Reversible Data Hiding in Encrypted Images Based on Progressive Recovery Source： IEEE Signal Processing Letters (Accepted)2016 Authors：Zhenxing Qian, Xinpeng Zhang, Guorui Feng Speaker ：Xiaozhu Xie Date ： 2016/09/29 The title I would like to present is “”. This paper has been accepted by “” this year, not yet final published . The authors are all from SHU, China. From the title, we may tell that the innovation of this paper is progressive recovery, which means , unlike other methods, it recovers the image step by step instead of in one step.

Outline Scenario Proposed schemes Experimental results Conclusions I will present in four parts. So first let us turn to scenario.

Scenario Embed additional messages(e.g.,lables, time stamps, category information etc.) KEMB KENC KENC [səˈnærioʊ] cloud storage is becoming more and more popular, I bet everyone here uses cloud storage more or less. However the security problem followed. For example, many photos of Hollywood actresses leaked from iCloud two years ago. So before upload the images, we would like to encrypt it. And on the server side, cloud server would like to embed some information for management, for example…. So

Proposed scheme(1/13)-Framework synthesis [ˈsɪnθɪsɪs] n. synthetize ['sɪnθəˌtaɪz] v The second part is proposed scheme. Before going into details, I would like to show the framework first. As we can see, there are three roles, content owner, data hider, and recipient. There is no difference in content owner part, just use the encryption key to encrypt the image. And the encrypted image will be separated into three sets, and then secret message will be embedded into each set respectively. After that, three sets synthetize to get the marked encrypted image.

Proposed scheme(2/13)- Data Embedding 125 76 155 62 90 50 57 24 96 118 138 78 60 82 97 211 114 216 49 148 66 89 160 144 56 41 44 33 21 14 125 76 155 62 90 50 57 24 96 118 138 78 60 82 97 211 114 216 49 148 66 89 160 144 56 41 44 33 21 14 Denoted as S1 (Sized MN/4) Denoted as S2 Sized MN/4 𝑋⊕ 𝐾 𝐸𝑁𝐶 divided into three sets Denoted as S3 Sized MN/2 Original image (Sized M*N) Now let us go to details~ we get an original image sized M by N. With an encryption key, we get the encrypted image. As JY has explained the process of encryption fully last time, I will just move on.. So the data-hider get the ..image. Encrypted image Three Sets of the encrypted Image

Proposed scheme(3/13)- Data Embedding 125 76 155 62 90 50 57 24 96 118 138 78 60 82 97 211 114 216 49 148 66 89 160 144 56 41 44 33 21 14 Seg1(1)=(125,155,90,50) Seg1(2)=(78,125,50,60) … Seg1(MN/4L1)=(…) Divide each set 𝑆 𝑖 into segments containing 𝐿 𝑖 pixels (L1>L2, and L1>L3) Take 𝑆 1 for example, let L1=4 Collect 3-LSBs in each segment Encrypted image Generate three binary matrices 𝐺 1 , 𝐺 2 , 𝐺 3 , for sets 𝑆 1 , 𝑆 2 , 𝑆 3 respectively 𝐵 1 1 =(1,0,1,0,1,1,0,1,0,0,1,0) 𝑇 size of 3 𝐿 1 ×1 𝐵 1 2 = 1,1,0,1,0,1,0,1,0,1,0,0 𝑇 … 𝐵 1 𝑀𝑁/ 4𝐿 1 =(…) before moving on, I would like to go into some details on the generation of G. compress 𝐾 𝐸𝑀𝐵

Proposed scheme(4/13)- Data Embedding 𝐼 is the identity matrix, Qi is the pseudo-randomly generated binary matrix controlled by KEMB, sized (3 𝐿 𝑖 −𝑃)×𝑃 compressed Vacate a spare room of P bits in each group. 𝑘 𝑖 ∈[1, 𝑅 𝑖 ] is the group index. G consist of two parts.

Proposed scheme(5/13)- Data Embedding 0 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 . 1 1 0 1 0 1 0 1 0 1 0 0 𝐶 1 2 = 𝐺 1 ∙ 𝐵 1 2 = =(1,0,0,1,1,0,0,0,0) 𝑇 Let us come back to the example… Take 𝐵 1 2 for example, let p=3.

