Challenge "Alberti Challenge - Part 2"

Challenge "Alberti Challenge - Part 2"  

  By: admin on May 14, 2022, 12:22 p.m.

A more difficult and even more interesting sequel to the "Alberti Challenge - Part 1". Will you manage to crack this puzzle this time as well?

Re: Challenge "Alberti Challenge - Part 2"  

  By: george4096 on May 27, 2022, 10:46 p.m.

Very nice challenge.

One comment: Although I solved the challenge (and found the underlying plaintext, through a combination of cryptanalysis and detective work), and entered the two Unicity Distance values, that were accepted as correct, actually, the assumed keyspace size (and unicity distance) when neither the inner disk nor the indicator are known, are actually wrong :-).

The answer expected here assumes that the key space size in this case is 26! (the number of possible inner disks) multiplied by 26 (the number of possible indicators). But there are equivalent combinations, that will provide the same encryptions/decryptions:

abcdefghijklmnopqrstuvwxyz with indicator a is eqivalent to bcdefghijklmnopqrstuvwxyza with indicator a cdefghijklmnopqrstuvwxyzab with indicator a .... zabcdefghijklmnopqrstuvwxy with indicator a

So the effective size of the keyspace is 26!, and not 26! x 26.

Great challenge, regardless of this minor issue!


Re: Challenge "Alberti Challenge - Part 2"  

  By: RandyWaterhouse on May 28, 2022, 11:53 a.m.

Dear George,

thanks for the nice comment and congratulations to being the first to solve the challenge!

Seems I messed up the unicity distance calculation, sorry for that!

Kind regards, Peter

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