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

3 Gray-Scale Image ．．． ．．． ．．． ． Pixel values

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

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

6 LSB Substitution, (permutation)

7 Original LSB Substitution,

8 LSB Substitution, transformation ？

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

11

12 Time consuming!

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 Square differences matrix Dynamic Programming Strategy

15 Square differences matrix Dynamic Programming Strategy

16 The Way to Produce Square Differences Matrix

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 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 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 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 min Dynamic Programming Strategy

22 min Dynamic Programming Strategy

23 min Dynamic Programming Strategy

24 Dynamic programming strategy Dynamic Programming Strategy

25 Drawback of the above LSB substitution methods LSB substitution Our method A pixel of Host-image Secret data Mse= ( ) 2 = ( ) 2 = Mse= ( ) 2 = ( ) 2 =9 A pixel of stego-image

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

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

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

29 Pixel adjustment Our method Case 4 A pixel of Host-imageSecret data It satisfied | |=5>2 3-1 and 111> and > Use pixel adjustment =

30 Pixel adjustment Our method Case 5 A pixel of Host-imageSecret data It satisfied | |=5>2 3-1 and Use pixel adjustment =

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

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 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 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 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 Our method Dynamic programming with Pixel adjustment min

39 Our method Dynamic programming with Pixel adjustment min

40 Our method Dynamic programming with Pixel adjustment Dynamic programming strategy

41 Experimental Results

42 Experimental Results

43 Experimental Results

44 Experimental Results

45 Experimental Results

46 Experimental Results

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

48 Q & A ?

49 Thank You !!