Perfect Synchronization
7 minutos de lectura
Se nos proporciona el código fuente en Python para cifrar la flag:
from os import urandom
from Crypto.Cipher import AES
from secret import MESSAGE
assert all([x.isupper() or x in '{_} ' for x in MESSAGE])
class Cipher:
def __init__(self):
self.salt = urandom(15)
key = urandom(16)
self.cipher = AES.new(key, AES.MODE_ECB)
def encrypt(self, message):
return [self.cipher.encrypt(c.encode() + self.salt) for c in message]
def main():
cipher = Cipher()
encrypted = cipher.encrypt(MESSAGE)
encrypted = "\n".join([c.hex() for c in encrypted])
with open("output.txt", 'w+') as f:
f.write(encrypted)
if __name__ == "__main__":
main()
Y también tenemos la salida del script (es una salida muy grande, no se muestra completamente aquí):
0fbf645baa0ecce12ed52071a4ed0d1d
ce2e2acd1155ac79105dcabcdb4fbbff
d78843699ad962a2a7c513d193d27ab4
9a2f91dbedaa39d9b53f8146c2301098
3ea6ab81fee5f5c718f48b86e8680732
d78843699ad962a2a7c513d193d27ab4
...
a952cfc1d886d6084113d5f3e13508f0
48654fae441fee7cd607d8cb90c1de44
3ea6ab81fee5f5c718f48b86e8680732
ddfdaa01baab48baf54067dd3a3de527
ddfdaa01baab48baf54067dd3a3de527
d3eae0ab73225a3d841241af5d8a0654
d78843699ad962a2a7c513d193d27ab4
Análisis de código fuente
En primer lugar, sabemos que los caracteres del mensaje son letras mayúsculas, guiones bajos, llaves o espacios:
assert all([x.isupper() or x in '{_} ' for x in MESSAGE])
Además, la función de cifrado toma un solo carácter, agrega un relleno aleatorio de 15 bytes y usa AES ECB para cifrarlo:
class Cipher:
def __init__(self):
self.salt = urandom(15)
key = urandom(16)
self.cipher = AES.new(key, AES.MODE_ECB)
def encrypt(self, message):
return [self.cipher.encrypt(c.encode() + self.salt) for c in message]
AES ECB
Revisemos cómo funciona AES ECB:
El punto aquí es que bloques iguales de texto claro producen bloques iguales de texto cifrados. De hecho, tenemos un mensaje que contiene un máximo de 30 caracteres diferentes (letras mayúsculas más el guion bajo, más las llaves más el espacio). Por lo tanto, tendremos como máximo 30 bloques de texto cifrados diferentes. Vamos a verlo:
$ python3 -q
>>> with open('output.txt') as f:
... cts = f.read().splitlines()
...
>>> len(set(cts))
30
Perfecto, tal y como esperábamos.
Enfoque de la solución
Dado que solo tenemos 30 símbolos diferentes en el mensaje y, por lo tanto, 30 bloques de cifrado diferentes, podemos aplicar un análisis de frecuencia con un poco de adivinación de palabras (en inglés) para encontrar letras y completar los huecos para formar palabras y frases que tengan sentido.
