Wpictf 2021
Baby’s First Reversing
Solved By : thewhiteh4t
- we have a python bytecode file
- we can de-compile it using
uncompyle6 - here is the resultant python script
def __main__(inp):
i = -4
for c in inp:
if i == 4:
if c != ' ':
exit(82)
else:
if i == -4:
if c != 'W':
exit(133)
else:
if i == -2:
if c != 'I':
exit(42069)
elif i == -1 and c != '{':
exit(11037)
if i == 10:
if c != '}':
exit(9001)
else:
if i == 1:
if c != '@':
exit(11037)
if i == 2 and c != '5':
exit(11037)
if i == 7 and c != 'P':
exit(11037)
if i == 3:
if c != 'E':
exit(11037)
else:
if i == 0:
if c != 'h':
exit(82)
if i == 5 and c != 'h':
exit(11037)
if i == -3 and c != 'P':
exit(133)
if i == 9:
if c != '!':
exit(133)
else:
if i == 6:
if c != '0':
exit(133)
if i == 8 and c != '3':
exit(133)
i += 1
else:
print(':)')
__main__(input('hi'))
- if we carefully observe we can see a pattern
- it is checking the value of
iandc - if the condition is not satisfied program exits with an exit code
if i == -4:
if c != 'W':
exit(133)
- if we look at the values of i and c we can see another pattern
i == -4 -> c = W
i == -3 -> c = P
i == -2 -> c = I
i == -1 -> c = {
i == 0 -> c = h
i == 1 -> c = @
i == 2 -> c = 5
i == 3 -> c = E
i == 4 -> c = SPACE
i == 5 -> c = h
i == 6 -> c = 0
i == 7 -> c = P
i == 8 -> c = 3
i == 9 -> c = !
i == 10 -> c = }
WPI{h@5E h0P3!}