The Thirteenth Labour

Card image

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.

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