Lo primero que podemos hacer es encontrar los bloques de texto cifrado para { y } (que aparecerán solo una vez):
>>> import json
>>> from collections import Counter
>>>
>>> print(json.dumps(Counter(cts), indent=2))
{
"0fbf645baa0ecce12ed52071a4ed0d1d": 31,
"ce2e2acd1155ac79105dcabcdb4fbbff": 89,
"d78843699ad962a2a7c513d193d27ab4": 142,
"9a2f91dbedaa39d9b53f8146c2301098": 7,
"3ea6ab81fee5f5c718f48b86e8680732": 41,
"b403c6f30eec7075d0643b5d4125de1b": 84,
"9e520d83ca02f81ab980ed7ff9a16526": 55,
"9b9233be563be442bf2edf0e3e42848d": 34,
"a952cfc1d886d6084113d5f3e13508f0": 230,
"00273fa04b9b836c553d876502e9a1e2": 101,
"d3eae0ab73225a3d841241af5d8a0654": 61,
"4ca1a8b8e8ca22b70d69ca257d79e5ed": 103,
"98c50ceddda78b850757dc37c1c0814d": 87,
"cb20a26ba8411b2e0072e7438ed67e54": 22,
"4065e6b94c150f4137af46b752e25204": 37,
"495c118c128a67d8a0f022e3f001775e": 75,
"d395fedae9dca912da1b1b50b0aca161": 100,
"c98f7e77e918e98833b8fa19de4b1653": 46,
"e3372f2164f66ea750b5b14f8166b0b9": 23,
"473a4eb8c699b9141c7d0bb4fa691674": 8,
"b1119ba67f26c30144e9cea6a6d7059c": 18,
"a33f02cf75488a76efb91511a0111982": 22,
"48654fae441fee7cd607d8cb90c1de44": 32,
"3321154f7ace2161b8cafb37c6307e6b": 9,
"4fdd58bb71f5f8c882eaa592d51cf647": 5,
"ddfdaa01baab48baf54067dd3a3de527": 8,
"8115f6662bacddd86db1ae8dc533c46c": 2,
"d4dde50295a8c036709603b3c216e44d": 1,
"97038e93767664edf41312c98285ba94": 2,
"25581e56ea570e8f68c9dab82f24f36e": 1
}
Entonces, sabemos que d4dde50295a8c036709603b3c216e44d y 25581e56ea570e8f68c9dab82f24f36e son los bloques de texto cifrado para { y }. Entonces, podemos adivinar cuatro caracteres más ( HTB) en la salida de texto cifrado:
...
9b9233be563be442bf2edf0e3e42848d
a952cfc1d886d6084113d5f3e13508f0 ·
c98f7e77e918e98833b8fa19de4b1653 H
d395fedae9dca912da1b1b50b0aca161 T
cb20a26ba8411b2e0072e7438ed67e54 B
d4dde50295a8c036709603b3c216e44d {
4ca1a8b8e8ca22b70d69ca257d79e5ed
98c50ceddda78b850757dc37c1c0814d
a33f02cf75488a76efb91511a0111982
48654fae441fee7cd607d8cb90c1de44
d3eae0ab73225a3d841241af5d8a0654
d78843699ad962a2a7c513d193d27ab4
97038e93767664edf41312c98285ba94
4ca1a8b8e8ca22b70d69ca257d79e5ed
3ea6ab81fee5f5c718f48b86e8680732
cb20a26ba8411b2e0072e7438ed67e54
4ca1a8b8e8ca22b70d69ca257d79e5ed
d395fedae9dca912da1b1b50b0aca161
98c50ceddda78b850757dc37c1c0814d
d395fedae9dca912da1b1b50b0aca161
3ea6ab81fee5f5c718f48b86e8680732
d395fedae9dca912da1b1b50b0aca161
98c50ceddda78b850757dc37c1c0814d
495c118c128a67d8a0f022e3f001775e
b403c6f30eec7075d0643b5d4125de1b
97038e93767664edf41312c98285ba94
9e520d83ca02f81ab980ed7ff9a16526
98c50ceddda78b850757dc37c1c0814d
48654fae441fee7cd607d8cb90c1de44
c98f7e77e918e98833b8fa19de4b1653
d78843699ad962a2a7c513d193d27ab4
ce2e2acd1155ac79105dcabcdb4fbbff
25581e56ea570e8f68c9dab82f24f36e }
a952cfc1d886d6084113d5f3e13508f0 ·
9e520d83ca02f81ab980ed7ff9a16526
...
