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I have been studying the Turing-Welchman Bombe for a bit over a year now and one question seems to find very inconsistent answers depending on where you look. In 2015, someone asked "under what conditions did a Bletchley bombe stop?" (referenced here). The well formed answer by a user name Geoff seems to suggest that the Bombe will stop whenever any wire in the test cable has no voltage.

Here is where the story gets a bit more complicated. Turing discusses in the Prof's Book what the expected number of stops the machine produces is for a given menu. He only describes the expected number of stops for what he refers to as "normal stops" this is the case where either 1 wire is live and 25 are dead, or 25 wires are live and 1 is dead. He claims this is the most common type of stop (though I can find no intuitive reason that this would be true) and this is a particularly useful stop since it immediately tells the checking Wrens a good deal of the plugboard settings since only one possible plugboard setting is deduced by the machine (save for the additional settings that may not be included in the menu). Turing notes that for a menu with no closures and only two letters we can expect that 92% of the settings we check will produce normal stops. This information, however, seems to be completely disconnected from the actual stops of the Bombe, for if the Bombe truly stops whenever any wire is dead, then a two letter menu should always stop -- even with a diagonal matrix. This is because feedback from the diagonal board could cause at most two wires to become live (well shy of the 26 needed to continue the stepping mechanism).

Further, every academic paper I have read on the subject of Turing's calculation of the H-M factor only considers these "normal stops".

Finally, in Geoffs edit to his answer he noted

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.

It would then seems the case that according to Turing's model of calculation, academic researchers on the subject, and Gordon Welchman, that the machine actually stops whenever 1 or 25 wires are live. On the other hand, when I asked the Bletchely team they explained "the test registers look only for 26 live wires. If that test is failed, the Bombe will stop." All in all, there seems to be a lot of conflicting descriptions of the Bombe and I can't seem to find one definitive primary source which describes the stopping mechanism of the machine. Turing seems to describe it one way, and Welchman another. I'm hoping someone with more history chops than me can help me track down a single answer to this question.

Daniel S
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Jonah Weinbaum
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1 Answers1

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First note that Turing says that the two letter menu will produce normal stops for 92% of values. These will be stops where exactly one wire is live. The other 8% of values will, as you note, produce stops, but these will be "abnormal" stops (in this case where exactly two wires are live).

Note also that Welchman's description the second does not technically preclude the case where there is more than one terminal that is not reached (though this is arguably a strained reading).

The actual stopping criterion for the Bombe is as described: that any of the wires should not be live. This definitely captures all of the desired causal settings, but (depending on the shape and size of the menu) could also detect false positives. A secondary check would be necessary to distinguish these whenever they occurred. In practice, the menus used would be such that the probability of such a false positive would be vanishingly small.

Daniel S
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