TUCoPS :: Crypto :: crcrev.txt

CRC and how to reverse it

Title           : CRC and how to Reverse it
Author          : Anarchriz/DREAD
Release Date    : 29 april 1999 (last modification 30 april 1999)
Difficulty      : beginner to advanced
Target          : CRC algorithm
Tools           : QEdit 2.1 (the best!) & Wordpad & some CRC progs
Description     : A CRC Tutorial & The c00l way to Reverse CRC

CRC and how to Reverse it


  This essay consists of a CRC tutorial and a way of how to reverse it. Many
Coders/Reversers don't know exactly how CRC works and almost no one knows how to
reverse it, while this knowledge could be very usefull. First the tutorial will
learn you how to calculate CRC in general, you can use it as data/code
protection. Second, the reverse part will learn you (mainly) how to reverse
CRC-32, you can use this to break certain CRC protections in programs or over
programs (like anti-virus). There seem to be utilities who can 'correct' CRCs
for you, but I doubt they also explain what they're doing.
  I'd like to warn you, since there is quite some math used in this essay. This
wont harm anyone, and will be well understood by the avarage Reverser or Coder.
Why? Well. If you dont know why math is used in CRC, I suggest that you click
that button with a X at the top-right of this screen. So I assume the reader has
knowledge of binair arithmetic.

Part 1: CRC Tutorial, what it is and how to calculate it

Cyclic Redundancy Code or CRC

  We all know CRC. Even if you don't recall, you will when you think of those
annoying messages RAR, ZIP and other compressors give you when the file is
corrupted due to bad connections or those !@#$% floppies. The CRC is a
value computed over a piece of data, for example for each file at the
time of compression. When the archiver is unpacking that file, it will read the
CRC and check it with the newly computed CRC of the uncompressed file. When
they match, there is a good chance that the files are identical. With CRC-32,
there is a chance of 1/2^32 of the check failing to recognize a change in data.

  A lot of people think CRC is short for Cyclic Redundancy Check. If indeed CRC
is short for Cyclic Redundancy Check then a lot of people use the term incorrect.
If it was you could not say 'the CRC of the program is 12345678'. People are also
always saying a certain program has a CRC check, not a Cyclic Redundancy Check 
check. Conclusion: CRC stands for Cyclic Redundancy Code and NOT for Cyclic
Redundancy Check.

  How is the calculation done? Well, the main idea is to see the file as one 
large string of bits divided by some number, which will leave you with a 
remainder, the CRC! You always have a remainder (can also be zero) which is at
most one bit less then the divisor (else it still has a divisor in it).
(9/3=3 remainder=0 ; (9+2)/3=3 remainder=2)
  Only here dividing with bits is done a little different. Dividing is repeatedly
substracting (x times) a number (divisor) from a number you want to divide, which
will leave you with the remainder. If you want the original number back you
multiply with the divisor or (idem) add x times the divisor with itself and
afterwards adding the remainder.
  CRC computation uses a special way of substracting and adding, i.e. a
new 'arithmetic'. While computing the carry for each bit calculation is
Lets look at 2 examples, number 1 is a normal substraction, 2&3 are special.
(1) 1101  (2) 1010  1010  (3) 0+0=0  0-0=0
    1010-     1111+ 1111-     0+1=1 *0-1=1
    ----      ----  ----      1+0=1  1-0=1
    0011      0101  0101     *1+1=0  1-1=0

  In (1), the second column from the right would evaluate to 0-1=-1, therefore
a bit is 'borrowed' from the bit next to it, which will give you this
substraction (10+0)-1=1. (this is like normal 'by-paper' decimal substraction)
The special case (2&3) 1+1 would normally have as answer 10, where the '1' is
the carry which 'transports' the value to the next bit computation. This value
is forgotten. The special case 0-1 would normally have as answer '-1', which
would have impact on the bit next to it (see example 1). This value is also
forgotten. If you know something about programming this looks like, or better,
it IS the XOR operation.
 Now look at an example of a divide:

In normal arithmetic:
1001/1111000\1101 13            9/120\13
     1001    -                    09  -|
     ----                         --   |
      1100                         30  |
      1001    -                    27  -
      ----                         --
       0110                         3 -> the remainder
       0000    -
        1001    -
         011 -> 3, the remainder

In CRC arithmetic:
1001/1111000\1110               9/120\14 remainder 6
     1001    -
      1001    -
       1001    -
        0000    -
         110 -> the remainder
(example 3)

  The quotient of a division is not important, and not efficient to remember,
because that would be only a couple of bits less than the bitstring where you
wanted to calculate the CRC from. What IS important is the remainder! That's
the thing that says something important over about the original file. That's
basicly the CRC!

