction and recovery, then save the extracted data in List-3 According to the indexes of the blocks where the message bits are extracted, reorder the message bits saved in List-1 and List-3 to form the list List-4 Push the message in List-2 first, then the message List-4 onto a stack. Repeat the above procedures until the message embedded in each embedding plane is completely extracted. Pop the data stored in the stack

28. 28 Experiments (1/5) Test images: (512*512) (a) Baboon; (b) Boat; (c) Lena; (d) Airplane; (e) Pepper; (f) Tiffany; (g) GoldHill; (h) Gradient complex smooth

29. 29 Experiments (2/5)

30. 30 Experiments (3/5)

31. 31

32. 32 Experiments (5/5)

33. 33 Conclusions For PSNR ≧ 30dB,the pure payload can be up to 1.32bpp,among the test images. The payload capacity of the proposed scheme is about six to ten times higher than that of Ni et al.’s algorithm. The proposed scheme can provide a high payload capacity and better image quality for applications that losslessly embed a message in a cover image.

34. 34 Comments Generalized  3-pexel block(2/3)  4-pexel block(3/4)  n- pexel block(n/n-1) Difference between two neighbor pixels  take the medium value as a reference value while calculating the difference values

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36. 36 Shifting histogram (1/6) 25310 42501 23446 33125 44442 histogram Cover image peak zero 26310 42601 23447 33126 44442 Shifted cover image histogram a = 4 b = 7

37. 37 Shifting histogram (2/6) Secret data ： 1010110 26310 42601 23447 33126 44442 26310 52601 23457 33126 45542 5  Embedding Stego-image Shifted cover image histogram Peak = 4 Zero = 7

38. 38 Shifting histogram (3/6)  Extraction 26310 52601 23457 33126 45542 Stego-image peakzero a = 4 b = 7 Secret data ： 1 010110 Recovered image 25 3 10 4 2501 23446 33 1 25 44442 histogram

39. 39 Shifting histogram (4/6) Total hiding capacity = # of peak point pixels = 7bits. Needs to keep a pair of pixel values on peak point and zero point for extraction and image recovery.

40. 40 maximum and minimum points (5/6) 25300 42516 23447 33125 33332 original image maximum point minimum point a=3 b=6 Record the coordinate of those pixels of b as overhead bookkeeping information hidden data=secret data + overhead data

41. 41 multiple pairs of maximum and minimum points (6/6) 25300 42516 23447 33125 33332 original image a1b1a2b2