# WU SEETF
## Crypto
### BabyRC4
chall.py:
```python=
from Crypto.Cipher import ARC4
from os import urandom
key = urandom(16)
flag = b'SEE{?????????????????????????????????}'[::-1]
def enc(ptxt):
cipher = ARC4.new(key)
return cipher.encrypt(ptxt)
print(f"c0 = bytes.fromhex('{enc(flag).hex()}')")
print(f"c1 = bytes.fromhex('{enc(b'a'*36).hex()}')")
"""
c0 = bytes.fromhex('b99665ef4329b168cc1d672dd51081b719e640286e1b0fb124403cb59ddb3cc74bda4fd85dfc')
c1 = bytes.fromhex('a5c237b6102db668ce467579c702d5af4bec7e7d4c0831e3707438a6a3c818d019d555fc')
"""
```
- Đây là dạng mã hóa RC4 và trong đoạn mã này sử dụng chung một key trong mục đích mã hóa. Chúng ta đã biết được một bản rõ `b'a' * 36` và hai đoạn ciphertext `c0` mã hóa flag, `c1` mã hóa `b'a' * 36`
- Vì vậy để giải chall này chúng ta sẽ đem xor bản rõ với c1 để tìm key và xor tiếp với c0 để ra flag.
solved.py:
```python=
from Crypto.Cipher import ARC4
from pwn import *
key="c4a356d7714cd709af271418a663b4ce2a8d1f1c2d695082111559c7c2a979b178b4349d"
key=bytes.fromhex(key)
def enc(ptxt):
cipher = ARC4.new(key)
return cipher.encrypt(ptxt)
def decrypt(ciphertext, key):
cipher = ARC4.new(key)
plaintext = cipher.decrypt(ciphertext)
return plaintext
# print(key)
a=b'a' * 36
c0 = bytes.fromhex('b99665ef4329b168cc1d672dd51081b719e640286e1b0fb124403cb59ddb3cc74bda4fd85dfc')
c1 = bytes.fromhex('a5c237b6102db668ce467579c702d5af4bec7e7d4c0831e3707438a6a3c818d019d555fc')
# a53f2e6a416a31b746d3cd5c3e4f0e74d644d3d01e5af1539d01f74170f9555c068c0738
# b50a2d636a2767c46165708acb816ec0
# print(enc(bytes.fromhex(a)).hex())
# b=xor(a,c1)
b=xor(c0,key)
print(b)
# b="1c54525953040700025b125412125418520a3e5522133e52543404133e132417520f1a24f83e"
# d=decrypt(c0,key)
# print(d)
#}5382efac:s5ss5y3k_4Cr_35Uer_rEv3n{E\x99_
```
- Vì lúc mã hóa đã đảo ngược lại flag nên chúng ta sẽ reverse lại flag:
> SEE{n3vEr_reU53_rC4_k3y5ss5s:cafe2835}
### Dumb Chall
chall.py:
```python=
import random
import time
from Crypto.Util.number import bytes_to_long, isPrime
from secret import FLAG
def fail():
print("You have disappointed the pigeon.")
exit(-1)
def generate_prime_number(bits: int = 128) -> int:
num = random.getrandbits(bits)
while not isPrime(num):
num += 1
return num
def generate_random_boolean() -> bool:
return bool(random.getrandbits(1))
def first_verify(g, p, y, C, w, r) -> bool:
assert w
return ((y * C) % p) == pow(g, w, p)
def second_verify(g, p, y, C, w, r) -> bool:
assert r
return pow(g, r, p) == C
p = generate_prime_number()
g = random.getrandbits(128)
x = bytes_to_long(FLAG.encode())
y = pow(g, x, p)
print(f"p = {p}")
print(f"g = {g}")
print(f"y = {y}")
print("Something something zero-knowledge proofs blah blah...")
print("Why not just issue the challenge and the verification at the same time? Saves TCP overhead!")
seen_c = set()
for round in range(30):
w, r = None, None
choice = generate_random_boolean()
if not choice:
w = int(input("Enter w: "))
C = int(input("Enter C: "))
if C in seen_c:
fail()
seen_c.add(C)
verify = first_verify
else:
r = int(input("Enter r: "))
C = int(input("Enter C: "))
if C in seen_c:
fail()
seen_c.add(C)
verify = second_verify
if not verify(g, p, y, C, w, r):
fail()
else:
print(f"You passed round {round + 1}.")
time.sleep(1)
print(
"You were more likely to get hit by lightning than proof correctly 30 times in a row, you must know the secret right?"