En Visual Studio Code u otro editor de texto, podemos seleccionar todas las ocurrencias de un bloque de texto cifrado y agregar el carácter de texto claro correspondiente (· para espacios) en toda la salida:
9b9233be563be442bf2edf0e3e42848d
a952cfc1d886d6084113d5f3e13508f0 ·
c98f7e77e918e98833b8fa19de4b1653 H
d395fedae9dca912da1b1b50b0aca161 T
cb20a26ba8411b2e0072e7438ed67e54 B
d4dde50295a8c036709603b3c216e44d {
4ca1a8b8e8ca22b70d69ca257d79e5ed
98c50ceddda78b850757dc37c1c0814d
a33f02cf75488a76efb91511a0111982
48654fae441fee7cd607d8cb90c1de44
d3eae0ab73225a3d841241af5d8a0654
d78843699ad962a2a7c513d193d27ab4
97038e93767664edf41312c98285ba94
4ca1a8b8e8ca22b70d69ca257d79e5ed
3ea6ab81fee5f5c718f48b86e8680732
cb20a26ba8411b2e0072e7438ed67e54 B
4ca1a8b8e8ca22b70d69ca257d79e5ed
d395fedae9dca912da1b1b50b0aca161 T
98c50ceddda78b850757dc37c1c0814d
d395fedae9dca912da1b1b50b0aca161 T
3ea6ab81fee5f5c718f48b86e8680732
d395fedae9dca912da1b1b50b0aca161 T
98c50ceddda78b850757dc37c1c0814d
495c118c128a67d8a0f022e3f001775e
b403c6f30eec7075d0643b5d4125de1b
97038e93767664edf41312c98285ba94
9e520d83ca02f81ab980ed7ff9a16526
98c50ceddda78b850757dc37c1c0814d
48654fae441fee7cd607d8cb90c1de44
c98f7e77e918e98833b8fa19de4b1653 H
d78843699ad962a2a7c513d193d27ab4
ce2e2acd1155ac79105dcabcdb4fbbff
25581e56ea570e8f68c9dab82f24f36e }
a952cfc1d886d6084113d5f3e13508f0 ·
9e520d83ca02f81ab980ed7ff9a16526
Ahora es el momento de usar la intuición para adivinar palabras. Por ejemplo esta:
4065e6b94c150f4137af46b752e25204
a952cfc1d886d6084113d5f3e13508f0 ·
cb20a26ba8411b2e0072e7438ed67e54 B
9b9233be563be442bf2edf0e3e42848d
a952cfc1d886d6084113d5f3e13508f0 ·
d395fedae9dca912da1b1b50b0aca161 T
c98f7e77e918e98833b8fa19de4b1653 H
d78843699ad962a2a7c513d193d27ab4
a952cfc1d886d6084113d5f3e13508f0 ·
00273fa04b9b836c553d876502e9a1e2
Es bastante obvio que debe ser ?·BY·THE·?, por lo que tenemos la Y y la E.
Entonces podemos encontrar esto:
9e520d83ca02f81ab980ed7ff9a16526
d395fedae9dca912da1b1b50b0aca161 T
a952cfc1d886d6084113d5f3e13508f0 ·
d395fedae9dca912da1b1b50b0aca161 T
c98f7e77e918e98833b8fa19de4b1653 H
00273fa04b9b836c553d876502e9a1e2
d395fedae9dca912da1b1b50b0aca161 T
a952cfc1d886d6084113d5f3e13508f0 ·
98c50ceddda78b850757dc37c1c0814d
Y aquí conseguimos la A de ?·THAT·?.
Lo siguiente debe ser ?·BOTH·THE·B?, por lo que encontramos la O:
98c50ceddda78b850757dc37c1c0814d
a952cfc1d886d6084113d5f3e13508f0 ·
cb20a26ba8411b2e0072e7438ed67e54 B
495c118c128a67d8a0f022e3f001775e
d395fedae9dca912da1b1b50b0aca161 T
c98f7e77e918e98833b8fa19de4b1653 H
a952cfc1d886d6084113d5f3e13508f0 ·
d395fedae9dca912da1b1b50b0aca161 T
c98f7e77e918e98833b8fa19de4b1653 H
d78843699ad962a2a7c513d193d27ab4 E
a952cfc1d886d6084113d5f3e13508f0 ·
cb20a26ba8411b2e0072e7438ed67e54 B
ce2e2acd1155ac79105dcabcdb4fbbff
Este es una suposición afortunada para ?E·AND·?:
e3372f2164f66ea750b5b14f8166b0b9
d78843699ad962a2a7c513d193d27ab4 E
a952cfc1d886d6084113d5f3e13508f0 ·
00273fa04b9b836c553d876502e9a1e2 A
b403c6f30eec7075d0643b5d4125de1b
4065e6b94c150f4137af46b752e25204
