MonthFebruary 2008

Crackme 4 - Linux Crackme

In this article I will explain how I solved lagalopex crackme from crackmes.de.You can download my keygen and its source code from here.

These are the rules and information:

Get a working keygen.
Allowed are only GPLed-tools.
Patching/Hijacking prohibited ;)

Difficulty: 3 - Getting harder
Platform: Unix/linux etc.
Language: C/C++

Ok, first let's execute it to see what's going on and if we are prompted for something:

daniel@gargamel:~/crackme/lagalopex$ ./cm3
Your name: daniel
Hello daniel, lets see what you've done so far...
daniel@gargamel:~/crackme/lagalopex$

Ok, we have done nothing so far so we cannot expect anything else. If we test the binary with the 'file' command we will see that it is not stripped so we might get some useful information from its 'nm' and 'strings' output:

daniel@gargamel:~/crackme/lagalopex$ nm cm3
0804bf58 T gcd1
0804bf92 T gcd2
0804bfcc T gcd3
0804bea6 T gcd_
U getline@@GLIBC_2.0
U getppid@@GLIBC_2.0
U getpwuid@@GLIBC_2.0
U getsid@@GLIBC_2.0
U getuid@@GLIBC_2.0
0804c007 T gs
0804c0ee T main
U memcpy@@GLIBC_2.0
U open@@GLIBC_2.0
U printf@@GLIBC_2.0
U pthread_attr_destroy@@GLIBC_2.0
U pthread_attr_init@@GLIBC_2.1
U pthread_attr_setdetachstate@@GLIBC_2.0
U pthread_create@@GLIBC_2.1
U pthread_exit@@GLIBC_2.0
U pthread_join@@GLIBC_2.0
U puts@@GLIBC_2.0
0804c03d T rad
0804bf04 T rad_
0804c086 T calc
0804bc34 t rmd160_final
0804be4a T rmd160_hash_buffer
080488b4 t rmd160_init
080488fe t rmd160_trans
0804bb15 t rmd160_write

daniel@gargamel:~/crackme/lagalopex$ strings cm3
[...]
Get a name first...
Hello %s, lets see what you've done so far...
%s/.key_%s
%s/.key_%s_%i
Nice, you got it!

I didn't list all of them but the interesting ones. We can see that there are threads involved which will probably make our reversing job harder. Also, we can see the presumably goodboy string: 'Nice, you got it' and some print format strings. I will not go much in detail with the reversing process and I will just show a flow diagram:

flowchart-2

As you can see in the flow diagram, it first computes a hash value of 20 bytes (RIPEMD-160) and then iterates over it through some calculations performed by 6 threads (hence, a total of 120 threads are created). I will describe briefly what these threads do:

  • gcd1
0804bf58 gcd1:
 804bf58:	55                   	push   ebp
 804bf59:	89 e5                	mov    ebp,esp
 804bf5b:	83 ec 08             	sub    esp,0x8
 804bf5e:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804bf61:	8b 50 04             	mov    edx,DWORD PTR [eax+4]
 804bf64:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804bf67:	8b 00                	mov    eax,DWORD PTR [eax]
 804bf69:	89 54 24 04          	mov    DWORD PTR [esp+4],edx
 804bf6d:	89 04 24             	mov    DWORD PTR [esp],eax
 804bf70:	e8 31 ff ff ff       	call   gcd_
 804bf75:	83 f8 01             	cmp    eax,0x1
 804bf78:	75 0c                	jne    804bf86
 804bf7a:	c7 04 24 01 00 00 00 	mov    DWORD PTR [esp],0x1
 804bf81:	e8 5e c7 ff ff       	call   pthread_exit
 804bf86:	c7 04 24 00 00 00 00 	mov    DWORD PTR [esp],0x0
 804bf8d:	e8 52 c7 ff ff       	call   pthread_exit

In this code you can see a pointer to a vector holding Xi, Yi and (Xi+Yi). So, as you can see this thread is computing the greatest common divisor of Xi and Yi. If this value is 1 (ie, they are coprimes), then the thread will return 1. Otherwise, the returned value will be zero.

