You have two tasks. You must design a 3-switch lamp and a 4-switch lamp. It is recommended that you try the 3-switch lamp first. Your design will include a lightbulb, wires, switches and power sources. The design must follow these rules:

1. You may only use 1 lightbulb for each lamp. The 3-switch lamp can only have one power source, and the 4-switch lamp must have exactly two. You may use any number of wires.
2. Every flip of a switch, no matter the previous positions, must turn the lamp from on to off or off to on.
3. Each wire may connect to any number of switches, power sources, and other wires, and to the lightbulb.
4. Each switch has two separate positions to which wires can connect. If the switch is up, then all the wires connected to position 1 are considered connected to each other. If the switch is down, all the wires connected to position 2 are considered connected to each other.
5. The lightbulb turns on if and only if there exists a complete circuit that includes both the lightbulb and at least one power source.
6. A circuit is a sequence of wires, power sources, and the lightbulb where each is connected to the next item in the sequence (the last is connected to the first). No such sequence may list the same wire, power source, or the lightbulb twice.

I recommend that you denote the different wires with letters like A, B, C, etc.

If the three switches are named A, B, C, then we must synthesize the function:
 F1(abc)=!a(b!c+!bc)+a!(b!c+!bc)
If D is the fourth switch then then we must synthesize the function:
F2(abcd)=!dF1(abc)+d!F1(abc)
Using 3-way switches for the first case the circuit is:

                      o-----o
                  B1!        !  C1
               ----- o      o--
               !                  !
               !      o-----o  !
              o                  !
            A!                   !
    M------o                   !--------N
                                   !
              o   B2       C2  !
               !      o      o    !
               !      ! \   / !    !
               -----o   X   o--!
                        /   \
                      o      o
The bulb, the power and the MN are in series. B1, B2 respectively C1, C2 are commuting simultaneously.
For the 4 switches and ONE power source we must double the above circuit, add the fourth switch and inverse the conections at A for one branch.
I don't understand for what would be the second power in the last case....
I feel that for the 4 switches case must be a more elegant solution.

Posted by Eugen on 2005-06-08 21:01:10