---
# System prepended metadata

title: WU SEETF

---

# WU SEETF

## Crypto

### BabyRC4
![](https://hackmd.io/_uploads/S1ideJPw3.png)
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 
![](https://hackmd.io/_uploads/Bk4eI1Pv2.png)
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.![](https://hackmd.io/_uploads/Skfd81DDh.png)

- 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
![](https://hackmd.io/_uploads/BkylikwDh.png)
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
![](https://hackmd.io/_uploads/rktMnyPvn.png)
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
37f606bfdc2004293292509364f154173a1c13aea5f6dd75be8b5cb0343e92c8efc405e7956b73132afa80d327521a8b
012a666c3dbf7a64760acf459fd4ad4de7a235ee8866c3810227c8f7cbcddda79130c46b820bff9521965dfe44de32af
ceb8fdf5acaf767a76f40ae6112b5c402a2e4e7713631e4c51e963fa61b193fd8ef0fbcbf088e5b525d1d808c5d094dd
71b1a7e5b7816b1de309daa88a8a3883a37d28650532ecbc3f462b005b353266011bdf3d642d78822c8c7299449f1f22
02caf72b6060b761d23c9bdfd4e9296a34bfd0a364eba5a81ae842a117bdd82c28ac8c692911b7d3eedbf3c543d0f48d
b959a90d2aab7a2be3684b69328d35ed9694d9df0eb4ea6e6938c806838d0db6b76718b9e7fc4268c14c17246f499af9
03b422faff6fd1de0dff181977560def1bef40ee307957d15039b06c202bc5827eba1d0e2c5027051fb5dc093bded056
fa5f0046485ea006d5917d353d3cba2010c3b5edbe87352f7450dd155c15e38de1d990be4754c5c0ff24d37a8f01af92
f027161939478e23ac9b99d56e37a0cdd76175a42892f8d7cb0dcc84964663d672c9f23812d016d445ff5c7589725c6d
30b8757465a5ec8801a5a190eb296d7d2c9ea3b1eb2c62f6062a0a8b5115f30c054cb0d5eeaba480dabfb3c5c70abef0
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9c76eae7ec44540bd41c06ba9c25b4734c9bbf68880494a38e0b118fca93677fa5770b98c0b982245a207880036e6445
2bd29f6aa55103be42b16e394a1190a03f8a22b7b7162a5d03c0423a19de47b4b8167bc330690f36db02ef5377624c80
c9b933108fad16ffd1275ea14e4b506d9c9c03df4287f99a71c09402c3a11c19932ad14af8468a84aedd8615a762d366
2e564b6e2730d220a65de2899ec7a99056c3d8e89f0178b5e974483bcc9b5949f69908777f7332bb4fe84d431e8a5a35
2f687205c31b07820a872e2a3d4e6897561ee3873086b445570a66b13232ade40e38397b648a5cac7a9a324664ece3bf
2dca9b0a9e96472b18119921a681d23d2ec3acacd1b9c0a011e23ca24407808be25e04edcbc736b9b11b6a167a2d297a
b2781278a3acd8768ca8ba1ad698a3b775c330df9a7837d18f9796cd838e5714e461f1bbfd1d3af9230c05ea33bc5555
191447109bb7de900a70e06170e492fd84d908fde7ccd2a5ed6a28c1860e655958d2f54e4638a54cbebbd35a61b528dc
54e0e6509a899baf44574eaa7d302bf00e5edfd8b0d0fa62f74524bf92bea63384773489c69e64b4b757afc9da0ecda4
941d9cb5676e366157d8bd6c92b7687539acb7a06bc611f8ad64871f0e6a598385957517aeb9972b719477bbdb31865f
1741225d735cc11377c31e8976a269473d66d8f788f34822cae720cd82f0e99af50807d1edb7513ba14efc097183d192
f5568fc82c2ca245465533f9b98bb537c9bc8e2e050d00cdf431e8acf6ed5c69b88f93760cb7fe85128a22db83242ba4
d00c1c38ff9cd6e15846aac2e821b9d60846a96d0688f86f40ea3178030868579300e2c0aa9cf9ae373ae9f817fe7ae5
76fb8669a5694795efaa9b4e38c0a306112054c891f9f871a80b6caec231f35b45c0c6ca8f64423861a5d5d09362af51
cffe394dd32e1bd6c71d300412f6551512a9eb71be7eb4b53c702caae94ae8673bf95cb7e76a0bb06e345e7e35305efa
