AVR Freaks

Surely, they would all have to be "made" at once - you always need each and every input character to be encoded to exactly one output character!

Oh, yeah-- I'd get some double-diode arrays to/from the rails at each I/O pin.  Along with at least a series resistor (and maybe also a tiny cap to Gnd between resistor and diodes/pin).

I guess I'd do what you mentioned for scanning -- excite each "source" pin one at a time, and read the "destination" ports.  Check that exactly one bit is made.  Repeat and build the source/destination masks.

26?  No numbers?  No spaces?  No period?  Gonna spell out all the attack coordinates?

[How many channels did Enigma have?  Wikipedia here I come...

Each rotor was a disc approximately 10 cm (3.9 in) in diameter made from hard rubber or bakelite with 26 ... typically the 26 letters A–Z ...

Hmmm--

Enigma's security came from using several rotors in series (usually three or four) and the regular stepping movement of the rotors, thus implementing a polyalphabetic substitution cipher.

Simple substitution ciphers are easy to break, aren't they?]

Right - so not actually like Enigma, then.

 Inside the body of the rotor, 26 wires connected each pin on one side to a contact on the other

Hence 26 'In' and 26 'Out'.

Indeed!

And, if the substitutions are constrained in pairs, it's even easier!

But Enigma wasn't just a simple substitution;  the clever bit was that the wheels moved during the message - so the substitution changed during the message.

Remember that you can't plug two patch cords into one patch receptacle.

Each pin is connected to two things.  The first is a receptacle.  The second is the cathode of a diode.  The anode of that diode is connected to a patch cable.  The end of that patch cable can be inserted into the receptacle of any unpopulated receptacle.

As the OP already knows, reading the state of the patch is done by scanning in 26 steps.  For each step, one of the 26 pins is made an output low while the remaining 25 pins are made inputs with internal pull-up.  If the pin's diode + patch cable are inserted into another pin's receptacle, that other pin will read low.  All other pins will read high.  If all pins read high, the output low pin is either not connected to anything, or is connected to itself.  In fact, the two states can be considered identical.

Invalid states can be identified after a complete scan.  If a pin has been scanned and determined to be connected either to itself or not connected, but that pin reads low while scanning another pin, this would be an illegal state and that fact can be indicated to the operator.

The use of the diodes will preclude any multiple circuit paths which would make scanning difficult or impossible.

Take the simple case of a 3-letter alphabet:

A--|<|--A

B--|<|--C

C--|<|--B

During the first scan step A is output low.  B & C read as high, implying A is patched to itself or unpatched (which are semantically the same).

During the 2nd scan step B is output low.  A reads high, but C reads low.

During the 3rd scan step C is output low.  A reads high, but B reads low.

Here's an 'illegal' state:

A--|<|-- (unconnected)

B--|<|--A

C--|<|--C

During the first scan step A is output low.  B & C read as high (as before), implying A is patched to itself or unpatched (which are semantically the same).

During the 2nd scan step B is output low.  A reads low, but C reads high.  This contradicts the results of the first scan step.  Illegal state:

If you examine all of the permutations, I think you'll find that there is a unique result to identify all legal states, and at least one result to identify all remaining (i.e. illegal) states.

I don't believe there is a way of accomplishing this with less than N pins where N is the number of symbols.  This is an N-factorial problem, not a 2^N problem.

EDIT: hang on, I might have spoken too soon, without sitting down with paper and pencil

+------- A ---|<|---+
|                   |
|   +---------------+
|   |
|   +--- B ---|<|---+
|                   |
|   +---------------+
|   |
|   +--- C ---|<|---+
|                   |
+-------------------+

The above patch will not be readable.  When A is output low, B will be pulled down by A's diode, but then C will be pulled down through A's and B's diodes in series.

It would be impossible to distinguish the above patch from this one:

+------- A ---|<|---+
|                   |
|   +---------------+
|   |
|   +--- C ---|<|---+
|                   |
|   +---------------+
|   |
|   +--- B ---|<|---+
|                   |
+-------------------+

There may be hope.  With one diode, the voltage at an input pin will be pulled to GND+Vf.  For normal silicon diodes, Vf will be about 0.7V.  This will read as low.  Two diode drops will result in about 1.4V.  This will still read as low, even if Vcc=3.3V.  You could run the mcu at 1.8V, but this might complicate the design of the rest of the circuit.

However, if each diode is placed in series with a resistor, the voltage at an input pin will be higher.  If the series resistor is carefully selected so that one 'hop' through one resistor/diode pair is below the input-low threshold, but two hops through two resistor/diode pairs is above the input-high threshold, this could work, even with Vcc=5V.  The selection of the resistor would depend upon the value of the internal pull-up resistor.  You would want to create a voltage divider such that the voltage at PIN here:

    Vcc
     |
     |
     /
     \
     / internal
     \ pull-up
     /
     |
     |
     +-------------PIN
     |
     |
   -----
   \   /
    \ /
   -----
     |
     |
     /
     \
     / R
     \
     /
     |
     |
    GND

... is less than the input-low threshold, but the voltage at PIN here:

    Vcc
     |
     |
     /
     \
     / internal
     \ pullup
     /
     |
     |
     +-------------PIN
     |
     |
   -----
   \   /
    \ /
   -----
     |
     |
     /
     \
     / R
     \
     /
     |
     |
   -----
   \   /
    \ /
   -----
     |
     |
     /
     \
     / R
     \
     /
     |
     |
    GND

... is greater than the input-high threshold.

Since the internal pullup won't be the same from device to device, or even from pin to pin, you might be best to use an external pullup instead.

26 pins

26 diodes

26 series resistors

26 external pullups

Even without self-reciprocity, you'd only need 26 pins.

With self-reciprocity, you'd also still need diodes, if only to detect illegal states (including those which violate self-reciprocity).