Wednesday 23 September 2015

Solving CSAW challenge Wyvern (rev-500) using afl-fuzz

We're given a Linux binary which asks for input:

 |    Welcome Hero       |

 [!] Quest: there is a dragon prowling the domain.
     brute strength and magic is our only hope. Test your skill.

 Enter the dragon's secret: xxxx_find_secret

 [-] You have failed. The dragon's power, speed and intelligence

     was greater.

Opening the binary in IDA shows that the main() function prints the above output. The input is processed by start_quest(std::string). If this function returns 0x1337 the level is solved.

Looking at the start_quest function, it first pushes 28 integers in a vector, does some "weird" if-tests, and finally calls sanitize_input. Interestingly, this function again contains a lot of similar and weird if-tests of the form:

    if ( y4 < 10 || (((_BYTE)x3 - 1) * (_BYTE)x3 & 1) == 0 )

It appears this is some kind of obfuscation. At this point I guessed (and hoped) that the input would be checked character by character. This means that if we guess a character correctly, the program will execute a (new) different path in the program. This is ideal for afl-fuzz, a fuzzer that detects "interesting fuzzing candidates" based on which path a program executes.

Compiling afl-fuzz with afl-qemu-trace support allows us to fuzz binaries without access to source code:

    $ mkdir indir
    $ echo "test" > indir/test1.bin
    $ afl-fuzz -Q -i indir -o sync_dir -M fuzzer01 ./wyvern
    $ afl-fuzz -Q -i indir -o sync_dir -S fuzzer02 ./wyvern
    $ afl-fuzz -Q -i indir -o sync_dir -S fuzzer07 ./wyvern  

While the fuzzer will not find crashes, it does find inputs that trigger new execution paths in the program! It took roughly half an hour of fuzzing on 7 cores to get some interesting files:

$ xxd id\:000006\,src\:000005\,op\:flip1\,pos\:4
00: 6472 3467 3034 3434 3434 3434 34cd 617c  dr4g044444444.a|
10: 6166 e803 666c 167c 6170 6116 47         af..fl.|apa.G

All generated files are 28 bytes long. That's the number of integers pushed into the vector! And it seems it must start with dr4g0n. The integers being put into the vector are:

    array = [0x64, 0xD6, 0x10A, 0x171, 0x1A1, 0x20F, 0x26E,
       0x2DD, 0x34F, 0x3AE, 0x41E, 0x452, 0x4C6, 0x538,
       0x5A1, 0x604, 0x635, 0x696, 0x704, 0x763, 0x7CC,
       0x840, 0x875, 0x8D4, 0x920, 0x96C, 0x9C2, 0xA0F]

What's interesting is that 0x64 is the ascii value of "d". Does this vector somehow encode the flag? We know the second letter is "r" with value 0x72. How can we get this value? By subtracting the first integer, 0x64, from the subsequent one: 0xD6 - 0x64 = 0x72! The solution is:

    flag = ""
    base = 0
    for num in array:
        flag += chr(num - base)
        base = num

    print flag

Flag: dr4g0n_or_p4tric1an_it5_LLVM

In hindsight it would've also been possible to use something like Intel PIN to count the number of executed instructions. The advantage of afl-fuzz is that it requires zero configuration: just give it some random start value and it works.