culling equivalent Enigma keys for short messages [closed]

Question

The following for-loop iterates over all possible keys for an M3 Enigma for three selected rotors:

for start_pos_left in A..Z:
    for start_pos_middle in A..Z:
        for ring_middle in 1..26:
            for start_pos_right in A..Z:
                for ring_right in 1..26:

For the middle and right-most rotors we need to know their start position and their ring setting. For the left-most rotor we need only know its start position as it never 'knocks' another wheel.

A computer-age attack on the Enigma (James J Gillogly. Ciphertext-only Cryptanalysis of Enigma. Cryptologia. 19(4) 405-412.) ignores the plugboard and iterates over the rotor settings, using Index-of-Coincidence to filter the likely settings. This is the attack I am trying to speed up. Here is my own implementation (using max freq rather than IC).

However, for short messages (real Enigma messages had a maximum of 320 letters for naval, and 250 for army and airforce messages), many of these keys are identical.

Here are some duplicate keys that turn a certain test message into identical ciphertext:

<A G,1 H,23> is dup of <A F,26 H,23>
<A G,1 I,24> is dup of <A F,26 I,24>
<A G,1 J,25> is dup of <A F,26 J,25>
<A G,1 K,26> is dup of <A F,26 K,26>
<A G,1 L,1> is dup of <A F,26 L,1>
<A G,1 M,2> is dup of <A F,26 M,2>
<A G,1 N,3> is dup of <A F,26 N,3>
<A G,1 O,4> is dup of <A F,26 O,4>
<A G,1 P,5> is dup of <A F,26 P,5>
<A G,1 Q,6> is dup of <A F,26 Q,6>
<A G,26 R,7> is dup of <A F,25 R,7>
<A G,26 S,8> is dup of <A F,25 S,8>
<A G,26 T,9> is dup of <A F,25 T,9>
<A G,26 U,10> is dup of <A F,25 U,10>
<A G,26 V,11> is dup of <A F,25 V,11>
<A H,1 R,7> is dup of <A F,25 R,7>
<A H,1 S,8> is dup of <A F,25 S,8>
<A H,1 T,9> is dup of <A F,25 T,9>
<A H,1 U,10> is dup of <A F,25 U,10>
<A H,1 V,11> is dup of <A F,25 V,11>
<A H,2 H,23> is dup of <A F,26 H,23>
<A H,2 I,24> is dup of <A F,26 I,24>
<A H,2 J,25> is dup of <A F,26 J,25>
<A H,2 K,26> is dup of <A F,26 K,26>
<A H,2 L,1> is dup of <A F,26 L,1>
<A H,2 M,2> is dup of <A F,26 M,2>
<A H,2 N,3> is dup of <A F,26 N,3>
<A H,2 O,4> is dup of <A F,26 O,4>
<A H,2 P,5> is dup of <A F,26 P,5>
<A H,2 Q,6> is dup of <A F,26 Q,6>
<A I,2 R,7> is dup of <A F,25 R,7>
<A I,2 S,8> is dup of <A F,25 S,8>
<A I,2 T,9> is dup of <A F,25 T,9>
<A I,2 U,10> is dup of <A F,25 U,10>
...

(In this example, the middle wheel is wheel II, which has a notch at E. The leftmost wheel is III and the rightmost wheel is I.)

How can you iterate over just canonical keys given you know the length of the message?