Xine - issue #4 - Phile 109

eZine lover (@eZine)

Published in 

Xine

 · 6 months ago

 
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                                        | Xine - issue #4 - Phile 109 | 
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Crypting datas in order to protect it is like an art, decrypting another. 
Crypting datas in order to avoid decryption had been headed for years, 
but I think each technique used is due to fail. Like CRC, for example, 
all ways used remember the password ( well, also encrypted ) so that it 
has already been - or it could be - hacked! But what about encryption that 
don't remember passwords? I'm thinking about ^ encryption. Imagine 
we make: 

for all bytes 
 byte = byte ^ n 
next bytes 

file: a b c d e f g h i j 
crpt: n n n n n n n n n n 
rslt: k l o p q r s t u v 
with k = a ^ n, l = b ^ n, o = c ^ n, ... 

I remember for those who don't very know XOR that a ^ a = 0 
=> b ^ a ^ a = b }=> c ^ a = b 
   c = b ^ a     } 

This mean that if u XOR "x" with a number "a", you xor the number u obtain 
again with "a" and obtain then "x" back. So, if u wanna decrypt the file, 
you only have to crypt it again with the same password... 

What's the main default of that ^ encryption? That's, if we know a byte of 
that file, we can know "n", so let be imaginative, use several numbers! 
Look that example: 

file: a b c d e f g h i j 
crpt: m n m n m n m n m n 
rslt: k l o p q r s t u v 
with k = a ^ m, l = b ^ n, o = c ^ m, ... 

So, we have to know 2 bytes of that file ( 1 for m and 1 for n, 2 for n is 
useless ). That's better but... Let's try with 10 numbers, or more! but all 
will have to be remembered! Imagine we use a sentence as password: 
"I'm so sick!", so n1 = asc("I"), n2 = asc("'"), n3 = asc("m"), ... 
Ok, but if someone know the first 12 bytes of that file! 

So, there is the problem: periodicity! 
How to avoid periodicity? I think the only way is to make a password longer 
than the file. Cool, I'll use my english lessons like passwords ;-). 
I won't explain ya how, but I'll show you... imagine we have 2 periodicity, 
one of 2 bytes and one of 3 bytes... the "total" periodicity will be of 
6 bytes... 

With this 2 passwords: mn (2) and opq (3) 

file: a b c d e f g h i j 
crpt: m n m n m n m n m n 
      o p q o p q o p q o 
rslt: k l r s t u v w x y 
with k = a ^ m ^ o, l = b ^ n ^ p, o = c ^ m ^ q, ... 

After, 2 periodicity, one of 2 bytes and one of 4 bytes will give a "total" 
of 4 bytes! That don't multiply perios, you have to make the LCM ( lower 
common multiple ) of all passwords used... 
That's easy to imagine a password of 7, one of 13 ( and why limit to 2 
pass? ), another of 11... that will give a perio of 1001 bytes ) 

So, let be imaginative, if e have a file longer that 1001 bytes: 
1st - you could add a 5 bytes long password, so that total perio is now 5005. 
2nd - imagine that a people have to know 1001 byte of youre file to decrypt 
        it! 
Yeah, he have to know 1001 byte, I tryed to find a system, with 2 passwords 
, one of 2 and one of 3, if I only have the total, I won't be able to find 
the 2 original pass, so if I miss some ( if I know even 900 of 1001, I won't 
be able to find the 101 left ), I'm also not able to know the total perio, 
perhaps could I guess it, but... 

Ok, here are the sources in C, made for Watcom11: 
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void Crypt( char* Data, int DataSize, char** Pws, int NPws ) 
{ 
 char* DataPtr, **PwsPtr; 
 int i, j; 

 PwsPtr = new char*[ NPws ]; 
 for( j=0; j<NPws; j++ ) 
  PwsPtr[j] = Pws[j]; 

 for( i=0, DataPtr=Data; i<DataSize; i++, DataPtr++ ) 
 { 
  for( j=0; j<NPws; j++ ) 
  { 
   *DataPtr ^= *(PwsPtr[j]++); 
   if( *(PwsPtr[j]) == 0 ) 
    PwsPtr[j] = Pws[j]; 
  } 
 } 
 delete PwsPtr; 
} 
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Notes: 
1st - That tutorial was made by n0ph on the 18-02-98, for IKX, for XINE4. 
2nd - If you find a way to hack that kind of password, write to 
        n0ph@HotMail.Com... 
3rd - Visit "members.xoom.com/n0ph" ! 
4th - That's all folks!