  • gcd2
0804bf92 gcd2:
 804bf92:	55                   	push   ebp
 804bf93:	89 e5                	mov    ebp,esp
 804bf95:	83 ec 08             	sub    esp,0x8
 804bf98:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804bf9b:	8b 50 08             	mov    edx,DWORD PTR [eax+8]
 804bf9e:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804bfa1:	8b 00                	mov    eax,DWORD PTR [eax]
 804bfa3:	89 54 24 04          	mov    DWORD PTR [esp+4],edx
 804bfa7:	89 04 24             	mov    DWORD PTR [esp],eax
 804bfaa:	e8 f7 fe ff ff       	call   gcd_
 804bfaf:	83 f8 01             	cmp    eax,0x1
 804bfb2:	75 0c                	jne    804bfc0
 804bfb4:	c7 04 24 01 00 00 00 	mov    DWORD PTR [esp],0x1
 804bfbb:	e8 24 c7 ff ff       	call   pthread_exit
 804bfc0:	c7 04 24 00 00 00 00 	mov    DWORD PTR [esp],0x0
 804bfc7:	e8 18 c7 ff ff       	call   pthread_exit

This thread will compute gcd(Xi,Xi+Yi), ie, the first and the third elements of our vector. If again they are coprimes, the thread will return 1; zero otherwise.

  • gcd3
0804bfcc gcd3:
 804bfcc:	55                   	push   ebp
 804bfcd:	89 e5                	mov    ebp,esp
 804bfcf:	83 ec 08             	sub    esp,0x8
 804bfd2:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804bfd5:	8b 50 08             	mov    edx,DWORD PTR [eax+8]
 804bfd8:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804bfdb:	8b 40 04             	mov    eax,DWORD PTR [eax+4]
 804bfde:	89 54 24 04          	mov    DWORD PTR [esp+4],edx
 804bfe2:	89 04 24             	mov    DWORD PTR [esp],eax
 804bfe5:	e8 bc fe ff ff       	call   gcd_
 804bfea:	83 f8 01             	cmp    eax,0x1
 804bfed:	75 0c                	jne    804bffb
 804bfef:	c7 04 24 01 00 00 00 	mov    DWORD PTR [esp],0x1
 804bff6:	e8 e9 c6 ff ff       	call   pthread_exit
 804bffb:	c7 04 24 00 00 00 00 	mov    DWORD PTR [esp],0x0
 804c002:	e8 dd c6 ff ff       	call   pthread_exit

This thread will compute gcd(Yi,Xi+Yi), ie, the second and the third elements of our vector. If again they are coprimes, the thread will return 1; zero otherwise.

  • gs
0804c007 gs:
 804c007:	55                   	push   ebp
 804c008:	89 e5                	mov    ebp,esp
 804c00a:	83 ec 08             	sub    esp,0x8
 804c00d:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c010:	8b 00                	mov    eax,DWORD PTR [eax]
 804c012:	85 c0                	test   eax,eax
 804c014:	7e 1b                	jle    804c031
 804c016:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c019:	8b 50 04             	mov    edx,DWORD PTR [eax+4]
 804c01c:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c01f:	8b 00                	mov    eax,DWORD PTR [eax]
 804c021:	39 c2                	cmp    edx,eax
 804c023:	7e 0c                	jle    804c031
 804c025:	c7 04 24 01 00 00 00 	mov    DWORD PTR [esp],0x1
 804c02c:	e8 b3 c6 ff ff       	call   pthread_exit
 804c031:	c7 04 24 00 00 00 00 	mov    DWORD PTR [esp],0x0
 804c038:	e8 a7 c6 ff ff       	call   pthread_exit

This thread will return 1 when Xi > Yi and zero otherwise.

  • rad
0804c03d rad:
 804c03d:	55                   	push   ebp
 804c03e:	89 e5                	mov    ebp,esp
 804c040:	83 ec 08             	sub    esp,0x8
 804c043:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c046:	8b 10                	mov    edx,DWORD PTR [eax]
 804c048:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c04b:	8b 40 04             	mov    eax,DWORD PTR [eax+4]
 804c04e:	0f af d0             	imul   edx,eax
 804c051:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c054:	8b 40 08             	mov    eax,DWORD PTR [eax+8]
 804c057:	0f af c2             	imul   eax,edx
 804c05a:	89 04 24             	mov    DWORD PTR [esp],eax
 804c05d:	e8 a2 fe ff ff       	call   rad_
 804c062:	89 c2                	mov    edx,eax
 804c064:	8b 45 08             	mov    eax,DWORD PTR [ebp+8]
 804c067:	8b 40 08             	mov    eax,DWORD PTR [eax+8]
 804c06a:	39 c2                	cmp    edx,eax
 804c06c:	7d 0c                	jge    804c07a
 804c06e:	c7 04 24 01 00 00 00 	mov    DWORD PTR [esp],0x1
 804c075:	e8 6a c6 ff ff       	call   pthread_exit
 804c07a:	c7 04 24 00 00 00 00 	mov    DWORD PTR [esp],0x0
 804c081:	e8 5e c6 ff ff       	call   pthread_exit

