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 gives 6,7,2,13,6,6 Click here to see
list of composites.
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
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.