Sharing secrets in stego images with authentication
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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

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

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


1 introduction 1 2

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.


1 introduction 2 2

1. Introduction(2/2)

  • Secret s for the (k, n)-threshold

    • Randomly choose m1, m2,…, mk-1

    • Secret share (xi, F (xi))


2 lin tsai s scheme 1 4

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)


2 lin tsai s scheme 2 4

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


2 lin tsai s scheme 3 4

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


2 lin tsai s scheme 4 4

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


3 yang et al s scheme 1 2

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


3 yang et al s scheme 2 2

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


4 the proposed scheme 1 5

4. The Proposed Scheme (1/5)

Diagram of the sharing and embedding procedure


4 the proposed scheme 2 5

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


4 the proposed scheme 3 5

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)


4 the proposed scheme 4 5

0001100P1

0001100P2

0001100P3

0001110P4

4. The Proposed Scheme (4/5)


4 the proposed scheme 5 5

4. The Proposed Scheme (5/5)


5 experimental results 1 5

5. Experimental Results (1/5)

The secret and cover images used in the first experiment.

(a) The secrete image, (b) the cover images.


5 experimental results 2 5

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


5 experimental results 3 5

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


5 experimental results 4 5

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


5 experimental results 5 5

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.


6 conclusions

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


7 comments yang et al s scheme 1 2

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)


7 comments yang et al s scheme 2 2

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)


7 comments chang et al s scheme 1 2

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)


7 comments chang et al s scheme 2 2

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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