)
print(f"A flag for your troubles - {FLAG}")
```
- Bài này chúng ta cần vượt qua 30 vòng xác minh, mỗi vòng xác minh sẽ ngẫu nhiên 1 trong 2 dạng xác minh sau.
- Với second_verify thì khá dễ ràng khi chúng ta biết được g và p rồi. Mình chọn r là một số bất kì và C = pow(g,r,p)
- Còn firse_verify để thỏa mãn `((y * C) % p) == pow(g, w, p)` thì mình chọn w=0 do đó C = y^-1. Để đảm bảo C mỗi vòng sẽ khác nhau mình sẽ thêm kp vào để không lặp lại.
solved.py:
```python=
from pwn import *
s = connect('win.the.seetf.sg', 3002)
s.recvuntil(b'p =')
p=int((s.recvuntil(b'\n').strip()).decode())
print(p)
s.recvuntil(b'g =')
g=int((s.recvuntil(b'\n').strip()).decode())
print(g)
s.recvuntil(b'y =')
y =int((s.recvuntil(b'\n').strip()).decode())
print(y)
for i in range(30):
da=(s.recv()).decode()
print(da)
if da.endswith('Enter w: '):# y*C % p = g^w % p
s.sendline(str(p-1).encode())
print(s.recv().decode())
s.sendline(str((pow(y,-1,p)+ p * i)).encode())
elif da.endswith('Enter r: '):
s.sendline(str(i+1).encode())# g^r % p = C =>
print(s.recv().decode())
s.sendline(str(pow(g,i+1,p)).encode())
s.interactive()
#SEE{1_571ll_h4v3_n0_kn0wl3d63}
```
> SEE{1_571ll_h4v3_n0_kn0wl3d63}
### OpenEndedRSA
chall.py
```python=
from Crypto.Util.number import *
from gmpy2 import iroot # this helps with super accurate square root calculations!
flag = b'????????????????????????'
m = bytes_to_long(flag)
e = 0x10001
pp = bytes_to_long(b'????????????????')
s = 1
assert isPrime(pp)
while not isPrime(s):
p = getPrime(512)
s = p**2 + pp**2
assert iroot(s-pp**2,2) == (p, True) # quick demo on how to use iroot()
assert s%2 == 1 # duh, s is a prime number after all!
q = getPrime(512)
n = p*q
c = pow(m,e,n)
print(f'n = {n}')
print(f'e = {e}')
print(f'c = {c}')
print(f's = {s}')
"""
n = 102273879596517810990377282423472726027460443064683939304011542123196710774901060989067270532492298567093229128321692329740628450490799826352111218401958040398966213264648582167008910307308861267119229380385416523073063233676439205431787341959762456158735901628476769492808819670332459690695414384805355960329
e = 65537
c = 51295852362773645802164495088356504014656085673555383524516532497310520206771348899894261255951572784181072534252355368923583221684536838148556235818725495078521334113983852688551123368250626610738927980373728679163439512668552165205712876265795806444660262239275273091657848381708848495732343517789776957423
s = 128507372710876266809116441521071993373501360950301439928940005102517141449185048274058750442578112761334152960722557830781512085114879670147631965370048855192288440768620271468214898335819263102540763641617908275932788291551543955368740728922769245855304034817063220790250913667769787523374734049532482184053
"""
```
- Chall này ta thấy `s= p**2 + pp**2` nên ta có thể tính được p bằng cách dùng tool https://www.alpertron.com.ar/FSQUARES.HTM để tính hoặc đơn giản hơn là ta lấy căn bậc 2 của `s`.
