| DT | PZ | CE |
| MG | RN | TQ |
| HL | UG | XU |
| UP | JL | JD |
| CK | BE | FO |
| AI | QX | RM |
| EJ | OD | NY |
| XV | TV | KB |
| BQ | MI | PV |
| NS | FW | WL |
| OR | AS | IG |
| YW | KH | HZ |
| FZ | YC | AS |
We must have V/A or V/S. If V/A we can identify the cycle (OHIWSADXRETZRZ) of the G5G2 with stecker, with the half compartment of the second box in this way
(OHIWSADXETGRZ)
(CHSWIVDXELGNZ)
i.e. we have to assume the stecker O/C, I/S, A/V, T/L, R/N, and that H, W, D, X, E, G, Z are unsteckered. This large number of unsteckered letters is a strong confirmation, and the repetition of the Stecker I/S is further confirmation. When we fit the rest of the boxes together we find that these five are all the Stecker.
There are other methods that can be applied, depending on the number of Stecker being small. The number of Stecker used in the Naval was 6 from 1931 to Nov. 1938 and possibly later. We might for instance have assumed that A and S were both unsteckered and therefore assumed that the constatation AS occurred in both the alphabets G3 and G6. With the Turing sheets we could find the possible positions for this, and then use a cyclometer to test the box shapes in those positions. This is naturally only worth while if we have no box-shape catalogue. Another possibility is to 'equate' the boxes, i.e. to find out from the permutations G4G1 etc what the original alphabets G1 and G4 were. In our case there are actually 13 different possibilities for G1, 13 for G2 and 12 for G3. There are two things we can do to distinguish between the correct and the incorrect possibilities. We can use known statistics about the list of admissible message settings, choosing that combination of alphabets that gives the greatest number of
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Back to Turing's Treatise on the Enigma. Chapter VII.