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 LinSpeaker: Chia-Chun Wu (吳佳駿 )Date: 2008/09/12 1

Outline1. Introduction2. Review of Lin & Tsai’s scheme3. Review of Yang et al.’s scheme4. The Proposed Scheme5. Experimental Results6. Conclusions7. Comments

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.

3

1. Introduction (2/2)

Secret s for the (k, n)-threshold

Randomly choose m1, m2,…, mk-1

Secret share (xi, F (xi))

11

221)(

kk xmxmxmsxF

11

221

121

222212

111

212111

)(

)(

)(

knknnn

kk

kk

xmxmxmsxF

xmxmxmsxF

xmxmxmsxF

4

2. Lin & Tsai’s scheme (1/4)

61 60

59 58

20 20

18 172827

3127

(2, 3)-threshold scheme

60

Secret s

Camouflage Image 1 Camouflage Image 2 Camouflage Image 3

F(xi) = s+ m1xi + m2xi2 + … + mk-1xi

k-1 mod p

F(xi) = s + m1xi mod p

F(27) = 60 + 2 × 27 mod 251 = 114 … F(x1)

F(61) = 60 + 2 × 61 mod 251 = 182 … F(x2)

F(20) = 60 + 2 × 20 mod 251 = 100 … F(x3)5

The Journal of Systems and Software, vol. 73, no. 3, pp. 405-414, Nov.-Dec., 2004.


2827

3127

Camouflage Image 1

F (27) = 60 + 2×27 mod 251 = 114 01110010(2)

Binary Bits

00011100

00011011

00011111

27

Even Parity Check

00011010

00011110

00011101

27

00011010

00011110

00011001

27

Camouflage Image 1 Hiding the values by LSB techniques

Embedding the watermark signal for the authentication

2630

2527

Stego-image 1 6


2630

2527

Stego Image 1

00011010

00011110

00011001

2701001000…..

1: odd0: even

Parity check bits

Secret key K as a seed for generating sequence of binary random number

Match False

Image Recovering Report Failure of Secret Recovery

7


2630

2527

Stego-image 1

6262

6261

Stego-image 2

2020

2120

Stego-image 3

F(27) = s + m1 × 27 mod 251 = 114 … F(x1)

F(20) = s + m1 × 20 mod 251 = 100 … F(x3)

m1 = 2

s = 608

00011010

00011110

00011001

27

01110010(2)=114 01100100(2)=100

00010100

00010100

00010101

20

60

Secret s

3. Yang et al.’s scheme (1/2)

2827

3127

Camouflage Image

F (26) = 60 + 2 × 26 mod GF(28) = 112 01110000(2)

Binary Bits

00011100

00011011

00011111

00011011

Hash Check

00011100

00011000

00011?11

00011001

2824

00011011

25

Camouflage Image

2824

2725

Stego Image

2824

?25

9

m1 = 2, xi = 00011010(2) = 26(2) 60

Secret s

The Journal of Systems and Software, vol. 80, no. 7, pp. 1070-1076, July, 2007.

3. Yang et al.’s scheme (2/2)

2827

3127

Camouflage Image

112(10)=01110000(2)

00011100

00011000

00011Pi

110001100

1

b = HK( 00011001||0001111||00011000||00011110||Block ID || Image ID )

Pi = XOR b = 0 10

00011100

00011011

00011111

00011011

2824

2725

Stego Image

4. The Proposed Scheme (1/5)

11

Diagram of the sharing and embedding procedure


2827

3127

Camouflage Image

F (3) = 5 + 250 × 3 mod 251 = 2 00000010(2)

Binary Bits

00011100

00011011

00011111

00011011

Check Bits

0001110?

0001100?

0001100?

0001100?

00011100

00011001

00011000

00011001

Camouflage Image

2825

2425

Stego Image

2824

?25

12

xi = 00011 (2) = 3(2) 5

Secret s

250


2827

3127

Camouflage Image

2(10)=00000010(2)

(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) 13

2825

2425

Stego Image

00011100

00011011

00011111

00011011

0001110P4

0001100P3

0001100P2

0001100P1


14

0001110P4

0001100P3

0001100P2

0001100P1

1 1

2 2

3 3

4 3

5

6

,

,

,

,

,

,

ij ijk k

ij ijk k

ij ijk k

ij ijk k

ijk

ijk

R X p

R V p

R W p

R Z p

R i

R j

1 (2) (10)

2 (2) (10)