Proposed scheme(6/13)- Data Embedding 𝐶 1 2 =(1,0,0,1,1,0,0,0,0) 𝑇 Embed additional bits 𝐴 1 2 = (0,1,0) 𝑇 𝐵 1 ` 2 =(1,0,0,1,1,0,0,0,0,0,1,0) 𝑇 Replaces 𝐵 1 2 with 𝐵 1 ` 2 Seg1`(2)=(76,126,48,58)

Proposed scheme(7/13)- Message Extraction 124 78 154 61 90 77 50 62 57 26 96 119 24 138 76 126 48 58 82 97 211 114 123 215 49 60 149 67 89 64 118 160 144 56 41 125 44 35 99 210 113 80 19 Seg1`(1)=(…) Seg1`(2)=(76,126,48,58) … Seg1`(MN/4L1)=(…) Divide each set 𝑆 𝑖 into segments containing 𝐿 𝑖 pixels Collect 3-LSBs in each segment Marked encrypted image The proposed scheme is separable. KEMB Case1: Embedded bits 010 𝐵 1 ` 2 =(1,0,0,1,1,0,0,0,0,0,1,0) 𝑇

Proposed scheme(8/13)- Directly decrypted image Case 2: K ENC ⊕ K ENC The marked encrypted image The directly decrypted image, with distortion of 3LSBs

Proposed scheme(9/13)- Message Extraction& progressive recovery KEMB Case 3: Directly decrypted image, with distortion of 3LSBs K ENC Original 𝐵 𝑖 𝑘 𝑖 can be achieved by calculating: 2 to the power of p 𝒂 is an arbitrary binary vector with P bits, and there are 2 𝑃 possible candidates for each group.

Proposed scheme(10/13)- Message Extraction& progressive recovery KEMB 𝐵 1 ` 2 =(1,0,0,1,1,0,0,0,0,0,1,0) 𝑇 𝐵 1 2 =(1,0,0,1,1,0,0,0,0,0,0,0) 𝑇 +𝑎∙ 𝐻 1 𝐻 1 = 𝑄 1 𝑇 𝐼 𝑃 = 0 1 0 0 1 1 0 1 0 1 0 0 1 1 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 1 2 3 possible candidates for 𝐵 1 2 𝑎= 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑥 𝑖 ∈ 0,1 ,𝑖∈[1,3] 2 3 possible candidates for 𝑎 When a=(1,0,0), we get original 𝐵 1 2 = 1,1,0,1,0,1,0,1,0,1,0,0 𝑇

Proposed scheme(11/13)- Message Extraction& progressive recovery

Proposed scheme(12/13)- Message Extraction& progressive recovery First round Step1: estimate pixel values within the Square set: 75 69 78 74 73 72 77 79 70 76 83 80 81 82 7 𝑝 3,3 = 76 8 + 75 8 + 74 8 +[ 80 8 ] 4 ∙8+4=78 Step 2: For each candidate vector 𝐵 1 ( 𝑘 1 ), put these bits into the original 3LSB-layers to construct an enciphered pixel segments, and then decipher the pixel segments using KENC , finally we get the pixel values 𝑡 𝑖,𝑗 . Directly decrypted image 2 𝑃 possible 𝐵 1 ( 𝑘 1 ) 2 𝑃 possible 𝑡 𝑖,𝑗

Proposed scheme(12/13)- Message Extraction& progressive recovery First round 75 69 78 74 73 72 77 79 70 76 83 80 81 82 7 Step 3: Calculate the difference D. 𝑡 𝑖,𝑗 that makes D smallest is the original pixel. 𝐺 1 𝑘 1 :The decrypted segment. Step 4: Update the reference image by substituting pixels in the square set with the original pixel 𝑡 𝑖,𝑗 . Directly decrypted image

Proposed scheme(13/13)- Message Extraction& progressive recovery Second round: Third round:

Experimental results (1/6) (a) original image (b) the marked encrypted image (c) the directly decrypted image (d) the losslessly recovered image Image sized 512×512. Fixed parameters P=5 and {L1=150, L2=125, L3=100} Hide 11350 bits (0.043 bpp) PSNR = 38.1dB

Experimental results (2/6) [10] X. Zhang, Separable reversible data hiding in encrypted image, IEEE Transactions Information Forensics and Security, 7(2): 826–832, 2012

Experimental results (3/6) [5] X. Zhang, Reversible data hiding in encrypted images, IEEE Signal Processing Letters, 18(4): 255–258, 2011 [6] W. Hong, T. Chen, and H. Wu, An improved reversible data hiding in encrypted images using side match, IEEE Signal Processing Letters, 19(4): 199–202, 2012 [7] M. Li, D. Xiao, A. Kulsoom, and Y. Zhang, Improved reversible data hiding for encrypted images using full embedding strategy, Electronic Letters, 51(9): 690-691, 2015 [10] X. Zhang, Separable reversible data hiding in encrypted image, IEEE Transactions Information Forensics and Security, 7(2): 826–832, 2012

Experimental results (4/6) Error rate in each round using different parameters

Experimental results (5/6) Error propagations, (a): errors in first round vs. the second, (b): errors in second round vs. the third

Experimental results (6/6) n represents the number of LSB-layers used for data embedding

CONCLUSIONS Extend the traditional recovery to the progressive based recovery. Provide a better prediction way for estimating the LSB-layers of the original image using three rounds. Improve the embedding rate.

Thank you!