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Where to 'S'tart (Posted on 2004-11-30) Difficulty: 4 of 5

1st pass encryption - MIF.CLGMFELCMLM.
2nd pass encryption - VO.IOSTSKQOPYY.
3rd pass encryption - K.XXHMMDBFEOX.

1. What answer would Zero pass encryption reveal, given that the encryption method does not change.
2. How much more difficult is it to get to zero level encryption if you don't have the level 1 values, only level 2?

See The Solution Submitted by Juggler    
Rating: 2.1667 (6 votes)

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Some Thoughts moderate spoiler on method; but not the answer | Comment 5 of 7 |

I think I see the pattern on how to get from 1st pass to 2nd pass etc.  Convert the letters to numbers:  A=1, B=2 etc.  Then add the first 2 letters in the first row (mod 26) to get the first letter of the second row.

eg  M+I = V  or 13 + 9 = 22
then I+F = O  etc.

So for the i-th "pass", the j-th character is given by:
Char(i,j)= Char(i-1, j) + Char(i-1, j+1)   again, mod 26

Haven't figured completely how to go "up" to the previous pass, but I'm thinking there will be a way to have n equations and n unknowns:  n-1 of the equations will be the Char(i,j) equations above, and then there should be one more equation based on the sum of all the characters in the message.   For the row above, adding each of the n characters once plus characters 2 through n-1 one more time, should be equal in some way to the sum of all the characters one row below.  But I'm not sure how the mod 26 business affects that sum.


  Posted by Larry on 2004-12-01 04:51:08
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