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bonanova
Daedalian Member

 Posted: Fri Mar 05, 2010 9:31 am    Post subject: 1 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
Daedalian Member

 Posted: Sat Mar 06, 2010 8:20 pm    Post subject: 2 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
Daedalian Member

 Posted: Sat Mar 06, 2010 9:03 pm    Post subject: 3 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
Tired of his old title

 Posted: Sat Mar 06, 2010 11:25 pm    Post subject: 4 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
Daedalian Member

Posted: Sun Mar 07, 2010 12:12 am    Post subject: 5

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

You can actually do slightly better with n=3 (but still worse on average than your answer of n=2):

You can ignore any flips that change all 3 bulbs. You're only interested in the ones that flip a single bulb.

Once you've found them label the switches and bulbs so that ABC=1 and ABD=2. Then A and B = 3, C = 1, D = 2.

The probability of finding the 2 interesting flips in the first 2 flips is 1/6, of needing 3 flips is 1/3, and of needing 4 flips is 1/2, for an average of 3 1/3 flips.
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