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Eye of Cain Cryptography Challenge - Win a Plex - No Entry Fee
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Anarielle Shahni
La Cage aux Folles
#1 - 2012-02-21 11:15:26 UTC  |  Edited by: Anarielle Shahni
Eye of Cain Cryptography Challenge (ECCC)


First challenge over - Congratulation Yuusef

Second challenge over - no winner

Third challenge started


Greetings,

Eye of Cain is willing to organize a series of cryptography challenges. Let me explain how we would proceed: The moment you enter the contest, We'll send you a message containing the position and password of a Small Secure Container. But the message will be encrypted. The first person to get to the item inside the container - a voucher with an identification code - and to send it back to us wins the prize. It's as simple as that. This is not a lottery, it's not about being lucky, it's about being smart.

You can (should) join "EC Crypto Challenge" ingame.

Challenges announcements will contain the following informations :

  • Difficulty : from newbie to expert, it will range from simple Vigenere ciphers to hardcore asymmetric key encryption algorithms. For the most difficult challenges, being able to write programs to help you might be convenient.
  • Entry fee : NO MORE ENTRY FEE.
  • Some additional information, if needed.

The message will, in case of hard challenges, contain some relevant information about the cypher used to encrypt the message. But keep in mind, Alice and Bob will not be happy about that.

A few guidelines :

  • You can join in even if the contest is already underway, except if it has already be won, of course.
  • Both the password and location of the container will be in the message. The SSC will ALWAYS be in high sec space. We do not want accidents.
  • There is no time limit. Once the SSC is anchored, it will stay there.
  • Quantum computers are strictly forbidden. Yes I know, it's harsh.


Finally, the winners will always be welcome to post a detailed explanation of how they solved the problem, if they want to.

First Challenge (26/02/2012) [OVER] :
Prize : Republic Fleet Firetail
Entry fee : Free, to celebrate the first challenge and give you an opportunity to discover cryptography.
Difficulty : easy

Second challenge (OVER) :
Prize : Cruor
Entry fee : 6M
Difficulty : Not so easy

The next challenge is not as the previous two. There is no container. You'll understand if you try it (and succeed).

Third challenge :
Prize : Plex
Entry Fee : Free
Difficulty : Moderate
Start :http://tyruiop.eu/contest/

Fly safe, and have fun.
Anarielle Shahni
La Cage aux Folles
#2 - 2012-02-25 22:47:37 UTC
The first challenge will begin tomorrow.

To motivate the people, the entry fee of the first challenge will be of 0.00 ISK. But you still win the prize, of course.
Yuusef
Strategic Exploration and Development Corp
Silent Company
#3 - 2012-02-26 13:32:37 UTC
Thanks very much to Anarielle for my shiny new Firetail, and also for running a fun and challenging contest!

The method and solution are as below:

Anarielle stated that the first challenge would be an easy one, so my first guess was that it was some form of letter substitution. After getting no promising results from Caesar ciphers and transposition ciphers, I decided to try Vigenere.

Now I'm in no way a cryptographer, merely someone with a slight curiosity in the art - as such, I didn't know what steps to take to reveal the key length, or if it were even a repeating key to begin with. Fortunately, when noone was making progress on the problem, Anarielle mailed a clue to the contestants stating the key length was 9.

From here, it was quite simple to run frequency analysis of the letters in each 'slice' of the cipher. The first attempt at freq. anal. gave very good results, and several discernable words appeared in the plaintext string. At this point, manual changes were made to the slices in order to complete words and phrases, thus revealing the entire plaintext.

The encrypted string: DIOUGTVHBOEMIGKOBXOSGCECPXYBCSNBEUMROGSLLKAHYXPPNAVBBXDIZFTESMSXURYLPGMOWDKFEVRLRESSUMYSIKCTGIKUKBVVCOUWDWKKLMO

The Vigenere key: KAOUTROTK

The plaintext string: THEANCHOREDCONTAINERWILLBEORBITINGTHEFIRSTMOONOFTHENINTHPLANETINTHESYSTEMCALLEDSHURIATHEPASSWORDWILLBEADMIRABLE

As promised, there was an anchered container with the password 'admirable', and the Firetail was delivered promptly after submitting the code contained within.

