Turing's Treatise on the Enigma

There are various things which could be done now. Of course one might put the whole data onto the spider, but at the time that this system was in force no such machine had been thought of. Another method, which was that principally used by the Poles is to have a permanent catalogue of the box shapes for G4G1, G5G2, G6G3 for every Grundstellung, assuming that there is not a T.O. between the first and last of the six alphabets. If we give some standard order or numbers to the box shapes, we can also put the possible series of three box shapes into an order, and can enter against each set of three box shapes the Grundstellungen for which this set is realised. To use this catalogue with our problem we should work out the box shapes viz. G4G1 is 26, G5G2 is 26, G6G3 is 24,2. The box shape 26 actually has the number 1 and 24,2 the number 2: they are the two commonest shapes as can be seen from the table p 19a. We then look up 1.1.2 in our catalogue, and find about 150 entries against it for each wheel order. Each of these will have to be tested out in some way or other. The most satisfactory method seems to be this. We form the permutation G6G3G5G2. It is
so that this permutation is of the class 20,4,1,1. For each possible Grundstellung it is possible to calculate the corresponding class for the unsteckered alphabets. This can fortunately be done mechanically by means of a form of 'cyclometer'. It would be as well to enter against each position the class of this permutation, and this might have been done at the time of construction of the catalogue. In the case in question the right Grundstellung is found to have the position 1,1,26 with order I, II, III (service machine, Umkehrwalze A). The corresponding boxes are


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