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I am trying to understand the conditions necessary for one of the Bletchley Park bombes to stop. Let me give an example.

I have been experimenting with Enigma machine and bombe simulators to try to understand better how the bombe works.

Using a standard three rotor Enigma machine set up as follows:

  • Rotors IV-I-III,
  • starting with QLH
  • rings set to AAA
  • with plugs EL GK US XZ WI NV FD BY MC PQ

The following plaintext WETTERVORHERSAGE becomes the following cryptotext RRHKJKNJXEUWNFOT (see enigma simulator) .

Using this as a crib, the menu that this produces has three loops (REHTK, EHT, RETK) and a click (loop of the two letters WR), which is pretty good as a crib as this means that the bombe then returns far fewer false stops.

When I plug this crib into the following bombe simulator (where all rings are set to A) I get a collection of about twenty stops, one of which is the correct one.

So my question: what is the condition that stops a bombe that is satisfied with respect to the crib menu?

My understanding is that the back of the bombe is set up to simulate the menu and then the bombe is switched on, stopping through a short circuit when, well, when what?

I tried to use the first false stop that the bombe simulator gave me to understand: 231-YAL. So in particular, my question can be applied to this situation. What is it about this set-up that would stop the bombe with the given crib-menu?

What I did notice in this configuration is that the click is correct with respect to the letters M and T (indicating that there should be plugs WM and RT) BUT the remainder of the loops do not match.

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2 Answers 2

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After a year I have managed to find a suitable solution to the problem. My understanding is based on the following picture that I created, based on a simplified version that I found online (at present I cannot find the reference, if I do I will edit it in).

enter image description here

It is supposed to be a bombe and can be understood in two parts.

  • The top contains 26 columns, one per letter of the alphabet. Each column is made of 26 connections one from each letter to each other. That what gives the triangular effect.
  • The bottom half are what would be the bombe connections depending on the menu derived from the crib. Each block that we can see spans and connects two columns are based on the crib and the corresponding position of the rotors.

The idea is that in the real bombe all these lines are wires and a current is introduced arbitrarily somewhere. Then the current flows through the diagram making lines 'live' and others not.

A live connection between two letters means that either there is a Steckerverbindung between them else they are not Steckered together. Although this sounds like a tautology, we do need to consider the picture as a whole. For example: if at the A column only the F connection is live, then it means that it is consistent that A and F are Steckered together; if at the A column more both F and G are live, then it means that A is not Steckered with either F or G; if at the A column all connections are live, then this means we have a definite contradiction, and that means the bombe position is false - and in such a case the bombe, I assume, would have just continued.

We can test the information above with such a diagram as above. The following diagrams are all close ups of the diagram above (the top right part of the bottom half):

  1. The bombe is in the correct position and the current is sent into a correct connection, as in the picture below.enter image description here Here the current is sent to the X connection in the R column. I indicated in red the live current. It flows through the diagram and as can be seen. Eventually we see that the L wire of the T column s live, as is the E wire of the V column. Nothing else in this block becomes live. So it is consistent that we have the three Steckerverbinugen RX, TL and VE.
  2. The bombe is in the correct position and the current is sent into a false connection. Again we can use the picture above but now we concentrate on the blue. Here I assumed the current was sent in at the A wire of the T column. In this case eventually every wire is live except the correct wires as given above (at least in the given part of the picture). When I say eventually, this means that one has to follow in and out of all the connections, and up through the top part of the diagram when necessary. Hence when the bombe is in a correct position and viewing the diagram as a graph, we would have (ideally) two connected components: the correct component (where there is at most one live wire per column) and the false component (where we would have at least one non-live wire pre column). I say ideally since there would be at least one extra component of information that is unobtainable from the rest (i.e. letters that are not part of the steckers we can find and don't appear in the crib).
  3. If the bombe is in a false position, then it shouldn't matter we the current is sent in. I did this in the following diagram, which is the same as the original but all rotors have been shifted by one position.enter image description here Here the current enters at the L wire of the T column. In this case, eventually, everything becomes live, which gives the contradiction.

