Efficient reversible data hiding based on the histogram
Efficient reversible data hiding based on the histogram modification of differences of pixel differences Source: Multimedia Tools and Applications, Apr. 2020. DOI: 10. 1007/s 11042 -02008623 -0 Author: Huang Delu and Wang Jianjun Speaker: Su Guodong Date: 2020/06/04 1
Outline • Introduction • Related works • Proposed scheme • Experimental results and analysis • Conclusions 2
Introduction Stego image Original image Data hiding (Difference based Histogram shifting) 3
Related works(1/4)---Pixel-difference based RDH scheme Original image xi 162 162 162 162 163 163 167 162 163 162 158 di 162 0 0 0 1 1 0 4 5 1 1 4 7 7 5 5 2 0 1 1 2 3 4 1 5 2 … p=0 di 0 1 2 3 4 5 1 6 … di 4
Related works(2/4)---Pixel-difference based RDH scheme Original image 162 162 163 163 167 162 163 162 158 p=0 d 7 = |xi-1 – xi| = |x 7 – x 7| = |162 – 163| = 1 (x 7 = 163) > (x 6 = 162) x 7 = 163 + 1 = 164 Shifted image 162 164 d 8 = |xi-1 – xi| = |x 7 – x 8| = |163 – 162| = 1 (x 8 = 162) < (x 7 = 163) x 8 = 162 - 1 = 161 162 0 0 0 1 1 1 0 4 5 1 1 4 di 161 164 168 161 164 161 157 5
Related works(3/4)---Pixel-difference based RDH scheme (Embedding) Shifted image Message: 0 1 1 0 0 0 1 … y 1 = x 1 + 0 = 162 162 163 y 2 = x 2 + 1 = 163 162 162 164 Stego image 162 162 164 y 3 = x 3 + 1 = 163 161 164 163 168 161 164 161 157 p=0 Original image 162 162 163 163 167 162 163 162 158 di 162 0 0 0 1 1 1 0 4 5 1 1 4 6
Related works(4/4)---Pixel-difference based RDH scheme (Extraction) Message: 0 1 1 0 0 0 1 di = yi – xi-1 Stego image Original image 162 163 162 d 1 = 0 m=0 x 1 = y 1 = 162 162 164 162 163 d 2 = 1 m=1 x 2 = y 2 - 1= 162 163 d 3 = 1 m=1 x 3 = y 3 - 1= 162 161 164 168 161 164 161 157 162 164 d 7 = 2 x 7 = y 7 - 1= 163 161 d 8 = 2 x 8 = y 8 + 1= 162 162 163 162 163 167 162 163 162 158 7
Proposed scheme (1/7)---Differences of Pixel differences (DPD) Original image 162 162 163 163 167 162 163 162 158 i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 xi 162 162 163 163 167 162 163 162 158 i = 12 i=2 163 167 162 162 d 12 = |x 12 – x 11| - | x 11 – x 10| = 5 – 4 =1 d 2 = |x 2 – x 1| - | x 1 – x 0| = 0 i 0 1 di 162 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 0 1 0 0 -1 4 1 -4 0 3 Comparison of three different histograms of “Lena” image 8 … 1 1 -4 -3 -2 -1 0 2 1 1 3 4 … di 8
Proposed scheme(2/7)---Flowchart 9
Proposed scheme(3/7)---Preprocess Location map (RLE compression) (a) original histogram; (b) shifted histogram 9
Proposed scheme(4/7)---First part of embedding Message: Original image Shifted image 0 1 1 0 0 0 1 1 1 st stego image 162 162 163 162 162 164 163 162 164 162 163 167 162 163 168 162 164 163 168 162 163 162 158 161 163 162 157 161 163 161 157 162 162 163 p=0 d 7 = |163 -162| - |162 -162| = 1 162 0 0 i 5 yi 162 0 0 0 1 0 0 -1 4 1 -4 0 3 6 163 164 162 163 y 7 = x 7 + 1 = 164 d’ 7 = 2 7 d 2 = 0 y 2 = x 2 + 0 = 162 d 3 = 0 d 4 = 0 d 5 = 0 y 3 = x 3 + 1 = 163 y 4 = x 4 + 1 = 163 y 5 = x 5 + 0 = 162 d 6 = 0 d 8 = 0 d 9 = 0 d 14 = 0 y 6 = x 6 + 0 = 162 y 8 = x 8 - 0 = 162 y 9 = x 9 + 1 = 164 y 14 = x 14 - 1 = 161 9
