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Sharing Secrets in Stego Images with Authentication

Sharing Secrets in Stego Images with Authentication. Source: Pattern Recognition, vol. 41, no. 10, pp. 3130-3137, October 2008. Authors: Chin-Chen Chang, Yi-Pei Hsieh, and Chia-Hsuan Lin Speaker: Chia-Chun Wu ( 吳佳駿 ) Date: 2008/09/12. Outline. 1. Introduction

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Sharing Secrets in Stego Images with Authentication

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  1. Sharing Secrets in Stego Images with Authentication Source: Pattern Recognition, vol. 41, no. 10, pp. 3130-3137, October 2008. Authors: Chin-Chen Chang, Yi-Pei Hsieh, and Chia-Hsuan Lin Speaker: Chia-Chun Wu (吳佳駿) Date: 2008/09/12

  2. Outline • 1. Introduction • 2. Review of Lin & Tsai’s scheme • 3. Review of Yang et al.’s scheme • 4. The Proposed Scheme • 5. Experimental Results • 6. Conclusions • 7. Comments

  3. 1. Introduction(1/2) • The proposed method based on a (k, n)-threshold scheme with the additional capabilities of steganography and authentication. • The fragile image watermarking based on Chinese remainder theorem (CRT) is adopted for image authentication during the secret sharing process.

  4. 1. Introduction(2/2) • Secret s for the (k, n)-threshold • Randomly choose m1, m2,…, mk-1 • Secret share (xi, F (xi))

  5. 60 Secret s The Journal of Systems and Software, vol. 73, no. 3, pp. 405-414, Nov.-Dec., 2004. 2. Lin & Tsai’s scheme (1/4) (2, 3)-threshold scheme 27 31 27 28 Camouflage Image 1 Camouflage Image 2 Camouflage Image 3 F(xi) = s+ m1xi + m2xi2 + … + mk-1xik-1 mod p F(xi) = s + m1xi mod p F(27) = 60 + 2 × 27mod 251 = 114 … F(x1) F(61) = 60 + 2 × 61 mod 251 = 182 … F(x2) F(20) = 60 + 2 × 20 mod 251 = 100 … F(x3)

  6. 27 31 27 28 27 00011111 27 27 00011001 00011101 Camouflage Image 1 00011011 00011100 00011110 00011110 00011010 00011010 27 25 30 26 Stego-image 1 2. Lin & Tsai’s scheme (2/4) 01110010(2) F (27) = 60 + 2×27 mod 251 = 114 Binary Bits Even Parity Check Camouflage Image 1 Hiding the values by LSB techniques Embedding the watermark signal for the authentication

  7. 27 00011001 01001000….. Parity check bits 00011110 00011010 1: odd 0: even 27 25 30 26 Stego Image 1 2. Lin & Tsai’s scheme (3/4) Secret key K as a seed for generating sequence of binary random number Match False Image Recovering Report Failure of Secret Recovery

  8. 20 21 20 20 60 Stego-image 2 Stego-image 3 Secret s 20 27 00010101 00011001 00010100 00011110 00010100 00011010 61 27 62 25 62 30 26 62 Stego-image 1 2. Lin & Tsai’s scheme (4/4) 01110010(2)=114 01100100(2)=100 F(27) = s + m1 × 27mod 251 = 114 … F(x1) F(20) = s + m1 × 20 mod 251 = 100 … F(x3) m1 = 2 s = 60

  9. 27 31 60 27 28 Secret s 00011011 00011111 00011001 25 00011?11 00011011 Camouflage Image 00011011 00011100 00011000 24 28 00011100 25 25 27 ? 24 24 28 28 Stego Image The Journal of Systems and Software, vol. 80, no. 7, pp. 1070-1076, July, 2007. 3. Yang et al.’s scheme (1/2) m1= 2, xi = 00011010(2) = 26(2) F (26) = 60 + 2 × 26 mod GF(28) = 112 01110000(2) Binary Bits Hash Check Camouflage Image

  10. 27 31 27 28 00011011 00011111 00011001 00011Pi11 00011011 00011100 00011000 00011100 25 27 24 28 Stego Image 3. Yang et al.’s scheme (2/2) 112(10)=01110000(2) Camouflage Image b = HK( 00011001||0001111||00011000||00011110||Block ID || Image ID ) Pi = XOR b = 0

  11. 4. The Proposed Scheme (1/5) Diagram of the sharing and embedding procedure

  12. 27 31 27 28 00011011 00011111 00011001 0001100? 00011000 0001100? Camouflage Image 00011011 00011100 0001100? 00011001 0001110? 00011100 25 25 24 ? 25 24 28 28 Stego Image 4. The Proposed Scheme (2/5) xi = 00011 (2) = 3(2) 5 250 F (3) = 5 + 250 × 3 mod 251 = 2 00000010(2) Secret s Binary Bits Check Bits Camouflage Image

