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1 An LSB Substitution base Information Hiding Technique 國立彰化師範大學 資訊工程學系教授兼系主任 蕭如淵 (Ju-Yuan Hsiao) 中華民國九十四年十二月十六日.

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Presentation on theme: "1 An LSB Substitution base Information Hiding Technique 國立彰化師範大學 資訊工程學系教授兼系主任 蕭如淵 (Ju-Yuan Hsiao) 中華民國九十四年十二月十六日."— Presentation transcript:

1 1 An LSB Substitution base Information Hiding Technique 國立彰化師範大學 資訊工程學系教授兼系主任 蕭如淵 (Ju-Yuan Hsiao) 中華民國九十四年十二月十六日

2 2 Outline Gray-Scale Image Information Hiding Least-Significant-Bit (LSB) Substitution Dynamic Programming Strategy Our method Experimental Results Future Works

3 3 Gray-Scale Image 124 126 134 128 ... 124 130 111 230 ... 98 99 121 111 ... . Pixel values

4 4 Information Hiding Host image Secret image Embed Extract Transmit Stego-image

5 5 Least-Significant-Bit (LSB) Substitution,  2 10011100  S

6 6 LSB Substitution, (permutation)

7 7 Original LSB Substitution,

8 8 LSB Substitution, transformation ?

9 9 Exhaustive LSB Substitution H H-R Z R Finding an optimal solution Replace and Attach Secret image (n-bits/pixel) Secret image (k-bits/pixel) Host image (n-bits/pixel) Residual image (k-bits/pixel) … Stego-image (n-bits/pixel) ((n-k)-bits/pixel) Positions transform

10 10

11 11

12 12 Time consuming!

13 13 Dynamic Programming Strategy H H-R Z R Finding an optimal solution Replace and Attach Secret image (n-bits/pixel) Secret image (k-bits/pixel) Host image (n-bits/pixel) Residual image (k-bits/pixel) Stego-image (n-bits/pixel) ((n-k)-bits/pixel)

14 14 Square differences matrix Dynamic Programming Strategy

15 15 Square differences matrix Dynamic Programming Strategy

16 16 The Way to Produce Square Differences Matrix

17 17 Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=9 Cost[3,{1}]=m[3][1]+Cost[4,{}]=4 Cost[3,{2}]=m[3][2]+Cost[4,{}]=1 Cost[3,{3}]=m[3][3]+Cost[4,{}]=0 Dynamic Programming Strategy

18 18 Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=9 Cost[3,{1}]=m[3][1]+Cost[4,{}]=4 Cost[3,{2}]=m[3][2]+Cost[4,{}]=1 Cost[3,{3}]=m[3][3]+Cost[4,{}]=0 Dynamic Programming Strategy

19 19 Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=9 Cost[3,{1}]=m[3][1]+Cost[4,{}]=4 Cost[3,{2}]=m[3][2]+Cost[4,{}]=1 Cost[3,{3}]=m[3][3]+Cost[4,{}]=0 Dynamic Programming Strategy

20 20 Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=9 Cost[3,{1}]=m[3][1]+Cost[4,{}]=4 Cost[3,{2}]=m[3][2]+Cost[4,{}]=1 Cost[3,{3}]=m[3][3]+Cost[4,{}]=0 Dynamic Programming Strategy

21 21 min Dynamic Programming Strategy

22 22 min Dynamic Programming Strategy

23 23 min Dynamic Programming Strategy

24 24 Dynamic programming strategy Dynamic Programming Strategy

25 25 Drawback of the above LSB substitution methods LSB substitution Our method A pixel of Host-image 10010001110 Secret data 10010110 Mse= ( 10010001 2 - 10010110 2 ) 2 = (145- 150) 2 =25 10001110 Mse= ( 10010001 2 - 10001110 2 ) 2 = (145- 142) 2 =9 A pixel of stego-image

26 26 Pixel adjustment Our method Case 1 A pixel of Host-imageSecret data 10010001010 It satisfied |001-010|=1 ≦ 2 3-1 Use traditional LSB

27 27 Pixel adjustment Our method Case 2 A pixel of Host-imageSecret data 11111110000 It satisfied |110-000|=6>2 3-1 and 000 255- 2 3-1 Use traditional LSB

28 28 Pixel adjustment Our method Case 3 A pixel of Host-imageSecret data 00000010111 It satisfied |010-111|=5>2 3-1 and 111> 2 3-1 and 00000010< 2 3-1 Use traditional LSB

29 29 Pixel adjustment Our method Case 4 A pixel of Host-imageSecret data 11010010111 It satisfied |010-111|=5>2 3-1 and 111> 2 3-1 and 11010010> 2 3-1 Use pixel adjustment 11010111-1000=11001111 3 5 210 215 207

30 30 Pixel adjustment Our method Case 5 A pixel of Host-imageSecret data 11010111000 It satisfied |111-010|=5>2 3-1 and 010 255-2 3-1 Use pixel adjustment 11010000+1000=11011000 1 7 216208 215

31 31 Our method Dynamic programming with Pixel adjustment Dynamic programming with LSB M[3][0]=(7- 4) 2 +(8-8) 2 +(0-0) 2 =9 Our method M[3][0]=(7-8) 2 +(8-8) 2 +(0-0) 2 =1

32 32 Our method Dynamic programming with Pixel adjustment Dynamic programming with LSB M[3][3]=(7-7) 2 +(8-11) 2 +(0-3) 2 =18 Our method M[3][3]=(7-7) 2 +(8-7) 2 +(0-3) 2 =10

33 33 Our method Dynamic programming with Pixel adjustment Dynamic programming with traditional LSB

34 34 Our method Dynamic programming with Pixel adjustment Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=1 Cost[3,{1}]=m[3][1]+Cost[4,{}]=6 Cost[3,{2}]=m[3][2]+Cost[4,{}]=5 Cost[3,{3}]=m[3][3]+Cost[4,{}]=10

35 35 Our method Dynamic programming with Pixel adjustment Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=1 Cost[3,{1}]=m[3][1]+Cost[4,{}]=6 Cost[3,{2}]=m[3][2]+Cost[4,{}]=5 Cost[3,{3}]=m[3][3]+Cost[4,{}]=11

36 36 Our method Dynamic programming with Pixel adjustment Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=1 Cost[3,{1}]=m[3][1]+Cost[4,{}]=6 Cost[3,{2}]=m[3][2]+Cost[4,{}]=5 Cost[3,{3}]=m[3][3]+Cost[4,{}]=11

37 37 Our method Dynamic programming with Pixel adjustment Cost[4,{}]=0 Cost[3,{0}]=m[3][0]+Cost[4,{}]=1 Cost[3,{1}]=m[3][1]+Cost[4,{}]=6 Cost[3,{2}]=m[3][2]+Cost[4,{}]=5 Cost[3,{3}]=m[3][3]+Cost[4,{}]=11

38 38 Our method Dynamic programming with Pixel adjustment min

39 39 Our method Dynamic programming with Pixel adjustment min

40 40 Our method Dynamic programming with Pixel adjustment Dynamic programming strategy

41 41 Experimental Results

42 42 Experimental Results

43 43 Experimental Results

44 44 Experimental Results

45 45 Experimental Results

46 46 Experimental Results

47 47 Future works Just-Noticeable Distortion (JND) Multiple Base

48 48 Q & A ?

49 49 Thank You !!


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