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Anarielle Shahni
Eye of Cain
0
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Posted - 2012.02.21 11:15:00 -
[1] - Quote
Eye of Cain Cryptography Challenge (ECCC)
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.
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 : In general, the entry fee will be around 10% of the prize's value.
- 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.
- To enter the contest, send the entry fee to the corp with GÇ£CryptographyGÇ¥ as the reason. You will then receive all the informations needed in an EVE Mail as soon as it begins, or as soon as we are online, if it has already begun.
- 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.
The first challenge will start this Sunday, February the 26th, at 12:00 Eve Time. Prize : Republic Fleet Firetail Entry fee : 1M Difficulty : easy
Fly safe, and have fun. |
Anarielle Shahni
Eye of Cain
1
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Posted - 2012.02.25 22:47:00 -
[2] - Quote
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
Suitcase Bomb Production Inc. Unfamiliar Presence
1
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Posted - 2012.02.26 13:32:00 -
[3] - Quote
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
Eye of Cain
1
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Posted - 2012.02.26 16:06:00 -
[4] - Quote
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
Pragmatic Kernel
0
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Posted - 2012.03.01 02:18:00 -
[5] - Quote
Hello, I didn't understand how was encrypted the message of the first challenge, is it possible to have some more explanation ? |
Anarielle Shahni
Eye of Cain
3
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Posted - 2012.03.01 10:16:00 -
[6] - Quote
Hello,
The first challenge was encrypted using the vig+¿nere 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
Pragmatic Kernel
0
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Posted - 2012.03.01 14:34:00 -
[7] - Quote
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 ? |
Anarielle Shahni
Eye of Cain
4
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Posted - 2012.03.05 21:54:00 -
[8] - Quote
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
Pragmatic Kernel
0
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Posted - 2012.03.06 13:19:00 -
[9] - Quote
oh ! ok I see =) thanks for the explanation ! |
Anarielle Shahni
Eye of Cain
4
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Posted - 2012.03.09 09:32:00 -
[10] - Quote
For those interested : the second challenge is still unsolved. And an easy -½ backup challenge -+ has been set to help. Entry fee is still 6M.
Fly safe. |
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Typical Ginger
Pragmatic Kernel
0
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Posted - 2012.03.20 21:20:00 -
[11] - Quote
maybe you should ask everybody if they're still searching ? concerning my-self, I resigned :D |
Anarielle Shahni
Eye of Cain
4
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Posted - 2012.04.06 14:29:00 -
[12] - Quote
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... |
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