Cryptographic History of Work on the German Naval Enigma

stretch of 21. If however it is wheel III AJD = AJY + 5 as there is a turnover in the 21 stretch but none in the 5 stretch. By comparing the scores in cases like this we can determine the correct wheel. The remaining letters of the alphabet can be fitted in by using other "good fits" if any, or by looking at the scores on the deciban sheets. E.g. in Para. 9 we saw there was a very good score for Q = B + 5. Q is L in our alphabet and 5 places back would give B/G which is still possible - so this score indicates B/G as right. Also at this stage we see if the alphabet "picks up" any four letter repeats (tetra). Going back to Para. 11 we see that we had a tetra CRS = CQT + 2.3; there are large numbers of tetras like this, mostly wrong, but looking at our alphabet we see that in fact S does equal T + 3. This has the double effect of slightly improving the chance of our alphabet and greatly improving the chance of the tetra being correct.

18. Now we can go on and tackle the "middle wheel" in the same way as we did the right hand wheel. Here (see the fit list) we have R = Y + 6, T = X + 3 (3, not 2, because there is a turnover in the stretch of 15 from U to P), R = X + 1, V = L + 1, L = A + 9, A = W + 1 and other values such as (Para. 15) R = Q + 2 from the tetras picked up by the right hand wheel alphabet. Thus some or all of the middle wheel alphabet can be fixed, the middle wheel determined and an alphabet or crib menu run off on the bombe - only a few of the 336 W.O's now being possible - to finish the job.

C  How it worked in practice.

19. The work outlined in Paras. 1 - 18 above was shared between the Banburists, the Big Room, the R.R. and the Hollerith section; we shall be chiefly concerned with the part played by Banburists and Big Room and I will deal more briefly with the other first.

20. The R.R. prepared the traffic for Freeborn - the Hollerith section was always known to us simply as Freeborn, the copies of the


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