OK, now I am thinking as an EE, about design of the plugboard. Clearly one of the pair designated for each letter is the output. And the other is the fed back input to the next part of the machine. This would match up to the exactly 26 pairs (52 total) jacks.
First, how could some of the jacks remain unconnected? If they remained unconnected, then their output side would go nowhere! It would not be left as its own code, it woud produce no code at all! Those banana jacks were not like our complex headphone jacks that have an inner switch--this is very simple hardware, appears to be nothing more than a plugboard of simple banana jacks in pairs.
(AND the logic requires that the jacks have outputs as well as inputs--otherwise how can they possibly map one letter to another at all--only by one letter's output mapping by a wire to another letter's input. Thus perfectly sensible 26 outputs, 26 inputs, exactly the 52 jacks on the board.)
Those look like ordinary banana jacks--I have many in my shop, a modern caryover from a bygone day. In fact they look like the paired banana jacks that I have many of, so paired cables could be used for ease of connection--just they constrain one end of the pair to go to the other end of the pair. And would of course swap--else connecting out to out, in to in, not working. Thus forming a swapping of the letter pair! Still could use 26 individual banana cables to map any letter to any letter.
This logic implies that there would be 26 wires total within all cables and jumper plugs as I wil discuss. (Not 10, not 13--though they could be paired cables for ease of connection in which case there are 13 double wire cables still equivalent to 26 wires just constrained in positions. There could even be 10 paired cables and 6 jumper plugs so 10 pairs swap letters as discussed, and then 6 jumpers mapped to the same letter--but this could also be generalized by 26 individual double-banana cables. Jumping the top row to the same letter's bottom row would map letter to itself.) The specifics could be a convention of how the cables were constructed as paired cables, and 10 paired cables and 6 jumper plugs would fill all positions and swap letters as discussed for exactly 10 pairs. If the "10" was derived from loosing 3 paired wires and there were no jumper plugs around, then the machine would not work! Is the "10" derived from coding sheet?
This might even make sense. The 6 jumper plugs used so that not all letters are remapped, thus NOT allowing codebreaking to assume no letter maps to itself. 10 paired cables to ease the operator connection method. (All reduced from the possability of any permutation mapping of 26 letters if 26 individual cables were provided.) 13 paired cables and no jumper cables would assure no character maps to itself, but allows any letter to be swapped with another and the next swap reduced by 2 recursively until all letters mapped.
Can't immediately figure a factorial notation, but 25*23*21*...*3*1 = 7.90585 * 10^12 ish combinations with 13 paired cables.
10 cables: 25*23*21*...*9*7=527.0569 * 10^9 ish combinations.
OK, now I am thinking as an EE, about design of the plugboard. Clearly one of the pair designated for each letter is the output. And the other is the fed back input to the next part of the machine. This would match up to the exactly 26 pairs (52 total) jacks.
First, how could some of the jacks remain unconnected? If they remained unconnected, then their output side would go nowhere! It would not be left as its own code, it woud produce no code at all! Those banana jacks were not like our complex headphone jacks that have an inner switch--this is very simple hardware, appears to be nothing more than a plugboard of simple banana jacks in pairs.
(AND the logic requires that the jacks have outputs as well as inputs--otherwise how can they possibly map one letter to another at all--only by one letter's output mapping by a wire to another letter's input. Thus perfectly sensible 26 outputs, 26 inputs, exactly the 52 jacks on the board.)
Those look like ordinary banana jacks--I have many in my shop, a modern caryover from a bygone day. In fact they look like the paired banana jacks that I have many of, so paired cables could be used for ease of connection--just they constrain one end of the pair to go to the other end of the pair. And would of course swap--else connecting out to out, in to in, not working. Thus forming a swapping of the letter pair! Still could use 26 individual banana cables to map any letter to any letter.
This logic implies that there would be 26 wires total within all cables and jumper plugs as I wil discuss. (Not 10, not 13--though they could be paired cables for ease of connection in which case there are 13 double wire cables still equivalent to 26 wires just constrained in positions. There could even be 10 paired cables and 6 jumper plugs so 10 pairs swap letters as discussed, and then 6 jumpers mapped to the same letter--but this could also be generalized by 26 individual double-banana cables. Jumping the top row to the same letter's bottom row would map letter to itself.) The specifics could be a convention of how the cables were constructed as paired cables, and 10 paired cables and 6 jumper plugs would fill all positions and swap letters as discussed for exactly 10 pairs. If the "10" was derived from loosing 3 paired wires and there were no jumper plugs around, then the machine would not work! Is the "10" derived from coding sheet?
This might even make sense. The 6 jumper plugs used so that not all letters are remapped, thus NOT allowing codebreaking to assume no letter maps to itself. 10 paired cables to ease the operator connection method. (All reduced from the possability of any permutation mapping of 26 letters if 26 individual cables were provided.) 13 paired cables and no jumper cables would assure no character maps to itself, but allows any letter to be swapped with another and the next swap reduced by 2 recursively until all letters mapped.
Can't immediately figure a factorial notation, but 25*23*21*...*3*1 = 7.90585 * 10^12 ish combinations with 13 paired cables.
10 cables: 25*23*21*...*9*7=527.0569 * 10^9 ish combinations.