Skip to content

Instantly share code, notes, and snippets.

@nbulischeck
Last active Mar 11, 2019
Embed
What would you like to do?
# Values from the white paper
EXAMPLE_PHDR = 'L2 F\' B2 U D2 B U D\' F2 L\' F U2 R U\' L\' R\' U\' B F D U\''.split(" ")
EXAMPLE_LHDR = 'B U2 D2 R\' F U2 B R L\' B L\' B L\' D F\' L U\' B2 R F2 L\' F2'.split(" ")
EXAMPLE_MSG = 'F L U2 L2 F\' B D\' B2 L\' B\' U L2 F\' R2 D\' B\' U\' L\' B R D2 L2 R\' B F2 D\' U R B D2 U2 R2 U\' F2 R2 F D F2 B2 D\' R\' D R\' U\' F\' B2 U F2 D R U L F U2 L2 D R B D\' B\' U L U\''.split(" ")
# Values from the challenge
CHALL_PHDR = 'B2 R U F\' R\' L\' B B2 L F D D\' R\' F2 D\' R R D2 B\' L R'.split(" ")
CHALL_LHDR = 'L\' L B F2 R2 F2 R\' L F\' B\' R D\' D\' F U2 B\' U U D\' U2 F\''.split(" ")
CHALL_MSG = 'L F\' F2 R B R R F2 F\' R2 D F\' U L U\' U\' U F D F2 U R U\' F U B2 B U2 D B F2 D2 L2 L2 B\' F\' D\' L2 D U2 U2 D2 U B\' F D R2 U2 R\' B\' F2 D\' D B\' U B\' D B\' F\' U\' R U U\' L\' L\' U2 F2 R R F L2 B2 L2 B B\' D R R\' U L'.split(" ")
notation1 = ["D", "B'", "L", "R'", "R", "L'", "F2", "U", "B2"]
notation2 = ["R2", "D'", "F", "U'", "B", "F'", "U2", "L2", "D2"]
default_table = [
[0, "L", "F"],
[1, "R", "B"],
[2, "U", "L2"],
[3, "D", "R2"],
[4, "F2", "U2"],
[5, "B2", "D2"],
[6, "L'", "F'"],
[7, "R'", "U'"],
[8, "B'", "D'"],
]
def find_encoded(i, seq):
if i in notation1:
f = lambda x: x[2] == i
elif i in notation2:
f = lambda x: x[3] == i
else:
quit(0)
for item in seq:
if f(item):
return item[1]
def find_default(i, seq):
if i in notation1:
f = lambda x: x[1] == i
elif i in notation2:
f = lambda x: x[2] == i
else:
quit(0)
for item in seq:
if f(item):
return item[0]
def get_enc_table(HEADER):
# Step 1: Map header to default table in base-9
ENCODED_P = ''.join([str(find_default(i, default_table)) for i in HEADER])
# Step 2: Convert first scramble into decimal
ENCODED_P = str(int(ENCODED_P, 9))
# Reverse engineering of 2.1.1 Embedding the permutation information
i = int(ENCODED_P[0])
sub1 = ENCODED_P[1:i+1]
P = ENCODED_P[i+1:i+10]
sub2 = ENCODED_P[i+10:]
# Use P to permute the default encoding table
enc_table = [default_table[int(x)] for x in P]
# Insert the "After" indexes
for idx, item in enumerate(enc_table):
item.insert(1, idx)
return enc_table
def get_length(enc_table, HEADER):
# Step 1: Map header to encoded table in base-9
HEADER_LENGTH = ''.join([str(find_encoded(i, enc_table)) for i in HEADER])
# Step 2: Convert first scramble into decimal
HEADER_LENGTH = str(int(HEADER_LENGTH, 9))
# Reverse engineering of 2.1.3 Embedding the length information
j = int(HEADER_LENGTH[0])
k = int(HEADER_LENGTH[1])
sub3 = HEADER_LENGTH[2:j+2]
length = int(HEADER_LENGTH[j+2:j+2+k])
sub4 = HEADER_LENGTH[j+2+k:]
return length
def decrypt_message(length, enc_table, MESSAGE):
# Step 1: Get m by taking length notations starting from third scramble
MESSAGE_PIECE = MESSAGE[0:length]
# Step 2: Map header to encoded table in base-9
MAPPED_MESSAGE = ''.join([str(find_encoded(i, enc_table)) for i in MESSAGE_PIECE])
# Step 2: Convert first scramble into decimal, then binary
MAPPED_MESSAGE = int(MAPPED_MESSAGE, 9)
BINARY_MESSAGE = f"{MAPPED_MESSAGE:b}"
# Step N: Pad
PADDED_MESSAGE = '0' * (8 - (151 % 8)) + BINARY_MESSAGE
# Step N: Chunk
CHUNKED_MESSAGE = [PADDED_MESSAGE[i:i+8] for i in range(0, len(PADDED_MESSAGE), 8)]
# Step N: Join
DECRYPTED_MESSAGE = ''.join([chr(int(x, 2)) for x in CHUNKED_MESSAGE])
print(DECRYPTED_MESSAGE)
enc_table = get_enc_table(CHALL_PHDR)
length = get_length(enc_table, CHALL_LHDR)
decrypt_message(length, enc_table, CHALL_MSG)
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment