# Anoraks Corner by Tony Sale A complete description of breaking this Enigma.

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Tony Sale's
Codes and Ciphers

## Scenario 1.

Itercept date: 15th March 1935. The following 40 ecipherments of message keys have been extracted.

AGL KYO . QDD DPG . COT FBU . FXN NKF
CUH FTD . SFZ TWA . EHT JSU . ZJA ZZQ
XLV BGB . GLW PGJ . RSR EIL . IQP ANM
ALG KGS . UEN IOF . RDM EPI . DHH YSD
VNT GVU . MKK LEP . YPQ HJZ . XBB BQX
QMC DRT . GTX PFW . BSS UIN . KEU OOE
HZU RHE . OUS XTN . WYE WCC . JWQ QAZ
LAY CXV . PBI SQK . BRO UDR . BII ULK
NDJ MPY . RAB EXX . FTV NFB . COT FBU
TNC VVT . FXN NKF . JCF QMH . IJI AZK

One partial message is:

OKH VON JSW 15/03 13.25 JYK QCP
WCKVE KOFLB UVQKI IEOAY UFA.....

The date of the intercepts implies a three rotor German Army/Air Force Enigma with six steckers. The two groups of three cipher letters in each message header inplies the repetion of the encipherment of the message key, thus Marian Rejewski's "characteristics" can be used to find the Enigma configuration.

To use Marian Rejewski's characteristics, the letter chains must first be deduced for the above set of enciphered message keys. Remember that in each set of six letters, 4 is an encipherment of the same message key letter as 1. Similarly 2 and 5 and 3 and 6. This links these letters together in pairs.
Start with 1 and 4 in each set. The first pair are AK. Now look for another occurance of K. It is KO. So now we have AKO. Continue this way until you get to I linked to A. This closes the cycle back to the first A. (AKOXBUIA) and is length 7. (only count A once).
The other sets are: (CFNMLC). (VGPSTV) . (ZZ) . (EJQDYHRE) . (WW)
So for 1,4 there are 6 sets length 7,7,5,5,1,1
Similarly for 2,5 there are just 2 sets legth 13,13 (note VU has to be deduced).
For 3,6 there are 6 of length 6,6,4,4,3,3.

In order to make the sizes of list manageble I have used wheel order 132 and revealed that the left hand indicator letter is F or G. This is so that you can use the already generated list of characteristics in VirtualBP.
Click here to see the list for 1,4 wheels 132, letters FAA to GZZ.
This gives many matches for 6,7,7,5,5,1,1
Now try the composite key: (Number of sets) (length of longest set)
for 1,4 then 2,5 then 3,6
This just shows two possibles, FSH and GHB.
Now you need an Enigma machine, get the 3 wheel Enigma from the toolbox. Use the function key F11 to get full screen size if needed.
Open the top lid and select wheels 132. Next set the rings to ZZZ. Close the lid and lift the wheel cover. Set the wheel to show FSH in the windows.
You are now looking for deducable message keys, either straights like AAA or keyboard patterns like ASD. Try decrypting AGL KYO. This just gives a jumble letters so reset the right hand wheel to H and try more groups.
The third group across the top hits the jackpot. COT FBU deciphers to PPP EPP. Obviously the E in postion 4 should be P so a stecker is needed. Restart at FSH and this time key in P at position 4. This gives A so A must be steckered to F. So do this. (See the instructions alongside the Enigma if required). Now resetting to FSH you now get PPP PPP. Confirming the stecker. One down five to go!
The first set on the 2nd line, CUH FTD gives PYD PYX. This implies D sould be X, since PYX is a keyboard pattern. So stecker D to X.
Work on through the sets of six letters and all the steckers come out.
Steckers: AF DX QR NV ZJ KM.
All revealed by the poor message key selection by the German operators!

Note that the alternative indicator from the list, GHB doesnt reveal any straights or patterns.

Now it gets harder. Lets decrypt the message fragment.
First reset to FSH and decrypt the six letters in the header of the message fragment. This gives UUU as the message key. But in order to decrypt the message we need to know the correct ring setting. The characteristics have assumed a ring setting of ZZZ. The correct ring setting can be found by exhaustive search using ANX ("to" space) as a guess of the first three letters of the message. (In 1935 this was a pretty solid guess!).
To do this first set the message key onto the rings. Now the Polish mathematicians would have had to start at AAA and go through all 17576 wheel start positions. I'll take pity on you (and your mouse!) and suggest you start at JAA. So now key into the Enigma machine, W, the first cipher letter of the message. And keep on keying it in until the A lamp lights. If it does try the next cipher letter, C, which should give N if you've found the right wheel position. If C doesn't give N just keep going on W.
Eventually, at JDY, WCK decrypts as ANX, so this is it? No not quite.
The combination of UUU on the rings and JDY on the wheels has to be transformed to UUU on the wheels and ??? on the rings. To do this remember the basic equation of: window position = core postion + ring position.
(treated as numeric values 1 - 26). We want a ring position which puts the core in the same position for the two cases. This comes out as (new ring letter) = 2x(key letter) - (letter just found).
Using this and UUU and JDY gives FLQ as the correct ring letters.
Now the moment of truth! Set the Enigma rings to FLQ, turn the wheels to UUU and key in the message cipher text revealing:

WCKVE KOFLB UVQKI IEOAY UFA
ANXOK HXDEX FSWYY DRING END

You have just broken a message enciphered on an Enigma machine which had a 10 million, million, million search space!! How? A combination of very clever Polish mathematics and very bad operator errors.

You may need to print this then go look for tools.

 This page was originally created by the late Tony Sale, the original curator of the Bletchley Park Museum,