unforum

[terano]

Promising: the decoded file for key 'NEEDFOOD' is shell archive or script for antique kernel text

[hairysocks]

I'm not sure if this has been mentioned - I didn't see it in the posts - but how do we know what order the bytes on the card should be in in the encrypted file? Everyone seems to be assuming it is left to right, top to bottom, but there are of course others:

L-R, B-T: BF EC 20 11 70 71 AA 22 A0 BB ....

R-L, T-B: 0F 56 53 48 8F 43 AD C5 FD 3A A9 69 C7 46 ....

R-L, B-T: 22 AA 71 70 11 20 EC BF D8 DE ....

Or even a spiral: 22 AA 71 70 11 20 EC BF A0 79 A2 2A ....

[hairysocks]

Here's another 13th Labour website. I'm surprised nobody has mentioned this before now. I haven't bothered reading past 11th July 2005 because the card was out by then. There aren't any "the password is xxx" phrases, but there are lots of 8 letter words in there:

http://the13labour.comicgen.com/d/20050112.html
[/url]

[Austin]

Is this the right place for a 'SPEC'?

There is always the possibility that the code does not need "de"crypting. If some meaningless rubbish has already been through the rc5 algorithm 12 times, why not encrypt it once more to make it up to 13? It could (just) explain Von's useless hint. (just "rccrypt")

So it could be "cat <binary file> | rccrypt -r 1 -k <key>" using the ricksoft rccrypt in linux.

(you can take a sensible message of the right length and 'decrypt' it to give you something that looks like it has been 64/12/8 encrypted, even using a zero key, I think) - this is not as simple as that though, if I typed it in correctly.

[BluesScale]

At the risk of sounding like a damned fool, could someone please tell me how I can view the output of the Windows version of rccypt?

I know it is a dumb question but perhaps not as dumb as not asking the question at all

BluesScale.

[x]

That is an interesting idea. 12 passes representing the current encrypted message. But 13 passes being the full decryption?

So perhaps our idea of decryption based on 12 passes isnt correct.

perhaps to complete the decryption, we go 64/1/8 we will find our answer?

[Uhtoff]

Two ideas have come to my mind -

Firstly, the last line of the text block just happens to be 16 characters of hex long...could that be the key? Have tried it in SteveC's decoder, but don't know enough about the encryption to know if that would work even if it was the right key, to decode the first lines.

Secondly, the only non-scary reason that this would need a collaborative effort that I can think of is that the key is somehow hidden in the 12 character code that you have to type in to get a solve. This would necessitate two people checking their codes against each others for similarities, thus a collaborative effort. To support this idea (to an extent) is that the card's code and the coe block are in a very similar font. If this was the case then the card would contain enough information on it to be solved by itself, but it would be easier to compare your code to someone elses and if 8 letters/numbers match then you're in gravy.

[pheebs]

tried nahl (as in the painter) who painted "the 13th labor" and got this:

Promising: the decoded file for key 'nahl' is Sendmail frozen configuration - version í‰)óׇf%©ížêw~`ü,G{sS§£pZ•æÊ)

does that mean anything to anyone? or is it just gobbledegook, as it is to me?

(this was probably a complete tangent but googling does lead to things you never knew!)

[BluesScale]

"unsquare" was considered promising by the website but yields nothing from the Windows version of rcscript - but then, nor did other keys marked as promising.

No offense at all intended to the author of this implementation of RCCrypt but does this look like user error, my failure to understand or a possible issue with the utility?

Thanks

Blues

[Hertspoppy]

Hi - this is my first post on here.

I've just started to try and solve this card and have scanned through all the posts so far so don't think this has been suggested....

Does the password have to be a word ? How about a number and taking the 64/12/8 as a sum like :

(64 / 12) / 8 = 0.666666667

I know it is more than 8 characters, but could this number mean anything ?

Boy is this hard ..... how are non computer experts supposed to solve something like this ? I would have thought the card should be solveable by anyone without the need for creating and running programs ?

[BluesScale]

A good thought but sadly one that I have tried.

I think that there is no reasonable doubt that this is an RC5 code and accordingly that software is required. One of the blues required QuickBasic to solve and the difficulty of this computing task is much harder as would befit a silver card. Indeed, it is hard to imagine a task that would be much harder and yet still possible.

Best regards

Blues

[Hertspoppy]

Thanks BluesScale. Oh well, it was an idea .....

I'll have to get the thinking hat on and see if I can come up with any other ideas around the password.

Is that all we are missing now ?

[Hertspoppy]

Me again

As there have been so many posts I can't remember all the passowrds mentioned.

Has anyone tried Sanguine as in

"Professor Mansoor was part of the team which created the Sanguine Algorithm, still used in many applications today"

from the Perplex City Academy ?

[Langley Moor]

[lownote]

The link from the title to the crypto puzzle is quite possibly Kerberos - the three-headed dog of hades that Hercules had to kidnap, and also the well-known cryptographic security implementation from MIT

The main implications are that Kerberos uses an initialization vector of zero for DES-encrypted messages, and has a specific model for constructing the cyphertext octets including confounders.

More detailed info here: ftp://ftp.isi.edu/in-notes/rfc1510.txt and here: http://web.mit.edu/kerberos/

It would normally take several thousand computers about 4 years to crack this by brute force, so I sort of hope this isn't right!