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Amazon.com Hosting Crypto-Contest
vic20 pointed us over to Amazon's crypto contest/promotional item. Crack the message, and you can get signed editions of books like Cryptonomicon, Applied Cryptography, The Code Book, and Between Silk and Cyanide as well as LEGO Mindstorms. They'll be giving more hints as the contest goes on, with a deadline of 11:59 PM PST Oct. 29.
For those who don't like random drawing/contest - you can bid for Signed copies of Cryptonomicon... Signed by Neal Stephenson or even Linus Torvalds.
:)
-Note: I'm not selling either.
Joseph Elwell.
Or at least I believe its the solution.
As someone pointed out, the first 4 groupings is the ISBN of the book. If you pull up the book in Amazon, each has a review by Amazon. The numbers are word offsets. The key thing to figure out is what they consider a word... I just did a preliminary scan through and came up with the following, if someone spends some time, they can undoubtably get it right...
Resonates from 918 tech to Alan
all ties brilliant coffers an sinks in
sent explains department cut and to men forget zoo would than
People need the bestselling conflict in the Information Age
The suitable specialist offers steps for secure source
As you can see, its not perfect. I was using Word, cutting text then using word count to get position. Things like hypenated words were screwing me up. But the last two sentences seem to well fitted to be chance.
if your asking why im posting this, well, this is some scam contest they are running for two weeks, with a "random drawing" at the end. Yeah right. Ive spent about as much time as I want to on it, Id be curious to the final solution once whoever it is puzzles it...
What Amazon has done is to give people the cypher-text and challenged them to recover the plain-text.
This is why crytanalysis is SOOO interesting. This is more of a "real-world" scenario. You get some stuff and have to find out how it was encrypted and what it says.
Amazon's challenge is much more difficult than the RSA challenges. RSA tells you what encryption algorythm is used and what the beginning of the plain-text message is.... all that's left is key-recovery. Here people will have to think about how to attack the problem first, then try breaking the code.
HINT: Presuming that Amazon's message is in English, and is most likely a cypher (be it block, substitution, output feedback) with standard math knowedge necessary, I can offer the following assumptions for all the budding crypoanalysts on
It is most likely a subtitution cypher
It is somewhat likely that these are not added modulus anything. (Due to the high range of values)
A good starting point would be to compare the recurring numbers within the cyphertext with common letters in the English language.
Most of all: Have fun...At least you don't do this for $$
The message is "this message is not yet here."
--
As has been pointed out, 0380973464 is the ISDN for Cryptonomicon, so
- 415-426
038-097-34-64-242-335-51-377-183-168
---ISDN------ C R Y P T O
038-097-34-64-380-330-115-289-273-189-56
---ISDN------ N O M I C O N
and further on,
068-486-42-23-87-434-10-468-151-345-150-494-376
---ISDN------ T H E C O D E B O O K
Kaa
Kaa
Kaa's Law: In any sufficiently large group of people most are idiots.
038-097-34-64-242-335-51-377-183-1681 5-426
Cryptonomicon
This is a truly strange deception
038-097-34-64-380-330-115-289-273-189-56
Cryptonomicon
funky protagonists are destined to want appendices
068-486-42-23-87-434-10-468-151-345-150-494-376-4
Between Silk and Cyanide : A Codemaker's War 1941-1945
He had great marks for the easiest execution of enemy explosions
038-549-53-15-1-193-121-29-109-66-28-160-106
The Code Book : The Evolution of Secrecy from Mary, Queen of Scots to Quantum Cryptography
People need the bestselling conflict in the information age
047-111-70-99-24-21-25-12-53-22-56-8
Applied Cryptography : Protocols, Algorithms, and Source Code in C
The suitable specialist offers steps for secure source
Well, not that I'm a crypto expert, but let's see how quickly a brute force decipher could take...
Okay, I obviously have to make a few assumptions. First off, I assume each number between dashes is a single alphanumeric. And since I said alphanumeric, you know I mean A-Z and 0-9.
There are 61 alphanumerics in this message. HOWEVER, I believe we can safely assume that each line is a phrase on its own. If so, we can solve each separately. The breakdown is 10, 11, 15, 13, and 12 (yes, that adds to 61).
So, we have 36^n combinations for each line (n being the # of alphanumerics in each line).
Sum(36^n for all n) = approx(2.1249 * 10^23)
Let's say that it takes 1 millisecond to calculate a single permutation. (I have *no* idea whether that is a fair estimate - I'm definitely no expert in cryptography).
This would make worst case time for solving the code would be 7.011 * 10^12 years. Of course, this doesn't rule out the possibility that there are multiple valid solutions (ie. the code interpreted differently still results in a valid English statement) and other such weird phenomena. I would guess the best chance to cracking this thing, at first, would be to find the weights of each of the numbers between the dashes, and compare that to the English alphabet's weight. At least, that might get you an E or something. I'll let someone who knows what he's talking about comment on that.
You should never take life too seriously - You'll never get out of it alive.