# What would happen if an Enigma machine's plugboard were used to map a letter to itself? [closed]

A famous characteristic of the Enigma machine was that its rotor design ensured/required that no letter ever map to itself. However, there was a plugboard on the front that could swap letter pairs. So let's say the machine were in a state where pressing A would make B light up. What would happen in that case if a plugboard connection were added swapping A and B?

## 2 Answers

In Enigma the wires followed from keyboard through rotors, then through plugboard, then back through the same rotors to lamps with letters. The resulting permutation would be (rotors)(plugboard)(rotors$$^{-1}$$). So you can't just swap A and B.

In this case B would still light up.

In Enigma encryption the Steckerbrett is passed through twice: once immediately after input and once immediately before output. In your example, the initial depression of the 'A' key is reinterpreted by the Steckerbrett as depressing a 'B' on an unsteckered Enigma. The unsteckered Enigma would then encrypt this as 'A' (because Enigma encryption is reciprocal) before returning to the Steckerbrett where the 'A' would be reinterpreted as 'B'. Thus the 'B' bulb would light up.

Note that the connectors on the Steckerbrett consisted of two wires with two connectors at each end:

one wire would connect the 'A' bulb/key to the 'B' Eintrittwaltz and the other would connect the 'B' bulb/key to the 'A' Eintrittwaltz so that in your case both wires would be live with current flowing in opposite directions.