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#1 | |
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Programming Guru
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I conceived a new kind of encryption method, based on the idea of number systems. Based on the concept of how hexidecimal, for example, can turn 4 numbers in to 1. But now ... imagine instead of encrypting in base 16 (hexidecimal), you were to encrypt in base 3.32*10^240. Yep, that's right:
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((width of surface * length of surface) * (R*G*B))^(height) So then from any y from this system can be thrown through a simple processing function to derive 24 values of R, G, B, X, Y=1..24 . Then those values are used to map a pixel to a 24/24 symbol. It works perfectly, processing lines of text over 100 characters in to just only 1 symbol!! Example: The two very small dots in the left corner by the border. That says Sane.This holds much potential, because I could still derive over 24 values with a more complex algorithm, I could pre-encrypt the unicode values of the text to shorten what needs to be encrypted. And best of all, it's fully capable of being reverse-engineered. Once all this is accomplished, as well as the implications in C++, I'm estimating the ratio will improve from 100:1 to 1000:1, making it the best encryption process anywhere. I'm looking for some people who'd be interested in perfecting this method and helping me publish it. It is programmed fully in Python, and I hope someone can convert it into C++ due to Python's inability to convert long to float (which I desperately need). I'm not showing the source code though because I'm that paranoid about it's safety of owner. ![]() Patent pending? ![]() Last edited by Sane; Sep 19th, 2005 at 9:58 PM. |
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#2 |
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Newbie
Join Date: Sep 2005
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Awesome Sane =D
I must do this as well: & ![]() |
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#3 |
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Professional Programmer
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I'd love to hear about the decryption process....
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% rc4 hexkey < input > output
#define S ,t=s[i],s[i]=s[j],s[j]=t /* rc4 hexkey <file */
unsigned char k[256],s[256],i,j,t;main(c,v,e)char**v;{++v;while(++i)s[
i]=i;for(c=0;*(*v)++;k[c++]=e)sscanf((*v)++-1,"%2x",&e);while(j+=s[i]
+k[i%c]S,++i);for(j=0;c=~getchar();putchar(~c^s[t+=s[i]]))j+=s[++i]S;} |
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#4 |
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Programming Guru
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Impressive.
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http://jasonpowers.net "There are a thousand hacking at the branches of evil to one who is striking at the root." |
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#5 |
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Expert Programmer
Join Date: May 2005
Location: East Lansing, MI
Posts: 663
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Nice! Keep us posted, i really want to know about decryption.
BTW, if you're using a lot of Python libraries, converting to c++ might be a pain. |
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#6 |
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Programming Guru
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this sounds really cool. good job!
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i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons. |
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#7 |
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Expert Programmer
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Knowing Sane, he hasn't used many libaries.
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Join us at #programmingforums @ irc.freenode.net! My software never has bugs. It just develops random features.
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#8 |
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Programming Guru
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I've only used Pygame. And that's to blit all the symbols together and save as a bitmap.
But I think I just noticed a flaw. In order to decrypt, the x is multiplied by 24*255*255*255*the number of pixels in the symbol. Then decoded in the same base to give you your characters. But the x has been rounded to an integer when placed on the bitmap, but it needs to be a precise float in order to get back to the original number. I think I have a solution though. It can work the float untill all the modulus' of each term returns true to the color values and X. So the decryption might be a bit more complicated then I thought ... Edit: And I think I might have explained it a little wierd in my post. Each term from the output of the conversion to base 3*10^240, becomes a big number. But since that number is too large to convert to ascii and whatnot, it is processed in to color and co-ordinate information until it becomes 0. So each symbol equals a term from base 3*10^240 (Just as every term from base 16 might become a letter or stay a number.). Last edited by Sane; Sep 20th, 2005 at 5:33 AM. |
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#9 |
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Programming Guru
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So how do you differentiate the order of the bitmap information to decode?
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#10 |
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Programming Guru
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From y 1 .. 24, and from every subsequent section in the sequence: (x/24)*24 .. end.
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