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.


1 Answer 1


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.

  • $\begingroup$ 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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