[Gus]

I probably missed something in your explanation, but I think there might be a problem with your solution: It appears to be very easy to 'double count' a prisonner: let's consider the following events:

Let's say the left switch is the 'odd' counter, the right one the 'even' counter. For each event, I give:

_ the name of the prisoner (A or B)
_ the number of visit the prisoner has made
_ the position of the switches when he enters (UU, UD...etc)
_ the position of the switches when he leaves
_ the 'odd' count of prisonner A
_ the 'even' count of prisoner B

A (1) DD -> UD 0 0
B (1) UD -> DD - -
A (2) DD -> DU 1 0
B (2) DU -> DD - -
B (3) DD -> UD - -
B (4) UD -> UU - -
A (3) UU -> DU 2 0
... etc

So after those events, the difference between A 'odd counter' and A 'even counter' is already at 2, even though only B has been coming to the room so far.

Let me know what you think.