We also learn from this set up the importance of the loops in the crib. This enables a lot of feedback, which is particularly important when the current is sent into the false wire.

Hence to answer my original question: when did the bombe stop? It seems that it should not stop when one column is completely live. And therefore perhaps it stopped when there was at least one non-live wire per column. If your crib had no loops, and therefore no feedback, then this condition would be easily satisfied, and there would be far too many bombe stops to deal with.

Edit after 4 years: I found in the book The Hut Six Story from Gordon Welchman first published in 1982 (although my copy seems to be first published in 1997) Appendix I p241 it says: Thus, in order to detect a "drop", the bombe needs to look for one of only two situations: a case in which current does not get back to any test register terminal other than A, or one in which there is some terminal other than A to which current does not get back.

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  • $\begingroup$ As the bombe is current mode loops, it probably just had a resistor at the end which created a voltage and caused the whole system to stop. It'd be electrically easy to use that to release the relay that started the motion. $\endgroup$
    – b degnan
    Dec 3, 2020 at 18:44
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Suppose it is an Army Enigma I. Choose reflector UKW-B or UKW-C. Choose three rotors from I, II, III, IV and V. For each of the 120 reflector/rotors permutations, there are 17,576 starting indicator positions to test.

Menu

Ciphertext $\tt RRHKJ \ KNJXE \ UWNFO \ T$ has 16 characters. Crib $\tt WETTE \ RVORH \ ERSAG \ E$ has 16 characters too. Align crib to ciphertext at position 1 and there is no clash.

          1     1 1
1   5     0     5 6
RRHKJ KNJXE UWNFO T <- ciphertext
WETTE RVORH ERSAG E <- crib

Assume turnover (t/o) position is at 13, 14, 15 or 16. We use letters at positions 1-12 and 16 to build a menu. It has 10 letters, 2 loops, 11 links of which one is a double link. There are two subsidiary letters and one subsidiary link.

Wetter menu positions 1-12, 16

Default ring setting is 26 26 26 (Z Z Z). Drum setting for character positions 1, 2, ..., 12 are Z Z A, Z Z B, ..., Z Z L and position 16 is Z A P. Bombe returns 32 stops. It is noted that Gordon Welchman called them "drops". Many Bletchley Park veterans called them stories. I like the term story. It is not to be trusted until verified.

Fri 29 Sep 10:39:30 BST 2023
UKW-B, I II IV,   offsets  2  3 15 (B C O), Cable E, wire o is a stop, drums C D Q, golden drums W V I : O
UKW-B, I II IV,   offsets  1  3 23 (A C W), Cable E, wire q is a stop, drums B D Y, golden drums X V A : Q
UKW-B, I III II,  offsets 14  6  5 (N F E), Cable E, wire j is a stop, drums O G F, golden drums K S T : J
UKW-B, I III II,  offsets 19 11 10 (S K J), Cable E, wire s is a stop, drums T L K, golden drums F N O : S
UKW-B, I III IV,  offsets 14  9 10 (N I J), Cable E, wire b is a stop, drums O J L, golden drums K P N : B
UKW-B, I III IV,  offsets  0 19 17 (Z S Q), Cable E, wire j is a stop, drums A T S, golden drums Y F G : J
UKW-B, I IV V,    offsets 22  6  8 (V F H), Cable E, wire s is a stop, drums W H K, golden drums C R O : S
UKW-B, I V II,    offsets 13 21 24 (M U X), Cable E, wire y is a stop, drums N X Y, golden drums L B A : Y
UKW-B, II III IV, offsets 15  4 14 (O D N), Cable E, wire z is a stop, drums P E P, golden drums J U J : Z
UKW-B, II V IV,   offsets  8 14 23 (H N W), Cable E, wire i is a stop, drums I Q Y, golden drums Q I A : I
UKW-B, III IV I,  offsets  5 20 25 (E T Y), Cable E, wire y is a stop, drums F V Z, golden drums T D Z : Y
UKW-B, III IV V,  offsets 25 24  1 (Y X A), Cable E, wire y is a stop, drums Z Z D, golden drums Z Z V : Y
UKW-B, IV I III,  offsets 16 11  7 (P K G), Cable E, wire l is a stop, drums R L H, golden drums H N R : L  <-- ture stop
UKW-B, IV III II, offsets  2  9 19 (B I S), Cable E, wire s is a stop, drums D J T, golden drums V P F : S
UKW-B, V II III,  offsets  8  7 19 (H G S), Cable E, wire l is a stop, drums K H T, golden drums O R F : L
UKW-B, V II III,  offsets  1 23 21 (A W U), Cable E, wire n is a stop, drums D X V, golden drums V B D : N
UKW-B, V II IV,   offsets 11  8 23 (K H W), Cable E, wire c is a stop, drums N I Y, golden drums L Q A : C
Fri 29 Sep 10:39:50 BST 2023 (17 stops, 20 seconds)

