Privacy-Preserving Reversible Watermarking for Data Exfiltration Prevention Through Lexicographic Permutations Source: IIH-MSP(2018): 330-339 Authors: Ching-Chun Chang and Chang-Tsun Li Speaker: Jiang-Yi Lin Date: 2019-08-08

Outline Introduction Related Works Proposed scheme Experimental results Conclusions

Introduction (1/3) – Reversible data hiding in encrypted image (RDHEI) Encryption Key 𝐾 𝑒 Embedding Key 𝐾 𝑟 Encrypt the image Embedding Original image m Marked image Key 𝐾 𝑒 Secret data extraction Key 𝐾 𝑟 Image recover & Secret data extraction 𝐾 𝑒 , 𝐾 𝑟

Introduction (2/3) – Vacating room before Encryption Encryption Key 𝐾 𝑒 Embedding Key 𝐾 𝑟 Encrypt the image Embedding Vacating room Original image m Marked image Key 𝐾 𝑒 Secret data extraction Key 𝐾 𝑟 Image recover & Secret data extraction 𝐾 𝑒 , 𝐾 𝑟

Introduction (3/3) – Vacating room after Encryption Encryption Key 𝐾 𝑒 Embedding Key 𝐾 𝑟 Vacating room Encrypt the image Embedding Original image m Marked image Key 𝐾 𝑒 Secret data extraction Key 𝐾 𝑟 Image recover & Secret data extraction 𝐾 𝑒 , 𝐾 𝑟

Related work(1/1)- Lexicographic Permutations 1. Given an original sequence: {12, 1, 7} 2. Permute the original sequence and sort permutation lexicons. 3. For watermark encoding, substitute {12,1,7} with a sorted lexicon {7, 1, 12} to encode the digit ‘2’. {1, 7, 12}: 0 {1, 12, 7}: 1 {7, 1, 12}: 2 {7, 12, 1}: 3 {12, 1, 7}: 4 {12, 7, 1}: 5

Proposed scheme (1/8)-embedding Create the original sequence by the low nybbles of a pixel pair from the encrypted image. 109 98 59 …………. 163 16 129 212 174 149 …………. 146 85 60 113 25 117 ………..... 115 48 241 174 = (10101110)2, low nybble: (1110)2 = 14 85 = (01010101)2, low nybble: (0101)2 = 5 Original sequence: {14, 5} from (174, 85) Permutation lexicons : {5, 14}: 0 {14, 5}: 1 Pixels at the black positions are modifiable in terms of their low nybbles, whereas those at the white positions are unmodifiable.

Proposed scheme (2/8) -embedding To embed bit 0, substitute the lexicon {14, 5} with {5, 14} by modifying the low nybbles of the pixel pair. 109 98 59 …………. 163 16 129 212 165 149 …………. 146 94 60 113 25 117 ………..... 115 48 241 Encrypted image 165 = (1010 0101)2, low nybble: (0101)2 = 5 94 = (0101 1110)2, low nybble: (1110)2 = 14 So the original (174, 85) has been changed to (165, 94).

Proposed scheme (3/8)-Extraction According to the Lexicographic order, the secret bit can be error-free extracted. Encrypted image 109 98 59 …………. 163 16 129 212 165 149 …………. 146 94 60 113 25 117 ………..... 115 48 241 Encrypted image 165 = (1010 0101)2, low nybble: (0101)2 = 5 94 = (0101 1110)2, low nybble: (1110)2 = 14 Lexicographic order: 0 The embedded bit: 0 Low nybbles: {5, 14}

Proposed scheme (4/8) -Extraction Step 1: Receiver restores two candidate encrypted images. 109 98 59 …………. 163 16 129 212 174 149 …………. 146 85 60 113 25 117 ………..... 115 48 241 Low nybbles: {14, 5} 174 = (10101110)2 85 = (01010101)2, Low nybbles: {5, 14} 165 = (1010 0101)2 94 = (0101 1110)2 Encrypted image 1 109 98 59 …………. 163 17 129 212 165 149 …………. 146 94 61 113 25 117 ………..... 116 49 241 Encrypted image 2

Proposed scheme (5/8) -Extraction Step 2: Decrypt two candidate images using the decryption key(same as the encryption key). 97 104 98 …………. 116 117 118 106 100 102 …………. 116 118 121 108 112 101 ………..... 120 119 115 97 104 98 …………. 116 117 118 106 91 102 …………. 116 127 121 108 112 101 ………..... 120 119 115

Proposed scheme (6/8) -Extraction Step 3: A content-adaptive estimation is designed for assisting host pixel recovery. 97 104 98 …………. 116 117 118 106 100 102 …………. 116 118 121 108 112 101 ………..... 120 119 115 Candidate image 1 Origin version Error = Sum(|{100,118} – {99,119}|) = Sum({1,1}) = 2 97 104 98 …………. 116 117 118 106 91 102 …………. 116 127 121 108 112 101 ………..... 120 119 115 Candidate image 2 Error = Sum(|{91,127} – {99,119}|) = Sum({8,8}) = 16

Proposed scheme (7/8)-Predictor

Proposed scheme (8/8)- Enhance distinction Low nybbles: {5, 4} 85 = (0101 0101)2 84 = (0101 0100)2, Low nybbles: {4, 5} 84 = (01010100)2 85 = (01010101)2 Low nybbles: {5, 4} 85 = (01010101)2 84 = (01010100)2 Extraction Pixels={85,84}. e={5, 4}, N=16. p={3,7},q={11,7}. Suppose secret bit w=0. pu is coprime to N e'={5*7, 4*7} (mod N) = {35, 28} (mod N) ={3, 12}. Since w=0, e'={3, 12} keep intact. In the extraction phase. e'={3, 12} => w=0. For any permutations of e’ G = {{3, 12}*q0, {3, 12}*q1}={{1,4},{5,4}} GB={{1,4},{5,4}} Thus, candidate pixels={81,84} or {85,84} G0={3, 12}*11 (mod N) ={33, 132} (mod N) ={1,4} G1={3, 12}*7 (mod N) ={21,84} (mod N) ={5,4} {4, 5}: 0 {5, 4}: 1 {3, 12}: 0 {12, 3}: 1

Experimental results (1/2)

Experimental results (2/2)

CONCLUSIONS Utilize the Lexicographic Permutations for RDHEI. A Content-Adaptive Estimator for Prediction. Enhance Distinction.

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