C 105: Two-Time Pad (20 pts + 65 extra)

Background

What You Need

Using a One-Time Pad

Unbreakable awesome secret secure system

Four score and seven

key = "Unbreakable awesome secret secure system"
plaintext = "Four score and seven"

i = 0
ciphertext = ""
for p in plaintext:
  ip = ord(p)
  k = key[i]
  ik = ord(k)
  inew = ip ^ ik
  ciphertext += chr(inew)
  print(p, hex(ip), k, hex(ik), hex(inew))
  i += 1
print(ciphertext.encode("ascii").hex())

Decrypting uses the same operation, since XOR reverses itself:

key = "Unbreakable awesome secret secure system"
ciphertext = bytes.fromhex("130117004512080e100945410f1345000a1b004e")

i = 0
plaintext = ""
for c in ciphertext:
  k = key[i]
  ik = ord(k)
  ip = c ^ ik
  p = chr(ip)
  print(c, k, ik, ip, p)
  plaintext += p
  i += 1
print(plaintext)

C 105.1: Using a Known Key (5 pts) Use this key: Wow, what a great key! Decrypt this ciphertext: 0e20220c67383c413d7440 The plaintext is the flag.

C 105.2: Crib Drag (15 pts) The key starts with: The Both these ciphertexts were encrypted with the same key: 1a0d134503550e0a1d4542071f56 0d0710071d19490d0e5607004f00074e47 To break this, start with a guess at the key, like this: The................................ This shows the first three letters of the two plaintext messages, as shown below. From the partial plaintexts, you can guess another letter or two, and use those letters to extend the key. Continue adding one or two letters at a time until you get the whole plaintext. This is called the crib drag attack. Submit either of the plaintext messages as the flag. Hint: Here's a script that lets you guess letters interactively.

C 105.3: Crib Drag (15 pts) Both these ciphertexts were encrypted with the same key: 13070b4954491b53114816450d1b101c100000090952480105044e4b000f4b4e5453 070d0a1e4c0c4e4d1d0f0d54431c10175311480a464604140a5b4e64373722233153 One of the plaintext messages contains the flag.

C 105.4: Crib Drag (50 pts) Both these ciphertexts were encrypted with the same key: 7e5e491142414350595441145c49435719454a1559425e5d4b114b49505c574e555d0016475e5757584f4651524054194149494c0c55595e144d4454195b494b1e19 734558115956551247545947510c5841195851500c535e544b544a58114844435b54495b190c665b540c545c514b1150460c7e76606a7962777c02 One of the plaintext messages contains the flag.