Imagine there is a 5x5 grid of lights, and only the middle light in the grid is on.

The lights are wired such that when you flip the switch for one light (from on to off or off to on) the others right next to it (not diagonally) flip as well.

Using this weird wiring of lights, what is the fewest number of switch changes it takes to turn all the lights off, and which lights should you switch? (Assume all the switches work in the manner explained, and there is 1 switch for each of the lights.)

After trying myself, this puzzle seems impossible so far. So, I need to find a way to prove it.

I think that to go about proving this, we have to keep in mind how many lights switch each time we flip a switch. Depending on which switch is flipped, 3,4, or 5 lights switch. We need to switch just one light total, in the middle.

It is crucial to recognize the fact that the possibilities of light combinations do not depend on the order that the switches are flipped. So, there are 2^25 possible combinations. This may match the number of total combinations, but some of the possibilities are probably the same.

The corner lights are important because only three switches change each of them. And we can only use two of the switches that switch the corner lights, or none at all. The lights on the side of the grid have only 4 corresponding switches. So, we can either use 4,2, or none of these switches. Similarly, with the rest of the lights that are off, 4,2,or none of the adjacent/same switches must be flipped. The middle light must have an odd number of changes.

If a combination of switches follows these rules, then this puzzle is possible. We must disprove (or prove) such a combination. I hope everyone is following me.

Posted by Tristan on 2003-11-02 15:56:24