LA CTF 2023 - crypto / one-more-time-pad

1. intro

2. code 및 분석

2.1.  code

from itertools import cycle
pt = b"Long ago, the four nations lived together in harmony ..."

key = cycle(b"lactf{??????????????}")

ct = ""

for i in range(len(pt)):
    b = (pt[i] ^ next(key))
    ct += f'{b:02x}'
print("ct =", ct)

#ct = 200e0d13461a055b4e592b0054543902462d1000042b045f1c407f18581b56194c150c13030f0a5110593606111c3e1f5e305e174571431e

 

2.2. 분석

xor으로 구성된 단순한 문제이다.

아래 ct와 같이 decrypt 된 flag가 있기에 xor의 특성을 이용해서 다시 한번 xor 해주면 된다.

 

3. exploit

ec = bytes.fromhex('200e0d13461a055b4e592b0054543902462d1000042b045f1c407f18581b56194c150c13030f0a5110593606111c3e1f5e305e174571431e')
pt = b"Long ago, the four nations lived together in harmony ..."

dc = ''

for i in range(len(pt)):
    dc += chr(pt[i]^ec[i])

print(dc)

 

┌[wyv3rn🐲]-(~)
└> python chall.py
ct = 200e0d13461a5850131f4b575a1f59504a4d1f511c18080c1a155b5356495a5b1f4b50585a4b575a4d5d050f431c0709525051461f111111
lactf{b4by_h1t_m3_0ne_m0r3_t1m3}lactf{b4by_h1t_m3_0ne_m0