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Simple LSB embedding (recap) CSM25 Secure Information Hiding Dr Hans Georg Schaathun University of Surrey Spring 2007 Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 1 / 34 Lesson Outcomes Chapter 2: LSB recap After this

  1. Simple LSB embedding (recap) CSM25 Secure Information Hiding Dr Hans Georg Schaathun University of Surrey Spring 2007 Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 1 / 34

  2. Lesson Outcomes Chapter 2: LSB recap After this session, everyone should be able to us Matlab to load an image and make simple changes handle binary information convert text to binary understand that any piece of information can be represented in numerous ways We may repeat material from CSM24 Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 2 / 34

  3. Review of LSB The image in the spatial domain Outline Review of LSB 1 The image in the spatial domain Numbers in Bits LSB of an image Merits and flaws Some steganalytic observations 2 The visual attack Histogramme Conclusions Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 3 / 34

  4. Review of LSB The image in the spatial domain A tiny grayscale example Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 4 / 34

  5. Review of LSB The image in the spatial domain A tiny grayscale example Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 4 / 34

  6. Review of LSB The image in the spatial domain The image file: pgm/pnm (spatial domain) P2 24 24 255 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 97 86 26 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 36 171 85 54 35 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 48 22 0 0 52 26 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 146 0 0 0 52 26 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 146 0 0 0 52 26 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 146 0 0 0 12 77 26 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 146 0 0 0 49 139 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 36 146 132 0 0 0 32 94 15 0 0 0 0 0 0 0 0 0 0 0 0 12 111 122 65 79 99 0 0 0 0 47 23 0 0 0 0 0 0 0 0 0 0 0 73 111 128 88 22 0 79 0 0 0 0 97 49 0 0 0 0 0 0 0 0 0 36 146 132 102 26 17 0 0 19 59 0 0 0 97 49 0 0 0 0 0 0 12 111 122 65 79 19 0 0 0 0 0 0 66 13 0 77 81 0 0 0 0 0 0 73 111 128 88 22 0 0 0 0 0 0 0 0 26 101 97 103 15 0 0 0 0 36 146 132 102 26 17 0 0 0 0 0 0 0 0 0 6 37 49 11 0 0 12 49 142 149 79 31 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 173 77 47 41 58 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12 37 6 57 99 71 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 128 137 70 45 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 171 49 0 45 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 36 0 0 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 45 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 22 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 5 / 34

  7. Review of LSB The image in the spatial domain Image formats pnm (netpbm) 8-bit grayscale (pgm) 24-bit RGB (ppm) Binary, i.e. 1-bit black/white (pbm) png (Portable Network Graphics) Kodak Photo-CD bmp Some file formats (e.g. TIFF) can carry different image formats. Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 6 / 34

  8. Review of LSB Numbers in Bits Outline Review of LSB 1 The image in the spatial domain Numbers in Bits LSB of an image Merits and flaws Some steganalytic observations 2 The visual attack Histogramme Conclusions Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 7 / 34

  9. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  10. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  11. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  12. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  13. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  14. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  15. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  16. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  17. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  18. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

  19. Review of LSB Numbers in Bits The coefficients in bits Take a number 217. Written Base 10 i.e. 217 = 7 · 10 0 + 1 · 10 1 + 2 · 10 2 Least significant digit: the last/the ones Same numer 217 10 = 11011001 2 Written in Base 2 11011001 2 = 1 · 2 0 + 0 · 2 1 + 0 · 2 2 + 1 · 2 3 + 1 · 2 4 + 0 · 2 5 + 1 · 2 6 + 1 · 2 7 Least significant bit: the last/the ones Any base could be used. (8 and 16 are common.) Dr Hans Georg Schaathun Simple LSB embedding (recap) Spring 2007 8 / 34

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