This thread will calculate the radical of (Xi*Yi*(Xi+Yi)) and if it's less than (Xi+Yi) it will return 1.

  • calc
0804c086 calc:
 804c086:	push   ebp
 804c087:	mov    ebp,esp
 804c089:	push   ebx
 804c08a:	sub    esp,0x14
 804c08d:	mov    eax,DWORD PTR [ebp+8]
 804c090:	mov    ecx,DWORD PTR [eax]
 804c092:	mov    DWORD PTR [ebp-8],0x6bca1af3
 804c099:	mov    eax,DWORD PTR [ebp-8]
 804c09c:	imul   ecx
 804c09e:	sar    edx,0x3
 804c0a1:	mov    eax,ecx
 804c0a3:	sar    eax,0x1f
 804c0a6:	mov    ebx,edx
 804c0a8:	sub    ebx,eax
 804c0aa:	mov    eax,DWORD PTR [ebp+8]
 804c0ad:	mov    ecx,DWORD PTR [eax+4]
 804c0b0:	mov    DWORD PTR [ebp-8],0xae4c415d
 804c0b7:	mov    eax,DWORD PTR [ebp-8]
 804c0ba:	imul   ecx
 804c0bc:	lea    eax,[edx+ecx]
 804c0bf:	mov    edx,eax
 804c0c1:	sar    edx,0x6
 804c0c4:	mov    eax,ecx
 804c0c6:	sar    eax,0x1f
 804c0c9:	mov    ecx,edx
 804c0cb:	sub    ecx,eax
 804c0cd:	mov    eax,ecx
 804c0cf:	lea    edx,[ebx+eax]
 804c0d2:	mov    eax,edx
 804c0d4:	sar    eax,0x1f
 804c0d7:	mov    ecx,eax
 804c0d9:	shr    ecx,0x18
 804c0dc:	lea    eax,[edx+ecx]
 804c0df:	and    eax,0xff
 804c0e4:	sub    eax,ecx
 804c0e6:	mov    DWORD PTR [esp],eax
 804c0e9:	call   pthread_exit

Now we have to find out what this function does since, at first glance, it's not as simple as the other ones.
It makes some operations with the first and second values of our vector, ie, Xi and Yi. And right before the call to pthread_exit instruction we can see that the returned value is AND'ed with 0xFF so the output of this function will be between 0x00 and 0xFF. I will not dig too much into all the shifts and the multiplications details; instead I will print its equivalent C source.

int calc(int *v)
{
	int x, y;
	x=v[0];
	y=v[1];

	return (   ((int)(x/19) + (int)(y/94)) % 256  );
}

Now let's have a look at the way the binary is collecting all this data and how it is processed:

 804c677:	call   80486c4
 804c67c:	test   eax,eax
 804c67e:	je     804c68f
 804c680:	mov    DWORD PTR [ebp-0x2510],0x3
 804c687:
 804c68a:	jmp    804c869
 804c68f:	mov    eax,DWORD PTR [ebp-0x360]
 804c695:	add    DWORD PTR [ebp-20],eax

This is the code which actually collects the data returned by the 'gcd1 threads'. Essentially it does the same than the 'gcd2', 'gcd3', 'gs', and 'rad' threads so I will just comment this one.

After the call to pthread_join, the returned value is at [ebp-0x360] address. If it is not zero (ie, if it is 1), the value is added to the value at [ebp-0x360] which is the counter shown in the flow chart.

