"DELAYED CODE"
technology
version 1.1
[*] Introduction
Let we wrote a virus. Avers will create antiviral code to detect it,
and after some time period all infected computers will be cured.
This article describes another technology of prolonging this time
period.
[*] Idea
We may write code which will change initial virus bytes (or any other
virus characteristics) after two months, for example.
If the virus will initially contain such modificating-code, antivirus
will just contain not one, but two or more checksums, or checksum
calculated at other, constant code range.
There are only one way to prohibit analysis of the modificating-code:
to hide it from avers. And there are the following ways to implement
this action:
1. At required time, download code from the Internet, or encrypt code
and wait for decryption-key.
2. Encrypt code with such method, that decryption will take
exactly required time. ("delayed code")
As you can see, last variant allows us to write completely automated
virus with hidden "delayed" features.
[*] Theory (sux)
Lets encrypt some random buffer A[] with some hashing algorithm so many
times N, so it will take us time period T.
After these calculations done, we have another random buffer B[] which
is used to encrypt/decrypt our "delayed" code.
There is no way to perform required N iterations using more than one
computer ('coz each time current buffer is encrypted), so minimal
decryption time is limited with maximal CPU speed.
If you will use computer which is fast enough, and use some time
to encrypt fucking random buffer, then you may be sure that
the same operation may not be done in a less time period.
So, each time the virus is active, it iterates N encryption cycles until
buffer A[] will be converted into B[].
After time T will be spent to decryption, virus will got buffer B[]
and use it to decrypt "delayed code".
[*] Theory (rulez)
Main trouble is that we dont want to wait for some months to encrypt
fucking data, 'coz in example showed above, encryption and decryption
both takes the same time period T.
This means that we will use RSA algorithm.
In the RSA algorithm, encryption and decryption keys are different.
So, to encrypt "delayed code" we will take random buffer A[], encrypt it
N times and got buffer B[].
But, in contrast to previously described hashing algorithm, our virus
will iterate not the same operation.
Our virus will decrypt buffer B[] back into A[].
In the encryption operation will be used small (low-bit) exponent,
and in the decryption operation we will use big exponent.
This means that encryption will take much less time than decryption.
We will encrypt our "delayed code" for some minutes/hours, but active
virus copies (as well as avers ;-) will decrypt it for some months.
[*] Encryption/decryption time interdependence
Now our task is to answere, how many times should we encrypt our
"delayed code" so it will be decrypted for some months.
As you know, RSA encryption means the following:
encryption: encr = (text ^ e) % m
decryption: text = (encr ^ d) % m
where {e,m} and {d,m} pairs are public and secret keys.
('^' sign means raising to a power, '%' means modulus,
remainder after division)
As a rule, e is a small number (with a low # of bits),
such as 3, 17, 50003 and so on.
And d is a big number, consisting of 1023 bits for example.
In our scheme there is no public/secret meaning at all.
Here is our scheme:
Encryption: Decryption:
~~~~~~~~~~ ~~~~~~~~~~
(encrypting "delayed code", (decrypting "delayed code",
at home ;-) on infected or on avers' PC)
■ A[] <-- random buffer/any data ■ x = B[]
■ encrypt "delayed code" with A[] N times: x = (x ^ d) % m
■ x = A[] A[] = x
N times: x = (x ^ e) % m ■ decrypt "delayed code" with A[]
B[] = x
■ store B[] in virus
Now lets consider our main 'modexp' subroutine.
// modexp: subroutine, raising to a power and returning modulus
// returns: x = (a ^ e) % m
void modexp(bignumber& x, // output
bignumber a, // input
bignumber e, // exponent
bignumber m, // modulus
{
x = 1;
bignumber t = a;
for (int i = 0; i <= MaxBit(e); i++)
{
if (e.GetBit[i] == 1)
x = (x * t) % m; // modmult()
t = (t * t) % m; // modmult()
}
}
As you can see, modexp() subroutine calls modmult() subroutine
(e.#bit + e.#bit1 - 1) times.
(-1 here means that last t*t multiplication is skipped;
i just simplified the code showed above)
This number of modmult() calls is a main difference between
encryption and decryption.
Decr.time = Encr.time * (D.#bit+D.#bit1 - 1) / (E.#bit1+E.#bit1 - 1) * K
where:
#bit == total number of bits
#bit1 == number of 1s
#bit + #bit1 - 1 == number of modmult() calls
K = 0.9+-10% -- Appeared because of variable modmult() time usage.
