WWII pigeon message stumps GCHQ decoders | Hacker News Reader

WWII pigeon message stumps GCHQ decoders

bbc.co.uk

102 pointssp864 comments

You have to look at the facts the message was sent by Sjt W Stot.

You do a search and you find this.

http://www.archieraf.co.uk/archie/1037zau.html

And you will find that there was a sgt H H Stott and his crew position is w/op Air Gunner so hence the sjt W Stot.

You will then find that On 27th April 1942, the crew of Halifax W1037 ZA-U from 10 Squadron failed to return hence the need to send a message.

You will also find that on the birds ID NURP.40.TW.194 and NURP.37.OK.76 - NURP” was the National Union of Racing Pigeons so it was british so this back up the fact it was send by the RAF. The number 40 relates to 1940 but you can't read to much into that if you see this website.

http://www.thebirdman.org/Index/Others/Others-Doc-Birds&...

You will see pigeons with number 38 was used in 1941 and 1942 and a pigeon with number 39 was used in 1945 and a pigeon with number 41 was used in 1942.

So this backs up the argument that the date would be between 1940 and 1945. So the date of the crash in 1942 wouldn't be far off.

All you have to do is find the codes used by Halifax W1037 ZA-U from 10 Squadron. The RAF would have records once you get that you can decode the message.

I thought the GCHQ had smart people working at it.

sgt H H Stott personal number is 1058698 take the first four numbers 1058 and use that as the ciper.

So the first letter A is plus 1 the second letter is O plus O the four letter is A and it is plus 5 which is G and the four letter K is plus 8 which is W so.

The first row is as follows.

BOG WT NHF LD KVG JB GJD IK

Which so far I have translated as.

BOMBER ON GROUND WITH TROOPS NEED HELF FAST LITTLE DEFENCE k___ v____ G_____ J___ B____ G____ J____ D____ I___ K___

Need help solving the rest.

On the other hand, 'W Stot' could well be different from 'H H Stott': it's a short surname, and not particularly rare:

http://www.192.com/atoz/people/surnames/stott/50/

Also, the article mentions that spelling 'sergeant' with a j, as in the abbreviation 'Sjt' was more common practice in the army, whereas the RAF would probably have used a g.

@SuperChihuahua posted a link to http://www.ciphermysteries.com/2012/11/02/dead-pigeon-sparks.... According to the author of that document it may have been Sergeant William Leslie Stott (508080 in the RAF, died in 1945 aged 35, buried in Chester’s Overleigh Cemetery) [1]. The spelling of "Sergeant" matches.

[1] http://www.cwgc.org/find-war-dead/casualty/2928300/STOTT,%20...

Assuming that this is the Stott who wrote the message and looking at the detailed nature of the mission to bomb the Tirpitz, could this information be used to programatically reverse engineer the key in a sensible amount of time? I.e. assuming certain keywords may be in the message, like "Tirpitz" or "down" etc...

>All you have to do is find the codes used by Halifax W1037 ZA-U from 10 Squadron. The RAF would have records once you get that you can decode the message.

Those records could very well have been destroyed, either through war time action or subsequent fire or flood or other catastrophe, or deliberately.

>I thought the GCHQ had smart people working at it.

It's possible that they've decrypted it but are declining to say so. Don't forget that they've only just this year (60 or so years later) released some of the work of Turing.

UK didnt even acknowledge the existing of GCHQ or the Cipher Schools at Bletchley until 70's and 80's.

This would explain why the AOAKN is at the start and end.

BOGWT - BOMBER ON GROUND WITH TROOPS

Because it is important.

This is very imaginative but it doesn't appear to have anything to do with actual code breaking.

Are you randomly guessing words based on their first character or is there any actual logic behind it?

KILLED VIOLENT GERMANS JUST BECAUSE GERMAN JUGGERNAUT DOESN'T INFER KETCHUP

Surely you must have miscalculated, if you are using the 1058 sequence then AOAKN would read BOFSO

According to this [1] article the message reads:

AOAKN HVPKD FNFJW YIDDC

RQXSR DJHFP GOVFN MIAPX

PABUZ WYYNP CMPNW HJRZH

NLXKG MEMKK ONOIB AKEEQ

WAOTA RBQRH DJOFM TPZEH

LKXGH RGGHT JRZCQ FNKTQ

KLDTS FQIRW AOAKN 27 1525/6

[1] http://www.telegraph.co.uk/news/uknews/defence/9697929/Wante...

This is the occurence of each leter in the message:

K = 10; N = 9; A = 9; H = 8; R = 8; F = 7; P = 7; O = 7; D = 6; Q = 6; M = 5; J = 5; G = 5; W = 5; T = 5; X = 4; E = 4; Z = 4; I = 4; C = 3; B = 3; L = 3; Y = 3; V = 2; S = 2; U = 1

Doesn't seem like a trully random (white-noise) distribution.