Fri 29 Sep 10:40:08 BST 2023
UKW-C, I III II,  offsets 18  0 13 (R Z M), Cable E, wire q is a stop, drums S A N, golden drums G Y L : Q
UKW-C, I IV V,    offsets  4 18 23 (D R W), Cable E, wire d is a stop, drums E T Z, golden drums U F Z : D
UKW-C, I V IV,    offsets 23 16  4 (W P D), Cable E, wire q is a stop, drums X S F, golden drums B G T : Q
UKW-C, II I III,  offsets  2 21 12 (B U L), Cable E, wire q is a stop, drums C V M, golden drums W D M : Q
UKW-C, II IV V,   offsets  2  7 21 (B G U), Cable E, wire a is a stop, drums C I X, golden drums W Q B : A
UKW-C, III II I,  offsets 16  5 17 (P E Q), Cable E, wire q is a stop, drums Q F R, golden drums I T H : Q
UKW-C, III II I,  offsets 21 19 19 (U S S), Cable E, wire v is a stop, drums V T T, golden drums D F F : V
UKW-C, IV II I,   offsets  1 16 11 (A P K), Cable E, wire e is a stop, drums C Q L, golden drums W I N : E
UKW-C, IV II III, offsets  8 12  3 (H L C), Cable E, wire g is a stop, drums J M D, golden drums P M V : G
UKW-C, IV III II, offsets  3 24 25 (C X Y), Cable E, wire d is a stop, drums E Y Z, golden drums U A Z : D
UKW-C, IV III V,  offsets 17 20  7 (Q T G), Cable E, wire y is a stop, drums S U J, golden drums G E P : Y
UKW-C, IV V I,    offsets 11 10 25 (K J Y), Cable E, wire f is a stop, drums M M Z, golden drums M M Z : F
UKW-C, IV V II,   offsets 24 19  9 (X S I), Cable E, wire d is a stop, drums Z V J, golden drums Z D P : D
UKW-C, V I III,   offsets 18 23 13 (R W M), Cable E, wire f is a stop, drums U X N, golden drums E B L : F
UKW-C, V II III,  offsets 10 16  8 (J P H), Cable E, wire c is a stop, drums M Q I, golden drums M I Q : C
Fri 29 Sep 10:40:28 BST 2023 (15 stops, 20 seconds)

Stop condition

Suppose test register is connected to Cable $\tt E$. "In order to detect a 'drop,' the bombe needs to look for one of the two situations: a case in which current does not get back to any test register terminal other than [its steckered letter, $\tt l$, in our case], or one in which there is vast majority of [in fact, all of] the register will fill up." (Welchman 2021, p.241)

Gordon Welchman (2021) The Hut Six Story: Breaking the Enigma Code. Cleobury Mortimer, Shropshire, UK: M & M Bladwin.