  13. 27 31 27 28 00011011 00011111 0001100P1 0001100P2 00011011 00011100 0001100P3 0001110P4 25 24 25 28 Stego Image 4. The Proposed Scheme (3/5) 2(10)=00000010(2) Camouflage Image (a1, a2, a3, a4)= (y1, y2, y3, y4) ⊕ (y5, y6, y7, y8) ⊕ … ⊕ (yn-3, yn-2, yn-1, yn) (p1, p2, p3, p4)=(a1, a2, a3, a4)⊕ (b1, b2, b3, b4)= (1, 0, 1, 0)

  14. 0001100P1 0001100P2 0001100P3 0001110P4 4. The Proposed Scheme (4/5)

  15. 4. The Proposed Scheme (5/5)

  16. 5. Experimental Results (1/5) The secret and cover images used in the first experiment. (a) The secrete image, (b) the cover images.

  17. 5. Experimental Results (2/5) Average = 39.1733 dB The experimental results for (2, 3)-threshold secret image sharing scheme. (a) The stego images generated by Lin & Tsai's method

  18. 5. Experimental Results (3/5) Average = 36.1833 dB The experimental results for (2, 3)-threshold secret image sharing scheme. (b) the stego images generated by Yang et al.'s method

  19. 5. Experimental Results (4/5) Average = 40.9533 dB The experimental results for (2, 3)-threshold secret image sharing scheme. (c) the stego images generated by the proposed method

  20. 5. Experimental Results (5/5) DR=NTPD/NTP, where DR means the detection ratio (DR) against the tampered region, NTP is the number of the tampered pixels, and NTPD is the number of the tampered pixels that are detected. An example of authenticating the obvious modified stego images. (a) The obvious modified stego image, (b) the authentication results.

  21. 6. Conclusions • The authentication is implemented by the concept of CRT • Authentication is improved • To prevent the participants from cheating • To improve the quality of stego-images • To improve the scheme to a lossless version

  22. 27 27 27 31 31 31 60 27 27 27 28 28 28 Secret s 00011011 00011011 00011011 00011111 00011111 00011111 Camouflage Image Camouflage Image Camouflage Image 00011011 00011011 00011011 00011100 00011100 00011100 7. Comments-Yang et al’s scheme (1/2) The Journal of Systems and Software, vol. 80, no. 7, pp. 1070-1076, July, 2007. (2, 3)-threshold scheme Camouflage Image Camouflage Image Camouflage Image xi = 00011010(2) = 26(2) xi = 00011010(2) = 26(2) xi = 00011010(2) = 26(2)

  23. 27 27 31 31 31 31 60 31 27 27 28 28 28 Secret s 00011111 00011011 00011011 00011111 00011111 00011111 Stego Image Stego Image Stego Image 00011011 00011111 00011011 00011100 00011100 00011100 7. Comments-Yang et al’s scheme (2/2) The Journal of Systems and Software, vol. 80, no. 7, pp. 1070-1076, July, 2007. (2, 3)-threshold scheme Camouflage Image Camouflage Image Camouflage Image xi = 00011010(2) = 26(2) xi = 00011011(2) = 27(2) xi = 00011110(2) = 30(2)

  24. 27 27 27 31 31 31 60 27 27 27 28 28 28 Secret s 00011011 00011011 00011011 00011111 00011111 00011111 Camouflage Image Camouflage Image Camouflage Image 00011011 00011011 00011011 00011100 00011100 00011100 7. Comments-Chang et al’s scheme(1/2) Pattern Recognition, vol. 41, no. 10, pp. 3130-3137, October 2008. (2, 3)-threshold scheme Camouflage Image Camouflage Image Camouflage Image xi = 00011 (2) = 3(2) xi = 00011 (2) = 3(2) xi = 00011 (2) = 3(2)

  25. 11 19 27 31 31 31 60 27 27 27 28 28 28 Secret s 00011011 00010011 00001011 00011111 00011111 00011111 Stego Image Stego Image Stego Image 00011011 00011011 00011011 00011100 00011100 00011100 7. Comments-Chang et al’s scheme(2/2) Pattern Recognition, vol. 41, no. 10, pp. 3130-3137, October 2008. (2, 3)-threshold scheme Camouflage Image Camouflage Image Camouflage Image xi = 00011 (2) = 3(2) xi = 00010 (2) = 2(2) xi = 00001 (2) = 1(2)

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