2
$\begingroup$

Rotor I with ring A would look like this:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
EKMFLGDQVZNTOWYHXUSPAIBRCJ

Inverse rotor I with ring A would look like this:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
UWYGADFPVZBECKMTHXSLRINQOJ

Rotor I with ring B would look like this:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
FLNGMHERWAOUPXZIYVTQBJCSDK

Am I right that inverse rotor I with ring B will look like this?

ABCDEFGHIJKLMNOPQRSTUVWXYZ
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
JUWYGADFPVZBECKMTHXSLRINQO
$\endgroup$
1
  • 1
    $\begingroup$ Checked only a few. Seems ok. It is just the inverse of the permutation. $\endgroup$
    – kelalaka
    Commented Jan 6, 2020 at 21:58

0

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