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K-Gen
Registerd on: 01/01/1970, 04:00:00
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Posts: 26
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Well I understand how this cryptography works ... It's a ve*en*rian cypher ... but It has too many keys to try (brute force) It can't be easily cryptanalised using the method for monoalphabetic encryption . I found and implemented an algorithm to find the possible key lenght and I got that most likley it is 21.. Is it really 21 or can i at least get a little hint here (like the key length)..?
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Use the source, Luke! |
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cp77fk4r
Global Admin
Registerd on: 01/01/1970, 04:00:00
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Posts: 621
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The length of the encryption key is the length of the string.
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Inj3ction
Registerd on: 01/01/1970, 04:00:00
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Posts: 102
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wtf?! i see somthing like a multiplie board. and i also noticed that thee isn't a "w". i don't have a clue what 2 do here. maybe if u could plz give me the name of the encription and than i'll pass it easily
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cp77fk4r
Global Admin
Registerd on: 01/01/1970, 04:00:00
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Posts: 621
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I saw it too and I fix that (the Y thing), but- no, try to do it by yourself.
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Inj3ction
Registerd on: 01/01/1970, 04:00:00
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Posts: 102
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oh so it's a mistake? i thought it was on purpose.
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pexalt
Registerd on: 01/01/1970, 04:00:00
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Posts: 11
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As did I, I thought W was a space or something like that, this explains why it only occurs once. None the less, wehn you say 'string' is that both lines or the first line is the string that is applied to the next line?
~Pexalt
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K-Gen
Registerd on: 01/01/1970, 04:00:00
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Posts: 26
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Wait a sec :) If the key is the same length as the string it's a one time pad !
One time pad is the only (matheaticaly prooven) unbrakable encryption !
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pexalt
Registerd on: 01/01/1970, 04:00:00
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Posts: 11
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Hmmm, Unbreakable encryption would be quite difficult to break I imagine. :P The first line is of a length that matches a very familar phrase used on this site quite frequently. I wondered then if this phrase was the string of which cp77fk4r had spoken. This would be a lot easier sliding the bar down from unbreakable and back into the medium range.
~Pexalt
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cp77fk4r
Global Admin
Registerd on: 01/01/1970, 04:00:00
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Posts: 621
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It's not Unbreakable encryption, there is a really cool way to decrypt it :)
google it :)
I think that i'll add scores for this challenge.
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K-Gen
Registerd on: 01/01/1970, 04:00:00
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Posts: 26
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well, i perfectly understood how this works ..
however if the key is the same length as the string it would be wrong to call it a substitution cipher.. since in a substitution cipher you change the letters accordingly to preveously changed letters. Here however you have 2 plain texts (assuming the the key is a real sentence of real words) that lies on the plaintext like in a one time pad, hence if you want to encrypt MYPASSWORD with ENCRYPTION it goes like that :
Like in a one time pad you can XOR the messages or you could use the given table (a-z*a-z) therefore:
M Y P A S S W O R D
E N C R Y P T I O N
--------------------
Q L R R Q H Q W F Q
As you see in the ciphertext the letter 'Q' repeats 4 times but No letter in the plaintext or key repeats more than twice. Hence it is only solvable by brute force given infinite computation power .... or I missed something :)
I think that you could really lower the key-space with brute force if you know that the key is made from real words .. however a simple calculation gives that if, say there are 100,000 words in english
and there are 62 characters in the cipher .. and say every word is average of 5-6 letters than :
62/6~=10 so you have an average of 10 spaces for 100,000 possibilities that means that the total ammount of possibilities is:
10^100000... A LOT !!
....Shortly I am brave enough to say that I have completely no Idea how to solve this :)
Edit by : K-Gen At 07/10/2005, 22:41:39
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pexalt
Registerd on: 01/01/1970, 04:00:00
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My theory is that the key can not be too difficult verging on being infintiely close to impossible for a medium challenge which means there is either a high chance of fluking the encryption or a logical means of deducing what it is. I haven't yet thought which method is going to be easiest to do. :)
~Pexalt
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K-Gen
Registerd on: 01/01/1970, 04:00:00
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Posts: 26
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Well i think that it is possible to do since we know the result of string1 XOR string 2. If both strings are made of words than if you test all 100,000 words once on the adequate part of the cipher you will get a word or some part of a word as well, so if the assumtion that both the key and the plaintext are of real dictionary words it is somehow possible to build a smart bruteforce algorithm that will deduce the key and the plaintext out of the cipher. And it will take a maximum of half an hour to calculate since if you have already found a word in the key that when you XOR it on a word of the plaintext and you get a correct part of the cipher , the chance of such a case to happen with different words is so low that there is no use to try and find all substitutes - hence you just write down what you found and continue the brutefoce from the point where you have stopped.
I will start developing an algorithm righ away :)
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superficial
Registerd on: 01/01/1970, 04:00:00
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Posts: 12
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hi, could i have a hint as to how to find the encryption key?
i found the method of cracking the code, as well as the key length, but i have no idea how to find the key itself...
like for example... is the key repeated? is it case sensitive?
Edit by : superficial At 04/11/2005, 17:35:39
Edit by : superficial At 21/11/2005, 10:49:20
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superficial
Registerd on: 01/01/1970, 04:00:00
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Posts: 12
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hello? could ihave some help pls?
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cp77fk4r
Global Admin
Registerd on: 01/01/1970, 04:00:00
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Posts: 621
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think think think!
start with guessing.
how all the start of the strings in the cryptographic challenges seems?
after that, check what is the key that you need to encrypt the start of the planetest and try to complet the key.
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