I'm bored today, so I present to you a simple encryption algorithm that I came up with a while back.
What the algorithm does is it splits each byte into 2 bytes, the first being the byte AND the mask, the second being the byte NOR the mask.
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Given mask M and byte array A,
Encrypt a new array B
for i = 1 -> A.length * 2
B[i * 2] = A[i] AND M
B[i * 2 + 1] = A[i] NOR M
This way, to decrypt it, you know where the byte definitely has 1's and where it definitely has 0's, any spots that you aren't defined in the AND or NOR, will be the inverse bits of the mask.
Now this by itself is unsafe because you can easily pin point where the mask has what bits. So, to make it more secure, the algorithm randomly salts any bits in the AND that will always be 0's, and any bits in the NOR that will always be 0's.
This is the Java code that encrypts a byte array according to this algorithm, given byte mXORMask = mask
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public byte [] transformBytes(byte [] vByteArray) {
byte [] nByteArray = new byte [vByteArray.length * 2];
nByteArray[0] = mXORMask;
for(int i = 0; i < vByteArray.length; i++) {
nByteArray[i * 2] = (byte)(mXORMask & vByteArray[i]);
nByteArray[i * 2 + 1] = (byte)(0xFF ^ (mXORMask | vByteArray[i]));
}
return nByteArray;
}
This is the Java code that salts the bytes given byte [] aSalterMask = {mXORMask ^ 0xFF, mXORMask};
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public byte[] transformBytes(byte[] vByteArray) {
byte [] nByteArray = new byte [vByteArray.length];
int tIndex = 0;
for(int i = 0; i < vByteArray.length; i++) {
nByteArray[i] = (byte)(((aSalterMask[tIndex] ^ 0xFF) & vByteArray[i]) | (aSalterMask[tIndex] & ByteLib.getRandomByte()));
tIndex++;
if(tIndex == mNumMasks) {
tIndex = 0;
}
}
return nByteArray;
}
The challenge to everyone is to find a way to crack this encryption without the mask available. Or possibly find the mask. A cookie to the first person to crack it without brute forcing.
Here's some encrypted 'hello world's for you:
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51 B3 04 9A 5C 91 0D 95 5E B2 69 DF 0F AE 17 14 1A AF 14 11 15 B9 29 DC 68 5E 39 78 78 5E 45 52 4A D3 39 17 7C 98 65 31 24 B7 3F 36 60 7B 66 2E 2E 30 63 AB 24 33 75 1B 60 58 68 5A 68 F8 05 D2 1A D5 5D 50 12 55 70 35 61 17 2D 1C 2D 1A 2D 91 25 13 2D 91 2C 15 67 B0 77 96 69 FB 3E 88 66 AC 3F 16 7E 14 7B 2F 2B 0B 35 31 74 35 75 9F 6C BB 28 5E 20 DA 29 F8 69 DC 31 5E 78 FC 79 5E 20 DE 04 F4 42 55 15 F4 5A 71 59 B1 3D 3C 15 37 6D 13 46 10 38 DF 17 A8 2E B4 02 2D 7D 13 4C 39 31 DC 39 6E 38 58 79 EC
-WallShadow <3
-- Thu Jul 19, 2012 11:51 pm --
Does no one have a solution for this?
I meant that, does the random number generator change the output? But I guess you answered that. My experience is that when using something multiple times in encryption, unless it's proven 'safe', will eventually weaken an algorithm. Inadvertently you will introduce a shortcut of some type. Otherwise all encryption would just use one thing over and over (yes, XOR and bit shifts are used a lot, but still). I'm sorry, I'm no expert on encryption. Quite the opposite actually, I did one course on it. I'll first have to translate your Java program into something more 'native' to me 
