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| tidmiste | Aug 15 2008, 02:33 AM |
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Just registered
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Ok. I'm going to release (just about) everything related to this cipher now, to make it a fair challenge, as you've insisted. Sorry that it's about a year late, but... I'd love to see someone solve it. Let me start with how it's encrypted. As I've said, it uses basic arithmetic, and to be more exact, it uses it in the form Ax (+ or -) B, where x is the numerical value of a given letter. The resulting number would be the number that appears in the code. Also, after encryption using one equation on one number, the equation shifts for the second. There could be between four and ten (even though that number can still be higher) equations that get shifted, and the order never changes. After you reach the last of the equations, you start back over on the first equation. The letters are numbered using a special system (which is also part of the key), but an example would like... 101=A, 102=B, 103=C, 104=D, 201=E, etc. That isn't what it is, but it's sort of like it. After I convert the numerical values of the letters, I combine them into 5-number groups. If there isn't enough numbers in the last group, I'll add as many zeroes as necessary to create a five number group. In addition, though I said it wasn't in the text I had originally posted, I could add nulls and fake digits, and the recipient of the message would know what they are, despite the arrangement of the digits. The system of encryption I created also creates a system that makes decryption in the prescence of unknown nulls still easy. Also, there could be billions of different variations because of it's math-based encryption process. The numbers created are definitely not just bigrams, but some are trigrams too. As I said above, because of the keys (the varying equations and numerical values), decrypting the values despite the varying bi- and trigrams is very easy. Hopefully, this helps a bit, jdege and Revelation. |
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Ancient chinese wise man once told me... SPOILER:GO AWAY!!!!! | |
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9:06 PM Nov 26