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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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2 Outline Gray-Scale Image Information Hiding Least-Significant-Bit (LSB) Substitution Dynamic Programming Strategy Our method Experimental Results Future Works
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3 Gray-Scale Image 124 126 134 128 ... 124 130 111 230 ... 98 99 121 111 ... . Pixel values
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4 Information Hiding Host image Secret image Embed Extract Transmit Stego-image
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5 Least-Significant-Bit (LSB) Substitution, 2 10011100 S
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6 LSB Substitution, (permutation)
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7 Original LSB Substitution,
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8 LSB Substitution, transformation ?
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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
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12 Time consuming!
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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)
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14 Square differences matrix Dynamic Programming Strategy
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15 Square differences matrix Dynamic Programming Strategy
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16 The Way to Produce Square Differences Matrix
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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
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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
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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
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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
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21 min Dynamic Programming Strategy
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22 min Dynamic Programming Strategy
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23 min Dynamic Programming Strategy
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24 Dynamic programming strategy Dynamic Programming Strategy
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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
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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
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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
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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
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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
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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
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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
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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
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33 Our method Dynamic programming with Pixel adjustment Dynamic programming with traditional LSB
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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
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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
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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
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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
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38 Our method Dynamic programming with Pixel adjustment min
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39 Our method Dynamic programming with Pixel adjustment min
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40 Our method Dynamic programming with Pixel adjustment Dynamic programming strategy
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41 Experimental Results
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42 Experimental Results
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43 Experimental Results
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44 Experimental Results
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45 Experimental Results
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46 Experimental Results
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47 Future works Just-Noticeable Distortion (JND) Multiple Base
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48 Q & A ?
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49 Thank You !!
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