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

Re: Challenge "Alberti Challenge - Part 2"  

  By: madness on Aug. 20, 2022, 11:12 p.m.

It's worse than that, because some parts of keys are irrelevant for short texts. E.g., with ciphertext 4, there are 7 letters in the inner alphabet that are undetermined, since they are not used to encrypt. So maybe you should divide by 7!.

Re: Challenge "Alberti Challenge - Part 2"  

  By: RandyWaterhouse on Aug. 21, 2022, 12:09 p.m.

Not sure I can follow that logic. If, for example, you have a simple monoalphabetic substitution but the message does not contain alle available letters from the plain alphabet (which, for short messages, typically is the case) one still has a keyspace of 26!. The unicity distance is, as far as I understand it, a feature of the encryption system/algorithm and not dependend on the specific messages one encrypts.

Re: Challenge "Alberti Challenge - Part 2"  

  By: madness on Sept. 1, 2022, 12:44 a.m.

I'm saying, e.g. for ciphertext 4, that there are 7! inner alphabets that will give the same solution.

Re: Challenge "Alberti Challenge - Part 2"  

  By: RandyWaterhouse on Sept. 1, 2022, 7:04 a.m.

I get that. But I don't think it's relevant, the unicity distance is a function of the cipher system/cipher algorithm, not of the specific message. As I said in my earlier post, you often have the situation that several keys deliver the same description, especially for short messages. That's true for many cipher systems.

Re: Challenge "Alberti Challenge - Part 2"  

  By: xiaozhuan on Jan. 20, 2024, 4:29 p.m.

Good challenge, fun to solve and I learned a lot. Thanks.

Re: Challenge "Alberti Challenge - Part 2"  

  By: RandyWaterhouse on Jan. 21, 2024, 7:42 a.m.

Thanks for the nice comment, I appreciate it!

Currently 185 guests and 0 members are online.
Powered by the CrypTool project
Contact | Privacy | Imprint
© 2009-2024 MysteryTwister team