On one side of the diagonal board, there are twenty-six 26-way cables. Each cable is equivalent to a letter in the menu. Each double-ended letchworth scrambler is equivalent to a link in the menu. Let's say, Cable I has 26 wires a, b, ..., i, j, ..., z and Cable J has a, b, ..., i, j, ..., z. The diagonal board connects wire Ij to wire Ji. Therefore, the are 351 distinct connections. It is the upper triangle of an 26x26 matrix.

Suppose each wire in the test register is a circuit on its own. Each wire has four possible states. Firstly, it connects to one circuit which is represented by an ${\tt 1}$. Secondly, it connects to 25 circuits which is represented by a ${\tt \sim}$. These two cases are the conditions we want for a stop. Thirdly, it connects to all 26 circuits, which means there are contradictions, and is represented by an ${\tt =}$. Fourthly, a wire is open circuit which is represented by a ${\tt 0}$. The last case happens when a letter is not in the menu and the diagonal board did not propagate to all its wires.

UKW-B, IV I III,  offsets 16 11  7 (P K G), Cable E, wire l is a stop, drums R L H, golden drums H N R : L
     wire  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z  wire
Cable A -  0  0  0  0  ~  0  0  ~  0  ~  ~  0  0  ~  ~  0  0  ~  0  ~  ~  ~  ~  ~  0  0  - A Cable
Cable B -     0  0  0  ~  0  0  ~  0  ~  ~  0  0  ~  ~  0  0  ~  0  ~  ~  ~  ~  ~  0  0  - B Cable
Cable C -        0  0  ~  0  0  ~  0  ~  ~  0  0  ~  ~  0  0  ~  0  ~  ~  ~  ~  ~  0  0  - C Cable
Cable D -           0  ~  0  0  ~  0  ~  ~  0  0  ~  ~  0  0  ~  0  ~  ~  ~  ~  ~  0  0  - D Cable
Cable E -              ~  ~  ~  ~  ~  ~  ~  1  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  - E Cable
Cable F -                 0  0  ~  0  ~  ~  0  0  ~  ~  0  0  ~  0  ~  ~  ~  ~  ~  0  0  - F Cable
Cable G -                    0  ~  0  ~  1  0  0  0  ~  0  0  ~  0  ~  ~  0  ~  ~  0  0  - G Cable
Cable H -                       1  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  - H Cable
Cable I -                          0  ~  ~  0  0  0  ~  0  0  ~  0  ~  ~  0  1  ~  0  0  - I Cable
Cable J -                             1  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  - J Cable
Cable K -                                ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  - K Cable
Cable L -                                   0  0  0  ~  0  0  ~  0  ~  ~  0  ~  ~  0  0  - L Cable
Cable M -                                      0  0  ~  0  0  ~  0  ~  ~  0  ~  ~  0  0  - M Cable
Cable N -                                         0  ~  ~  ~  ~  0  ~  ~  0  ~  ~  ~  0  - N Cable
Cable O -                                            1  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  ~  - O Cable
Cable P -                                               0  0  ~  0  ~  ~  ~  ~  ~  0  0  - P Cable
Cable Q -                                                  0  ~  0  ~  ~  ~  ~  ~  0  0  - Q Cable
Cable R -                                                     1  ~  ~  ~  ~  ~  ~  ~  ~  - R Cable
Cable S -                                                        0  ~  1  0  ~  ~  0  0  - S Cable
Cable T -                                                           1  ~  ~  ~  ~  ~  ~  - T Cable
Cable U -                                                              ~  ~  ~  ~  ~  ~  - U Cable
Cable V -                                                                 0  ~  ~  ~  0  - V Cable
Cable W -                                                                    ~  ~  ~  ~  - W Cable
Cable X -                                                                       ~  ~  1  - X Cable
Cable Y -                                                                          0  0  - Y Cable
Cable Z -                                                                             0  - Z Cable
     wire  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z  wire

    XZ EL KG WI US maybe steckered
    steckered pair - EL KG US WI XZ - 
    self-steckered - H J O R T - 
           unknown - A B C D F M N P Q V Y

Now, apply current to connection Cable $\tt E$ wire $\tt l$. It energises itself but no other wires in the test register. We know that $\tt E$ is steckered to $\tt L$. Using the checking machine, by the wiring alone, it can be found that $\tt GK, IW, SU, ZX$ are also steckered and $\tt H, J, O, R, T$ are self-steckered.