Those counts look fairly typical from a uniform distribution to me. There are fancy tests you could do, but to get a feel for it, I suggest simulation and eyeballing. Add up the counts after drawing 135 characters from a uniform character distribution 20 times, and compare the sorted lists of counts to this set of numbers after sorting.

Somehow the really random distributions look less random than ones biased towards flattening-out. Within 500 truly random points in a square you'll see suspiciously looking clusters of points, choose 5 in each of the 10x10 subsquares and suddenly it looks much more "random".

Edit: A bookmarklet to display some random points in a square, tested in Firefox 10 only:

javascript:"<html><body><canvas id=\"tutorial\" width=\"150\" height=\"150\">foo</canvas><script>var canvas = document.getElementById('tutorial');var ctx = canvas.getContext('2d');ctx.fillStyle = \"rgb(000,0,0)\";for (var i=0;i<150;i++) {ctx.fillRect (Math.random() * 150,Math.random() * 150, 2, 2); };</script></body></html>"

...and one showing true random and flattened random side by side:

javascript:"<html><body><canvas id=\"tutorial\" width=\"200\" height=\"200\">foo</canvas><-><canvas id=\"tutorial2\" width=\"200\" height=\"200\">foo</canvas><script>var canvas = document.getElementById('tutorial');var ctx = canvas.getContext('2d');ctx.fillStyle = \"rgb(000,0,0)\";for (var i=0;i<400;i++) {ctx.fillRect (Math.random() * 200,Math.random() * 200, 2, 2); };</script><script>var canvas = document.getElementById('tutorial2');var ctx = canvas.getContext('2d');ctx.fillStyle = \"rgb(000,0,0)\";for (var i=0;i<20;i++) for(var j=0;j<20;j++) for(k=0;k<1;k++) {ctx.fillRect (i * 10 + Math.random() * 10, j * 10 + Math.random() * 10, 2, 2); };</script></body></html>"

Wow, you're right. The random distribution looks far more "meaningful".

You might want to look into the "incidence of coincidence", which is the probability that letters from two randomly chosen text's are the same. It can be changed to measure a single message, and gives a distinctive difference for english vs. random texts.

"Hitler is evil take him out" Nailed it

Hi there :-)

I read the message slightly differently (I believe some of the supposed 'w's are 'u's and one of your 'F' s is I believe a 'G'. As follows:

AOAKN HVPKD FNFJW YIDDC RQXSR DJHFP GOVFN MIAPX PABUZ WYYNP CMPNW HJRZH NLXKG MEMKK ONOIB AKEEQ UAOTA RBQRH DJOFM TPZEH LKXGH RGGHT JRZCQ FNKTQ KLDTS GQIRW AOAKN

Also do you know who 'Mason' is? Maybe the person who sent in the photo? Is this a recruiting sergeant for GCHQ? :-))

Not being a code breaker myself what relevance does AOAKN have? Beginning and end of message.

This pigeon was found in a chimney - what can we learn from that? What was the procedure when a message was sent but not received? This can be solved :-)

One time pads are mathematically provably secure. So, if it's a one time pad then we may not be able to solve this. There are some possibilities. If there's enough information with the pigeon and the message to find the original other half of the pad the message may be decrypted.

i'd be consistent and consider FNFJW to be actually FNFJU ;)

When the news first broke, redditors did a pretty impressive analysis: http://en.reddit.com/r/cryptography/comments/12jipi/ww2_pige...

I'd place my money on this comment: "Poem code with AOAKN being the indicator group, 27 representing the block count and 1525/6 being a page or line number."

Interesting point about the indicator-group at the start and end. But I thought poem codes were mainly used for under-cover situations where having a physical copy of the key could be compromising. Why then is this message written on an official-looking message pad with "Pigeon Service" printed at the top? That seems incongruous to me.

The chaos of WWII can easily be enough of an explanation. I guess I would argue that the repeat of AOAKN is a good indication of a poem code, more than the official pad can dissuade me.

AOAKN seems like greetings or best wishes, in some languages we use same initial and ending for greetings.

The template and the font used in the paper does not seems like typewriter print but LaserJet printer. specially the font used in the words on the top "PIGEON SERVICE". does not seems like printed with a typewriter.

The war's secret message are supposed not to use a single english readable words. Here we see some english words like "TO", FORM. Date, Reply, originator, copies, sender, signature. Why these words are not encrypted.

when a paper is folded in a 1 1/8 cylinder for 70 years there should be wrinkles marks on the paper.

A capsule was owned by the family of the soldier who brought it home in 1918

http://www.interbug.com/pigeon/messaging/photos/wwi_british_...

AOA could mean Asalam-O-Aliquem kn=koran hi or hello in English

I don't thing FNFJW is correct, I think it's a U. There is another very definite W written in the note, and it looks nothing like the square-ish looking U in a couple of other places.

I think Tirpitz,Gough,Gulag are mentioned.Frigates also. CQ=help(before current SOS signal) KN=in the King's Name but the Arabic possible. The flat bottomed U may refer to a 'reed'-read Also St Nazaire France was a base for carrier pigeons in WWII. Operation Chariot?

