Conference on theoretical and applied computer science

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Conference on theoretical and applied computer science

  1. 1. Conference on Theoretical and Applied Computer Science (TACS 10) Oklahoma State University Stillwater, OK November 5th, 2010 Recursive Information Hiding in Visual Cryptography Sandeep Katta Computer Science Department Oklahoma State University Stillwater, OK Email: sandeep.katta@okstate.edu
  2. 2. Agenda  Traditional Secret Sharing  Abstract  Introduction  k out of n sharing problem  Model  Properties of 3 out of n scheme  Recursive Information Hiding in 3 out of 5 scheme  General k out of k scheme  Applications
  3. 3. Traditional Secret Sharing  5 bank employees share a bank vault  Management don’t trust a single individual  They assume there will be no collusion between more than 2 of them.  The management split up the password for the vault in such a way that:  Any 3 or more employees working together can have access to vault, but NOT < 3.
  4. 4. Abstract: Visual Cryptography is a secret sharing scheme that uses the human visual system to perform computations.  In this study, a new recursive hiding scheme for 3 out of 5 is proposed.  The idea used is to hide smaller secrets in the shares of a larger secret without an expansion in the size of the latter.
  5. 5. Introduction Visual cryptography (VC) was introduced by Moni Naor and Adi Shamir at EUROCRYPT 1994.  It is used to encrypt written material (printed text, handwritten notes, pictures, etc) in a perfectly secure way. The decoding is done by the human visual system directly, without any computation cost.
  6. 6. VC – How it works  Divide image into two parts:  Key: a transparency  Cipher: a printed page  Separately they are random noise  Combination reveals an image Simple Example
  7. 7. VC – How it works  Every single pixel is split into subpixels.  Human vision still perceives them as one pixel.  The 2 out of 2 method uses:  2 foils  1 pixel with 4 subpixels  This overlay results in black, so the original pixel was also black.
  8. 8. k out of n sharing problem  For a set of n participants, a secret image s is encoded into n shadow images called shares.  Each participant get one share.  The original message is visible if any k or more of them are stacked together, but totally invisible if fewer than k transparencies are stacked together.
  9. 9. Model  Assume the message consists of a collection of black and white pixels and each pixel is handled separately.  Each share is a collection of m black and white subpixels.  The resulting picture can be thought as a [nxm] Boolean matrix S = [si,j]  si,j = 1 if the j-th subpixel in the i-th share is black.  si,j = 0 if the j-th subpixel in the i-th share is white.
  10. 10. Model  Pixels are split:  n shares per pixel: m m n Share 1 Share 2 Share n
  11. 11. Definition  A solution to the k out of n visual secret sharing scheme consists of two collections of n×m Boolean matrices C0 and C1.  To share a white pixel, one of the matrices in C0 is randomly chosen.  For a black pixel, one of the matrices in C1 is randomly chosen.  The chosen matrix defines the color of the m subpixels in each one of the n transparencies.
  12. 12. Definition  The solution is considered valid if the following three conditions are met.  Contrast  For S in C0 (White): H(V) ≤ d – αm  For S in C1 (Black): H(V) ≥ d  Security  The two collections of q×m (1≤q ≤k) matrices formed by restricting n×m matrices in C0 and C1 to any q rows, are indistinguishable.
  13. 13. Preliminary Notation  n → Group Size  k → Threshold  m → Pixel Expansion  α → Relative contrast  C0 → Collection of n × m Boolean matrices for shares of white pixel.  C1 → Collection of n × m Boolean matrices for shares of black pixel.  V → OR’ed k rows  H(V) → Hamming weight of V  d → number in [1,m]  r → Size of collections C0 and C1
