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## PowerPoint Slideshow about 'Cubic cipher' - Gideon

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Presentation Transcript

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

- Each pair has two directions
- E.g. COL or LO
- So, total of 13X2 = 26 pairs

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

decryption

- For our case, basically repeat the above for all 3 planes (xy, xz, yz) of the cube

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?

Thanks!

All Hail the Fu

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