Back in 2012, a man in the South of England was apparently renovating his chimney when he came across the skeleton of a pigeon.
One leg bone had a capsule attached, which turned out to contain what looked very much like a coded message dating back to World War 2.
The British made extensive use of homing pigeons during the war as a way to get messages back from Nazi-occupied Europe to England.
The system was not without peril, both to the birds (apparently the majority didn’t make it back) and to the message-sender releasing the bird, who faced the death penalty for espionage if caught.
With 70 years having passed, little more to go on than “likely place of release – 1940s France”, and no indication of what code or cipher had been used, even the UK’s code-breaking boffins at Bletchley Park warned that the message would probably never be cracked:
One-time pad
If the sender was a field agent in occupied France, he may well have had a one-time pad, a sort of cipher that uses a randomly generated key that is as long as the message.
Only two copies of the key ever exist: the agent takes one key, or more usually a code-pad consisting of numerous sheets of daily keys, and the agent’s handler keeps the other.
By destroying one page of the code-pad each day, whether a message was sent or not, the field agent can ensure that each key is only ever used once, or not at all, and the handler can keep in synchronisation.
If each key really is used for one message only; if both copies of the key are handled securely; and if the key is truly random (i.e. no algorithm was used that could possibly be repeated), a one-time pad is completely and provably secure.
That’s because every possible string of characters is equally likely to have been the key, so there is no way to choose between different plausible decryptions.
Here’s an example to demonstrate, assuming that our encrypted message is XARYOPT .
Imagine that decryption involves shifting each encrypted letter back by the amount denoted by the alphabetic position of each key letter.
In other words, a key of A means “shift by 0” (M remains M), B means “shift by 1” (E becomes D), M means “shift by 13” (Z becomes M), and so on.
You can pick any possible key from AAAAAAA to ZZZZZZZ, so you can “decrypt” the message XARYOPT however you choose:
Code book
On the other hand, if decryption involved a code book, for example so that HVPKG meant ARTILLERY and FNFJU meant MOVEMENTS, then in the absence of the codebook you would be stuck.
AOAKN, which is repeated at the start and the end of the message, might be some kind of delimiter, but for the rest of the message, you could come up with any decoding you like and claim that “your” code book was the right one.
No-one would be able to prove otherwise.
Interestingly, even though every codeword in the encoded message is unique, except for AOAKN, that doesn’t mean there are no repeated words in the original message.
Codebooks often provide more than one encoding option for common words, so you can avoid telltale repetition in your coded text.
The first “solution”
In late 2012, a Canadian named Gord Young claimed to have decrypted the message easily, using his “great-uncle’s aerial observers’ notes, the WW1 code formula, and WW1 coding as a basis.”
Young chose to read off the code groups top-to-bottom, left-to-right, rather than starting left-to-right as they seem to have been written, and came up with this:
AOAKN - Artillery Observer At "K" Sector, Normandy.
RQXSR - Requested [Head] Quaters Supplement Report
PABLIZ - Panzer Attack - Blitz
NLXKG - Now loading [e] X [tra] {sector] "K" Guns
WAOTA - West Artillery Observer Tracking Attack
LKXGH - Lt. Knows [that] (e) X [tra] Guns [are] Here
. . .
DJHFP - Determined Jerry's Headquarters Front Posts
. . .
Criticisms of Young’s claim quickly appeared, pointing out details such as:
- Top-to-bottom reading is unlikely.
- Panzer (German tank) would not be in a WW1 codebook.
- Jerry (German soldier) would not be in a WW1 codebook.
- PABLIZ is wrongly transcribed and should be PABUZ.
But the most cutting criticism of Young’s decryption is this: it’s drivel.
No British observer in Normandy would have wasted a precious pigeon to send such nonsense.
Cracked at last?
And that was that for the Dead Pigeon code until last week, when a 22-year-old Spaniard, Dídac Sánchez, announced he’d cracked it at last.
He says that he started on the task when the news first broke about the message back in in 2012, and that he has been working on it every since.
According to The Telegraph in the UK, Sánchez claimed to have spent a whopping €1,500,000 (about $1.7m/£1.1m) to solve the puzzle.
By this point, I’m sure you’re dying to know, “What was the pigeon trying to say?”
Here’s comes the suspicious part, I’m afraid.
Sánchez isn’t telling.
You’ll have to AOAKN ERFLP QSQMD WWEBN AOAKN 5 for the full story.
Image of pigeon postie courtesy of Shutterstock.
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