Going over to the real CRC computation

  To perform a CRC calculation we need to choose a divisor, we call it the
'poly' from now on. The width W of a poly is the position of the highest bit,
so the width of poly 1001 is 3, and not 4. Note that the highest bit is always
one, when you have chosen the width of the poly you only have to choose a value
for the lower W bits.

  If we want to calculate the CRC over a bitstring, we want to make sure all
the bits are processed. Therefore we need to add W zero bits to the end of the
bitstring. In the case of example 3, we could say the bitstring was 1111.
Look at a little bigger example:

Poly                = 10011, width W=4
Bitstring + W zeros = 110101101 + 0000

10011/1101011010000\110000101 (we don't care about the quotient)
      10011|||||||| -
       10011|||||||  -
        00000||||||   -
         00000|||||    -
          00000||||     -
           00000|||      -
            10011||       -
             00000|        -
              10011         -
               1111 -> the remainder -> the CRC!
(example 4)

There are 2 important things to state here:
1.Only when the highest bit is one in the bitstring we XOR it with the poly,
 otherwise we only 'shift' the bitstring one bit to the left.
2.The effect of XORring is, that it's XORed with the lower W bits, because the
 highest bit always gives zero.

Going over to a Table-Driven Algorithm

  You all should understand that an algorithm based on bitwise calculation will
be very slow and inefficient. It would be far more efficient if you could
calculate it on a per-byte basis. But then we can only accept poly's with a
width of a multiple of 8 bits (that's a byte ;). Lets visualize it in a example
poly with a width of 32 (W=32):

          3   2   1   0    byte
Pop! <--|   |   |   |   |<-- bitstring with W zero bits added, in this case 32
       1<--- 32 bits ---> this is the poly, 4*8 bits

(figure 1)

  This is a register you use to store the temporary result of the CRC, I call
it the CRC register or just register from now on. You are shifting bits from
the bitstring in at the right side, and bits out at the left side. When the bit
just shifted out at the left side is one, the whole register is XORred by the
lower W bits of the poly (in this case 32). In fact, we are doing exactly the
same thing as the divisions above.
What if (as I said) we would shift in & out a whole group of bits at once.
Look at an example of 8 bit CRC with 4 bits at once shifted in & out:

The register just before the shift : 10110100
Then 4 bits (at the top) are shifted out at the left side while shifting 4 new
bits in at the right side. In this example 1011 is shifted out and 1101 (new)
is shifted in.

Then the situation is this:
8 bits currently CRC/Register   : 01001101
4 top bits just shifted out     : 1011
We use this poly                : 101011100, width W=8

Now we calculate just as usual the new value of the register.

Top  Register
---- --------
1011 01001101  the topbits and the register
1010 11100   + (*1) Poly is XORred on position 3 of top bits (coz there is a one)
0001 10101101 result of XORring

Now we still have a one on bit position 0 of topbits:
0001 10101101  previous result
   1 01011100+ (*2) Poly is XORred on position 0 of top bits (coz there is a one)
0000 11110001 result of second XORring
Now there are all zero's in the topbits, so we dont have to XOR with the poly
anymore for this sequence of topbits.

The same value in the register you get if you first XOR (*1) with (*2) and the
result with the register. This is because of the standard XOR property:
(a XOR b) XOR c = a XOR (b XOR c)

1010 11100     poly on position 3 of top bits
   1 01011100+ poly XORred on position 0 of top bits
1011 10111100  (*3) result of XORring

The result (*3) is XORred with the register
1011 10111100
1011 01001101+ the top bits and the register
0000 11110001

You see? The same result! Now (*3) is important, because with the top bits 1010
is always the value (*3)=10111100 (only the lower W=8 bits) bound (under the
stated conditions, of course) This means you can precompute the XOR values for
each combination of top bits. Note that top bits always become zero after one
iteration, this must be because the combination of XORring leads to it.

  Now we come back to figure 1. For each value of the top byte (8 bits) just
shifted out, we can precompute a value. In this case it would be a table
consisting of 256 (2^8) entries of double words (32bit). (the CRC-32 table is
in the appendix)
In pseudo-language our algoritm now is this:
  While (byte string is not exhausted)
    Top = top_byte of register ;
    Register = Register shifted 8 bits left ORred with a new byte from string ;
    Register = Register XORred by value from precomputedTable at position Top ;

The direct Table Algorithm

  The algorithm proposed above can be optimized. The bytes from the byte string
don't need to travel through the whole register before they are used. With
this new algorithm we can directly XOR a byte from a byte string with the byte
shifted out of the register. The result points to a value in the precomputed
table which will be XORred with the register.
  I don't know exactly why this gives the same result (it has to do with a XOR
property), but it has the Big advantage you don't have to append zero
bytes/bits to your byte string. (if you know why, pleaz tell me :)