As it seems, when N-->oo, K-->1
[*] Example
Lets calculate, how many times (N) should we encrypt the message,
so it will be decrypted for 10 minutes.
Note, that all these tests were performed on a slow pc, so
not the numbers, but their proportionality has a meaning.
At first, we must create RSA key.
Let it be 1024-bit key with low exponent e = 3.
Executing: 'KEYGEN.EXE KEY\DPGN 1024 3 3'
Key parameters are: 1024-bit N, E==3, D=1023-bit/519*'1'
Theoretical time ratio: (1023+519-1)/(2+2-1)*0.9 = 462+-10%
But we want higher ratio precision, so lets find real time ratio,
kinda "calibrate" a key.
Executing: 'DGPGN.EXE e 100'
result: 100 iterations done, encryption time = 815 ms
Executing: 'DGPGN.EXE d 100'
result: 100 iterations done, decryption time = 360228 ms
'Real' ratio: 360228/815 = 441
Now we may calculate N for 10-min decryption.
If 100 decryption iterations used 360228 ms, and N iterations will
use 10*60*1000 ms (10 minutes), then
N = 60*10*1000 * 100 / 360228 = 167
If 167 decryption iterations will use 60*10*1000 ms, then
encryption time is 60*10*1000 / 441 = 1360 ms.
So, about one second of encryption will result in 10 mins of decryption.
Executing: 'DPGN.EXE e 167'
encryption time: 1268 ms
Executing: 'DPGN.EXE d 167'
decryption time: 600477 ms = 10 minutes + 0.5 seconds
Lets show here some other calculation results:
Decryption: Encryption: N (1024-bit key)
K5-100 Celeron-500
10 min 1.3 sec 167 950
1 hour 7.8 sec 1000 ...
1 day 3 min 24000
1 week 21 min 168000
1 month 1.5 hour 672000
1 year 18 hours 8064000
16 months 1 day
10 years 1 week
40 years 1 month
8-( )
[*] Increasing speed of calculations
Decryption algorithm 'text = (encr ^ d) % m' may be also represented
as follows:
a = ((encr % p) ^ dp) % p // dp = d % (p-1)
b = ((encr % q) ^ dq) % q // dq = d % (q-1)
if (b < a) b += q
text = a + p * (((b - a) * u) % q) // u: (u * p) % q = 1
In such algorithm, decryption time should be faster by some times.
But, i dont know if it is possible to find p and q knowing d,e and m.
[*] Slowing down speed of calculations
Because of d is not unique key, and d variants may be calculated
using formula:
d' = d + (p-1)*(q-1) * t, где t=0,1,2,...,
then d length may be increased as we want,
that will slow down calculations by many times.
But, if p and q will be found (knowing d'), then original d can
also be found, and then somebody will be able to perform
DPGN calculations many times faster, that is bad.
[*] Other bad stuff
1. In a real life, there is no need to publish N number
(# of iterations), just any good CRC will be enough.
So, nobody will know what is IT and when IT will be decrypted and
executed. |->
Also, it is good to add some random part to N number and do not
make resulting time T day- or hour- aligned.
2. I'm not sure, but - maybe - {N times: x = (x ^ d) % m} operation
may be changed to something more fast (using some perverted math).
To avoid such shit, you may do some additional encryption between
'modexp' calls, for example as following:
N times:
{
x = (x ^ d) % m;
x = x xor ;
}
DPGN.EXE just xors some dwords within the x number on each iteration.
[*] "Delayed code" usage (which code to encrypt?)
■ As was said above, entrypoint-modificating code, i.e. virus checksum
changer. Imagine: your virus has been spreaded, av has been written,
and now 99% of infected PCs are cured. But, that 1% that remains
infected, after some time, changes own checksum and new virus begins
to spread starting not from one-two PCs (as it was with "host"
modification), but from thousands of infected computers.
■ We all are thinking about downloading viral plugins from the Internet.
"Delayed code" technology allows your virus to contain any fixed urls,
those will be hidden from avers, until right time.
■ Instead of deleting user's data you can quickly encrypt it, leaving
a chance to decryption... in some years ;-)
■ Recursive encryption. I mean that decrypted "delayed code" can
contain another "surprise package".
(x) 2000 Z0MBiE
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