Let's now analyze the 'pthread_join' to the 'calc' threads:

 804c7f8:	call   pthread_join
 804c7fd:	test   eax,eax
 804c7ff:	je     804c80d
 804c801:	mov    DWORD PTR [ebp-0x2510],0x3
 804c808:
 804c80b:	jmp    804c869
 804c80d:	mov    eax,DWORD PTR [ebp-24]
 804c810:	movzx  eax,BYTE PTR [ebp+eax-0x24fc]
 804c817:
 804c818:	movsx  eax,al
 804c81b:	movzx  edx,al
 804c81e:	mov    eax,DWORD PTR [ebp-0x360]
 804c824:	cmp    edx,eax
 804c826:	je     804c834
 804c828:	mov    DWORD PTR [ebp-0x2510],0x5
 804c82f:
 804c832:	jmp    804c869
 804c869:	mov    eax,DWORD PTR [ebp-0x2510]
 804c86f:	add    esp,0x252c
 804c875:	pop    ecx
 804c876:	pop    ebx
 804c877:	pop    edi
 804c878:	leave
 804c879:	lea    esp,[ecx-4]
 804c87c:	ret

This code gets the value returned from the 'calc' thread and compares it to the H[i] which holds the i-th hash value. If they don't match, the program will exit with code 5. Otherwise, it will continue picking up the values returned by the threads. So for the sake of writing the keygen we now know that the (Xi,Yi) values read from the file must satisfy this constraint. If not, the program will just exit without printing out the goodboy message.

After all the exit values are collected and processed we will find the following code:

 804c841:	8b 5d ec             	mov    ebx,DWORD PTR [ebp-20]
 804c844:	83 eb 64             	sub    ebx,0x64
 804c847:	e8 78 bf ff ff       	call   rand
 804c84c:	0f af c3             	imul   eax,ebx
 804c84f:	85 c0                	test   eax,eax
 804c851:	75 0c                	jne    804c85f
 804c853:	c7 04 24 b4 c9 04 08 	mov    DWORD PTR [esp],0x804c9b4
 804c85a:	e8 55 bf ff ff       	call   puts

The 'result' value is loaded into ebx, then 0x64 is substracted from it and multiplied by a random value.
If the multiplication returns zero, the goodboy string is moved onto the stack and printed out through puts function:

if( (counter-0x64) * rand()  ==  0)
   puts("Nice, you got it!");

The way to go is making 'counter' to be 0x64. In every iteration, counter is incremented by 5 if the conditions are satisfied; as there are 20 iterations, counter will be 100 (0x64) if (Xi,Yi) pairs are correctly chosen. And this is what our keygen should do.

Putting it all together:

eq-forall

eq-calc

eq-rad

eq-constraints

To get a working keygen, we must implement a sort of intelligent brute-force keeping in mind all the constraints listed above.

  • ABC Conjecture

This conjecture was first put forward in 1980 by Joseph Oesterle of the University of Paris and David Masser of the Mathematics Institute of the University of Basel in Switzerland, which is now considered one of the most important unsolved problems in number theory. This conjecture deals with numbers that have no common factors. For example, if A=3 and B=7 then we'll have C=10. These 3 numbers are relative primes but rad(A*B*C) > C. Occasionally this is not always true; for instance, if A=1 and B=8, then C=9, rad(A*B*C)=1*2*3=6 < 9.

From here we can read:

If rad(a*b*c) can be less than c, just how much less? Can we find an example where rad(abc) is so much smaller that the square of rad(abc) is also less than c? No one knows of such a triple, but there are some examples where [rad(abc)]α < c for some exponent α greater than 1 though less than 2. Take the triple {2, 6436341, 6436343}. Here b is equal to 310 × 109 and c is equal to 235, and so rad(abc) = 2 × 3 × 23 × 109 = 15042. And 15042α < 6436343 for any α < 1.6299.

In 1985 Joseph Oesterlé of the University of Paris and David W. Masser of the University of Basel conjectured that there are only finitely many such exceptional triples. Given a positive number ε that can be made arbitrarily small, [rad(abc)]1 + ε can be less than c only in a finite number of cases (which will depend on ε). That’s the abc conjecture.

First of all we should try only the (Xi,Yi) pairs so that calc(Xi,Yi) = H(i); to achieve this we need a way to solve that equation:

int solve_calc(int x, int k)
{
	int d = k - (int)(x/19);
	return  d*94;
}

We will get the Yi candidates by solving the equation with some Xi and the H[i] value for each iteration. The 'solve_calc' function will return a valid Yi but it's not the only one which will satisfy the 'calc' condition. Valids Yi will range from the returned value to the returned value+93 and we can add them n*256 (for any positive n) since the calc function performs a modulo 256 operation.