  Lets visuallize this algorithm:

  +----< byte string (or file)
  v       3   2   1   0    byte
  |     +---+---+---+---+
 XOR---<|   |   |   |   |  Register
  |     +---+---+---+---+
  |             |
  |            XOR
  |             ^
  v     +---+---|---+---+
  |     |   |   |   |   |  Precomputed table
  |     +---+---+---+---+
  +--->-:   :   :   :   :
        |   |   |   |   |
(figure 2)

The 'reflected' direct Table Algorithm

  To make things more complicated there is a 'reflected' version of this
algorithm. A Reflected value/register is that it's bits are swapped around
it's centre. For example 0111011001 is the reflection of 1001101110.

  They came up with this because of the UART (chip that performs serial IO),
which sends each byte with the least significant bit (bit 0) first and the most
significant bit (bit 7) last, this is the reverse of the normal situation.
  Instead then of reflecting each byte before processing, every else is
reflected. An advantage is that it gives more compact code in the
implementation. So, in calculating the table, bits are shifted to the right and
the poly is reflected. In calculating the CRC the register is shifted to the
right and (of course) the reflected table is used.

  byte string (or file) -->---+
                              |    1. In the table each entry is reflected
    byte  3   2   1   0       V    2. The initial register is reflected
        +---+---+---+---+     |    3. The bytes from the byte string aren't
        |   |   |   |   |>---XOR       reflected, because all the rest is.
        +---+---+---+---+     |
                |             |
               XOR            V
                ^             |
        +---+---|---+---+     |
        |   |   |   |   |     |   Precomputed table
        +---+---+---+---+     |
        :   :   :   :   : <---+
        |   |   |   |   |
(figure 3)

  Our algorithm is now:
1. Shift the register right by one byte
2. XOR the top byte just shifted out with a new byte from the byte string
    to yield an index into the table ([0,255])
3. XOR the table value into the register
4. Goto 1 if there are more bytes to process

Some implementations in Assembly

To get everything settled here's the complete CRC-32 standard:
Name            : "CRC-32"
Width           : 32
Poly            : 04C11DB7
Initial value   : FFFFFFFF
Reflected       : True
XOR out with    : FFFFFFFF

As a bonus for you curious people, here's the CRC-16 standard: :)
Name            : "CRC-16"
Width           : 16
Poly            : 8005
Initial value   : 0000
Reflected       : True
XOR out with    : 0000

'XOR out with' is the value that is XORred with the final value of the register
before getting (as answer) the final CRC.
There are also 'reversed' CRC poly's but they are not relevant for this
tutorial. Look at my references if you want to know more about that.

  For the assembly implementation I use 32 bit code in 16 bit mode of DOS...
so you will see some mixing of 32 bit and 16 bit code... it is easy to convert
it to complete 32 bit code. Note that the assembly part is fully tested to be
working correctly, the Java or C code is derived from that.
Ok. Here is the assembly implementation for computing the CRC-32 table:

        xor     ebx, ebx   ;ebx=0, because it will be used whole as pointer
        xor     eax, eax   ;eax=0 for new entry
        mov     al, bl     ;lowest 8 bits of ebx are copied into lowest 8 bits of eax

        ;generate entry
        xor     cx, cx
        test    eax, 1
        jz     no_topbit
        shr     eax, 1
        xor     eax, poly
        jmp    entrygoon
        shr     eax, 1
        inc     cx
        test    cx, 8
        jz     entryLoop

        mov     dword ptr[ebx*4 + crctable], eax
        inc     bx
        test    bx, 256
        jz     InitTableLoop

Notes:  - crctable is an array of 256 dwords
        - eax is shifted to the right because the CRC-32 uses reflected Algorithm
        - also therefore the lowest 8 bits are processed...

In Java or C (int is 32 bit):

for (int bx=0; bx<256; bx++){
  int eax=0;
  eax=eax&0xFFFFFF00+bx&0xFF;      // the 'mov al,bl' instruction
  for (int cx=0; cx<8; cx++){
    if (eax&&0x1) {
    else eax>>=1;

The implementation for computing CRC-32 using the table:

        xor     ebx, ebx
        xor     al, [si]
        mov     bl, al
        shr     eax, 8
        xor     eax, dword ptr[4*ebx+crctable]
        inc     si
        loop   computeLoop

        xor     eax, 0FFFFFFFFh

Notes:  - ds:si points to the buffer where the bytes to process are
        - cx contains the number of bytes to process
        - eax contains current CRC
        - crctable is the table computed with the code above
        - the initial value of the CRC is in the case of CRC-32: FFFFFFFF
        - after complete calculation the CRC is XORred with: FFFFFFFF
           which is the same as NOTting.

