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Old November 1st, 2012, 09:09 AM   #1

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Can you crack the pigeon code?


The remains of a WW2 homing pigeon with previously unopened message attached have been recovered in Britain.
For some weeks now the code has been with the successors to Betchley Park, GCHQ and so far, they have not been able to read the message. Fame beyond your wildest imaginings beckons ( or at least 15 seconds of fame).
Skeleton of hero World War II carrier pigeon found down a chimney with a secret coded message still attached to its leg | Mail Online

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Old November 1st, 2012, 09:12 AM   #2
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Is there any cash prize?
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Old November 1st, 2012, 09:15 AM   #3

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I could try. But it is gonna take some time.
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Old November 1st, 2012, 09:36 AM   #4

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Nurp nurp.
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Old November 1st, 2012, 10:12 AM   #5

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Quote:
Originally Posted by Naomasa298 View Post
Nurp nurp.
NURP was a code for a specific bird. A couple of the pigeons awarded the Dicken medal are know only by their NURP numbers. This suggests to me that two birds were released. As XO2 was Bomber Command HQ, the implication is that two birds were released by a bomber to confirm SOMETHING. Now if the bird's ring still exists and it should, somewhere the number will show up in old RAF records as being signed for by some W/T Operator. If the other bird arrived safe, the message will exist in its decoded form---maybe.
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Old November 1st, 2012, 12:03 PM   #6

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Looks like some kind of a substitution cipher. Shouldn't be too hard to decrypt in this day and age.

Last edited by Enkidu; November 1st, 2012 at 12:20 PM.
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Old November 1st, 2012, 12:27 PM   #7

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I tried couple of decryption applications with no result. What makes this extra hard is that I can't be certain what some of the letters are.
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Old November 1st, 2012, 01:11 PM   #8

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Is this correct?

AOAKN HVPKD FNFJU YIDDC
RQXDR DJHFP GOVFN MIAPX
PABUZ WYYNP OMPNW HJRZH.
NLXKG MEMEK ONOIB AREEQ
UAOTA. RBQRH DJOFM TPZEH
LKXEH RGGHT JRZCQ FNKTQ
KLDTS GQIRU AOAKN

Curious facts:

Starts and ends with AOAKN.

Pairs of D, E, G, K, and Y. This might be helpful assuming this is substitution cipher, and only if they really are "paired".

27 x 5 = 135 letters in the cipher.

Last edited by Enkidu; November 1st, 2012 at 01:23 PM.
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Old November 1st, 2012, 01:57 PM   #9
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Quote:
Originally Posted by Enkidu View Post
Is this correct?

AOAKN HVPKD FNFJU YIDDC
RQXDR DJHFP GOVFN MIAPX
PABUZ WYYNP OMPNW HJRZH.
NLXKG MEMEK ONOIB AREEQ
UAOTA. RBQRH DJOFM TPZEH
LKXEH RGGHT JRZCQ FNKTQ
KLDTS GQIRU AOAKN

Curious facts:

Starts and ends with AOAKN.

Pairs of D, E, G, K, and Y. This might be helpful assuming this is substitution cipher, and only if they really are "paired".

27 x 5 = 135 letters in the cipher.
Copycat. Well someone else reads the ATS forums ;p

I've tried at this for quite a while with no luck. Hope someone figures it out soon!
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Old November 1st, 2012, 02:17 PM   #10

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Letter frequencies don't look like a single substitution - 9 each for R, N and A, 8 for K & H - too smooth? All 26 letters used so not a Playfair encryption.
Needs better maths than I've got, IMO.
Time of origin and time of liberation - middle of the day, so probably not a major bombing raid, over an hour apart so no apparent urgency.
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