a952cfc1d886d6084113d5f3e13508f0 ·
4ca1a8b8e8ca22b70d69ca257d79e5ed
Aquí hay otra suposición afortunada para ·FREQUENCY·ANALYSIS·:
a952cfc1d886d6084113d5f3e13508f0 ·
0fbf645baa0ecce12ed52071a4ed0d1d
ce2e2acd1155ac79105dcabcdb4fbbff
d78843699ad962a2a7c513d193d27ab4 E
9a2f91dbedaa39d9b53f8146c2301098
3ea6ab81fee5f5c718f48b86e8680732
d78843699ad962a2a7c513d193d27ab4 E
b403c6f30eec7075d0643b5d4125de1b N
9e520d83ca02f81ab980ed7ff9a16526
9b9233be563be442bf2edf0e3e42848d Y
a952cfc1d886d6084113d5f3e13508f0 ·
00273fa04b9b836c553d876502e9a1e2 A
b403c6f30eec7075d0643b5d4125de1b N
00273fa04b9b836c553d876502e9a1e2 A
d3eae0ab73225a3d841241af5d8a0654
9b9233be563be442bf2edf0e3e42848d Y
4ca1a8b8e8ca22b70d69ca257d79e5ed
98c50ceddda78b850757dc37c1c0814d
4ca1a8b8e8ca22b70d69ca257d79e5ed
a952cfc1d886d6084113d5f3e13508f0 ·
Aquí tenemos ·LANGUAGE·CRYPTANALYSIS·:
a952cfc1d886d6084113d5f3e13508f0 ·
d3eae0ab73225a3d841241af5d8a0654 L
00273fa04b9b836c553d876502e9a1e2 A
b403c6f30eec7075d0643b5d4125de1b N
e3372f2164f66ea750b5b14f8166b0b9
3ea6ab81fee5f5c718f48b86e8680732 U
00273fa04b9b836c553d876502e9a1e2 A
e3372f2164f66ea750b5b14f8166b0b9
d78843699ad962a2a7c513d193d27ab4 E
a952cfc1d886d6084113d5f3e13508f0 ·
98c50ceddda78b850757dc37c1c0814d I
b403c6f30eec7075d0643b5d4125de1b N
a952cfc1d886d6084113d5f3e13508f0 ·
9e520d83ca02f81ab980ed7ff9a16526 C
ce2e2acd1155ac79105dcabcdb4fbbff R
9b9233be563be442bf2edf0e3e42848d Y
48654fae441fee7cd607d8cb90c1de44
d395fedae9dca912da1b1b50b0aca161 T
00273fa04b9b836c553d876502e9a1e2 A
b403c6f30eec7075d0643b5d4125de1b N
00273fa04b9b836c553d876502e9a1e2 A
d3eae0ab73225a3d841241af5d8a0654 L
9b9233be563be442bf2edf0e3e42848d Y
4ca1a8b8e8ca22b70d69ca257d79e5ed S
98c50ceddda78b850757dc37c1c0814d I
4ca1a8b8e8ca22b70d69ca257d79e5ed S
a952cfc1d886d6084113d5f3e13508f0 ·
Esto es ·CIPHERTEXT·THE·METHOD·:
a952cfc1d886d6084113d5f3e13508f0 ·
9e520d83ca02f81ab980ed7ff9a16526 C
98c50ceddda78b850757dc37c1c0814d I
48654fae441fee7cd607d8cb90c1de44 P
c98f7e77e918e98833b8fa19de4b1653 H
d78843699ad962a2a7c513d193d27ab4 E
ce2e2acd1155ac79105dcabcdb4fbbff R
d395fedae9dca912da1b1b50b0aca161 T
d78843699ad962a2a7c513d193d27ab4 E
4fdd58bb71f5f8c882eaa592d51cf647
d395fedae9dca912da1b1b50b0aca161 T
a952cfc1d886d6084113d5f3e13508f0 ·
d395fedae9dca912da1b1b50b0aca161 T
c98f7e77e918e98833b8fa19de4b1653 H
d78843699ad962a2a7c513d193d27ab4 E
a952cfc1d886d6084113d5f3e13508f0 ·
a33f02cf75488a76efb91511a0111982
d78843699ad962a2a7c513d193d27ab4 E
d395fedae9dca912da1b1b50b0aca161 T
c98f7e77e918e98833b8fa19de4b1653 H
495c118c128a67d8a0f022e3f001775e O
4065e6b94c150f4137af46b752e25204 D
a952cfc1d886d6084113d5f3e13508f0 ·
Estas son ·BREAKING· t ·SOLVING·:
a952cfc1d886d6084113d5f3e13508f0 ·
cb20a26ba8411b2e0072e7438ed67e54 B
ce2e2acd1155ac79105dcabcdb4fbbff R
d78843699ad962a2a7c513d193d27ab4 E
00273fa04b9b836c553d876502e9a1e2 A
3321154f7ace2161b8cafb37c6307e6b
98c50ceddda78b850757dc37c1c0814d I
b403c6f30eec7075d0643b5d4125de1b N
e3372f2164f66ea750b5b14f8166b0b9 G
a952cfc1d886d6084113d5f3e13508f0 ·
...