In Java or C it is like this:

 for (int cx=0; cx<bytesOfFile.length; cx++){
   int ebx=0;
   eax^=bytesOfFile[cx]; // only the lowest byte should be XORred

So now we landed at the end of the first part: The CRC tutorial
If you want to make a little deeper dive in CRC I suggest reading the document
I did, you will find the URL at the end of this document.
Ok. On to the most interresting part of this document: Reversing CRC!

Part 2: Reversing CRC

  When I was thinking of a way to reverse it... I got stuck several times. I
tried to 'deactivate' the CRC by thinking of such an sequence of bytes that it
then shouldn't matter anymore what bytes you would place behind it. I couldn't
do it... Then I realized it could NEVER work that way, because CRC algorithm is
build in such a way it wouldn't matter which _bit_ you would change, the
complete CRC _always_ (well always... almost) changes drasticly. Try that
yourself (with some simple CRC programs)... :)
  I realized I only could 'correct' the CRC _after_ the bytes I wanted to
change. So I could make such a sequence of bytes, that would 'transform' the
CRC into whatever I wanted!
Lets visualize the idea:

Bunch of bytes: 01234567890123456789012345678901234567890123456789012
You want to change from  ^  this byte to  ^  this one.
Thats position 9 to 26.
We also need 4 extra bytes (until position 30 ^) for the sequence of bytes which
will change the CRC back to its original value after the patched bytes.

  When you are calculating the CRC-32 it goes fine until the byte on position 9,
in the patched bunch of bytes the CRC radically changes from that point on.
Even when pass position 26, from where the bytes are not changed, you never get
the original CRC back. NOT! When you read the rest of this essay you know how.
In short you have do this when patching a certain bunch of bytes while
maintainting the CRC:
1. Calculate the CRC until position 9, and save this value.
2. Continue calculating until position 27 and 4 extra bytes, save the resulting
3. Use the value of 1 for calculating the CRC of the 'new' bytes and the extra
 4 bytes (this should be 27-9+4=22 bytes) and save the resulting value.
4. Now we have the 'new' CRC value, but we want the CRC to be the 'old' CRC
 value. We use the reverse algorithm to compute the 4 extra bytes.

We can to point 1 to 3, below you learn to do point 4.

Reversing CRC-16

  I thought, to make it more easy for you, first to calculate the reverse of
CRC-16. Ok. We are on a certain point after the patched code where you want to
change the CRC back to its original. We know the original CRC (calculated before
patching the data) and the current CRC register. We want to calculate the
2-bytestring which changes the current CRC register to the original CRC.
First we calculate 'normally' the CRC with the unknown 2 bytes naming them X
and Y, for the register I take  a1 a0 , the only non-variable is zero (00). :)
Look again at our latest CRC algorithm, figure 3, to understand better what im
Ok, here we go:

Take a 2-bytestring 'X Y'. Bytes are processed from the left side.
Take for register  a1 a0.
For a XOR operation I write '+' (as in the CRC tutorial)

Processing first byte, X:
a0+X            this is the calculated topbyte (1)
b1 b0           sequence in table where the topbyte points at
00 a1           to right shifted register
00+b1 a1+b0     previous 2 lines XORred with eachother

Now the new register is: (b1) (a1+b0)

Processing second byte, Y:
(a1+b0)+Y       this is the calculated topbyte (2)
c1 c0           sequence in table where the topbyte points at
00 b1           to right shifted register
00+c1 b1+c0     previous 2 lines XORred with eachother

Now the final register is: (c1) (b1+c0)

I'll show it a little different way:

a0 + X      =(1)  points to b1 b0 in table
a1 + b0 + Y =(2)  points to c1 c0 in table
     b1 + c0=d0  new low byte of register
          c1=d1  new high byte of register
    (1)  (2)

Wow! Let this info work out on you for a while... :)
Don't be afraid, a real value example is coming soon.
  What if you wanted the register to be some d1 d0 (the original CRC) and you
know the value of the register before the transformation (so a1 a0)... what 2
bytes or what X and Y would you have to fed through the CRC calculation?
  Ok. We will begin working from the back to the front. d0 must be b1+c0 and
d1 must be c1... But how-the-hell, I hear you say, can you know the value of
byte b1 and c0??? ShallI remember you about the Table? You can just lookup
the value of the word C0 C1 in the Table because you know C1. Therefore you
need to make a 'lookup' routine. If you found the value, be sure to remember
the index to the value because that's the way to find the unknown topbytes e.g.
  So now you found c1 c0, how to get b1 b0? If b1+c0=d0 then b1=d0+c0! Now you
use the lookup routine to lookup the b1 b0 value too. Now we know everything
to calculate X & Y ! Cool huh?
 a1+b0+Y=(2) so Y=a1+b0+(2)
 a0+X=(1)    so X=a0+(1)

