Case 41 · cipher · Medium · About 10–15 minutes
A conservatory messenger uses a repeating keyword to change the alphabet shift from letter to letter. The keyword has been left in plain sight. The message is secure only against a visitor who skips the operating instructions.
Recover the instruction exactly. No plant identification, historical cipher knowledge, or interpretation of the illustration is required.
Write the alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ with values A=0 through Z=25.
For encryption, write the keyword MOSS repeatedly underneath the letters of a spaceless message. Add each message-letter value to the key-letter value beneath it. If the total is 26 or more, subtract 26. Convert the result back to a letter.
For example, A encrypted with key B becomes B, because 0+1=1. Z with key C becomes B, because 25+2=27, and 27−26=1.
The complete encrypted message is:
RCDDAKJWPHZJQOV
The key begins at M beneath the first ciphertext letter and continues M–O–S–S–M–O–S–S without resetting. The plaintext has fifteen letters. After decrypting, put spaces at word lengths 6, 3, 6.
What instruction was sent? Every ciphertext character is correct. The example pairs are separate demonstrations and do not use up any part of the repeated keyword.
Hint 1 - reveal when you want a nudge
To undo addition, subtract the key value. When the result is negative, add 26 before converting back to a letter.
Hint 2 - reveal when you want a nudge
The first four key letters are MOSS, with values 12,14,18,18. R−M gives F, C−O wraps to O, and the next two letters each decode to L.
Reveal the solution and explanation
The instruction is FOLLOW RED THREAD.
Repeat the key as MOSSMOSSMOSSMOS. Subtracting its values from RCDDAKJWPHZJQOV, with alphabet wraparound, gives FOLLOWREDTHREAD. The specified lengths divide it into FOLLOW, RED, and THREAD.
The first letter is a direct subtraction: R has value 17 and M has value 12, leaving 5, which is F. The second crosses the alphabet boundary: C has value 2 and O has value 14;2−14=−12, and adding 26 gives 14, which is O. The key does not start over at RED. It remains aligned with the continuous fifteen-character string.
This is a fully keyed cipher. A solver is not expected to infer MOSS by frequency analysis or guess a password from a picture. Once the key alignment is fixed, each position is a reversible addition modulo 26 and has one possible plaintext letter.
A forward check adds the same key values to the recovered message. It recreates all fifteen printed ciphertext letters, including the wraparound positions. The spacing check is separate:6+3+6 accounts for every letter exactly once.
Solve another: The Cabinet of Fifty Mysteries · Browse all 50 cases