When it is not a stop, e.g. UKW-B, IV I III, 16 11 8, apply current to any one wire of Cable $\tt E$ energises all 26 wires in the test register.

UKW-B, IV I III,  offsets 16 11  8 (P K H) is not a stop
     wire  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z  wire
Cable A -  0  0  0  0  =  0  0  =  0  =  =  0  0  =  =  0  0  =  0  =  =  =  =  =  0  0  - A Cable
Cable B -     0  0  0  =  0  0  =  0  =  =  0  0  0  =  0  0  =  0  =  =  0  =  =  0  0  - B Cable
Cable C -        0  0  =  0  0  =  0  =  =  0  0  =  =  0  0  =  0  =  =  0  =  =  0  0  - C Cable
Cable D -           0  =  0  0  =  0  =  =  0  0  0  =  0  0  =  0  =  =  0  =  =  0  0  - D Cable
Cable E -              =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  - E Cable
Cable F -                 0  0  =  0  =  =  0  0  0  =  0  0  =  0  =  =  0  =  =  0  0  - F Cable
Cable G -                    0  =  0  =  =  0  0  0  =  0  0  =  0  =  =  0  =  =  0  0  - G Cable
Cable H -                       =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  - H Cable
Cable I -                          0  =  =  0  0  0  =  0  0  =  0  =  =  0  =  =  0  0  - I Cable
Cable J -                             =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  - J Cable
Cable K -                                =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  =  - K Cable
Cable L -                                   0  0  0  =  0  0  =  0  =  =  0  =  =  0  0  - L Cable
Cable M -                                      0  =  =  0  0  =  0  =  =  =  =  =  0  0  - M Cable
Cable N -                                         0  =  0  =  =  =  =  =  =  =  =  =  0  - N Cable
Cable O -                                            =  =  =  =  =  =  =  =  =  =  =  =  - O Cable
Cable P -                                               0  0  =  0  =  =  0  =  =  0  0  - P Cable
Cable Q -                                                  0  =  0  =  =  =  =  =  0  0  - Q Cable
Cable R -                                                     =  =  =  =  =  =  =  =  =  - R Cable
Cable S -                                                        0  =  =  =  =  =  0  0  - S Cable
Cable T -                                                           =  =  =  =  =  =  =  - T Cable
Cable U -                                                              =  =  =  =  =  =  - U Cable
Cable V -                                                                 =  =  =  =  0  - V Cable
Cable W -                                                                    =  =  =  =  - W Cable
Cable X -                                                                       =  =  =  - X Cable
Cable Y -                                                                          0  0  - Y Cable
Cable Z -                                                                             0  - Z Cable
     wire  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z  wire

Turnover at position 16

At t=0, initial ring setting is 26 26 26 (Z Z Z) making the middle rotor to turnover at t=16. It suggests that DG is a steckered pair. Since GK is a pair, there is a contradiction. So, the possibility of middle rotor turnover at position 16 is eliminated.

Turnover at position 15

Change ring setting to 26 26 01 (Z Z A) making middle rotor to turnover at t=15. And hence at t=0, starting indicator is P K H (16 11 08). At t=15, middle rotor turns over and yet again indicator is P L W.

    steckered pair - EL GK IW SU ZX - 
    self-steckered - H J O R T - 
           unknown - A B C D F M N P Q V Y

X  -       RRHKJ KNJXE UWNFO T
X1 -       RRHGJ G-JZL SI--O T

X8 -       ILTTL R-ORH LR--K L
Y  -       WETTE R-ORH ER--G E
crib       WETTE RVORH ERSAG E
diff              ^      ^^   
t/o                        ~ ~