Non-variable example for CRC-16

Lets look at an example with real values:
-register before: (a1=)DE (a0=)AD
-wanted register: (d1=)12 (d0=)34
Look up the entry beginning with 12 in the CRC-16 table in the appendix.
-This is entry 38h with value 12C0. Try to find another entry beginning with 12.
You can't find another because we calculated each entry for each possible value
of the topbyte and that's 256 values, remember!
Now we know (2)= 38, c1= 12 and c0= C0, so b1= C0+34=F4, now look up the entry
of B1 beginning with F4.
-This is entry 4Fh with value F441.
Now we know (1)= 4F, b1= F4 and b0= 41. Now all needed values are known, to
compute X and Y we do:
X=a0+(1)   =AD+4F   =E2

Conclusion: to change the CRC-16 register from DEAD to 1234 we need the bytes
 E2 A7 (in that order).

You see, to reverse CRC you have to 'calculate' your way back, and remember the
values along the way. When you are programming the lookup table in assembly,
remember that intel saves values backwards in Little-Endian format.
Now you probably understand how to reverse CRC-16.... now CRC-32

Reversing CRC-32

Now we had CRC-16, CRC-32 is just as easy (or as difficult). You now work with
4 bytes instead of 2. Keep looking and comparing this with the 16bit version
from above.

Take a 4-bytestring X  Y  Z  W  , bytes are taken from the LEFT side
Take for register  a3 a2 a1 a0
Note that a3 is the most significant byte and a0 the least.

Processing first byte, X:
a0+X                    this is the calculated topbyte (1)
b3    b2    b1    b0    sequence in table where the topbyte points at
00    a3    a2    a1    to right shifted register
00+b3 a3+b2 a2+b1 a1+b0 previous 2 lines XORred with eachother

Now the new register is: (b3) (a3+b2) (a2+b1) (a1+b0)

Processing second byte, Y:
(a1+b0)+Y                       this is the calculated topbyte (2)
c3    c2    c1       c0         sequence in table where the topbyte points at
00    b3    a3+b2    a2+b1      to right shifted register
00+c3 b3+c2 a3+b2+c1 a2+b1+c0   previous 2 lines XORred with eachother

Now the new register is: (c3) (b3+c2) (a3+b2+c1) (a2+b1+c0)

Processing third byte, Z:
(a2+b1+c0)+Z                    this is the calculated topbyte (3)
d3    d2    d1       d0         sequence in table where the topbyte points at
00    c3    b3+c2    a3+b2+c1    to right shifted register
00+d3 c3+d2 b3+c2+d1 a3+b2+c1+d0 previous 2 lines XORred with eachother

Now the new register is: (d3) (c3+d2) (b3+c2+d1) (a3+b2+c1+d0)

Processing fourth byte, W:
(a3+b2+c1+d0)+W                 this is the calculated topbyte (4)
e3    e2    e1       e0         sequence in table where the topbyte points at
00    d3    c3+d2    b3+c2+d1    to right shifted register
00+e3 d3+e2 c3+d2+e1 b3+c2+d1+e0 previous 2 lines XORred with eachother

Now the final register is: (e3) (d3+e2) (c3+d2+e1) (b3+c2+d1+e0)

I'll show it a little different way:
a0 + X                  =(1)  points to  b3 b2 b1 b0  in table
a1 + b0 + Y             =(2)  points to  c3 c2 c1 c0  in table
a2 + b1 + c0 + Z        =(3)  points to  d3 d2 d1 d0  in table
a3 + b2 + c1 + d0 + W   =(4)  points to  e4 e3 e2 e1  in table
     b3 + c2 + d1 + e0  =f0
          c3 + d2 + e1  =f1
               d3 + e2  =f2
                    e3  =f3
    (1)  (2)  (3)  (4)
(figure 4)

This is reversed in the same way as the 16bit version. I shall give an example
with real values. For the table values use the CRC-32 table in the appendix.
Take for CRC register before, a3 a2 a1 a0 -> AB CD EF 66
Take for CRC register after,  f3 f2 f1 f0 -> 56 33 14 78 (wanted value)
Here we go:

First byte of entries            entry   value
e3=f3                     =56 -> 35h=(4) 56B3C423 for e3 e2 e1 e0
d3=f2+e2      =33+B3      =E6 -> 4Fh=(3) E6635C01 for d3 d2 d1 d0
c3=f1+e1+d2   =14+C4+63   =B3 -> F8h=(2) B3667A2E for c3 c2 c1 c0
b3=f0+e0+d1+c2=78+23+5C+66=61 -> DEh=(1) 616BFFD3 for b3 b2 b1 b0

