challenge info  
Name: Impossible
Category: Cryptography
Difficulty: Medium
Points: 50




Is this challenge impossible?

nc 32200


source code

def fake_psi(a, b):
    return [i for i in a if i in b]

def zero_encoding(x, n):
    ret = []
    for i in range(n):
        if (x & 1) == 0:
            ret.append(x | 1)
        x >>= 1
    return ret

def one_encoding(x, n):
    ret = []
    for i in range(n):
        if x & 1:
        x >>= 1
    return ret

print("Supply positive x and y such that x < y and x > y.")
x = int(input("x: "))
y = int(input("y: "))

if len(fake_psi(one_encoding(x, 64), zero_encoding(y, 64))) == 0 and x > y and x > 0 and y > 0:


we need to satisfy 3 conditions to get the flag

  1. x must be greater than y.
  2. Both x and y must be positive
  3. The length of the list returned by fake_psi should be 0.
    fake_psi(one_encoding(x, 64), zero_encoding(y, 64))


We can set y to 1, which means that zero_encoding(y, 64) returns a list of 64 ones, where there are 64 ones in total.

To satisfy the condition that fake_psi should return an empty list, we need to find a value of x such that one_encoding(x, 64) returns an empty list. The one_encoding function only appends to the list if the least significant bit of x is 1, so we need to set the 64 least significant bits of x to 0.

We can set x to 1 << 64, which means shifting the binary representation of 1 64 bits to the left, effectively adding 64 zeros to the end. Therefore, x is equal to 18446744073709551616.

So, to obtain the flag, you should enter x as 18446744073709551616 and y as 1.


└─$ nc 32200
Supply positive x and y such that x < y and x > y.
x: 18446744073709551616
y: 1



Flag: actf{se3ms_pretty_p0ssible_t0_m3_7623fb7e33577b8a}