The Thirteenth Labour

Season 1 Card #251 - The Thirteenth Labour

Card text

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

The absolute basics

• We assume a block of text has been scrambled (encrypted) so that we cannot read it
• the RC5.64 standard refers to how it has been scrambled
• This scrambling requires a 'Key' to unlock it - we are searching for that key.
• So there are two camps
  • one camp is supplying thousands of words randomly as keys to see if that produces anything reasonable
  • The other camp is trying to figure out the appropriate word (perhaps using the Labours of Hercules as a clue) then trying that word
  • http://www.connected.ltd.uk/pxc/

• Of course, if we are wrong about the encryption standard, then we are fairly fubarred, but it would appear that all roads point to it.

• Of course, there was some speculation that the 12 referred to the fact that there were 12 tasks of Hercules, and 12 RC5 challenges posted by RSA security
  • This was therefore the '13th Challenge'
  • Also, if this does need a brute force or grid computing attack - this would be a 'Herculean Challenge'

The distributed assault is currently underway, see below and at the Assault on 13th Labour site.

Cowheads and 64/12/8

• The cows represent the symbol of distributed.net, who first cracked the RC5.64 protocol, possibly indicating that this card requires a simple RC5.64 crack as well

• However, the card doesn't say RC5.64, it says 64/12/8. The assumption is that:
  • the 5 cow heads is a pointer to the RC5 standards, as worked on by the folks at www.distributed.net
  • The 64 would therefore be the encryption standard (RC5.64)
  • The 12 is supposedly the number of rounds, and the key is 8 characters long
  • That is why the brute force using 8 character words and phrases

Distributed Effort

Recently, work has been done by chimera245 to set up a coordinated Grid Computing solve on this Card. There is more information on this here kindly maintained by GuiN. The Grid Client is running nicely and is available here.

Earlier efforts included some work by Scott who did some math, and observed how long it would take to work out an answer here. See: Rccrypt Work Unit checkout for the original proposal for a combined effort.

Update:

As of 23-SEP-06, the client has tried every single combination of letters and numbers. That is, every single 8 character combination of the characters: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789. Nothing has come of this. Basically, this means that the key to unlocking the code HAS to contain at least one symbol. Chimera245 has updated the program, adding 5 more characters to test - <space> . , ? and ! - This will double the size of the possible keyspace.

Please see Brian Enigma's impressive Card Catalog for a sorted, comprehensive listing of cards.