Now we have all needed values, then
X=(1)+         a0=         DE+66=B8
Y=(2)+      b0+a1=      F8+D3+EF=C4
Z=(3)+   c0+b1+a2=   4F+2E+FF+CD=53
(final computation)

Conclusion: to change the CRC-32 register from ABCDEF66 to 56331478 we need
 this sequence of bytes: B8 C4 53 8E

The reverse Algorithm for CRC-32

  If you look at the by-hand computation of the sequence of bytes needed to
change the CRC register from  a3 a2 a1 a0  to  f3 f2 f1 f0  its difficult to
transform this into a nice compact algorithm.

Look at an extended version of the final computation:
X =(1) +                a0     0
Y =(2) +           b0 + a1     1
Z =(3) +      c0 + b1 + a2     2
W =(4) + d0 + c1 + b2 + a3     3
f0= e0 + d1 + c2 + b3          4
f1= e1 + d2 + c3               5
f2= e2 + d3                    6
f3= e3                         7
(figure 5)

  It is just the same as figure 4, only some values/bytes exchanged. This view
will help us to get a compact algorithm. What if we take a buffer of 8 bytes
that is, for every line you see in figure 5 one byte is reserved. Bytes 0 to
3 are filled with a0 to a3, bytes 4 to 7 are filled with f0 to f3. As before,
we take the last byte e3 which is equal to f3 and lookup the complete value in
the CRC table. Then we XOR this value (e3 e2 e1 e0) on position 4 (as in figure
5). Then we automatically know what the value of d3 is, because we already
XORred f3 f2 f1 f0 with e3 e2 e1 e0, and f2+e2=d3. Because we now already know
what the value of (4) is (the entry number), we can directly XOR the value into
position 3. Now we know d3 use this to lookup the value of d3 d2 d1 d0 and XOR
this on one position earlier, that is position 3 (look at the figure!). XOR the
found entry number (3) for the value on position 2. We now know c3 because we
have the value f1+e1+d2=c3 on position 5.
  We go on doing this until we XORred b3 b2 b1 b0 on position 1. Et voila!
Bytes 0 to 3 of the buffer now contains the needed bytes X to W!

Summarized is here the algorithm:
1. Of the 8 byte buffer, fill position 0 to 3 with a0 to a3 (the start value of
 the CRC register), and position 4 to 7 with  f0 to f3 (wanted end value of CRC
2. Take the byte from position 7 and use it to lookup the complete value.
3. XOR this value (dword) on position 4
4. XOR the entry number (byte) on position 3
5. Repeat step 2 & 3 three more times while decreasing the positions each time
 by one.

Implementation of the Reverse Algorithm

  Now its time for some code. Below are the implementation of the reverse
algorithm for CRC-32 in Assembly (it is not difficult to do this for other
languages and/or CRC standards). Note that in assembly (on PC's) dwords are
written to and read from memory in reverse order.

crcBefore       dd (?)
wantedCrc       dd (?)
buffer          db 8 dup (?)

        mov     eax, dword ptr[crcBefore] ;/*
        mov     dword ptr[buffer], eax
        mov     eax, dword ptr[wantedCrc] ; Step 1
        mov     dword ptr[buffer+4], eax  ;*/

        mov     di, 4
        mov     al, byte ptr[buffer+di+3] ;/*
        call   GetTableEntry              ; Step 2 */
        xor     dword ptr[buffer+di], eax ; Step 3
        xor     byte ptr[buffer+di-1], bl ; Step 4
        dec     di                        ;/*
        jnz    computeReverseLoop         ; Step 5 */

-Registers eax, di bx are used

Implementation of GetTableEntry

crctable        dd 256 dup (?)       ;should be defined globally somewhere & initialized of course

        mov     bx, offset crctable-1
        add     bx, 4                ;points to (crctable-1)+k*4 (k:1..256)
        cmp     [bx], al             ;must always find the value somewhere
        jne     getTableEntryLoop

        sub     bx, 3
        mov     eax, [bx]
        sub     bx, offset crctable
        shr     bx, 2


 On return eax contains a table entry, bx contains the entry number.


  Well... your reached the end of this essay. If you now think: wow, all those
programs which are protected by CRC can say 'bye, bye'. Nope. It is very easy
to make an anti-anti-CRC code. To make a succesfull CRCreverse you have to
know exactly from what part of the code the CRC is calculated and what CRC
algorithm is used. A simple countermeasure is using 2 different CRC algorithms,
or combination with another dataprotection algorithm.
  Anywayz... I hope all this stuff was interesting and that you enjoyed reading
it as I enjoyed writing it.