A very fun challenge, and I look forward to seeing what comes next!
Anarielle Shahni
La Cage aux Folles
#4 - 2012-02-26 16:06:39 UTC  |  Edited by: Anarielle Shahni
Crypto Challenge - 2


Ladies and gentlemen, I require your attention.

I've been informed by the Amarr secret services that a Blood Raider informant has been captured near the Empress palace. Unfortunatly, he was able to send a message to his contact before that. The message was heavily encrypted and the Amarrian forces weren't able to break it.

They require your assistance. Suspicions are that the spy was trying to send one of the Amarrian covert ops communication codes to his contact, and we can't warn them without revealing their position. We need to intercept the message before its too late !

Good luck.

Second challenge 26-02-2012 20:00 Eve Time :
Prize : Cruor
Entry fee : 6M
Difficulty : Not so easy
Shusakurwa
Imperial Shipment
Amarr Empire
#5 - 2012-03-01 02:18:09 UTC
Hello,
I didn't understand how was encrypted the message of the first challenge, is it possible to have some more explanation ?
Anarielle Shahni
La Cage aux Folles
#6 - 2012-03-01 10:16:40 UTC
Hello,

The first challenge was encrypted using the vigènere cipher ( http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher ). The idea is the following, imagine you want to encrypt "cryptography is awesome" with the key "crypto" :

You first remove the spaces and split the message in blocks of 6 characters (the length of the key) :
"crypto graphy isawes ome"

Then, you apply the key to each block, summing numerical values of the caracters (starting to 0, not 1, i.e. : a = 0, z = 25).
e.g. : crypto is [2, 17, 24, 15, 14], then applying the key "crypto" give us : [4, 34, 48, 30, 28], and finally, we get the values mod 26 :
[4, 12, 22, 4, 2] which is emwec.

Now you repeat this procedure for each block and you've got your encrypted message.

I hope that helped you.

Anarielle
Shusakurwa
Imperial Shipment
Amarr Empire
#7 - 2012-03-01 14:34:56 UTC  |  Edited by: Shusakurwa
yeah that's what I was thinking, but "THE" with "KAO" should be "DHS":

K->10 T->19 19+10 mod 26 = 3 -> D
A->0 H->7 7+0 mod 26 = 7 -> H
O->14 E->4 14+4 mod 26 = 18 -> S

although the first 3 letters of the encrypted string are "DIO", what am I missing here ?

edit: and in your ex. I think 34 mod 26 is 8, not 12. so the result is eiwec, not emwec right ?
Anarielle Shahni
La Cage aux Folles
#8 - 2012-03-05 21:54:54 UTC
Yes, sorry, it was 8 not 12.

Originally, for the first challenge the key I used was "eveonline" (yes I know, it's not very original). But Yuusef's key kind of worked with the little program I used to. I said kind of because the result was... well.. only partial.

But more than enough to guess the password.
Shusakurwa
Imperial Shipment
Amarr Empire
#9 - 2012-03-06 13:19:21 UTC
oh ! ok I see =)
thanks for the explanation !
Anarielle Shahni
La Cage aux Folles
#10 - 2012-03-09 09:32:00 UTC
For those interested : the second challenge is still unsolved. And an easy « backup challenge » has been set to help. Entry fee is still 6M.

Fly safe.
Typical Ginger
The Scope
Gallente Federation
#11 - 2012-03-20 21:20:58 UTC
maybe you should ask everybody if they're still searching ? concerning my-self, I resigned :D
Anarielle Shahni
La Cage aux Folles
#12 - 2012-04-06 14:29:10 UTC
I think I can say "time's up" now. I'll reimburse every participant to the challenge, and make a new one asap.

I might have underestimated the difficulty of this one...