As well, for each Xi value multiple of 19, we will get the same 'calc' value for the next 18 ones due to the integer division.

With these in mind we now have to bruteforce the obtained pairs to find the ones who satisfy the co-primality and radical restrictions. The way I will deal with the co-primality restrictions is by computing the GCD with the Extended Euclidean Algorithm. However, the hardest part is computing the radical of the numbers since a factorization is needed.

To speed the factorization process up I will use the Sieve of Eratosthenes to compute a table with all the primes up to a limit.

Please, have a look at the source code of the keygen for a further understanding of the process I have described. In a Pentium 4 3.0 GHz a valid license is generated in less than 1 second but it also depends on the hash value.


daniel@gargamel:~/crackme/lagalopex$ time ./keygen
Username: daniel
Sieve of Eratosthenes Prime Generation...OK
100% completed
Writing license info into /home/daniel/.key_daniel

real 0m1.079s
user 0m0.420s
sys 0m0.030s

As you can see it took about 0.45 seconds to generate the numbers and write the license file. The rest of the time is how much it took me to write my username :-).


daniel@gargamel:~/crackme/lagalopex$ time ./cm3
Your name: daniel
Hello daniel, lets see what you've done so far...
Nice, you got it!

real 0m1.711s
user 0m1.020s
sys 0m0.030s

Voila ! We got a working keygen and it takes less than half the time in the bruteforcing process than the crackme itself in verifying our license.

Notes:

Regarding the string to be hashed I'd like to point out that the final part of it is a number of value: getppid()-getsid(). This number should be zero since the parent PID and the session ID are the same if the program is running from a shell. However if you are debugging the crackme (gdb, strace, ltrace, ...) the parent process ID will be the PID of the debugger and this assumption is not true anymore. To debug the program with the data generated by my KeyGen I made a little trick based on LD_PRELOAD so that getppid and getsid always return 0:

int getppid() { return 0; }
int getsid(int pid) { return 0; }

gcc -shared -o fakeids.so fakeids.c
export LD_PRELOAD=$PWD/fakeids.so

daniel@gargamel:~/crackme/lagalopex$ ./testuids
ppid()=0
getsid()=0

This way, the debugging process becomes easier.

Hope you found this article interesting.

Regards,

Daniel

X-PIC Development System

robot-arm-xpic

X-PIC-3The X-PIC Development System was designed by my colleague and friend Alberto Calvo and me during our studies at University. It is composed of two main boards: X-PIC and X-BOT which I will describe briefly. Our goal was to design a microcontroller based system powerful enough for our Robotics projects. However, we realized that it could be a good idea to design a more generic system which could be used as a small development platform without the need to have a big amount of LED's, buttons, LCD, switches, jumpers, etc. as most boards use to have. So it is suitable for any projects which require a microcontroller.

  • X-PIC

X-PIC-1 This is the main board. It has got a PIC16F877A microcontroller at a clock frequency of 20MHz. There's also an EEPROM serial memory which can be interfaced both externally and through the microcontroller. All the GPIO pins can be easily accessed through the 10-Pin connectors, all of them including VCC and GND signals.
For testing purposes, the board's got a general purpose button and one LED. It can be programmed through its ICSP connector or through its RS232 interface if a bootloader is previously loaded into its internal Flash memory.

  • X-BOT

X-BOT-1

This board was designed specifically for Robotics applications and can be easily connected to the X-PIC board and mounted together. It's got eight connectors which allow the microcontroller to read their value through selectable digital or analog IO pins. Also it includes two DC Motor drivers to drive up to 4 motors (1A per channel) which make this board suitable for a huge number of Robotics and Electronics applications.

X-TRK-1As an extension to this system we designed the X-TRK sensor board specially aimed to Sniffer Robots. It can drive up to 8 IR sensors with just 4 IO pins thanks to its 3 to 8 line decoder. In the picture you can see a 3D version of the board, which has not been yet manufactured (just two home-made protoypes).

You can download all the documentation and schematics in Spanish (sorry for the inconvenience):

X-PIC User Guide

X-BOT User Guide

X-PIC Schematics

X-BOT Schematics

Regards,

Daniel