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
encryption
• Basic idea: for a cleartext letter, ciphertext = pair of letters around it on a straight line
• E.g. FEG and VRY COL EFG QRS
UMB H J TVW
IAD KNP XYZ
• Since there’re multiple adjacent pairs, there’re multiple ways to encrypt.
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. CLO is also a valid encryption
LCOL EFG QRS
UMB H J TVW
IAD KNP XYZ
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
Encryption
• But 2 of them involve the empty space in the middle, so drop them.
• Overall, 24 possible encryptions for each letter.
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
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.
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
decryption
• Case 2: lie on different x-coordinates
• That means cleartext lies on the third, unoccupied x-coordinate
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
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’ ‘
nulls
• Some examples:
MVEPVMVMPEVMMVEP VMMVVMMVOLMVMVMVC
PEEPLOPEEPOLPEEPCC
• The cipher sprinkles nulls at random throughout the ciphertext
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)
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
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);
}
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)
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)
Weaknesses
• subject to bigraph frequency analysis, although nulls and 1-to-24 help a little
• too simple?
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