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Cubic cipher

Cubic cipher. key generation. First, form a Keystring like Playfair E.g. Keyword=COLUMBIA Keystring=COLUMBIADEFGHJKNPQRSTVWXYZ. key generation. Then wrap keystring around a 3X3X3 cube, but leave center empty. 2D and 3D views: z=0 z=1 z=2 COL EFG QRS UMB H J TVW IAD KNP XYZ.

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Cubic cipher

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  1. Cubic cipher

  2. key generation First, form a Keystring like Playfair E.g. Keyword=COLUMBIA Keystring=COLUMBIADEFGHJKNPQRSTVWXYZ

  3. key generation • Then wrap keystring around a 3X3X3 cube, but leave center empty. • 2D and 3D views: z=0 z=1 z=2 COL EFG QRS UMB H J TVW IAD KNP XYZ

  4. encryption • Basic idea: for a cleartext letter, ciphertext = pair of letters around it on a straight line • E.g. FEG and VRY COL EFG QRS UMB H J TVW IAD KNP XYZ • Since there’re multiple adjacent pairs, there’re multiple ways to encrypt.

  5. Encryption • Adjacent pairs can go beyond the cube and “wrap back” to it (or think of the cube as being surrounded by copies of itself) • E.g. CLO is also a valid encryption LCOL EFG QRS UMB H J TVW IAD KNP XYZ

  6. encryption • Two other examples: CIU CDM

  7. encryption • A less intuitive example: CPV COL EFG QRS UMB H J TVW IAD KNP XYZ

  8. encryption • 13 such symmetric pairs (above/below, left/right, in front/behind, in front-above/behind-below, …) • Mathematically… (x, y, z = coordinates of letter) 1. x+1 and x-1 2. y+1 and y-1 3. z+1 and z-1 4. x+1, y-1 and x-1, y+1 5. x+1, y+1 and x-1, y-1 6. x+1, z-1 and x-1, z+1 7. x+1, z+1 and x-1, z-1 8. y+1, z-1 and y-1, z+1 9. y+1, z+1 and y-1, z-1 10. x+1, y+1, z+1 and x-1, y-1, z-1 11. x+1, y+1, z-1 and x-1, y-1, z+1 12. x+1, y-1, z+1 and x-1, y+1, z-1 13. x+1, y-1, z-1 and x-1, y+1, z+1

  9. encryption • Each pair has two directions • E.g. COL or LO • So, total of 13X2 = 26 pairs

  10. Encryption • But 2 of them involve the empty space in the middle, so drop them. • Overall, 24 possible encryptions for each letter.

  11. encryption • Which of the 24 encryptions to pick? • Use a distribution function to decide • Right now, distribution function simply returns random integer from 0 to 24 All 24 encryptions have equal chance • More on this later.

  12. decryption • Basic idea: take the two ciphertext characters, find the letter that forms a straight line with them. • Technical term? Zeph: “It’s called Cubic projection… or something like that.” • Very simple, but not sure how to describe in English.

  13. decryption • Use 2D as example. Only 2 possibilities • Case 1: lie on the same x-coordinate • That means cleartext lies on the same coordinate also

  14. decryption • Case 2: lie on different x-coordinates • That means cleartext lies on the third, unoccupied x-coordinate

  15. decryption • For our case, basically repeat the above for all 3 planes (xy, xz, yz) of the cube

  16. nulls • Nulls = meaningless characters in the ciphertext to confuse cryptanalyst • In my cipher, you can produce nulls by encrypting the empty space in the middle

  17. nulls • E.g. EP/PE and MV/VM are adjacent pairs of the empty space. COL EFG QRS COL EFG QRS UMB H J TVW UMB H J TVW IAD KNP XYZ IAD KNP XYZ • So, when decrypted, EP, PE, MV, and VM conveniently become the empty character and disappear: EP’ ‘

  18. nulls • Some examples: MVEPVMVMPEVMMVEP VMMVVMMVOLMVMVMVC PEEPLOPEEPOLPEEPCC • The cipher sprinkles nulls at random throughout the ciphertext

  19. nulls • (Note: since the adjacent pairs of the empty space cannot contain the empty space itself, all adjacent pairs are valid. So there are 26 possible nulls, not just 24.) • (Another note: because of nulls, ciphertext is in general more than twice as long as cleartext)

  20. Sample encryptions • Cleartext: “If one examines dialectic materialism, one is faced with a choice: either accept textual neocultural theory or conclude that narrative is created by the masses. Many narratives concerning the role of the observer as reader exist.” (From Postmodernism Generator) • Ciphertext: YUNENIQMBITVHESQKOJOHEXFHGFNNZPXZAWYXJQOTVRNQIHEUYTXGELWCXVPBIATRVTRMTXFCFBKNIUXEHOUPVZAQXDVBVNRYXIDNKYWIGTQRACIHMPZGNZDXFOWWYAQUOWGOSSHEHZYANUOBQSDOWYMDJTOEHMLARANMGPDLMQWWLUKWVPSGRKZAULBFBDIYVGYJRSLZIGXLMRWBICWTAHZZDWYVIHEUAXBXUIHGXJCMZCBPSHSWYQFWLPOPLSDMDPVHWIZQIFXMXUOZKAZEPRNGONFMFPYSHWKPVYQEIHZSHRXTSNFLTOKGBIDGBMGSOQXGEUCTJEHHZXUEQGOUTKYKOHSBEHOFNYJYMJMJTPZZRTAVPTJNOZSSBVECIYXUTWHUEDHWKEUSCCXIDAMHETVQLYNQXIBHWSUXRMGMDWLDALSYXGFPXWKBQSPUNQGVRKIIQ

  21. Further development? • As I mentioned, right now all 24 (26 for null) ciphers for each letter have equal probability. • Dist.Func. just returns random int in [0, 25] int distFunc(char X) { return rand.nextInt(26); }

  22. Further development? • But can make it more complicated by limiting the choices for some letters to less than 24 • E.g. frequency balancing for homophony • Or the opposite: “unbalance” it (make the ciphertext for ‘Z’ the most frequency bigraph) to throw people off (Shane’s suggestion)

  23. Further development? • Encrypt the ciphertext again using the same key or some key derived from it? • Instead of adjacent pairs, use weirder pairs (or perhaps determine what to use by looking at the keyword)

  24. Weaknesses • subject to bigraph frequency analysis, although nulls and 1-to-24 help a little • too simple?

  25. Thanks! All Hail the Fu

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