Proposed scheme(5/7)---Second part of embedding Message: 1 0 1 2 st stego image 1 st stego image Shifted image 162 162 162 163 162 161 163 162 164 162 164 163 168 162 164 163 169 161 163 161 157 163 162 164 p’ = -1 < p = 0 162 0 1 -1 2 0 0 -1 4 2 -5 0 2 d’ 13 = |162 -161| - |161 -168| = -5 i 11 12 z 2 = y 2 - 1 = 161 d 6 = -1 z 4 = y 4 + 0 = 163 d 10 = -1 z 8 = y 8 - 1 = 161 13 169 168 161 163 yi 168 z 11 = y 11 + 1 = 169 d’’ 13 = -6 d 4 = -1 9
Proposed scheme(6/7)---Extraction 162 161 163 d’’ 15 = 2 d’’ 14 = 0 1 st stego image y 13 = z 13 = 163 y 13 = z 13 = 161 162 162 163 162 164 p’ = -1 164 163 169 161 163 161 157 1 0 1 163 161 157 2 st stego image Message: (2) (1) 162 164 163 168 (3) (6) 161 163 161 157 169 161 163 161 164 163 d’’ 13 = -6 < (p’ = -1) d’’ 8 = -2 < (p’ = -1) y 11 = z 11 – 1 = 168 y 8 = z 8 + 1 = 162 9
Proposed scheme(7/7)---Extraction (1) 2 st stego image 162 162 163 Message: 0 1 1 0 0 0 1 1 (2) 162 162 162 163 d’ 2 = 0 > (p = 0) d’ 3 = 1 > (p = 0) x 2 = y 2 = 162 163 162 164 x 3 = y 3 – 1 = 162 162 163 p=0 162 164 163 168 161 163 161 157 Original image 162 163 167 (6) 162 163 162 158 (9) 162 164 162 163 d’ 7 = 2 > (p =0) d’ 10 = -1 < (p = 0) x 7 = y 7 – 1 = 163 x 10 = y 10 = 163 9
Experiments(1/4) 15
Experiments(2/4)-Comparisons [2] Anushiadevi R, Praveenkumar P, Rayappan JBB, Amirtharajan R (2017) Reversible secret data hiding based on adjacency pixel difference. J Artif Intel 10: 22– 31. [10] Hung K-M, Hsieh C-T, Chen T-W, Chen L-M (2016) Reversible data hiding based on improved multilevel histogram modification of pixel differences. J Appl Sci Eng 19(4): 489– 495. [15] Li X, Zhang W, Gui X, Yang B (2013) A novel reversible data hiding scheme based on two-dimensional difference-histogram modification. IEEE Trans Inf Forens Security 8(7): 1091– 1100. [33] Zhao Z, Lou H, Lu Z-M, Pan J-S (2011) Reversible data hiding based on multilevel histogram modification and sequential recovery. AEU Int J Electron Commun 65(10): 814– 826 16
Experiments(3/4)-Comparisons [10] Hung K-M, Hsieh C-T, Chen T-W, Chen L-M (2016) Reversible data hiding based on improved multilevel histogram modification of pixel differences. J Appl Sci Eng 19(4): 489– 495. [11] Kamstra L, Heijmans HJAM (2005) Reversible data embedding into images using wavelet techniques and sorting. IEEE Trans Image Process 14(12): 2082– 2090 [18] Li F, Mao Q, Chang C-C (2018) Reversible data hiding scheme based on the Haar discrete wavelet transform and interleaving prediction method. Multimed Tools Appl 77: 5149– 4168 [23] Ou B, Li X, Zhang W, Zhao Y (2019) Improving pairwise PEE via hybrid-dimensional histogram generation and adaptive mapping selection. IEEE Trans Circuits Syst Video Technol 29(7): 2176– 2190 17
Experiments(4/4)---Comparisons Results of comparison of the proposed extended method with other methods 18
Conclusions ØA novel RDH method based on DPD ØDPD-based histogram has a higher peak point, which results in more embeddable pixels and higher visual quality of the marked images. ØThe proposed method is extended by histogram modification techniques to achieve a higher EC. 19
Thanks! 20
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