- Ta thu được `p=11336109240426199545380121772211758030643077084354723510396623108046045877936239312932896062461038721613541632921592699626408876519614365179680368296156007`
- Đã có được p ta giải mã theo mật mã RSA và thu được flag.
solved.py:
```python=
from Crypto.Util.number import *
from gmpy2 import iroot
e = 65537
n = 102273879596517810990377282423472726027460443064683939304011542123196710774901060989067270532492298567093229128321692329740628450490799826352111218401958040398966213264648582167008910307308861267119229380385416523073063233676439205431787341959762456158735901628476769492808819670332459690695414384805355960329
e = 65537
c = 51295852362773645802164495088356504014656085673555383524516532497310520206771348899894261255951572784181072534252355368923583221684536838148556235818725495078521334113983852688551123368250626610738927980373728679163439512668552165205712876265795806444660262239275273091657848381708848495732343517789776957423
s = 128507372710876266809116441521071993373501360950301439928940005102517141449185048274058750442578112761334152960722557830781512085114879670147631965370048855192288440768620271468214898335819263102540763641617908275932788291551543955368740728922769245855304034817063220790250913667769787523374734049532482184053
p=11336109240426199545380121772211758030643077084354723510396623108046045877936239312932896062461038721613541632921592699626408876519614365179680368296156007
q=n//p
phi=(p-1)* (q-1)
d=inverse(e,phi)
m=long_to_bytes(pow(c,d,n))
print(m)
SEE{0dd_3vEN:deadbeef}
```
> SEE{0dd_3vEN:deadbeef}
### Semaphore
chall.py:
```python=
import ecdsa # https://pypi.org/project/ecdsa/
import os
flag = os.environ.get('FLAG', 'SEE{not_the_real_flag}').encode()
sk = ecdsa.SigningKey.generate()
for nibble in flag.hex():
signature = sk.sign(flag + nibble.encode())
print(signature.hex())
# Code ends here. The output is printed below.
'''
04a22664841747a475192a7f9b6461404c0f85c46810d00d1a18249338e5bb58fe32e6580e6aab74dad040e1e7ae0a2d
4992c6691ae1e326f5038d050bce13d5007625c3c694a86dc9a75748f395ae80b5d904b6464f0e17b2df17e6b9f540ee
836b146003baaf00b28403ca52b7df2d949350cf4e57b7e9f981b6ee9293c93d44977f102b5bd4000d93890ebfff2d28
5e7d2389cb92df48845cf3d70a172e458de9456743964568c2532f080ea5b4ba70301b27f5d9019f5d0ac4aafd2147af
1cf28514e763cf6494814e0d64d32bd20ce99a983e88dc66225aaa4246c58ab6f33a2bfe441e023582b47937ac8d15a4
1a1980d999797b5af80d4f550d741708352f738c137f8ca6e1fd9dd3b70fa75f800f865b6b5098fb95c62ab40a80594f
83b70fc96da176c34f616a48be43ebc22b090545a52b972cc784d55c79996ef842493514a6969db7c5d34b9ef6fcd28e
99c5d66a1f7f79d5b194397af66a347b7cba9dc5d327424aea71bc82fba1bbf5d50acd57e6eea4fa366a542f1a91e350
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012a666c3dbf7a64760acf459fd4ad4de7a235ee8866c3810227c8f7cbcddda79130c46b820bff9521965dfe44de32af
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cc82ce7885a5ac4bc08bdbd13976e3cd3b2a3f439180a329e826275e3dee8386720bf28d75596c289b5bc0b9d364eef9
4f4eb4b9bb6c6deaf1173763dbd1518079096d8863a4140dca9c01693ae4bfaab6f125e0ed5a5f20200b8c1a86e58ef8
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71b8b89846f6849a17cd6258695790f7c71945e0c6b8fd3f297809ad61ddd9f73f8ec4538e45d1d420f328fefd71fd39
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ddcfe5abc02ba60668d6d87420fb1d74fc54b97707ff01cf2a0200ddf2eb708e40e9d9ae89841df4bb0e5c1aea8b6885
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'''
```
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