3 (2) (10)

4 (2) (10)

5 (10)

6 (10)

0001100 12 ,

0001100 12 ,

0001100 12 ,

0001110 14 ,

1 ,

1 ,

ijk

ijk

ijk

ijk

ijk

ijk

R

R

R

R

R

R

1 (10)

2 (10)

3 (10)

4 (10)

5 (10)

6 (10)

131 ,

137 ,

139 ,

149 ,

521 ,

523 ,

ijk

ijk

ijk

ijk

ijk

ijk

M

M

M

M

M

M

6

1

(10)

131 137 139 149 521 523

101282017477111 ,

ij ijk kn

n

M M


15

1 1 1

2 2 2

3 3 3

4 4 4

5 5 5

6 6 6

mod ,

mod ,

mod ,

mod ,

mod ,

mod ,

ij ijk k

ij ijk k

ij ijk k

ij ijk k

ij ijk k

ij ijk k

Y M R

Y M R

Y M R

Y M R

Y M R

Y M R

1

2

3

4

5

6

mod 131 12,

mod 137 12,

mod 139 12,

mod 149 14,

mod 521 1,

mod 523 1,

ijk

ijk

ijk

ijk

ijk

ijk

Y

Y

Y

Y

Y

Y

-11 1 1

-12 2 2

-13 3 3

-14 4 4

-15 5 5

-16 6 6

( / ) mod ,

( / ) mod ,

( / ) mod ,

( / ) mod ,

( / ) mod ,

( / ) mod ,

ij ij ij ijk k k k

ij ij ij ijk k k k

ij ij ij ijk k k k

ij ij ij ijk k k k

ij ij ij ijk k k k

ij ij ij ijk k k k

I M M M

I M M M

I M M M

I M M M

I M M M

I M M M

6

(10)1

101282017477111 ,ij ijk kn

n

M M

6

1

/ ,ij ij ij ij ijk kn k kn kn

n

Y R M M I

5. Experimental Results (1/5)

16

The secret and cover images used in the first experiment. (a) The secrete image, (b) the cover images.


17

The experimental results for (2, 3)-threshold secret image sharing scheme. (a) The stego images generated by Lin & Tsai's method

Average = 39.1733 dB


18

The experimental results for (2, 3)-threshold secret image sharing scheme. (b) the stego images generated by Yang et al.'s method



19

The experimental results for (2, 3)-threshold secret image sharing scheme. (c) the stego images generated by the proposed method



20

An example of authenticating the obvious modified stego images. (a) The obvious modified stego image, (b) the authentication results.

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.

6. Conclusions

The authentication is implemented by the concept of CRTAuthentication is improvedTo prevent the participants from cheatingTo improve the quality of stego-imagesTo improve the scheme to a lossless version

21

7. Comments-Yang et al’s scheme (1/2)


60

Secret s

22


2827

3127

Camouflage Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011010(2) = 26(2)

2827

3127

Camouflage Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011010(2) = 26(2)

2827

3127

Camouflage Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011010(2) = 26(2)

7. Comments-Yang et al’s scheme (2/2)


60

Secret s

23


2827

3127

Stego Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011010(2) = 26(2)

00011100

00011111

00011111

00011011

Camouflage Image

xi = 00011011(2) = 27(2)

00011100

00011011

00011111

00011111

Camouflage Image

xi = 00011110(2) = 30(2)

2831

3127

Stego Image

2827

3131

Stego Image

7. Comments-Chang et al’s scheme (1/2)


60

Secret s

24

Pattern Recognition, vol. 41, no. 10, pp. 3130-3137, October 2008.

2827

3127

Camouflage Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011 (2) = 3(2)

2827

3127

Camouflage Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011 (2) = 3(2)

2827

3127

Camouflage Image

00011100

00011011

00011111

00011011

Camouflage Image

xi = 00011 (2) = 3(2)

7. Comments-Chang et al’s scheme (2/2)


60

Secret s

25

Pattern Recognition, vol. 41, no. 10, pp. 3130-3137, October 2008.

2827

3127

Stego Image

Camouflage Image Camouflage Image Camouflage Image

2827

3119

Stego Image

2827

3111

Stego Image

00011100

00011011

00011111

00011011

xi = 00011 (2) = 3(2)

00011100

00011011

00011111

00010011

xi = 00010 (2) = 2(2)

00011100

00011011

00011111

00001011

xi = 00001 (2) = 1(2)

Sharing Secrets in Stego Images with Authentication

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