cc82ce7885a5ac4bc08bdbd13976e3cd3b2a3f439180a329e826275e3dee8386720bf28d75596c289b5bc0b9d364eef9
4f4eb4b9bb6c6deaf1173763dbd1518079096d8863a4140dca9c01693ae4bfaab6f125e0ed5a5f20200b8c1a86e58ef8
9cad001be5d25df58a5a804ca97d9e6c7f32d6779fc020bf68727a83548382c5dc4e12359e0914cf7fca879c60fce06d
1f7091b011dfdd1ea0dcb6071f85d5c75642e37517d5263d4b468490496f90abb26e02307aa54d3d4dfde6d80626d933
7a05a981041f8e406fc69a5569c5aa2c2fa8a4c6445e3bd974a19f9fb3720cb5218c79e5eb586bfc3915d0e7ca7a9638
71b8b89846f6849a17cd6258695790f7c71945e0c6b8fd3f297809ad61ddd9f73f8ec4538e45d1d420f328fefd71fd39
d8397c87bd54c91a060d0565ff0f6edc0af0287889cd7a03ac321ea9ae1999f370ee8881b2a01b924f4fbc199cf1d868
5b848e0310b6d652310686af9cbbc6d08f42e37eb83c3c0cec22e643f859c30fc7bbeb8c1fe4e16c8dcf9e325a685a7a
4cc8713a14a5a074ecc112e22192bb162e1d5ea303939dbf2bbb9620da0ff9eb326fb5b30c7c133283a7949230ee7177
fa7e00b2da873ad104a50df8639ee6e8e702e256fd4a8ebbb25bde28988750eefcef91d4ec891458f3e700fef132071f
1fdfd4432de9d3647a194cf8d66b3d8a5e3eac5bb1e832b366ed37b4b7875fa6346b1b65bb3f068c61358ad7b7e5ee5e
0d660198e6c811a34136dc087693e81ebbd5877c722a7624ddb682043e5d24c8cf892f13b2cf2383f476f12e68bd94b5
e8f9d1f42b54c76836bae42b05bba24ffcdd0dd935380b615223a0c6ca1569a448d74cbb78e47bca635dade3e802c8a5
04a21057c034f7cb9eff39935f464aa5d4dc027b9d5da05ee5f097f12f40a4046f87683ad1638e1e05e61246ee178fbd
cabf588910bce6d41552ac5addb00f196f90c809dc6856f8e8c7490338b98bc423d2cf42c3e3faa73ab171a5a8544433
d61eeaa195eaa6970dc99e4e0ed779ee0e450f4b73917b6fd00f4bd99b26a0bb05313cdc8915d17d7b0d9942410658af
c758b4cf5847593c4cb4e356ec865e8797d74a564cf9ad1549feb4d8439e8598d0b69cd44c3267d36c1d7317303e2aa1
bd21290aee61dd300f3850b567114c8f5a1b6f8b170711b8b79f225a04961c5c39903d100a8725025da33922beba9098
a832c1f4b54699f59adff2dc5d9401888ef48eb1e3dee75d6b24334db5d8cdbe17cc87131fcfeb2fc59a084a817047a1
9251d8c0e198b84bb52e656388a4a5fc7e487dd777effa80cce8f0a9feab8b5a96bffeb73f36984c16fbb1214e2cfe65
8e2bc98948c3d0989ef1fcb26620a383a41414e55b5844e6e61e50335c18c097c1850c78acd776ffe4d35194227fb358
f18f9004efa7d5c6201faa99481df0870403f0952beffcecf9bae9eb05be06e777c83bd3fe30ba8666d72292c9f6c093
a07f86f2a093ca8d6fdbc63cb56d3527e6c85225fb77f86efea0dc1550333e7b5abcb49368c60a70ef4cf8456600b73e
a520893f94bc749b07d25d8bd1185105c001f12f4e0004733f6b1ff2197925941891f343f3eaa97bfcd461776b46217d
739c2f2c68b83569cccfb8d9178ac5e5299723fedae515d6df245ad5a0964b91f9a5a6e869c7ef57598fd33757db51a3
cb59854d3b20c9f5d432e205af140bfbd34ccee0b579f67e6ceefd91e64815df1e84969a9ee2a57bd6e559d7ef9ec3c9
1b9556ba0691912615c25a08eebb0b56e46ee639c56ef2b979322e5cbc797c9c83fa59b83c5e9e45b9b7eb51eaa1ed77
0f49b18563052f9b64ebbc1a9e0fa1e5685e09c8bd266e4231e42fd25ac74f7bea9d4decd17bd9b5c9b8c808a3751612
77f08e003539e15a5a5f860cb8e9cd42ae239c9c2c7bb7d7e094b15834c86f70e4b8bb22fde71ddb4a6916b2e9725d00
5073b22c6ac1c3171e01bfaaf0ab55b2d8be72fc259efe5c36bc1cdfa2a15db2e682411d4289bd66090417aa9893c46a
a11eadfd3a5c22820b1bc815088bebf74804b8479a77a8b72641a2743f6a1401038006cb90c777d7f29f42f519ef1545
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'''
```