a952cfc1d886d6084113d5f3e13508f0 ·
4ca1a8b8e8ca22b70d69ca257d79e5ed S
495c118c128a67d8a0f022e3f001775e O
d3eae0ab73225a3d841241af5d8a0654 L
473a4eb8c699b9141c7d0bb4fa691674
98c50ceddda78b850757dc37c1c0814d I
b403c6f30eec7075d0643b5d4125de1b N
e3372f2164f66ea750b5b14f8166b0b9 G
a952cfc1d886d6084113d5f3e13508f0 ·
Aquí podemos ver ·WORLD·WAR·II·:
a952cfc1d886d6084113d5f3e13508f0 ·
b1119ba67f26c30144e9cea6a6d7059c
495c118c128a67d8a0f022e3f001775e O
ce2e2acd1155ac79105dcabcdb4fbbff R
d3eae0ab73225a3d841241af5d8a0654 L
4065e6b94c150f4137af46b752e25204 D
a952cfc1d886d6084113d5f3e13508f0 ·
b1119ba67f26c30144e9cea6a6d7059c
00273fa04b9b836c553d876502e9a1e2 A
ce2e2acd1155ac79105dcabcdb4fbbff R
a952cfc1d886d6084113d5f3e13508f0 ·
98c50ceddda78b850757dc37c1c0814d I
98c50ceddda78b850757dc37c1c0814d I
a952cfc1d886d6084113d5f3e13508f0 ·
Lo siguiente representa ·CROSSWORD·PUZZLES·IN·MAJOR·NEWSPAPERS·:
e3372f2164f66ea750b5b14f8166b0b9 G
a952cfc1d886d6084113d5f3e13508f0 ·
9e520d83ca02f81ab980ed7ff9a16526 C
ce2e2acd1155ac79105dcabcdb4fbbff R
495c118c128a67d8a0f022e3f001775e O
4ca1a8b8e8ca22b70d69ca257d79e5ed S
4ca1a8b8e8ca22b70d69ca257d79e5ed S
b1119ba67f26c30144e9cea6a6d7059c W
495c118c128a67d8a0f022e3f001775e O
ce2e2acd1155ac79105dcabcdb4fbbff R
4065e6b94c150f4137af46b752e25204 D
a952cfc1d886d6084113d5f3e13508f0 ·
48654fae441fee7cd607d8cb90c1de44 P
3ea6ab81fee5f5c718f48b86e8680732 U
ddfdaa01baab48baf54067dd3a3de527
ddfdaa01baab48baf54067dd3a3de527
d3eae0ab73225a3d841241af5d8a0654 L
d78843699ad962a2a7c513d193d27ab4 E
4ca1a8b8e8ca22b70d69ca257d79e5ed S
a952cfc1d886d6084113d5f3e13508f0 ·
98c50ceddda78b850757dc37c1c0814d I
b403c6f30eec7075d0643b5d4125de1b N
a952cfc1d886d6084113d5f3e13508f0 ·
a33f02cf75488a76efb91511a0111982 M
00273fa04b9b836c553d876502e9a1e2 A
8115f6662bacddd86db1ae8dc533c46c
495c118c128a67d8a0f022e3f001775e O
ce2e2acd1155ac79105dcabcdb4fbbff R
a952cfc1d886d6084113d5f3e13508f0 ·
b403c6f30eec7075d0643b5d4125de1b N
d78843699ad962a2a7c513d193d27ab4 E
b1119ba67f26c30144e9cea6a6d7059c W
4ca1a8b8e8ca22b70d69ca257d79e5ed S
48654fae441fee7cd607d8cb90c1de44 P
00273fa04b9b836c553d876502e9a1e2 A
48654fae441fee7cd607d8cb90c1de44 P
d78843699ad962a2a7c513d193d27ab4 E