  14. 14. Basic 2 out of 2 scheme (2 subpixels)
  15. 15. Basic 2 out of 2 scheme (4 subpixels) Horizontal shares Vertical shares Diagonal shares
  16. 16. Properties of 3 out of n scheme, (n ≥ 3)  Pixel Expansion, m = 2n – 2.  Relative Contrast, α = .  Let B be the black n×(n-2) matrix which contains only 1’s.  Let I be the Identity n × n matrix which contains 1’s on the diagonal and 0’s elsewhere.  BI is an n×(2n-2) concatenated matrix.  C(BI) is the complement of BI.
  17. 17. Properties of 3 out of n scheme, (n ≥ 3)  C0 contains matrices obtained permuting columns of C(BI).  C1 contains matrices obtained permuting columns of BI.  Any single share contains an arbitrary collection of (n-1) black & (n-1) white subpixels.  Any pair of shares has (n-2) common black & two Individual black subpixels.  Any stacked triplet of shares from C0 has n black subpixels.  Any stacked triplet of shares C1 has (n+1) black subpixels.
  18. 18. Matrix design for 3 out of 5 scheme  B = n×(n-2)→5×(5-2) = 5×3  I = n×n = 5×5
  19. 19. Matrix design for 3 out of 5 scheme  Two collections of n × m Boolean matrices C0 and C1  m = 2n – 2 → 2(5) – 2 = 8  Hamming weight of C(BI) i.e. is White H(V) = 5  Hamming weight of BI i.e. is Black H(V) = 8
  20. 20. Matrix design for 3 out of 5 scheme  C0 = {all the matrices obtained by permuting the columns of C(BI)}  C1 = {all the matrices obtained by permuting the columns of BI}  If the columns are not permuted then there is a possibility to reveal the secret information in any single share and therefore the process fails.  For example here are few different permutations  {[1] [8] [2] [7] [3] [6] [4] [5]}  {[2] [4] [6] [8] [1] [3] [5] [7]}  {[3] [2] [1] [8] [7] [6] [4] [5]}
  21. 21. 3 out of 5 scheme (9 subpixels)
  22. 22. 3 out of 5 scheme (9 subpixels)  Superimposition of white and black pixels gives  Need to observe, any single share contains 4 black and 4 white subpixels.  So to make it a complete square array without distorting their aspect ratio, we need to add one more pixel.  It should be either black or white.
  23. 23. Recursive Information Hiding in 3 out of 5 Scheme  RI hiding is a technique where certain additional secret information can be hidden in one of the shares of the original secret image.  Secret information which we are going to hide is taken according to their sizes.  Small images to larger  Shares are distributed at each consecutive level.
  24. 24. Recursive Information Hiding in 3 out of 5 Scheme  Original secret image under consideration is of size 5×5  First secret image is of size 1×1  Second secret image is of size 5×1  Shares of second secret image is designed by seeing the shares of first & original second secret image.
  25. 25. Interpretation of the process  Shares of original secret image are constructed by using the shares of second secret image & placed in diff. levels.  The process gets repeated. Finally when we combine any 3 shares the original secret information is revealed.
  26. 26. Results
  27. 27. General k out of k Scheme  Matrix size = k×  S0 : handles the white pixels  All 2k-1 columns have an even number of 1’s  No two k rows are the same  S1 : handles the black pixels  All 2k-1 columns have an odd number of 1’s  No two k rows are the same  C0/C1 : all the permutation of columns in S0/S1 2k-1
  28. 28. Applications  Remote Electronic voting  Banking Customer Identification  Message Concealment  Key Management
  29. 29. References  Naor, M. and Shamir, A. 1995. Visual Cryptography. Advances in Cryptography-Eurocrypt, 950: 1-12.  Shamir, A. 1979. How to Share a Secret. Communications of the ACM. 22: 612-613.  Gnanaguruparan, M. and Kak, S. 2002. Recursive Hiding of Secrets in Visual Cryptography. Cryptologia 26: 68-76.  Parakh, A. and Kak, S. 2008. A Recursive Threshold Visual Cryptography Scheme. Cryptology ePrint Archive, Report 2008/535.
  30. 30. Questions?

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