Fnx go out to the beta-testers Douby/DREAD and Knotty Dread for the good
comments on my work which made it even better!

For a sample CRC-32 correcting patcher program visit my webpages:
    http://surf.to/anarchriz  -> Programming -> Projects
(it's still a preview but will give you a proof of my idea)

For more info on DREAD visit http://dread99.cjb.net

If you still have questions you can mail me at anarchriz@hotmail.com,
or try the channels #DreaD, #Win32asm, #C.I.A and #Cracking4Newbies (in that
order) on EFnet (on IRC).

CYA ALL! - Anarchriz

"The system makes its morons, then despises them for their ineptitude, and
rewards its 'gifted few' for their rarity." - Colin Ward


CRC-16 Table

  00h   0000 C0C1 C181 0140 C301 03C0 0280 C241
  08h   C601 06C0 0780 C741 0500 C5C1 C481 0440
  10h   CC01 0CC0 0D80 CD41 0F00 CFC1 CE81 0E40
  18h   0A00 CAC1 CB81 0B40 C901 09C0 0880 C841

  20h   D801 18C0 1980 D941 1B00 DBC1 DA81 1A40
  28h   1E00 DEC1 DF81 1F40 DD01 1DC0 1C80 DC41
  30h   1400 D4C1 D581 1540 D701 17C0 1680 D641
  38h   D201 12C0 1380 D341 1100 D1C1 D081 1040

  40h   F001 30C0 3180 F141 3300 F3C1 F281 3240
  48h   3600 F6C1 F781 3740 F501 35C0 3480 F441
  50h   3C00 FCC1 FD81 3D40 FF01 3FC0 3E80 FE41
  58h   FA01 3AC0 3B80 FB41 3900 F9C1 F881 3840

  60h   2800 E8C1 E981 2940 EB01 2BC0 2A80 EA41
  68h   EE01 2EC0 2F80 EF41 2D00 EDC1 EC81 2C40
  70h   E401 24C0 2580 E541 2700 E7C1 E681 2640
  78h   2200 E2C1 E381 2340 E101 21C0 2080 E041

  80h   A001 60C0 6180 A141 6300 A3C1 A281 6240
  88h   6600 A6C1 A781 6740 A501 65C0 6480 A441
  90h   6C00 ACC1 AD81 6D40 AF01 6FC0 6E80 AE41
  98h   AA01 6AC0 6B80 AB41 6900 A9C1 A881 6840

  A0h   7800 B8C1 B981 7940 BB01 7BC0 7A80 BA41
  A8h   BE01 7EC0 7F80 BF41 7D00 BDC1 BC81 7C40
  B0h   B401 74C0 7580 B541 7700 B7C1 B681 7640
  B8h   7200 B2C1 B381 7340 B101 71C0 7080 B041

  C0h   5000 90C1 9181 5140 9301 53C0 5280 9241
  C8h   9601 56C0 5780 9741 5500 95C1 9481 5440
  D0h   9C01 5CC0 5D80 9D41 5F00 9FC1 9E81 5E40
  D8h   5A00 9AC1 9B81 5B40 9901 59C0 5880 9841

  E0h   8801 48C0 4980 8941 4B00 8BC1 8A81 4A40
  E8h   4E00 8EC1 8F81 4F40 8D01 4DC0 4C80 8C41
  F0h   4400 84C1 8581 4540 8701 47C0 4680 8641
  F8h   8201 42C0 4380 8341 4100 81C1 8081 4040

CRC-32 Table

  00h   00000000 77073096 EE0E612C 990951BA
  04h   076DC419 706AF48F E963A535 9E6495A3
  08h   0EDB8832 79DCB8A4 E0D5E91E 97D2D988
  0Ch   09B64C2B 7EB17CBD E7B82D07 90BF1D91

  10h   1DB71064 6AB020F2 F3B97148 84BE41DE
  14h   1ADAD47D 6DDDE4EB F4D4B551 83D385C7
  18h   136C9856 646BA8C0 FD62F97A 8A65C9EC
  1Ch   14015C4F 63066CD9 FA0F3D63 8D080DF5

  20h   3B6E20C8 4C69105E D56041E4 A2677172
  24h   3C03E4D1 4B04D447 D20D85FD A50AB56B
  28h   35B5A8FA 42B2986C DBBBC9D6 ACBCF940
  2Ch   32D86CE3 45DF5C75 DCD60DCF ABD13D59

  30h   26D930AC 51DE003A C8D75180 BFD06116
  34h   21B4F4B5 56B3C423 CFBA9599 B8BDA50F
  38h   2802B89E 5F058808 C60CD9B2 B10BE924
  3Ch   2F6F7C87 58684C11 C1611DAB B6662D3D

