# I Hate Python

## Description

I hate Python, and now you will too. Find the password.
import random
def do_thing(a, b):
return ((a << 1) & b) ^ ((a << 1) | b)
x = input("What's the password? ")
if len(x) != 25:
print("WRONG!!!!!")
else:
random.seed(997)
k = [random.randint(0, 256) for _ in range(len(x))]
a = { b: do_thing(ord(c), d) for (b, c), d in zip(enumerate(x), k) }
b = list(range(len(x)))
random.shuffle(b)
c = [a[i] for i in b[::-1]]
print(k)
print(c)
kn = [47, 123, 113, 232, 118, 98, 183, 183, 77, 64, 218, 223, 232, 82, 16, 72, 68, 191, 54, 116, 38, 151, 174, 234, 127]
valid = len(list(filter(lambda s: kn[s[0]] == s[1], enumerate(c))))
if valid == len(x):
else:
print("WRONG!!!!!!")

## Solution

Okay, let's work this backwards.
kn = [47, 123, 113, 232, 118, 98, 183, 183, 77, 64, 218, 223, 232, 82, 16, 72, 68, 191, 54, 116, 38, 151, 174, 234, 127]
valid = len(list(filter(lambda s: kn[s[0]] == s[1], enumerate(c))))
if valid == len(x):
else:
print("WRONG!!!!!!")
We see that c is checked against kn, and they must be the same in order for our password to be correct.
random.seed(997)
k = [random.randint(0, 256) for _ in range(len(x))]
a = { b: do_thing(ord(c), d) for (b, c), d in zip(enumerate(x), k) }
b = list(range(len(x)))
random.shuffle(b)
c = [a[i] for i in b[::-1]]
This part is a little confusing. The first thing to notice is that the RNG is seeded, so the values of k and b are always the same.
Since we know the value that c must be, and the value of b after random.shuffle() is known, we can recover a.
c = kn
print("Need c =", c)
a = [None for _ in range(len(b[::-1]))]
for i in range(len(b[::-1])):
a[b[::-1][i]] = c[i]
print("Need a =", a)
Now, we need to work out what the value of x must be. Notice that every character in x is passed through the do_thing() function, with the corresponding value in k.
a = { b: do_thing(ord(c), d) for (b, c), d in zip(enumerate(x), k) }
What we need to do is to recover the value of each character in x, knowing the corresponding values in k. To do that, we need to understand the do_thing() function.
def do_thing(a, b):
return ((a << 1) & b) ^ ((a << 1) | b)
We can consider the two cases, where the bit
$b_i$
is either 0 or 1.
Notice that if
$b_i=0$
, then this simplifies to 0 ^ (a << 1) = (a << 1), and if
$b_i=1$
, then this simplifies to 1 ^ (a << 1) = !(a << 1).
So this operation flips every bit in (a << 1), where the corresponding bit in b is 1. This is the same as (a << 1) ^ b.
Hence, to undo this operation and recover the flag, we simply perform the following:
def undo_thing(a, b):
return (a ^ b) >> 1
Here's the full solver script to obtain the password.
def undo_thing(a, b):
return (a ^ b) >> 1
x = 'a' * 25
random.seed(997)
k = [random.randint(0, 256) for _ in range(len(x))]
print("k =", k)
a = { b: do_thing(ord(c), d) for (b, c), d in zip(enumerate(x), k) }
b = list(range(len(x)))
random.shuffle(b)
c = [a[i] for i in b[::-1]]
kn = [47, 123, 113, 232, 118, 98, 183, 183, 77, 64, 218, 223, 232, 82, 16, 72, 68, 191, 54, 116, 38, 151, 174, 234, 127]
valid = len(list(filter(lambda s: kn[s[0]] == s[1], enumerate(c)))) # i.e. c = kn
print("---")
c = kn
print("Need c =", c)
a = [None for _ in range(len(b[::-1]))]
for i in range(len(b[::-1])):
a[b[::-1][i]] = c[i]
print("Need a =", a)
undo_a = { b: chr(undo_thing(a[b], d)) for (b, c), d in zip(enumerate(x), k) }
print(''.join(undo_a[i] for i in range(25)))
The flag is MetaCTF{yOu_w!N_th1\$_0n3}.