Rating:
# Potentially Eazzzy
1. Given a python file, let's see the script
```
#!/usr/bin/env python3
# Potentially Eazzzy
# Author: chainsaw10
# If I screwed up and you can't generate a key for your email, I tested this
# program with [email protected] and it should work with that email and an
# appropriate key. Still yell at me in Discord though ;)
try:
FLAG = open("flag.txt", "r").read()
except:
FLAG = "DogeCTF{Flag is different on the server}"
import itertools
ALPHABET = [chr(i) for i in range(ord("*"), ord("z")+1)]
def print_flag():
print("Generating flag...")
print(FLAG)
a = lambda c: ord(ALPHABET[0]) + (c % len(ALPHABET))
o = lambda c: ord(c)
oa = lambda c: a(o(c))
def indexes(s, needle):
a = 0
for idx, c in enumerate(s):
if c == needle:
a += idx
return a
def m(one, two, three, four):
d = len(ALPHABET)//2
s = ord(ALPHABET[0])
s1, s2, s3 = o(one) - s, o(two) - s, o(three) - s
return sum([s1, s2, s3]) % d == four % d
def validate(email, key):
email = email.strip()
key = key.strip()
if len(key) != 32:
return False
email = email[:31].ljust(31, "*")
email += "*"
for c in itertools.chain(email, key):
if c not in ALPHABET:
return False
if email.count("@") != 1:
return False
if key[0] != "Z":
return False
dotcount = email.count(".")
if dotcount < 0 or dotcount >= len(ALPHABET):
return False
if a(dotcount) != o(key[1]):
return False
if o(key[3]) != a(o(key[1])%30 + o(key[2])%30) + 5:
return False
if o(key[2]) != a(indexes(email, "*") + 7):
return False
if o(key[4]) != a(sum(o(i) for i in email)%60 + o(key[5])):
return False
if o(key[5]) != a(o(key[3]) + 52):
return False
if o(key[6]) != a((o(key[7])%8)*2):
return False
if o(key[7]) != a(o(key[1]) + o(key[2]) - o(key[3])):
return False
if o(key[8]) != a((o(key[6])%16) / 2):
return False
if o(key[9]) != a(o(key[6]) + o(key[4]) + o(key[8]) - 4):
return False
if o(key[10]) != a((o(key[1])%2) * 8 + o(key[2]) % 3 + o(key[3]) % 4):
return False
if not m(email[3], key[11], key[12], 8):
return False
if not m(email[7], key[13], key[4], 18):
return False
if not m(email[9], key[14], key[3], 23):
return False
if not m(email[10], key[15], key[10], 3):
return False
if not m(email[11], key[13], key[16], 792):
return False
if not m(email[12], key[17], key[4], email.count("d")):
return False
if not m(email[13], key[18], key[7], email.count("a")):
return False
if not m(email[14], key[19], key[8], email.count("w")):
return False
if not m(email[15], key[20], key[1], email.count("g")):
return False
if not m(email[16], email[17], key[21], email.count("s")):
return False
if not m(email[18], email[19], key[22], email.count("m")):
return False
if not m(email[20], key[23], key[17], 9):
return False
if not m(email[21], key[24], key[13], 41):
return False
if not m(email[22], key[25], key[10], 3):
return False
if not m(email[23], key[26], email[14], email.count("1")):
return False
if not m(email[24], email[25], key[27], email.count("*")):
return False
if not m(email[26], email[27], key[28], 7):
return False
if not m(email[28], email[29], key[29], 2):
return False
if not m(email[30], key[30], email[18], 4):
return False
if not m(email[31], key[31], email[4], 7):
return False
return True
def main():
print("Welcome to Flag Generator 5000")
print()
print("Improving the speed quality of CTF solves since 2020")
print()
print("You'll need to have your email address and registration key ready.")
print("Please note the support hotline is closed for COVID-19 and will be")
print("unavailable until further notice.")
print()
email = input("Please enter your email address: ")
key = input("Please enter your key: ")
if validate(email, key):
print_flag()
else:
print("License not valid. Please contact support.")
main()
```
As we can see, we have to send a email and key that can be verified to pass the constraints.
2. Luckily, the constraints seem to be not so hard. That is, there are many solutions for this challenge. So just write the [python script](sol.py) with z3 and we can get the flag.
`DawgCTF{h0pe_th15_w@snt_t00_eaz^3y_4_u}`
