bulbs and switches - the ones in my kitchen

[bonanova]

Here's a piece of fluff for the back of a napkin at lunch time.
Might not even need the napkin...

In my kitchen there are 3 ceiling lights and 4 wall switches.
Here's the deal:

1. Each switch controls 1 and only 1 light.
2. Each light is controlled by at least one switch.
3. When you enter the kitchen [you might not want to open the fridge] you find the switches positioned randomly up or down.
4. You are to determine how the lights and switches are connected.
5. You may [repeatedly, but using the same value of n each time] simultaneously change any n [0 < n < 4] of the switches and note what happens.

Here's the question:
If you want to minimize the number of times you execute Step 5 above, is there a preferred value of n?
_________________
_{Vidi, vici, veni.}

[Trojan Horse]

Okay, so one of the lights is controlled by exactly 2 switches.

Question: do we know exactly how those 2 switches affect that light? Do we know for sure that the 2 switches are connected in an "XOR" fashion; that is, flipping either switch always flips the light? Or is it possible that the switches control the light in some other way? Maybe the light turns on only if both switches are in the up position, or if at least one switch is in the down position, or whatever.

(I once lived in an apartment where the only way to turn the hall light on was to flip BOTH of two switches to the down position. So I'm not assuming anything here.)

[bonanova]

In all cases if you flip a switch, one and only one bulb will change its state.

In my entrance hall, I have the case you talked about.
I have not been able to fix those connectons.
Anyone have a clue on that issue, I'd be glad to hear it.
_________________
_{Vidi, vici, veni.}

[Zag]

Then the answer is clearly n=2, which can always be solved in 3 turns.

obviously, n=0 and n=4 are useless.

if n=1, then your first three tries might each change a different bulb, and you will need a fourth try to know which bulb is double-switched.

if n=3, this is actually the same as n=1.

with n=2, you switch AB, BC, CD.

One of these situations must be the case (once you number the bulbs to fit).

AB=no change, BC=12, CD=23 (A and B = 1, C=2, D=3)
AB=12, BC=12, CD=13 (A and C = 1, B=2, D=3)
AB=12, BC=23, CD=13 (A and D = 1, B=2, C=3)
AB=12, BC=no change, CD=23 (A=1, B and C=2, D=3)
AB=12, BC=23, CD=23 (A=1, B and D = 2, C=3)
AB=12, BC=23, CD=no change (A=1, B=2, C and D = 3)

[ChickenMarengo]