Or else a sophisticated plug for 'Call of Duty'?

You are all being so clever. What is wrong with our suggestion?

Point: There was a proposed D Day in August 1942 which was called off - hence OK for Operation Chariot March 28th 1942. Also CMPWN - Campbeltown? See Call of Duty Failed D Day. It is a SPOOF - albeit a good one. Not to say it is indecipherable.

Wow. I have in my possession some really similar looking crypted pages form an Italian partisan of the WWII (prepared and parachuted by the English forces).

It seems to me that the situations might be analog. How can I contact them? I see that they are asking for help but I don't see contact info.

I know some folks at GCHQ. Email me via my profile here and I will put you in contact.

Post the scans or transcripts please!

If this were from 2012, the message would be in plaintext in some DB, and probably easily crackable.

WW2 crypto FTW

And if this were from 2012 said DB would be on an encrypted hard drive that can only be unlocked by typing "password".

Which would reveal a filesystem with the plausible looking content, while having typed a different password would have revealed the true contents of the message.

The real password is "Swordfish".

And the weight of the hard drive would clearly be the cause of the pigeon getting pulled down into the chimney and unable to get out.

love the friday humor guys ;-)

It's very curious that the first and last groups are identical: AOAKN. That would not be expected from a one time pad.

Perhaps it is the identifier of the pad used, and doubles as a kind of EOF signal?

1 in 11 million probability of the first and last blocks being the same by pure random chance. I'd guess they must be some sort of pre-agreed message sentinel, but even 1 in 11 million chances happen sometimes!

Right -- as improbable as that sounds, it pales in comparison to the likelihood of renovating your chimney and finding a pigeon with a coded message from WWII!

As suggested in the comments on the Telegraph article, it would would seem likely that this message was encrypted with a one-time pad, making it essentially unbreakable. Unless the key is found in some MoD vault (and released to the public).

Probably correct. Another option is that the message was produced using a cipher machine, which were in use by the allied forces at that time. If a machine and records of the pre-arranged keys still exist then decryption might be possible.

If we can find records of how that machine worked, then we should be able to build a program to break the encryption.

This message was sent by a homeland "Y Station", most likely from the RAF, to a decoding facility downstream, in the SIGINT community. At Bletchley or elsewhere. Y Stations were monitoring German military communications from the UK. It is why what we can now read on paper is the verbatim retranscription of a procedure that was heard on a radio receiver by a radio intelligence operator, --ie the presence of the standard "checksum" and EOM signature comprising the reiteration of the header group followed by the total number of 5-groups sent -- 27 in the instance. The origin time "1522" is the ZULU at which the actual transmission started while the end of transmission time is written after the checksum "1525/6". Taking pauses into account, this means the morse code operator was transmitting the groups at 10wpm, a credible speed for this type of operation. I hope this helps. Greetings from rainy Montreal David Thorne-Alexander http://en.wikipedia.org/wiki/Beaumanor_Hall

By the signature, in the video clip, there's a box labelled 'Number of copies sent', filled in with a number 2. So, assuming the other copy went on another pigeon, it could well have already been received and deciphered back in WWII.

Here's the still from the video showing that and the signature: http://i.imgur.com/KTp15.png

If you're interested in this you might be interested in a another unsolved message. http://en.wikipedia.org/wiki/Taman_Shud_Case

Also written up recently at http://www.smh.com.au/lifestyle/riddle-on-the-sands-20121119...

Unfortunately the evidence is that decryption might depend on either a unavailable first edition of the Enlgish translation of http://en.wikipedia.org/wiki/Rubaiyat_of_Omar_Khayyam and/or information from a person who died in 2007.

While much has been published about the German Enigma system I've not seen much about the British equivalents during WW2.

If this message was produced using a one time pad - which is practical for short messages - then it may be very difficult to decypher.

More reading here: http://www.ciphermysteries.com/2012/11/02/dead-pigeon-sparks...

This has some good ideas and research, but the author seems fixed on the idea that the message was produced by an SOE agent using a poem cypher. This seems unlikely to me because the cyphertext is written on an official-looking form. Blank message pads and quantities of homing pigeons are presumably not something that an under-cover agent would want to carry around with them.

And I'm just reading Cryptonomicon by Neal Stephenson. Amazing book.

Me too, really made the article stand out for me.

Running low on #pigeons. Please send more #WWII

1525/6 could be a time and day. The Normandy landings took place on 6th June 1944.

Could be a page/offset in a one-time pad.

The article missed some of the message, which read:

IDKFA IDDQD IDCLIP IDCLEV

IDSPISPOPD

If you count you will find that there are 26 diferent codes. This tells me that each 5 letter code represents a letter of the alphbet. This is not a message but it is the code itself.

I still say this could be from the underground you have to stop and think . The underground use many forms of code to talk to one another

Reply on news.ycombinator.com