  40h   76DC4190 01DB7106 98D220BC EFD5102A
  44h   71B18589 06B6B51F 9FBFE4A5 E8B8D433
  48h   7807C9A2 0F00F934 9609A88E E10E9818
  4Ch   7F6A0DBB 086D3D2D 91646C97 E6635C01

  50h   6B6B51F4 1C6C6162 856530D8 F262004E
  54h   6C0695ED 1B01A57B 8208F4C1 F50FC457
  58h   65B0D9C6 12B7E950 8BBEB8EA FCB9887C
  5Ch   62DD1DDF 15DA2D49 8CD37CF3 FBD44C65

  60h   4DB26158 3AB551CE A3BC0074 D4BB30E2
  64h   4ADFA541 3DD895D7 A4D1C46D D3D6F4FB
  68h   4369E96A 346ED9FC AD678846 DA60B8D0
  6Ch   44042D73 33031DE5 AA0A4C5F DD0D7CC9

  70h   5005713C 270241AA BE0B1010 C90C2086
  74h   5768B525 206F85B3 B966D409 CE61E49F
  78h   5EDEF90E 29D9C998 B0D09822 C7D7A8B4
  7Ch   59B33D17 2EB40D81 B7BD5C3B C0BA6CAD

  80h   EDB88320 9ABFB3B6 03B6E20C 74B1D29A
  84h   EAD54739 9DD277AF 04DB2615 73DC1683
  88h   E3630B12 94643B84 0D6D6A3E 7A6A5AA8
  8Ch   E40ECF0B 9309FF9D 0A00AE27 7D079EB1

  90h   F00F9344 8708A3D2 1E01F268 6906C2FE
  94h   F762575D 806567CB 196C3671 6E6B06E7
  98h   FED41B76 89D32BE0 10DA7A5A 67DD4ACC
  9Ch   F9B9DF6F 8EBEEFF9 17B7BE43 60B08ED5

  A0h   D6D6A3E8 A1D1937E 38D8C2C4 4FDFF252
  A4h   D1BB67F1 A6BC5767 3FB506DD 48B2364B
  A8h   D80D2BDA AF0A1B4C 36034AF6 41047A60
  ACh   DF60EFC3 A867DF55 316E8EEF 4669BE79

  B0h   CB61B38C BC66831A 256FD2A0 5268E236
  B4h   CC0C7795 BB0B4703 220216B9 5505262F
  B8h   C5BA3BBE B2BD0B28 2BB45A92 5CB36A04
  BCh   C2D7FFA7 B5D0CF31 2CD99E8B 5BDEAE1D

  C0h   9B64C2B0 EC63F226 756AA39C 026D930A
  C4h   9C0906A9 EB0E363F 72076785 05005713
  C8h   95BF4A82 E2B87A14 7BB12BAE 0CB61B38
  CCh   92D28E9B E5D5BE0D 7CDCEFB7 0BDBDF21

  D0h   86D3D2D4 F1D4E242 68DDB3F8 1FDA836E
  D4h   81BE16CD F6B9265B 6FB077E1 18B74777
  D8h   88085AE6 FF0F6A70 66063BCA 11010B5C
  DCh   8F659EFF F862AE69 616BFFD3 166CCF45

  E0h   A00AE278 D70DD2EE 4E048354 3903B3C2
  E4h   A7672661 D06016F7 4969474D 3E6E77DB
  E8h   AED16A4A D9D65ADC 40DF0B66 37D83BF0
  ECh   A9BCAE53 DEBB9EC5 47B2CF7F 30B5FFE9

  F0h   BDBDF21C CABAC28A 53B39330 24B4A3A6
  F4h   BAD03605 CDD70693 54DE5729 23D967BF
  F8h   B3667A2E C4614AB8 5D681B02 2A6F2B94
  FCh   B40BBE37 C30C8EA1 5A05DF1B 2D02EF8D


> A painless guide to CRC error detection algorithm
  url: ftp://ftp.adelaide.edu.au/pub/rocksoft/crc_v3.txt
  (I bet this 'painless guide' is more painfull then my 'short' one ;)
> I also used a random source of a CRC-32 algorithm to understand the algorithm
> Link to crc calculation progs... hmmm search for 'CRC.ZIP' or 'CRC.EXE' or something
 alike at ftpsearch (http://ftpsearch.lycos.com?form=advanced)

Copyright (c) 1998,1999 by Anarchriz
(this is REALLY the last line :)

TUCoPS is optimized to look best in Firefox® on a widescreen monitor (1440x900 or better).
Site design & layout copyright © 1986-2024 AOH