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The enigma machine permuted the words by using a plugboard, three rotors, and a reverting drum.
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The plugboard, the three rotors, and the reverting drum performed permutations on the incoming key.
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The permutation can be given by $$SNMLRL^{-1}M^{-1}N^{-1}S^{-1}$$ Now, when the operator of the enigma pressed the key, the right-most rotor is rotated once.
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So, to accommodate this change in our permutation, we added a permutation P that takes an alphabet and spits out the next alphabet.
$$P = (a\ b\ c\ d\ e\ f\ g\ h\ i\ j\ k\ l\ m\ n\ o\ p\ q\ r\ s\ t\ u\ v\ w\ x\ y\ z)$$ So, now the net permutation becomes, $$SPNP^{-1}MLRL^{-1}M^{-1}PN^{-1}P^{-1}S^{-1}$$

I got till this, but this is a special case when all the rotors are lined up initially i.e. "a" of the right rotor matched with "a" of the plugboard and "a" of the middle rotor, similarly for the others.

But what if the rotors are not lined up i.e. "a" of the right rotor matched with say "p" of the plugboard. Or in other words, there is an initial offset in all the rotors.
Will the above permutation work in the general case when the rotors have some offsets?

The images are taken from: Marian Rejewski. "An application of the theory of permutations in breaking the Enigma cipher." Applicationes Mathematicae 16.4 (1980): 543-559. http://eudml.org/doc/264403

Sumit Gupta
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The permutation P permutes all letters by one i.e. a goes to b, b to c ... z to a. You can easily show that the permutation P^2 permutes all letters by 2, so a goes to c, b to d etc In general, P^x permutes all letters by x.

So following your example, if a of the right-hand rotor was aligned with p of the input commutator (which is connected directly to the plugboard output using the identity permutation - Note that when Rejewski was initially trying to work out the Enigma hardware, he did not know this and so was using the permutation found on commercial Enigma machines, which was along the lines of the qwerty keyboard, and so assuming the identity permutation was an inspired guess) then you would need to insert the appropriate P^x which transforms p to a (and of course, q to b, r to c ...). I think x=11 would do the trick.

Tim Flack
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