ce2e2acd1155ac79105dcabcdb4fbbff R
4ca1a8b8e8ca22b70d69ca257d79e5ed S
a952cfc1d886d6084113d5f3e13508f0 ·
Y finalmente, solo quedan los guiones bajos:
a952cfc1d886d6084113d5f3e13508f0 ·
c98f7e77e918e98833b8fa19de4b1653 H
d395fedae9dca912da1b1b50b0aca161 T
cb20a26ba8411b2e0072e7438ed67e54 B
d4dde50295a8c036709603b3c216e44d {
4ca1a8b8e8ca22b70d69ca257d79e5ed S
98c50ceddda78b850757dc37c1c0814d I
a33f02cf75488a76efb91511a0111982 M
48654fae441fee7cd607d8cb90c1de44 P
d3eae0ab73225a3d841241af5d8a0654 L
d78843699ad962a2a7c513d193d27ab4 E
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a952cfc1d886d6084113d5f3e13508f0 ·
Para resumir, este es el mensaje completo:
FREQUENCY ANALYSIS IS BASED ON THE FACT THAT IN ANY GIVEN STRETCH OF WRITTEN
LANGUAGE CERTAIN LETTERS AND COMBINATIONS OF LETTERS OCCUR WITH VARYING
FREQUENCIES MOREOVER THERE IS A CHARACTERISTIC DISTRIBUTION OF LETTERS THAT
IS ROUGHLY THE SAME FOR ALMOST ALL SAMPLES OF THAT LANGUAGE IN CRYPTANALYSIS
FREQUENCY ANALYSIS ALSO KNOWN AS COUNTING LETTERS IS THE STUDY OF THE
FREQUENCY OF LETTERS OR GROUPS OF LETTERS IN A CIPHERTEXT THE METHOD IS
USED AS AN AID TO BREAKING CLASSICAL CIPHERS FREQUENCY ANALYSIS REQUIRES
ONLY A BASIC UNDERSTANDING OF THE STATISTICS OF THE PLAINTEXT LANGUAGE AND
SOME PROBLEM SOLVING SKILLS AND IF PERFORMED BY HAND TOLERANCE FOR EXTENSIVE
LETTER BOOKKEEPING DURING WORLD WAR II BOTH THE BRITISH AND THE AMERICANS
RECRUITED CODEBREAKERS BY PLACING CROSSWORD PUZZLES IN MAJOR NEWSPAPERS AND
RUNNING CONTESTS FOR WHO COULD SOLVE THEM THE FASTEST SEVERAL OF THE CIPHERS
USED BY THE AXIS POWERS WERE BREAKABLE USING FREQUENCY ANALYSIS FOR EXAMPLE
SOME OF THE CONSULAR CIPHERS USED BY THE JAPANESE MECHANICAL METHODS OF
LETTER COUNTING AND STATISTICAL ANALYSIS GENERALLY HTB{SIMPLE_SUBSTITUTION_CIPHER}
CARD TYPE MACHINERY WERE FIRST USED IN WORLD WAR II POSSIBLY BY THE US ARMYS
SIS TODAY THE HARD WORK OF LETTER COUNTING AND ANALYSIS HAS BEEN REPLACED BY
COMPUTER SOFTWARE WHICH CAN CARRY OUT SUCH ANALYSIS IN SECONDS WITH MODERN
COMPUTING POWER CLASSICAL CIPHERS ARE UNLIKELY TO PROVIDE ANY REAL
PROTECTION FOR CONFIDENTIAL DATA PUZZLE PUZZLE PUZZLE
Flag
Y la flag es: HTB{SIMPLE_SUBSTITUTION_CIPHER}.