At t=7, scrambler $S_7 = {\tt (AJ) (BX) (CU) (DK) (ET) (FR) (GM) (HY) (IL) (NV) (OQ) (PW) (SZ)}$. It is clear that NV is a steckered pair making N > V > N > V. At t=13, scrambler $S_{13} = {\tt (AG) (BR) (CZ) (DI) (EJ) (FQ) (HT) (KS) (LN) (MX) (OY) (PW) (UV)}$, it follows that N > V > U > S. Let's make two wild guesses that FD is a pair and A is self-steckered. We have,

    steckered pair - EL GK IW SU ZX - NV FD
    self-steckered - H J O R T - A
           unknown - B C M P Q Y

X  -       RRHKJ KNJXE UWNFO T
X1 -       RRHGJ GVJZL SIVDO T

X8 -       ILTTL RNORH LRUAK L
Y  -       WETTE RVORH ERSAG E
crib       WETTE RVORH ERSAG E
diff                          
t/o                        ~ ~

Therefore, we can conclude that ring setting is ?? ?? 01 (? ? A), starting indicator is ? ? H (?? ?? 08), EL GK IW SU ZX NV FD are steckered pairs and H J O R T A are self-steckered. Also, indicator/ring offset for the left and the middle rotors are 16 11. Any settings satisfy the above conditions will yield the correct ciphertext/plaintext pair.

Alternative solutions for turnover at position 15

There is no dispute that NV is a steckered pair. However, it is equally valid to make two alternative wild guesses by saying that F is self-steckered and QA is a pair. Given that ring setting is ?? ?? 01 (? ? A), starting indicator is ? ? H (?? ?? 08), any settings that satisfy that EL GK IW SU ZX NV QA are steckered pairs, H J O R T F are self-steckered and indicator/ring offset for the left and the middle rotors are 16 11 will return the correct ciphertext/plaintext pair too.

    steckered pair - EL GK IW SU ZX - NV QA
    self-steckered - H J O R T - F
           unknown - B C D M P Y

X  -       RRHKJ KNJXE UWNFO T
X1 -       RRHGJ GVJZL SIVFO T

X8 -       ILTTL RNORH LRUQK L
Y  -       WETTE RVORH ERSAG E
crib       WETTE RVORH ERSAG E
diff                          
t/o                        ~ ~

Turnover at position 14

There is at least one contradiction. Therefore, the possibility of middle rotor turnover at position 14 is eliminated.

Turnover at position 13

There is at least one contradiction. Therefore, the possibility of middle rotor turnover at position 13 is eliminated.

Stop for UKW-C, I IV V, 4 18 23, Cable E, wire d

There is at least one contradiction at each of turnover positions at 13, 14, 15 and 16. Therefore, the possibility this stop can be rejected altogether.

Insufficient information

First, the ciphertext is too short, we cannot determine the rest of steckered pairs and self-steckered letters. Second, there is no message preamble, we are unable to determine ring setting for the left and the middle rotors.

End note

(1) The Turing-Welchman Bombe actually spins the left drums (at the first row of each bank) fastest, then the middle drums (at the second row) and the right drums (at the third row) slowest. It doesn't affect the stops themselves but only the order that they are coming out from the Bombe. (2) For an unknown reason, the markings on drums I, II, III, IV and V are shifted by 1, 1, 1, 2 and 3 places respectively comparing to the Enigma machines. Magnus Ekhall and Fredrik Hallenberg describe this oddity in the Turing Bombe Tutorial for their very comprehensive Bombe Simulator. I presume it is the same in the Checking Machines.

Frank Carte (2010) The Turing Bombe. The Rutherford Journal 3. http://rutherfordjournal.org/article030108.html

Magnus Ekhall and Fredrik Hallenberg (2020) Turing Bombe Tutorial. https://www.lysator.liu.se/~koma/turingbombe/TuringBombeTutorial.pdf

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